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6200	https://www.chegg.com/homework-help/questions-and-answers/derive-nodal-temperature-equation-case-exterior-corner-node-one-adjacent-side-insulated-on-q80746640	Solved Derive the nodal temperature equation for the case of \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Engineering Mechanical Engineering Mechanical Engineering questions and answers Derive the nodal temperature equation for the case of an exterior corner node with one adjacent side insulated and one adjacent side to a convective heat transfer, as shown in Fig bellow. (10 marks) B: Two large parallel plate are at T1=800K, ϵ1=0.3, T2=400K, ϵ2=0.7. Calculate the heat transfer rate between the two plans. where σ =5.67X10-8 W/m2.K4 Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Derive the nodal temperature equation for the case of an exterior corner node with one adjacent side insulated and one adjacent side to a convective heat transfer, as shown in Fig bellow. (10 marks) B: Two large parallel plate are at T1=800K, ϵ1=0.3, T2=400K, ϵ2=0.7. Calculate the heat transfer rate between the two plans. where σ =5.67X10-8 W/m2.K4 Derive the nodal temperature equation for the case of an exterior corner node with one adjacent side insulated and one adjacent side to a convective heat transfer, as shown in Fig bellow. (10 marks) B: Two large parallel plate are at T1=800K, ϵ1=0.3, T2=400K, ϵ2=0.7. Calculate the heat transfer rate between the two plans. where σ =5.67X10-8 W/m2.K4 Show transcribed image text Here’s the best way to solve it.Solution 100%(1 rating) Share Share Share done loading Copy link View the full answer Previous questionNext question Transcribed image text: 04:- A: Derive the nodal temperature equation for the case of an exterior corner node with one adjacent side insulated and one adjacent side to a convective heat transfer, as shown in Fig bellow. m-1.n (10 marks) h.T. dy m.n dx m.11-1 B: Two large parallel plate are at Ty=800K, 6=0.3, T2=400K, €2=0.7. Calculate the heat transfer rate between the two plans, where 6 -5.67X10-8 W/m²K (10 marks) Not the question you’re looking for? Post any question and get expert help quickly. 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6201	https://stats.stackexchange.com/questions/107610/what-is-the-reason-the-log-transformation-is-used-with-right-skewed-distribution	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams What is the reason the log transformation is used with right-skewed distributions? Ask Question Asked Modified 3 years ago Viewed 116k times 42 $\begingroup$ I once heard that log transformation is the most popular one for right-skewed distributions in linear regression or quantile regression I would like to know is there any reason underlying this statement? Why is the log transformation suitable for a right-skewed distribution? How about a left-skewed distribution? regression distributions data-transformation skewness Share Improve this question edited Jun 3, 2017 at 15:55 whuber♦ 342k6666 gold badges823823 silver badges1.4k1.4k bronze badges asked Jul 11, 2014 at 14:50 user3269user3269 5,3421010 gold badges5050 silver badges5656 bronze badges $\endgroup$ 0 Add a comment \| 5 Answers 5 Reset to default 50 $\begingroup$ Economists (like me) love the log transformation. We especially love it in regression models, like this: \begin{align} \ln{Y_i} &= \beta_1 + \beta_2 \ln{X_i} + \epsilon_i \end{align} Why do we love it so much? Here is the list of reasons I give students when I lecture on it: It respects the positivity of $Y$. Many times in real-world applications in economics and elsewhere, $Y$ is, by nature, a positive number. It might be a price, a tax rate, a quantity produced, a cost of production, spending on some category of goods, etc. The predicted values from an untransformed linear regression may be negative. The predicted values from a log-transformed regression can never be negative. They are $\widehat{Y}_j=\exp{\left(\beta_1 + \beta_2 \ln{X_j}\right)} \cdot \frac{1}{N} \sum \exp{\left(e_i\right)}$ (See an earlier answer of mine for derivation). The log-log functional form is surprisingly flexible. Notice: \begin{align} \ln{Y_i} &= \beta_1 + \beta_2 \ln{X_i} + \epsilon_i \ Y_i &= \exp{\left(\beta_1 + \beta_2 \ln{X_i}\right)}\cdot\exp{\left(\epsilon_i\right)}\ Y_i &= \left(X_i\right)^{\beta_2}\exp{\left(\beta_1\right)}\cdot\exp{\left(\epsilon_i\right)}\ \end{align} Which gives us:That's a lot of different shapes. A line (whose slope would be determined by $\exp{\left(\beta_1\right)}$, so which can have any positive slope), a hyperbola, a parabola, and a "square-root-like" shape. I've drawn it with $\beta_1=0$ and $\epsilon=0$, but in a real application neither of these would be true, so that the slope and the height of the curves at $X=1$ would be controlled by those rather than set at 1. As TrynnaDoStat mentions, the log-log form "draws in" big values which often makes the data easier to look at and sometimes normalizes the variance across observations. The coefficient $\beta_2$ is interpreted as an elasticity. It is the percentage increase in $Y$ from a one percent increase in $X$. If $X$ is a dummy variable, you include it without logging it. In this case, $\beta_2$ is the percent difference in $Y$ between the $X=1$ category and the $X=0$ category. If $X$ is time, again you include it without logging it, typically. In this case, $\beta_2$ is the growth rate in $Y$---measured in whatever time units $X$ is measured in. If $X$ is years, then the coefficient is annual growth rate in $Y$, for example. The slope coefficient, $\beta_2$, becomes scale-invariant. This means, on the one hand, that it has no units, and, on the other hand, that if you re-scale (i.e. change the units of) $X$ or $Y$, it will have absolutely no effect on the estimated value of $\beta_2$. Well, at least with OLS and other related estimators. If your data are log-normally distributed, then the log transformation makes them normally distributed. Normally distributed data have lots going for them. Statisticians generally find economists over-enthusiastic about this particular transformation of the data. This, I think, is because they judge my point 8 and the second half of my point 3 to be very important. Thus, in cases where the data are not log-normally distributed or where logging the data does not result in the transformed data having equal variance across observations, a statistician will tend not to like the transformation very much. The economist is likely to plunge ahead anyway since what we really like about the transformation are points 1,2,and 4-7. Share Improve this answer edited Apr 13, 2017 at 12:44 CommunityBot 1 answered Jul 11, 2014 at 18:51 BillBill 8,0743333 silver badges3434 bronze badges $\endgroup$ 1 9 $\begingroup$ These are standard points but it's very good to have them brought together concisely. Many accounts cover only some of these points. Small point: I think your contrast between economists' attitudes and statisticians' attitudes is a little overdone. For example, the importance of link over error family runs through generalised linear model literature, although it could do with more trumpeting. Keene, Oliver N. 1995. The log transformation is special. Statistics in Medicine 14: 811-819. DOI:10.1002/sim.4780140810 is another example. $\endgroup$ Nick Cox – Nick Cox 2015-02-21 10:09:32 +00:00 Commented Feb 21, 2015 at 10:09 Add a comment \| 48 $\begingroup$ First let's see what typically happens when we take logs of something that's right skew. The top row contains histograms for samples from three different, increasingly skewed distributions. The bottom row contains histograms for their logs. You can see that the center case ($y$) has been transformed to something close to symmetry, while the more mildly right skew case ($x$) is now somewhat left skew. One the other hand, the most skew variable ($z$) is still (slightly) right skew, even after taking logs. If we wanted our distributions to look more symmetric, and perhaps more normal, the transformation clearly improved the second and third case. We can see that this might help at least sometimes to reduce the amount of right-skewness. [However, a note of caution; in many cases you may be better not trying to achieve symmetry but rather in considering a more suitable model for your variables. Sometimes log transformations make sense regardless of distributional considerations, and often when that's the case, you happen to get greater symmetry at the same time, which is nice but rarely the central goal.] So why does it work? Note that when we're looking at a picture of the distributional shape, we're not considering the mean or the standard deviation - that just affects the labels on the axis. So we can imagine looking at some kind of "standardized" variables (while remaining positive, all have similar location and spread, say) Taking logs "pulls in" more extreme values on the right (high values) relative to the median, while values at the far left (low values) tend to get stretched back, further away from the median. In the first diagram, $x$, $y$ and $z$ all have means near 178, all have medians close to 150, and their logs all have medians near 5. When we looks at the original data, a value at the far right - say around 750 - is sitting far above the median. In the case of $y$, it's 5 interquartile ranges above the median. But when we take logs, it gets pulled back toward the median; after taking logs it's only about 2 interquartile ranges above the median. Meanwhile a low value like 30 (only 4 values in the sample of size 1000 are below it) is a bit less than one interquartile range below the median of $y$. When we take logs, it's again about two interquartile ranges below the new median. It's no accident that the ratio of 750/150 and 150/30 are both 5 when both log(750) and log(30) ended up about the same distance away from the median of log(y). That's how logs work - converting constant ratios to constant differences. It's not always the case that the log will help noticeably. For example if you take say a lognormal random variable and shift it substantially to the right (i.e. add a large constant to it) so that the mean became large relative to the standard deviation, then taking the log of that would make very little difference to the shape. It would be less skew - but barely. But other transformations - the square root, say - will also pull large values in like that. Why are logs in particular, more popular? I touched on one reason just at the end of the previous part - constant ratios tend to constant differences. This makes logs relatively easy to interpret, since constant percentage changes (like a 20% increase to every one of a set of numbers) become a constant shift. So a decrease of $-0.162$ in the natural log is a 15% decrease in the original numbers, no matter how big the original number is. A lot of economic and financial data behaves like this, for example (constant or near-constant effects on the percentage scale). The log scale makes a lot of sense in that case. Moreover, as a result of that percentage-scale effect. the spread of values tends to be larger as the mean increases - and taking logs also tends to stabilize the spread. That's usually more important than normality. Indeed, all three distributions in the original diagram come from families where the standard deviation will increase with the mean, and in each case taking logs stabilizes variance. [This doesn't happen with all right skewed data, though. It's just very common in the sort of data that crops up in particular application areas.] There are also times when the square root will make things more symmetric, but it tends to happen with less skewed distributions than I use in my examples here. We could (fairly easily) construct another set of three more mildly right-skew examples, where the square root made one left skew, one symmetric and the third was still right-skew (but a bit less skew than before). What about left-skewed distributions? If you applied the log transformation to a symmetric distribution, it will tend to make it left-skew for the same reason it often makes a right skew one more symmetric - see the related discussion here. Correspondingly, if you apply the log-transformation to something that's already left skew, it will tend to make it even more left skew, pulling the things above the median in even more tightly, and stretching things below the median down even harder. So the log transformation wouldn't be helpful then. See also power transformations/Tukey's ladder. Distributions that are left skew may be made more symmetric by taking a power (greater than 1 -- squaring say), or by exponentiating. If it has an obvious upper bound, one might subtract observations from the upper bound (giving a right skewed result) and then attempt to transform that. Sometimes, transformation just doesn't seem to help. Why not? Here's two common issues, but they're not exhaustive. (i) The impact of shifting Sometimes taking logs (for example) seems to work quite well on a right skewed distribution but another time it doesn't seem to work at all with a distribution that's not even as skewed as the first one. This may seem counterintuitive given the diagrams above. Here's the kicker. While adding a constant to a variable doesn't change its skewness, it very much changes the impact of a power-type transformation (such as those on the Tukey-ladder), including the log-transform. The more you shift it up the less the effect of a transformation like log or square root. Because of this sort of effect, you can easily have two variables that have exactly the same skewness, and find that taking logs will work nicely on one and barely improve things at all on the other. The same goes for square roots, and so forth. (ii) Discreteness With a discrete variable, a transformation can move the probability spikes around, but the values that are together will always stay the same (all the values at 1 go to whatever 1 transforms to). A monotonic transformation, including log and square root, will leave them in the same order, to boot. For example, if you had say $70\%$ of the distribution at $1$, $20\%$ at $2$ and the rest spread across higher values, then no matter which monotonic transformation you apply, the two lowest values would still have $70\%$ and $20\%$ of the values respectively. Squashing up the $10\%$ that's in the far right tail just won't help much. I will take as read that you don't use transformations that "lose" values, for what I hope are obvious reasons. So no $\log(0)$'s for example. Share Improve this answer edited Sep 7, 2022 at 23:41 answered Jul 12, 2014 at 5:10 Glen_bGlen_b 296k3737 gold badges674674 silver badges1.1k1.1k bronze badges $\endgroup$ 1 $\begingroup$ Thank you Glen_b for this excellent answer. You give us empirical data to illustrate and then give an intuitive explanation for why/how this transformation works. Much appreciated. $\endgroup$ Ram – Ram 2018-10-08 19:18:13 +00:00 Commented Oct 8, 2018 at 19:18 Add a comment \| 9 $\begingroup$ The log function essentially de-emphasizes very large values. Look at the image below which shows $y = ln(x)$. Notice how large values on the $x$-axis are relatively smaller on the y-axis. Now, in a right-skewed distribution you have a few very large values. The log transformation essentially reels these values into the center of the distribution making it look more like a Normal distribution. Share Improve this answer answered Jul 11, 2014 at 15:06 TrynnaDoStatTrynnaDoStat 8,28933 gold badges3030 silver badges4343 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ All of these answers are sales pitches for the natural log transformation. There are caveats to its use, caveats that are generalizable to any and all transformations. As a general rule, all mathematical transformations reshape the PDF of the underlying raw variables whether acting to compress, expand, invert, rescale, whatever. The biggest challenge this presents from a purely practical point of view is that, when used in regression models where predictions are a key model output, transformations of the dependent variable, Y-hat, are subject to potentially significant retransformation bias. Note that natural log transformations are not immune to this bias, they're just not as impacted by it as some other, similar acting transformations. There are papers offering solutions for this bias but they really don't work very well. In my opinion, you're on much safer ground not messing with trying to transform Y at all and finding robust functional forms that allow you to retain the original metric. For instance, besides the natural log, there are other transformations that compress the tail of skewed and kurtotic variables such as the inverse hyperbolic sine or Lambert's W. Both of these transformations work very well in generating symmetric PDFs and, therefore, Gaussian-like errors, from heavy-tailed information, but watch out for the bias when you try to bring the predictions back into the original scale for the DV, Y. It can be ugly. Share Improve this answer edited Jun 11, 2015 at 9:53 answered Jun 11, 2015 at 1:40 user78229user78229 10.9k22 gold badges2626 silver badges4848 bronze badges $\endgroup$ 15 3 $\begingroup$ This seems to end focused on what to do with heavy-tailed distributions (by kurtotic you mean possessing high kurtosis). I think you need to spell out how that relates to the question. Similarly, how Lambert's $W$ relates to the question isn't clear. I don't get how transformation bias is less of a possible problem for the logarithmic transformation than for related transformations (which ones?) as in this respect and in others the logarithmic behaves as you would expect as a member of a wider family, for example in being intermediate in effect between the square root and the reciprocal. $\endgroup$ Nick Cox – Nick Cox 2015-06-11 07:28:15 +00:00 Commented Jun 11, 2015 at 7:28 3 $\begingroup$ We all chafe at various aspects of the rules, but many of us continue to interact here because we have come to see the wisdom of them and have found constructive ways to work around the apparent restrictions. This rule is fundamental: a post that does not answer a question doesn't belong. It tends to keep each thread coherent, limited, clean, and on-topic. It is key to creating material that tends to be more useful and interesting than you will find on any other Q&A site. $\endgroup$ whuber – whuber ♦ 2015-06-11 15:40:24 +00:00 Commented Jun 11, 2015 at 15:40 3 $\begingroup$ You have tinkered with this but in my view it remains very problematic as an answer. 1. You're broadening the question in several ways, e.g. by bringing in heavy-tailed distributions as well. That can be a reasonable thing to do in some threads, but here is a well-focused thread with high-quality answers and the extra answer here is by and large muddying the waters. When there are existing good answers to a question there has to be a really good reason for a new answer. $\endgroup$ Nick Cox – Nick Cox 2015-06-11 23:46:16 +00:00 Commented Jun 11, 2015 at 23:46 4 $\begingroup$ 2. The assertions about transformation bias remain arm-waving; there's no technical precision to the answer that matches the claims, including the mysterious assertion that the log is less problematic than other similar transformations. $\endgroup$ Nick Cox – Nick Cox 2015-06-11 23:46:54 +00:00 Commented Jun 11, 2015 at 23:46 4 $\begingroup$ 3. The detail about Lambert's $W$ remains cryptic. More broadly, the message is that transformations are dubious except that asinh and Lambert's can be good. This seems contradictory and isn't well explained. You are clearly very knowledgeable but this needs a straighter expository style to be valuable. Hence I can't upvote this in good conscience. Your earlier decision to remove it was better in my view. Here and elsewhere I don't think you're quite catching CV style: there isn't a rigid prescription but answers have to be focused; chatty, discursive posts don't usually fit well. $\endgroup$ Nick Cox – Nick Cox 2015-06-11 23:51:25 +00:00 Commented Jun 11, 2015 at 23:51 \| Show 10 more comments 1 $\begingroup$ Many interesting points have been made. A few more? 1) I would suggest that another issue with linear regression is that the 'left hand side' of the regression equation is E(y) : the expected value. If the error distribution is not symmetrical, then merits for the study of the expected value are weak. The expected value is not of central interest when the errors are asymmetrical. One could explore quantile regression instead. Then the study of, say, the median, or other percentage points might be worthy even if the errors are asymmetrical. 2) If one elects to transform the response variable, then one may wish to transform one of more of the explanatory variables with the same function. For example, if one has a 'final' outcome as response, then one might have a 'baseline' outcome as an explanatory variable. For interpretation, it makes sense the transform 'final' and 'baseline' with the same function. 3) The main argument for transforming an explanatory variable is often around the linearity of the response - explanatory relationship. These days, one can consider other options like restricted cubic splines or fractional polynomials for the explanatory variable. There is certainly often a certain clarity if linearity can be found though. Share Improve this answer answered Jun 10, 2019 at 23:03 ghfickghfick 9122 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions regression distributions data-transformation skewness See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 1 What would a suitable justification be to use 'log(exports)' as a dependent variable instead of solely 'exports'? Why transform distribution could help model accuracy? If $X$ is normally distributed, can $\log(X)$ also be normally distributed? Back transformation of an MLR model Does a log transform always bring a distribution closer to normal? Transform normal distribution to skewed distribution without changing its support 5 Strictly positive response in regression: what should my "default" model be? 0 When does taking the log transformation of a univariate not remove skew? 0 Probability distribution & Interpretation 1 Why use the square transform to reduce left skew? See more linked questions Related 1 Transforming highly left skewed x with log((-1)x+max(x)) in linear regression 5 How do I transform an extremely skewed distribution to use it for linear regression? 3 Choosing appropriate regression model for a dependent variable with many zeros and highly right skewed distribution Hot Network Questions Switch between math versions but without math versions Why include unadjusted estimates in a study when reporting adjusted estimates? RTC battery and VCC switching circuit How do trees drop their leaves? What meal can come next? What happens if you miss cruise ship deadline at private island? 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6202	https://cp-algorithms.com/geometry/circle-line-intersection.html	Circle-Line Intersection¶ Given the coordinates of the center of a circle and its radius, and the equation of a line, you're required to find the points of intersection. Solution¶ Instead of solving the system of two equations, we will approach the problem geometrically. This way we get a more accurate solution from the point of view of numerical stability. We assume without loss of generality that the circle is centered at the origin. If it's not, we translate it there and correct the $C$ constant in the line equation. So we have a circle centered at $(0,0)$ of radius $r$ and a line with equation $Ax+By+C=0$. Let's start by find the point on the line which is closest to the origin $(x_0, y_0)$. First, it has to be at a distance Second, since the vector $(A, B)$ is perpendicular to the line, the coordinates of the point must be proportional to the coordinates of this vector. Since we know the distance of the point to the origin, we just need to scale the vector $(A, B)$ to this length, and we'll get: The minus signs are not obvious, but they can be easily verified by substituting $x_0$ and $y_0$ in the equation of the line. At this stage we can determine the number of intersection points, and even find the solution when there is one or zero points. Indeed, if the distance from $(x_0, y_0)$ to the origin $d_0$ is greater than the radius $r$, the answer is zero points. If $d_0=r$, the answer is one point $(x_0, y_0)$. If $d_0<r$, there are two points of intersection, and now we have to find their coordinates. So, we know that the point $(x_0, y_0)$ is inside the circle. The two points of intersection, $(a_x, a_y)$ and $(b_x, b_y)$, must belong to the line $Ax+By+C=0$ and must be at the same distance $d$ from $(x_0, y_0)$, and this distance is easy to find: Note that the vector $(-B, A)$ is collinear to the line, and thus we can find the points in question by adding and subtracting vector $(-B,A)$, scaled to the length $d$, to the point $(x_0, y_0)$. Finally, the equations of the two points of intersection are: Had we solved the original system of equations using algebraic methods, we would likely get an answer in a different form with a larger error. The geometric method described here is more graphic and more accurate. Implementation¶ As indicated at the outset, we assume that the circle is centered at the origin, and therefore the input to the program is the radius $r$ of the circle and the parameters $A$, $B$ and $C$ of the equation of the line. Practice Problems¶
6203	https://www.newark.com/decimal-fraction-table	Skip to main content We will no longer be supporting Internet Explorer, to ensure you have the best possible experience we recommend using a modern browser. Your quotes are ready to view Log InRegister Hi My Account My Orders Order History & Tracking Order Preferences Buying Tools Quotes BOM Upload Saved Basket Favorites My Profile Account Summary Profile Information Log outLog in Close This item has been added to your cart Newark Part No. Manufacturer Quantity Pack Size Unit Price Re-reeling Charge Newark Part No. Manufacturer Quantity Unit Price items have been added to your basket Review your cart for details on availability and shipping Continue Shopping Go To Checkout 0 items $0.00 Go to Cart Your Cart is Empty \| Description \| Newark Part No. \| Manufacturer \| Quantity \| Total \| --- --- Only the last 10 items added are shown Show all Fraction and Decimal Conversion Table Newark's decimal and fraction conversion chart gives you the decimal equivalent for commonly used fractions along with other fractions that express the same value (2/4 and 3/6, for example) as well as lowest common denominators. Decimal and Fraction Conversion Decimal and Fraction Conversion Chart \| Fraction \| Equivalent Fractions \| \| \| \| \| \| \| \| \| \| \| Decimal \| --- --- --- --- --- --- \| 1/2 \| 2/4 \| 3/6 \| 4/8 \| 5/10 \| 6/12 \| 7/14 \| 8/16 \| 9/18 \| 10/20 \| 11/22 \| 12/24 \| .5 \| \| 1/3 \| 2/6 \| 3/9 \| 4/12 \| 5/15 \| 6/18 \| 7/21 \| 8/24 \| 9/27 \| 10/30 \| 11/33 \| 12/36 \| .333 \| \| 2/3 \| 4/6 \| 6/9 \| 8/12 \| 10/15 \| 12/18 \| 14/21 \| 16/24 \| 18/27 \| 20/30 \| 22/33 \| 24/36 \| .666 \| \| 1/4 \| 2/8 \| 3/12 \| 4/16 \| 5/20 \| 6/24 \| 7/28 \| 8/32 \| 9/36 \| 10/40 \| 11/44 \| 12/48 \| .25 \| \| 3/4 \| 6/8 \| 9/12 \| 12/16 \| 15/20 \| 18/24 \| 21/28 \| 24/32 \| 27/36 \| 30/40 \| 33/44 \| 36/48 \| .75 \| \| 1/5 \| 2/10 \| 3/15 \| 4/20 \| 5/25 \| 6/30 \| 7/35 \| 8/40 \| 9/45 \| 10/50 \| 11/55 \| 12/60 \| .2 \| \| 2/5 \| 4/10 \| 6/15 \| 8/20 \| 10/25 \| 12/30 \| 14/35 \| 16/40 \| 18/45 \| 20/50 \| 22/55 \| 24/60 \| .4 \| \| 3/5 \| 6/10 \| 9/15 \| 12/20 \| 15/25 \| 18/30 \| 21/35 \| 24/40 \| 27/45 \| 30/50 \| 33/55 \| 36/60 \| .6 \| \| 4/5 \| 8/10 \| 12/15 \| 16/20 \| 20/25 \| 24/30 \| 28/35 \| 32/40 \| 36/45 \| 40/50 \| 44/55 \| 48/60 \| .8 \| \| 1/6 \| 2/12 \| 3/18 \| 4/24 \| 5/30 \| 6/36 \| 7/42 \| 8/48 \| 9/54 \| 10/60 \| 11/66 \| 12/72 \| .166 \| \| 5/6 \| 10/12 \| 15/18 \| 20/24 \| 25/30 \| 30/36 \| 35/42 \| 40/48 \| 45/54 \| 50/60 \| 55/66 \| 60/72 \| .833 \| \| 1/7 \| 2/14 \| 3/21 \| 4/28 \| 5/35 \| 6/42 \| 7/49 \| 8/56 \| 9/63 \| 10/70 \| 11/77 \| 12/84 \| .143 \| \| 2/7 \| 4/14 \| 6/21 \| 8/28 \| 10/35 \| 12/42 \| 14/49 \| 16/56 \| 18/63 \| 20/70 \| 22/77 \| 24/84 \| .286 \| \| 3/7 \| 6/14 \| 9/21 \| 12/28 \| 15/35 \| 18/42 \| 21/49 \| 24/56 \| 27/63 \| 30/70 \| 33/77 \| 36/84 \| .429 \| \| 4/7 \| 8/14 \| 12/21 \| 16/28 \| 20/35 \| 24/42 \| 28/49 \| 32/56 \| 36/63 \| 40/70 \| 44/77 \| 48/84 \| .571 \| \| 5/7 \| 10/14 \| 15/21 \| 20/28 \| 25/35 \| 30/42 \| 35/49 \| 40/56 \| 45/63 \| 50/70 \| 55/77 \| 60/84 \| .714 \| \| 6/7 \| 12/14 \| 18/21 \| 24/28 \| 30/35 \| 36/42 \| 42/49 \| 48/56 \| 54/63 \| 60/70 \| 66/77 \| 72/84 \| .857 \| \| 1/8 \| 2/16 \| 3/24 \| 4/32 \| 5/40 \| 6/48 \| 7/56 \| 8/64 \| 9/72 \| 10/80 \| 11/88 \| 12/96 \| .125 \| \| 3/8 \| 6/16 \| 9/24 \| 12/32 \| 15/40 \| 18/48 \| 21/56 \| 24/64 \| 27/72 \| 30/80 \| 33/88 \| 36/96 \| .375 \| \| 5/8 \| 10/16 \| 15/24 \| 20/32 \| 25/40 \| 30/48 \| 35/56 \| 40/64 \| 45/72 \| 50/80 \| 55/88 \| 60/96 \| .625 \| \| 7/8 \| 14/16 \| 21/24 \| 28/32 \| 35/40 \| 42/48 \| 49/56 \| 56/64 \| 63/72 \| 70/80 \| 77/88 \| 84/96 \| .875 \| \| 1/9 \| 2/18 \| 3/27 \| 4/36 \| 5/45 \| 6/54 \| 7/63 \| 8/72 \| 9/81 \| 10/90 \| 11/99 \| 12/108 \| .111 \| \| 2/9 \| 4/18 \| 6/27 \| 8/36 \| 10/45 \| 12/54 \| 14/63 \| 16/72 \| 18/81 \| 20/90 \| 22/99 \| 24/108 \| .222 \| \| 4/9 \| 8/18 \| 12/27 \| 16/36 \| 20/45 \| 24/54 \| 28/63 \| 32/72 \| 36/81 \| 40/90 \| 44/99 \| 48/108 \| .444 \| \| 5/9 \| 10/18 \| 15/27 \| 20/36 \| 25/45 \| 30/54 \| 35/63 \| 40/72 \| 45/81 \| 50/90 \| 55/99 \| 60/108 \| .555 \| \| 7/9 \| 14/18 \| 21/27 \| 28/36 \| 35/45 \| 42/54 \| 49/63 \| 56/72 \| 63/81 \| 70/90 \| 77/99 \| 84/108 \| .777 \| \| 8/9 \| 16/18 \| 24/27 \| 32/36 \| 40/45 \| 48/54 \| 56/63 \| 64/72 \| 72/81 \| 80/90 \| 88/99 \| 96/108 \| .888 \| Calculator & Conversion Charts Menu Technical Resources Articles, Solution Guides, Webinars, and more. 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6204	http://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f19/www/scribes/lecture03.pdf	15-859FF: Coping with Intractability CMU, Fall 2019 Lecture #3: Reductions for Problems in FPT September 11, 2019 Lecturer: Venkatesan Guruswami Scribe: Anish Sevekari 1 Vertex Cover on Hypergraphs We deﬁne a d-regular hypergraph H = (V, F) where F is a family of subsets of V of size d. The vertex cover problem on hypergraphs is to ﬁnd the smallest set S such that every edge X ∈F intersects S at least once. We parameterize the vertex cover problem by size of the optimum solution. Then the parameterized version of the problem is given a hypergraph H and an integer k, does H have a vertex cover of size k? We will now give a dkpoly(n) algorithm that solves the vertex cover problem and a O(d2d!kd) kernel for d-regular Hypergraph Vertex Cover problem. 1.1 An algorithm for Hypergraph Vertex Cover The algorithm is a simple generalization of the 2kpoly(n) algorithm for vertex cover over graphs. We pick any edge, then branch on the vertex on this edge which is in the vertex cover. Algorithm 1 VertexCover(H = (V, F), k) 1: if k = 0 and \|F\| > 0 then 2: return ⊥ 3: end if 4: Pick any edge e ∈F. 5: for v ∈e do 6: H′ = (V \ {v}, F \ {f ∈F : v ∈f}) 7: if VertexCover(H′, k) ̸=⊥then 8: return {v} ∪VertexCover(H′, k −1) 9: end if 10: end for 11: return ⊥ Note that the recursion depth of this algorithm is k, and we branch on at most d points inside each call. Therefore, total size of the recurrence tree is dk. Hence, this algorithm terminates in dkpoly(n) time, where n = \|V \| + \|F\| is size of the input. We think of d as a ﬁxed constant, and hence \|F\| ∈poly(\|V \|). Further, if this algorithm returns a set S, then the set F \ {f ∈F : f ∩S ̸= ∅} is empty. And therefore, S hits all the edges. Therefore, if an algorithm returns a set, then it is a vertex cover of size less than k. Now we need to show that if H has a vertex cover of size ≤k, then the algorithm above ﬁnds it. We will do this by induction. The base case k = 0 is handled in the algorithm. Assume that the algorithm works correctly for all t < k. Let S be a vertex cover of H with \|S\| ≤k. Then for any edge e ∈F, S ∩e ̸= ∅. Let v ∈S ∩e. Therefore, the graph H′ = (V \ {v}, F \ {f ∈F : c ∈f}) has vertex cover S {v} of size less than k −1. By induction hypothesis, VertexCover(H′, k −1) 1 Figure 3.1: A (k + 1)-sunﬂower. correctly ﬁnds a vertex cover of H′ of size at most k −1. Since the algorithm returns {v} ∪ VertexCover(H′, k −1) or something better, it ﬁnds a vertex cover of size at most k for H correctly. 1.2 Kernel for Hypergraph Vertex Cover The kernel presented here is based on Sunﬂower Lemma. First, we deﬁne what a sunﬂower is Deﬁnition 3.1. A family of sets S1, . . . , Sk+1 is called a k + 1-sunﬂower if and only if for all i, j Si ∩Sj = k+1 \ t=1 St Note that if Tk+1 t=1 St is empty, then k + 1-sunﬂower is set of k + 1 disjoint sets/edges. This in particular implies that F doesn’t have a vertex cover of size k. Lemma 3.2 (Sunﬂower Lemma). [ER60] Let k ≥2 and F be a family of sets such that each element of F has size d. Then if \|F\| > d! · kd, then F contains a k + 1 sunﬂower. Proof. We will induct on d. If d = 1, then we want to ﬁnd k + 1 distinct sets in F, since any two sets in F would be disjoint. This holds when F > k. This covers the base case. Greedily pick disjoint sets from F. If we can ﬁnd k + 1 such sets, then we get a k + 1-sunﬂower. Therefore, suppose this process stops after picking l ≤k sets, say T1, . . . , Tl. Then Sl i=1 Ti = T has size at most ld ≤kd, and every set in F intersects T. Since \|F\| ≥d! · kd, there is an element v which is contained in at least \|F\|/ld > (d −1)! · kd−1 sets. Let G = {X {v} : X ∈F, v ∈X}. Note that each set in G has size d−1. Since \|G\| > (d−1)!·kd−1 G contains a k + 1-sunﬂower by induction hypothesis. Let S1, . . . , Sk+1 be a k + 1-sunﬂower in G, then S1 ∩{v}, . . . , Sk+1 ∩{v} is a k + 1-sunﬂower in G, which completes the induction step. Note that we can compute T and G in O(k\|F\|) time. Then we have to solve a smaller problem where value of d is dropped by 1. Therefore, we can actually ﬁnd a sunﬂower in O(dk\|F\|) time. This also proves the lemma below. Lemma 3.3. If \|F\| > d! · kd then we can ﬁnd k + 1-sunﬂower in poly(\|V \|, \|F\|) time. 2 The known lower bound for sunﬂower lemma is cdkd for some constant c where d is size of the sets in the family F. Erd˝ os and Rado conjecture in [ER60] that upper bound should in fact be c(k)d for some constant c(k) that depends only on k. A recent improvement on sunﬂower lemma managed to get the upper bound down. Lemma 3.4 (Improved Sunﬂower Lemma [ALWZ19]). In the same setting as the Sunﬂower Lemma, if \|F\| > (log d)d((k + 1) · log log d)O(d) then F contains a sunﬂower. Now we use the sunﬂower lemma to get a kernel for Hypergraph Vertex Cover problem Theorem 3.5 ([CFK+15], Theorem 2.26). d-regular Hypergraph Vertex Cover admits a kernel with at most d! · kd sets and at most d! · kd · d2 elements. Proof. Let H = (V, F) be the hypergraph. We will relax the condition that each set in F has size d, and only insist that each set in F has size ≤d. First we observe that if F contains a sunﬂower {S1, . . . , Sk+1}, then every vertex cover T of size ≤k must intersect the core Y = Tk+1 t=1 St. If T does not intersect S, then it must intersect St \ Y for all t. Since these sets are disjoint, that would imply that \|T\| ≥k + 1, which is a contradiction. For the reduction, we do the following: If F > d! · kd, then we ﬁnd a k + 1-sunﬂower using the sunﬂower lemma. Let S1, . . . , Sk+1 be the sunﬂower with core Y . Then we return the instance H′ = (V ′, F′) where V ′ = V \ (Sk+1 t=1 Si \ Y ) and F′ = (F ∪{Y }) \ {S1, . . . , Sk+1}. This process continues as long as \|F\| > d! · kd. The argument in paragraph above implies that the old instance had a vertex cover of size ≤k if the new instance has a vertex cover of size ≤k, and any vertex cover of the new instance is a vertex cover of the old instance. Therefore, the two instances are equivalent. At each step, we are decreasing \|F\|, as long as k ≥2. Therefore, the process eventually terminates. Suppose process ends with H = (V, F), then by sunﬂower lemma, number of sets with size exactly d′ ≤d in F is less than d′! · kd′. And hence \|F\| ≤d! · kd · d. Now we throw away all the vertices v that are not in any edge. Since each element in F contains at most d elements, number of elements left is at most d! · kd · d2. Hence, the ﬁnal instance has size at most O(d! · kd · d2), giving us the required kernel. 2 Hitting Set and FPT reductions The Set Cover or Hitting Set problem is very similar to Hypergraph Vertex Cover prob-lem. Only diﬀerence is that now we remove the constraint that all edges have ﬁxed constant size d. More precisely, the problem is Given a set V and F ⊂P(V ), is there a set S ⊂V such that \|S\| ≤k and S ∩X ̸= ∅for all X ∈F. The branching algorithm may need to branch into n diﬀerent branches at each point, and hence takes nk time, which is worse than n k runtime of the brute force algorithm. In fact, nothing much better is known. We strongly believe that there are no FPT algorithms to solve the Hitting Set problem. The main reason for this is that Clique can be reduced to Hitting Set by an FPT reduction. Clique ≤FPT Hitting Set In fact, there is a whole hierarchy of complexity classes which “classiﬁes” the hardnesss of the problem in FPT sense. These classes are called W, W, . . .. The class W contains all the problems that are solvable in FPT time. We will formally deﬁne all the classes later in the course. 3 It turns out that Clique is the W-complete, and Hitting Set is W-complete. Therefore, and FPT algorithm for Hitting Set would imply W = W = W, which we believe to be false. This belief is actually independent (rather not known to be dependent) of the belief that P ̸= NP, that is we have not proved P ̸= NP implies W ̸= W. Deﬁnition 3.6 (FPT reduction). φ : Σ∗× N →Σ∗× N is called a FPT reduction from A to B (A ≤FPT B) if and only if 1. (x, k) ∈A ⇔φ((x, k)) ∈B 2. There exists a computable function g : N →N such that if φ((x, k)) = (y, k′) then k′ < g(k) for all x, k. 3. φ can be computed in f(k)\|⟨x, k⟩\|c time. Where \|⟨x, k⟩\|, the length concatenation of repre-sentation of x and k, is the total input size. (iii) and (ii) imply that if A ≤FPT B and B ∈FPT then A ∈FPT. Suppose we have an algorithm that solves instance (y, k′) of B in time h(k′)\|⟨y, k′⟩\|c′, then we can solve an instance of A in time h(g(k))(\|⟨x, k⟩\|c + f(k)c′\|⟨x, k⟩\|cc′ ≤h(g(k))f(k) · \|⟨x, k⟩\|cc′+O(1) by converting it to an instance of B using the reduction. Therefore, A ∈FPT. Remark 3.7. Vertex Cover and IndSet (Independent Set) have a trivial polynomial time reduction, since a graph has vertex cover of size k if and only if its complement has an independent set of size n −k, but this is not an FPT reduction, since n −k is not a bounded by any function of k and depends on n as well. Theorem 3.8. (Theorems 13.5 to 13.9 in [CFK+15]) Clique ≤FPT Multicolored Clique ≤FPT Hitting Set The Multicolored Clique problem is deﬁned as follows: Given a k-partite graph, does it have a k-clique? We parameterize the problem by k. Note that since graph is k-partite, we must have one vertex in every part. Proof. First, we will give the reduction Clique ≤FPT Multicolored Clique. The reduction map is given by ⟨G = (V, E), k⟩7→⟨H = (V1 ⨿. . . ⨿Vk), e E, k⟩ where Vi = V × {i}, and {(v1, i), (v2, j)} ∈e E if and only if {v1, v2} ∈E and i ̸= j. Here ⨿denotes the disjoint union. The map can clearly be computed in poly(n, k) time and k′ = k. Therefore, this reduction satisﬁes the requirements (ii) and (iii). We only need to show that (i) holds as well. If G has as clique {v1, . . . , vk} then (vi, i) and (vj, j) are connected by an edge by deﬁnition of e E. Therefore, the set {(v1, 1), . . . , (vk, k)} is a clique in H. On the other hand, if {(v1, 1), . . . , (vk, k)} is a clique in H, then {vi, vj} ∈E for all i ̸= j. This implies that {v1, . . . , vk} forms a k-clique in G. Hence, two instances of corresponding problems are equivalent and hence (i) also holds, making this a valid FPT reduction. Now we will reduce Multicolored Clique into Dominating Set, which is a special case of the Hitting Set problem. The Dominating Set problem is deﬁned as follows: Given a graph G = (V, E) is there a set S ⊂V of size k such that for all v / ∈S, ∃u ∈S such that {u, v} ∈E? 4 Note that a set S ⊂V satisﬁes this condition if and only if S ∩(N(v) ∪{v}) ̸= ∅for all vertices v ∈V , where N(v) is the set of neighbors of v. Hence, Dominating Set is a special case of Hitting Set problem. That is, Dominating Set ≤FPT HItting Set. Hence, it suﬃces to show that Multicolored Clique ≤FPT Dominating Set. The reduction map is given by ⟨G = (V1 ⨿. . . Vk, E), k⟩7→(H, k′ = k) where H is constructed from empty graph as follows: 1. For each i, add a copy of Vi to H, and add every edge inside this set to H. 2. For each i, add two new vertices pi, qi and connect them to every vertex in Vi. 3. For each pair e = {u, v} / ∈E, where u ∈Vi and v ∈Vj add a vertex we, and add edges {we, w} for all w ∈Vi \ {u} and for all w ∈Vj \ {v}. The graph H can be constructed from G in poly(n) time. Further, k′ = k. Therefore, (ii) and (iii) hold. We just need to show the correctness of the reduction. Suppose there is a k-clique {v1, . . . , vk} ⊂G. Then each vi must belong to a diﬀerent Vj. Let’s assume that vi ∈Vi. Then {v1, . . . , vk} is a dominating set for H. For every u ∈Vi \ {vi}, {u, vi} is an edge in H. For each i, {pi, vi} and {qi, vi} are edges in H. And for any we ∈H, where e = {u, v} with u ∈Vi and v ∈Vj, either vi ̸= u or vj ̸= v since e is a non-edge. Suppose u ̸= vi, and hence we is connected to vi. Therefore, the set {v1, . . . , vk} dominates all the vertices in H, and is a valid dominating set of size k. On the other hand, if S is a dominating set in H of size k, then S either contains a vertex from Vi or S contains both pi, qi as it needs to hit the neighbor set of pi and qi. Therefore, S has size at least k. But, since S has size exactly k, it cannot pick both pi, qi for any i. Hence, S must contain exactly one element from each Vi. Further, if there are u, v ∈S such that u ∈Vi, v ∈Vj and e = {u, v} is a non-edge in G, then neither u nor v are connected to we in H. As we is connected only to the vertices in Vi and Vj, this implies that we is not dominated by S. Therefore, for all u, v ∈S, {u, v} is an edge in G. Therefore, S forms a k-clique in G. This proves the other direction for the correctness of the reduction. Hence this is a valid FPT reduction. References [ALWZ19] Ryan Alweiss, Shachar Lovett, Kewen Wu, and Jiapeng Zhang. Improved bounds for the sunﬂower lemma, 2019. 3.4 [CFK+15] Marek Cygan, Fedor V. Fomin, Lukasz Kowalik, Daniel Lokshtanov, Marx D´ aniel, Marcin Pilipczuk, Micha l Pilipczuk, and Saket Saurabh. Parametrized algorithms. Springer, 2015. 3.5, 3.8 [ER60] Paul Erd˝ os and Richard Rado. Intersection theorems for systems of sets. Journal of the London Mathematical Society, 35(1):85–90, 1960. 3.2, 1.2 5
6205	https://www.mindgems.com/info/file-download-time-calculator/	Download Time Calculator Calculate the Time Needed to Download/Upload a File How long will it take to complete a file download given a specific Internet download speed? Use the Download Time Calculator to accurately calculate the time needed to download a file for multiple Internet connection speeds. Enter the downloaded file size in the field below and specify the file size measure units in the drop-box. Check also the free online File Size calculator and converter. Test your download speed at SpeedTest.NET. Downloading 100 Gigabytes at 10 Mbps takes 22h 13m 20s \| \| \| --- \| \| Connection Speed \| Time (DD:HH:MM:SS) \| \| 9.6 Kbps \| \| \| 14.4 Kbps \| \| \| 28.8 Kbps \| \| \| 33.6 Kbps \| \| \| 56 Kbps \| \| \| 64 Kbps (ISDN) \| \| \| 128 Kbps (ISDN-2) \| \| \| 256 Kbps \| \| \| 512 Kbps \| \| \| 1.024 Mbps \| \| \| 1.544 Mbps (DS1, T1) \| \| \| 2.048 Mbps (E1, ISDN-32) \| \| \| \| \| --- \| \| Connection Speed \| Time (DD:HH:MM:SS) \| \| 10 Mbps (10Base-T) \| \| \| 25.6 Mbps (ATM25) \| \| \| 34 Mbps (E3) \| \| \| 45 Mbps (DS3, T3) \| \| \| 51 Mbps (OC1) \| \| \| 100 Mbps (100Base-T) \| \| \| 155 Mbps (OC3) \| \| \| 622 Mbps (OC12) \| \| \| 1 Gbps (1000Base-T) \| \| \| 2.4 Gbps (OC48) \| \| \| 10 Gbps (OC192) \| \| \| 100 Gbps \| \| Folder Size is a free drive size analyzer tool that will list the sizes of all folder and files on your computer's hard drives, USB drives, CDROMs, Floppy disks and network shares. Download this freeware tool and check the distribution of your disk space. Find largest files and folders, find empty files and folders, find longest file names and paths, print folders, print files list, and much more. DOWNLOAD NOW Compatible with Windows 11/10/8.1/8/7 (Both 32 & 64 Bit) Low on disk space? Duplicate File Finder FREE can solve this easily. Remove duplicate files and save disk space. Find duplicates, find similar documents, find files with similar or duplicate names, find files with the same sizes. DOWNLOAD NOW Compatible with Windows 11/10/8.1/8/7 (Both 32 & 64 Bit) Table of contents: Definition of "Download" What is Download Time? What is Download Speed? How to Calculate Download Time Automatically? How to Calculate Download Time Manually? How Long to Download 100GB Game? How Long Does it Take to Download 1 GB Internet Connection Types How Much Internet Speed do I Need? Definition of "Download" To download means to receive a file from the Internet. Such files typically are downloaded from a web server such as HTTP, FTP, mail server, or other. In contrast, upload means transferring a file to a remote server. The term "download" may also refer to a file that is offered for downloading, has been downloaded or the actual process of transferring the file.E.g. "A download is available on the product page.", "This is a very large download.", "A download is running so I will make a coffee." Downloading transfers files for local use and storage. Do not mistake this with streaming, where the data is used immediately, while the transmission is in progress. Such data usually is not stored long-term. Streaming usually is used to display video or audio in a web browser or application. For example, YouTube is a video streaming service. Such streaming services even prevent the storage and use of the downloaded data. Please note that downloading is not the same as data transfer. The term "data transfer" is used to denote moving or copying data between storage devices. Receiving data from the Internet is downloading. The duration of the download is known as download time. What is Download Time? Download time is the time needed to transfer a file from the Internet to a local computer, phone, tablet, or another Internet-connected device. The download time is determined by the connection speed between the two devices and the size of the transferred file. The connection speed is determined by the hardware capabilities of the two devices and also the ISP (Internet Service Provider) that they use. A faster connection speed will result in a faster download and shorter download time. Of course, a smaller file will require a shorter download time too. What is Download Speed? Download speed is the amount of data that is transferred per second between two Internet-connected devices. It may be expressed in various units which are multiples of a BIT - the smallest unit in computers. Those are bits, Kilobits, Megabits, Gigabits, and so on. To convert Kilobits to Megabits you have to divide them by 1024. To convert the Megabits to Gigabits you have to divide them into 1024 again and so on. The download speed is determined by your service provider and the hardware limits of the connection. For example, the highest download speed of a 100Base-T connection is 100 Megabits. Of course, that speed is the maximum that can be reached on such hardware and may be slower based on the Internet plan that you have signed for. The download and upload speeds are usually different. In most cases, the upload speed is much slower than the download speed. If the two are equal we call such connection symmetric. If the connection is not "shaped" (limited) by an ISP it is symmetric by default. Please note that the above test is not always correct. IS providers usually detect where your computer connects and provide higher speeds for tests if they detect that you are trying to test your connection speed. Use the Download Time Calculator above to enter your file size and calculate the time needed for the transfer. Multiple Internet connection speeds are listed and you can see the ones that are close to yours. How to Calculate Download Time Automatically To calculate the download time using the download time calculator, enter the file size in the edit field and select the measure units from the drop-down box on the right - bits, bytes, kilobytes, megabytes, gigabytes, terabytes, petabytes, exabytes, zettabytes, or yottabytes. The time needed to download or upload the file will be calculated automatically for multiple connection speeds. You can view all the download times in the table below the download time calculator. How to Calculate Download Time Manually To calculate download time manually you have to divide the file size by the connection speed. The more difficult part is that you have to convert the speed and time units to the desired ones. All the download speeds are in bits per second e.g. Mbps is Megabits per second, Gbps is Gigabits per second. In the above calculation those seconds are converted to days, hours, minutes and seconds for easier reading. To convert the seconds to minutes you have to divide them by 60 (seconds in a minute) and the whole part will be minutes and the fractional part that is left is the seconds. How Long to Download 100GB Game The time it takes to download a 100GB game depends on several factors, including your Internet connection speed. The download speed is typically measured in megabits per second (Mbps) or gigabits per second (Gbps). Here's a rough estimate: Download Speed: Download Speed 1 Mbps: Approximately 115 hours Download Speed 10 Mbps: Approximately 12 hours Download Speed 50 Mbps: Approximately 2.5 hours Download Speed 100 Mbps: Approximately 1.25 hours Download Speed 500 Mbps: Approximately 15 minutes Download Speed 1 Gbps: Approximately 7.5 minutes These are just rough download time estimates, and actual download times may vary. Additionally, other factors like network congestion, the efficiency of the server distributing the game, and any throttling imposed by your Internet service provider can also affect download speeds. It's important to note that these download estimates assume a consistent and uninterrupted download speed, which may not always be the case. If your Internet connection fluctuates or if there are interruptions during the download, it may take longer. To get a more accurate estimate, you can use online tools that measure your current Internet speed. Keep in mind that download speeds are usually expressed in megabits per second (Mbps), so you may need to convert the game size to megabits (divide the size in gigabytes by 0.125) to compare with your Internet speed. How Long Does it Take to Download 1 GB The time it takes to download a specific amount of data is influenced by the speed of your Internet connection. Internet speeds are measured in megabits per second (Mbps) or gigabits per second (Gbps), and they play a crucial role in determining the download duration. To provide a sense of scale, downloading 1 GB of data at varying speeds results in significantly different time-frames. A basic understanding of these estimates can help users gauge how long they might need to wait for downloads to complete based on their Internet connection. From sluggish connections taking hours to lightning-fast gigabit speeds reducing the wait to mere seconds, the download time is a direct reflection of the efficiency and capacity of one's Internet infrastructure. Download Speed: 1 Mbps: Approximately 2.27 hours Download Speed: 10 Mbps: Approximately 13.68 minutes Download Speed: 50 Mbps: Approximately 2.74 minutes Download Speed: 100 Mbps: Approximately 1.37 minutes Download Speed: 500 Mbps: Approximately 16.96 seconds Download Speed: 1 Gbps: Approximately 8.48 seconds These are rough estimates and actual download times may vary based on various factors such as network conditions and server responsiveness. Internet Connection Types Dial-up: Uses a standard telephone line and a modem. Speed: Up to 56 Kbps (kilobits per second). DSL (Digital Subscriber Line): Utilizes existing telephone lines. Speed: Ranges from 256 Kbps to 100 Mbps. Cable: Delivers Internet access over cable television lines. Speed: Ranges from 10 Mbps to 1 Gbps. Fiber-optic: Transmits data using light signals through thin strands of glass or plastic. Speed: Ranges from 100 Mbps to 10 Gbps or more. Satellite: Relies on satellite signals to provide Internet access. Speed: Varies, but can range from 1 Mbps to 100 Mbps. Fixed Wireless: Uses radio signals between towers and antennas. Speed: Varies, typically ranging from a few Mbps to several Mbps. Mobile Data (3G, 4G LTE, 5G): Provides Internet access through cellular networks. Speed varies across generations: 3G (Few hundred Kbps to few Mbps), 4G LTE (Several Mbps to 100 Mbps or more), 5G (Potentially several Gbps). Wi-Fi: Allows devices to connect wirelessly within a limited range. Speed: Depends on the Wi-Fi standard, ranging from Mbps to Gbps. Ethernet: Wired connection using Ethernet cables. Speed: Typically ranges from 10 Mbps (older standards) to 1 Gbps or more (modern standards). This list provides a clear structure for understanding different Internet connection types and their associated speed ranges. How Much Internet Speed do I Need? The appropriate Internet speed for you depends on your specific needs and usage patterns. Different online activities require varying amounts of bandwidth. Here are some general guidelines to help you determine how much Internet speed you might need: Basic Browsing and Email Speed Recommendation: 1 to 5 Mbps Suitable for basic web browsing, sending emails, and light online activities. Streaming Standard Definition (SD) Video Speed Recommendation: 3 to 10 Mbps Suitable for streaming standard-definition video on platforms like Netflix, Hulu, or YouTube. Streaming High Definition (HD) Video Speed Recommendation: 5 to 15 Mbps Suitable for streaming high-definition video content. Online Gaming Speed Recommendation: 10 to 50 Mbps Suitable for online gaming with a reasonable ping and minimal lag. Working from Home (Video Conferencing) Speed Recommendation: 5 to 20 Mbps Suitable for smooth video conferencing and file uploads/downloads. Multiple Devices Simultaneously Consider higher speeds if multiple devices in your household will be using the Internet simultaneously for various activities. 4K Video Streaming Speed Recommendation: 25 Mbps or higher Suitable for streaming ultra-high-definition (4K) video content. It's essential to note that these recommendations are approximate and can vary based on the number of users, devices, and specific online applications you use. If you're unsure about your Internet needs, you can contact your Internet service provider (ISP) for personalized advice or consider starting with a mid-tier plan and adjusting based on your experiences. Also, consider factors like data caps, reliability, and customer support when choosing an Internet plan. Keep in mind that advancements in technology or changes in your usage patterns may influence your Internet speed requirements over time.
6206	https://www.impan.pl/~pmh/teach/algebra/additional/formal.pdf	Formal power series From Wikipedia, the free encyclopedia In mathematics, formal power series are a generalization of polynomials as formal objects, where the number of terms is allowed to be infinite; this implies giving up the possibility to substitute arbitrary values for indeterminates. This perspective contrasts with that of power series, whose variables designate numerical values, and which series therefore only have a definite value if convergence can be established. Formal power series are often used merely to represent the whole collection of their coefficients. In combinatorics, they provide representations of numerical sequences and of multisets, and for instance allow giving concise expressions for recursively defined sequences regardless of whether the recursion can be explicitly solved; this is known as the method of generating functions. Contents 1 Introduction 2 The ring of formal power series 2.1 Definition of the formal power series ring 2.1.1 Ring structure 2.1.2 Topological structure 2.1.3 Alternative topologies 2.2 Universal property 3 Operations on formal power series 3.1 Multiplying series 3.2 Power series raised to powers 3.3 Inverting series 3.4 Dividing series 3.5 Extracting coefficients 3.6 Composition of series 3.6.1 Example 3.7 Composition inverse 3.8 Formal differentiation of series 4 Properties 4.1 Algebraic properties of the formal power series ring 4.2 Topological properties of the formal power series ring 5 Applications 6 Interpreting formal power series as functions 7 Generalizations 7.1 Formal Laurent series 7.1.1 Formal residue 7.2 The Lagrange inversion formula 7.3 Power series in several variables 7.3.1 Topology 7.3.2 Operations 7.3.3 Universal property 7.4 Non-commuting variables 7.5 Replacing the index set by an ordered abelian group 8 Examples and related topics 9 Notes Formal power series - Wikipedia, the free encyclopedia 1 of 16 10/11/2012 19:27 10 References Introduction A formal power series can be loosely thought of as an object that is like a polynomial, but with infinitely many terms. Alternatively, for those familiar with power series (or Taylor series), one may think of a formal power series as a power series in which we ignore questions of convergence by not assuming that the variable X denotes any numerical value (not even an unknown value). For example, consider the series If we studied this as a power series, its properties would include, for example, that its radius of convergence is 1. However, as a formal power series, we may ignore this completely; all that is relevant is the sequence of coefficients [1, −3, 5, −7, 9, −11, ...]. In other words, a formal power series is an object that just records a sequence of coefficients. It is perfectly acceptable to consider a formal power series with the factorials [1, 1, 2, 6 , 24, 120, 720, 5040, … ] as coefficients, even though the corresponding power series diverges for any nonzero value of X. Arithmetic on formal power series is carried out by simply pretending that the series are polynomials. For example, if then we add A and B term by term: We can multiply formal power series, again just by treating them as polynomials (see in particular Cauchy product): Notice that each coefficient in the product AB only depends on a finite number of coefficients of A and B. For example, the X5 term is given by For this reason, one may multiply formal power series without worrying about the usual questions of absolute, conditional and uniform convergence which arise in dealing with power series in the setting of analysis. Once we have defined multiplication for formal power series, we can define multiplicative inverses as follows. The multiplicative inverse of a formal power series A is a formal power series C such that AC = 1, provided that such a formal power series exists. It turns out that if A has a multiplicative inverse, it is unique, and we denote it by A −1. Now we can define division of formal power series by defining B / A to be the product B A −1, provided that the inverse of A exists. For example, one can use the definition of multiplication above to verify the familiar formula Formal power series - Wikipedia, the free encyclopedia 2 of 16 10/11/2012 19:27 An important operation on formal power series is coefficient extraction. In its most basic form, the coefficient extraction operator for a formal power series in one variable extracts the coefficient of Xn, and is written e.g. [Xn] A, so that [X2] A = 5 and [X5] A = −11. Other examples include and Similarly, many other operations that are carried out on polynomials can be extended to the formal power series setting, as explained below. The ring of formal power series The set of all formal power series in X with coefficients in a commutative ring R form another ring that is written R, and called the ring of formal power series in the variable X over R. Definition of the formal power series ring One can characterize R abstractly as the completion of the polynomial ring R[X] equipped with a particular metric. This automatically gives R the structure of a topological ring (and even of a complete metric space). But the general construction of a completion of a metric space is more involved than what is needed here, and would make formal power series seem more complicated than they are. It is possible to describe R more explicitly, and define the ring structure and topological structure separately, as follows. Ring structure As a set, R can be constructed as the set RN of all infinite sequences of elements of R, indexed by the natural numbers (taken to include 0). Designating a sequence whose term at index n is an by , one defines addition of two such sequences by and multiplication by This type of product is called the Cauchy product of the two sequences of coefficients, and is a sort of discrete convolution. With these operations, RN becomes a commutative ring with zero element (0, 0, 0, ...) and multiplicative identity (1, 0, 0,...). The product is in fact the same one used to define the product of polynomials in one indeterminate, which suggests using a similar notation. One embeds R into R by sending any (constant) a ∈ R to the sequence (a, 0, 0, ...) and designates the sequence (0, 1, 0, 0, ...) by X; then using the above definitions every sequence with only finitely many nonzero terms can be expressed in terms of these special elements as Formal power series - Wikipedia, the free encyclopedia 3 of 16 10/11/2012 19:27 these are precisely the polynomials in X. Given this, it is quite natural and convenient to designate a general sequence an by by the formal expression , even though the latter is not an expression formed by the operations of addition and multiplication defined above (from which only finite sums can be constructed). This notational convention allows reformulation the above definitions as and which is quite convenient, but one must be aware of the distinction between formal summation (a mere convention) and actual addition. Topological structure Having stipulated conventionally that one would like to interpret the right hand side as a well-defined infinite summation. To that end, a notion of convergence in RN is defined and a topology on RN is constructed. There are several equivalent ways to define the desired topology. We may give RN the product topology, where each copy of R is given the discrete topology. We may give RN the I-adic topology, where I = (X) is the ideal generated by X, which consists of all sequences whose first term a0 is zero. The desired topology could also be derived from the following metric. The distance between distinct sequences (an) and (bn) in RN, is defined to be where k is the smallest natural number such that ak ≠ bk; the distance between two equal sequences is of course zero. Informally, two sequences (an) and (bn) become closer and closer if and only if more and more of their terms agree exactly. Formally, the sequence of partial sums of some infinite summation converges if for every fixed power of X the coefficient stabilizes: there is a point beyond which all further partial sums have the same coefficient. This is clearly the case for the right hand side of (1), regardless of the values an, since inclusion of the term for i = n gives the last (and in fact only) change to the coefficient of Xn. It is also obvious that the limit of the sequence of partial sums is equal to the left hand side. This topological structure, together with the ring operations described above, form a topological ring. This is called the ring of formal power series over R and is denoted by R. The topology has the useful property that an infinite summation converges if and only if the sequence of its terms converges to 0, which just means that any fixed power of X occurs in only finitely many terms. Formal power series - Wikipedia, the free encyclopedia 4 of 16 10/11/2012 19:27 The topological structure allows much more flexible use of infinite summations. For instance the rule for multiplication can be restated simply as since only finitely many terms on the right affect any fixed Xn. Infinite products are also defined by the topological structure; it can be seen that an infinite product converges if and only if the sequence of its factors converges to 1. Alternative topologies The above topology is the finest topology for which always converges as a summation to the formal power series designated by the same expression, and it often suffices to give a meaning to infinite sums and products, or other kinds of limits that one wishes to use to designate particular formal power series. It can however happen occasionally that one wishes to use a coarser topology, so that certain expressions become convergent that would otherwise diverge. This applies in particular when the base ring R already comes with a topology other than the discrete one, for instance if it is also a ring of formal power series. Consider the ring of formal power series then the topology of above construction only relates to the indeterminate Y, since the topology that was put on has been replaced by the discrete topology when defining the topology of the whole ring. So converges to the power series suggested, which can be written as ; however the summation would be considered to be divergent, since every term affects the coefficient of Y (which coefficient is itself a power series in X). This asymmetry disappears if the power series ring in Y is given the product topology where each copy of is given its topology as a ring of formal power series rather than the discrete topology. As a consequence, for convergence of a sequence of elements of it then suffices that the coefficient of each power of Y converges to a formal power series in X, a weaker condition that stabilizing entirely; for instance in the second example given here the coefficient of Y converges to , so the whole summation converges to . This way of defining the topology is in fact the standard one for repeated constructions of rings of formal power series, and gives the same topology as one would get by taking formal power series in all inderteminates at once. In the above example that would mean constructing , and here a sequence converges if and only if the coefficient of every monomial XiYj stabilizes. This topology, which is also the I-adic topology, where I = (X,Y) is the ideal generated by X and Y, still enjoys the property that a summation converges if and only if its terms tend to 0. The same principle could be used to make other divergent limits converge. For instance in the limit Formal power series - Wikipedia, the free encyclopedia 5 of 16 10/11/2012 19:27 does not exist, so in particular it does not converge to . This is because for i≥2 the coefficient of Xidoes not stabilize as n goes to infinity. It does however converge in the usual topology of R, and in fact to the coefficient of exp(X). Therefore, if one would give the product topology of RN where the topology of R is the usual topology rather than the discrete one, then the above limit would converge to exp(X). This more permissive approach is not however the standard when considering formal power series, as it would lead to convergence considerations that are as subtle as they are in analysis, while the philosophy of formal power series is on the contrary to make convergence questions as trivial as they can possibly be. With this topology it would not be the case that a summation converges if and only if its terms tend to 0. Universal property The ring R may be characterized by the following universal property. If S is a commutative associative algebra over R, if I is an ideal of S such that the I-adic topology on S is complete, and if x is an element of I, then there is a unique Φ : R → S with the following properties: Φ is an R-algebra homomorphism Φ is continuous Φ(X) = x. Operations on formal power series One can perform algebraic operations on power series to generate new power series. Multiplying series The product of two series is given by where The sequence is the Cauchy product of the sequences and . Power series raised to powers If n is a natural number we have where Formal power series - Wikipedia, the free encyclopedia 6 of 16 10/11/2012 19:27 for . (This formula can only be used if and are invertible in the ring of scalars.) In the case of formal power series with complex coefficients, the complex powers are well defined at least for series with constant term equal to . In this case, can be defined either by composition with the binomial series , or by composition with the exponential and the logarithmic series, , or as the solution of the differential equation with constant term , the three definitions being equivalent. The rules of calculus and easily follow. Inverting series The series in R is invertible in R if and only if its constant coefficient a0 is invertible in R. This condition is necessary, for the following reason: if we suppose that A has an inverse then the constant term of is the constant term of the identity series, i.e., it is 1. This condition is also sufficient; we may compute the coefficients of the inverse series B via the explicit recursive formula An important special case is that the geometric series formula is valid in R: If R=K is a field, then a series is invertible if and only if the constant term is non-zero, i.e., if and only if the series is not divisible by X. This says that is a discrete valuation ring with uniformizing parameter X. Dividing series The computation of a quotient f/g = h assuming the denominator is invertible (that is, is invertible in the ring of scalars), can be performed as a product f and the inverse of g, or directly equating the coefficients in f = gh: Formal power series - Wikipedia, the free encyclopedia 7 of 16 10/11/2012 19:27 Extracting coefficients The coefficient extraction operator applied to a formal power series in is written and extracts the coefficient of , so that Composition of series Given formal power series and one may form the composition where the coefficients cn are determined by "expanding out" the powers of f(X): Here the sum is extended over all (k,j) with and with A more explicit description of these coefficients is provided by Faà di Bruno's formula, at least in the case where the coefficient ring is a field of characteristic 0. A point here is that this operation is only valid when f(X) has no constant term, so that the series for g(f(X)) converges in the topology of R. In other words, each cn depends on only a finite number of coefficients of f(X) and g(X). Formal power series - Wikipedia, the free encyclopedia 8 of 16 10/11/2012 19:27 Example Assume that the ring R has characteristic 0. If we denote by exp(X) the formal power series then the expression makes perfect sense as a formal power series. However, the statement is not a valid application of the composition operation for formal power series. Rather, it is confusing the notions of convergence in R and convergence in R; indeed, the ring R may not even contain any number e with the appropriate properties. Composition inverse Any formal series with has a composition inverse provided is an invertible element of R. The coefficients are found recursively from the above formula for the coefficients of a composition, equating them with those of the composition identity X (that is 1 at degree 1 and 0 at every degree greater than 1) . In the case when the coefficient ring is a field of characteristic 0, the Lagrange inversion formula provides a powerful tool to compute the coefficients of g, as well as the coefficients of the (multiplicative) powers of g. Formal differentiation of series Given a formal power series in R, we define its formal derivative, denoted Df or , by The symbol D is called the formal differentiation operator. The motivation behind this definition is that it simply mimics term-by-term differentiation of a polynomial. This operation is R-linear: for any a, b in R and any f, g in R. Additionally, the formal derivative has many of the properties of the usual derivative of calculus. For example, the product rule is valid: Formal power series - Wikipedia, the free encyclopedia 9 of 16 10/11/2012 19:27 and the chain rule works as well: whenever the appropriate compositions of series are defined (see above under composition of series). Thus, in these respects formal power series behave like Taylor series. Indeed, for the f defined above, we find that where Dk denotes the kth formal derivative (that is, the result of formally differentiating k times). Properties Algebraic properties of the formal power series ring R is an associative algebra over R which contains the ring R[X] of polynomials over R; the polynomials correspond to the sequences which end in zeros. The Jacobson radical of R is the ideal generated by X and the Jacobson radical of R; this is implied by the element invertibility criterion discussed above. The maximal ideals of R all arise from those in R in the following manner: an ideal M of R is maximal if and only if M ∩ R is a maximal ideal of R and M is generated as an ideal by X and M ∩ R. Several algebraic properties of R are inherited by R: if R is a local ring, then so is R if R is Noetherian, then so is R; this is a version of the Hilbert basis theorem if R is an integral domain, then so is R if R = K is a field, then K is a discrete valuation ring. Topological properties of the formal power series ring The metric space (R, d) is complete. The ring R is compact if and only if R is finite. This follows from Tychonoff's theorem and the characterisation of the topology on R as a product topology. Applications Formal power series can be used to solve recurrences occurring in number theory and combinatorics. For an example involving finding a closed form expression for the Fibonacci numbers, see the article on Examples of generating functions. One can use formal power series to prove several relations familiar from analysis in a purely algebraic setting. Consider for instance the following elements of Q: Formal power series - Wikipedia, the free encyclopedia 10 of 16 10/11/2012 19:27 Then one can show that and as well as (the latter being valid in the ring Q). In algebra, the ring K (where K is a field) is often used as the "standard, most general" complete local ring over K. Interpreting formal power series as functions In mathematical analysis, every convergent power series defines a function with values in the real or complex numbers. Formal power series can also be interpreted as functions, but one has to be careful with the domain and codomain. If f = ∑an Xn is an element of R, S is a commutative associative algebra over R, I is an ideal in S such that the I-adic topology on S is complete, and x is an element of I, then we can define This latter series is guaranteed to converge in S given the above assumptions on X. Furthermore, we have and Unlike in the case of bona fide functions, these formulas are not definitions but have to be proved. Since the topology on R is the (X)-adic topology and R is complete, we can in particular apply power series to other power series, provided that the arguments don't have constant coefficients (so that they belong to the ideal (X)): f(0), f(X2−X) and f( (1 − X)−1 − 1) are all well defined for any formal power series f∈R. With this formalism, we can give an explicit formula for the multiplicative inverse of a power series f whose constant coefficient a = f(0) is invertible in R: Formal power series - Wikipedia, the free encyclopedia 11 of 16 10/11/2012 19:27 If the formal power series g with g(0) = 0 is given implicitly by the equation where f is a known power series with f(0) = 0, then the coefficients of g can be explicitly computed using the Lagrange inversion formula. Generalizations Formal Laurent series A formal Laurent series over a ring R is defined in a similar way to a formal power series, except that we also allow finitely many terms of negative degree (this is different from the classical Laurent series), that is series of the form where for all but finitely many negative indices n. Multiplication of such series can be defined. Indeed, similarly to the definition for formal power series, the coefficient of of two series with respective sequences of coefficients and is which sum is effectively finite because of the assumed vanishing of coefficients at sufficiently negative indices, and which sum zero for sufficiently negative k for the same reason. For a non-zero formal Laurent series, the minimal integer n such that an≠0 is called the order of f, denoted ord(f). (The order of the zero series is +∞.) The formal Laurent series form the ring of formal Laurent series over R, denoted by R((X)). It is equal to the localization of R with respect to the set of positive powers of X. It is a topological ring with the metric . If R = K is a field, then K((X)) is in fact a field, which may alternatively be obtained as the field of fractions of the integral domain K. One may define formal differentiation for formal Laurent series in a natural way (term-by-term). Precisely, the formal derivative of the formal Laurent series f above is which is again an element of K((X)). Notice that if f is a non-constant formal Laurent series, and K is a field of characteristic 0, then one has Formal power series - Wikipedia, the free encyclopedia 12 of 16 10/11/2012 19:27 However, in general this is not the case since the factor n for the lowest order term could be equal to 0 in R. Formal residue Assume that R is a field K of characteristic 0. Then the map is a K-derivation that verifies The latter shows that the coefficient of X−1 in ƒ is of particular interest; it is called formal residue of ƒ and denoted Res(ƒ). The map is K-linear, and by the above observation one has an exact sequence Some rules of calculus. As a quite direct consequence of the above definition, and of the rules of formal derivation, one has, for any ƒ and g in K((X)) i. ii. iii. iv. v. Property (i) is part of the exact sequence above. Property (ii) follows from (i) as applied to (ƒg)' = ƒg' + ƒ'g. Property (iii): any ƒ can be written in the form ƒ = xm g, with m = ord(ƒ) and ord(g) = 0: then ƒ'/ƒ = mX −1 + g'/g. Since ord(g) = 0, the element g is invertible in K ⊂ im(D) = ker(Res), whence Res(ƒ'/ƒ) = m. Property (iv): Since ker(Res) ⊂ im(D), we can write ƒ = ƒ−1 X −1 + F' , with F ∈ K((X)). Consequently, (ƒ g) g' = f−1 g−1 g' + (F' g) g' = ƒ−1 g'/g + (F g)' and (iv) follows from (i) and (iii). Property (v) is clear from the definition. The Lagrange inversion formula As mentioned above, any formal series with and has a composition inverse . The following relation between the coefficients of and holds ("Lagrange inversion formula"): In particular, for n = 1 and all k ≥ 1, Formal power series - Wikipedia, the free encyclopedia 13 of 16 10/11/2012 19:27 Since the proof of the Lagrange inversion formula is a very short computation, it is worth reporting it here. By the above rules of calculus, Generalizations. One may observe that the above computation can be repeated plainly in more general settings than : a generalization of the Lagrange inversion formula is already available working in the -modules , where is a complex exponent. As a consequence, if f and g are as above, with , we can relate the complex powers of f/X and g/X: precisely, if and are non-zero complex numbers with negative integer sum, , then . For instance, this way one finds the power series for complex powers of the Lambert function. Power series in several variables Formal power series in any number of indeterminates (even infinitely many) can be defined. If I is an index set and XI is the set of indeterminates Xi for i∈I, then a monomial Xα is any finite product of elements of XI (repetitions allowed); a formal power series in XI with coefficients in a ring R is determined by any mapping from the set of monomials Xα to a corresponding coefficient cα, and is denoted . The set of all such formal power series is denoted R, and it is given a ring structure by defining and Topology The topology on R is such that a sequence of its elements converges only if for each monomial Xα the corresponding coefficient stabilizes. This is the J-adic topology, where J is the ideal of R generated by all the indeterminates in XI. As remarked above, the topology on a repeated formal power series ring like R is usually chosen in such a way that it becomes isomorphic as a topological ring to R. Operations Formal power series - Wikipedia, the free encyclopedia 14 of 16 10/11/2012 19:27 All of the operations defined for series in one variable may be extended to the several variables case. A series is invertible if and only if its constant term is invertible in R. The composition f(g(X)) of two series f and g is defined if f is a series in a single indeterminate, and the constant term of g is zero. For a series f in several indeterminates a form of "composition" can similarly be defined, with as many separate series in the place of g as there are indeterminates. In the case of the formal derivative, there are now separate partial derivative operators, which differentiate with respect to each of the indeterminates. They all commute with each other. Universal property In the several variables case, the universal property characterizing R becomes the following. If S is a commutative associative algebra over R, if I is an ideal of S such that the I-adic topology on S is complete, and if x1, ..., xr are elements of I, then there is a unique Φ : R → S with the following properties: Φ is an R-algebra homomorphism Φ is continuous Φ(Xi) = xi for i = 1, ..., r. Non-commuting variables The several variable case can be further generalised by taking non-commuting variables Xi for i∈I,where I is an index set and then a monomial Xα is any word in the XI; a formal power series in XI with coefficients in a ring R is determined by any mapping from the set of monomials Xα to a corresponding coefficient cα, and is denoted . The set of all such formal power series is denoted R«XI», and it is given a ring structure by defining addition pointwise and multiplication by where · denotes concatenation of words. These formal power series over R form the Magnus ring over R. Replacing the index set by an ordered abelian group Main article: Hahn series Suppose G is an ordered abelian group, meaning an abelian group with a total ordering "<" respecting the group's addition, so that a < b if and only if a + c < b + c for all c. Let I be a well-ordered subset of G, meaning I contains no infinite descending chain. Consider the set consisting of for all such I, with ai in a commutative ring R, where we assume that for any index set, if all of the ai are zero then Formal power series - Wikipedia, the free encyclopedia 15 of 16 10/11/2012 19:27 the sum is zero. Then R((G)) is the ring of formal power series on G; because of the condition that the indexing set be well-ordered the product is well-defined, and we of course assume that two elements which differ by zero are the same. V arious properties of R transfer to R((G)). If R is a field, then so is R((G)). If R is an ordered field, we can order R((G)) by setting any element to have the same sign as its leading coefficient, defined as the least element of the index set I associated to a non-zero coefficient. Finally if G is a divisible group and R is a real closed field, then R((G)) is a real closed field, and if R is algebraically closed, then so is R((G)). This theory is due to Hans Hahn, who also showed that one obtains subfields when the number of (non-zero) terms is bounded by some fixed infinite cardinality. Examples and related topics Bell series are used to study the properties of multiplicative arithmetic functions Formal groups are used to define an abstract group law using formal power series Notes ^ Sec 0.313, I.S. Gradshteyn (И.С. Градштейн), I.M. Ryzhik (И.М. Рыжик); Alan Jeffrey, Daniel Zwillinger, editors. Table of Integrals, Series, and Products, seventh edition. Academic Press, 2007. ISBN 978-0-12-373637-6. Errata. (Several previous editions as well.) 1. ^ Ivan Niven, "Formal Power Series", American Mathematical Monthly, volume 76, number 8, October 1969, pages 871–889. 2. ^ Koch, Helmut (1997). Algebraic Number Theory. Encycl. Math. Sci.. 62 (2nd printing of 1st ed.). Springer-Verlag. p. 167. ISBN 3-540-63003-1. Zbl 0819.11044 ( format=complete) . 3. ^ Moran, Siegfried (1983). The Mathematical Theory of Knots and Braids: An Introduction. North-Holland Mathematics Studies. 82. Elsevier. p. 211. ISBN 0-444-86714-7. Zbl 0528.57001 ( /zmath/en/search/?q=an:0528.57001&format=complete) . 4. References Nicolas Bourbaki: Algebra, IV , §4. Springer-V erlag 1988. Retrieved from " Categories: Abstract algebra Ring theory Enumerative combinatorics Mathematical series This page was last modified on 7 November 2012 at 21:29. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. See Terms of Use for details. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Formal power series - Wikipedia, the free encyclopedia 16 of 16 10/11/2012 19:27
6207	https://www.khanacademy.org/science/biology/human-biology/neuron-nervous-system/a/depolarization-hyperpolarization-and-action-potentials	Depolarization, hyperpolarization & neuron action potentials (article) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Biology archive Course: Biology archive>Unit 24 Lesson 2: The neuron and nervous system Anatomy of a neuron Overview of neuron structure and function The membrane potential Electrotonic and action potentials Saltatory conduction in neurons Neuronal synapses (chemical) The synapse Neurotransmitters and receptors Q & A: Neuron depolarization, hyperpolarization, and action potentials Overview of the functions of the cerebral cortex Science> Biology archive> Human biology> The neuron and nervous system © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Q & A: Neuron depolarization, hyperpolarization, and action potentials Google Classroom Microsoft Teams Answer to #AskKhanAcademy Fall Finals 2015 question. Question: What causes the hyperpolarization and depolarization of membrane potential, and how does change in membrane potential trigger graded and action potentials for the transmission of signals? – Vincent Tse Answer: Hi Vincent, That’s a great question! Here is a written explanation, with links to some videos that may also help you. Hyperpolarization and depolarization At rest, a typical neuron has a resting potential (potential across the membrane) of −60‍ to −70‍ millivolts. This means that the interior of the cell is negatively charged relative to the outside. See videos: Neuron resting potential description, Neuron resting potential mechanism Hyperpolarization is when the membrane potential becomes more negative at a particular spot on the neuron’s membrane, while depolarization is when the membrane potential becomes less negative (more positive). Depolarization and hyperpolarization occur when ion channels in the membrane open or close, altering the ability of particular types of ions to enter or exit the cell. For example: The opening of channels that let positive ions flow out of the cell (or negative ions flow in) can cause hyperpolarization. Examples: Opening of channels that let K+‍ out of the cell or Cl−‍ into the cell. The opening of channels that let positive ions flow into the cell can cause depolarization. Example: Opening of channels that let Na+‍ into the cell. The opening and closing of these channels may depend on the binding of signaling molecules such as neurotransmitters (ligand-gated ion channels), or on the voltage across the membrane (voltage-gated ion channels). Graded potentials A hyperpolarization or depolarization event may simply produce a graded potential, a smallish change in the membrane potential that is proportional to the size of the stimulus. As its name suggests, a graded potential doesn’t come in just one size – instead, it comes in a wide range of slightly different sizes, or gradations. Thus, if just one or two channels open (due to a small stimulus, such as binding of a few molecules of neurotransmitter), the graded potential may be small, while if more channels open (due to a larger stimulus), it may be larger. Graded potentials don’t travel long distances along the neuron’s membrane, but rather, travel just a short distance and diminish as they spread, eventually disappearing. See videos: Electrotonic and action potentials, Neuron graded potential description, Neuron graded potential mechanism Action potential Alternatively, a large enough depolarization event, perhaps resulting from multiple depolarizing inputs that happen at the same time, can lead to the production of an action potential. An action potential, unlike a graded potential, is an all-or-none event: it may or may not occur, but when it does occur, it will always be of the same size (is not proportional to the size of the stimulus). Image modified from "How neurons communicate: Figure 3," by OpenStax College, Biology (CC BY 3.0). An action potential begins when a depolarization increases the membrane voltage so that it crosses a threshold value (usually around−55‍mV‍). At this threshold, voltage-gated Na+‍ channels in the membrane open, allowing many sodium ions to rush into the cell. This influx of sodium ions makes the membrane potential increase very rapidly, going all the way up to about +40‍mV‍. After a short time, the sodium channels self-inactivate (close and become unresponsive to voltage), stopping the influx of sodium. A set of voltage-gated potassium channels open, allowing potassium to rush out of the cell down its electrochemical gradient. These events rapidly decrease the membrane potential, bringing it back towards its normal resting state. The voltage-gated potassium channels stay open a little longer than needed to bring the membrane back to its resting potential. This results in a phenomenon called “undershoot,” in which the membrane potential briefly dips lower (more negative) than its resting potential. Eventually, the voltage-gated potassium channels close and the membrane potential stabilizes at resting potential. The sodium channels return to their normal state (remaining closed, but once more becoming responsive to voltage). The action potential cycle may then begin again. See videos: Electrotonic and action potentials, Neuron action potential description, Neuron action potential mechanism Transmission of a signal by action potentials The cycle above is described for just one patch of membrane. However, an action potential can travel down the length of a neuron, from the axon hillock (the base of the axon, where it joins the cell body) to the tip of the axon, where it forms a synapse with the receiving neuron. See video: Anatomy of a neuron This directional transmission of the signal occurs for two reasons: First, when one patch of membrane (say, right at the axon hillock) undergoes an action potential, lots of Na+‍ ions rush into the cell through that patch. These ions spread out laterally inside the cell and can depolarize a neighboring patch of membrane, triggering the opening of voltage-gated sodium channels and causing the neighboring patch to undergo its own action potential. Second, the action potential can only travel in one direction – from the cell body towards the axon terminal – because a patch of membrane that has just undergone one action potential is in a “refractory period” and cannot undergo another. The refractory period is primarily due to the inactivation of voltage-gated sodium channels, which occurs at the peak of the action potential and persists through most of the undershoot period. These inactivated sodium channels cannot open, even if the membrane potential goes above threshold. The slow closure of the voltage-gated potassium channels, which results in undershoot, also contributes to the refractory period by making it harder to depolarize the membrane (even once the voltage-gated sodium channels have returned to their active state). The refractory period ensures that an action potential will only travel forward down the axon, not backwards through the portion of the axon that just underwent an action potential. Image credit: "How neurons communicate: Figure 4," by OpenStax College, Biology (CC BY 3.0). When the action potential reaches the end of the axon (the axon terminal), it causes neurotransmitter-containing vesicles to fuse with the membrane, releasing neurotransmitter molecules into the synaptic cleft (space between neurons). When the neurotransmitter molecules bind to ligand-gated ion channels on the receiving cell, they may cause depolarization of that cell, causing it to undergo its own action potential. (Some neurotransmitters also cause hyperpolarization, and a single cell may receive both types of inputs.) See video: Neuronal synapses (chemical) I hope that helps! Good luck with your finals studying! Best wishes, Emily (Khan Academy Biology) References References: Animation: Voltage-gated channels and the action potential. (2015). In Essentials of anatomy & physiology: Online learning center. Retrieved from OpenStax College, Biology. (2015). How neurons communicate. In OpenStax CNX. Retrieved from Refractory period. (2015, July 8). Retrieved December 9, 2015 from Wikipedia: Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Neurons, synapses, and signaling. In Campbell biology (10th ed., pp. 1061-1078). San Francisco, CA: Pearson. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Jimmy 10 years ago Posted 10 years ago. Direct link to Jimmy's post “Why does people with one ...” more Why does people with one sort of epilepsy suffer from seizures while others don't? Is it because some sort of abnormality in the membrane potential? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ivana - Science trainee 6 years ago Posted 6 years ago. Direct link to Ivana - Science trainee's post “The two are not the same....” more The two are not the same. Epilepsy is defined by a state of recurrent, spontaneous seizures. If one seizure occurs in an individual, it may not necessarily mean that they have epilepsy because the seizure may have been provoked and that individual may never have a seizure again. The concept of epileptogenesis refers to the development of the state of epilepsy. seizures arise when there is a disruption of mechanisms that normally create a balance between excitation and inhibition. Thus, normally there are controls that keep neurons from excessive action potential discharge, but there are also mechanisms that facilitate neuronal firing so the nervous system can function appropriately. Normally a high concentration of potassium exists inside a neuron and there is a high extracellular sodium concentration, as well as additional ions, leading to a net transmembrane potential of −60 mV . If the balance is perturbed (eg, if potassium is elevated in the extracellular space), this can lead to depolarization that promotes abnormal activity in many ways : terminals may depolarize, leading to transmitter release, and neurons may depolarize, leading to action potential discharge. Excessive discharge alone does not necessarily cause a seizure. Synchronization of a network of neurons is involved. Therefore, how synchronization occurs becomes important to consider. There are many ways neurons can synchronize In broad terms, there are 1) seizures due to genetic causes, 2) seizures due to developmental disorders or malformations, and 3) seizures that develop after a progression of changes in response to an insult or injury. an initial insult or injury leads to a period of time without evidence of overt seizures, and then recurrent seizures begin. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Bob Boblin a year ago Posted a year ago. Direct link to Bob Boblin's post “Thanks for helping me.” more Thanks for helping me. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer bio fida 10 years ago Posted 10 years ago. Direct link to bio fida's post “are there any other types...” more are there any other types of neurons? If so then how many Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer YEBrandy 10 years ago Posted 10 years ago. Direct link to YEBrandy's post “Many different types, bro...” more Many different types, broadly categorized with respect to their shape or their function. Motor neurons, interneurons (AKA relay neurons) and sensory neurons are the traditional classifications with respect to function. Motor neurons transmit a signal to an 'effector' of some kind (a muscle or a gland perhaps), interneurons transmit signals between surrounding neurons, and sensory neurons 'receive' stimuli (interpreting the stimulus and integrating it). Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Vanessa Palmer 7 years ago Posted 7 years ago. Direct link to Vanessa Palmer's post “When would a nicotinic re...” more When would a nicotinic receptor depolarize? Does it hyperpolarize and how does this happen? I dont undertstand the connection between the receptors binding to channels such as nicotinic and muscarinic and how they depolarize or hyperpolarize the cell? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ivana - Science trainee 6 years ago Posted 6 years ago. Direct link to Ivana - Science trainee's post “At the synapse of a motor...” more At the synapse of a motor neuron and striated muscle cell, binding of acetylcholine to nicotinic acetylcholine receptors triggers a rapid increase in permeability of the membrane to both Na+ and K+ ions, leading to depolarization, an action potential, and then contraction. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more patelvrajemail 2 years ago Posted 2 years ago. Direct link to patelvrajemail's post “At what voltage do the vo...” more At what voltage do the voltage-gated sodium and potassium channels open and close? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “I do not remember the exa...” more I do not remember the exact values but the range I remember is around -100 mV to about + 50 mV Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more abbeileena 8 years ago Posted 8 years ago. Direct link to abbeileena's post “What is an overshoot? and...” more What is an overshoot? and what is the difference between an undershoot and an overshoot? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Maj Master 8 years ago Posted 8 years ago. Direct link to Maj Master's post “What happens to the sodiu...” more What happens to the sodium from the action potential in the axon terminal? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Phuong Russ 9 years ago Posted 9 years ago. Direct link to Phuong Russ's post “How does a neuron get fro...” more How does a neuron get from hyperpolarization back to resting potential? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer SpinosaurusRex 9 years ago Posted 9 years ago. Direct link to SpinosaurusRex's post “Hyperpolarization is ofte...” more Hyperpolarization is often caused by efflux of K+ (a cation) through K+ channels, or influx of Cl– (an anion) through Cl– channels. On the other hand, influx of cations, e.g. Na+ through Na+ channels or Ca2+ through Ca2+ channels, inhibits hyperpolarization. If a cell has Na+ or Ca2+ currents at rest, then inhibition of those currents will also result in a hyperpolarization. This voltage-gated ion channel response is how the hyperpolarization state is achieved. In neurons, the cell enters a state of hyperpolarization immediately following the generation of an action potential. While hyperpolarized, the neuron is in a refractory period that lasts roughly 2 milliseconds, during which the neuron is unable to generate subsequent action potentials. Sodium-potassium ATPases redistribute K+ and Na+ ions until the membrane potential is back to its resting potential of around –70 millivolts, at which point the neuron is once again ready to transmit another action potential. Check out this website as well: Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Mack Galante 7 years ago Posted 7 years ago. Direct link to Mack Galante's post “What would happen if the ...” more What would happen if the Na and K gates open simultaneously in a cell? Would the change in flux timing not allow the cell to depolarize so no action potential would be produced, or would it just impact the timing of the refractory period? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ivana - Science trainee 6 years ago Posted 6 years ago. Direct link to Ivana - Science trainee's post “Depolarization wouldn't b...” more Depolarization wouldn't be effective since repolarization would almost immediately happen... Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Haley 8 years ago Posted 8 years ago. Direct link to Haley's post “How long is one cycle of ...” more How long is one cycle of depolarization to depolarization and hyperbolarization Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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6208	https://www.nist.gov/document/nistir-xxxx-selected-procedures-length-calibration-final-draft-20141015pdf	Selected Laboratory and Measurement Practices and Procedures for Length Calibrations (2014 Ed) Jose A. Torres Ferrer Office of Weights and Measures (contractor) Physical Measurement Laboratory Georgia L. Harris, editor Office of Weights and Measures Physical Measurement Laboratory NISTIR XXXX This page is CODEN page on back of title page or Blank. NISTIR XXXX Selected Laboratory and Measurement Practices and Procedures for Length Calibrations (2014 Ed) Jose A. Torres Ferrer Office of Weights and Measures (contractor) Physical Measurement Laboratory Georgia L. Harris, editor Office of Weights and Measures Physical Measurement Laboratory This publication is available free of charge from: October 2014 U.S. Department of Commerce Penny Pritzker, Secretary National Institute of Standards and Technology Willie May, Acting Under Secretary of Commerce for Standards and Technology and Acting Director This page is CODEN page on back of title page or Blank. October NISTIR XXXX i Foreword This NIST IR of Selected Publications was been compiled as an update for a number of Good Laboratory Practices (GLP), Good Measurement Practices (GMP), Standard Operating Procedures (SOP), Statistical Techniques and Reference Tables. This is the fourth in a series of updates that also include: NISTIR 5672 (2014), NISTIR 6969 (2014), NISTIR 7383 (2013). This publication covers those GMPs and SOPs required for length calibrations as used in weights and measures applications. Most of these procedures are updates to procedures that were originally published in NBS Handbook 145, Handbook for the Quality Assurance of Metrological Measurements, in 1986, by Henry V. Oppermann and John K. Taylor. The major changes incorporated 1) uncertainty analyses that comply with current international methods, 2) measurement assurance techniques using check standards or replicate measurements, and 3) . No substantive changes were made to core measurement processes or equations. The following Procedure is new in the 2014 publication, but was prepared by Ken Fraley (retired, OK) and first made available as a draft for laboratories to follow in 2001: Standard Operating Procedures for: Calibration of Pi Tapes (23) October NISTIR XXXX ii This page intentionally made blank. NISTIR XXXX page iii Table of Contents Foreword i Good Measurement Practices 2 Reading the Center of Graduations When Using a Microscope GMP 2-1 8 Reporting Tape Calibrations GMP 8-1 9 Equations for Metallic Tapes GMP 9-1 Standard Operating Procedures 10 Calibration of Rigid Rules SOP 10-1 11 Calibration of Metal Tapes – Bench Method SOP 11-1 12 Calibration of Metal Tapes – Tape-to-Tape Method SOP 12-1 23 Calibrations of PI Tapes – Bench Method SOP 23-1 NISTIR XXXX page iv This page intentionally made blank. Good Measurement Practices This page intentionally made blank. October 2014 GMP 2 Page 1 of 2 GMP 2 Good Measurement Practice for Reading the Center of Graduations when using a Microscope The microscope furnished to State laboratories with the length bench has a reticle on which each graduation represents 0.002 inch which must have a suitable calibration from an accredited supplier to ensure metrological traceability to the International System of Units (SI). The difference in length between two tapes is measured from the center of the graduations of interest. The tape should be placed on the length bench such that the reading edge of the tape partially covers the graduations marked on the length bench. Place the microscope so that both the edges of the tape and the edge of the bench graduations are in the field-of-vision of the microscope, preferably near the center, and within reading range of the graduated reticle. Position the microscope so that part of the graduations on the reticle overlaps the bottom portion of the graduation on the tape and part overlaps the graduation on the length bench. See Figure 1 for illustration. Do not move or refocus the microscope once the comparison between the unknown, X, and standard, S, has begun. Figure 1. Graduated reticle overlapping tape and bench graduations. Determine the center of a graduation by first determining the width of the graduation, dividing the width by two and adding this value to the reticle reading for the left edge of the graduation. In the illustration above, let the length bench be the standard, S, and the tape be the unknown, X. The left edge of the tape graduation is at 206 (0.206 inch); the right edge is at 216 (0.216 inch). The width of the graduation is 0.010 inch. Adding half this value to 206 (0.206 inch), gives the center of the graduation at 211 (0.211 inch). Following the same procedure for the length bench graduation shows the left edge of the graduation at 190 (0.190 inch) and the right edge at 197 (0.197 inch). The center of the graduation is 193.5 (0.193 5 inch). The difference between the two tapes is 211 – 193.5, which is 17.5 (0.211 inch – 0.193 5 inch = 0.017 5 inch). Since X is longer than S, d is positive. Using the equation X – S = d , X – S = 0.017 5 inch or X = S + 0.017 5 inch October 2014 GMP 2 Page 2 of 2 If it is necessary to calibrate a rigid rule to the end of the rule, the edge of ends of the rule are reference points for the measurement. In these cases, there will be only a single value read from the microscope reticle for each end of the rule since the center of graduation is assumed to be the end of the rules. October 2014 GMP 8 Page 1 of 4 GMP 8 Good Measurement Practice for Reporting Tape Calibrations Tape calibration results should be reported as described in SOP 1, Preparation of Calibration Certificates. The use of the Calibration Certificate and Test Report Review Checklist, (SOP 1, Appendix B), is highly recommended. Information pertinent to this type of calibration should include: 1. A note that reported lengths may be converted to catenary suspension support lengths by use of the “Equations for Metallic Tapes” (GMP 9); 2. Description of the procedure, and the position of the test artifact relative to the observer; 3. Tension applied to the tape during the test; 4. Type of support, e.g., continuous, catenary (include information regarding spacing); 5. The assumed linear coefficient of expansion for the tape material; 6. The relationship between the International System of Units (SI) and US customary units; 7. The degrees of freedom (effective) associated with the coverage factor k; AE value, if requested by customer; and 9. Weight per Unit Length, if requested by customer. An example of a Calibration Certificate follows in the Appendix. October 2014 GMP 8 Page 2 of 4 Appendix – Example COMPLIANT CALIBRATION LABORATORY 123 Some Ave. City, State 12312-1231 CALIBRATION CERTIFICATE FOR 25 foot Iron-Nickel Alloy Tape Maker: LUFSTARTOYO Lab Test No.: TI-14-056 Serial No.: C917R NMI Test No.: 822/1234 SUBMITTED BY YOUR CUSTOMER, INC. Customer’s Address City, State This tape has been compared to the standards of Compliant Calibration Laboratory. The horizontal distances between the centers of graduations of the indicated intervals have the following lengths at 20 °C (68 °F) when the tape is subjected to 10 pound horizontally applied tension and supported on a horizontal flat surface. The terminal points of the indicated intervals are the centers of the graduations at the edge of the tape ribbon nearest the observer when the zero graduation is to the left. Interval (feet) Length (inches) Uncertainty (inches) Interval (feet) Length (inches) Uncertainty (inches) 0 to 1 12.000 3 0.001 0 0 to 14 167.999 4 0.001 0 0 to 2 24.000 5 0.001 0 0 to 15 180.000 1 0.001 0 0 to 3 36.000 3 0.001 0 0 to 16 192.000 4 0.001 0 0 to 4 48.000 2 0.001 0 0 to 17 204.000 3 0.001 0 0 to 5 59.999 8 0.001 0 0 to 18 216.000 2 0.001 0 0 to 6 71.999 6 0.001 0 0 to 19 220.000 4 0.001 0 0 to 7 84.000 2 0.001 0 0 to 20 239.999 8 0.001 0 0 to 8 96.000 1 0.001 0 0 to 21 252.000 4 0.001 0 0 to 9 108.000 2 0.001 0 0 to 22 264.000 1 0.001 0 0 to 10 120.000 4 0.001 0 0 to 23 275.999 7 0.001 0 0 to 11 132.000 4 0.001 0 0 to 24 287.999 6 0.001 0 0 to 12 144.000 2 0.001 0 0 to 25 300.000 7 0.001 0 0 to 13 155.999 6 0.001 0 The data in the above table only applies to those items specifically listed on this report. This document does not represent or imply endorsement by Compliant Calibration Laboratory, NMI, or any agency of the State and/or national governments. This document may not be reproduced, except in full, without the written permission of Compliant Calibration Laboratory. Page 1 of 3 October 2014 GMP 8 Page 3 of 4 Reported lengths may be converted to catenary suspension support lengths by use of the “Equations for Metallic Tapes” (GMP 9). The exact relationship between the International System of Units (SI) and the US customary units of length is one foot equals 0.304 8 meter. Uncertainty statement: The combined standard uncertainty includes the standard uncertainty reported for the standard, the standard uncertainty for the measurement process, the standard uncertainty of the coefficient of expansion, the standard uncertainty of the tension weights, the standard uncertainty of the graduated reticle, and a component of uncertainty to account for any observed deviations from NIST values that are less than surveillance limits. The combined standard uncertainty is multiplied by a coverage factor of 2.07, for 35 effective degrees of freedom, to provide an expanded uncertainty, which defines an interval having a level of confidence of approximately 95 percent. The expanded uncertainty presented in this report is consistent with the ISO/IEC Guide to the Expression of Uncertainty in Measurement (2008). The expanded uncertainty is not to be confused with a tolerance limit for the user during application. Traceability statement: The Standards of the Compliant Calibration Laboratory are traceable to the International System of Units (SI) through the National Institute of Standards and Technology, and are part of a comprehensive measurement assurance program for ensuring continued accuracy and measurement traceability within the level of uncertainty reported by this laboratory. The laboratory test number identified above is the unique report number to be used in referencing measurement traceability for artifacts identified in this report only. Supplemental Information Description of artifacts submitted for testing: 25 foot, iron-nickel alloy (invar), tape. A linear coefficient of thermal expansion of 0.000 000 4 /°C (0.000 000 22 /°F) was assumed. Conditions of artifacts submitted for testing: Artifact showed evidence of proper handling. No bents or kinks were observed. A light film of oil covered the tape ribbon. Treatment of artifacts prior to testing: Artifact was cleaned with cheesecloth and alcohol to remove the protective oil film. The artifact was kept in the length laboratory, next to the standard, for 36 h to allow for thermal equilibrium. Equipment & Standards Standard: Length bench Calibrated: August 2014 Tension weights: 10 pound, Class F Microscope: B&L, 0.002 inch graduated reticle This document does not represent or imply endorsement by Compliant Calibration Laboratory, NMI, or any agency of the State and/or national governments. This document may not be reproduced, except in full, without the written permission of Compliant Calibration Laboratory. Page 2 of 3 October 2014 GMP 8 Page 4 of 4 Procedure used: NISTIRXXXX, SOP 11, Calibration of Metal Tapes, Bench Method Environmental conditions at time of test: Temperature: 20.1 C to 20.2 C Barometric Pressure: 752.7 mmHg Relative Humidity: 43.35 % to 43.40 % Date artifact was received: September 15, 2014 Date of test: September 17, 2014 Due date per customer’s request: September 17, 2016 Date of report preparation: September 18, 2014 Josh Balani, Jr. Test performed by: Josh Balani, Jr. Metrology Expert Member: NCSLI NCWM ASQ ASTM This document does not represent or imply endorsement by Compliant Calibration Laboratory, NMI, or any agency of the State and/or national governments. This document may not be reproduced, except in full, without the written permission of Compliant Calibration Laboratory. Page 3 of 3 October 2014 GMP 9 Page 1 of 2 GMP 9 Good Measurement Practice for Equations for Metallic Tapes Table 1. Symbols used in this practice. Symbol Description L Horizontal straight-line distance LS Calibrated length of the tape interval on a flat surface at TS and PS Ln Designated nominal length of the tape interval P Applied tension PS Standard tension applied to the tape interval for LS AE Average cross-sectional area times Young’s Modulus of Elasticity T Temperature TS Standard temperature of the tape interval for LS, 20 °C (68 °F) α Coefficient of thermal expansion of the tape ribbon W Average weight per unit length of the tape ribbon N Number of equidistant catenary suspensions PO Tension of accuracy (Tension of accuracy is defined as that tension which must be applied to the tape interval to produce its designated nominal length at the observed temperature of the tape.) PC Tension of accuracy while the tape is supported in catenary suspensions The horizontal straight-line distance, L, of a tape interval can be computed by the following equation for an applied tension, P, and temperature, T, when the tape is supported for N, number of equidistant catenary suspensions. When both a standard tape and a tape to be calibrated are placed in a catenary suspension condition, the horizontal straight line distance for the compared intervals for each tape must be calculated. Intervals can be compared only at the supported points along the tape length. 24 )()()( 2 PNLWLTTLAE PPLLL nnSnSnS  (1) For simplicity, it is recommended to entirely support the tape on a horizontal flat surface for calibration. When the tape is supported entirely on a horizontal flat surface, N = ∞, and the general equation is reduced to: )()( SnSnS TTLAE PPLLL  (2) The distance, L, of the tape interval can be set to the designated nominal length, Ln, for determining the tension of accuracy, PO, while the tape is supported on a flat surface, by writing equation (2) as follows: October 2014 GMP 9 Page 2 of 2 )()( SnSOnSn TTLAE PPLLL  (3) from which the following equations are developed: )()( SnSnSO TTAE LLLAE PP  (4) or )()( SnSnOS TTAE LLLAE PP  (5) Substituting equation (5) for PS in the general equation (1), we have 24 )()( 2 CnnOnn PNLWLAE PPLLL  (6) The distance, L, of the tape interval again can be set to the designated nominal length, Ln, for determining the tension of accuracy, PC, while the tape is supported in catenary suspensions, by writing this equation as follows 24 )()( 2 CnnOCnnn PNLWLAE PPLLL  (7) from which: 24 )()(22 NLWAE PPP nOCC  (8) or 24 )()()( 22 NLWAE TTAE LLLAE PPP nSnSnSCC   (9) The value of PC can be solved by first determining the right side of the equal signs in equations (8) or (9), then substituting various values for PC until the left side approaches the right side within the desired limits. If the value is greater than the right side, reduce the value of PC.Standard Operating Procedures This page intentionally made blank. October 2014 SOP 10 Page 1 of 10 SOP No. 10 Recommended Standard Operating Procedure for Calibration of Rigid Rules Introduction 1.1. Purpose This SOP describes the procedure to be followed for the calibration of rigid rules by comparison to the 18 inch metal rule issued to each State as the State reference standard which is calibrated in the interval from 1 inch to 13 inches. Equations provided here assume that the rigid rules being compared have the same coefficient of expansions. The maximum length of rule that can be directly compared to the standard rule is 12 inches. However, longer lengths can be calibrated in segments of 12 inches with reference to the standard rule. 1.2. Prerequisites 1.2.1. Valid calibration certificates with appropriate values and uncertainties must be available for all of the standards used in the calibration. All standards must have demonstrated metrological traceability to the international system of units (SI), which may be to the SI through a National Metrology Institute such as NIST. 1.2.2. The ocular microscope used in measuring differences in lengths must be in good operating condition and must be equipped with a graduated reticle having established traceability. 1.2.3. The operator must be trained and experienced in precision measuring techniques with specific training in GMP 2, GMP 8, GMP 9, SOP 10, and SOP 29. 1.2.4. Laboratory facilities must comply with following minimum conditions to meet the expected uncertainty possible with this procedure. Equilibration of supporting surface, standard rule and rule to be calibrated requires environmental stability of the laboratory within the stated limits for a minimum of 24 hours before a calibration. October 2014 SOP 10 Page 2 of 10 Table 1. Environmental conditions. Temperature Requirements During a Calibration Relative Humidity (%) Lower and upper limits: 18 °C to 22 °C Maximum changes: < ± 1 °C / 24 h and ± 0.5 °C / 1 h 40 to 60 ± 10 / 4 h 2. Methodology 2.1. Scope, Precision, Accuracy The accuracy of calibration of standard rules is possible to within 0.000 1 inch, provided suitable calibration of standards are obtained. The precision of intercomparison and the accuracy of the standard limit the uncertainty of calibration to 0.002 inch under optimum conditions. 2.2. Summary A rigid rule (test rule) is calibrated by comparing intervals on it with intervals of the standard rule. A reticle eye piece (ocular micrometer) is used for this purpose. Test rules longer than the standard rule may be calibrated in segments, using the last calibrated graduation as the zero graduation mark for the succeeding segments. Deviations from nominal that are calculated for each successive interval are cumulative. 2.3. Apparatus/Equipment Required 2.3.1. Length bench or similar flat surface on which to lay the test rule and the standard rule. 2.3.2. Calibrated standard rule. 2.3.3. Microscope with calibrated graduated reticle having graduations spaced at intervals no greater than 0.002 inch. October 2014 SOP 10 Page 3 of 10 2.4. Symbols Table 2. Symbols used in this procedure. Symbol Description A, B, C, D Calculated centers of graduations S Standard X Unknown X0L Left edge of zero on unknown X0R Right edge of zero on unknown S0L Left edge of zero on standard S0R Right edge of zero on standard Xm Center of graduation of unknown Sm Center of graduation of standard di Difference between X and S. The subscript i, designates the trial number, 1 or 2. LX Calibrated length of test rule 2.5. Procedure 2.5.1. Both the test rule and the standard rule must be in temperature equilibrium with the environment of the length laboratory. 2.5.2. Place the test rule and the standard rule on the length bench or similar flat surface, parallel to one another with the reading edges adjacent. It is not necessary to have the “zero” graduations in coincidence. Ordinarily, this will require that one rule reads left-to-right (increasing units) while the other reads right-to-left (decreasing units). In this case, for convenience of calculation, the standard rule is placed in the right-to-left position. A worksheet to reverse the calibration on the standard rule is provided at the end of this SOP. Place shims under the rules as necessary so that the upper surfaces of both are in the same plane. 2.5.3. Place the microscope on the bench in the vicinity of the zero position and align it so that its reticle scale is parallel to the scales under test. (See GMP No. 2 for instructions on reading graduations with the microscope.) 2.5.3.1. Observe readings of left and right sides of the corresponding graduation of the standard rule and record to the nearest 0.001 inch. The average of these readings will give a value for A. 2.5.3.2. Observe readings of left and right sides of the corresponding graduation of the test rule and record to the nearest 0.001 inch. The average of these readings will give a value for B. If the October 2014 SOP 10 Page 4 of 10 “zero” graduation is the end of the rule, only the reading for the end of the rule is taken. 2.5.4. Move the microscope successively to each graduation to be calibrated and record readings as described in 2.5.3.1 and 2.5.3.2 identifying the readings for the standard rule as C and for the unknown as D. CAUTION! Be certain that the rules are not disturbed during movement of the microscope. 2.5.5. Return the microscope to the zero graduation and repeat readings (2.5.3.1. and 2.5.3.2.) to verify that the rules have not been disturbed. Accept all previous data if the initial zero readings do not disagree with the final zero readings by more than 0.002 inch; otherwise, discard all previous data and repeat entire sequence of readings until a satisfactory set of zero readings are obtained. 2.5.6. Repeat 2.5.3. thru 2.5.5. for every additional segment of the test rule requiring calibration. This will require repositioning the rules, aligning the last measured interval graduation on the test rule with the initial graduation of the standard rule. Deviations from nominal for successive segments are cumulative. 2.5.7. Move the rules and reposition, making a second set of measurements as directed in 2.5.3, 2.5.4, and 2.5.5 for Trial 2. .3. Calculations 3.1. Calculate the difference between the Initial (A, B) and Final (A 1 , B 1 ) zero measurements to ensure that the initial and final readings agree within 0.002 inch. (2.5.5.) \|A – B – A 1 + B 1 \| (1) 3.2. Calculate the center of graduation for unknown, Xm, and standard, Sm , for each set of measurements and each scale interval. The center of the “starting” graduation on the standard rule is A; for the test rule it is B. The subsequent centers of graduations for the standard and the test rule are recorded as C and D, respectively, for each interval measured. For each trial, the values for A and B will remain constant for all measured intervals on that trial, and will be used to compute the measured differences between the test rule and the standard. CASSS RLm ,2)(  (2) October 2014 SOP 10 Page 5 of 10 DBXXX RLm ,2)(  (3) 3.3. Calculate the differences, d1, and d2 between X and S for each scale interval and each trial. (The subscript i is associated with the trial number.) iii DCBAd  (4) 3.4. Obtain the calibrated length of the standard, LS, for the measured interval from the calibration certificate for the standard rule. 3.5. Calculate the length of the test rule, LX, for each interval measured using the mean of the measured differences.   SX LddL  221 (5) 4. Measurement Assurance 4.1. Duplicate the process with a suitable check standard or have a suitable range of check standards for the laboratory. See NISTIR 7383 SOP 17, SOP 20 and NISTIR 6969 SOP 30. Plot the check standard length and verify it is within established limits OR a t-test may be incorporated to check the observed value against an accepted value. The mean of the check standard observations is used to evaluate bias and drift over time. Check standard observations are used to calculate the standard deviation of the measurement process which contributes to the Type A uncertainty components. 4.2 If a standard deviation chart is used for measurement assurance, the standard deviation of each combination of Trial 1 and Trial 2 is calculated and the pooled (or average) standard deviation is used as the estimate of variability in the measurement process. Note: the pooled or average standard deviation over time will reflect varying conditions of test items that are submitted to the laboratory. A standard deviation chart will be needed for each interval calibrated so that the variability resulting from transfers will be measured (the number of charts may be adjusted through analysis using F-tests). 5. Assignment of Uncertainty 5.1. The limits of expanded uncertainty, U, include estimates of the standard uncertainty of the length standards used, u s, estimates of the standard deviation of the measurement process, s p, and estimates of the effect of other components associated with this procedure, uo. These estimates should be combined using the root-sum-squared method (RSS), and the expanded uncertainty, U, reported with October 2014 SOP 10 Page 6 of 10 a coverage factor to be determined based on degrees of freedom, which if large enough will be 2, ( k = 2), to give an approximate 95 percent level of confidence. See NISTIR 6969, SOP 29 (Standard Operating Procedure for the Assignment of Uncertainty) for the complete standard operating procedure for calculating the uncertainty. 5.1.1 The expanded uncertainty for the standard, U, is obtained from the calibration report. The combined standard uncertainty, uc, is used and not the expanded uncertainty, U, therefore the reported uncertainty for the standard will usually need to be divided by the coverage factor k. When transfers are used, u s for values after the transfer are dependent and cumulative. See NISTIR 6969, SOP 29 for handling of dependent uncertainties. 5.1.2. The standard deviation of the measurement process, s p, is taken from a control chart for a check standard or standard deviation charts. See NISTIR 7383, SOP 17, SOP 20, and NISTIR 6969, SOP 30. 5.1.3. Uncertainty associated with bias, ud . Any noted bias that has been determined through analysis of control charts and round robin data must be less than limits provided in SOP 29 and included if corrective action is not taken. See SOP 29 for additional details 5.1.4. Other standard uncertainties usually included at this calibration level include uncertainties associated with the ability to read the graduated reticle, only part of which is included in the process variability due to parallax and visual capabilities, and uncertainties associated with the graduations of the reticle. Table 3. Example uncertainty budget table. Component Description Reference us Standard uncertainty for standards Calibration report, divide by k s p Standard uncertainty for the process Measurement assurance process; range charts ugr Standard uncertainty for graduated reticle Must be assessed experimentally or from a calibration certificate ud Standard uncertainty for disparity due to drift/bias Rectangular distribution and reasons, 0.577 d, 0.29 d; SOP 29 (NISTIR 6969) ures Standard uncertainty due to resetting of the rules Must be assessed experimentally; if an interval based standard deviation chart is used u res will be included in the control chart standard deviation value. uo Standard uncertainty for other factors 6. Report October 2014 SOP 10 Page 7 of 10 Report results as described in SOP No. 1 Preparation of Calibration Certificates. October 2014 SOP 10 Page 8 of 10 Appendix A Rigid Rule Calibration Data Sheet Date Environmental parameters Metrologist Before After Unc/ability to measure Test No. Temperature °C °C s p in Pressure mmHg mmHg df Humidity % %Based on NISTIR 6969, SOP 29, Appendix A at 95.45 % probability distribution: k factor ID Range S X Trial 1 Trial 2 Rule Graduation Left Right Center of graduation Left Right Center of graduation S A = A = X B = B = S 1 C1 = C1 = X1 D 1 = D 1 = d 1 = A – B – C 1 + D 1 = d 2 = A – B – C 1 + D 1 = Average d = ( d1 + d 2) / 2 = L S = Length of X, LX = average d + L S = S 2 C2 = C2 = X2 D 2 = D 2 = d 1 = A – B – C 2 + D 2 = d 2 = A – B – C 2 + D 2 = Average d = ( d1 + d 2) / 2 = L S = Length of X, LX = average d + L S = S 3 C3 = C3 = X3 D 3 = D 3 = d 1 = A – B – C 3 + D 3 = d 2 = A – B – C 3 + D 3 = Average d = ( d1 + d 2) / 2 = L S = Length of X, LX = average d + L S = S A 1 = A 1 = X B1 = B1 =Trial 1: Absolute difference between initial and final zero measurements = \|A – B – A 1 + B 1\|= Is result ≤ 0.002 inch? Yes/No If YES, accept all data. Trial 2: Absolute difference between initial and final zero measurements =\|A – B – A 1 + B 1\|= Is result ≤ 0.002 inch? Yes/No If YES, accept all data. October 2014 SOP 10 Page 9 of 10 October 2014 SOP 10 Page 10 of 10 Form A-2. Worksheet to Reverse the Calibration of the Rigid Rule Standard Nominal Length of Interval (inches) Calculations to be performed Length of Interval (inches) 1 L1 = C 13 – C 12 = -2 L2 = C 13 – C 11 = -3 L3 = C 13 – C 10 = -4 L4 = C 13 – C 9 = -5 L5 = C 13 – C 8 = -6 L6 = C 13 – C 7 = -7 L7 = C 13 – C 6 = -8 L8 = C 13 – C 5 = -9 L9 = C 13 – C 4 = -10 L10 = C 13 – C 3 = -11 L11 = C 13 – C 2 = -12 L12 = C 13 – C 1 = -To reverse the calibration on the State standard rigid rule, subtract the lengths for each calibrated interval from the length from the 1 inch to the 13 inch graduation. Example: Length from 13 to 12 = Length from 1 to 13 – Length from 1 to 12 Length from 13 to 12 = 11.999 8 – 10.999 7 = 1.000 1 inch October 2014 SOP 11 Page 1 of 15 SOP No. 11 Recommended Standard Operating Procedure for Calibration of Metal Tapes Bench Method Introduction 1.1. Purpose Metal tapes are used by contractors, surveyors, and others for building layouts, measurement of land areas, establishment of land boundaries, and similar purposes. Inaccuracies in such measurements can cause structural misalignments, boundary controversies, and other problems. The test method described here provides a procedure to calibrate such tapes to four decimal places in the case of measurements in customary units (inches), and to 6 decimal places for measurements expressed in SI units (meters). The calibrated length bench is used as the standard. 1.2. Prerequisites 1.2.1. Valid calibration certificates with appropriate values and uncertainties must be available for all of the standards used in the calibration. All standards must have demonstrated metrological traceability to the international system of units (SI), which may be to the SI through a National Metrology Institute such as NIST. 1.2.2. The ocular microscope used in measuring differences in lengths must be in good operating condition and must be equipped with a calibrated graduated reticle having established traceability. 1.2.3. The operator must be trained and experienced in precision measuring techniques with specific training in GMP 2, GMP 8, GMP 9, SOP 11, and SOP 29. 1.2.4. Laboratory facilities must comply with following minimum conditions to meet the expected uncertainty possible with this procedure. Equilibration of length bench and tapes requires environmental stability of the laboratory within the stated limits for a minimum of 24 hours before a calibration. October 2014 SOP 11 Page 2 of 15 Table 1. Environmental conditions. Temperature Requirements During a Calibration Relative Humidity (%) Lower and upper limits: 18 °C to 22 °C Maximum changes: < ± 1 °C / 24 h and ± 0.5 °C / 1 h 40 to 60 ± 10 / 4 h 2. Methodology 2.1. Scope, Precision, Accuracy The precision and accuracy attainable depend upon the care exercised in aligning the tape on the length bench, and the skill acquired in the use of an optical micrometer to measure scale differences. 2.2. Summary The tape to be calibrated is laid over the bench scale and sufficient tension is applied to insure that it lies flat on the bench. Differences between the graduation on the tape and that of the bench scale are measured using an ocular micrometer. The temperature of the tape is observed, and corrections applied, to calculate the length at the reference temperature of 20 °C using formulae or tables which are in Appendix B. Tapes longer than the length bench may be calibrated in segments, using the last calibrated graduation as the zero graduation mark for the succeeding segments. Typical lengths to be tested on a 100 foot tape when using the 16 foot length bench are: every foot through 10 feet and then 15, 20, 30, 40, 45, 50, 60, 70, 75, 80, 90, and 100 foot lengths. 2.3. Apparatus/Equipment Required 2.3.1. Calibrated thermometers with sufficiently small resolution, stability, and uncertainty, capable of indicating temperatures in the range of 15 °C to 30 °C, and accurate to ± 0.5 °C. 2.3.2. Calibrated 5 meter (16 foot) length bench. 2.3.3. Microscope with calibrated graduated reticle having graduations spaced at intervals no greater than 0.002 inch. 2.3.4. Clamps, fabric tape in lengths required to connect tension weights to the tape being calibrated, and weights to apply an appropriate tension to the tape under calibration. (See C.3.) October 2014 SOP 11 Page 3 of 15 2.4. Symbols Table 2. Symbols used in this procedure. Symbol Description S Standard X Unknown d0i Initial zero difference d0f Final zero difference X0L Left edge of zero on unknown X0R Right edge of zero on unknown S0L Left edge of zero on standard S0R Right edge of zero on standard Xm Center of graduation of unknown Sm Center of graduation of standard d Difference of interval between X and S t Average of initial and final temperatures t corr Temperature correction Ln Nominal length of tape interval under test in inches LX Calibrated length of tape interval under test in inches α Linear coefficient of thermal expansion for the standard bench (0.00001063 / ºC) β Linear coefficient of thermal expansion for the tape (0.00001160 / ºC) AE Cross-sectional area times Young’s modulus of elasticity Q0 Lower load (tension) applied to the tape Q1 Higher load (tension) applied to the tape L0 Length of tape under load Q0 L1 Length of tape under load Q1 2.5. Procedure 2.5.1. Inspect the tape to ensure that no kinks, dents, or other damage are present that will affect the accuracy of the tape 2.5.2. Clean the tape and bench by first wiping with a soft cloth, and then with a soft cloth saturated with alcohol to remove protective oil film. (See Appendix C, Section C.1.) 2.5.3. Place the tape, under the clamp, at the zero end of the bench, and adjust so the starting mark on the tape is near the zero graduation on the length bench. When the ring is part of the measuring tape, a special clamp must be used. See Appendix C.5 for further information on this matter. 2.5.4. Lay the tape flat on the bench. The tape should extend well beyond the end roller of the bench to permit tension to be applied. Slide one end of a fabric strap onto another tape clamp. Fasten this tape clamp to the tape on the portion that extends below the end roller. Hang the tension weight from the October 2014 SOP 11 Page 4 of 15 bottom of the fabric strap. Check to see that the tape is lying straight on the bench and parallel to the bench scale. Adjust, if necessary. Apply tension using a weight of 10 pounds, unless other tension is required (See Appendix C.3.) When a tape is shorter than the 5 m length bench, a fabric strap is used to effectively lengthen the tape so that the tension weight is properly suspended below the rollers of the length bench. 2.5.5. Lay two thermometers (See 2.3.1.) on the bench and tape at intervals of 1/3 and 2/3 of the length of the bench to determine its initial and final temperatures at the time of the test. 2.5.6. Adjust the tape clamp on the zero end of the bench so that the starting graduation on the tape coincides with the center of the zero graduation of the bench. Note: Some tapes are calibrated using the one foot mark as the reference point, rather than the zero graduation. In this case, align the 1 foot mark of the tape with the zero graduation mark of the bench. 2.5.7. Check all alignments and coincidence of the graduations down the length of the bench before proceeding with calibration. Use the lateral adjustments at the left end of the bench to facilitate alignment. Caution! Take care that tape is not touched or disturbed during the following sequence of measurements. Record all observations on a suitable data sheet. 2.5.8. Record the temperatures indicated by the two thermometers. 2.5.9. Place the microscope on the bench in the vicinity of the zero position and align it so that its reticle scale is parallel to the tape under test. (See GMP No. 2 for instructions on reading graduations with the microscope.) 2.5.9.1. Observe readings of left and right sides of zero graduation of tape and record to the nearest 0.001 inch. 2.5.9.2. Observe readings of left and right sides of zero graduation of bench and record to the nearest 0.001 inch. 2.5.10. Move microscope successively to each graduation that needs to be calibrated and record readings similarly as in 2.5.9.1 and 2.5.9.2. 2.5.11. Return microscope to the zero graduation and repeat readings (2.5.9.1. and 2.5.9.2.) to verify that the tape has not moved. Accept all previous data if the zero readings have not changed by more than 0.001 inch; otherwise, discard all previous data and repeat entire sequence of readings until a satisfactory set of zero readings are obtained. 2.5.12. Loosen the tape by removing the tension weights, move the tape, hang the October 2014 SOP 11 Page 5 of 15 tension weights back on the fabric strap, and realign the zero marks on the tape and the bench to coincidence. 2.5.13. Repeat 2.5.8 thru 2.5.12 for every 15 foot (5 meter) section of the tape requiring calibration. This will require repositioning the tape, aligning the last measured interval graduation on the tape with the zero graduation of the bench. 2.5.14. Make a second set of measurements as directed in 2.5.8 to 2.5.13. .2.5.15. After the final measurement is taken and accepted, record the temperatures indicated by the two thermometers. 2.5.16. After all measurements are completed; apply a thin film of oil to the tape. 3. Calculations 3.1. Calculate the Initial Zero Difference, d0i (2.5.8.) 2)( 00000 RLRLi SSXXd  (1) 3.2. Calculate the Final Zero Difference, d0f (2.5.10.) 2)( 00000 RLRLf SSXXd  (2) 3.3. Calculate the center of graduation for unknown, Xm, and standard, Sm , for each set of measurements and each scale interval. 2)( RLmXXX  (3) 2)( RLmSSS  (4) 3.4. Calculate the difference, d, between X and S for each scale interval. mm SXd  (5) 3.5. Obtain the correction to the standard, CS, for the measured interval from the calibration certificate for the length bench scale. October 2014 SOP 11 Page 6 of 15 3.6. Calculate the temperature correction, t corr .   )(20    tLt ncorr (6) 3.7. Calculate a correction, CX, for each trial and each scale interval. corr SX tCdC  (7A) Be sure to include cumulative errors from previous intervals when transfers are made. (For example, corrections for two prior intervals added for the case when a third interval is calibrated.) "' xxcorr SX CCtCdC  (7B) 3.8. Calculate and report the mean, XC of the two corrections for each interval. 3.9. Calculate the length of tape under the 10 pound load. XnX CLL  (8) Measurement Assurance 4.1. Duplicate the process with a suitable check standard or have a suitable range of check standards for the laboratory. See NISTIR 7383 SOP 17, SOP 20 and NISTIR 6969 SOP 30. Plot the check standard length and verify it is within established limits OR a t-test may be incorporated to check the observed value against an accepted value. The mean of the check standard observations is used to evaluate bias and drift over time. Check standard observations are used to calculate the standard deviation of the measurement process which contributes to the Type A uncertainty components. 4.2 If a standard deviation chart is used for measurement assurance, the standard deviation of each combination of Trial 1 and Trial 2 is calculated and the pooled (or average) standard deviation is used as the estimate of variability in the measurement process. Note: the pooled or average standard deviation over time will reflect varying conditions of test items that are submitted to the laboratory. A standard deviation chart will be needed for each interval calibrated (at least initially) so that the variability resulting from transfers will be measured. 5. Assignment of Uncertainty 5.1. The limits of expanded uncertainty, U, include estimates of the standard uncertainty of the length standards used, u s, estimates of the standard deviation of the measurement process, s p, and estimates of the effect of other components associated with this procedure, uo. These estimates should be combined using the October 2014 SOP 11 Page 7 of 15 root-sum-squared method (RSS), and the expanded uncertainty, U, reported with a coverage factor to be determined based on degrees of freedom, which if large enough will be 2, ( k = 2), to give an approximate 95 percent level of confidence. See NISTIR 6969, SOP 29 (Standard Operating Procedure for the Assignment of Uncertainty) for the complete standard operating procedure for calculating the uncertainty. 5.1.1 The expanded uncertainty for the standard, U, is obtained from the calibration report. The combined standard uncertainty, u c, is used and not the expanded uncertainty, U, therefore the reported uncertainty for the standard will usually need to be divided by the coverage factor k. When transfers are used, u s for values after the transfer are dependent and cumulative. See NISTIR 6969, SOP 29 for handling of dependent uncertainties. 5.1.2. The standard deviation of the measurement process, s p, is taken from a control chart for a check standard or standard deviation charts. (See SOP 17, SOP 20, and SOP 30) 5.1.3. Uncertainty associated with bias, ud . Any noted bias that has been determined through analysis of control charts and round robin data must be less than limits provided in SOP 29 and included if corrective action is not taken. See SOP 29 for additional details 5.1.4. Uncertainty associated with temperature correction includes values for the linear coefficients of expansion for the Standard and Unknown, the accuracy of temperature measurements, and factors associated with potential gradients in measuring the temperature on the length bench and tape. 5.1.5. Other standard uncertainties usually included at this calibration level include uncertainties associated with the ability to read the graduated reticle, only part of which is included in the process variability due to parallax and visual capabilities, and uncertainties associated with the graduations of the reticle. October 2014 SOP 11 Page 8 of 15 Table 3. Example uncertainty budget table. Component Description Reference us Standard uncertainty for standards Calibration report, divide by k s p Standard uncertainty for the process Measurement assurance process; range charts ugr Standard uncertainty for graduated reticle Must be assessed experimentally or from a calibration certificate utc Standard uncertainty for temperature correction Calibration certificate ut Standard uncertainty for temperature HB 143 accuracy guideline, 0.1 Culce ( ,) Standard uncertainty for linear coefficient of expansion 5 % to 10 % of the coefficient of expansion value utw Standard uncertainty for tension weights Calibration certificate ud Standard uncertainty for disparity due to drift/bias Rectangular distribution and reasons, 0.577 d, 0.29 d; SOP 29 (NISTIR 6969) ures Standard uncertainty due to resetting of the tape Must be assessed experimentally; may be included in the control chart standard deviation where present uo Standard uncertainty for other factors 6. Report Report results as described in SOP No. 1 Preparation of Calibration Certificates. October 2014 SOP 11 Page 9 of 15 Appendix A Bench Method Data Sheet Date Environmental parameters Metrologist Before After Unc/ability to measure Test No. Temperature °C °C s p in Pressure mmHg mmHg df Humidity % %Based on NISTIR 6969, SOP 29, Appendix A at 95.45 % probability distribution: k factor ID Range Linear coefficient of thermal expansion S Bench 16 ft (5 m) α 0.000 010 63 /°C X Tape β 0.000 011 60 /°C Tension lb Support Initial temperatures Final temperatures t1 °C t3 °C Average temperature °C t2 °C t4 °C Interval, ft Trial Initial Zero Graduations Bench Tape Tape Bench 0 01 X0L X0R S 0L S 0R Initial zero, difference between X and S, d 0i in 2 X0L X0R S 0L S 0R Initial zero, difference between X and S, d 0i in Intervals after zero Tape Bench X1L X1R Xm S 1L S 1R S m d CS tcorr CX Range 12Average CX in X1L X1R Xm S 1L S 1R S m d CS tcorr CX Range 12Average CX in Final Zero Graduations Tape Bench 0 01 X0L X0R S 0L S 0R Final zero, difference between X and S, d 0f in Difference between Initial Zero and Final Zero, d 0i - d 0f in \| d 0i - d 0f \| ≤ 0.001 in? If YES, accept previous data. 2 X0L X0R S 0L S 0R Final zero, difference between X and S, d 0f in Difference between Initial Zero and Final Zero, d 0i - d 0f in \| d 0i - d 0f \| ≤ 0.001 in? If YES, accept previous data.   )(20    tLt ncorr corr SX tCdC  XnX CLL  "'xxcorr SX CCtCdC October 2014 SOP 11 Page 10 of 15 Appendix B Table B-1. Temperature Correction Factors for Calibration of Steel Tapes graduated in feet when linear coefficient of thermal expansion for the tape is 0.00001160 / ºC. All values in inches x 10 -4 L nFeet Temperature, °C 19 20 21 22 23 24 25 26 27 28 29 30 1 0.1 0.0 -0.1 -0.2 -0.3 -0.5 -0.6 -0.7 -0.8 -0.9 -1.0 -1.2 2 0.2 0.0 -0.2 -0.5 -0.7 -0.9 -1.2 -1.4 -1.6 -1.9 -2.1 -2.3 3 0.3 0.0 -0.3 -0.7 -1.0 -1.4 -1.7 -2.1 -2.4 -2.8 -3.1 -3.5 4 0.5 0.0 -0.5 -0.9 -1.4 -1.9 -2.3 -2.8 -3.3 -3.7 -4.2 -4.7 5 0.6 0.0 -0.6 -1.2 -1.7 -2.3 -2.9 -3.5 -4.1 -4.7 -5.2 -5.8 6 0.7 0.0 -0.7 -1.4 -2.1 -2.8 -3.5 -4.2 -4.9 -5.6 -6.3 -7.0 7 0.8 0.0 -0.8 -1.6 -2.4 -3.3 -4.1 -4.9 -5.7 -6.5 -7.3 -8.1 8 0.9 0.0 -0.9 -1.9 -2.8 -3.7 -4.7 -5.6 -6.5 -7.4 -8.4 -9.3 9 1.0 0.0 -1.0 -2.1 -3.1 -4.2 -5.2 -6.3 -7.3 -8.4 -9.4 -10.5 10 1.2 0.0 -1.2 -2.3 -3.5 -4.7 -5.8 -7.0 -8.1 -9.3 -10.5 -11.6 15 1.7 0.0 -1.7 -3.5 -5.2 -7.0 -8.7 -10.5 -12.2 -14.0 -15.7 -17.5 20 2.3 0.0 -2.3 -4.7 -7.0 -9.3 -11.6 -14.0 -16.3 -18.6 -21.0 -23.3 30 3.5 0.0 -3.5 -7.0 -10.5 -14.0 -17.5 -21.0 -24.4 -27.9 -31.4 -34.9 40 4.7 0.0 -4.7 -9.3 -14.0 -18.6 -23.3 -27.9 -32.6 -37.2 -41.9 -46.6 45 5.2 0.0 -5.2 -10.5 -15.7 -21.0 -26.2 -31.4 -36.7 -41.9 -47.1 -52.4 50 5.8 0.0 -5.8 -11.6 -17.5 -23.3 -29.1 -34.9 -40.7 -46.6 -52.4 -58.2 60 7.0 0.0 -7.0 -14.0 -21.0 -27.9 -34.9 -41.9 -48.9 -55.9 -62.9 -69.8 70 8.1 0.0 -8.1 -16.3 -24.4 -32.6 -40.7 -48.9 -57.0 -65.2 -73.3 -81.5 75 8.7 0.0 -8.7 -17.5 -26.2 -34.9 -43.7 -52.4 -61.1 -69.8 -78.6 -87.3 80 9.3 0.0 -9.3 -18.6 -27.9 -37.2 -46.6 -55.9 -65.2 -74.5 -83.8 -93.1 90 10.5 0.0 -10.5 -21.0 -31.4 -41.9 -52.4 -62.9 -73.3 -83.8 -94.3 -104.8 100 11.6 0.0 -11.6 -23.3 -34.9 -46.6 -58.2 -69.8 -81.5 -93.1 -104.8 -116.4 For example: For L n = 20 feet at 25 °C, the temperature correction is -0.001 16 inches. October 2014 SOP 11 Page 11 of 15 Table B-2. Temperature Correction Factors for Calibration of Steel Tapes graduated in meters when linear coefficient of thermal expansion for the tape is 0.00001160 / ºC. All values in meters x 10 -6 LnMeters Temperature, °C 19 20 21 22 23 24 25 26 27 28 29 30 0.1 0.1 0.0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1.0 0.2 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4 -1.6 -1.7 -1.9 0.3 0.3 0.0 -0.3 -0.6 -0.9 -1.2 -1.5 -1.7 -2.0 -2.3 -2.6 -2.9 0.4 0.4 0.0 -0.4 -0.8 -1.2 -1.6 -1.9 -2.3 -2.7 -3.1 -3.5 -3.9 0.5 0.5 0.0 -0.5 -1.0 -1.5 -1.9 -2.4 -2.9 -3.4 -3.9 -4.4 -4.8 0.6 0.6 0.0 -0.6 -1.2 -1.7 -2.3 -2.9 -3.5 -4.1 -4.7 -5.2 -5.8 0.7 0.7 0.0 -0.7 -1.4 -2.0 -2.7 -3.4 -4.1 -4.8 -5.4 -6.1 -6.8 0.8 0.8 0.0 -0.8 -1.6 -2.3 -3.1 -3.9 -4.7 -5.4 -6.2 -7.0 -7.8 0.9 0.9 0.0 -0.9 -1.7 -2.6 -3.5 -4.4 -5.2 -6.1 -7.0 -7.9 -8.7 1 1.0 0.0 -1.0 -1.9 -2.9 -3.9 -4.8 -5.8 -6.8 -7.8 -8.7 -9.7 2 1.9 0.0 -1.9 -3.9 -5.8 -7.8 -9.7 -11.6 -13.6 -15.5 -17.5 -19.4 3 2.9 0.0 -2.9 -5.8 -8.7 -11.6 -14.6 -17.5 -20.4 -23.3 -26.2 -29.1 4 3.9 0.0 -3.9 -7.8 -11.6 -15.5 -19.4 -23.3 -27.2 -31.0 -34.9 -38.8 5 4.8 0.0 -4.8 -9.7 -14.6 -19.4 -24.3 -29.1 -34.0 -38.8 -43.6 -48.5 10 9.7 0.0 -9.7 -19.4 -29.1 -38.8 -48.5 -58.2 -67.9 -77.6 -87.3 -97.0 15 14.6 0.0 -14.6 -29.1 -43.6 -58.2 -72.7 -87.3 -101.9 -116.4 -131.0 -145.5 20 19.4 0.0 -19.4 -38.8 -58.2 -77.6 -97.0 -116.4 -135.8 -155.2 -174.6 -194.0 25 24.3 0.0 -24.3 -48.5 -72.7 -97.0 -121.3 -145.5 -169.8 -194.0 -218.3 -242.5 30 29.1 0.0 -29.1 -58.2 -87.3 -116.4 -145.5 -174.6 -203.7 -232.8 -261.9 -291.0 For example: For L n = 4 meters at 25 °C, the temperature correction is -0.000 019 4 meters. October 2014 SOP 11 Page 12 of 15 Appendix C Supplemental Information C.1. Cleaning To clean a steel tape before calibration, first wipe the tape with a soft cloth; then with a soft cloth saturated with alcohol to remove the film of oil used to protect the tape. After calibration, a thin film of light oil, such as sewing machine oil, should be applied to the tape for protection. C.2. Tolerances The tolerances for measuring tapes are those stated below. Table C-1. Tolerances for a 30 meter tape. Length Interval Tolerances 0 through 15 meters 1.27 mm (0.050 inch) 15 through 22 meters 1.91 mm (0.075 inch) 22 through 30 meters 2.54 mm (0.100 inch) C.2.1. The inaccuracy in the length of the ribbon, when supported on a horizontal surface with a tension of 10 pounds at a temperature of 20 °C (68 °F) shall not exceed 0.050 inch for the 75 foot length, and 0.100 inch for the 100 foot length. C.3. Tension Specifications The length of a tape will be affected by the temperature of the tape, the tension applied to the tape, and the manner in which the tape is supported. The tape will stretch when tension is applied and will return to its normal length when the tension is removed, provided the tape has not been permanently deformed when it was stretched. The tensions, at which steel tapes are to be calibrated, expressed in terms of the load in kilograms (or pounds) to be applied to obtain the tension, are stated below. The loads should be accurate within 45 g (0.01 lb). Table C-2. Tension to be applied, in terms of load. Length Interval Tension Less than 10 m (25 ft) 2 kg (3.5 lb) 10 m through 30 m (25 ft through 100 ft) 5 kg (10 lb) Greater than 30 m (100 ft) 10 kg (20 lb) C.4. Methods of Support and Tension Considerations Tapes calibrated in a State laboratory are normally supported on a horizontal surface October 2014 SOP 11 Page 13 of 15 throughout the entire length of the tape. Also, tapes may be calibrated and used when supported in catenary types of suspension. In these cases, the tape is supported at equidistant points because the weight of the tape affects its length. The weight of the tape increases the tension and the “sag” causes the horizontal length to be shorter than when the tape is supported throughout its length. Equations are given in GMP No. 10 to compute the horizontal straight-line distance of a tape supported at N number of equidistant catenary suspensions and for computing the tension of accuracy, defined as the tension that must be applied to the tape interval to produce its designated nominal length at the observed temperature of the tape. It is sufficient to provide the user of a steel tape with the calibrated length of the tape under standard temperature and tension conditions, the weight per unit-length of the tape, and the AE value for the tape, as might be requested. This information will enable the user to compute the values desired using the equations cited above. C.5. Zero Reference Point Metal measuring tapes submitted to a State laboratory for calibration normally will be made of steel. Generally, these tapes will have a ring on the end of the tape. For maximum calibration and measurement accuracy, a tape should have a blank end between the ring and the zero graduation. The zero graduation is then more precisely defined and more easily referenced for calibration and use. Tapes that have the ring as part of the measuring portion of the tape are more difficult to calibrate than a tape with a blank end. When the ring is part of the measuring portion of the tape, the zero reference point shall be the outside end of the ring unless otherwise specified. It is more difficult to obtain a good zero reference setting on the ring due to its curvature and to parallax in reading the edge of the ring against a reference mark. Additionally, the ring may become permanently deformed in use and change the length of the tape. For these tapes the NIST normally calibrates from the 1-foot mark over the length of the tape, and then calibrates from the ring to the 1-foot mark. These values are reported separately so the user can obtain maximum measurement accuracy by using the 1-foot graduation as the zero reference point. When the ring is part of the measuring range of the tape, a special holder for the ring is needed to clamp the tape to the length bench. A strap with an open area in the middle is needed to permit the end of the ring to be seen. The strap is slipped through the ring and the strap is clamped to the length bench. See example below. Figure C-1. Strap, holder, and tape ring. The edge of the tape to be calibrated (the reading edge) is the edge nearest the observer October 2014 SOP 11 Page 14 of 15 when the zero graduation is to the observer’s left. When viewed through a microscope, some graduations will appear to have irregular edges. The portion of the graduation to be used for calibration is the portion of the graduation at the bottom of the reading edge of the tape. This provides a reference point that can be repeated and referenced by others. Do not attempt to estimate the ‘best overall’ edge of a graduation because this is not easily repeatable and cannot be accurately reproduced by other. If the graduations to be calibrated do not reach to the edge of the tape, the tape should not be calibrated . C.6. Temperature Considerations The reference temperature for length calibrations is 20 °C. The length of the tape can be determined at any other temperature T by using the following equation.  )20 (120  TLLT  (C-1) Since α is always positive, it can be seen that for temperatures above 20 °C, the tape is longer than it is at 20 °C. For temperatures below 20 °C, the tape is shorter than it is at 20 °C. If two length standards have different coefficients of expansion because they are made of different materials, the lengths of the tapes will change at different rates as the temperature changes. If two tapes are being compared at a temperature other than 20 °C, these lengths must be corrected back to 20 °C for calibration. If two length standards are being compared and they have the same coefficient of expansion, then as the temperature changes the lengths will change by the same amount. Hence, if the standards are compared at a temperature other than 20 °C, the relationship between the two standards will be the same as if they were being compared at 20 °C; thus, no temperature correction is needed. Table C-3. Coefficient of expansion of various length standards. Type /°C /°F Steel tape 11.60 x 10 -6 6.45 x 10 -6 Length bench 10.63 x 10 -6 5.91 x 10 -6 Invar tape 4.0 x 10 -7 2.2 x 10 -7 C.7. Invar tapes Invar is an alloy of nickel and steel. Invar tapes are used to obtain measurements of greater accuracy than can be made with steel tapes, because invar has a very low coefficient of expansion. It has the added benefit of being very slow to tarnish from exposure to the atmosphere. However, invar tapes require very careful handling to prevent twists and kinks. The load to be applied to an invar tape to maintain the desired tension is normally 20 lb. A load of 40 lb is used for the higher tension to determine the AE value (see C.8.). For October 2014 SOP 11 Page 15 of 15 metric tapes, the normal load is 5 kg. A load of 10 kg is used to determine the AE value. C.8. AE Value The AE Value (area elongation value) for a tape is determined by first calibrating the tape under its normal tension. The load is then increased by10 lb or 20 lb and one length interval is recalibrated to determine the length of the tape under the increased tension. The AE factor is computed with the following equation. n LLLQQAE 0101  (C-2) For example, a 100 foot tape is calibrated from 0 feet to 100 feet with a load of 10 lb applied to the tape with a resulting length of 99.992 feet. The load is increased to 20 lb and the new length is found to be 100.004 feet. The AE value is: lb feet feet lb feet lb lb AE 83333 012 .01000 100 )992 .99 004 .100 ()10 20 (  It is recommended that the AE value be determined over the longest interval that is convenient to measure. This minimizes the error in the AE value because of the better readability of the change in length C.9 Weight per Unit Length The weight per unit length of a tape can be determined as follows: C.9.1. Weigh the tape and reel (or case). C.9.2. Remove the tape from the reel or case and weigh the empty reel (or case). C.9.3. Measure the length of any blank ends on the tape and add this to the measuring length. C.9.4. Correct for the weight of the loop on the tape. The weight of the loop that is normally used on steel tapes is approximately 2.5 grams. The weight per unit length is the computed as follows. ends blank of length tape of length loop of weight reel empty of weight reel loaded of Weight Length Unit per Weight This page intentionally made blank. October 2014 SOP 12 Page 1 of 12 SOP No. 12 Recommended Standard Operating Procedure for Calibration of Metal Tapes 1Tape-to-Tape Method I. Introduction 1.1. Purpose Metal tapes are used by contractors, surveyors, and others for building layouts, measurement of land areas, establishment of land boundaries, and similar purposes. The accuracy of the measurement often must be provable to 0.000 1 meter on a 100 meter tape (0.000 1 foot on a 100 foot tape), sometimes in a court of law. A significant feature of such proof is the knowledge of the accurate length of the tape used. Inaccuracies in such measurements can cause structural misalignments, boundary controversies, and other problems. The test method described here provides a procedure to calibrate such tapes by comparison to a calibrated standard tape. 1.2. Prerequisites 1.2.1. Valid calibration certificates with appropriate values and uncertainties must be available for all of the standards used in the calibration. All standards must have demonstrated metrological traceability to the international system of units (SI), which may be to the SI through a National Metrology Institute such as NIST. 1.2.2. The operator must be trained and experienced in precision measuring techniques with specific training in GMP 8, GMP 9, SOP 12, and SOP 29. 1.2.3. Laboratory facilities must comply with following minimum conditions to meet the expected uncertainty possible with this procedure. Equilibration of tapes requires environmental stability of the laboratory within the stated limits for a minimum of 24 hours before a calibration. 1 Many references are provided in in-pound units due to common use in U.S. weights and measures requirements. The International System of Units, SI, is the official system of units for metrological traceability. October 2014 SOP 12 Page 2 of 12 Table 1. Environmental conditions. Temperature Requirements During a Calibration Relative Humidity (%) Lower and upper limits: 18 °C to 22 °C Maximum changes: < ± 1 °C / 24 h and ± 0.5 °C / 1 h 40 to 60 ± 10 / 4 h 2. Methodology 2.1. Scope, Precision, Accuracy The method is applicable to the calibration of metal tapes such as used by surveyors, builders, and contractors. The overall length and specified intermediate lengths may be checked by the technique. The accuracy is limited by the accuracy of the calibration of the standard tape and by the precision of intercomparison. The latter should be within ± 0.001 foot, corresponding to 10 parts per million in a 100 foot tape. The method is limited to calibration of steel tapes (because the tension is specified as the result of a 10 pound load). 2.2. Summary This procedure is based upon the method developed by C. Leon Carroll Jr., National Bureau of Standards, NBSIR 74-451, “Field Comparisons of Steel Surveyors’ Tapes.” The tape to be calibrated is stretched out parallel to a standard tape on a reasonably flat surface. Paper scales (graph paper), graduated in millimeters are used at the zero and at each specified interval of calibration to measure any difference between the two tapes. The length of the tape undergoing calibration is computed from the known length of the standard tape and the observed differences between the test tape and the standard. Calibrations are usually made at each 1 foot interval for the first 10 feet, and at each 10 foot interval to the full length of the tape. 2.3. Apparatus/Equipment Required 2.3.1. Standard tape, calibrated to within ± 0.001 foot. 2.3.2. Pieces of graph paper (10 x 10 to the centimeter, i.e., millimeter graduations), approximately 5 cm in width by 15 cm in height. Number the horizontal centimeter graduations, 0, 1, 2, etc.. Evaluate and document the range of use to within 0.1 mm using a calibrated rigid rule or a microscope with calibrated reticles to ensure that the graph paper has sufficient accuracy. October 2014 SOP 12 Page 3 of 12 2.3.3. Equipment, see Figure 1, to apply tension to the tapes under test, consisting of the following. 2.3.3.1. Spring scales (two); one capable of indicating a load of 10 pounds and the other capable of indicating a load of 20 pounds. The scales should be calibrated with an accuracy of ± 0.1 lb. This may be done by the arrangement shown in Figure 2 with results internally documented. 2.3.3.2. Turnbuckles, suitable for adjusting tension on the tapes. 2.3.3.3. Swivel connectors to prevent axial twisting of the tapes. 2.3.3.4. Magnifying glass to aid in reading the graph paper values. 2.3.3.5. Small weights (< 5 lb) to hold the tapes down flat to the paper. Figure 1. Experimental Arrangement. October 2014 SOP 12 Page 4 of 12 Figure 2. Calibration of Spring Scales. 2.4. Symbols Table 2. Symbols used in this procedure. Symbol Description S Length of the Standard Tape at the calibration interval L X Length of the Test Tape at the calibration interval cf Conversion factor, 0.032 808 ft/cm for tapes graduated in feet 0.010 000 m/cm for tapes graduated in meters L n Nominal length of tape interval under test AE Cross-sectional area times Young’s modulus of elasticity Q0 Lower load (tension) applied to the tape Q1 Higher load (tension) applied to the tape L 0 Length of tape under load Q0 L 1 Length of tape under load Q1 2.5. Procedure 2.5.1. Inspect the tapes to ensure that no kinks, dents, or other damage are present which will affect the accuracy of the calibration. 2.5.2. Clean the tapes by first wiping with a soft cloth, and then with a soft cloth saturated with alcohol to remove protective oil film. October 2014 SOP 12 Page 5 of 12 2.5.3. Stretch the test tape and standard tape parallel to each other on areasonably flat surface. The evenness of the surface is less important than the parallelism of the tapes. The two tapes should be separated by a constant distance of about 1 to 3 centimeters. The zero and test intervals of the two tapes should not be in coincidence but rather displaced by one or two centimeters, as indicated in the detail of Figure 1. 2.5.4. Use the turnbuckles to apply equal loads of 10 pounds to the two tapes as indicated by the spring scales. (Note the use of swivels to prevent axial twisting.) 2.5.5. Place a piece of graph paper under the zero interval and each interval to be calibrated as shown in the detail in Figure 1. Adjust the tapes and the paper so that the former are precisely aligned with the lateral rulings of the paper. It is convenient but not necessary for these to be the bold centimeter rulings of the paper. Note the amount of separation of the tapes at the zero interval and make corresponding adjustments at each calibration interval of interest. In this way, parallelism of the two tapes is easily verified. 2.5.6. Make final adjustment of tensions on the tapes and recheck for parallelism at all test points before taking the readings described in 2.5.7. Do not disturb during the measurement sequence. 2.5.7 Read the distances A, B, C, and D as indicated in the detail of Figure 1. Note that A and B are for the zero (or first) interval and are the same for all test intervals. C and D have subscripts 1, 2, etc. corresponding to the interval, i, calibrated. Make all readings to the center of the graduation mark tested and estimate to the closest 0.1 mm. Record all readings in centimeters. 2.5.8. Record all measurements as Trial 1. 2.5.9. Release the tension to the tapes and reapply it. 2.5.10. Displace each piece of graph paper a few millimeters, then readjust the load, check for parallelism, and record a second series of measurements as Trial 2. 2.5.11. Release the tension to the tapes and reapply it. 2.5.12. Displace each piece of graph paper a few millimeters, then readjust the load, check for parallelism, and record a second series of measurements as Trial 3. 2.5.13. After all measurements are completed; apply a thin film of oil to the tapes. October 2014 SOP 12 Page 6 of 12 3. Calculations 3.1. Calculate and record A – B – C + D for each trial, then record the value of R, the range of these values (difference of highest and lowest) for each scale interval. The range should not exceed 0.15 cm. Sum the values for A, B, C, D for the three trials to use when calculating the length, L, of each interval. 3.2. The value obtained from ƩA – ƩB – ƩC + ƩD must equal the sum of the column A – B – C + D, otherwise an error has been made in the calculations. 3.3. Calculate the length of the test tape at each calibration interval according to the following equation. )(3 DCBAcf SL X  (1) 4. Measurement Assurance 4.1. Duplicate the process with a suitable check standard or have a suitable range of check standards for the laboratory. See NISTIR 7383 SOP 17, SOP 20 and NISTIR 6969 SOP 30. Plot the check standard length and verify it is within established limits OR a t-test may be incorporated to check the observed value against an accepted value. The mean of the check standard observations is used to evaluate bias and drift over time. Check standard observations are used to calculate the standard deviation of the measurement process which contributes to the Type A uncertainty components. 4.2 If a standard deviation chart is used for measurement assurance, the standard deviation of each combination of 3 Trials is calculated and the pooled (or average) standard deviation is used as the estimate of variability in the measurement process. Note: the pooled or average standard deviation over time will reflect varying conditions of test items that are submitted to the laboratory. A standard deviation chart will be needed for each interval calibrated (at least initially) so that the variability resulting from transfers will be measured. 5. Assignment of Uncertainty 5.1. The limits of expanded uncertainty, U, include estimates of the standard uncertainty of the length standards used, u s, estimates of the standard deviation of the measurement process, s p, and estimates of the effect of other components associated with this procedure, uo. These estimates should be combined using the root-sum-squared method (RSS), and the expanded uncertainty, U, reported with a coverage factor to be determined based on degrees of freedom, which if large enough will be 2, ( k = 2), to give an approximate 95 percent level of confidence. See NISTIR 6969, SOP 29 (Standard Operating Procedure for the Assignment of October 2014 SOP 12 Page 7 of 12 Uncertainty) for the complete standard operating procedure for calculating the uncertainty. 5.1.1 The expanded uncertainty for the standard, U, is obtained from the calibration report. The combined standard uncertainty, u c, is used and not the expanded uncertainty, U, therefore the reported uncertainty for the standard will usually need to be divided by the coverage factor k. 5.1.2. The standard deviation of the measurement process, s p, is taken from a control chart for a check standard or standard deviation charts. (See SOP 17, SOP 20, and SOP 30) 5.1.3. Uncertainty associated with bias, ud . Any noted bias that has been determined through analysis of control charts and round robin data must be less than limits provided in SOP 29 and included if corrective action is not taken. See SOP 29 for additional details 5.1.4. Other standard uncertainties usually included at this calibration level include uncertainties associated with the ability to read the graph paper, only part of which is included in the process variability due to parallax and visual capabilities, and uncertainties associated with the graduations of the graph paper. Table 3. Example uncertainty budget table. Component Description Reference us Standard uncertainty for standards Calibration report, divide by k s p Standard uncertainty for the process Measurement assurance process; range charts ugp Standard uncertainty for graph paper Must be assessed experimentally utw Standard uncertainty for the spring scales Must be assessed experimentally or from a calibration certificate ud Standard uncertainty for disparity due to drift/bias Rectangular distribution and reasons, 0.577 d, 0.29 d; SOP 29 (NISTIR 6969) ures Standard uncertainty due to resetting of the tapes Must be assessed experimentally uo Standard uncertainty for other factors 6. Report Report results as described in SOP No. 1 Preparation of Calibration Certificates. October 2014 SOP 12 Page 8 of 12 Appendix A Tape-to-Tape Method Data Sheet Date Environmental parameters Metrologist Before After Unc/ability to measure Test No. Temperature °C °C s p in Pressure mmHg mmHg df Humidity % %Based on NISTIR 6969, SOP 29, Appendix A at 95.45 % probability distribution: k factor ID Range S X Tension lb Interval Trial A B C D A-B-C+D 1)(3 DCBAcf  S L23 Ʃ Range 1)(3 DCBAcf  S L23 Ʃ Range 1)(3 DCBAcf  S L23 Ʃ Range 1)(3 DCBAcf  S L23 Ʃ Range 1)(3 DCBAcf  S L23 Ʃ Range 1)(3 DCBAcf  S L23 Ʃ Range October 2014 SOP 12 Page 9 of 12 Appendix B Supplemental Information B.1. Cleaning To clean a steel tape before calibration, first wipe the tape with a soft cloth; then with a soft cloth saturated with alcohol to remove the film of oil used to protect the tape. After calibration, a thin film of light oil, such as sewing machine oil, should be applied to the tape for protection. B.2. Tolerances The tolerances for measuring tapes are those stated below. Table B-1. Tolerances for a 30 meter tape. Length Interval Tolerances 0 through 15 meter 1.27 mm (0.050 inch) 15 through 22 meter 1.91 mm (0.075 inch) 22 through 30 meter 2.54 mm (0.100 inch) B.2.1. The inaccuracy in the length of the ribbon, when supported on a horizontal surface with a tension of 10 pounds at a temperature of 20 °C (68 °F) shall not exceed 0.050 inch for the 75 foot length, and 0.100 inch for the 100 foot length. B.3. Tension Specifications The length of a tape will be affected by the temperature of the tape, the tension applied to the tape, and the manner in which the tape is supported. The tape will stretch when tension is applied and will return to its normal length when the tension is removed, provided the tape has not been permanently deformed when it was stretched. The tensions, at which steel tapes are to be calibrated, expressed in terms of the load in kilograms (or pounds) to be applied to obtain the tension, are stated below. The loads should be accurate within 45 g (0.01 lb). Table B-2. Tension to be applied, in terms of load. Length Interval Tension Less than 10 m (25 ft) 2 kg (3.5 lb) 10 m through 30 m (25 ft through 100 ft) 5 kg (10 lb) Greater than 30 m (100 ft) 10 kg (20 lb) B.4. Methods of Support and Tension Considerations Tapes calibrated in a State laboratory are normally supported on a horizontal surface October 2014 SOP 12 Page 10 of 12 throughout the entire length of the tape. Also, tapes may be calibrated and used when supported in catenary types of suspension. In these cases, the tape is supported at equidistant points because the weight of the tape affects its length. The weight of the tape increases the tension and the “sag” causes the horizontal length to be shorter than when the tape is supported throughout its length. Equations are given in GMP No. 10 to compute the horizontal straight-line distance of a tape supported at N number of equidistant catenary suspensions and for computing the tension of accuracy, defined as the tension that must be applied to the tape interval to produce its designated nominal length at the observed temperature of the tape. It is sufficient to provide the user of a steel tape with the calibrated length of the tape under standard temperature and tension conditions, the weight per unit-length of the tape, and the AE value for the tape, as might be requested. This information will enable the user to compute the values desired using the equations cited above. B.5. Zero Reference Point Metal measuring tapes submitted to a State laboratory for calibration normally will be made of steel. Generally, these tapes will have a ring on the end of the tape. For maximum calibration and measurement accuracy, a tape should have a blank end between the ring and the zero graduation. The zero graduation is then more precisely defined and more easily referenced for calibration and use. Tapes that have the ring as part of the measuring portion of the tape are more difficult to calibrate than a tape with a blank end. When the ring is part of the measuring portion of the tape, the zero reference point shall be the outside end of the ring unless otherwise specified. It is more difficult to obtain a good zero reference setting on the ring due to its curvature and to parallax in reading the edge of the ring against a reference mark. Additionally, the ring may become permanently deformed in use and change the length of the tape. For these tapes the NIST normally calibrates from the 1 foot mark over the length of the tape, and then calibrates from the ring to the 1 foot mark. These values are reported separately so the user can obtain maximum measurement accuracy by using the 1 foot graduation as the zero reference point. When the ring is part of the measuring range of the tape, a special holder for the ring is needed to clamp the tape to the length bench. A strap with an open area in the middle is needed to permit the end of the ring to be seen. The strap is slipped through the ring and the strap is clamped to the length bench. See example below. Figure B-1. Strap, holder, and tape ring. The edge of the tape to be calibrated (the reading edge) is the edge nearest the observer October 2014 SOP 12 Page 11 of 12 when the zero graduation is to the observer’s left. When viewed through a microscope, some graduations will appear to have irregular edges. The portion of the graduation to be used for calibration is the portion of the graduation at the bottom of the reading edge of the tape. This provides a reference point that can be repeated and referenced by others. Do not attempt to estimate the ‘best overall’ edge of a graduation because this is not easily repeatable and cannot be accurately reproduced by other. If the graduations to be calibrated do not reach to the edge of the tape, the tape should not be calibrated . B.6. Temperature Considerations No temperature correction is required, provided the test tape and the standard tape are at the same temperature and of the same material. This will be the case when the measurements are made inside a building. Tapes of the same color would be expected to attain the same temperature, even in sunlight. However, black and white tapes have shown temperature differences of as much as 8 °C when exposed to direct sunlight. In such cases, the temperature differences, even when measured, would be uncertain due to variability of exposure along the length of the tape. Accordingly, calibrations in the laboratory are preferred, when possible. B.7. Invar tapes Invar is an alloy of nickel and steel. Invar tapes are used to obtain measurements of greater accuracy than can be made with steel tapes, because invar has a very low coefficient of expansion. It has the added benefit of being very slow to tarnish from exposure to the atmosphere. However, invar tapes require very careful handling to prevent twists and kinks. The load to be applied to an invar tape to maintain the desired tension is normally 20 lb. A load of 40 lb is used for the higher tension to determine the AE value (see C.8.). For metric tapes, the normal load is 5 kg. A load of 10 kg is used to determine the AE value. B.8. AE Value The AE Value (area elongation value) for a tape is determined by first calibrating the tape under its normal tension. The load is then increased by10 lb or 20 lb and one length interval is recalibrated to determine the length of the tape under the increased tension. The AE factor is computed with the following equation. n LLLQQAE 0101  (B-1) For example, a 100 foot tape is calibrated from 0 feet to 100 feet with a load of 10 lb applied to the tape with a resulting length of 99.992 feet. The load is increased to 20 lb and the new length is found to be 100.004 feet. The AE value is: October 2014 SOP 12 Page 12 of 12 lb feet feet lb feet lb lb AE 333 83 012 .0000 1100 )992 .99 004 .100 ()10 20 (  It is recommended that the AE value be determined over the longest interval that is convenient to measure. This minimizes the error in the AE value because of the better readability of the change in length B.9 Weight per Unit Length The weight per unit length of a tape can be determined as follows: B.9.1. Weigh the tape and reel (or case). B.9.2. Remove the tape from the reel or case and weigh the empty reel (or case). B.9.3. Measure the length of any blank ends on the tape and add this to the measuring length. B.9.4. Correct for the weight of the loop on the tape. The weight of the loop that is normally used on steel tapes is approximately 2.5 grams. The weight per unit length is the computed as follows. ends blank of length tape of length loop of weight reel empty of weight reel loaded of Weight Length Unit per Weight October 2014 SOP 23 Page 1 of 11 SOP 23 Recommended Standard Operations Procedure for Calibrations of PI Tapes Bench Method Introduction 1.1 Purpose PI tapes are circumferential measuring tapes that are wrapped around a circular or cylindrical object and when read, indicate the diameter of the object. The tapes are manufactured for various diameter ranges and for either outside diameters (OD) or inside diameters (ID). The test method described here provides a procedure to calibrate such tapes to four decimal places in inches (diameter). Acalibrated length bench (transfer standard, calibrated using SOP 11) is used as the standard. 1.2 Prerequisites 1.2.1 Valid calibration certificates with appropriate values and uncertainties must be available for all of the standards used in the calibration. All standards must have demonstrated metrological traceability to the international system of units (SI), which may be to the SI through a National Metrology Institute such as NIST. 1.2.2. The ocular microscope used in measuring differences in lengths must be in good operating condition and must be equipped with a graduated reticle having established traceability. 1.2.3. The operator must be trained and experienced in precision measuring techniques with specific training in GMP 2, GMP 8, GMP 9, SOP 11, SOP 23, and SOP 29. 1.2.4. Laboratory facilities must comply with following minimum conditions to meet the expected uncertainty possible with this procedure. Equilibration of length bench and tapes requires environmental stability of the laboratory within the stated limits for a minimum of 24 hours before a calibration. Table 1. Environmental conditions. Temperature Requirements During a Calibration Relative Humidity (%) Lower and upper limits: 18 °C to 22 °C Maximum changes: < ± 1 °C / 24 h and ± 0.5 °C / 1 h 40 to 60 ± 10 / 4 h October 2014 SOP 23 Page 2 of 11 Methodology 2.1. Scope, Precision, Accuracy The precision and accuracy attainable depend upon the care exercised in aligning the tape on the length bench, and the skill acquired in the use of an optical micrometer to measure scale differences. 2.2 Physical Characteristics 2.2.1. PI tapes are manufactured with corrections built into the tape where they will indicate the proper diameter without using extensive mathematical corrections. These corrections will have to be taken into consideration when calibrating a PI tape on a flat horizontal surface. OD PI tapes are referenced to 20 °C with 2.25 kg (5 lb) tension (within ± 0.5 percent of the load, i.e., Class F mass standards are more than adequate.) applied. There is concern about the proper tension to apply to an ID PI tape since they are normally used under some degree of compression. Most tapes are made of 1095 spring steel, which has a coefficient of thermal expansion of 0.00000633 / °F. Figure 1. Schematic drawing of Pi tape As the tape is wrapped around the circumference of an object, the reading portion of the tape and the vernier come into alignment. The smallest graduation on the main body equals 0.5 mm (0.025 in). The line on the October 2014 SOP 23 Page 3 of 11 vernier scale which comes closest to lining up with a graduation on the main body is the value to be added to the reading. The vernier allows readings to be made to the nearest 0.001 in, as shown in Figure 2. Figure 2. Reading the Vernier scale on a Pi tape. 2.3. Fundamental Considerations 2.3.1. The relationship of circumference, C, diameter, D, and pi ( ) is C = (D), D = C/ . See Figure 3 where  = 3.141592654 Figure 3. Circumference, diameter, pi ( ). 2.3.2. The thickness of the tape will affect the measurements in two locations, as shown in Figure 4. L1 is the outside portion of the tape, which becomes longer when the tape is bent. L2 is the center of the tape (neutral axis). It remains the same length when the tape is bent. L3 is the inside portion of the tape and it compresses and becomes shorter when the tape is bent. Figure 4. Side view of tape. October 2014 SOP 23 Page 4 of 11 2.3.3 Another point that must be taken into consideration is that the tape was tested for accuracy on a flat horizontal surface but now it is curved around an object. The tape becomes temporarily deformed but the amount of distortion is predictable using the modulus of elasticity. As shown in the diagram (Figure 4.), the outside portion of the tape stretches and the inside compresses but there is a point midway through the thickness of the tape which does not change in length and is called the neutral axis. For all homogeneous beams with the same modulus of elasticity in tension and compression, the neutral axis passes through the center of the section. If the tape has a thickness of 0.010 inch, the distance from the neutral axis to the surface of the object is 0.005 inch. Therefore, the reading we obtained is not the distance on the outside surface of the tape but is actually the distance along the neutral axis of the tape. When we calculate a diameter, this is actually the diameter of the neutral axis of the tape. Since this thickness affects the diameter in two locations, the amount of correction that must be taken into account is 0.010 inches. 2.4. Summary The tape to be calibrated is laid over the bench scale and tension is applied to ensure that it lies flat on the bench. Differences between the graduation on the tape and that of the bench scale are measured using an ocular micrometer. The temperature of the tape is observed and corrections applied to the reference temperature of 20 °C. Tapes longer than the length bench may be calibrated in segments, using the last calibrated graduation as the zero graduation mark for the succeeding segments. PI tape graduations represent object diameters and may not directly coincide with length bench graduations. 2.5. Apparatus/Equipment Required 2.5.1. Calibrated thermometers with sufficiently small resolution, stability, and uncertainty, capable of indicating temperatures in the range of 15 °C to 30 °C, and accurate to ± 0.5 °C. 2.5.2. Calibrated 5 meter (16 foot) length bench. 2.5.3. Microscope with graduated reticle having graduations spaced at intervals no greater than 0.002 inch. 2.5.4. Clamps, fabric tape in lengths required to connect tension weights to the tape being calibrated, and weights to apply an appropriate tension to the tape under calibration. (See "Tension Specifications", in Appendix C.3 in SOP 11). October 2014 SOP 23 Page 5 of 11 2.6. Symbols Table 2. Symbols used in this procedure. Symbol Description S Standard X Unknown d0i Initial zero difference d0f Final zero difference X0L Left edge of zero on unknown X0R Right edge of zero on unknown S0L Left edge of zero on standard S0R Right edge of zero on standard Xm Center of graduation of unknown Sm Center of graduation of standard d Difference of interval between X and S t Average of initial and final temperatures t corr Temperature correction Ln Nominal length of tape interval under test in inches LX Calibrated length of tape interval under test in inches α Linear coefficient of thermal expansion for the standard bench (0.00001063 / ºC) β Linear coefficient of thermal expansion for the tape (0.00001160 / ºC) Ld Diameter length of interval π 3.141592654 2.7. Procedure 2.7.1. Inspect the tape to ensure that no kinks, dents, or other damage are present which will affect the accuracy of the tape. 2.7.2. Clean the tape by first wiping with a soft cloth, and then with a soft cloth saturated with alcohol to remove the protective oil film. (See Appendix C, Section C.1 for SOP 11). 2.7.3. Place the tape, under the clamp, at the zero end of the bench, and adjust so the starting mark on the tape is near the zero graduation on the length bench. 2.7.4. Lay the tape flat on the bench. The tape should extend well beyond the end roller of the bench to permit tension to be applied. Slide one end of a fabric strap onto another tape clamp. Fasten this tape clamp to the tape on the portion that extends below the end roller. Hang the tension weight from the bottom of the fabric strap. Check to see that the tape is lying straight on the bench and parallel to the bench scale. Adjust, if necessary. Apply tension using a weight of 2.25 kg (5 lb) unless other tension is specified. (See Appendix C.3 in SOP 11.) When a tape is shorter than the October 2014 SOP 23 Page 6 of 11 5 m length bench, fabric strap will be be used to effectively lengthen the tape so that the tension weight is properly suspended below the rollers of the length bench. 2.7.5. Lay two thermometers (see 2.4.1.) on the bench at intervals of one-third and two-thirds of the length of the bench to determine its temperature at the time of test. 2.7.6. Adjust the tape clamp on the zero end of the bench so that the tape zero graduation coincides with the center of the zero graduation of the bench. 2.7.7. Check all alignments and coincidence of zero graduations before proceeding with the calibration. Use the lateral adjustments of the left end of the bench to facilitate alignment. Caution! Take care that the tape is not touched or disturbed during the following sequence of measurements. Record all observations on a suitable data sheet. (See SOP 11, Appendix A.) 2.7.8. Record the temperatures indicated by the two thermometers. 2.7.9. Place the ocular microscope on the bench in the vicinity of the zero position and align it so that its scale is parallel to the tape under test. (See GMP No. 2 for instructions on how to make readings.) 2.7.9.1.Observe readings of left and right sides of zero graduation of tape and record to the nearest 0.001 inch. The mean of these values will be used to determine the center of the tape graduation. 2.7.9.2.Observe readings of left and right sides of zero graduation of bench and record to the nearest 0.001 inch. The mean of these values will be used to determine the center of the bench graduation. 2.7.10. Move the ocular microscope successively to each graduation that is to be calibrated and record readings similarly as in 2.7.9.1 and 2.7.9.2. 2.7.11. Return the ocular microscope to the zero graduation and repeat readings to verify that the tape has not moved. Accept all previous data if the initial zero reading does not disagree with the final zero reading by more than 0.001 inch; otherwise, discard previous data and repeat entire sequence of readings until a satisfactory set is obtained. 2.7.12. Loosen the tape by removing the tension weights, move the tape, hang the tension weights back on the fabric strap, and realign the zero marks on the tape and bench to coincidence. October 2014 SOP 23 Page 7 of 11 2.7.13. Repeat 2.7.9. thru 2.7.12. for every 15 foot (5 meter) section of the tape requiring calibration. This will require repositioning the tape, aligning the last measured interval graduation on the tape with the zero graduation of the bench. 2.7.14. Make a second set of measurements as directed in 2.7.9 through 2.7.13. 2.7.15. After the final measurement is taken and accepted, record the temperatures indicated by the two thermometers. 2.7.16. After all measurements are completed, apply a thin film of oil to the tape. 3. Calculations 3.1. Calculate the Initial Zero Difference, d 0i (2.6.9.) 2)( 00000 RLRLi SSXXd  (1) 3.2. Calculate the Final Zero Difference, d0f (2.6.11.) 2)( 00000 RLRLf SSXXd  (2) 3.3. Calculate the center of graduation for unknown, Xm, and standard, Sm , for each set of measurements and each scale interval. 2)( RLmXXX  (3) 2)( RLmSSS  (4) 3.4. Calculate the difference, d, between X and S for each scale interval. mm SXd  (5) 3.5. Obtain the correction to the standard, CS, for the measured interval from the calibration certificate for the length bench scale. 3.6. Calculate the temperature correction, t corr .   )(20    tLt ncorr (6) 3.7. Calculate a correction, CX, for each trial and each scale interval. October 2014 SOP 23 Page 8 of 11 corr SX tCdC  (7) 3.8. Calculate and report the mean, XC , of the two corrections for each interval. 3.9. Calculate the length of tape under the 10 pound load. XnX CLL  (8) 3.10. As appropriate, convert the flat length of the tape to an outside diameter (OD) measurement using the following formula: )( inches in thickness tape LL XOD  (9) 3.11. As appropriate, convert the flat length of the tape to an inside diameter (ID) measurement using the following formula: )( inches in thickness tape LL XID  (9) 4. Measurement Assurance 4.1. Duplicate the process with a suitable check standard or have a suitable range of check standards for the laboratory. See NISTIR 7383 SOP 17, SOP 20 and NISTIR 6969 SOP 30. Plot the check standard length and verify it is within established limits OR a t-test may be incorporated to check the observed value against an accepted value. The mean of the check standard observations is used to evaluate bias and drift over time. Check standard observations are used to calculate the standard deviation of the measurement process which contributes to the Type A uncertainty components. 4.2 If a standard deviation chart is used for measurement assurance, the standard deviation of each combination of Trial 1 and Trial 2 is calculated and the pooled (or average) standard deviation is used as the estimate of variability in the measurement process. Note: the pooled or average standard deviation over time will reflect varying conditions of test items that are submitted to the laboratory. A standard deviation chart will be needed for each interval calibrated (at least initially) so that the variability resulting from transfers will be measured. 5. Assignment of Uncertainty 5.1. The limits of expanded uncertainty, U, include estimates of the standard uncertainty of the length standards used, u s, estimates of the standard deviation of the measurement process, s p, and estimates of the effect of other components associated with this procedure, uo. These estimates should be combined using the root-sum-squared method (RSS), and the expanded uncertainty, U, reported with a coverage factor to be determined based on degrees of freedom, which if large October 2014 SOP 23 Page 9 of 11 enough will be 2, ( k = 2), to give an approximate 95 percent level of confidence. See NISTIR 6969, SOP 29 (Standard Operating Procedure for the Assignment of Uncertainty) for the complete standard operating procedure for calculating the uncertainty. 5.1.1 The expanded uncertainty for the standard, U, is obtained from the calibration report. The combined standard uncertainty, u c, is used and not the expanded uncertainty, U, therefore the reported uncertainty for the standard will usually need to be divided by the coverage factor k. 5.1.2. The standard deviation of the measurement process, s p, is taken from a control chart for a check standard or standard deviation charts. (See SOP 17, SOP 20, and SOP 30) 5.1.3. Uncertainty associated with bias, ud . Any noted bias that has been determined through analysis of control charts and round robin data must be less than limits provided in SOP 29 and included if corrective action is not taken. See SOP 29 for additional details 5.1.4. Uncertainty associated with temperature correction includes values for the linear coefficients of expansion for the Standard and Unknown, the accuracy of temperature measurements, and factors associated with potential gradients in measuring the temperature on the length bench and tape. 5.1.5. Other standard uncertainties usually included at this calibration level include uncertainties associated with the ability to read the graduated reticle, only part of which is included in the process variability due to parallax and visual capabilities, and uncertainties associated with the graduations of the reticle. October 2014 SOP 23 Page 10 of 11 Table 3. Example uncertainty budget table. Component Description Reference us Standard uncertainty for standards Calibration report, divide by k s p Standard uncertainty for the process Measurement assurance process; range charts ugr Standard uncertainty for graduated reticle Must be assessed experimentally or from a calibration certificate utc Standard uncertainty for temperature correction Calibration certificate ut Standard uncertainty for temperature Handbook 143 accuracy guideline, 0.1 Culce ( ,) Standard uncertainty for linear coefficient of expansion 5 % to 10 % of the coefficient of expansion value utw Standard uncertainty for tension weights Calibration certificate ud Standard uncertainty for disparity due to drift/bias Rectangular distribution and reasons, 0.577 d, 0.29 d; SOP 29 (NISTIR 6969) ures Standard uncertainty due to resetting of the tape Must be assessed experimentally uo Standard uncertainty for other factors 6. Report Report results as described in SOP No. 1 Preparation of Calibration Certificates. October 2014 SOP 23 Page 11 of 11 Appendix A Bench Method Data Sheet Date Environmental parameters Metrologist Before After Unc/ability to measure Test No. Temperature °C °C s p in Pressure mmHg mmHg df Humidity % %Based on NISTIR 6969, SOP 29, Appendix A at 95.45 % probability distribution: k factor ID Range Linear coefficient of thermal expansion S Bench 16 ft (5 m) α 0.000 010 63 /°C X Tape β 0.000 011 60 /°C Tension lb Support Initial temperatures Final temperatures t1 °C t3 °C Average temperature °C t2 °C t4 °C Interval, ft Trial Initial Zero Graduations Bench Tape Tape Bench 0 01 X0L X0R S 0L S 0R Initial zero, difference between X and S, d 0i in 2 X0L X0R S 0L S 0R Initial zero, difference between X and S, d 0i in Intervals after zero Tape Bench X1L X1R Xm S 1L S 1R S m d CS tcorr CX Range 12Average CX in X1L X1R Xm S 1L S 1R S m d CS tcorr CX Range 12Average CX in Final Zero Graduations Tape Bench 0 01 X0L X0R S 0L S 0R Final zero, difference between X and S, d 0f in Difference between Initial Zero and Final Zero, d 0i - d 0f in \| d 0i - d 0f \| ≤ 0.001 in? If YES, accept previous data. 2 X0L X0R S 0L S 0R Final zero, difference between X and S, d 0f in Difference between Initial Zero and Final Zero, d 0i - d 0f in \| d 0i - d 0f \| ≤ 0.001 in? If YES, accept previous data.   )(20    tLt ncorr corr SX tCdC  XnX CLL This page intentionally made blank.
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Let their radii r and their midpoints respectively be O 1 and O 2. Find the greatest possible value of the area of the rectangle O 1 A O 2 B. 1995 Estonia Open Junior 1.4 The midpoint of the hypotenuse A B of the right triangle A B C is K. The point M on the side B C is taken such that B M=2⋅M C. Prove that ∠B A M=∠C K M. 1995 Estonia Open Junior 2.1 A rectangle, whose one sidelength is twice the other side, is inscribed inside a triangles with sides 3 cm, 4 cm and 5 cm, such that the long sides lies entirely on the long side of the triangle. The other two remaining vertices of the rectangle lie respectively on the other two sides of the triangle. Find the lengths of the sides of this rectangle. 1996 Estonia Open Junior 1.4 In a trapezoid, the two non parallel sides and a base have length 1, while the other base and both the diagonals have length a. Find the value of a. 1996 Estonia Open Junior 2.4 A pentagon (not necessarily convex) has all sides of length 1 and its product of cosine of any four angles equal to zero. Find all possible values of the area of such a pentagon. 1997 Estonia Open Junior 1.3 Juku invented an apparatus that can divide any segment into three equal segments. How can you find the midpoint of any segment, using only the Juku made, a ruler and pencil? 1998 Estonia Open Junior 1.3 Two non intersecting circles with centers O 1 and O 2 are tangent to line s at points A 1 and A 2, respectively, and lying on the same side of this line. Line O 1 O 2 intersects the first circle at B 1 and the second at B 2. Prove that the lines A 1 B 1 and A 2 B 2 are perpendicular to each other. 1998 Estonia Open Junior 2.5 The points E and F divide the diagonal B D of the convex quadrilateral A B C D into three equal parts, i.e. \|B E\|=\|E F\|=\|F D\|. Line A E interects side B C at X and line A F intersects D C at Y. Prove that: a) if A B C D is parallelogram then X,Y are the midpoints of B C,D C, respectively, b) if the points X,Y are the midpoints of B C,D C, respectively , then A B C D is parallelogram 1999 Estonia Open Junior 1.2 Two different points X and Y are chosen in the plane. Find all the points Z in this plane for which the triangle X Y Z is isosceles. 1999 Estonia Open Junior 2.3 On the plane there are two non-intersecting circles with equal radii and with centres O 1 and O 2, line s going through these centres, and their common tangent t. The third circle is tangent to these two circles in points K and L respectively, line s in point M and line t in point P. The point of tangency of line t and the first circle is N. a) Find the length of the segment O 1 O 2. b) Prove that the points M,K and N lie on the same line 2000 Estonia Open Junior 1.3 Consider a shape obtained from two equal squares with the same center. Prove that the ratio of the area of this shape to the perimeter does not change when the squares are rotated around their center. 2000 Estonia Open Junior 1.5 Find the total area of the shaded area in the figure if all circles have an equal radius R and the centers of the outer circles divide into six equal parts of the middle circle. 2000 Estonia Open Junior 2.4 In the plane, there is an acute angle ∠A O B . Inside the angle points C and D are chosen so that ∠A O C=∠D O B. From point D the perpendicular on O A intersects the ray O C at point G and from point C the perpendicular on O B intersects the ray O D at point H. Prove that the points C,D,G and H are conlyclic. 2001 Estonia Open Junior 1.3 Consider points C 1,C 2 on the side A B of a triangle A B C, points A 1,A 2 on the side B C and points B 1,B 2 on the side C A such that these points divide the corresponding sides to three equal parts. It is known that all the points A 1,A 2,B 1,B 2,C 1 and C 2 are concyclic. Prove that triangle A B C is equilateral. 2001 Estonia Open Junior 2.2 In a triangle A B C, the lengths of the sides are consecutive integers and median drawn from A is perpendicular to the bisector drawn from B. Find the lengths of the sides of triangle A B C. 2002 Estonia Open Junior 1.1 A figure consisting of five equal-sized squares is placed as shown in a rectangle of size 7×8 units. Find the side length of the squares. 2002 Estonia Open Junior 1.4 Consider a point M inside triangle A B C such that triangles A B M,B C M and C A M have equal areas. Prove that M is the intersection point of the medians of triangle A B C. 2002 Estonia Open Junior 2.3 In a triangle A B C we have \|A B\|=\|A C\| and ∠B A C=α. Let P≠B be a point on A B and Q a point on the altitude drawn from A such that \|P Q\|=\|Q C\|. Find ∠Q P C. 2003 Estonia Open Junior 1.2 Circles with centres O 1 and O 2 intersect in two points, let one of which be A. The common tangent of these circles touches them respectively in points P and Q. It is known that points O 1,A and Q are on a common straight line and points O 2,A and P are on a common straight line. Prove that the radii of the circles are equal. 2003 Estonia Open Junior 1.4 Mari and Juri ordered a round pizza. Juri cut the pizza into four pieces by two straight cuts, none of which passed through the centre point of the pizza. Mari can choose two pieces not aside of these four, and Juri gets the rest two pieces. Prove that if Mari chooses the piece that covers the centre point of the pizza, she will get more pizza than Juri. 2003 Estonia Open Junior 2.2 The shape of a dog kennel from above is an equilateral triangle with side length 1 m and its corners in points A,B and C, as shown in the picture. The chain of the dog is of length 6 m and its end is fixed to the corner in point A. The dog himself is in point K in a way that the chain is tight and points K,A and B are on the same straight line. The dog starts to move clockwise around the kennel, holding the chain tight all the time. How long is the walk of the dog until the moment when the chain is tied round the kennel at full? 2003 Estonia Open Junior 2.4 Consider the points A 1 and A 2 on the side A B of the square A B C D taken in such a way that \|A B\|=3\|A A 1\| and \|A B\|=4\|A 2 B\|, similarly consider points B 1 and B 2,C 1 and C 2,D 1 and D 2 respectively on the sides B C, C D and D A. The intersection point of straight lines D 2 A 1 and A 2 B 1 is E, the intersection point of straight lines A 2 B 1 and B 2 C 1 is F, the intersection point of straight lines B 2 C 1 and C 2 D 1 is G and the intersection point of straight lines C 2 D 1 and D 2 A 1 is H. Find the area of the square E F G H, knowing that the area of A B C D is 1. 2004 Estonia Open Junior 1.2 Diameter A B is drawn to a circle with radius 1. Two straight lines s and t touch the circle at points A and B, respectively. Points P and Q are chosen on the lines s and t, respectively, so that the line P Q touches the circle. Find the smallest possible area of the quadrangle A P Q B. 2004 Estonia Open Junior 2.3 Circles c 1 and c 2 with centres O 1 and O 2, respectively, intersect at points A and B so that the centre of each circle lies outside the other circle. Line O 1 A intersects circle c 2 again at point P 2 and line O 2 A intersects circle c 1 again at point P 1. Prove that the points O 1,O 2,P 1,P 2 and B are concyclic 2005 Estonia Open Junior 1.3 In triangle A B C, the midpoints of sides A B and A C are D and E, respectively. Prove that the bisectors of the angles B D E and C E D intersect at the side B C if the length of side B C is the arithmetic mean of the lengths of sides A B and A C. 2005 Estonia Open Junior 2.3 The vertices of the square A B C D are the centers of four circles, all of which pass through the center of the square. Prove that the intersections of the circles on the square A B C D sides are vertices of a regular octagon. 2006 Estonia Open Junior 1.3 Let A B C D be a parallelogram, M the midpoint of A B and N the intersection of C D with the angle bisector of A B C. Prove that C M and B N are perpendicular iff A N is the angle bisector of D A B 2006 Estonia Open Junior 2.3 Two non-intersecting circles, not lying inside each other, are drawn in the plane. Two lines pass through a point P which lies outside each circle. The first line intersects the first circle at A and A′ and the second circle at B and B′, here A and B are closer to P than A′ and B′, respectively, and P lies on segment A B. Analogously, the second line intersects the first circle at C and C′ and the second circle at D and D′. Prove that the points A,B,C,D are concyclic if and only if the points A′,B′,C′,D′ are concyclic. 2007 Estonia Open Junior 1.2 The sides A B,B C,C D and D A of the convex quadrilateral A B C D have midpoints E,F,G and H. Prove that the triangles E F B,F G C,G H D and H E A can be put together into a parallelogram equal to E F G H. 2007 Estonia Open Junior 1.4 Call a scalene triangle K disguisable if there exists a triangle K′ similar to K with two shorter sides precisely as long as the two longer sides of K, respectively. Call a disguisable triangle integral if the lengths of all its sides are integers. (a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter. (b) Let K be an arbitrary integral disguisable triangle for which no smaller integral disguisable triangle similar to it exists. Prove that at least two side lengths of K are perfect squares. 2007 Estonia Open Junior 2.2 The center of square A B C D is K. The point P is chosen such that P≠K and the angle ∠A P B is right . Prove that the line P K bisects the angle between the lines A P and B P. 2008 Estonia Open Junior 1.3 Let M be the intersection of the medians A B C of the triangle and the midpoint of the side B C. A line parallel to side B C and passing through point M intersects sides A B and A C at points X and Y respectively. Let the point of intersection of the lines X C and M B be Q and let P intersection point of the lines Y B and M C be P . Prove that the triangles D P Q and A B C are similar. 2008 Estonia Open Junior 2.2 In a right triangle A B C, K is the midpoint of the hypotenuse A B and M such a point on the B C that \|B M\|=2\|M C\|. Prove that ∠M A B=∠M K C. 2009 Estonia Open Junior 1.2 The feet of the altitudes drawn from vertices A and B of an acute triangle A B C are K and L, respectively. Prove that if \|B K\|=\|K L\| then the triangle A B C is isosceles. 2009 Estonia Open Junior 2.1 A Christmas tree must be erected inside a convex rectangular garden and attached to the posts at the corners of the garden with four ropes running at the same height from the ground. At what point should the Christmas tree be placed, so that the sum of the lengths of these four cords is as small as possible? 2009 Estonia Open Junior 2.4 The triangle A B C is \|B C\|=a and \|A C\|=b. On the ray starting from vertex C and passing the midpoint of side A B , choose any point D other than vertex C. Let K and L be the projections of D on the lines A C and B C, respectively, K and L. Find the ratio \|D KD L\|. 2010 Estonia Open Junior 1.2 Given a convex quadrangle A B C D with \|A D\|=\|B D\|=\|C D\| and ∠A D B=∠D C A, ∠C B D=∠B A C, find the sizes of the angles of the quadrangle. 2010 Estonia Open Junior 2.3 On the side B C of the equilateral triangle A B C, choose any point D, and on the line A D, take the point E such that \|B A\|=\|B E\|. Prove that the size of the angle A E C is of does not depend on the choice of point D, and find its size. 2011 Estonia Open Junior 1.3 Consider a parallelogram A B C D. a) Prove that if the incenter of the triangle A B C is located on the diagonal B D, then the parallelogram A B C D is a rhombus. b) Is the parallelogram A B C D a rhombus whenever the circumcenter of the triangle A B C is located on the diagonal B D? 2011 Estonia Open Junior 2.3 Consider the diagonals A 1 A 3,A 2 A 4,A 3 A 5,A 4 A 6,A 5 A 4 and A 6 A 2 of a convex hexagon A 1 A 2 A 3 A 4 A 5 A 6. The hexagon whose vertices are the points of intersection of the diagonals is regular. Can we conclude that the hexagon A 1 A 2 A 3 A 4 A 5 A 6 is also regular? 2012 Estonia Open Junior 1.3 A rectangle A B E F is drawn on the leg A B of a right triangle A B C, whose apex F is on the leg A C. Let X be the intersection of the diagonal of the rectangle A E and the hypotenuse B C of the triangle. In what ratio does point X divide the hypotenuse B C if it is known that \|A C\|=3\|A B\| and \|A F\|=2\|A B\|? 2012 Estonia Open Junior 1.5 A hiking club wants to hike around a lake along an exactly circular route. On the shoreline they determine two points, which are the most distant from each other, and start to walk along the circle, which has these two points as the endpoints of its diameter. Can they be sure that, independent of the shape of the lake, they do not have to swim across the lake on any part of their route? 2012 Estonia Open Junior 2.3 Two circles c and c′ with centers O and O′ lie completely outside each other. Points A,B, and C lie on the circle c and points A′,B′, and C lie on the circle c′ so that segment A B∥A′B′, B C∥B′C′, and ∠A B C=∠A′B′C′. The lines A A′,B B', and C C′ are all different and intersect in one point P, which does not coincide with any of the vertices of the triangles A B C or A′B′C′. Prove that ∠A O B=∠A′O′B′. 2012 Estonia Open Junior 2.5 Is it possible that the perimeter of a triangle whose side lengths are integers, is divisible by the double of the longest side length? 2013 Estonia Open Junior 1.4 Inside a circle c with the center O there are two circles c 1 and c 2 which go through O and are tangent to the circle c at points A and B crespectively. Prove that the circles c 1 and c 2 have a common point which lies in the segment A B. 2013 Estonia Open Junior 2.3 In an isosceles right triangle A B C the right angle is at vertex C. On the side A C points K,L and on the side B C points M,N are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point P inside the triangle A B C such that ∠K P L=∠M P N=45 o. 2014 Estonia Open Junior 1.5 In a triangle A B C the midpoints of B C,C A and A B are D,E and F, respectively. Prove that the circumcircles of triangles A E F,B F D and C D E intersect all in one point. 2014 Estonia Open Junior 2.2 In a scalene triangle one angle is exactly two times as big as another one and some angle in this triangle is 36 o. Find all possibilities, how big the angles of this triangle can be. 2014 Estonia Open Junior 2.5 In the plane there are six different points A,B,C,D,E,F such that A B C D and C D E F are parallelograms. What is the maximum number of those points that can be located on one circle? 2015 Estonia Open Junior 1.5 Let A B C be an acute triangle. The arcs A B and A C of the circumcircle of the triangle are reflected over the lines AB and A C, respectively. Prove that the two arcs obtained intersect in another point besides A. 2015 Estonia Open Junior 2.5 Let A B C be an acute-angled triangle, H the point of intersection of its altitudes , and A A′ the diameter of the circumcircle of triangle A B C. Prove that the quadrilateral H B A′C is a parallelogram. 2016 Estonia Open Junior 1.5 A right triangle A B C has the right angle at vertex A. Circle c passes through vertices A and B of the triangle A B C and intersects the sides A C and B C correspondingly at points D and E. The line segment C D has the same length as the diameter of the circle c. Prove that the triangle A B E is isosceles. 2016 Estonia Open Junior 2.4 Let d be a positive number. On the parabola, whose equation has the coefficient 1 at the quadratic term, points A,B and C are chosen in such a way that the difference of the x-coordinates of points A and B is d and the difference of the x-coordinates of points B and C is also d. Find the area of the triangle A B C. 2016 Estonia Open Junior 2.5 On the plane three different points P,Q, and R are chosen. It is known that however one chooses another point X on the plane, the point P is always either closer to X than the point Q or closer to X than the point R. Prove that the point P lies on the line segment Q R. 2017 Estonia Open Junior .1.5 Find all possibilities: how many acute angles can there be in a convex polygon? 2018 Estonia Open Junior 1.5 Let M be the intersection of the diagonals of a cyclic quadrilateral A B C D. Find the length of A D, if it is known that A B=2 mm , B C=5 mm, A M=4 mm, and C D C M=0.6. 2018 Estonia Open Junior 2.5 Medians A D,B E, and C F of triangle A B C intersect at point M. Is it possible that the circles with radii M D,M E, and M F a) all have areas smaller than the area of triangle A B C, b) all have areas greater than the area of triangle A B C, c) all have areas equal to the area of triangle A B C? 2019 Estonia Open Junior 1.5 Point M lies on the diagonal B D of parallelogram A B C D such that M D=3 B M. Lines A M and B C intersect in point N. What is the ratio of the area of triangle M N D to the area of parallelogram A B C D? 2019 Estonia Open Junior 2.1 A pentagon can be divided into equilateral triangles. Find all the possibilities that the sizes of the angles of this pentagon can be. 2019 Estonia Open Junior 2.5 Different points C and D are chosen on a circle with center O and diameter A B so that they are on the same side of the diameter A B. On the diameter A B is chosen a point P different from the point O such that the points P,O,D,C are on the same circle. Prove that ∠A P C=∠B P D. 2020 Estonia Open Junior 1.5 A circle c with center A passes through the vertices B and E of a regular pentagon A B C D E. The line B C intersects the circle c for second time at point F. Prove that the lines D E and E F are perpendicular. 2020 Estonia Open Junior 2.5 The circle ω 2 passing through the center O of the circle ω 1, is tangent to the circle ω 2 at the point A. On the circle ω 2, the point C is taken so that the ray A C intersects the circle ω 1 for second time at point D, the ray O C intersects the circle ω 1 at point E and the lines D E and A O are parallel. Find the size of the angle D A E. Senior 1993 Estonia Open p5 Within an equilateral triangle A B C, take any point P. Let L,M,N be the projections of P on sides A B,B C,C A respectively. Prove that A P N L=B P L M=C P M N. 1994 Estonia Open 1.4 Prove that if A C B C=A B+B C A C in a triangle A B C , then ∠B=2∠A . 1994 Estonia Open 2.2 The two sides B C and C D of an inscribed quadrangle A B C D are of equal length. Prove that the area of this quadrangle is equal to S=1 2⋅A C 2⋅sin∠A 1995 Estonia Open Senior 1.3 We call a tetrahedron a "trirectangular " if it has a vertex (we call this is called a "right-angled" vertex) in which the planes of the three sides of the tetrahedron intersect at right angles. Prove the "three-dimensional Pythagorean theorem": The square of the area of the opposite face of the "right-angled" vertex of the ""trirectangular " tetrahedron is equal to the sum of the squares of the areas of three other sides of the tetrahedron . 1995 Estonia Open Senior 2.4 Find all points on the plane such that the sum of the distances of each of the four lines defined by the unit square of that plane is 4. 1996 Estonia Open Senior 1.4 A unit square has a circle of radius r with center at it's midpoint. The four quarter circles are centered on the vertices of the square and are tangent to the central circle (see figure). Find the maximum and minimum possible value of the area of the striped figure in the figure and the corresponding values of r such these, the maximum and minimum are achieved. 1996 Estonia Open Senior 2.4 The figure shows a square and a circle with a common center O, with equal areas of striped shapes. Find the value of cos a. 1997 Estonia Open Senior 1.4 Let H,K,L be the feet from the altitudes from vertices A,B,C of the triangle A B C, respectively. Prove that\|A K\|⋅\|B L\|⋅\|C H\|=\|H K\|⋅\|K L\|⋅\|L H\|=\|A L\|⋅\|B H\|⋅\|C K\|. 1997 Estonia Open Senior 1.5 Is it possible to fill space with regular tetrahedrons so that the peak of one tetrahedron does not coincide with another tetrahedron at a point other than the top? 1997 Estonia Open Senior 2.3 The figure shows a square and three circles of equal radius tangent to each other and square passes. Find the radius of the circles if the square length is 1. 1998 Estonia Open Senior 1.2 Prove that the parallelogram A B C D with relation ∠A B D+∠D A C=90 o, is either a rectangle or a rhombus. 1998 Estonia Open Senior 2.1 Circles C 1 and C 2 with centers O 1 and O 2 respectively lie on a plane such that that the circle C 2 passes through O 1. The ratio of radius of circle C 1 to O 1 O 2 is 2+3–√−−−−−−√. a) Prove that the circles C 1 and C 2 intersect at two distinct points. b) Let A,B be these points of intersection. What proportion of the area of circle is C 1 is the area of the sector A O 1 B ? 1998 Estonia Open Senior 2.5 The plane has a semicircle with center O and diameter A B. Chord C D is parallel to the diameter A B and ∠A O C=∠D O B=7 16 (radians). Which of the two parts it divides into a semicircle is larger area? 1999 Estonia Open Senior 1.5 On the side B C of the triangle A B C a point D different from B and C is chosen so that the bisectors of the angles A C B and A D B intersect on the side A B. Let D′ be the symmetrical point to D with respect to the line A B. Prove that the points C,A and D′ are on the same line 1999 Estonia Open Senior 2.3 Two right triangles are given, of which the incircle of the first triangle is the circumcircle of the second triangle. Let the areas of the triangles be S and S′ respectively. Prove that S S′≥3+2 2–√ 1999 Estonia Open Senior 2.5 Inside the square A B C D there is the square A′B′C′D′ so that the segments A A′,B B′,C C′ and D D′ do not intersect each other neither the sides of the smaller square (the sides of the larger and the smaller square do not need to be parallel). Prove that the sum of areas of the quadrangles A A′B′B and C C′D′D is equal to the sum of areas of the quadrangles B B′C′C and D D′A′A. 2000 Estonia Open Senior 1.3 In the plane, the segments A B and C D are given, while the lines A B and C D intersect. Prove that the set of all points P in the plane such that triangles A B P and C D P have equal areas , form two lines intersecting at the intersection of the lines A B and C D. 2000 Estonia Open Senior 2.4 The diagonals of the square A B C D intersect at P and the midpoint of the side A B is E. Segment E D intersects the diagonal A C at point F and segment E C intersects the diagonal B D at G. Inside the quadrilateral E F P G, draw a circle of radius r tangent to all the sides of this quadrilateral. Prove that r=\|E F\|−\|F P\|. 2001 Estonia Open Senior 1.1 Points A,B,C,D,E and F are given on a circle in such a way that the three chords A B,C D and E F intersect in one point. Express angle E F A in terms of angles ABC and CDE (find all possibilities). 2001 Estonia Open Senior 2.3 Let us call a convex hexagon A B C D E F boring if ∠A+∠C+∠E=∠B+∠D+∠F. a) Is every cyclic hexagon boring? b) Is every boring hexagon cyclic? 2002 Estonia Open Senior 1.2 The sidelengths of a triangle and the diameter of its incircle, taken in some order, form an arithmetic progression. Prove that the triangle is right-angled. 2002 Estonia Open Senior 1.4 In a triangle A B C we have ∠B=2⋅∠C and the angle bisector drawn from A intersects B C in a point D such that \|A B\|=\|C D\|. Find ∠A. 2002 Estonia Open Senior 2.3 Let A B C D be a rhombus with ∠D A B=60 o. Let K,L be points on its sides A D and D C and M a point on the diagonal A C such that K D L M is a parallelogram. Prove that triangle B K L is equilateral. 2003 Estonia Open Senior 1.2 Four rays spread out from point O in a 3-dimensional space in a way that the angle between every two rays is a. Find cos a. 2003 Estonia Open Senior 2.4 Consider the points D,E and F on the respective sides B C,C A and A B of the triangle A B C in a way that the segments A D,B E and C F have a common point P. Let \|A P\|\|P D\|=x,\|B P\|\|P E\|=y and \|C P\|\|P F\|=z. Prove that x y z−(x+y+z)=2. 2004 Estonia Open Senior 1.3 a) Does there exist a convex quadrangle A B C D satisfying the following conditions (1) A B C D is not cyclic; (2) the sides A B,B C,C D and D A have pairwise different lengths; (3) the circumradii of the triangles A B C,A D C,B A D and B C D are equal? b) Does there exist such a non-convex quadrangle? 2004 Estonia Open Senior 1.5 Find the smallest real number x for which there exist two non-congruent triangles with integral side lengths having area x. 2004 Estonia Open Senior 2.4 On the circumcircle of triangle A B C, point P is chosen, such that the perpendicular drawn from point P to line A C intersects the circle again at a point Q, the perpendicular drawn from point Q to line A B intersects the circle again at a point R and the perpendicular drawn from point R to line B C intersects the circle again at the initial point P. Let O be the centre of this circle. Prove that ∠P O C=90 o. 2005 Estonia Open Senior 1.2 Two circles c 1 and c 2 with centres O 1 and O 2, respectively, are touching externally at P. On their common tangent at P, point A is chosen, rays drawn from which touch the circles c 1 and c 2 at points P 1 and P 2 both different from P. It is known that ∠P 1 A P 2=120 o and angles P 1 A P and P 2 A P are both acute. Rays A P 1 and A P 2 intersect line O 1 O 2 at points G 1 and G 2, respectively. The second intersection between ray A O 1 and c 1 is H 1, the second intersection between ray A O 2 and c 2 is H 2. Lines G 1 H 1 and A P intersect at K. Prove that if G 1 K is a tangent to circle c 1, then line G 2 A is tangent to circle c 2 with tangency point H 2. 2005 Estonia Open Senior 2.4 Three rays are going out from point O in space, forming pairwise angles α,β and γ with 0 o<α≤β≤γ<180 o. Prove that sin α 2+sin β 2>sin γ 2. 2006 Estonia Open Senior 1.3 Let A B C be an acute triangle and choose points A 1,B 1 and C 1 on sides B C,C A and A B, respectively. Prove that if the quadrilaterals A B A 1 B 1,B C B 1 C 1 and C A C 1 A 1 are cyclic then their circumcentres lie on the sides of A B C. 2006 Estonia Open Senior 2.3 Four points A,B,C,D are chosen on a circle in such a way that arcs A B,B C, and C D are of the same length and the arc D A is longer than these three. Line A D and the line tangent to the circle at B intersect at E. Let F be the other endpoint of the diameter starting at C of the circle. Prove that triangle D E F is equilateral. 2007 Estonia Open Senior 1.2 Three circles with centres A, B, C touch each other pairwise externally, and touch circle c from inside. Prove that if the centre of c coincideswith the orthocentre of triangle ABC, then ABC is equilateral. 2007 Estonia Open Senior 2.3 Tangents l 1 and l 2 common to circles c 1 and c 2 intersect at point P, whereby tangent points remain to different sides from P on both tangent lines. Through some point T, tangents p 1 and p 2 to circle c 1 and tangents p 3 and p 4 to circle c 2 are drawn. The intersection points of l 1 with lines p 1,p 2,p 3,p 4 are A 1,B 1,C 1,D 1, respectively, whereby the order of points on l 1 is: A 1,B 1,P,C 1,D 1. Analogously, the intersection points of l 2 with lines p 1,p 2,p 3,p 4 are A 2,B 2,C 2,D 2, respectively. Prove that if both quadrangles A 1 A 2 D 1 D 2 and B 1 B 2 C 1 C 2 are cyclic then radii of c 1 and c 2 are equal. 2007 Estonia Open Senior 2.5 Consider triangles whose each side length squared is a rational number. Is it true that (a) the square of the circumradius of every such triangle is rational; (b) the square of the inradius of every such triangle is rational? 2008 Estonia Open Senior 1.2 Let O be the circumcentre of triangle A B C. Lines A O and B C intersect at point D. Let S be a point on line B O such that D S∥A B and lines A S and B C intersect at point T. Prove that if O,D,S and T lie on the same circle, then A B C is an isosceles triangle. 2008 Estonia Open Senior 2.3 Two circles are drawn inside a parallelogram A B C D so that one circle is tangent to sides A B and A D and the other is tangent to sides C B and C D. The circles touch each other externally at point K. Prove that K lies on the diagonal A C. 2009 Estonia Open Senior 1.3 Three circles in a plane have the sides of a triangle as their diameters. Prove that there is a point that is in the interior of all three circles. 2009 Estonia Open Senior 1.5 Let any point D be chosen on the side B C of the triangle A B C. Let the radii of the incircles of the triangles A B C,A B D and A C D be r 1,r 2 and r 3. Prove that r 1<r 2+r 3. 2009 Estonia Open Senior 2.4 a) An altitude of a triangle is also a tangent to its circumcircle. Prove that some angle of the triangle is larger than 90 o but smaller than 135 o. b) Some two altitudes of the triangle are both tangents to its circumcircle. Find the angles of the triangle. 2010 Estonia Open Senior 1.4 Circle c passes through vertices A and B of an isosceles triangle ABC, whereby line A C is tangent to it. Prove that circle c passes through the circumcenter or the incenter or the orthocenter of triangle A B C. 2010 Estonia Open Senior 2.1 The diagonals of trapezoid A B C D with bases A B and C D meet at P. Prove the inequality S P A B+S P C D>S P B C+S P D A, where S X Y Z denotes the area of triangle X Y Z. 2011 Estonia Open Senior 1.3 Consider an acute-angled triangle A B C and its circumcircle. Let D be a point on the arc A B which does not include point C and let A 1 and B 1 be points on the lines D A and D B, respectively, such that C A 1⊥D A and C B 1⊥D B. Prove that \|A B\|≥\|A 1 B 1\|. 2011 Estonia Open Senior 1.5 Given a triangle A B C where \|B C\|=a,\|C A\|=b and \|A B\|=c, prove that the equality 1 a+b+1 b+c=3 a+b+c holds if and only if ∠A B C=60 o. 2011 Estonia Open Senior 2.1 A square A B C D lies in the coordinate plane with its vertices A and C lying on different coordinate axes. Prove that one of the vertices B or D lies on the line y=x and the other one on y=−x. 2011 Estonia Open Senior 2.3 Let A B C be a triangle with integral side lengths. The angle bisector drawn from B and the altitude drawn from C meet at point P inside the triangle. Prove that the ratio of areas of triangles A P B and A P C is a rational number. 2012 Estonia Open Senior 1.3 Let A B C be a triangle with median AK. Let O be the circumcenter of the triangle A B K. a) Prove that if O lies on a midline of the triangle A B C, but does not coincide with its endpoints, then A B C is a right triangle. b) Is the statement still true if O can coincide with an endpoint of the midsegment? 2013 Estonia Open Senior 1.4 Inside a circle c there are circles c 1,c 2 and c 3 which are tangent to c at points A,B and C correspondingly, which are all different. Circles c 2 and c 3 have a common point K in the segment B C, circles c 3 and c 1 have a common point L in the segment C A, and circles c 1 and c 2 have a common point M in the segment A B. Prove that the circles c 1,c 2 and c 3 intersect in the center of the circle c. 2013 Estonia Open Senior 2.3 Circles c 1,c 2 with centers O 1,O 2, respectively, intersect at points P and Q and touch circle c internally at points A 1 and A 2, respectively. Line P Q intersects circle c at points B and D. Lines A 1 B and A 1 D intersect circle c 1 the second time at points E 1 and F 1, respectively, and lines A 2 B and A 2 D intersect circle c 2 the second time at points E 2 and F 2, respectively. Prove that E 1,E 2,F 1,F 2 lie on a circle whose center coincides with the midpoint of line segment O 1 O 2. 2014 Estonia Open Senior 1.4 In a plane there is a triangle A B C. Line A C is tangent to circle c A at point C and circle c A passes through point B. Line B C is tangent to circle c B at point C and circle c B passes through point A. The second intersection point S of circles c A and c B coincides with the incenter of triangle A B C. Prove that the triangle A B C is equilateral. 2014 Estonia Open Senior 2.3 The angles of a triangle are 22.5 o,45 o and 112.5 o. Prove that inside this triangle there exists a point that is located on the median through one vertex, the angle bisector through another vertex and the altitude through the third vertex. 2015 Estonia Open Senior 1.5 Let A B C be a triangle. Let K,L and M be points on the sides B C,A C and A B, respectively, such that \|A M\|\|M B\|⋅\|B K\|\|K C\|⋅\|C L\|\|L A\|=1. Prove that it is possible to choose two triangles out of A L M,B M K,C K L whose inradii sum up to at least the inradius of triangle A B C. 2015 Estonia Open Senior 2.5 The triangle K 2 has as its vertices the feet of the altitudes of a non-right triangle K 1. Find all possibilities for the sizes of the angles of K 1 for which the triangles K 1 and K 2 are similar. 2016 Estonia Open Senior 1.5 The bisector of the angle A of the triangle A B C intersects the side B C at D. A circle c through the vertex A touches the side B C at D. Prove that the circumcircle of the triangle A B C touches the circle c at A. 2016 Estonia Open Senior 2.5 The circumcentre of an acute triangle A B C is O. Line A C intersects the circumcircle of A O B at a point X, in addition to the vertex A. Prove that the line X O is perpendicular to the line B C. 2017 Estonia Open Senior 1.5 On the sides B C,C A and A B of triangle A B C, respectively, points D,E and F are chosen. Prove that 1 2(B C+C A+A B)<A D+B E+C F<3 2(B C+C A+A B). 2017 Estonia Open Senior 2.5 The bisector of the exterior angle at vertex C of the triangle A B C intersects the bisector of the interior angle at vertex B in point K. Consider the diameter of the circumcircle of the triangle B C K whose one endpoint is K. Prove that A lies on this diameter. 2018 Estonia Open Senior 1.1 Is there an equilateral triangle in the coordinate plane, both coordinates of each vertex of which are integers? 2018 Estonia Open Senior 1.5 The midpoints of the sides B C,C A, and A B of triangle A B C are D,E, and F, respectively. The reflections of centroid M of A B C around points D,E, and F are X,Y, and Z, respectively. Segments X Z and Y Z intersect the side A B in points K and L, respectively. Prove that A L=B K. 2018 Estonia Open Senior 2.2 The lengths of all sides of a right triangle are integers. The length of one leg is an odd prime p. Find the lengths of the other two sides of this triangle in terms of p. 2018 Estonia Open Senior 2.5 Let A′ be the result of reflection of vertex A of triangle ABC through line B C and let B′ be the result of reflection of vertex B through line A C. Given that ∠B A′C=∠B B′C, can the largest angle of triangle A B C be located: a) At vertex A, b) At vertex B, c) At vertex C? 2019 Estonia Open Senior 1.1 Juri and Mari play the following game. Juri starts by drawing a random triangle on a piece of paper. Mari then draws a line on the same paper that goes through the midpoint of one of the midsegments of the triangle. Then Juri adds another line that also goes through the midpoint of the same midsegment. These two lines divide the triangle into four pieces. Juri gets the piece with maximum area (or one of those with maximum area) and the piece with minimum area (or one of those with minimum area), while Mari gets the other two pieces. The player whose total area is bigger wins. Does either of the players have a winning strategy, and if so, who has it? 2019 Estonia Open Senior 1.5 Polygon A 0 A 1...A n−1 satisfies the following: ∙A 0 A 1≤A 1 A 2≤...≤A n−1 A 0 and ∙∠A 0 A 1 A 2=∠A 1 A 2 A 3=...=∠A n−2 A n−1 A 0 (all angles are internal angles). Prove that this polygon is regular. 2019 Estonia Open Senior 2.5 The plane has a circle ω and a point A outside it. For any point C, the point B on the circle ω is defined such that A B C is an equilateral triangle with vertices A,B and C listed clockwise. Prove that if point B moves along the circle ω, then point C also moves along a circle. 2020 Estonia Open Senior 1.5 A circle c with center A passes through the vertices B and E of a regular pentagon A B C D E . The line B C intersects the circle c for second time at point F. The point G on the circle c is chosen such that \|F B\|=\|F G\| and B≠G. Prove that the lines A B,E F and D G intersect at one point. 2020 Estonia Open Senior 2.5 The bisector of the interior angle at the vertex B of the triangle A B C and the perpendicular line on side B C passing through the vertex C intersects at D. Let M and N be the midpoints of the segments B C and B D, respectively, with N on the side A C. Find all possibilities of the angles of the triangles A B C, if it is known that \|A M\|\|B C\|=\|C D\|\|B D\|. source: Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest No comments: Post a Comment Home Subscribe to: Posts (Atom) aops olympiad geo articles lits of articles with sources google drive link containing all of them geometry problems not existing in aops contest collections all collected here international contests + TSTs national contests regional contests from Belarus regional contests from China regional contests from Russia regional contests from Ukraine regional contests not from above geometry olympiads not added at this page yet Cabri Clubs 2024 Sharygin Geometry Olympiad 2024 Oral Moscow Geometry Olympiad 2024 Iranian Geometry Olympiad (IGO) 2023 Yasinsky Geometry Olympiad 2023 Sharygin Geometry Olympiad 2023 Oral Moscow Geometry Olympiad 2023 Novosibirsk Oral Olympiad in Geometry 2023 Iranian Geometry Olympiad (IGO) 2023 DJGO = Discord Junior Geometry Olympiad 2023 DGO = Discord Geometry Olympiad 2023 Adygea Teachers' Geometry Olympiad 2022 Yasinsky Geometry Olympiad 2022 Novosibirsk Oral Olympiad in Geometry 2022 DJGO = Discord Junior Geometry Olympiad 2022 DGO = Discord Geometry Olympiad 2022 Adygea Teachers' Geometry Olympiad 2023 BMoEG ΙII = Buratino's Mock of Euclidean Geometry IIΙ /// Closing date: 23:59 (GMT+7) November problem's pdf original aops thread aops post collection 2022 BMoEG II = Buratino's Mock of Euclidean Geometry II problem's pdf problems with official solutions' pdf original aops thread aops post collection 2021 BMoEG = Buratino's Mock of Euclidean Geometry I problems' pdf problems with official solutions' pdf original aops thread aops post collection geometry olympiads posted in aops, not collected in this page yer Cabri Clubs (Argentina) Open Regional Geometry Olympiad Anischenko (Russia) geometry problems from not final rounds of NMO posted in aops, not collected in this page yet Argentina Contests Brazil Contests Kazakhstan Contests Mexico Contests Peru Contests Ukraine NMO - geometry a few collection links inside aops 2022 Olympiad Geometry from Contests a few geometry problem collections inside aops ( + geo olympiads) Aops Geo Mocks and Geometry from Mocks (>=JMO), contests collections combinatorial geometry from Russian and Ukrainian Tournaments forum collections not yet in Aops' contest collections USA MS HS contests + tournaments' team rounds, besides aops contest collections my aops forum collections 2022 Olympiad Geometry from Contests Contest Collections / Geo Junior Geometry USA Contests besides aops contest collections latest updated collections with more than 10 problems German Federal 1974 - 2022 (BWM) Hungarian Kurschák Competition 1947 - 2022 Japan Junior Math Olympiad 2009-22 (JJMO) new contests in this blogspot created by: Takis Chronopoulos (parmenides51) from Greece contact email: parmenides51 # gmail.com if any link is incorrect / broken, please let me know 14642Olympiad Geometry problems with Art Of Problem Solving links 270 high school math contests collected, 68 of them with solutions aops = artofproblemsolving.com geometry problemsfrom mathematical olympiads, contests, competitions, tournaments total views 1,685,367 info parmenides51View my complete profile 2019 Imogeometry review part 1, International + Regional, version 2 part 2, IMO shortlist, TSTs and unofficial my english public groups + forums Geometry Mathley Old High School Olympiads - 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Repêchage 2009-21 (CMOQR) 24p Chile 1989 - 2020 levels 1-2 and TST 66p (uc) China Girls 2012-21 (CGMO) 43p China 1986 - 2021 (CMO) 43p China Northern 2005-09, 2016-20 (CNMO) 20p (-08) China 2nd Round T2 1983 - 2020 42p China South East 2005-21 51p China Western 2001-19 (CWMI) 38p Croatia Junior 2017-20 (HJMO) 8p Croatia 2010-20 (HMO) 22p Czech & Slovak MO, III A 1991 - 2021 61p Denmark 1991 - 2021 (Mohr) 54p Ecuador 2016-19 L2 L3 (OMEC) 14p Estonia 1996 - 2007 IX-XII 59p Estonia Open 1993 - 2020 128p France 2011-20 (Fall Animath Cup) 16p Finland 1997 - 2019 (FHSMC) 27p Germany 2015-21 (De_MO) 12p Germany Federal 1995 - 2020 (BWM) 55p Greece 1995 - 2020 26p Greece Junior 1996 - 2021 27p Hungary 1985 - 2020 (Kürschák) 29p Hungary OKTV Iceland 2010-22 (-18) 16p India 1986 - 2021 (INMO) 69p Indian IOQM 2021-22 22p India Postal Coaching 2004-5,2008-11,2014-16 76p India pre-Regional 2012-19 (PRMO) 55p (-15) India Regional 1990 - 2019 (RMO) 111p Indonesia MO Province 2002-20 (OSP SMA) 24p Indonesia 2002-20 (INAMO) (OSN) + SHL 2014-15 47p Iran MO 2nd Round 1983 - 2021 72p Iran MO 3rd Round 1996 - 2020 120p (-99,01) Ireland 1988 - 2021 (IRMO) 75p Israel 2011-19 (Gillis) 17p Italy 1985 - 2021 (ITAMO) 78p Japan Junior Finals 2018-20 (JJMO) Japan Finals 1991 - 2021 (JMO) 33p Kazakhstan 1999 - 2020 IX-XI 115p Korea S. Junior 2005-18 (KJMO) 28p Korea S. 2nd Round 2004-20 (KMO) 33p Korea S. Final 1993 - 2021 (FKMO) 49p Kosovo 2011-21 IX-XII 26p (-12,-14,-15,-18) Macedonia North 2006-21 20p Macedonia North Juniors 2020-21 3p Malaysia Juniors SHL 2015 8p Malaysia IMONST R2 2020 p2 Mexico 1987- 2020 (OMM) 68p Moldova 2009, 2014-20 VII+ 106p Netherlands (Dutch) 1990 - 2020 (DMO) 50p New Zealand 2010-20 (NZOMC, NZMO) 32p Norway 1993 - 2021 (Abel) 43p Paraguay 2001-20 (OMAPA) 38p (-16-17,-18) Peru 2004-19 levels 1-3 (ONEM) 26p Philippines 2008-21 (PMO) 12p (-14,-15) Philippines Area II 2008-20 (PMO) 11p (-11) Poland Junior 2006-21 113p Poland 2nd Round 1993 - 2021 59p Poland Finals 1985 - 2021 76p Portugal 1994 - 2020 51p (OPM) (-04) Romania 2000-19 VII+ 102p Romania District 2001-19 VII+ 97p Serbia 2007-21 (SMO) 26p Singapore 2nd Round (SMO) 1995- 2019 70p South Africa 1995 - 2021 (SAMO) 47p Spain 1984 - 2021 55p Sweden 1962 - 2020 78p Switzerland 2004-21 (SMO) 45p Taiwan 1992 - 2002 16p Thailand 2004-20 (TMO) + SHL 2011-12,-14 54p Turkey 1993 - 2020 47p Turkey Junior 1996 - 2020 24p Ukraine 1991 - 2021 VIII-XI 223p UK MO for Girls 2011-20 (UKMOG) 14p USAJMO 2010-21 21p USAMO 1972 - 2021 66p Vietnam 1962 - 2021 (VMO) 93p JBMO TST olympiad geometry problems - JTST JBMO 1997- SL 130p Bosnia & Herzegovina JBMO TST 2003-22 21p Bulgaria JBMO TST 2007-22 31p (-09,-15) Greece JBMO TST 1998 - 2022 26p France JBMO TST 2013-22 34p Macedonia North Juniors / JBMO TST 2012-22 17p Moldova JBMO TST 1999 - 2022 (-09) 45p Romania JBMO TST 1998 - 2019, 2021-22 131p (+ ROMOP '11) Saudi Arabia JBMO TST 2015-22 25p (-18, -20) Serbia JBMO TST 2007-22 20p (-12,-13) Turkey JBMO TST 2012-22 15p (-19,-20) regional olympiad geometry problems (not from Australia, Canada, Latin America, UK, USA, ex-USSR) Dürer 2008-20 (Hungary) 47p Hanoi Open MC 2006-18 (HOMC) (Vietnam) 71p IFYM 2010-19 X-XII (Bulgaria) (-13R1, -17R2, -16) India STEMS 2021 5p India IITB Mathathon 2019, 2021 5p Lotfi Zadeh Olympiad 2021 1p (Aze) LuMaT 2018-19 6p (Indonesia) Mathley 80p Maths Beyond Limits 2017-20 (Poland) 45p Memorial "Aleksandar Blazhevski-Cane" 2020-21 (N. Macedonia) 3p QEDMO 2005-12, 2015-17 35p (1st-11th,14th-15th) (Germany) Australia contests, olympiad geometry problems Auckland MO 2015-21 (New Zealand) 13p Australia 1979 - 2020 (AMO) 88p New Zealand 2010-22 (NZOMC, NZMO) 39p UNSW School MC 1964 - 2019, -21 (Australia) 79p UQ/QAMT PS Competition 2002-16 (Australia) 51p (-14) Canada contests, olympiad geometry problems Alberta HS MC 2007-22 (AHSMC) (Canada) 21p AMQ Concours 2004-21 (Canada) 65p Canadian MO Qual. Repêchage 2009-21 (CMOQR) 24p Canada 1969 - 2022 (CMO) 74p Canada Junior 2020-22 (CJMO) 4p Canada IMO Training Latin Americans, olympiad geometry problems (posted also elsewhere above) Argentina 1995 - 2020 levels 1-3 (OMA) 103p Argentina TST 1996 - 2021 (IMO - CSMO - OIM) 108p Argentina Training Lists (OIM, OMCS, OMR) 101p Brazil 1979 - 2020 levels 2,3 (OBM) 104p Brazil TST 1997 - 2020 (IMO - OMCS - OIM) 72p Brazil Training Lists OMCS 1999 - 2020 219p Centroamerican 1999 - 2021 (OMCC) 42p CentroAmerican Shortlist (OMCC SHL) 76p Chile 1989 - 2020 levels 1-2 and TST 74p (uc) COMATEQ 2017-21 (Puerto Rico) 13p Cono Sur / Southern Cone 1989 - 2020 (OMCS) + SHL 70p Cono Sur Shortlist (OMCS SHL) 56p Cordoba pre-TST 2014-16 (Argentina) 9p Ecuador 2016-19 L2 L3 (OMEC) 14p Ecuador TST 2006-21 48p El Número de Oro 1997 - 2019 (Arg) 60p (-99,-06,-07,-08) Grupo MATE OIM 2021 (Peru) 4p Iberoamerican 1985 - 2020 (OIM) 65p IberoAmerican Shortlist (OIM SHL) 153p Istmo Centroamericano 2017-19 3p Lusophon / Portuguese Language 2011-20 (OMCPLP) 16p May Olympiad 1995 - 2021 levels 1-2 (Mayo) 59p Mexican Geometry Olympiad 2014, 2020-21 26p Mexico 1987- 2020 (OMM) 68p Mexico OMMock 2017-21 8p Mexico Regional Center Zone 2007-21 26p (-13) Mexico Regional Southeast 2015-21 13p OIFMAT I-III 2010-13 (Fmat Chile) 7p Olympic Revenge 2002-20 (Brazilian) 23p (-06,-08) Omaforos Contests OFO, FOFO, COFFEE 2015-22 57p (Arg) PAGMO 2021 Paraguay 2001-20 (OMAPA) 38p (-16-17,-18) Peru 2004-19 levels 1-3 (ONEM) 26p Peru TST 2006-19 (IMO - CSMO - OIM - EGMO) 75p Puerto Rico TST 2007-21 32p Quarantine MO 2020 (Mexican, Global) 5p Rioplatense 1990 - 2019 levels 1-3 (OMR) 128p Rio de Janeiro 1998 - 2020 (OMERJ) (Brazil) 63p Romania contests, olympiad geometry problems Clock-Tower School Juniors 2007-14 (Romania) 14p Danube / Donova 2005-19 (Romania) 24p IMAC Arhimede 2007-14 (Romania) 17p IMAR 2003-19, 2022 (Romania) 18p (-04) Romania District 2001-19 VII+ 97p Romania 2000-19 VII+ 102p Romania JBMO TST 1998 - 2019, 2021-22 131p (+ ROMOP '11) Romania TST 1987 - 2019, 2021 133p Romanian Master of Mathematics 2008-21 (RMM) + SHL 2016-19 29p Science ON 2021 (Romania) 8p Stars Of Mathematics 2007-22 (Romania) 26p UK contests, olympiad geometry problems British 1966 - 2020 (BrMO BMO1 BMO2 FIST) 155p British TST 1985 - 2015 (UK FST, NST) 62p UK MO for Girls 2011-20 (UKMOG) 14p USA contests, olympiad geometry problems Bay Area MO 1999 - 2019 (BAMO) 28p Carnegie Mellon Informatics & MC 2016-20 (CMIMC) 77p ELMO 2009- SL 71p Harvard-MIT Mathematics Tournament 2014-20 (HMMT) 70p HMMT Invitational Competition (HMIC) 2013-21 8p Math Prize for Girls 2009-19 (MPfG) 54p MOP Homework 2005-06 (MOSP) 39p NIMO Summer & Winter Contests 2011-17 29p NIMO Monthly Contests 2012-17 65p Online Math Open 2012-20 (OMO) 137p Princeton University MC 2007-19 (PUMaC) 185p Purple Comet (MS-HS) 2003-20 San Diego Power Contest 2018-21 (SDPC) 13p US Math Competition Association 2019-20 (USMCA) 17p USAJMO 2010-21 21p USAMO 1972 - 2021 66p USAMTS 2011-16 USA TST 2000-20 (IMO - EGMO) 41p USA TSTST 2011-19 20p US Ersatz Math Olympiad 2020 (USEMO) 4p ex - USSR, olympiad geometry problems Adygea Teachers' Geometry 2017-21 20p AESC MSU Internet MO 2014-21 VII-X (Russia) 33p All - Russian 1993 - 2019 IX-XI (ARO) 149p All - Siberian Open 2007-21 VII+ (Russia) 194p All - Soviet Union (ASU) 1961-92 155p Almaty City MO 2008-21 (Kazakhstan) 16p Armenia 2009-21 96p Belarus 1995 - 2019 VIII-XI 192p (-96,-13,-16) Belarus TST 1995,'98,'00,'09-19 69p (-13) Caucasus 2015-21 (Russia) 22p Champions Tournament 2001-19 (Ukraine) 33p (-14) Chisinau City MO 1949-56, 1973-79 (Moldova) 63p Estonia 1996 - 2007 IX-XII 59p Estonia Open 1993 - 2020 128p Euler Olympiad 2009-21 VIII (Russia) 106p Euler Teachers' MO 2007-20 62p European Math Tournament 2018-21 VII-VIII (Belarus) 49p Final Mathematical Cup 2019-20 (FMC) 3p Formula of Unity / Third Millennium 2013-21 (FdI) 56p Georgia TST 2005 4p High Standards 2009-21 VII+ (Высшая проба) (Russia) 46p Innopolis University Open MO 2015-21 (Russia) 16p Izumrud Olympiad Emerald - UrFU 2016-21 (Russia) 30p Kazakhstan 1999 - 2020 IX-XI 115p Kharkiv City MO 2012-21 VIII-XI (Ukraine) 37p Kharkiv Lyceum No.27 2014-21 (Ukraine) 34p Kharkiv Masters Tournament 2018-19 (Ukraine) 56p Kukin MO - OmSU 2008-21 VII+ (Russia) 80p Kurchatov 2013-21 (Russia) 48p Kvanta MO 2018-21 VI-VIII (Ukraine) 6p Kyiv City MO 1994-83 & 2003-22 VIII-XI (Ukraine) 231p Kyiv Math Festival 2005-21 VIII-X (Ukraine) 29p Kyiv TST 2005-21 (Ukraine) 141p (-07) Latvia TST 2019-20 (BW) 8p League of Winners Tournament 2014, 2016-17 (Russia) 23p Lithuania: Grand Duchy 2009-21 (BW TST) 13p Lithuania TST 2005-19 (IMO - MEMO) 26p (-07) Lomonosov Tournament 1978 - 2021 (Russia) 89p Math. Multiathlon Tournament 2008-17 (Russia) 192p Matol Online Olympiad 2021-22 (Kazakshtan) 8p Metropolises 2016-21 (IOM) (Russia) 10p Minsk City Internet MO 2014-20 VI-IX (Belarus) 18p MIPT Metropolitan 2014-20 (Russia) 22p (-15) Moldova 2009, 2014-20 VII+ 106p Moldova JBMO TST 1999 - 2020 (-09) 40p Moldova TST 2002-20 (IMO - EGMO) 45p (-04,-12) Moscow City MO 1935-56 (MMO) 136p Moscow Oral Geo 2003-21 216p Moscow City Oral MO 2005-21 VI-VII 24p Moscow City Oral Team MO 1999 - 2021 VIII-XI 248p Moscow Correspondence Competition VI-VIII Novosibirsk Oral Geometry 2016-21 VII-IX (Russia) 58p Rusanovsky Lyceum, Kyiv 2001-06, 2010-21 (Ukraine) 140p Savin Competition VI-IX 1991 - 2020 (Russia) 157p Sharygin 2005-21 757p Silk Road 2002-21 (SRMC) (Kazakhstan) 19p St. Petersburg State School MO 2010-20 (SPbU) (Russia) 153p St. Petersburg City MO 2008-21 IX-XI (Russia) 72p Tournament of Towns 1980-87 (ToT) 37p Tournament of Towns 1998 - 2013,2020 (ToT) 162p Tuymaada 1994 - 2021 (Russia) 93p Ukraine Correspodence MO 2004-20 V+ 52p Ukraine From Tasks to Tasks 2010-16 V-IX 22p Ukraine TYM I-XXIII 126p Ukraine TST 2007-20 28p Ukraine 1991 - 2021 VIII-XI 223p VU MIF Olympiad 2016-19 (Lithuania) 10p Yasinsky 2017-21 98p Youth Mathematical School 2005-21 (Russia) 64p Zhautykov 2005-22 (IZhO) (Kazakhstan) 37p Zhautykov City 2001-21 VII-IX (Kazakhstan) 96p 239 Open MO 1999 - 2021 (St.Petersburg) (-05,-07,-18,-20) 69p Sel-Geo.ru unofficial olympiad geometry problems AoPS Problem Making Contest 2016 +SL (APMC) 8p Cyberspace MC 2020 (CMC) 2p DeuX MO 2020 5p (aops) ELMO 2009- SL 71p EMMO 2016 (India) 8p Functional Equations Online Olympiad 2020 (FEOO) 4p (aops) QEDMO 2005-12, 2015-17 (Germany) 35p (-13th) InfinityDots MO 2017-19 6p IMEO 2017-20 5p IMOC 2017-21 (Taiwan) 34p International Olympic Revenge 2017-18 (IMOR) 2p Kvanta MO 2018-21 VI-VIII (Ukraine) 6p Mathley 80p Mathlinks Contest 2002-08, 2020 51p Metrix MO 2020 8p Mexico OMMock 2017-21 8p MO Discord Server 2020 (MODS) (uc) MOP Homework 2005-06 (MOSP) 39p NIMO Monthly Contests 2012-17 65p NIMO Summer & Winter Contests 2011-17 29p Olympic Revenge 2002-21 (Brazil) 24p (-06,-08) Online Math Open 2012-20 (OMO) 137p Purple Comet (MS-HS) 2003-20 Quarantine MO 2020 (Mexican, Global) 5p Real IMO Shortlist 2018-19 (Monster's) 16p US Ersatz Math Olympiad 2020-21 (USEMO) 6p geometry olympiads Adygea Teachers' Geometry 2017-21 20p Discord Geometry Olympiad 2021 (DGO) 10p Geolympiad 2015 (Aops) 28p Iranian 2014-20 (IGO) 91p Mexican Geometry Olympiad 2014, 2020-21 26p Moscow Oral Geo 2003-21 216p Novosibirsk Oral Geometry 2016-21 VII-IX (Russia) 58p Sharygin 2005-21 757p Ukrainian Geometry 2017,2020 (2) 35p Yasinsky 2017-21 98p geo + mock oly geo shortlists Junior only contests, olympiad geometry problems (posted also elsewhere above,besides JBMO TST) Austria Junior Regional 2019-20 2p Belgium Flanders Juniors 2002-21 (JWO) 31p Belgium OMB Juniors 2004-21 23p Canada Junior 2020-22 (CJMO) 4p COMATEQ 2017-21 (Puerto Rico) 13p Croatia Junior 2017-20 (HJMO) 8p Czech-Polish-Slovak Junior Match 2012-19 (CPSJ) 32p Euler Olympiad 2009-21 VIII (Russia) 106p European Math Tournament 2018-21 VII-VIII (Belarus) 49p Greece Junior 1996 - 2021 27p Japan Junior Finals 2018-20 (JJMO) JBMO 1997- SL 125p Istmo Centroamericano 2017-19 3p Korea S. Junior 2005-18 (KJMO) 28p Kvanta MO 2018-21 VI-VIII (Ukraine) 6p Macedonia North Juniors 2020-21 3p Malaysia Juniors SHL 2015 8p May Olympiad 1995 - 2021 levels 1-2 (Mayo) 59p Minsk City Internet MO 2014-22 VI-IX (Belarus) 19p Moscow City Oral MO 2005-21 VI-VII 24p Moscow Correspondence Competition VI-VIII Novosibirsk Oral Geometry 2016-21 VII-IX (Russia) 58p Poland Junior 2006-21 113p Savin Competition VI-IX 1991 - 2020 (Russia) 157p Turkey Junior 1996 - 2020 24p Ukraine From Tasks to Tasks 2010-16 V-IX 22p USAJMO 2010-21 21p Zhautykov City 2001-19 VII-IX (Kazakhstan) 91p correspodence only / internet olympiads (posted also elsewhere above) AESC MSU Internet MO 2014-21 VII-X (Russia) 33p IMEO 2017-20 5p Cyberspace MC 2020 (CMC) 2p Discord Geometry Olympiad 2021 (DGO) 10p Mathley 80p India Postal Coaching 2004-5,2008-11,2014-16 76p Mathlinks Contest 2002-08, 2020 51p Matol Online Olympiad 2021 (Kazakshtan) 3p Mexico OMMock 2017-21 8p Minsk City Internet MO 2014-20 VI-IX (Belarus) 18p Moscow Correspondence Competition VI-VIII MO Discord Server 2020 (MODS) (uc) NIMO Monthly Contests 2012-17 65p NIMO Summer & Winter Contests 2011-17 29p OIFMAT I-III 2010-13 (Fmat Chile) 7p Oliforum Contest I-V 2008-17 (Italy) 9p Omaforos Contests OFO, FOFO, COFFEE 2015-22 57p (Arg) Online Math Open 2012-20 (OMO) 137p Purple Comet (MS-HS) 2003-20 Quarantine MO 2020 (Mexican, Global) 5p Savin Competition VI-IX 1991 - 2020 (Russia) 157p Ukraine Correspodence MO 2004-20 V+ 52p Ukraine From Tasks to Tasks 2010-16 V-IX 22p USAMTS 2011-16 VMEO I-IV 2004-06, 2015 (VMF Vietnam) 21p revenge olympiads (posted also elsewhere above) International Olympic Revenge (IMOR) Olympic Revenge (Brazilian) forum olympiads (posted also elsewhere above) Oliforum Contest I-V 2008-17 (Italy) 9p OIFMAT I-III 2010-13 (Fmat Chile) 7p VMEO I-IV 2004-06, 2015 (VMF Vietnam) 21p Omaforos Contests OFO, FOFO, COFFEE 2015-22 57p (Arg) girls only contests, olympiad geometry problems (posted also elsewhere above) Brazilian Girls in Mathematics Tournament 2019 (TM^2) 2p China Girls 2012-21 (CGMO) 43p European Girls 2012-21 (EGMO) 19p Math Prize for Girls 2009-19 (MPfG) 54p PAGMO 2021 UK MO for Girls 2011-20 (UKMOG) 14p teachers' olympiads (posted also elsewhere above) Adygea Teachers' Geometry 2017-21 20p El Número de Oro 1997 - 2019 60p (Arg) (-99,-06,-07,-08) Euler Teachers' MO 2007-20 62p city olympiads (posted also elsewhere above) Almaty City MO 2008-21 (Kazakhstan) 16p Chisinau City MO 1949-56, 1973-79 (Moldova) 63p Kharkiv City MO 2012-21 VIII-XI (Ukraine) 37p Kyiv City MO 1994-83 & 2003-22 VIII-XI (Ukraine) 231p Minsk City Internet MO 2014-22 VI-IX (Belarus) 19p Moscow City MO 1935-56 (MMO) 136p Moscow Oral Geo 2003-22 228p Moscow City Oral MO 2005-22 VI-VII 27p Moscow City Oral Team MO 1999 - 2021 VIII-XI 248p St. Petersburg City MO 2008-21 IX-XI (Russia) 72p school olympiads (posted also elsewhere above) 239 Open MO 1994 - 2021 (-97,-98,-07,-18,-20) 84p Kharkiv Lyceum No.27 2014-21 (Ukraine) 34p Rusanovsky Lyceum, Kyiv 2001-06, 2010-21 (Ukraine) 140p ended contests, olympiad geometry problems (posted also elsewhere above) Austrian-Polish 1978 - 2006 (APMC) 53p Clock-Tower School Juniors 2007-14 (Romania) 14p Mathley 80p Hungary - Israel 1990 - 2009 21p (-04) IMAC Arhimede 2007-14 (Romania) 15p League of Winners Tournament 2014, 2016-17 (Russia) 23p Math. Multiathlon Tournament 2008-17 (Russia) 192p Moscow Correspondence Competition VI-VIII NIMO Monthly Contests 2012-17 65p NIMO Summer & Winter Contests 2011-17 29p OIFMAT I-III 2010-13 (Fmat Chile) 7p Oliforum Contest I-V 2008-17 (Italy) 9p Ukraine From Tasks to Tasks 2010-16 V-IX 22p UNESCO Competition 1995-96 VMEO I-IV 2004-06, 2015 (VMF Vietnam) 21p possible new problem collections from Argentina Brazil Bulgaria Colombia Cuba Cyprus Czech Estonia Germany Hungary Latvia Lithuania Mexico Moldova Russia Slovakia Ukraine under construction, olympiad geometry problems MO Discord Server 2020 (MODS) (uc) Moscow Teachers' ΜΟ (uc) Moscow Correspondence Competition VI-VIII Hungary OKTV Romantics of Geometry facebook group ENGLISH / GREEK readers solved within this page ----------------------- 1959 IMO Problem 4 (HUN) 1959 IMO Problem 5 (ROM) 1959 IMO Problem 6 (CZS) 1960 IMO Problem 5 (CZS) 1984 BMO Problem 2 (ROM) 1985 BMO Shortlist 2 (GRE) 2010 JBMO Shortlist G1 2011 JBMO Shortlist G1 2011 JBMO Shortlist G2 2014 JBMO Shortlist G1 my JBMO Geometry Shortlists' solutions old IMO Plane Geometry solutions Kostas Dortsios' old IMO 3D Geometry solutions my English geometry problem collections --------------------------------------- 45 Selected Geometry Problems in Greek Exams 1964 - 1980 EN with fb links 50 geometry problems from fb group JUVENTUD MATEMÁTICA 127 geometry problems from fb group JUVENTUD GEÓMETRA 259 geometry writings by Christopher J. Bradley 345 geometry problems from 8floxes.com 397 geometry problems from Sobregeometrias SX SG Mathematical Olympiad problems in pdf with solutions by John Scholes (kalva) ------------------------------------------------------------------------------- All Soviet Union MO 1961-1992 with solutions by John Scholes (kalva) Asian Pacific MO (APMO) 1989 - 2004 EN with solutions by John Scholes (kalva) CentroAmerican MO (OMCC) 1999 - 2003 EN with solutions by John Scholes (kalva) Iberoamerican MO (OIM) 1985-2003 EN with solutions by John Scholes (kalva) IMO problems 1959 - 2003 EN with solutions by John Scoles (kalva) Russian Mathematical Olympiad 1995-2002 with partial solutions by John Scholes (kalva) my geometry problem collections from mags inside aops ----------------------------------------------------- In the World of Mathematics, part I In the World of Mathematics, part II Mathematical Excalibur Mathematical Reflections - Junior Mathematical Reflections - Olympiad Mathematical Reflections - Senior Mathematical Reflections - Undergraduate Quantum - English Edition my problem collections from geometry Shortlists inside aops ----------------------------------------------------------- Cono Sur Olympiad Shortlist IMOC Shortlist (Taiwan) Indonesia MO Shortlist (INAMO OSN) Real IMO Shortlist (Monster's) Romanian Master in Mathematics Shortlist Thailand MO Shortlist (TMO) aops various shortlists (not IMO, JBMO, BMO, ELMO) -------------------------------------------------- Indonesia Shortlist 2014 Insonesia Shortlist 2015 Pan African Shortlist 2017 Pan African Shortlist 2018 Romanian Master of Mathematics SHL 2016 Romanian Master of Mathematics SHL 2017 Romanian Master of Mathematics SHL 2018 Romanian Master of Mathematics SHL 2019 geometry articles from ---------------------- Canada IMO Training Crux Mathematicorum International Tournament of Towns Summer Consferences Jean-Louis Ayme Mathematical Excalibur Mathematical Reflections Revista Escolar de la Olimpíada Iberoamericana de Matemática geometry problems from ---------------------- In the World of Mathematics (U sviti matematyky) Mathematical Excalibur Mathematical Reflections Quantum Magazine (English) problem collections from ------------------------ Mathematical Reflections Romanian Mathematical Magazine a beloved competition quote by BOGTRO in aops --------------------------------------------- I'll give you some multiple choice question practice: Which of the below is most likely to improve your computational math ability? A) Insisting you are bad at computational math B) Avoiding computational math C) Complaining about some competitions emphasizing computational math D) Practicing computational math a few notations, line related ----------------------------- (A B) is an open segment [A B] a closed segment A B¯¯¯¯¯¯¯¯ an orientated segment (A B a ray [A B a closed ray A B a line my greek geometry page για τους ρομαντικούς της γεωμετρίας ------------------------------------------------------------- tabs in Greek ------------- IMO 1959-1962 JBMO Shortlist 2009- Balkan MO 1984- recent comments --------------- parmenides51thanks, fixed Anonymous2018 2 and 4 are swapped parmenides51Thanks, just fixed it AnonymousThe 2019 BMO Shortlist G1 link is incorrect. AnonymousGreat source helped me very much as i couldn't copy the text from other site… AnonymousProbably then u must read all the data carefully or know what is IOQM really is … AnonymousRonaldo 777777777 suiiii rustyasspotahtoBro you a real onegod bless ya AnonymousHORRIBLE !!!! i tried to find something but couldn't. What kind of informati… Anonymousmagdiragheb2007@gmail.com I much appreciate this excellent effort and say God bl… Widget byHelplogger Euclidean Geometry in Mathematical Olympiads by Evan Chen, Aops Solutions Forum ------------------------------------------------------------------------------- math forums - besides aops -------------------------- Dxdy (Russia) Fmat (Chile) Forum.portal.edu.ro (Romania) Matematika (Bulgaria) Mathematica (Greece) Matholympiad (Bangladesh) Mathscope (Vietnam) Oliforum (Italy) OMA foros (Argentina) Pir2.forumeiros (Brazil) Tutorbrasil (Brazil) VMF Diendantoanhoc (Vietnam) links to sites collecting Mathematical Olympiads material --------------------------------------------------------- A Collection of Math Olympiad Problems- hvernaev Alekdimitrov.com (BUL) Animath (FR) Archives.math.utk.edu (links) Artofproblemsolving.com / Contests section Cone Sul Seleção (BR) European Mathematical Society Members Examenemate (ROM) Geometrykanal (telegram) IMO official page IMO unofficial old page (olympiads.win.tue.nl) Imoibero (BR) Imomath Ivyleaguecenter - List kangourou countries Klasirane (BUL) Mategl (ROM) Math Archives Math-bg (BUL) Mathematical Circles Library Mathizen Mathus.ru Mexico IMO training blog MG Belgrade (High School ) Molympiad - wiki contests list Molympiad (VIET) Moscow Math Circles Old MO links Pregătire Matematică Olimpiade Juniori (ROM) Russian links Russian links more Selinaeducation - links collection Sergey Negoda (UKR) Shapovalov Links (RUS) Societatea de Ştiinţe Matematice din România (ROM) sqing' links Torneo de las Cuencas - Torneo Geométricos (ARG) Training Camps (ARG) Treinamentoconesul (BR) ViitoriOlimpici.ro (ROM) www.diary.ru (RU) Zaba.ru (RU) Олимпиадная геометрия (zen.yandex.ru) John Scholes (www.kalva.demon.co.uk) online backups --------------------------------------------------- webee.technion.ac.il (full) www.cs.cornell.edu (partial) mks.mff.cuni.cz (partial) Olympiad Mathematics Handouts ----------------------------- Alexander Remorov Berkeley Math Circle Canada IMO Training Evan Chen Po-Shen Loh Yufei Zhao USA math constests, another link collections -------------------------------------------- Math problems, by National Association of Math Circles Mathcounts Other Math Contests by SBU Math Contest Regional Contests, a list by Mu Alpha Theta Summer Math Camps by AMS Russian webpages ---------------- All-Russian MO Moscow MO Russian Olympiads St. Petersburg MO Links 0 Aops Forums ------------------- Compilation of useful posts by BOGTRO Evan Chen's EGMO study group Lemmas in Olympiad Geometry - active forum Old High School Olympiads Old Russian Math Olympiads Tran Quang Hung's geometry group Links 1 Geometry Journals - Magazines ------------------------------------- Forum Geometricorum Global journal of advanced research on classical and modern geometries International Journal of Computer Discovered Mathematics International Journal of Geometry Journal of Classical Geometry Sangaku Journal of Mathematics Links 2 Ancient Greek Geometry ------------------------------ All figures from Apollonius Conica in Sketchpad Ancient Mathematicians and Astronomers, Greek Ancient Mathematicians and Astronomers, Internationally Euclid's Elements online, with html search Evangelos Stamatis' bibliography, contains Euclid, Apollonius, Archimedes online proofs of Pythagorean Theorem using Geogebra Links 3 Groups -------------- Euclid / Geometry Research Mailing List FB Group \ Geometria y Trigonometria FB Group \ GEOMETRY FB Group \ Juventud Geometra (inactive) FB Group \ Peru Geometrico FB Group \ Romantics of Geometry FB Group \ Sobregeometrias SG - SX (inactive) FB Group\ GEOMETRIA (ASESORIA TOTAL) Yahoo Groups \ Advanced Plane Geometry (inactive) Yahoo Groups \ Hyacinthos (inactive) Links 4 Personal Pages ---------------------- Christopher Bradley Darij Grinberg (new page) Darij Grinberg (old inactive page) Dick Klingens Ercole Suppa (Italy) Francisco Javier García Capitán Hatzipolakis Antreas (Greece) Jean-Louis Ayme (France) Lorenzo Roi Nguyen Van Linh (Vietnam) Pamfilos Paris (Greece) Paul Kunkel (Whistler Alley) Paul Yiu Psychas Vaggelis (Greece) Quang Duong Telv Cohl Tran Quang Hung (Vietnam) Tsintsifas George (Greece) Virgil Nicula (Aops) Wilson Stothers Xah Lee Links 5 Geometry of (Triangle) Sites ------------------------------------ Cubics in triangle plane (Bernard Gibert) Encyclopedia of Polygon Figures (Chris Van Tienhoven) Encyclopedia of Quadri Figures (Chris Van Tienhoven) Encyclopedia of Triangle Centers (Clark Kimberling) Hechos Geométricos en el Triángulo (Angel Montesdeoca) Lugares Geometricos (locus) Online catalogue of Archimedean circles (Floor van Lamoen) Todo Triangulos Web (Quim Castellsaguer) Links 6 Geometry Problem Sites ------------------------------ 8 Foxes (inactive) Artofproblemsolving Cut The Knot \ Geometry (Alexander Bogomolny) Gallery Geometrikon (Paris Pamfilos) Geometry from the Land of Incas (contains also site Gogeometry) (Antonio Gutierrez) Geometry Junkyard Geometry.ru (Russian) Gogeometry Leo, problems & solutions (Leonad Giugiuc) Mathpages \ Geometry Peru Geometrico (inactive) SobreGeometrias (Spanish) (Milton Donaire) (inactive) Wolfram Mathworld \ Geometry (Eric Weisstein) Lemmas in Olympiad Geometry by Titu Andreescu,Sam Korsky, Cosmin Pohoata Aops Solutions Forum ----------------------------------------------------------------------------------------------- USA, national & international ----------------------------- math olympiads abbreviations aops extraordinary geometry posts --------------------------------- Examples of using harmonic divisions and polarity collecting my aops' geometry solutions -------------------------------------- geometry problems synthetically solved top olympiad sites (besides aops) --------------------------------- IMO official site Imomath Olimpiada.ru Pregătire Matematică Olimpiade Juniori youtube suggestions ------------------- Geometría exámenes de admisión by Cristian Tello (solved problems with commentary) Lines&Circles by zuss77 (visualisations of problems) Olympiad Geometry by Michael Greenberg (solved problems with commentary) missing problems before 2000 ---------------------------- El Número de Oro 1997 - 2019 (Arg) 60p (-99,-06,-07,-08) Moscow City MO 1935-56 (MMO) 136p Poland Finals 1985 - 2021 76p Portugal 1994 - 2020 51p (OPM) (-04) Romania 2000-19 VII+ 102p Romania TST 1987 - 2019, 2021 128p Serbia 2007-21 (SMO) 26p Tournament of Towns 1980-87 (ToT) 37p missing problems after 2000 --------------------------- Balkan BMO SL 2007- 99p Belarus 1995 - 2019 VIII-XI 192p (-96,-13,-16) Belarus TST 1995,'98,'00,'09-19 69p (-13) Bulgaria TST 2003-08, 2012-15, 2020 25p El Número de Oro 1997 - 2019 60p (Arg) (-99,-06,-07,-08) Estonia 1996 - 2007 IX-XII 59p France TST 2000-07, 2012-20 45p (-01) Germany 2015-21 (De_MO) 12p Harvard-MIT Mathematics Tournament 2014-20 (HMMT) ... 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6210	http://page.mi.fu-berlin.de/szabo/PDF/transversals.pdf	Extremal problems for transversals in graphs with bounded degree Tibor Szab´ o ∗ G´ abor Tardos † Abstract We introduce and discuss generalizations of the problem of independent transversals. Given a graph property R, we investigate whether any graph of maximum degree at most d with a vertex partition into classes of size at least p admits a transversal having property R. In this paper we study this problem for the following properties R: “acyclic”, “H-free”, and “having connected components of order at most r”. We strengthen a result of . We prove that if the vertex set of a d-regular graph is partitioned into classes of size d + ⌊d/r⌋, then it is possible to select a transversal inducing vertex disjoint trees on at most r vertices. Our approach applies appropriate triangulations of the simplex and Sperner’s Lemma. We also establish some limitations on the power of this topological method. We give constructions of vertex-partitioned graphs admitting no independent transversals that partially settles an old question of Bollob´ as, Erd˝ os and Szemer´ edi. An extension of this construction provides vertex-partitioned graphs with small degree such that every transversal contains a ﬁxed graph H as a subgraph. Finally, we pose several open questions. 1 Introduction Let G be a graph and let P be a partition of V (G) into sets V1, . . . , Vn. A transversal (of P) is a subset T of V (G) for which \|T ∩Vi\| = 1 for each i = 1, . . . , n. The starting point of our discussion is the following theorem of Haxell. Theorem 1.1 Let G be a graph of maximum degree d and V1 ∪. . . ∪Vn = V (G) be a partition of its vertex set with \|Vi\| ≥2d. Then there is a transversal T which is an independent set in G. This theorem seems to have appeared ﬁrst explicitly in Haxell , although it is also a conse-quence of a more general result of Meshulam and implicitly, even earlier, of Haxell . The result has two proofs: one combinatorial and another via combinatorial topology ; it is not clear ∗Department of Computer Science, ETH Z¨ urich, 8092 Switzerland. Email address: szabo@inf.ethz.ch. Research supported by the joint Berlin/Zurich graduate program Combinatorics, Geometry, Computation, ﬁnanced by the German Science Foundation (DFG) and ETH Z¨ urich. †Alfr´ ed R´ enyi Institute of Mathematics, Pf 127, H-1364 Budapest, Hungary. Email address: tardos@renyi.hu 1 how closely these two arguments are related. The statement also has several applications in diﬀerent problems of graph theory, see [4, 7, 11]. In the present paper we set to study the generalizations of Theorem 1.1 in two diﬀerent directions. 1.1 Acyclic transversals with bounded components The ﬁrst generalization we consider here was introduced in in order to improve on a result of Alon, Ding, Oporowski and Vertigan . For a ﬁxed degree d and component size r, let us deﬁne p(d, r) to be the smallest integer such that any graph with maximum degree at most d partitioned into classes of size at least p(d, r) has a transversal that induces components of size at most r. Theorem 1.1 can then be rephrased as p(d, 1) ≤2d. In a generalization of the combinatorial argument of implied that p(d, r) ≤d + ⌊d/r⌋for any d and r. This was known to hold with equality only when r = 1 and d is a power of 2 [14, 18]. In Corollary 3.4 we establish p(d, 1) = 2d for every d. Unfortunately, for r > 1 even the asymptotical truth escapes us. The estimate is not tight in general: p(2, 2) = 2 as shown in . In fact, the best lower bound known for r > 1 is p(d, r) ≥d and even p(d, 2) = d is possible at the moment. There was hope that a generalization of the “topological” argument could provide stronger upper bounds. In the present paper we provide this missing proof via Sperner’s Lemma and appropriate triangulations of the simplex. Alas, we end up with the exact same result which follows from the combinatorial counterpart. For the proof we construct triangulations which generalize the ones of Aharoni, Chudnovsky and Kotlov and then ﬁnish along the lines of Aharoni and Haxell applying Sperner’s Lemma for our appropriately deﬁned colored triangulation. In fact, in Corollary 2.6 we obtain a slightly stronger statement. We prove that if the class sizes are at least d + ⌊d/r⌋, then a transversal could be selected inducing connected components which are trees on at most r vertices. One of our main tools for this strengthening is a triangulation from . In Corollary 2.4 we obtain a bound p(d, forest) ≤d on the minimum class size p(d, forest) such that any graph of maximum degree at most d partitioned into classes of size p(d, forest) has a transversal inducing a forest. We allow multigraphs in this deﬁnition. With Construction 3.3 we show that this bound is optimal for even d. For simple graphs the analogous value might be somewhat lower though. Our best construction here is Construction 3.7 for H = K3. For an even d this construction gives a maximum degree d graph Gd whose vertices are partitioned into classes of size 3 4d −1 and no transversal of Gd is triangle-free. In Corollary 2.9 we note that our proof implies that the r-component complex Kr(G) of a d-regular graph G with many (more than (m + 1)(d −1 + (d + 1)/r)) vertices is m-connected. Here Kr(G) denotes the simplicial complex deﬁned on the vertex set of G, where a subset forms a simplex if all connected components of the induced subgraph is of order at most r. Unfortunately our proof does not decide the asymptotics of p(d, r) for r > 1. Thus it is natural 2 to investigate “how good” such proofs could become with a possibly more clever choice of colored triangulations. With Construction 3.2 of Section 3 we ﬁnd that for r = 2, where the truth is between d and 3 2d, there is an intrinsic limit of 5 4d to where such type of arguments could improve the upper bound. In particular, for d = 2, the combinatorial proof of p(2, 2) = 2 cannot be substituted by a topological argument. 1.2 H-free transversals The second direction we intend to generalize Theorem 1.1 is about H-free-transversals. Given a ﬁxed graph H, let p(d, H) be the smallest integer such that any graph of maximum degree at most d partitioned into classes of size at least p(d, H) admits a transversal with no subgraph isomorphic to H. Theorem 1.1 can be phrased as p(d, K2) ≤2d. We ﬁnd the case of H = Kk particularly interesting, but at this point we are only able to provide a lower bound which we conjecture to be best possible. For any r-regular graph H on n vertices and for any d divisible by r, in Corollary 3.8 we prove that p(d, H) ≥ n (n−1)rd. The special case of the same construction establishes p(d, 1) = p(d, K2) = 2d for every d. Earlier this was only known for powers of 2. (See Jin and Yuster .) Construction 3.3 is a modiﬁed version of the above construction and provides a partial solution for a problem of Bollob´ as, Erd˝ os and Szemer´ edi studied by several researchers [4, 5, 14, 18, 6]. Let ∆(r, n) be the largest integer such that any r-partite graph Gr(n) with vertex classes Vi of size n each and of maximum degree less than ∆(r, n) contains an independent transversal, i.e., an independent set containing one vertex from each Vi. Deﬁne ∆r = limn→∞∆(r, n)/n, where the limit is easily seen to exist. Trivially ∆(2, n) = n, thus ∆2 = 1. Graver (c.f. ) showed ∆3 = 1. Bollob´ as, Erd˝ os and Szemer´ edi proved that 2 r ≤∆r ≤1 2 + 1 r −2, thus establishing µ = limr→∞∆r ≤1/2. Alon showed ∆r ≥1/(2e) for every r. This was improved to ∆r ≥1/2 by Haxell thus eventually settling a conjecture of and establishing µ = 1/2. Exact values of ∆r were known only when r = 3, 5 or a power of 2. Alon observed that a theorem of Aharoni and Haxell also gives ∆r ≥ l r 2(r−1) m . This can be paired with the constructions of Jin to provide ∆r = r 2r−1 for r = 2p. For other integers r, the upper bounds of Jin were somewhat improved by Alon, but exact results were not known. Here we extend the above for every even r. More precisely, in Corollary 3.6 we prove that for every r ≥2 even and for every n, ∆(r, n) = rn 2(r −1) . 3 2 Transversals spanning bounded connected components Given a graph G, a simplicial complex K, and a mapping l : V (K) →V (G), the pair (K, l) is called a G-labeled simplicial complex. If l is clear from the context we simply use K to denote the G-labeled complex. A 1-dimensional simplex {x, y} of K is called ruined if l(x) and l(y) are adjacent in G. An r-dimensional simplex S of K is called ruined if the graph of ruined 1-dimensional faces of S is connected and spans all r + 1 vertices of S. The m-dimensional solid ball is denoted by B m; its boundary, the (m −1)-dimensional sphere is Sm−1. S−1 is just the empty set. The link of a simplex σ in a simplicial complex is deﬁned as lkK(σ) := {τ ∈K : τ ∩σ = ∅, τ ∪σ ∈ K}. The join of two simplicial complexes with disjoint vertex sets is deﬁned as K ∗K′ := {τ ∪τ ′ : τ ∈K, τ ′ ∈K′}. For a more detailed discussion of the topological concepts we refer the reader to the excellent survey of Bj¨ orner . Throughout this paper when talking about a subdivision of a complex, we mean PL-subdivision, when we speak of a triangulation of the sphere Sm or ball Bm we mean PL-triangulation, where PL stands for piecewise linear. This technical property is needed to assure key properties like the following: the link of a simplex of a PL-triangulated sphere or ball is itself a PL-triangulated sphere unless the simplex is in the boundary of the ball . We mention here that one can avoid PL-triangulations and triangulated balls and spheres by using an alternative homologic approach, and speaking about boundaries of chains. This alternative treatment is somewhat less intuitive, and requires a homologic version of Sperner’s theorem, but it avoids most technical diﬃculties. A triangulation T ′ of Bm is called a ﬁlling of the triangulation T of Sm−1 if T is the boundary of T ′. A G-labeled triangulation (T ′, l′) of Bm is called a ﬁlling of the G-labeled triangulation (T , l) of Sm−1, if T ′ is a ﬁlling of T and l′\|V (T ) = l. We say that a simplex σ is multi-colored if all its vertices are assigned distinct colors. We use Sperner’s Lemma . It states that an appropriately colored subdivision of a multi-colored simplex contains a multi-colored simplex. Lemma 2.1 Let T be a triangulation of the n dimensional simplex σ. Suppose that the vertices of T are colored by n + 1 colors, such that (1) each vertex of σ receives a diﬀerent color (i.e., σ is multi-colored) and (2) the vertices of T on any face τ of σ are colored by the colors of the vertices of τ. Then there exists a multi-colored n-dimensional simplex in T . As a warm-up let us discuss forest transversals. The next theorem is our tool to construct forest transversals in graphs. Theorem 2.2 Let m ≥0 and d ≥0 be arbitrary integers. Let G be a graph of maximum degree d and W be a designated subset of the vertices V (G), \|W\| > md. Every G-labeled triangulation (T , l) of Sm−1 has a G-labeled ﬁlling (T ′, l′), such that 4 (i) l′(v) ∈W for every v ∈V (T ′) \ V (T ). (ii) Every cycle of ruined edges is contained in T . (iii) Every path of ruined edges with both endpoints in T is fully contained in T . For the proof we use the following triangulation constructed by Aharoni, Chudnovsky and Kotlov . Lemma 2.3 [1, Lemma 1.2] Given a triangulation T of Sm−1, there is a ﬁlling T ′ of T and an ordering of the new vertices V (T ′)\V (T ) = {v1, . . . , vs} such that for all i, the vertex vi is connected to at most 2m vertices of V (T ) ∪{v1, . . . , vi−1}. □ Proof of Theorem 2.2. The triangulation of Lemma 2.3 provides a ﬁlling of T such that we add the vertices one by one and each new vertex is connected to at most 2m older ones. Since \|W\| > md we can ensure that the label of each new vertex is chosen such that there is only at most one ruined edge from that vertex going to an older vertex. Thus we avoid the creation of cycles of ruined edges and also paths of ruined edges connecting vertices of T . □ An immediate application of Theorem 2.2 is for forest-transversals. Let us recall that p(d, forest) is the smallest integer, such that any d-regular graph partitioned into classes of size at least p(d, forest) has a cycle-free transversal. Corollary 2.4 p(d, forest) ≤d. Proof. Suppose G is a graph of maximum degree d and V (G) = V1 ∪. . . ∪Vn, \|Vi\| ≥d. We consider the (n −1)-dimensional simplex σ with vertex set {v1, . . . , vn}. We create a G-labeled triangulation (T , l) of σ, such that • for every vertex x of the triangulation and face τ of σ containing x, l(x) ∈∪i:vi∈τVi and • the graph of ruined edges induces a forest. We proceed by cell-induction, i.e., subdivide and label the faces of σ in an arbitrary nondecreasing order of their dimension. We start by labeling each vertex vi by an arbitrary element l(vi) of Vi. Let τ be an m-dimensional face of σ, m > 0, whose boundary is subdivided and G-labeled. Let W = ∪i:vi∈τVi. As \|W\| ≥(m + 1)d > md, we can apply Theorem 2.2 to obtain an appropriate labeled subdivision of τ. Notice that we do not create a cycle of ruined edges disjoint from the boundary, because of condition (ii). Cycles of ruined edges intersecting the boundary could not be created either because of condition (iii). Eventually the whole simplex σ is subdivided without creating a cycle of ruined edges. If each vertex of this triangulation is colored with the index of the class of its label, then the assumptions of Sperner’s Lemma are satisﬁed and the existence of a full-dimensional multi-colored simplex is guaranteed. The labels of this multi-colored simplex determine a transversal with no cycle. □ For multigraphs the bound in Corollary 2.4 is tight for even d, as it is witnessed by doubling the edges of any family of graphs which provide p(d/2, 1) = d. These graphs are given in Construction 3.3. 5 Our best example for simple graphs is weaker. Construction 3.7 for H = K3 shows that the analogous value ps(d, forest) for simple graphs satisﬁes ps(d, forest) ≥3 4d for even d. Next we prove a strengthening of the bounded component transversal result from . Our main tool is the following theorem about the existence of labeled ﬁllings with certain properties. Theorem 2.5 Let m ≥0 and r ≥1 be arbitrary integers. Let G be a graph of maximum degree d ≥r −1 and designated subset W ⊆V (G) of size \|W\| > m(d −1 + (d + 1)/r). Then a G-labeled triangulation (T , l) of Sm−1 admits a G-labeled ﬁlling (T ′, l′) satisfying the following properties: (a) l′(v) ∈W for every v ∈V (T ′) \ V (T ). (b) Every cycle of ruined edges of (T ′, l′) is contained in T . (c) There is no ruined edge between V (T ) and V (T ′) \ V (T ). (d) The ruined r-simplices of (T ′, l′) are contained in T . For a vertex w of a graph G, N(w) denotes the set of vertices adjacent to w. Proof. We prove the theorem by induction on m. For m = 0 the statement is trivial. Suppose m > 0. We construct T ′ in three phases. First we apply the “excising technique” of to create an inner “crust” which contains no ruined edges going to the boundary. We excise the vertices of T one by one from the inner boundary and use the induction hypothesis for (m −1) in each step. In the second phase we ﬁll (the inner boundary of) the crust constructed in the ﬁrst phase. We use Theorem 2.2 here. We obtain a ﬁlling S0 of T satisfying properties (a)-(c). But we may create a number of ruined r-simplices in this phase. Finally, in the third phase we remove the ruined r-simplices constructed in the second phase. We remove them one by one and ﬁll up the resulting “holes”. We use the induction hypothesis for (m −1) in each step. Let us start with the ﬁrst phase. We take the vertices u1, . . . , ut of V (T ) and excise them one by one. We do this by creating an increasing sequence T = T0 ⊆T1 ⊆· · · ⊆Tt of labeled complexes. The complex Ti satisﬁes properties (a)-(d), furthermore it has an “inner boundary” T ′ i with V (T ′ i )∩V (T ) = {ui+1, . . . , ut}, such that T ′ i is a triangulation of Sm−1 and the union of Ti with any ﬁlling of T ′ i is a ﬁlling of T . We start with T0 = T ′ 0 = T . For i > 0 consider the vertex ui and its link lkT ′ i−1({ui}), which is an (m −2)-sphere. (Note that we talk about PL-triangulations.) By induction there is a G-labeled ﬁlling ˜ Ti of this link, with subset ˜ Wi = W\N(l(ui)) satisfying properties (a)-(d). (We naturally assume here, and later in this proof, that the set of new vertices introduced in a ﬁlling is disjoint from the set of old vertices, that is we have (V ( ˜ Ti)\V (lkT ′ i−1({ui})))∩V (Ti−1) = ∅.) Observe that \| ˜ Wi\| ≥\|W\| −d > (m −1)(d −1 + (d + 1)/r). We then create Ti by adding the join of ˜ Ti with ui to Ti−1. This operation excises ui from the interior boundary of Ti. More formally, let Ti = Ti−1∪( ˜ Ti∗{∅, {ui}}) and we obtain the inner boundary T ′ i = (T ′ i−1 ( ˜ Ti ∗{∅, {ui}}))∪˜ Ti. By the choice of ˜ Wi, we do not add a ruined edge going to ui. By property (c) of the induction hypothesis 6 there are no ruined edges added going to other vertices of V (T ). In conclusion, the newly introduced ruined edges are “separated” from the old ones, that is there are no ruined edges between V (Ti−1) and V (Ti) \ V (Ti−1). Thus properties (a)-(d) hold for Ti by the induction hypothesis. Eventually all vertices of V (T ) excised from the inner boundary. Hence Tt is a crust having properties (a)-(d) and its inner boundary T ′ t is disjoint from T . As the second phase of the construction, we apply Theorem 2.2 to ﬁll T ′ t such that the new labels are from W. This is possible, because our extra condition on the maximum degree ensures that m(d −1 + (d + 1)/r) ≥md. Let S0 be the union of Tt and this ﬁlling of T ′ t . Clearly, S0 is a ﬁlling of T satisfying properties (a), (b), and (c). In the third phase of our construction we get rid of any ruined r-simplices in S0 \ T that we may have created in the second phase. For m < r no such simplices are created, so property (d) is automatically satisﬁed. For m ≥r our plan is to modify S0 to get rid of all ruined r-simplices one by one. We will change our triangulation locally and be careful not to spoil properties (a), (b), and (c). In one step we remove a ruined r-simplex σ / ∈T , together with all the simplices containing it, thus creating an m-dimensional “hole” in Bm. Then we ﬁll up this hole diﬀerently, such that we do not create new ruined r-simplices, properties (a)-(c) are still satisﬁed, while σ is gone. Since the number of ruined simplices was ﬁnite to begin with, after applying this operation ﬁnitely many times we will have a triangulation T ′ with no ruined r-simplices outside of T . Suppose that Si is the ﬁlling of T we obtained after getting rid of the ith ruined r-simplex in S0. Fix an arbitrary ruined r-simplex σi+1 of Si \ T . In fact σi+1 ∈S0 \ Tt. The link lkSi(σi+1) of the r-simplex σi+1 in Si is a triangulated Sm−r−1. We ﬁnd a ﬁlling ˜ Si of lkSi(σi+1) using the induction hypothesis for the designated subset Wi ⊆W containing the non-neighbors of the labels of the vertices of σi+1. Formally, let Nσi+1 be the set of vertices of G that are neighbors to the label of some vertex of σi+1 and let Wi = W \ Nσi+1. Then \|Nσi+1\| ≤d(r + 1) −(r −1), since σi+1 is a ruined r-simplex. Hence \|Wi\| ≥ \|W\| −\|Nσi+1\| > m(d −1 + (d + 1)/r) −dr −d + r −1 = (m −r)(d −1 + (d + 1)/r), i.e., we can indeed use the induction hypothesis. Now we are ready to deﬁne Si+1. First we remove all simplices from Si which contain σi+1. This of course creates an m-dimensional “hole” in Bm. Then in order to ﬁll it, we add all simplices of the form σ′ ∪˜ σ, where σ′ ⫋σi+1 and ˜ σ ∈˜ Si. That is, we add ˜ Si ∗δσi+1, where δσi+1 is the boundary complex of σi+1. Starting from the ﬁlling Si of T we replaced a subcomplex with another, both of them triangulated Bm and having lkSi(σi+1) ∗δσi+1 as their boundary, so the resulting complex Si+1 is also a ﬁlling of T . There are no ruined edges between V (σi+1) and V (Si+1) \ V (Si) because of the choice of Wi. All other edges between V (Si) and V (Si+1) \ V (Si) are also edges between lkSi(σi+1) and V ( ˜ Si)\lkSi(σi+1), hence not ruined by property (c) of the induction hypothesis. In conclusion, there are no ruined edges between the newly introduced vertices and the “old” vertices. Thus, we did not 7 spoil properties (b) or (c), and did not create any new ruined r-simplices. Clearly, property (a) is also maintained. The ruined simplex σi+1 is gone, so the number of ruined r-simplices decreased by one. After ﬁnitely many steps we obtain a ﬁlling T ′ of T satisfying properties (a)-(d). □ The following corollary is immediate. Corollary 2.6 Let r be an arbitrary positive integer. Let G be a graph of maximum degree d, and let P be a partition V1 ∪. . . ∪Vm = V (G) such that \|Vi\| ≥d + ⌊d/r⌋for i = 1, . . . , m. Then there exists a transversal T of P such that the connected components of the induced subgraph G\|T are trees on at most r vertices. Proof. It is enough to prove the statement for r ≤d + 1 (for higher values of r the statement of the Corollary is weaker than the one for r = d + 1). We construct a transversal T of P such that the connected components of the induced subgraph G\|T are trees on at most r vertices. Let us denote the vertices of the (n −1)-dimensional simplex σ by v1, . . . , vn. Our goal is to deﬁne a G-labeled subdivision of the complex consisting of the faces of σ such that • for every vertex x of the subdivision of a face τ of σ we have l(x) ∈∪i:vi∈τVi • the subdivision has no cycle of ruined edges and • has no ruined r-simplex. We, again, proceed by cell-induction. As a start, we label each vertex vi of σ by an arbitrary vertex l(vi) ∈Vi. Suppose we are given an m ≥1-dimensional face τ of σ with a labeled subdivision of its boundary. Then, by the previous theorem, it is possible to extend this triangulation to the interior of τ without creating ruined r-simplices and cycles, such that the labels are from the set ∪i:vi∈τVi. We just note that \|∪i:vi∈supp(τ)Vi\| ≥(d+⌊d/r⌋)(m+1) ≥(d+d/r−(r−1)/r)(m+1) > (d−1+(d+1)/r)m. Eventually, the whole simplex σ has such a labeled triangulation. Assigning color i to the vertices with label from Vi we obtain a colored triangulation respecting the assumptions of Sperner’s Lemma. Thus, a multi-colored simplex could be found. The labels of the vertices of this multi-colored simplex form a transversal having the desired property. □ We also obtained a new proof of the following statement on ﬁnding transversals with bounded connected components (which are not necessarily acyclic). Corollary 2.7 [13, Theorem 4.1] For arbitrary positive integers r and d, p(d, r) ≤d + d r . For a graph G let Kr(G) denote the simplicial complex deﬁned on the vertices of G, which contains all simplices inducing connected components of size at most r in G. In particular K1(G) is called the independent set complex of G, it consists of the the independent sets of G. We refer to K2(G) as the 8 induced matching complex of G. Using this notation Theorem 2.5 could be stated in the language of topology. A simplicial complex K is said to be m-connected if its body \|\|K\|\| (the corresponding topological space) is m-connected, i.e., every continuous f : Si →\|\|K\|\| can be extended to a continuous map Bi+1 →\|\|K\|\| for −1 ≤i ≤m (in other words f is nullhomotopic). The m-connectedness of Kr(G) can be described using ﬁllings of G-labeled triangulations. In the remainder of this section a G-labeled simplex is called multi-labeled if the labels of its vertices are all distinct. Proposition 2.8 For a graph G, and m, r ≥0 the complex Kr(G) is m-connected if and only if the following holds for all −1 ≤i ≤m: Every G-labeled triangulation of S i without a ruined, multi-labeled r-simplex has a ﬁlling without a ruined, multi-labeled r-simplex. Proof. Notice that for any complex K the map l : V (K) →V (G) is a simplicial map l : K → Kr(G) if and only if the G-labeled complex (K, l) has no ruined, multi-labeled r-simplex. Indeed, the image under l of a ruined, multi-labeled r-simplex is a set of r + 1 distinct vertices spanning a connected subgraph in G, and such a set is not a simplex of Kr(G). To see the reverse direction assume S is a simplex of K but its image under l is not a simplex of Kr(G). Then l(S) contains r + 1 distinct vertices spanning a connected subgraph, and taking inverse images of these we ﬁnd a ruined, multi-labeled, r-dimensional face of S. Both directions of the proposition is a simple consequence of the above observation and the simplicial approximation theorem. Assume ﬁrst that the ﬁlling property is satisﬁed. We need to show that every continuous map f : Si →\|\|Kr(G)\|\| is nullhomotopic for i ≤m. By the simplicial approximation theorem, there exist a triangulation T of Si and a simplicial map l : T →Kr(G) such that its aﬃne extension \|\|l\|\| : Si →\|\|Kr(G)\|\| is homotopic to f. (In fact, any ﬁne enough triangulation T will do here.) Therefore, it is enough to to show that \|\|l\|\| is nullhomotopic by ﬁnding a continuous extension to Bi. As (T , l) has no ruined, multi-labeled r-simplex it has a ﬁlling (T ′, l′) that has no ruined, multi-labeled r-simplex. Thus l′ : T ′ →Kr(G) is a simplicial map and its aﬃne extension \|\|l′\|\| is a continuous function extending \|\|l\|\| to the ball Bi+1. For the reverse implication assume Kr(G) is m-connected. Let i ≤m, and let (T , l) be a G-labeled triangulation of Si without ruined, multi-labeled r-simplices. Now l is simplicial map from T to Kr(G) and its aﬃne extension \|\|l\|\| is a continuous map from \|\|T \|\| ∼ = Si to \|\|Kr(G)\|\|. Therefore it can be extended to a continuous map f : Bi+1 →\|\|Kr(G)\|\|. We use simplicial approximation for f and ﬁnd a suitable triangulation T ′ of Bi+1 and get a simplicial map l′ : T ′ →Kr(G) approximating f. As f\|Si = \|\|l\|\| we can make sure that the boundary of the complex T ′ is T and l and l′ agree on T . This means that l′ is a ﬁlling of l with no ruined, multi-labeled r-simplices. □ 9 Corollary 2.9 Let m ≥0 and r ≥1 be arbitrary integers. If G is a graph on more than m(d −1 + (d + 1)/r) vertices with maximum degree d ≥r −1, then Kr(G) is (m −1)-connected. Proof. By Proposition 2.8 we need to show that a G-labeled triangulation of S i without ruined, multi-labeled simplices has a ﬁlling still without ruined, multi-labeled simplices for i < m. Theo-rem 2.5 applies here with W = V (G) and states the existence of a ﬁlling without any new ruined r-simplices, much less ruined, multi-labeled simplices. □ 3 Constructions 3.1 Non-ﬁllable labeled triangulations First we give bounds on the connectedness of Kr(G) for r = 1 and 2. The examples are very similar to each other. The graphs constructed are disjoint unions of smaller graphs and we use the simple observation that for the disjoint union of two graphs G and G′ the complex Kr(G ∪G′) is the join of the complexes Kr(G) and Kr(G′). Our ﬁrst example shows that Corollary 2.9 (and thus also Theorem 2.5) is best possible when r = 1. Note that independent set complexes are widely studied and several lower bound on their connectedness and acyclicity is known. The one closest to our result is Proposition 3.1 in a special case of which claims that K1(G) is (⌈˜ γ(G)/2⌉−2)-acyclic over the reals for any graph G. Here ˜ γ(G) is the total domination number, the cardinality of the smallest set S of vertices in G such that every vertex of G is a neighbor of a vertex in S. Using the obvious bound ˜ γ(G) ≥n/d, where n is the number of vertices of G and d is the maximum degree we obtain the same bound on the acyclicity of K1(G) as Corollary 2.9 gives for its connectedness. We remark that in the r = 1 case our argument also naturally generalizes to show that K1(G) is (⌈˜ γ(G)/2⌉−2)-connected. Construction 3.1 Take G1 to be the disjoint union of m copies of Kd,d. It has 2md vertices, one too few for Corollary 2.9 to show that the independent set complex K1(G1) is (m −1)-connected. We show that K1(G) is homotopy equivalent to Sm−1 and therefore it is not (m −1)-connected. The complex K1(Kd,d) consists of two disjoint (d −1)-simplices, so it is homotopy equivalent to S0. As G1 consists of m copies of Kd,d, the complex K1(G1) is the m-fold join of K1(Kd,d) and therefore homotopy equivalent to Sm−1 as claimed. □ Although Corollary 2.9 is best possible in general for r = 1, the independence complex K1(G) can be arbitrarily more connected for particular graphs G, than what is guaranteed by this result. For the cycle Corollary 2.9 gives that K1(Cn) is (⌈n/4⌉−2)-connected. Kozlov determined the homotopy type of of the independent set complex of cycles and his result implies the stronger statement that K1(Cn) is ⌊(n + 1)/3⌋−2 connected. 10 For r > 1 we don’t know whether Corollary 2.9 is tight. The following construction for r = 2 shows that in this case the lower bound of 3 2md −1 2m on the size of the graph in Corollary 2.9 cannot be lowered below 5 4md. For r = d = 2 the construction gives 5 2m, which is tight. Clearly, this construction bounds also how far the lower bound on the size of W in Theorem 2.5 can be lowered. In a combinatorial argument establishes that p(2, 2) = 2. Notice that the standard topological proof of the same fact (similar to the proofs of Corollaries 2.4 and 2.6) through the method of Aharoni and Haxell is impossible. It would require a strengthening of Theorem 2.5 for the r = d = 2 case, which is impossible by the example below. Construction 3.2 Let d be even. A blown-up ﬁve-cycle Hd/2 is a graph on the vertex set ∪4 j=0Aj, \|Aj\| = d/2, where x ∈Aj and y ∈Al are connected if and only if j −l ≡±1 modulo 5. Let G2 be the disjoint union of k copies of the blown-up ﬁve-cycle Hd/2. We claim that K2(G2) is homotopy equivalent to S2k−1 and therefore it is not (2k −1)-connected. As in the previous construction it is enough to prove that for a single blown-up ﬁve cycle Hd/2 the complex K2(Hd/2) is homotopy equivalent to the circle S1. This implies that K2(G2) is homotopy equivalent to the k-fold join of S1, which is S2k−1. The maximal simplices of the independent set complex K1(Hd/2) are the sets Aj ∪Aj+2 for 0 ≤j ≤4. Here (and later in this construction) the indices are understood modulo 5. This complex is easily seen to be homotopy equivalent with the cycle S1. The maximal simplices of K2(Hd/2) are the same simplices together with the simplices {x, y}∪Aj, where x ∈Aj−2, y ∈Aj+2 and 0 ≤j ≤4. Any 1-simplex spanning an edge in Hd/2 is contained in a unique maximal simplex of K2(Hd/2). If a non-maximal simplex of a simplicial complex is contained in a unique maximal simplex then one can collapse this face, i.e., remove all simplices containing it and the remaining complex is homotopy equivalent to the one before the collapse. Therefore we can collapse any 1-simplex of K2(Hd/2) which spans an edge in Hd/2,d/2 and the remaining complex is homotopy equivalent to K2(Hd/2). As the maximal simplices containing these 1-simplices are distinct we can collapse all the 1-simplices spanning an edge simultaneously and the remaining complex is still homotopy equivalent to (in fact a strong deformation retract of) K2(Hd/2). Notice, that the complex remaining after collapsing all the 1-simplices corresponding to edges of Hd/2 is exactly K1(Hd/2). Therefore K2(Hd/2) is homotopy equivalent to K1(Hd/2) and to S1. □ This example shows that with parameters m = 2k, r = 2 an d even, \|V (G)\| = 5(d/2)k = 5 4dm, the statement of Corollary 2.9 is not true. The question remains open, whether the topological proof for r = 2 could be strengthened from 3 2dm or the counterexample improved from 5 4dm. 11 3.2 Partitioned graphs without independent transversals Let n, d, k ≥1 be integers such that d ≥kn/(2k −1). In this section we construct a graph Gk,n,d of maximum degree at most d, together with a vertex set partition into 2k disjoint subsets V1, . . . , V2k of size \|Vi\| = n, i = 1, . . . , 2k, such that there exists no independent transversal with respect to this partition, i.e., every subset T ⊆V (G) with the property \|T ∩Vi\| = 1, i = 1, . . . , 2k, spans at least one edge. Construction 3.3 If n ≤d, then Gk,n,d could be chosen to be the disjoint union of k ≥1 complete bipartite graphs Kn,n, the bipartite classes forming the vertex partition into 2k parts. Thus we can assume d < n and by our condition n ≤2d −d k < 2d. Let i = 2d −n, q = ⌈d−i i ⌉and r = d −qi. We have 1 ≤r ≤i ≤d −1 and 1 ≤q ≤k −1. The graph Gk,n,d is the disjoint union of 2q + 1 complete bipartite graphs Hi with vertex sets Ai ∪Bi, i = 1, . . . , 2q + 1 and an independent set W of 2(k −q −1)n points. The graph Hq+1 is isomorphic to Kd−i+r,d−i+r and all other graphs Hi are isomorphic to Kd,d. The partition classes are deﬁned as follows. For i = 1, . . . q, Vi = Ai ∪B′ i+1, where B′ j ⊆Bj is an arbitrary (d −i)-element subset of Bj. Symmetrically, for i = 1, . . . q, Vq+1+i = Bq+1+i ∪A′ q+i, where A′ j ⊆Aj is an arbitrary (d −i)-element subset of Aj. The leftover elements are divided into two classes: Vq+1 = B1 ∪(∪q+1 j=2(Bj \ B′ j)) and V2q+2 = A2q+1 ∪(∪2q j=q+1(Aj \ A′ j)). This way all the classes are of size 2d −i = n. In case q < k −1, then W ̸= ∅and we create the required 2k classes by arbitrarily partitioning the independent set W. V1 V2 V4 V5 V3 V6 B4 B5 B3 B2 B1 A1 A2 A3 A4 A5 Figure 1: The partitioned graph G3,5,3 Suppose for a contradiction that there exists an independent transversal T of Gk,n,d. If T ∩Bi ̸= ∅ for some index i ≤q, then T ∩Ai = ∅because T is independent. Therefore T ∩Bi+1 ̸= ∅as well, since T is a transversal. Thus, eventually, T ∩Bq+1 ̸= ∅, since Vq+1 ⊆∪q+1 j=1Bj ensures that there is at least one index i ≤q + 1 with T ∩Bi ̸= ∅. For symmetric reasons T ∩Aq+1 ̸= ∅, which provides the contradiction sought after. □ 12 Generalization of this construction will be presented in the next subsection. We prefer to discuss the important special case of independent transversals in this formulation, because we ﬁnd it more transparent, than the (more intuitive) way of Construction 3.7. We remark that if k = d is a power of 2 and n = 2d −1, then our graphs are the same as the one used by Yuster , but our vertex partitions are diﬀerent. Taking the parameters d = k, and n = 2d −1 our construction establishes p(d, 1) = 2d for every d. Previously this only was known for powers of 2 [14, 18]. Corollary 3.4 For every integer d ≥1, p(d, 1) = 2d. □ The same construction partially answers a question of Bollob´ as, Erd˝ os, and Szemer´ edi , which was studied extensively by a number of researchers. Let us recall that ∆(r, n) denotes the largest integer such that any r-partite graph Gr(n) with vertex classes Vi of size n each and of maximum degree less than ∆(r, n) contains an independent transversal, i.e., an independent set containing one vertex from each Vi. The limit ∆r = limn→∞∆(r, n)/n is easily seen to exist. Haxell showed µ = limr→∞∆r = 1/2, but until very recently the exact values of ∆r were known only for r = 2, 3, , and r = 4, 5 . Alon observed that the method of actually implies ∆r ≥ l r 2(r−1) m . Thus Jin’s construction is optimal and for powers of 2 one has ∆r = r 2(r−1). Here we extend the above result for all even r and determine not only ∆r, but all the values ∆(r, n) in this case. The following Proposition appears in and is an immediate consequence of a theorem of Aharoni and Haxell . Proposition 3.5 [6, Proposition 5.2] ∆(r, n) ≥ rn 2(r −1) . For each r even and for arbitrary n our construction provides graphs of maximum degree d = ⌈ rn 2(r−1)⌉with no independent transversal. Hence we have Corollary 3.6 For every integer n ≥1 and r ≥2 even, ∆(r, n) = rn 2(r −1) . Therefore for every r even we have ∆r = r 2(r −1). 13 3.3 Partitioned graphs without H-free transversals Construction 3.7 Let H be an r-regular graph on n vertices and let d be a multiple of r. We prove a lower bound on p(d, H) (see deﬁnition in Section 1.2) by giving an inductive construction. For every c < n n−1 · d r we construct a graph of maximum degree at most d with a vertex partition into classes of size c which does not admit an H-free transversal. We proceed by induction on c. For a positive integer j, let H(j) be a blow-up of H, such that each vertex is replaced with j independent vertices and each edge is replaced with a copy of Kj,j. For c ≤d/r, one can take the blow-up H(c) of H, with the independent sets being the classes of its vertex partition. Now let d r < c < n n−1 · d r and suppose we have a graph ˜ G with vertex partition ˜ V1 ∪. . . ∪˜ Vm, and class size \| ˜ Vi\| = nc −nd/r containing no H-free transversal. Such a partitioned graph exists by our induction hypothesis since our assumption on c guarantees that nc −nd/r < c. Our graph G will contain a copy of ˜ G and m copies H1, . . . , Hm of H(d/r). The vertex partition of G will consist of mn classes of size c. For each i and j, 1 ≤i ≤m, 1 ≤j ≤n, we deﬁne a class V j i = W j i ∪˜ W j i of G, where W j i is the set of d/r independent vertices corresponding to vertex j of H in Hi and ˜ W 1 i ∪. . . ∪˜ W n i is an arbitrary partition of ˜ Vi into parts of size \| ˜ W j i \| = c −d/r. Note that the size of ˜ Vi is n(c −d/r) and that the size of each class V j i is c. It remains to see why there is no H-free transversal. Suppose there is one, denoted by T. Since T is H-free, for each copy Hi of H(d/r) there is a j, 1 ≤j ≤n, such that T ∩W j i = ∅. That is T ∩˜ W j i ̸= ∅, since T is a transversal with respect to the V j i . Thus for each i, 1 ≤i ≤m, T ∩˜ Vi ̸= ∅. So T contains a transversal of ˜ G with respect to the sets ˜ Vi, which cannot be H-free, a contradiction. □ Corollary 3.8 Let H be an r-regular graph on n vertices and d be a multiple of r. Then we have p(d, H) ≥ n (n −1)r d. Note that for H = K3 this corollary gives p(d, K3) ≥3 4d. 4 Remarks and open problems Given an arbitrary graph property R, deﬁne p(d, R) to be the smallest integer p such that any graph G of maximum degree d with a vertex partition into classes of size p admits a transversal spanning a subgraph having property R. We propose the general question of determining p(d, R) for various graph properties R. In this paper we investigated this function when R is “H-free”, “acyclic”, or “having connected components of order at most r”. 14 The most interesting open question regarding H-free transversals is the case of cliques, in par-ticular triangle-free transversals. Currently we only know 3 4d ≤p(d, K3) ≤d. For regular H we conjecture that our construction is optimal. For non-regular H we don’t even have a conjecture. Our other most important problem is the asymptotic determination of p(d, r) for any ﬁxed r, when d tends to inﬁnity. This problem is already open for r = 2. The lone existing lower bound makes even p(d, 2) = d a possibility. This in fact was shown to be true for d = 2 (). The smallest unknown case is p(3, 2) which is either 3 or 4. The question is whether every partition of the vertex set of a 3-regular graph into subsets of size 3 allows for a transversal inducing only a matching. An interesting line of research is to investigate the limits of the triangulation-method of Aharoni and Haxell more thoroughly, i.e., to decide whether Corollary 2.9 (or Theorem 2.5) is optimal. Let us formulate a special case of this problem more precisely. Suppose C is a constant, 5/4 ≤C < 3/2. Given any G-labeled triangulation of Sm−1 containing no ruined 2-simplex, does there exist an extension into a Bm with no ruined 2-simplex, provided \|V (G)\| > Cmd? In other words, what is the smallest number of vertices in a d-regular graph, which guarantees the m-connectedness of the induced matching complex. We know that p(d, forest) = d but the graphs showing the lower bound have parallel edges. It would be interesting to ﬁnd (at least asymptotically) the minimum class size for a vertex partition of simple d-regular graphs that ensures the existence of a cycle-free transversal. This value is between ⌈3 4d⌉and d. The numbers ∆r for odd r ≥7 are extremely intriguing. Currently it is known that r 2(r −1) ≤∆r ≤∆r−1 = r −1 2(r −2). The fact that ∆2 = ∆3 (or ∆4 = ∆5) means that the freedom of an extra class of size n besides the ﬁrst two (or the ﬁrst four) does not help to prevent an independent transversal. It would be very interesting to decide whether this phenomenon is just an artifact of the parameters being too small or there is something deeper going on implying ∆2l = ∆2l+1 for every l. We vote for the latter. Acknowledgement. We would like to thank P´ eter Csorba and G´ abor Moussong for fruitful dis-cussions. P´ eter Csorba greatly simpliﬁed our ﬁrst proof of the bound on the connectedness of the complexes discussed in Constructions 3.1 and 3.2. Note added in proof. The very last conjecture of the last section, namely that ∆2l = ∆2l+1 for every l, was proved by P. Haxell and the ﬁrst author . References R. Aharoni, M. Chudnovsky, A. Kotlov, Triangulated spheres and colored cliques, Discrete and Computational Geometry 28 (2) (2002), 223–229. 15 R. Aharoni, P. Haxell, Hall’s Theorem for hypergraphs, Journal of Graph Theory 35 (2) (2000), 83–88. R. Aharoni, P. Haxell, Systems of disjoint representatives, to appear. N. Alon, The linear arboricity of graphs, Israel Journal of Mathematics 62 (1988), 311–325. N. Alon, The strong chromatic number of a graph, Random Structures and Algorithms 3 (1992), 1–7. N. Alon, Problems and results in extremal combinatorics, Part I, Discrete Mathematics 273 (2003), 31–53. N. Alon, G. Ding, B. Oporowski, D. Vertigan, Partitioning into graphs with only small compo-nents, Journal of Combinatorial Theory, Series B 87 (2003), 231–243. A. Bj¨ orner, Topological Methods, in Handbook of Combinatorics (R. Graham, M. Gr¨ otschel, and L. Lov´ asz, eds.), Elsevier and MIT Press (1995). B. Bollob´ as, P. Erd˝ os, E. Szemer´ edi, On complete subgraphs of r-chromatic graphs, Discrete Mathematics 13 (1975), 97–107. P. Haxell, A condition for matchability in hypergraphs, Graphs and Combinatorics 11 (3) (1995), 245–248. P. Haxell, A note on vertex list colouring, Combinatorics, Probability and Computing 10 (2001), 345–348. P. Haxell, T. Szab´ o, Odd transversals are odd, submitted. P. Haxell, T. Szab´ o, G. Tardos, Bounded size components—partitions and transversals, Journal of Combinatorial Theory, Series B 88 (2) (2003), 281–297. G. P. Jin, Complete subgraphs of r-partite graphs, Combinatorics, Probability and Computing, 1 (3) (1992), 241–250. D. Kozlov, Complexes of directed trees, Journal of Combinatorial Theory, Series A 88 (1) (1999), 112–122. R. Meshulam, The clique complex and hypergraph matchings, Combinatorica 21 (1) (2001), 89–94. E. Sperner, Neuer Beweis f¨ ur die Invarianz der Dimensionzahl und des Gebietes, Abhandlungen aus dem Mathematischen Seminar der Universit¨ at Hamburg 6 (1928), 265-272. 16 R. Yuster, Independent transversals in r-partite graphs, Discrete Mathematics 176 (1997), 255– 261. 17
6211	https://mitocw.ups.edu.ec/high-school/mathematics/exam-prep/taylor-series/maclaurin-taylor-series	Maclaurin & Taylor Series Analytic Functions Algorithm for Computing Taylor Series More Examples of Taylor Series Taylor Approximations and Power Series Analytic Functions Analytic functions and Taylor series are defined, including the concept of an infinitely differentiable function. 18.01 Single Variable Calculus, Fall 2005 Prof. Jason Starr Course Material Related to This Topic: Read lecture notes, section 2 on pages 2–4 Back to Top Algorithm for Computing Taylor Series Step-by-step method for computing a Taylor series, with example of finding the Taylor series expansion of f(x) = (1-x)-1 about x = 0. 18.01 Single Variable Calculus, Fall 2005 Prof. Jason Starr Course Material Related to This Topic: Read lecture notes, section 3, pages 4–5 Back to Top More Examples of Taylor Series Taylor series expansions of (1-x)-1, ex, sin(x), and cos(x) about any point x=a. 18.01 Single Variable Calculus, Fall 2005 Prof. Jason Starr Course Material Related to This Topic: Read lecture notes, section 4, pages 5–9 Back to Top Taylor Approximations and Power Series Five questions which involve Taylor series approximations for sine, cosine, exponential, logarithmic, and other functions, as well as finding error bounds on these approximations. 18.01 Single Variable Calculus, Fall 2006 Prof. David Jerison Course Material Related to This Topic: Complete exam problems 7C-1 to 7C-5 on page 44 Check solution to exam problems 7C-1 to 7C-5 on pages 100–2 Back to Top
6212	https://ucmp.berkeley.edu/cnidaria/cnidaria.html	\| \| \| Introduction to CnidariaJellyfish, corals, and other stingers \| \| Cnidarians are incredibly diverse in form, as evidenced by colonial siphonophores, massive medusae and corals, feathery hydroids, and box jellies with complex eyes. Yet, these diverse animals are all armed with stinging cells called nematocysts. Cnidarians are united based on the presumption that their nematocysts have been inherited from a single common ancestor. The name Cnidaria comes from the Greek word "cnidos," which means stinging nettle. Casually touching many cnidarians will make it clear how they got their name when their nematocysts eject barbed threads tipped with poison. Many thousands of cnidarian species live in the world's oceans, from the tropics to the poles, from the surface to the bottom. Some even burrow. A smaller number of species are found in rivers and fresh water lakes. There are four major groups of cnidarians: Anthozoa, which includes true corals, anemones, and sea pens; Cubozoa, the amazing box jellies with complex eyes and potent toxins; Hydrozoa, the most diverse group with siphonophores, hydroids, fire corals, and many medusae; and Scyphozoa, the true jellyfish. \| \| \| Click on the buttons below to learn more about Cnidaria. \| Visit the Cnidaria WWW Server for information and links dealing with all manner of cnidarians. Visit the home page of the Hydrozoan Society. \|
6213	https://www.ikonixusa.com/whitepapers/true-power-vs-apparent-power	Whitepapers - True Power vs. Apparent Power: Understanding the Difference The store will not work correctly when cookies are disabled. My Account Sign In Create an Account Ensuring Electrically Safe Products.® Toggle Nav Customer Resources Product Catalogs Careers Menu Products Accessories Software Consulting Support Calibration Repair Submit an RMA Trade-in Instrument Drivers Technical Question Education Videos Whitepapers Safety Agency Resources System Integration Resources Glossary About Who We Are Customer Benefits Contact Us International Locations Customer Resources Product Catalogs Careers Account My Account Sign In Create an Account Quick Links Send in Your Tester Trade in Search Search Categories Products See All Pages Popular Suggestions Sorry, no results were found! Search Change Sign In Create an Account My Cart My Cart 0 CloseYou have no items in your shopping cart. Home Whitepapers \| True Power vs. Apparent Power: Understanding the Difference Download PDF Version of True Power vs. Apparent Power: Understanding the Difference > True Power vs. Apparent Power: Understanding the Difference Introduction AC power sources are essential pieces of equipment for providing flexible and reliable power generation. However, the concepts of power generation and power loss can be daunting for end users and sales personnel alike. This is especially true when it comes to talking about the different types of power quantities. What is the difference between true power and apparent power? Why is power referred to in both watts and volt-amps (VA)? This paper attempts to answer these questions and provide an understanding on the various aspects of power. Energy and Power In order to fully understand power, it is critical to grasp the concept of energy and how it relates to power. Energy is a force multiplied by a distance. An example is shown in Fig. 1. A force of 50 Newtons is applied to the block at point A in order to move it 10m to point B. The total energy expended is E = 50N 10m = 500 Joules. The term "Joule" is the quantity used to define energy. If the same force were used to push the block only 5m, the energy expended would be E = 50N 5m = 250 Joules. This makes sense since more energy would be required to push a block 10m rather than moving it 5m. Fig. 1: Force Block Diagram So what does energy have to do with power? Power is defined as the amount of energy expended per unit time. Looking back at the block example from Fig. 1, say it takes 5 seconds to move the block 10m from point A to point B. Then the total power, P = 500 Joules/5 sec = 100 Joules/sec = 100 Watts. So essentially, power is a force multiplied by a rate. Using this concept, it can be applied to electrical quantities. This dynamic is the same when dealing with power in electronics. Voltage or "potential" is the force necessary to move electrons (rather than blocks). Current is actually a rate of flow of charges (in this case the charge of electrons) per second through a material with that voltage applied. By taking that force (voltage) multiplied by a rate (current), the end result is energy expended over time. This quantity is power. Thus electrical power is voltage multiplied by current. P = V I where power (P) is in watts, voltage (V) is in volts and current (I) is in amps The voltage and current of an AC circuit are sinusoidal in nature (see Fig. 2). This means that the amplitude of the current and voltage of an AC circuit will constantly change over time. Since power is voltage times current (P = VI), power is maximized when the voltage and current are "lined up" with one another. The peaks and zero points on the waveform for voltage and current occur at the same time (Fig. 3). When this happens, the two signals are said to be "in phase" with one another. Two waveforms are said to be "out of phase" or "phase shifted" when the two signals do not match up from point to point. So when are the voltage and current in phase? What causes them to shift so that they would be out of phase? This has to do with the circuit components involved in the system. Due to the behavior of voltage and current in AC circuits, power is a quantity that comes in several different flavors. Fig. 2: AC Waveform Fig. 3: In Phase Voltage and Current Signals Circuit Components: To Shift or Not To Shift? The three main circuit components are the resistor, the capacitor and the inductor. Each component has a different effect on the phase shift between voltage and current. A resistor causes no shift between current and voltage (a zero degree shift - shown in Fig. 3). Power that is the result of a purely resistive load is known as "true" power. It is the power that performs work in the circuit and it is measured in Watts (W). This is the power that will drive a circuit board or a motor. It is the desired outcome of an electrical system. A capacitor and inductor will cause a 90 degree phase shift between voltage and current (Fig. 4). The resulting power will have a value of zero every time the voltage or current has a zero value since the two quantities are multiplied to get power (Fig. 5). This is not a desirable model because although the source is generating power, no work is being done at the load. This type of power is known as "reactive" power because it counteracts the effects of true power. Reactive power is the lazy brother of true power. True power is doing all the work while reactive power is actually taking away from the power in the system making the true power work harder to get the job done. It is measured in what is known as reactive volt-amps (VAr). A capacitive load will introduce negative reactive power and the voltage "lags" the current waveform by 90° (Fig. 6). An inductive load will introduce positive reactive power and the voltage "leads" the current waveform by 90° (Fig. 7). This is because capacitors "generate" reactive power and inductors "consume" reactive power. Fig. 5: Voltage, Current and Resulting Power waveforms Fig. 6: Negative Reactive Power - Voltage Lags Current Fig. 7: Positive Reactive Power - Voltage Leads Current Why are these values important? The reason is because all practical electrical applications will contain a combination of resistive, capacitive and inductive elements. Therefore, any practical AC circuit or system will have a combination of both true and reactive power which will vary the phase angle between voltage and current. The desired outcome is to maximize true power while limiting reactive power. Taking both into account, the end result of the tug of war battle between true and reactive power is called complex power. It is measured in volt-amps (VA). The Power Triangle The power triangle is shown in Fig. 8. This gives a graphical representation of how all these quantities are related. Fig. 8: The Power Triangle True Power (P): power that performs work measured in Watts (W) Reactive Power (Q): power that does not perform work (sometimes called "wattless power") measured in VA reactive (VAr) Complex Power (S): the vector sum of the true and reactive power measured in volt amps (VA) Apparent Power (ISI): the magnitude of the complex power measured in volt amps (VA). Φ = phase angle. This is the angle used to describe the phase shift between the voltage and current. The larger the phase angle, the greater the reactive power generated by the system. In this case, the horizontal leg is the true power, the vertical leg is the reactive power and the hypotenuse is the complex power (Fig. 8). True power is the work we want out of the system. As a result we are going to say that it is on the "true" axis (horizontal line). Since reactive power essentially "leeches" off of true power and does not contribute to powering the load, we'll define it separately and say it is on the "imaginary axis (vertical line whose value indicated by the letter "I" before the quantity). The apparent power is calculating by simply taking the magnitude of S (ISI -> magnitude of a value shown by the value with a vertical line on either side). For example, if the true power in a system is 4W and the reactive power is 3VAr, the complex power would be S = sqrt(4A2 + JA2) = 5VA This method of calculation is known as a vector summation. Putting the Pieces Together The important quantity to pull from this is the apparent power. The apparent power is the vector summation of both the true power and reactive power. That basically means that both the true and reactive power elements will affect the apparent power value. This will depend upon the magnitude of the true power, the magnitude of the reactive power and the phase shift between the voltage and the current. Apparent power is used to define the total power delivered to a load. This is also how ratings are often described on power sources, including the EEC series. For example, the power source is a l000VA unit. That VA quantity refers to the apparent power. Fig. 9: Black Box Example Take a look at the image in Fig. 9. This is a black box example. There are three parts to this diagram: the input, the system and the output. The user has to input a certain amount of power, the system performs the work and the end result is that there is output power. In a perfect world with no losses, the input power would be equal to the output power. However, there will always be losses on a system. Say that the required output to run a motor is 1000VA. It is known that 200VA of power are "lost" in the system due to the reactive power components. This means that the user will need to input at least 1200VA of power to run the motor. Since 200VA is lost in the "work" portion of the system, the output power = 1200VA - 200VA = 1000VA. Otherwise, the motor will not have enough power to operate. It is important to know the phase shift between the voltage and current because this gives a value for reactive power. This value needs to be known so that enough power can be applied at the input to get the desired output. The above example outlines the importance of calculating losses in a system. Without knowing how much power is lost on a system, it is possible that the source would not have the necessary power to supply the load. The end result would either be that the source would be unable to supply the load or the system would operate at a highly inefficient rate. Therefore it is necessary to calculate the power consumption of the entire system rather than just the load so that the source can supply the necessary amount of power. Conclusion The apparent power is a combination of both reactive power and true power. True power is a result of resistive components and reactive power is a result of capacitive and inductive components. Almost all circuitry on the market will contain a combination of these components. Since reactive power takes away from true power, it must be considered in a system to ensure that the apparent power output from a system is sufficient to supply the load. This is a critical aspect of understanding AC power sources because the source must be capable of supplying the necessary volt-amp power for a given application. As with any product, understanding the needs and specifications of the end user will ensure a successful application. 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6214	https://collegedunia.com/exams/dichromate-chemistry-articleid-1954	Dichromate (Cr2O72-): Structure, Chemical and Physical Properties, Applications Select Goal & City Select Goal Search for Colleges, Exams, Courses and More.. 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This compound acts as a strong oxidizing agent and is also primarily used as the standard solutionfor volumetric analysis. In aqueous solutions these dichromate ions are interconvertible. Potassium Dichromate is one of the most common examples of Dichromates. Potassium dichromate is an orange crystalline solid that gives chromic oxide and potassium chromate when it readily decomposes. Dichromate’s molecular formula is Cr 2 O 7 2- and its density is 2.68 grams per centimetre cube. The molecular weight weighs 294.185 g/mol over molar mass. This compound also has a boiling point of 500 C and it also has a melting point of 398 C. Table of Content What is Dichromate? Dichromate: Structure Chemical Properties of Dichromate Physical Properties of Dichromate Uses of Dichromate Things To Remember Sample Questions Keyterms:Dichromate, Anion, Oxidizing agent, Solution, Aqueous solution, Chromic oxide, Crystalline solid, Potassium chromate What is Dichromate? [Click Here for Sample Questions] Dichromate is an anion with the chemical formula Cr 2 O 7 2-. It is a strong oxidizing agent in organic chemistry and is used as a primary standard solution in volumetric analysis. Chromate and Dichromate ions are inconvertible in aqueous solutions. Dichromate \| Cr 2 O 7 2- \| Dichromate \| \| Density \| 2.68 g/cm³ \| \| Molecular Weight/ Molar Mass \| 294.185 g/mol \| \| Boiling Point \| 500 °C \| \| Melting Point \| 398 °C \| \| Chemical Formula \| Cr 2 O 7 2- \| Also Read: \| Related Articles \| \| Hydrochloric Acid \| Neodymium \| Promethium \| \| Atomic Orbitals \| Transuranium Element \| Atomic Radius \| Dichromate: Structure [Click Here for Sample Questions] The structure of the dichromateion is shown in the following diagrams. Dichromate Physical Properties of Dichromate [Click Here for Sample Questions] It is odourless. Its Appearance is a Red-orange crystalline solid. Its Valency is 2. Its pH level is 4. Its Oxidation state is +6. It is appreciably soluble in hot water and moderately in cold water. Potassium Dichromate Chemical Properties of Dichromate [Click Here for Sample Questions] A deep blue colour with peroxide due to the formation of [CrO(O 2)2] is formed when an acidified solution of Dichromate. Cr 2 O 7 2- + 4H 2 O 2 + 2H+ → 2CrO 5 + 5H 2 O It reacts with hydrogen sulphide and oxidises it to sulphur It oxidises sulphites to sulphates. It oxidises chlorides to chlorine. It oxidises nitrites to nitrates. It oxidises thiosulphates to sulphates. It oxidises sulphur to stannic salts. Cr 2 O 7 2- + 3H2S + 8H+ → 2Cr3+ + 3S + 7H2O Potassium Dichromate Uses of Dichromate [Click Here for Sample Questions] There are various uses of Dichromate, some of these uses are listed below: It is for the hardening of gelatin film in photography. It is used in the leather industry for chrome tanning. It is also used as a mordant in dyeing fixatives like Cr(OH)3. It is used in the volumetric estimation of iodine, sulphite and ferrous salts. It is used for the preparation of compounds containing chromium such as chrome yellow, chrome red and chrome alum. It is used for chrome plating which helps in protecting metals from corroding and improves the adhesion of paint. The heavy metal, alkaline earth and lanthanide salts of Dichromate are used as pigments. Things To Remember The use of lead-containing pigment chrome yellow stopped due to environmental regulations. Dichromate is an anion and its chemical formula is Cr 2 O 7 2-. This compound acts as a strong oxidizing agent and is also primarily used as the standard solution for volumetric analysis. In aqueous solutions these dichromate ions are interconvertible. Potassium Dichromate is one of the most common examples of Dichromates. Potassium dichromate is an orange crystalline solid that gives chromic oxide and potassium chromate when it readily decomposes. Dichromate’s molecular formula is Cr2O72- and its density is 2.68 grams per centimetre cube. The molecular weight weighs 294.185 g/mol over molar mass. This compound also has a boiling point of 500 C and it also has a melting point of 398 C. Also Read: \| Related Articles \| \| Amphoteric Oxides \| 2 4 Dinitrophenylhydrazine \| Diazotization Reaction \| \| Sulphonation Mechanism \| Acetate \| What is the Unit of Sound? \| \| d and f block elements MCQ \| D and F Block Important Questions \| d and f Block Elements Ncert Solutions \| Sample Questions Ques: How do I prepare a 0.1 N potassium dichromate solution? (4 marks) Ans:Normality = (mass of solute x 1000) ÷ (mass of solute equivalent to grams x volume of solution (ml)). Molar mass in grams KCr 2 O 7 = 294.18 grams. To find the equivalent of potassium dichromate, it is necessary to examine the changes in the oxidation state of Cr in the reactants and products, and in the acidic medium. (Cr 2 O 7 2- → 2Cr 3). Here, the oxidation state changed (the change was 6 because it changed from 2 × 6 to 2 × 3) to the equivalent of KCr 2 O 7 = 294.18 / 6 = 49.03. To prepare one regular solution, one gram of KCr 2 O 7, or 49.03 grams, must be dissolved in one litre of solution. To prepare a regular solution of 0.1, dissolve 4.903 grams in water to make 1 litre of solution. Ques: How do I calculate the equivalent weight of potassium dichromate? (3 marks) Ans:The potassium dichromatein an acid medium is a strong oxidant. This means that electrons earn during the redox reaction. The potassium dichromatein acid solution leads to: KCr 2 O 7 14H 6E- → 2K 2 CR 3 7H 2 O equivalent weight of KCr 2 O 7 = 294.26 / 6 = 49 Ques: What happens to chromium hybridization with dichromate ion (Cr 2 O 7 2-)? (3 marks) Ans:Cr 2 O 7 2- is a special compound. The oxidation state of Cr is 6. There are no electrons in the Cr 6 valence shell. That is, there are empty 4s and 3d subshells. Three 3D orbitals, ie. 3dxy, 3dyz, 3dzx, and One4s orbits are used. If you're wondering why you can't see the other d-orbitals, it's because they're in high energy levels (CFT theory) and easy to link to them. Therefore, this results in hybridization d3. Ques: What are the uses of dichromate compounds? (2 marks) Ans: Dichromatecompounds have many uses. They are used as oxidants and are also used in the preparation of various elements such as waxes, paints and adhesives. However, because potassium dichromateis a hexavalent chromium product, it is carcinogenic and highly toxic. Ques: What is the dichromate test? (2 marks) Ans:Potassium dichromatepaper can be used to look for sulfur dioxide as it clearly changes from orange to white. This is typical of all redox reactions in which hexavalent chromium is reduced to trivalent chromium. Ques: Is Sodium Dichromate an Oxidizer? (2 marks) Ans:Sodium dichromateis a very powerful oxidant. Sodium dichromatein sulfuric acid is used to oxidize primary alcohols but is severely restricted to the corresponding acid by peroxidation with hydrated aldehydes. Ques: Why doesn't potassium dichromate act as a self-indicating drug? (2 marks) Ans:Cr 2 O 7 is orange and the first drop must be yellow, so it is supposed to be used as a self-index, but Cr is 3 ((reduced form of Cr 2 O 7))'green, so yellow is visible. not. This makes it difficult to display and visualize. Ind should be used as diphenylamine sulfonic acid as it was detected. Ques: What is the use of potassium dichromate? (2 marks) Ans:Potassium dichromatehas a wide range of uses as an oxidizer in many chemical and industrial applications, as well as in leather colour, colour and tanning. It is also used medically as an external disinfectant or astringent. Ques: What happens if I add potassium dichromate to hydrochloric acid? (3 marks) Ans:Potassium dichromateis a powerful oxidant and is strong enough to oxidize HCl to Cl2 (chlorine gas; hopefully this experiment was done in succession). In this reaction, Cr2O7 (2-) dichromate, which has an oxidation number of Cr 6, is reduced to Cr (3). The half-reaction equation is: 2Cl- → Cl 2 2 e- Cr 2 O 7 2- 6 e-14H → 2Cr 3 7H 2 O The combination of these gives a reaction equation. 6 Cl- Cr 2 O7 2-14H → 3 Cl 2 2Cr 3 7H 2 O. 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6215	https://en.wikipedia.org/wiki/TP53BP1	Jump to content TP53BP1 Cymraeg Galego Татарча / tatarça Українська Edit links From Wikipedia, the free encyclopedia Protein-coding gene in the species Homo sapiens \| TP53BP1 \| \| \| \| --- --- \| \| \| \| \| \| \| \| Available structures \| \| \| \| --- --- \| \| PDB \| Ortholog search: PDBe RCSB \| \| \| \| List of PDB id codes \| \| --- \| \| 1GZH, 1KZY, 1XNI, 2G3R, 2IG0, 2LVM, 2MWO, 2MWP, 3LGF, 3LGL, 3LH0, 4CRI, 4RG2, 4X34, 5ECG \| \| \| \| \| \| \| \| Identifiers \| \| \| \| \| Aliases \| TP53BP1, 53BP1, p202, p53BP1, TP53, TDRD30, tumor protein p53 binding protein 1 \| \| \| \| External IDs \| OMIM: 605230; MGI: 1351320; HomoloGene: 4137; GeneCards: TP53BP1; OMA:TP53BP1 - orthologs \| \| \| \| \| Gene location (Human) \| \| \| \| --- --- \| \| \| \| \| \| \| Chr. \| Chromosome 15 (human) \| \| \| \| \| \| \| \| \| Band \| 15q15.3 \| Start \| 43,403,061 bp \| \| End \| 43,510,728 bp \| \| \| \| \| \| \| Gene location (Mouse) \| \| \| \| --- --- \| \| \| \| \| \| \| Chr. \| Chromosome 2 (mouse) \| \| \| \| \| \| \| \| \| Band \| 2\|2 E5 \| Start \| 121,023,762 bp \| \| End \| 121,101,888 bp \| \| \| \| \| \| \| RNA expression pattern \| \| \| \| --- --- \| \| Bgee \| \| Human \| Mouse (ortholog) \| --- \| \| \| \| \| Top expressed in \| \| pituitary gland anterior pituitary gonad ventricular zone stromal cell of endometrium Achilles tendon right uterine tube ganglionic eminence middle temporal gyrus sural nerve \| \| \| \| \| Top expressed in \| \| primary oocyte ascending aorta substantia nigra secondary oocyte ciliary body fossa renal corpuscle Rostral migratory stream aortic valve condyle \| \| \| More reference expression data \| \| \| \| \| \| BioGPS \| \| \| \| \| \| --- --- \| \| \| \| \| \| \| More reference expression data \| \| \| \| \| \| \| \| \| \| \| Gene ontology \| \| \| \| --- --- \| \| Molecular function \| DNA binding transcription coregulator activity p53 binding protein binding ubiquitin modification-dependent histone binding damaged DNA binding methylated histone binding telomeric DNA binding sequence-specific DNA binding histone binding \| \| Cellular component \| cytoplasm nucleoplasm chromosome telomere chromosome, centromeric region kinetochore nuclear body nucleus replication fork site of double-strand break DNA repair complex \| \| Biological process \| regulation of transcription, DNA-templated cellular response to X-ray DNA damage checkpoint signaling positive regulation of DNA-binding transcription factor activity transcription, DNA-templated cellular response to DNA damage stimulus positive regulation of transcription, DNA-templated positive regulation of transcription by RNA polymerase II DNA repair protein homooligomerization double-strand break repair via nonhomologous end joining positive regulation of isotype switching negative regulation of double-strand break repair via homologous recombination \| \| Sources:Amigo / QuickGO \| \| \| \| \| \| \| \| \| \| Orthologs \| \| \| \| --- --- \| \| Species \| Human \| Mouse \| \| Entrez \| \| \| \| 7158 \| \| \| \| \| 27223 \| \| \| Ensembl \| \| \| \| ENSG00000067369 \| \| \| \| \| ENSMUSG00000043909 \| \| \| UniProt \| \| \| \| Q12888 \| \| \| \| \| P70399 \| \| \| RefSeq (mRNA) \| \| \| \| NM_001141979 NM_001141980 NM_005657 NM_001355001 \| \| \| \| \| NM_001290830 NM_013735 \| \| \| RefSeq (protein) \| \| \| \| NP_001135451 NP_001135452 NP_005648 NP_001341930 \| \| \| \| \| NP_001277759 NP_038763 \| \| \| Location (UCSC) \| Chr 15: 43.4 – 43.51 Mb \| Chr 2: 121.02 – 121.1 Mb \| \| PubMed search \| \| \| \| \| \| \| \| Wikidata \| \| \| \| \| \| \| \| \| \| --- --- \| \| View/Edit Human \| \| View/Edit Mouse \| \| \| \| \| \| Tumor suppressor p53-binding protein 1 also known as p53-binding protein 1 or 53BP1 is a protein that in humans is encoded by the TP53BP1 gene. Clinical significance [edit] 53BP1 is underexpressed in most cases of triple-negative breast cancer. DNA repair [edit] DNA double-strand breaks (DSBs) are cytotoxic damages that can be repaired either by the homologous recombinational repair (HR) pathway or by the non-homologous end-joining (NHEJ) pathway. NHEJ, although faster than HR, is less accurate. The early divergent step between the two pathways is end resection, and this step is regulated by numerous factors. In particular, BRCA1 and 53BP1 play a role in determining the balance between the two pathways. 53BP1 restricts resection and promotes NHEJ. Age-associated deficient repair [edit] Ordinarily during the G1 phase of the cell cycle, when a sister chromatid is unavailable for HR, NHEJ is the predominant pathway for repairing DNA double-strand breaks (DSBs). However, as individuals age, recruitment of 53BP1 to DSBs during G1 becomes deficient. The absence of 53BP1 at such DSBs appears to promote the alternative error-prone repair process Alt-EJ. This repair process, also referred to as microhomology-mediated end joining, is highly inaccurate and likely contributes to the aging process. Interactions [edit] 53BP1 has been shown to physically interact with: Histone H4 dimethylated or monomethylated at Lysine 20 Histone H2A or Histone H2A.X ubiquitinated at Lysine 15 p53 DYNLL1 TIRR (NUDT16L1) References [edit] ^ Jump up to: a b c GRCh38: Ensembl release 89: ENSG00000067369 – Ensembl, May 2017 ^ Jump up to: a b c GRCm38: Ensembl release 89: ENSMUSG00000043909 – Ensembl, May 2017 ^ "Human PubMed Reference:". National Center for Biotechnology Information, U.S. National Library of Medicine. ^ "Mouse PubMed Reference:". National Center for Biotechnology Information, U.S. National Library of Medicine. ^ Iwabuchi K, Bartel PL, Li B, Marraccino R, Fields S (Jun 1994). "Two cellular proteins that bind to wild-type but not mutant p53". Proceedings of the National Academy of Sciences of the United States of America. 91 (13): 6098–102. Bibcode:1994PNAS...91.6098I. doi:10.1073/pnas.91.13.6098. PMC 44145. PMID 8016121. ^ Iwabuchi K, Li B, Massa HF, Trask BJ, Date T, Fields S (Oct 1998). "Stimulation of p53-mediated transcriptional activation by the p53-binding proteins, 53BP1 and 53BP2". The Journal of Biological Chemistry. 273 (40): 26061–8. doi:10.1074/jbc.273.40.26061. PMID 9748285. ^ "Entrez Gene: TP53BP1 tumor protein p53 binding protein 1". ^ Bouwman P, Aly A, Escandell JM, Pieterse M, Bartkova J, van der Gulden H, Hiddingh S, Thanasoula M, Kulkarni A, Yang Q, Haffty BG, Tommiska J, Blomqvist C, Drapkin R, Adams DJ, Nevanlinna H, Bartek J, Tarsounas M, Ganesan S, Jonkers J (Jun 2010). "53BP1 loss rescues BRCA1 deficiency and is associated with triple-negative and BRCA-mutated breast cancers". Nature Structural & Molecular Biology. 17 (6): 688–95. doi:10.1038/nsmb.1831. PMC 2912507. PMID 20453858. ^ Panier S, Boulton SJ (2014). "Double-strand break repair: 53BP1 comes into focus". Nat. Rev. Mol. Cell Biol. 15 (1): 7–18. doi:10.1038/nrm3719. PMID 24326623. S2CID 3752325. ^ Li J, Xu X (2016). "DNA double-strand break repair: a tale of pathway choices". Acta Biochim. Biophys. Sin. (Shanghai). 48 (7): 641–6. doi:10.1093/abbs/gmw045. PMID 27217474. ^ Jump up to: a b Anglada T, Genescà A, Martín M (December 2020). "Age-associated deficient recruitment of 53BP1 in G1 cells directs DNA double-strand break repair to BRCA1/CtIP-mediated DNA-end resection". Aging. 12 (24): 24872–24893. doi:10.18632/aging.202419. PMC 7803562. PMID 33361520. ^ Botuyan MV, Lee J, Ward IM, Kim JE, Thompson JR, Chen J, Mer G (Dec 2006). "Structural basis for the methylation state-specific recognition of histone H4-K20 by 53BP1 and Crb2 in DNA repair". Cell. 127 (7): 1361–73. doi:10.1016/j.cell.2006.10.043. PMC 1804291. PMID 17190600. ^ Fradet-Turcotte A, Canny MD, Orthwein A, Leung CC, Huang H, Landry MC, Kitevski-LeBlanc J, Noordermeer SM, Sicheri F, Durocher D (Jun 2013). "53BP1 is a reader of the DNA-damage-induced H2A Lys 15 ubiquitin mark". Nature. 499 (7456): 50–4. Bibcode:2013Natur.499...50F. doi:10.1038/nature12318. PMC 3955401. PMID 23760478. ^ Derbyshire DJ, Basu BP, Serpell LC, Joo WS, Date T, Iwabuchi K, Doherty AJ (Jul 2002). "Crystal structure of human 53BP1 BRCT domains bound to p53 tumour suppressor". The EMBO Journal. 21 (14): 3863–72. doi:10.1093/emboj/cdf383. PMC 126127. PMID 12110597. ^ Ekblad CM, Friedler A, Veprintsev D, Weinberg RL, Itzhaki LS (Mar 2004). "Comparison of BRCT domains of BRCA1 and 53BP1: a biophysical analysis". Protein Science. 13 (3): 617–25. doi:10.1110/ps.03461404. PMC 2286730. PMID 14978302. ^ Joo WS, Jeffrey PD, Cantor SB, Finnin MS, Livingston DM, Pavletich NP (Mar 2002). "Structure of the 53BP1 BRCT region bound to p53 and its comparison to the Brca1 BRCT structure". Genes & Development. 16 (5): 583–93. doi:10.1101/gad.959202. PMC 155350. PMID 11877378. ^ Derbyshire DJ, Basu BP, Date T, Iwabuchi K, Doherty AJ (Oct 2002). "Purification, crystallization and preliminary X-ray analysis of the BRCT domains of human 53BP1 bound to the p53 tumour suppressor". Acta Crystallographica Section D. 58 (Pt 10 Pt 2): 1826–9. Bibcode:2002AcCrD..58.1826D. doi:10.1107/S0907444902010910. PMID 12351827. ^ Lo KW, Kan HM, Chan LN, Xu WG, Wang KP, Wu Z, Sheng M, Zhang M (Mar 2005). "The 8-kDa dynein light chain binds to p53-binding protein 1 and mediates DNA damage-induced p53 nuclear accumulation". The Journal of Biological Chemistry. 280 (9): 8172–9. doi:10.1074/jbc.M411408200. PMID 15611139. ^ Drané P, Brault ME, Cui G, Meghani K, Chaubey S, Detappe A, Parnandi N, He Y, Zheng XF, Botuyan MV, Kalousi A, Yewdell WT, Münch C, Harper JW, Chaudhuri J, Soutoglou E, Mer G, Chowdhury D (Mar 2017). "TIRR regulates 53BP1 by masking its histone methyl-lysine binding function". Nature. 543 (7644): 211–6. doi:10.1038/nature21358. PMC 5441565. PMID 28241136. ^ Botuyan MV, Cui G, Drané P, Oliveira C, Detappe A, Brault ME, Parnandi N, Chaubey S, Thompson JR, Bragantini B, Zhao D, Chapman JR, Chowdhury D, Mer G (Jul 2018). "Mechanism of 53BP1 activity regulation by RNA-binding TIRR and a designer protein". Nat Struct Mol Biol. 25 (7): 591–600. doi:10.1038/s41594-018-0083-z. PMC 6045459. PMID 29967536. Further reading [edit] Choubey D, Lengyel P (Mar 1995). "Binding of an interferon-inducible protein (p202) to the retinoblastoma protein". The Journal of Biological Chemistry. 270 (11): 6134–40. doi:10.1074/jbc.270.11.6134. PMID 7890747. Choubey D, Li SJ, Datta B, Gutterman JU, Lengyel P (Oct 1996). "Inhibition of E2F-mediated transcription by p202". The EMBO Journal. 15 (20): 5668–78. doi:10.1002/j.1460-2075.1996.tb00951.x. PMC 452311. PMID 8896460. Datta B, Li B, Choubey D, Nallur G, Lengyel P (Nov 1996). "p202, an interferon-inducible modulator of transcription, inhibits transcriptional activation by the p53 tumor suppressor protein, and a segment from the p53-binding protein 1 that binds to p202 overcomes this inhibition". The Journal of Biological Chemistry. 271 (44): 27544–55. doi:10.1074/jbc.271.44.27544. PMID 8910340. Wen Y, Yan DH, Spohn B, Deng J, Lin SY, Hung MC (Jan 2000). "Tumor suppression and sensitization to tumor necrosis factor alpha-induced apoptosis by an interferon-inducible protein, p202, in breast cancer cells". Cancer Research. 60 (1): 42–6. PMID 10646849. Xia Z, Morales JC, Dunphy WG, Carpenter PB (Jan 2001). "Negative cell cycle regulation and DNA damage-inducible phosphorylation of the BRCT protein 53BP1". The Journal of Biological Chemistry. 276 (4): 2708–18. doi:10.1074/jbc.M007665200. PMID 11042216. Joo WS, Jeffrey PD, Cantor SB, Finnin MS, Livingston DM, Pavletich NP (Mar 2002). "Structure of the 53BP1 BRCT region bound to p53 and its comparison to the Brca1 BRCT structure". Genes & Development. 16 (5): 583–93. doi:10.1101/gad.959202. PMC 155350. PMID 11877378. Derbyshire DJ, Basu BP, Serpell LC, Joo WS, Date T, Iwabuchi K, Doherty AJ (Jul 2002). "Crystal structure of human 53BP1 BRCT domains bound to p53 tumour suppressor". The EMBO Journal. 21 (14): 3863–72. doi:10.1093/emboj/cdf383. PMC 126127. PMID 12110597. Lai H, Lin L, Nadji M, Lai S, Trapido E, Meng L (2002). "Mutations in the p53 tumor suppressor gene and early onset breast cancer". Cancer Biology & Therapy. 1 (1): 31–6. doi:10.4161/cbt.1.1.37. PMID 12170762. Derbyshire DJ, Basu BP, Date T, Iwabuchi K, Doherty AJ (Oct 2002). "Purification, crystallization and preliminary X-ray analysis of the BRCT domains of human 53BP1 bound to the p53 tumour suppressor". Acta Crystallographica Section D. 58 (Pt 10 Pt 2): 1826–9. Bibcode:2002AcCrD..58.1826D. doi:10.1107/S0907444902010910. PMID 12351827. Wang B, Matsuoka S, Carpenter PB, Elledge SJ (Nov 2002). "53BP1, a mediator of the DNA damage checkpoint". Science. 298 (5597): 1435–8. Bibcode:2002Sci...298.1435W. doi:10.1126/science.1076182. PMID 12364621. S2CID 30076227. Richie CT, Peterson C, Lu T, Hittelman WN, Carpenter PB, Legerski RJ (Dec 2002). "hSnm1 colocalizes and physically associates with 53BP1 before and after DNA damage". Molecular and Cellular Biology. 22 (24): 8635–47. doi:10.1128/MCB.22.24.8635-8647.2002. PMC 139863. PMID 12446782. DiTullio RA, Mochan TA, Venere M, Bartkova J, Sehested M, Bartek J, Halazonetis TD (Dec 2002). "53BP1 functions in an ATM-dependent checkpoint pathway that is constitutively activated in human cancer". Nature Cell Biology. 4 (12): 998–1002. doi:10.1038/ncb892. PMID 12447382. S2CID 7430614. Fernandez-Capetillo O, Chen HT, Celeste A, Ward I, Romanienko PJ, Morales JC, Naka K, Xia Z, Camerini-Otero RD, Motoyama N, Carpenter PB, Bonner WM, Chen J, Nussenzweig A (Dec 2002). "DNA damage-induced G2-M checkpoint activation by histone H2AX and 53BP1". Nature Cell Biology. 4 (12): 993–7. doi:10.1038/ncb884. PMID 12447390. S2CID 12380387. Peng A, Chen PL (Mar 2003). "NFBD1, like 53BP1, is an early and redundant transducer mediating Chk2 phosphorylation in response to DNA damage". The Journal of Biological Chemistry. 278 (11): 8873–6. doi:10.1074/jbc.C300001200. PMID 12551934. Stewart GS, Wang B, Bignell CR, Taylor AM, Elledge SJ (Feb 2003). "MDC1 is a mediator of the mammalian DNA damage checkpoint". Nature. 421 (6926): 961–6. Bibcode:2003Natur.421..961S. doi:10.1038/nature01446. PMID 12607005. S2CID 4410773. Yan DH, Abramian A, Li Z, Ding Y, Wen Y, Liu TJ, Hunt K (Mar 2003). "P202, an interferon-inducible protein, inhibits E2F1-mediated apoptosis in prostate cancer cells". Biochemical and Biophysical Research Communications. 303 (1): 219–22. doi:10.1016/S0006-291X(03)00320-6. PMID 12646190. Kao GD, McKenna WG, Guenther MG, Muschel RJ, Lazar MA, Yen TJ (Mar 2003). "Histone deacetylase 4 interacts with 53BP1 to mediate the DNA damage response". The Journal of Cell Biology. 160 (7): 1017–27. doi:10.1083/jcb.200209065. PMC 2172769. PMID 12668657. Caldwell RB, Braselmann H, Schoetz U, Heuer S, Scherthan H, Zitzelsberger H (July 4, 2016). "Positive Cofactor 4 (PC4) is critical for DNA repair pathway re-routing in DT40 cells". Sci. Rep. 6: 28890. Bibcode:2016NatSR...628890C. doi:10.1038/srep28890. PMC 4931448. PMID 27374870.{{cite journal}}: CS1 maint: article number as page number (link) External links [edit] Overview of all the structural information available in the PDB for UniProt: Q12888 (TP53-binding protein 1) at the PDBe-KB. \| PDB gallery \| \| --- \| \| 1gzh: CRYSTAL STRUCTURE OF THE BRCT DOMAINS OF HUMAN 53BP1 BOUND TO THE P53 TUMOR SUPPRESSOR 1kzy: Crystal Structure of the 53bp1 BRCT Region Complexed to Tumor Suppressor P53 1ssf: Solution structure of the mouse 53BP1 fragment (residues 1463-1617) 1xni: Tandem Tudor Domain of 53BP1 2g3r: Crystal Structure of 53BP1 tandem tudor domains at 1.2 A resolution 2ig0: Structure of 53BP1/methylated histone peptide complex \| \| Retrieved from " Category: Genes on human chromosome 15 Hidden categories: Articles with short description Short description matches Wikidata CS1 maint: article number as page number
6216	https://im.kendallhunt.com/MS_ACC/teachers/1/6/5/preparation.html	Illustrative Mathematics \| Kendall Hunt Illustrative Mathematics \| Kendall Hunt Skip to main content IM CurriculumAbout UsContact UsGoogle ClassroomEducators Register/Log in Accelerated Grade 6 Middle School AcceleratedAccelerated Grade 6Accelerated Grade 7 Unit 6 Accelerated Grade 6Unit 1Unit 2Unit 3Unit 4Unit 5Unit 6Unit 7Unit 8Unit 9 Lesson 123456789101112 Lesson 5 Percent Increase and Decrease with Equations PreparationLessonPractice View Student Lesson Lesson Narrative In this lesson, students represent situations involving percent increase and percent decrease using equations. They write equations like y = 1.06x to represent growth of a bank account, and use the equation to answer questions about different starting amounts. They write equations like t -0.25t = 12 or 0.75t=12 to represent the initial price t of a T-shirt that was $12 after a 25% discount. The focus of this unit is writing equations and understanding their connection to the context. In a later unit on solving equations the focus will be more on using equations to solve problems about percent increase and percent decrease. When students repeatedly apply a percent increase to a quantity and see that this operation be expressed generally by an equation, they engage in MP8. Learning Goals Teacher Facing Explain (orally and in writing) how to calculate the original amount given the new amount and a percentage of increase or decrease. Generate algebraic expressions that represent a situation involving percent increase or decrease, and justify (orally) the reasoning. Student Facing Let’s use equations to represent increases and decreases. Learning Targets Student Facing I can solve percent increase and decrease problems by writing an equation to represent the situation and solving it. CCSS Standards Building On 5.NF.B Addressing 7.RP.A.3 Building Towards 7.RP.A.3 Print Formatted Materials Teachers with a valid work email address canclick here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials. Student Task Statementspdfdocx Cumulative Practice Problem Setpdfdocx Cool DownLog In Teacher GuideLog In Teacher Presentation Materialspdfdocx Additional Resources Google SlidesLog In PowerPoint SlidesLog In Privacy Policy \| Accessibility Information IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations and additions to create IM 6–8 Math Accelerated are copyright 2020 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.
6217	https://www.youtube.com/watch?v=Z5mA_l-ptqk	Speed and Average Speed \| Middle School Science \| Khan Academy Khan Academy India - English 549000 subscribers 1 likes Description 241 views Posted: 9 Aug 2025 In this video, we explore the concept of speed – how fast something is moving – and learn how to calculate it using distance and time. We also dive into average speed, the total distance covered divided by total time taken, even when the speed changes during the journey. Timestamps: Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month. Support Us: Created by Shankar R Transcript: Intro All right, you might have heard of the storrist and hair story. It's one of the stales we grew up with, right? But here's the fun question. Why do tortoises win race in stories? I mean, come on. In real life, we know tortoise is much much much slower than a hair. Like, it's not even close. But still, HMA always in this particular story, the tortoise wins. Why? You know the moral of the story? Slow and steady wins the race. It's wise. It's catchy. But wait a second. Do we actually know the physics behind it? Do we know the science behind this whole thing? Let's find out if we can explain why exactly the tortoise won this race scientifically. Let's break it down. Speed qualitatively Now, if you look at speed qualitatively, we always talk about how fast or slow things go. That's how we experience motion in real life. Imagine a race between a snail, a bicycle, and a rocket blasting off into the space. While lined up, right? They can all cover up same distance eventually, but what makes them feel so different is how fast each each one eats up the distance. That's what speed captures. But here's the interesting part. The phrase how fast is most often used to describe how quickly something is moving at a speed at a specific instant. Speed is quite simply how quickly distance changes with time. Let's look at another example. Remember that rays we saw when we were measuring time? Yeah, the red guy, the blue guy, and the yellow guy, right? So if you look at them right now, you can feel it. Blue is the fastest, red's the slowest. And instinctively, we just know. We watch the move and say, "Yep, red is the slowest, blue is the fastest, and yellow, yeah, yellow is just slow." That's the character of the motion. That's what we mean when we are looking at motion qualitatively based on how it feels. But here is the thing in this example. Sure, blue is fast, faster, red is slowest, and yellow is somewhere in between. But how much faster? How much slower? Feelings are great but are they enough for science? Not really. Science need numbers. So that's what we are going to do now. Let's look at speed quantitatively. All right. So how do we Speed quantitatively do that? So we are going to measure the distance traveled by each of these races and we are going to calculate the time. Right? We have got our trusty digital stopwatch ready to measure exactly how long each one takes. And here we go. We have the data already. So we can see that all of them are covering the same distance of 20 m. Uh the red takes 10 seconds, blue takes 4 seconds and yellow takes 8 seconds. Now that we have got some data that is distance and time, here is where it gets interesting. Scientists have figured out a wonderfully simple trick. If you take the distance an object travels and divide it by time it takes, you get speed. Exactly. Quantitatively, that's how we measure fast and slow. So technically speaking, speed is just how quickly or how much distance an object covers in certain amount of time. And when we put that into the math form, we get speed is equal to distance upon time. Now let's take a look at the device measuring speed. All right. So uh Speedomemeter we got an F1 car here and this one belongs to Louis Hamilton. Yeah, it's a Mercedes pretty sleek car right now. What you see on screen that number that's the speed at that particular instant which is why we call it instantaneous speed. And that number you are seeing that's what we call the speedometer reading. The device doing all this magic that is measuring the speed it's called the speedomemeter. Right here is how it works. Inside the car there are these tiny sensors that track how much the wheel has traveled the distance and it also records time. It divides that distance by that tiny split second of a time and boom. This classic distance divided by time calculation is done instantly by car's computer or the speedometer. And just like that you get the speed reading in uh real time. Okay. Now let's watch the speed. Okay. It's increasing 339 340. It's touching uh and then it's slowing down. Yeah, it's a corner. Uh yeah, speed is dropping. Picking up again. Yeah, it's picking up again. See, the driver's speed isn't constant. It's always changing. Speeding up. Speeding now. Now speeding up again. Got it? That's how we measure and understand instantaneous speed. Average Speed Now, here is another question to think about. If the car took 1 minute to cover full circuit of 5 km, how fast was the car? Pretty simple question, right? Now, obviously, the speed wasn't the same entire time. You just saw that the speed kept changing, right? It was speeding up, slowing down the corners. So, we are not talking, we're not going to talk about instantaneous speed anymore. If somebody asks you what is the speed of the car when it completes 5 km track in 1 minute, we are looking at a different kind of speed called the average speed. Average speed is simply the total distance divided by the total time taken. It gives us the average of all the different speeds over the entire journey. Whether it's speeding up, slowing down, it doesn't matter. What matters is the total how much distance it has traveled and how much time it has taken. While earlier we were zooming in on one particular moment with instantaneous speed, now we are zooming out to look at the whole picture. So in our F1 situation here, the car covers 5 kilometer in 1 minute. Since 1 minute is 1x 60 hour, we'll do this conversion thing later in another video. All right, stay tuned for that. But since we want to calculate the speed in kilometer per hour, I'm converting it into 1x60th of an hour. Now do the math and we get 300 km per hour as the average speed. Again, remember this is not the speed at every moment. Around corners, it was definitely going slower. On straight stretches, it probably went faster than 300. But this 300 km is the overall average for the whole lab. And that brings us back to our good old story of totis and hair that we started with. Right? Now, we know their speed. All right? The tortoise moves at 0.2 km/h and the hair at 56 km/h. By this speed what we mean is the instantaneous speed. All right. So what actually happened in the race? We have to find the average speed to figure out what really happened in the race. So they had a let's say 1,000 m race that is distance is equal to 1 kilometer and let's say the time taken by the tus is 4 hour and time taken by the hair is 5 hours. Why did hair take five hour? Because remember in the story hair slept hair thought I am lightning fast. I can even beat him in a in a matter of a second. But he slept off and the overall distance he covered took 5 hours. If you look at their average speed, it is very evident that the tortoise covered 1 kilometer in 4 hours and his speed is 0.25 km/h. Whereas the air covered 1 kilometer in 5 hours and his average speed, okay, it's not the speed, it's average speed is 0.2 km/h. Clearly the average speed of turtles is greater than average speed of hair. So that's the moral of the story. The one with higher average speed wins the race.
6218	https://mathworld.wolfram.com/PopulationGrowth.html	Population Growth -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Applied Mathematics Population Dynamics History and Terminology Disciplinary Terminology Biological Terminology History and Terminology Disciplinary Terminology Political Terminology MathWorld Contributors Stover More...Less... Population Growth The differential equation describing exponential growth is (1) This can be integrated directly (2) to give (3) where . Exponentiating, (4) This equation is called the law of growth and, in a much more antiquated fashion, the Malthusian equation; the quantity in this equation is sometimes known as the Malthusian parameter. Consider a more complicated growth law (5) where is a constant. This can also be integrated directly (6) (7) (8) Note that this expression blows up at . We are given the initial condition that , so . (9) The in the denominator of (◇) greatly suppresses the growth in the long run compared to the simple growth law. The (continuous) logistic equation, defined by (10) is another growth law which frequently arises in biology. It has solution (11) See also Gompertz Curve, Growth, Law of Growth, Life Expectancy, Logistic Map, Lotka-Volterra Equations, Makeham Curve, Malthusian Parameter, Survivorship Curve Portions of this entry contributed by Christopher Stover Explore with Wolfram\|Alpha More things to try: population growth population growth history world population growth rank by country References Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp.290-295, 1999. Referenced on Wolfram\|Alpha Population Growth Cite this as: Stover, Christopher and Weisstein, Eric W. "Population Growth." From MathWorld--A Wolfram Resource. Subject classifications Applied Mathematics Population Dynamics History and Terminology Disciplinary Terminology Biological Terminology History and Terminology Disciplinary Terminology Political Terminology MathWorld Contributors Stover More...Less... About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
6219	https://stackoverflow.com/questions/26711407/what-should-be-the-optimal-way-of-solving-recurrence-relation-for-really-huge-nu	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog What should be the optimal way of solving Recurrence relation for really Huge number greater than Integer maximum value Ask Question Asked Modified 10 years, 9 months ago Viewed 489 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I want to find the Nth number of the Recurrence Equation ``` T(n)=T(n-1)+3T(n-2)+3T(n-3)+(n-4),T(1)=T(4)=1,T(2)=T(3)=3 ``` so if suppose you entered 2,5,9 as input, output should be T(2)=3,T(5)=20,T(9)=695 what I did is create an array of size equal to maximum of all input value and storing solution of T(i) at index i.Then look up into the array for specific index. eg array for T(3),array for T(5),etc The code worked fine till maximum number is not greater than maximum integer value system can hold i.e ``` Integer.MAXValue. ``` Because the index of array can only be integer then if number is n=1855656959555656 what should be the best way to find the solution of T(1855656959555656)? as clearly I cant create an array of size=1855656959555656.. I have even tried BigInteger from java.Math but with no success. I have to find some other approach.please suggest some ideas.. Thanks java arrays recurrence Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Nov 3, 2014 at 12:43 Cyclotron3x3 asked Nov 3, 2014 at 9:37 Cyclotron3x3Cyclotron3x3 2,22911 gold badge2929 silver badges4141 bronze badges 5 Yoy want precalculate all values? maxvalue array will be 16GB size. It isn't best use of space. (because you need at least longs to store value. – talex Commented Nov 3, 2014 at 9:46 Yes exactly.This isn't the best of Algo to solve this kind of relations.But I am not able to think of other than this as of now. @talex – Cyclotron3x3 Commented Nov 3, 2014 at 9:54 There is an answer already. But you run into another problem. Value grows is exponential. You will have memory problem trying to store T(MAX_INT) value – talex Commented Nov 3, 2014 at 10:02 Have you tried writing in file the intermediate output – Mujtaba Hasan Commented Nov 3, 2014 at 11:12 @ Mujtaba Hasan yes I have tried..every logic seems working until number gets really big...eg. n=1855656959555656..execution gets almost stuck and timeout.. – Cyclotron3x3 Commented Nov 3, 2014 at 12:25 Add a comment \| 6 Answers 6 Reset to default This answer is useful 4 Save this answer. Show activity on this post. you do not need to store every T(i), you only need to store 3 values T(i-1), T(i-2), T(i-3). While looping over i, check if the current i should be part of your output, if so put it out immediately or save it to an "output"-array. edit: this part is quite inefficient. You check in every iteation EVERY needed output. ``` for (int k = 0; k < arr.length; ++k) { if (count == arr[k]) T[k] = temp[i]; else if (arr[k] == 1) T[k] = 1; else if (arr[k] == 2) T[k] = 3; else if (arr[k] == 3) T[k] = 3; else if (arr[k] == 4) T[k] = 1; } ``` so your code runs in time (maxarr.length) you can reduce it to only (max). Use a HashMap with key=neededPosition (=count) value=position in arr Init the map like this: ``` Map map = new HashMap(); for (int i = 0; i < arr.length; i++) { map.put(arr[i], i); } if (map.containsKey(count)) { T[map.get(count)] = temp[i] } ``` check the values 1-4 just once after the whole thing! Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Nov 3, 2014 at 15:48 answered Nov 3, 2014 at 9:49 MichaelMichael 81511 gold badge66 silver badges2323 bronze badges 6 sounds like a good alternative.let me write it and i 'll get back to you. – Cyclotron3x3 Commented Nov 3, 2014 at 10:08 While I think this solves the actual problem, it may not help people getting here searching for large arrays (reading the title of the post....) – Stefan Haustein Commented Nov 3, 2014 at 10:33 @Michael code is working but with no extra benefit. when I entered n=1855656959555656..It is taking years to execute..any further modification You can suggest.otherwise we are still stuck at same place. – Cyclotron3x3 Commented Nov 3, 2014 at 12:18 can you post your code please? I see if we can speed up the performance a little bit – Michael Commented Nov 3, 2014 at 12:23 @Michael How will you initialize the map.We can't initialize it with every possible count value and its respected array position during initializing.It will be more convenient if you just post your map initialization logic.. – Cyclotron3x3 Commented Nov 3, 2014 at 15:43 \| Show 1 more comment This answer is useful 1 Save this answer. Show activity on this post. Not possible. The array size can be a maximum of Integer.MAX_VALUE (minus something usually 5 or 8, depending on the JVM capabilities). Why?. The index for an Array should be an integer thats a limitation. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Nov 3, 2014 at 9:38 TheLostMindTheLostMind 36.3k1212 gold badges7272 silver badges109109 bronze badges 2 yes i know that.But there must be a way to solve this recurrence relation using programming techniques. @TheLostMind – Cyclotron3x3 Commented Nov 3, 2014 at 9:48 @Ashish - Create a List and start removing objects as soon as their use is over.. You might have to call the GC explicitly in the middle. Don't keep values of intermediate steps. – TheLostMind Commented Nov 3, 2014 at 9:50 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. It can't be done. So you need to solve the problem by introducing a sharding mechanism. The simplest way would be to just have arrays of arrays with a fixed length. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Nov 3, 2014 at 9:41 Sean Patrick FloydSean Patrick Floyd 300k7272 gold badges480480 silver badges597597 bronze badges 3 How array of array can solve the issue?I didn't understand?can you just elaborate.thanks – Cyclotron3x3 Commented Nov 3, 2014 at 9:46 @Ashish - Array of Arrays will help you in working around the max size issue. Although you might get an OutOfMemoryError. You could escape that as well by increasing your heap space. – TheLostMind Commented Nov 3, 2014 at 9:48 you need a formula for where to look up what value. for value n, you will divide n by the max array length and you know the array offset to look in, and the modulo will tell you the position – Sean Patrick Floyd Commented Nov 3, 2014 at 9:49 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Edit: You really do not need this much storage for your problem at hand (as pointed out in another answer; this code fragment avoids arrays altogether to avoid bounds checks / indirection): ``` public void t(long n) { if (n < 5) { return (n == 2 \|\| n == 3) ? 3 : 1; } long i = 5; // Initialize variables for n == 5; long tn_1 = 1; // T(n-1) = T(4) = 1; long tn_2 = 3; // T(n-2) = T(3) = 3; long tn_3 = 1; // T(n-3) = T(2) = 1; long tn_4 = 3; // T(n-4) = T(1) = 3; while (true) { long tn = tn_1 + 3tn_2 + 3tn_3 + tn_4; if (i++ == n) { return tn; } tn_4 = tn_3; tn_3 = tn_2; tn_2 = tn_1; tn_1 = tn; } } ``` To answer the question in the title anyway: If your array is sparse, use a map (TreeMap or HashMap) of Long or BigInteger: ``` Map t = new TreeMap() ``` The memory consumption of sparse arrays depends on the number of elements actually stored, so you may want to delete values from the map that are no longer needed. If your array is not sparse, use a 2-level array (memory consumption will depend on the pre-allocated size only): ``` public class LongArray { static final long BLOCK_SIZE = 0x40000000; long[][] storage; public LongArray(long size) { long blockCount = (size + BLOCK_SIZE - 1) / BLOCK_SIZE; storage = new long[][(int) blockCount]; for (long i = 0; i < blockCount; i++) { if (i == blockCount - 1) { storage[i] = new long[(int) size - BLOCK_SIZE (blockCount - 1)]; } else { storage[i] = new long[(int) BLOCK_SIZE]; } } } public long get(long index) { return storage[(int) (index / BLOCK_SIZE)][(int) (index % BLOCK_SIZE)]; } public void put(long index, long value) { storage[(int) (index / BLOCK_SIZE)][(int) (index % BLOCK_SIZE)] = value; } } ``` In both cases, use t.get(index) and t.put(index, value) instead of t[index] to access your array (if t is the name of the array). Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Nov 5, 2014 at 10:28 answered Nov 3, 2014 at 10:05 Stefan HausteinStefan Haustein 18.8k33 gold badges4040 silver badges5454 bronze badges 3 It looks like a good option .but can you elaborate how to calculate T(i) using Map. – Cyclotron3x3 Commented Nov 3, 2014 at 10:12 You use t.get(index) / t.put(index, value) instead of t[index] (if t is your array) – Stefan Haustein Commented Nov 3, 2014 at 10:27 p.s. As pointed out elsewhere, you don't need a map or array, see edit. – Stefan Haustein Commented Nov 5, 2014 at 10:29 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. You can do one thing. Check if the value of n is equal to 1855656959555656 in the beginning or if its multiple. Suppose, the value of n is twice of 1855656959555656. Then you can create two arrays and link them together virtually. This should solve your problem but it will involve a lot of overhead. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Nov 3, 2014 at 9:53 hchawlahchawla 12688 bronze badges 5 I have tried something similar (linking two arrays virtually) but stuck at same place.I even wrote separate code for number greater than Integer.MAXValue.if(num>Integer.MAXValue){ two arrays linked solution} else {one array solution} – Cyclotron3x3 Commented Nov 3, 2014 at 10:15 @Ashish Probably, you would need to refine your logic to make it work :) Also, if you are aware of ArrayList or Linked List, you can create a global Array List, and increase its size one by one whenever required. Thus, the last element of the Array List shall give you the required answer to the formula you are using. It might not even require recursion. – hchawla Commented Nov 3, 2014 at 10:23 problem is not the size now but the performance degrades exponentially with very large number.It almost stuck for minutes. – Cyclotron3x3 Commented Nov 3, 2014 at 15:47 @Ashish A recursive call for such a large number will definitely take a lot of time and memory. If you wish to execute just this program then, I would suggest hard-coding the values once and for all, at least till n=1855656959555656. Later, you can conduct an efficient search method to get the result for the required value of n. – hchawla Commented Nov 4, 2014 at 5:58 Further,for numbers greater than 1855656959555656, you can start implementing the logic and it will take lesser time as you already have values ready up to n=1855656959555656. This will reduce the time and memory consumption for further runs of the code as values are readily available and need not be calculated.I would'nt say that this is a good programming approach, but for this program and your requirements, I think it will be a proper approach. Eventually, you will have to limit the value of n that can be taken as an input for the program. – hchawla Commented Nov 4, 2014 at 5:59 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Use recursive call: ``` int T(int n){ if (n==1 \|\| n==4){ return 1; } else if (n==2 \|\| n==3){ return 3; } else { return T(n-1)+3T(n-2)+3T(n-3)+T(n-4); } } ``` Edit: Time consumming. Won't work with large numbers Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Nov 3, 2014 at 10:08 answered Nov 3, 2014 at 9:49 jhamonjhamon 3,69144 gold badges2929 silver badges3939 bronze badges 5 This is very slow code. Loop is faster. Because recursive version calculate same value multiple times. – talex Commented Nov 3, 2014 at 9:50 1 @jhamon - each method call will be in its own stack frame.. So, more memory is consumed in recursion. – TheLostMind Commented Nov 3, 2014 at 9:52 T(35) takes 8174ms, T(36) 15690ms, you never go to stack overflow with this code :) – talex Commented Nov 3, 2014 at 10:00 @TheLostMind, hmm, yea you're right. I didn't think of that. – jhamon Commented Nov 3, 2014 at 10:02 This solution is not good for very large number since it is stack based.It will throw StackOverflow error when you enter a number 1554894. @jhamon – Cyclotron3x3 Commented Nov 3, 2014 at 10:05 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions java arrays recurrence See similar questions with these tags. 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6220	https://gre.magoosh.com/definitions/gloomy	Definition for gloomy – Thickly shaded; cheerlessly obscure; shadowy; dark; somber. – Affected with, characterized by, or expressing gloom; wearing the aspect of sorrow; depressed or depressing; melancholy; doleful: as, a gloomy countenance; a gloomy prospect. – Synonyms Dim, dusky, cloudy, cheerless, lowering. See darkness. – adjective – Imperfectly illuminated; dismal through obscurity or darkness; dusky; dim; clouded. adjective – Affected with, or expressing, gloom; melancholy; dejected. adjective – Imperfectly illuminated; dismal through obscurity or darkness; dusky; dim; clouded. 4 more defintions on Wordnik More definitions and example sentences on Wordnik
6221	https://teachchemistry.org/classroom-resources/preparation-and-evaluation-of-buffers	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Skip to main content Preparation and Evaluation of Buffers Mark as Favorite (34 Favorites) LESSON PLAN in Acid & Base Theories, Buffers, Unlocked Resources. Last updated October 03, 2024. Summary In this lesson students will use multiple methods to calculate and prepare buffered solutions with a desired pH. Upon preparation of the solutions, the students will explore differing aspects of buffers including buffering capacity and predominant form. Grade Level High School (AP Chemistry) AP Chemistry Curriculum Framework This activity supports the following unit, topics and learning objectives: Unit 8: Acids and Bases Topic 8.4: Acid-Base Reactions and Buffers 8.4.A: Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases. Topic-8.8:Properties of Buffers 8.8.A: Explain the relationship between the ability of a buffer to stabilize pH and the reactions that occur when an acid or a base is added to a buffered solution. Topic-8.9:Henderson-Hasselbalch Equation 8.9.A:Identify the pH of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer. Topic 8.10:Buffer Capacity 8.10.A: Explain the relationship between the buffer capacity of a solution and the relative concentrations of the conjugate acid and conjugate base components of the solution. Objectives By the end of this lesson, students should be able to Prepare a buffered solution with a desired pH from a weak acid and its salt. Prepare a buffered solution with a desired pH by partially neutralizing a weak acid with a strong base. Compare the buffering capacity between two buffered solutions. Evaluate the predominant form of an acid in a solution of a specific pH. Chemistry Topics This lesson supports students’ understanding of Acids & Bases Acid – Base Theory Buffers pH Buffering capacity Time Teacher Preparation: 30 minutes Lesson: 90-120 minutes Materials per group 100 mL 1.0 M HC2H3O2 100 mL 0.50 M NaOH pH meter or pH paper Approximately 5 g Sodium Acetate, NaC2H3O2 50 ml Graduated cylinders 50 ml Graduated cylinders (50 mL and 100 mL) 250 mL beaker 1- 5.00 mL pipette Balance capable of measuring to 0.01 g Weighing boats Safety Always wear safety goggles when handling chemicals in the lab. Students should wash their hands thoroughly before leaving the lab. When students complete the lab, instruct them how to clean up their materials and dispose of any chemicals. When working with acids, if any solution gets on students’ skin, they should immediately alert you and thoroughly flush their skin with water. When diluting acids, always add acid to water. Teacher Notes Lesson Outline: Day 1 (45-60 minutes): Use the short PowerPoint provided to cover all of the concepts that the students will be addressing during the lab. There is also a formative quiz that can be used with students to determine the level of readiness to proceed with the lab. Also the Pre-lab section of the lab should be completed and discussed by the lab groups. The Pre-lab section does not appear to be very time consuming at first glance, however, these calculations are some of the most troubling and challenging that the students will encounter. Ensuring that they can successfully complete them is worth dedicating class time, rather than assigning them for homework. Day 2 (45-60 minutes): The lab activity should be completed, post lab questions should be answered and a consensus should be reached within each group. This lab is designed to help students understand one of the more challenging ideas in AP Chemistry: buffers and buffering capacity. It was designed with free response question 3 from the 2017 administration of the AP Chemistry Exam in mind. Some parts of this FRQ would be an excellent way to determine if students understood the concepts from this lab. The teacher will need to either purchase standardized solutions of 1.0 M HC2H3O2 and 0.50 M NaOH solutions or prepare and standardize the solutions. In this lab students will use two different methods to prepare buffered solutions with the same pH. Buffer 1 is prepared using a weak acid, acetic acid, and its salt, sodium acetate. Buffer 2 is prepared by partially neutralizing a weak acid, acetic acid, with a strong base, sodium hydroxide. Student lab groups of 3 can be assigned varying target pH values to promote each lab group to complete their own calculations. This can be done by varying the assigned pH values as follows: Group 1 pH = 4.50, Group 2 pH = 4.55, Group 3 pH = 4.60, etc. The pKa of acetic acid is 4.75, so you may or may not want to assign a group the value of 4.75. Download the Excel spreadsheet for this resource to calculate the mass of sodium acetate needed for Buffer 1 and the volume of sodium hydroxide needed for Buffer 2. This lab should be completed once students are comfortable with all of the AP essential knowledge regarding buffer calculations, and the concept of buffering capacity which are outlined below. Buffer Background Information: Essential Knowledge 6.C.2: The pH is an important characteristic of aqueous solutions that can be controlled with buffers.Comparing pH to pKa allows one to determine the protonation state of a molecule with a labile proton. The first point goal of teaching buffers is recognizing what a buffer is composed of. In order for a solution to be classified as a buffer it must contain both members of a conjugate acid-base pair.This allows any added base to react with conjugate acid and any added acid to react with conjugate base. By comparing pH to pKaof any acid in solution, the ratio between the acid form and base form can be determined (protonation state). If pH < pKa the acid form has a higher concentration than the base form and if pH > pKa the base form has a higher concentration when compared to the acid form. The pH of a buffer is related to both pKa as well as the ratio of acid and base forms (evidenced by the Henderson-Hasselbalch equation). The buffer capacity is related to absolute concentration of the acid and base forms.Therefore, it is possible for two buffers of equal pH to respond differently to the addition of a strong acid, or strong base, therefore have a differing buffer capacity. Past Free-Response Questions Relating to These Concepts: An old FRQ to get the students used to the calculations involved in the process would be the 2002 Form B #1. It is pretty straight forward but does get them used to the equations and processes needed. Good predominant form question 2013 secure practice exam #7. This question does get convoluted with interparticle force ideas, but the predominant form ideas are well done.Since it is on the secure exam a link cannot be provided. Another great predominant form question can be found on the 2016 released exam question number 4. According to the Student Performance Q & A, most common predominant form mistake made by students was to assume at any pH greater than 7 the acid would be in its basic form. One of the better buffer capacity questions can be found on the 2007 AP Chemistry Exam Form B Question#5 ci-iii. The 2011 FRQ #1 was a really good question for this concept, but part c is now in an exclusion statement.This does really do a good job of addressing what a buffer is. For the Student Lesson Background In this experiment, we will use two different methods to prepare buffered solutions with the same assigned pH. Buffer 1 will be prepared using acetic acid, HC2H3O2, and sodium acetate, NaC2H3O2.Buffer 2 will be prepared using acetic acid, HC2H3O2, and sodium hydroxide, NaOH. Both buffers will have a target pH of ________. Acetic acid has a Ka = 1.76 x 10-5 and a pKa = 4.75. Pre-lab Questions Determine what mass of sodium nitrite, NaNO2, would be required to prepare a buffer, Buffer A, with a pH of 3.13 from 50.0 mL of 1.0 M nitrous acid, HNO2.The Ka and pKa for nitrous acid is 4.0 x 10-4 and 3.40 respectively. What is the predominant form of the acid in Buffer A? Explain your answer. If the pH of Buffer A were 3.40, what would the predominant form of the acid be? A second buffered solution, Buffer B, is prepared with twice the number of moles of acid and base, but the same ratio of acid to base as Buffer A. Is the pH of Buffer B greater than, less than, or equal to 3.13?Justify your answer. Which buffered solution, Buffer A or Buffer B, would be more resistant to pH change when a strong acid or a strong base is added? Justify your answer. Find step 3 from the Buffer 1 procedure. Perform the calculations required to determine the mass of sodium acetate necessary to produce the buffer of the pH assigned to your lab group. Find step 3 from the Buffer 2 procedure. Perform the calculations required to determine the volume of sodium hydroxide necessary to produce the buffer of the pH assigned to your lab group. Objective The purpose of this lab is to use two different methods to prepare buffered solutions with the same assigned pH value. Materials 100 mL 1.0 M HC2H3O2 100 mL 0.50 M NaOH pH meter 50.0 g Sodium Acetate, NaC2H3O2 50 mL Graduated cylinder 00 mL Graduated cylinder 250 mL beaker Balance capable of measuring 0.01 g Weigh boat Stirring rod Disposable pipettes Safety Always wear safety goggles when handling chemicals in the lab. Wash your hands thoroughly before leaving the lab. Follow the teacher’s instructions for cleanup of materials and disposal of chemicals. When working with acids and bases, if any solution gets on your skin immediately rinse the area with water. When diluting acids, always add acid to water. Procedure for Buffer 1 Using the graduated cylinder, measure 50.0 mL of 1.0 M HC2H3O2. You may want to use a disposable pipette to get an exact volume. Pour HC2H3O2 into a 250 ml beaker. Mass out the appropriate number of grams of solid NaC2H3O2 required to reach your assigned pH. Add the solid NaC2H3O2 to the small beaker and stir with a clean stirring rod until completely dissolved. Measure the resulting pH using a pH meter and record. Procedure for Buffer 2 Using the graduated cylinder, measure 50.0 mL of 1.0 M HC2H3O2. You may want to use a disposable pipette to get an exact volume. Pour HC2H3O2 into a 250 ml beaker. Measure out the appropriate volume of 0.50 M NaOH required to reach your assigned pH using a clean graduated cylinder. Add the NaOH to the small beaker and stir with a clean stirring rod. Measure the resulting pH using a pH meter and record. Data \| \| \| \| --- \| \| Buffer 1 \| Buffer 2 \| \| Assigned pH \| \| \| \| Volume of HC2H3O2 (mL) \| \| \| \| Mass of NaC2H3O2 (g) \| \| N/A \| \| Volume of NaOH (mL) \| N/A \| \| \| Actual pH \| \| \| Calculations Calculate your percent error using the assigned pH and your actual pH for both Buffer 1 and Buffer 2. Analysis If either percent error is greater than 5%, provide a reasonable source of error and explain precisely how the error would affect the actual pH. A different lab group attempts to make a buffer with an acid and a salt.They choose HCl and NaCl. Explain why their attempt will not produce a solution that is resistant to pH change. Justify your answer using net-ionic equations. Circle the beaker below that best represents the particulate diagram for Buffer 1. Explain your reasoning referencing the Henderson-Hasselbalch equation. It is observed that Buffer 2 changes more significantly than Buffer 1 upon the addition of strong acid.Explain this observation.
6222	https://www.youtube.com/watch?v=ov-sMWBpMUE	Range Equation Derivation Engineering Futures 674 subscribers 2 likes Description 90 views Posted: 12 Jan 2021 Deriving the range equation with simple algebra and trigonometry. This is a crucial skill for a Physics, AP Physics, or college physics class. ENJOY If you like this video and want more study materials or teaching resources, please feel free to visit my Teachers Pay Teachers (TPT) store: 2 comments Transcript: hey how's it going i'm anthony todd and today i'm going to show you how to to derive the range equation for your ap physics 1 or ap physics c class or even in if you're in college this will help you okay so how this works is we understand that we have some projectile being launched with some initial velocity v naught at some angle above the horizontal and as it flies through it's going to follow a parabolic arch until it hits the ground okay now in order for us to understand this this is what we're trying to look for we are looking for this displacement in the x direction so how far does this object go okay now we understand using some simple kinematics that the displacement of this object will just be equal to the velocity in the x component times the time parameter okay and with some trigonometry and some vectors we know this that if you have an object being launched at v naught with some angle theta that's the velocity in the x direction down here so the velocity in the x component will just be v naught cosine theta and the velocity in the y direction so the velocities y component will be equal to v naught sine of theta okay and that's just using pythagorean's theorem and some sohcahtoa to get that so now that we know this component we can rewrite this as delta x is just equal to v naught cosine theta times t okay and we're going to go ahead and solve for t well we can just kind of leave it here if we want to now this is the x component so this is x now in the y direction so in the y we know now this this range equation only works when your vertical displacement is 0 meters so what that means is this object has to be fired off from some spot in the y direction and it has to land at the exact same spot on the other side so pretty much it has to be level ground okay so this will not work if you have any displacement that is not zero now if we know this and we know the velocity initial or the velocity of the y component initially will just be v naught e naught sine of theta and we know that gravity is always a constant okay it's a constant and the question is well we have to kind of correlate these two together so we got to take our x components and our y components or our y equations and kind of smush them together so here we notice we have time we don't have displacement the x we don't have you know velocity and x over here the only thing that you know is the same in both parameters is this time so that's kind of a key so we're going to look for an equation that has time in it and it has all of these components and we're going to look for a kinematic equation well the kinematic equation that satisfies that is this one the displacement in the y direction is equal to the velocity initial times time i should say the y direction plus g t squared all over 2. okay so we now know that the vertical displacement is zero and we know that this is just going to be v naught sine of theta times t plus g t squared all over two and what we can see here is we can actually factor out a t so we get this v naught sine of theta plus g t all over two and then we can actually you know you know take away this right here because anything divided by zero so we still have this v naught sine of theta plus g t all over 2. so what we are going to do now is we're going to actually solve for t okay so we'll move this over becomes negative g t over 2 equals v naught sine of theta and then solving for time we actually get this time is equal to v naught sine of theta divided by negative g oops sorry this is 2 up here 2 sorry 2 v not sine of theta divided by negative g now it is negative but honestly when you plug this into your values you know gravity is actually negative 9.81 so you have a negative and negative so you can actually just go ahead and write that as a positive okay so we now know our time parameter in the y direction and we know that time has to be the same in both the x and y components so this is a very important equation and this is a very important equation remember we are looking for the range okay so we are looking for the range of this and the range is actually just the you know fancy word for the horizontal displacement so here's our range equation sometimes you'll see this with r so now what we're going to do is we're going to take these two equations and we're going to combine them together to make one massive equation and then when we finish it up and factor some things out we should actually see the final range equation come in so now we have this our displacement in the x direction is equal to v naught cosine of theta times t well we don't really know what t is numerically but we know what it is algebraically i'm going to switch this up real quick this is a better color forgive me i was trying to be all fancy and use different colors times t but we do know what this t is we're going to take this equation and we're going to plug it in right there so we get this times 2 v naught and that is sine of theta all over g okay so now trying to simplify this a little bit you get this the displacement in the x direction is equal to v naught squared um that's that's actually two right here sorry two b naught squared cosine theta times sine of theta all over g now there actually is this is not done yet okay so example you could plug this into your calculator and it would get you you know everything right but if you want to simplify this there is notice here cosine theta times sine theta and there's a 2 right here that's actually a trig identity okay so example that's something you might get into more advanced class so 2 cosine theta times sine of theta actually gives you sine of 2 theta okay and it's a very complex process to get that i'm not going to go into that but if we know this we can substitute this in right here to make this look a little more simpler okay so we have 2 cosine sine theta right here so we can go ahead and substitute that to that sine 2 so delta x is equal to is equal to v naught v naught squared times sine 2 theta all over d rational constant and there it is there is the range equation in all of its glory okay so this is a great video and a great skill to know especially for any ap regular physics or heck even in college physics class to be able to derive the arranged equation and you do it with just some simple kinematics and vectors i hope this video helped if so give me a thumbs up and subscribe for more physics content i hope you'll have a great day thank you
6223	https://promova.com/antonyms-of/support	What is the opposite of support? \| Antonyms support \| Promova For Business Group lessons English tutors App For Business Group lessons English tutors App Get the App: Free for Ukrainians Sign Up Free for Ukrainians Sign Up What is another word for support Type your word here Try:Lovely Find antonyms of words, enhance your English language skills support /səˈpɔːrt/ Assistance provided to someone to help them cope or succeed; to agree with and give encouragement to someone or something. Antonyms of support Verb Strongest matches: oppose hinder undermine obstruct resist Weak matches: discourage dissuade thwart contradict refute Noun Strongest matches: opposition resistance obstruction hindrance antagonism Weak matches: discouragement deterrence defiance disapproval contradiction Synonyms assist, aid, help, back, uphold, sustain, bolster, reinforce, boost, encourage, advocate, champion, endorse, promote, further, foster, facilitate, accommodate, stand by, second Usage examples: The tech team offered 24/7 support to ensure that all users could navigate the new software without any hitches. With the support of the sturdy beams, the ancient bridge has stood for centuries, defying time and weather. Emotional support from friends and family played a crucial role in her recovery from the surgery, showing how important a strong network can be during challenging times. 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6224	https://study.com/academy/lesson/effective-annual-rate-formula-calculations.html	Effective Annual Rate \| Formula, Calculations & Examples - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Business Courses Business 110: Business Math Effective Annual Rate \| Formula, Calculations & Examples Contributors: Cynthia Helzner, Yuanxin (Amy) Yang Alcocer Author Author: Cynthia Helzner Show more Instructor Instructor: Yuanxin (Amy) Yang Alcocer Show more Know the definition of the effective annual rate (EAR), see the formula for calculating the effective annual rate, and explore some examples on how to calculate the effective annual rate. Updated: 11/21/2023 Table of Contents Effective Annual Rate (EAR) Effective Annual Rate (EAR) Examples Why is the Effective Annual Rate (EAR) Important? Effective Annual Rate (EAR) vs Annual Percentage Rate (APR) Lesson Summary Show FAQ What is the difference between APR and effective annual rate? APR does not take into account the effect of compounding interest multiple times per year, whereas EAR does take that compounding into account. Thus, EAR is a more accurate reflection of the interest that will be earned or owed. How do you calculate effective annual rate? The formula for EAR is: EAR = (1 + i/n)^n - 1 where i is the stated interest rate as a decimal and n is the number of interest payments per year. The stated interest rate is typically given as a percentage so remember to divide that percentage by 100 to get the decimal version. What is effective interest rate with example? The effective interest rate is the actual percent interest that a borrower pays on their loan or earns on their investment. The formula for effective interest rate is EAR = {(1 + i/n)^n - 1} 100, where i is the nominal rate as a decimal and n is the number of compounding periods per year. For example, 2% interest compounded quarterly would have an EAR of {(1 + 0.02/4)^4 - 1} 100 = 2.015% Create an account LessonTranscript VideoQuizCourse Click for sound 4:20 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? 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Your next lesson will play in 10 seconds 0:02 Interest Rate 0:44 Effective Annual Rate 1:43 Formula 2:58 Example 3:42 Lesson Summary QuizCourseView Video Only Save Timeline 84K views Recommended lessons and courses for you Related LessonsRelated Courses ##### How to Calculate Present Value of an Investment: Formula & Examples 4:06 ##### Nominal vs. Real Interest Rate \| Differences & Examples 7:13 ##### Financial Mathematics \| Definition, Equations & Problems 6:37 ##### Present Value of Annuity \| Overview, Formula & Examples 4:26 ##### Rule of 72 Definition, Formula & Examples 3:58 ##### Real Interest Rate \| Definition, Formula & Calculation 3:01 ##### Simple Interest Definition, Formula & Examples 8:46 ##### Principal Amount \| Definition, Formula & Examples 6:20 ##### Discounted Cash Flow, Net Present Value & Time Value of Money 5:35 ##### What is Compound Interest? - Definition, Formula & Examples 6:32 ##### Present & Future Values of Multiple Cash Flows 4:41 ##### Future Value Definition Formula & Examples 4:55 ##### Types of Interest Rates & Investments 7:45 ##### Fair Value \| Definition, Formula & Method 3:55 ##### Financial Markets & Goods Markets 7:36 ##### Capital Market Efficiency & Price Behavior 5:12 ##### Nominal Rate \| Definition, Formula & Examples 4:25 ##### Zero Coupon Bond \| Overview, Formula & Examples 4:48 ##### Arithmetic vs. Geometric Return \| Definition & Calculation 5:13 ##### Present Value of Annuity \| Overview, Formula & Examples 4:26 ##### Investing: Help & Tutorials ##### Money Management: Help & Review ##### Business 100: Intro to Business ##### Intro to Excel: Essential Training & Tutorials ##### Contemporary Math ##### UExcel Introduction to Macroeconomics: Study Guide & Test Prep ##### UExcel Organizational Behavior: Study Guide & Test Prep ##### UExcel Business Law: Study Guide & Test Prep ##### UExcel Business Ethics: Study Guide & Test Prep ##### UExcel Workplace Communications with Computers: Study Guide & Test Prep ##### Effective Communication in the Workplace: Certificate Program ##### Introduction to Public Speaking: Certificate Program ##### DSST Business Mathematics Study Guide and Test Prep ##### Investing: Help & Tutorials ##### Money Management: Help & Review ##### Business 100: Intro to Business ##### Intro to Excel: Essential Training & Tutorials ##### Contemporary Math ##### UExcel Introduction to Macroeconomics: Study Guide & Test Prep ##### UExcel Organizational Behavior: Study Guide & Test Prep Effective Annual Rate (EAR) --------------------------- When someone earns interest on an investment, such as an interest-bearing bank account, or pays interest on a loan, such as a mortgage, the advertised interest rate from the bank or lender is called the nominal rateor the stated interest rate. However, that rate does not take into account the effect of compounding, i.e. having multiple interest payments per year. Effective annual rate (EAR), in contrast, does include the effect of compounding as is thus a more accurate way to calculate the amount of interest that will accrue. What is Effective Annual Rate? Effective annual rate (EAR), otherwise known as the effective annual interest rate is the actual percent interest that a borrower pays on their loan or that an investor earns on their investment. EAR depends on (and is thus calculated from) the nominal rate and the number of compounding periods per year. EAR does not depend on the amount invested. Why Does Compounding Matter? Compounding increases the effective rate of interest, whether that be interest earned or interest paid. That increase occurs because the interest snowballs i.e. interest is earned (or owed) on interest. For example, if someone invests $100 at 5% interest, they were initially earning 5% on the $100 that they invested. In contrast, once they earn $5 of interest, they will then earn 5% on $105 rather than on $100. When this snowballing occurs, more interest is accrued than what the nominal rate would imply. Effective Annual Rate Formula When choosing an investment or loan, it is helpful to know how to find effective annual rate. The effective annual rate formula for the decimal version of EAR is: E A R=(1+i n)n−1 where i represents the nominal rate as a decimal and n represents the number of compounding periods per year i.e. the number of times that interest is paid per year. To convert percentages to decimals, simply divide the percentage by 100. For example, if a bank account earns 3% interest compounded monthly, then i would be 0.03 because 3% as a decimal is 3/100 = 0.03, and n would be 12 because there are 12 monthly payments per year. The effective interest rate formula as a percentage is 100 times the decimal version of the formula: E A R=[(1+i n)n−1]⋅100 As before, i represents the nominal rate as a decimal and n represents the number of compounding periods per year. To unlock this lesson you must be a Study.com memberCreate an account Effective Annual Rate (EAR) Examples ------------------------------------ The examples below demonstrate how to calculate EAR using the effective annual rate formula. Carlos takes out a loan to pay for his car. The stated interest rate of the loan is 6%. If the interest on the loan is compounded quarterly, what is the effective annual rate as a decimal? Explanation: The stated interest rate as a percentage is 6% so i is 6/100 = 0.06. Next, n is 4 because there are 4 quarterly payments per year. Plugging those values into the decimal formula for EAR yields E A R=(1+0.06 4)4−1 which equals approximately 0.06136. Yeonmi invests $2,000 at 4% interest compounded daily. What is the effective annual rate in 2023 as a percentage? Explanation: Because the stated interest rate is 4%, i is 4/100 = 0.04. n is 365 because there are 365 daily payments per year, as 2023 is not a leap year. Plugging those values into the percentage formula for EAR yields E A R=[(1+0.04 365)365−1]⋅100 which equals approximately 4.081%. Note that this calculation is independent of the amount of money invested. Alex's checking account earns 0.2% interest compounded monthly. What is the EAR of that account? Explanation: Since the nominal rate is 0.2%, i is 0.2/100 = 0.002. n is 12 because there are 12 monthly payments per year. Plugging those values into the decimal formula for EAR yields E A R=(1+0.002 12)12−1 which equals approximately 0.002002. We could use the percentage formula for EAR to find the percentage from scratch but, since we already know it as a decimal, it is faster to simply multiply our decimal answer by 100: E A R=0.002002⋅100, which equals 0.2002%. Amani invests $1,000 in a savings account that earns 6% interest compounded semiannually. How much interest will he earn in a year? Explanation: Since the interest rate is 6%, i is 6/100 = 0.06. n is 2 because semiannual means twice per year. Plugging those values into the decimal formula for EAR yields E A R=(1+0.06 2)2−1 which equals 0.0609. Now, to find the interest earned in one year, we multiply the investment by the EAR as a decimal: $1,000⋅0.0609=$60.90 EAR is important in choosing the best investment option. Dana could invest in a savings account that earns 5% interest compounded annually or in a CD that offers 4.9% interest compounded daily. Which is the better option? Explanation: For the savings account, the stated interest rate is 5%, so i is 5/100 = 0.05. n is 1 for interest compounded annually. Plugging those values into the decimal formula for EAR yields E A R=(1+0.05 1)1−1 which equals 0.05. For the CD, the stated interest rate is 4.9%, so i is 4.9/100 = 0.049. n is 365 for a non-leap year and n is 366 for a leap year. Plugging those values into the decimal formula for EAR yields E A R=(1+0.049 365)365−1 and E A R=(1+0.049 366)366−1, both of which equal approximately 0.05022. That value is higher than the 0.05 EAR for the savings account. So, even though the nominal rate is lower for the CD, the CD actually yields more interest than the savings account does. To unlock this lesson you must be a Study.com memberCreate an account Why is the Effective Annual Rate (EAR) Important? ------------------------------------------------- The primary importance of the effective annual rate (EAR) to borrowers and investors is that it is a more accurate representation of how much interest will be owed on a loan or earned on an investment. Furthermore, EAR can help investors to figure out which investment option will be more profitable. For example, an investment that offers a higher stated interest rate may end up having a lower EAR than one that offers a slightly lower interest rate that is compounded more often. Similarly, borrowers can use EAR to pick the loan option on which they will owe the least interest. Compared to the interest rate stated in a loan contract, the EAR is a more accurate reflection of how much a loan will actually cost the borrower. To unlock this lesson you must be a Study.com memberCreate an account Effective Annual Rate (EAR) vs Annual Percentage Rate (APR) ----------------------------------------------------------- Another term that you may have heard when it comes to borrowing or investing is annual percentage rate (APR). APR is an annualized interest rate so it treats the account as if interest only accrues once per year (in our EAR formulas, that corresponds to n = 1). So, unlike the effective annual rate (EAR), the APR does not take into account the effect of compounding multiple times per year. The formula for calculating the annual percentage rate is: A P R=t o t a l i n t e r e s t+f e e s p r i n c i p a l⋅365 l e n g t h o f t h e l o a n i n d a y s⋅100 where principal means the stated amount or face value of the loan (e.g. $10,000 for a borrower who borrowed $10,000 and will then need to pay it back with interest). For situations or math problems in which there are no fees, only interest, you can treat the APR as the nominal rate for purposes of calculating EAR. To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- The nominal rateor stated interest rate is the advertised interest rate and does not take into account the effect of compounding (accruing interest on interest). Similarly, annual percentage rate (APR) expresses the interest in an annualized fashion and thus does not take into account the effect of having multiple interest payments per year. Thus, the nominal rate and APR are lower than the actual interest that will accrue on a loan or investment, and so a more accurate representation is useful. Unlike the nominal rate or the APR, the effective annual rate (EAR) i.e. effective annual interest rate does take into account the effect of compounding. The EAR is derived from the nominal rate, i, and the number of compounding periods per year, n, whereas it does not depend on the principal. The formula for EAR is: E A R=(1+i n)n−1 where i represents the nominal rate as a decimal and n represents the number of compounding periods per year i.e. the number of times that interest is paid per year. This version of the formula gives the EAR as a decimal. Multiply it by 100 to get the rate as a percentage. When calculating how much interest will accrue, multiply the decimal form of EAR by the principal (i.e. the initial investment or loan amount). To find the total balance at the end of one year, add the interest to the principal (i.e. if an investment of $1,000 earns $44 in interest, the final balance would be $1,000 + $44 = $1,044). To unlock this lesson you must be a Study.com memberCreate an account Video Transcript Interest Rate Almost everyone has a credit card nowadays. There are lots of different credit card companies out there. They all want more customers, more people to carry their credit card. The way credit card companies work is that you spend money that you don't have, then you pay a small portion every month until you pay it off. However, each credit card has a stated interest rate that you also have to pay based on how much you owe. The credit card companies use this interest rate to calculate how much extra you have to pay each month. This interest rate is how credit card companies make their money. You will see the interest rate listed as an APR (Annual Percentage Rate). You will see such numbers as 10.99%, 23.99%, etc. Effective Annual Rate However, this number that you see is not the effective annual rate, the actual interest rate when the calculations are done more than once a year, because the interest rate that they show you is the interest rate if the calculation is done just once a year. With credit card companies, the calculations are done on a monthly basis. Let me show you the difference when the calculation is done once a year versus once a month. Say we have an annual interest rate of 11%. If we have $500 in the account, then at the end of the year, we will have $555 if we make the calculation just once a year. Now, if we make the calculation on a monthly basis, at the end of the year we will have $557.86. We have slightly more here than when our calculation was done just once a year. This tells us that if we calculate the interest more than once a year, then we need to adjust our effective annual rate to account for this. Formula The good news is that we already have a formula in place to do just that. In this formula, the i stands for the interest rate that is given to you by the company. The n is the number of times that calculations are made in a year. So, if calculations are done on a monthly basis, the n is 12. When using this formula, we change our percentage to decimal form. Let's use this formula for our 11% annual interest rate and see what our effective annual rate is when our calculations are done on a monthly basis. Plugging in 0.11 for i and 12 for n, we have (1 + 0.11 / 12)^12 - 1 = 0.1157 for a percentage of 11.57. So, our effective annual rate is 11.57%. Is this correct? If we make just one calculation using this effective annual rate on our $500 balance, at the end of year, we will have $557.85. We are 1 penny off because we have rounded our effective annual rate to just two decimal places. So, our effective annual rate is correct. Example See if you can do this problem on your own: Calculate the effective annual rate for an interest rate of 15% when the calculations are done on a monthly basis. Our interest rate is 15, so our i is 15. Our n is 12 since the calculations are done on a monthly basis. So, plugging in this information into our formula, we have (1 + 0.15 / 12)^12 - 1 = 0.16075. So, our effective annual rate is 16.07%. Is this what you got? Lesson Summary Let's review what we've learned now. The effective annual rate is the actual interest rate when the calculations are done more than once a year. The formula for the effective annual rate is: The little i stands for the interest rate that is given to you and the n is the number of times a year that calculations are made. Using this formula will give you the effective annual rate if the calculations are done more than once a year. Most often the n is 12 for monthly calculations. Learning Outcomes Measure your ability to do the following after watching this video lesson: Provide specifics about the effective annual rate Distinguish between the effective annual rate and the annual percentage rate Illustrate the formula for calculating the effective annual rate Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. 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6225	https://www.uptodate.com/contents/specific-phobia-in-adults-epidemiology-clinical-manifestations-course-and-diagnosis	Specific phobia in adults: Epidemiology, clinical manifestations, course, and diagnosis - UpToDate Subscribe Sign in English Deutsch Español 日本語 Português Why UpToDate? Product Editorial Subscription Options How can UpToDate help you?Select the option that best describes you Medical Professional Resident, Fellow, or Student Hospital or Institution Group Practice Subscribe Specific phobia in adults: Epidemiology, clinical manifestations, course, and diagnosis View Topic Share Font Size Small Normal Large Bookmark Rate Feedback Tools Formulary drug information for this topic No drug references linked in this topic. Find in topic Formulary Print Share Feedback Font Size Small Normal Large Outline SUMMARY INTRODUCTION SPECIFIC PHOBIA SPECIFIERS EPIDEMIOLOGY Prevalence Comorbidities PATHOGENESIS Genetic factors Neurobiologic factors Personality factors Cognitive factors Social and environmental factors Evolutionary pathway CLINICAL MANIFESTATIONS COURSE OF ILLNESS DIAGNOSIS Assessment for symptoms and associated behaviors Diagnostic criteria Differential diagnosis - Agoraphobia - Panic disorder - Social anxiety disorder - Posttraumatic stress disorder - Separation anxiety disorder - Illness anxiety disorder - Eating disorders - Other diagnostic considerations - Specific phobia in children ASSESSMENT TOOLS SOCIETY GUIDELINE LINKS SUMMARY REFERENCES GRAPHICS Tables - Self-report measures for specific phobia by phobia subtype RELATED TOPICS Acute procedural anxiety and specific phobia of clinical procedures in adults: Treatment overview Acute procedural anxiety in adults: Epidemiology, clinical features, assessment, and diagnosis Agoraphobia in adults: Epidemiology, pathogenesis, clinical manifestations, course, and diagnosis Anxiety disorders in children and adolescents: Epidemiology, pathogenesis, clinical manifestations, and course Eating disorders: Overview of epidemiology, clinical features, and diagnosis Generalized anxiety disorder in adults: Epidemiology, pathogenesis, clinical manifestations, course, assessment, and diagnosis Illness anxiety disorder: Epidemiology, clinical presentation, assessment, and diagnosis Major depression in adults: Epidemiology Overview of fears and phobias in children and adolescents Panic disorder in adults: Epidemiology, clinical manifestations, and diagnosis Posttraumatic stress disorder in adults: Epidemiology, pathophysiology, clinical features, assessment, and diagnosis Risky drinking and alcohol use disorder: Epidemiology, clinical features, adverse consequences, screening, and assessment Social anxiety disorder in adults: Epidemiology, clinical features, assessment, and diagnosis Society guideline links: Anxiety and anxiety disorders in adults Specific phobia in adults: Cognitive-behavioral therapy Specific phobia in adults: Treatment overview Specific phobia in adults: Epidemiology, clinical manifestations, course, and diagnosis Author:Randi E McCabe, PhDSection Editor:Murray B Stein, MD, MPHDeputy Editor:Michael Friedman, MD Literature review current through:Aug 2025. This topic last updated:May 16, 2025. INTRODUCTION Specific phobia is an anxiety disorder characterized by clinically significant fear of a particular object or situation that typically leads to avoidance behavior. Phobic fears include animals, insects, heights, water, enclosed places, driving, flying, seeing blood, getting an injection, and choking or vomiting. The phobic anxiety may be triggered by anticipation of the stimulus, actual exposure to the stimulus, and even hearing the stimulus name spoken aloud (eg, hearing the word spider for an individual with spider phobia). The focus of fear may include disgust, danger of harm, and/or the experience of physical symptoms in the phobic situation . This topic describes the epidemiology, pathogenesis, clinical manifestations, course, and diagnosis of specific phobia in adults. Specific phobias relating to clinical procedures (eg, blood-injection-injury phobia) and other manifestations of acute procedural anxiety; specific phobias and other anxiety disorders in children; and pharmacotherapy and psychotherapy for specific phobia in adults are discussed separately. ●(See "Anxiety disorders in children and adolescents: Epidemiology, pathogenesis, clinical manifestations, and course".) ●(See "Overview of fears and phobias in children and adolescents".) To continue reading this article, you must sign in with your personal, hospital, or group practice subscription. Subscribe Sign in Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. 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6226	https://www.maths.dur.ac.uk/users/anna.felikson/Geometry/Geometry20/hints7-8.pdf	Dr. Anna Felikson, Durham University Geometry, 23.11.2020 Hints 7-8 7.2. Use aﬃne transformation to simplify the question. 7.3. Apply aﬃne transformation to map three vertices of the pentagon to the vertices of the regular pentagon. Use the fact that aﬃne maps preserve parallelism to conclude about the images of the other points. (You will probably also need some continuity argument). 7.4. Apply the sine law to three diﬀerent triangles you can ﬁnd in the picture. 8.3. If it is not clear how to ﬁnd the cross-ratio of points on the lineat inﬁnity , we could ﬁrst send the line at inﬁnity to any other nicer line. Another option would be to recall that the cross-ratio of points in RP2 is the cross-ratio of the corresponding lines in R3. 1
6227	https://www.quora.com/Someone-please-help-Suppose-that-a-b-is-on-the-circle-x-2-y-2-r-2-Show-that-the-line-ax-by-r-2-is-tangent-to-the-circle-at-a-b-I-have-this-math-problem-to-solve-and-I-dont-understand-it-Can-someone-help-me	Something went wrong. Wait a moment and try again. Tangent Lines Coordinate Plane Help Needed Equation of Lines Tangents and Secants Math Geometry 5 Someone please help! Suppose that (a,b) is on the circle x^2 + y^2=r^2. Show that the line ax+by=r^2 is tangent to the circle at (a,b). I have this math problem to solve and I don't understand it. Can someone help me? 6 Answers Haresh Sagar Studied Science & Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views · 4y Circle:x2+y2=r2 Point on circle : P=(a,b) Take derivative of circle equation ⟹ 2x+2yy′=0 ⟹ y′=−xy Slope of tangent at P ⟹ m=−ab Equation of tangent at P ⟹ y−b=−ab(x−a) ⟹ by−b2=−ax+a2 ⟹ ax+by=a2+b2 As we know that circle is centred at origin and radius of circle is r . P is on circumference of circle so, r=√(a−0)2+(b−0)2 ⟹ r=√a2+b2 ⟹ r2=a2+b2 Hence proved that given line is tangent to given circle. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Greg Weidman B.E. in Electrical Engineering & Mathematics, Villanova University (Graduated 1992) · Author has 379 answers and 10.7M answer views · 4y The way to approach this starts with looking at the radius line from the origin to the point (a,b). You know the slope of this line is b/a. You also know that given a line of slope m, any line perpendicular to it has slope -1/m. Next, realize that any tangent to a circle is perpendicular to the radius line at the tangent point. So, the slope of your line is -a/b. You also have a point on the tangent line, (a,b). You can now use the point-slope formula to find the equation for the tangent line. Some simple algebraic manipulation will give you what you are looking for. Good luck! Graham Dolby Author has 3.2K answers and 1.7M answer views · 4y Suppose that (a,b) is on the circle x2+y2=r2. Show that the line ax+by=r2 is tangent to the circle at (a,b). The circle x2+y2=r2 has radius r and centre (0,0). If (a,b) is on the circle then by substitution a2+b2=r2. The radial line through (0,0) and (a,b) is bx−ay=0. This is normal to the circle at (a,b). The perpendicular line to the normal at (a,b) is ax+by=a×a+b×b which is ax+by=r2. The line perpendicular to the normal at (a,b) is the tangent at (a,b), and this is ax+by=r2, as aleady stated. Related questions Can anyone solve this maths question? The perpendicular bisector of a chord XY cuts XY at N and the circle at P. If XY=16 cm and NP=2cm, calculate he radius of the circle. The answer to this question is 17cm, but I can't solve it. A circle C1 is tangent to y axis, on which the centre of another circle C2 lies. What are the circles if their common tangent is √3y = x+2? In the xy'-plane below, the equation of the circle is x^2+ y^2 = 13. Line l is tangent to the circle at (-2,3). What is the slope of line l? When is a 2 + b 2 = x 2 + y 2 ? Can you answer this math problem? X²+y²=10x+6y-2=0. How would you identify the center and the radius of the circle? Allen Ries B.Ed. from University of Alberta (Graduated 1974) · Author has 25.1K answers and 9.7M answer views · 4y x²+y²=r² y’= -x/y Now using this slope (y’, the derivative of the circle equation) and the point (a,b) we look for the tangent line using the slope-point method: y-b = (-x/y)(x-a) y-b=-a/b(x-a) by-b²=-ax+a² ax+by=a²+b² Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. 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Find the slope of the radius at the point : The radius from the center of the circle (the origin ( 0 , 0 ) ) to the point ( a , b ) has a slope given by: slope of radius = b − 0 a − 0 = b a 3. Rearrange the line equation to find its slope : The line a x + b y = r 2 can be rearranged into slope-intercept form y = m x + c : b y = − a x + r 2 ⇒ y = − To show that the line a x + b y = r 2 is tangent to the circle x 2 + y 2 = r 2 at the point ( a , b ) , we can follow these steps: Confirm the point lies on the circle : Since ( a , b ) is on the circle, we have: a 2 + b 2 = r 2 2. Find the slope of the radius at the point : The radius from the center of the circle (the origin ( 0 , 0 ) ) to the point ( a , b ) has a slope given by: slope of radius = b − 0 a − 0 = b a 3. Rearrange the line equation to find its slope : The line a x + b y = r 2 can be rearranged into slope-intercept form y = m x + c : b y = − a x + r 2 ⇒ y = − a b x + r 2 b The slope of this line is − a b . 4. Check the relationship between the slopes : For the line to be tangent to the circle at the point ( a , b ) , the slopes of the radius and the tangent line must be negative reciprocals of each other. Thus, we need to check: slope of radius ⋅ slope of tangent line = − 1 Substituting the slopes: b a ⋅ ( − a b ) = − 1 This equation holds true, confirming that the line is perpendicular to the radius at point ( a , b ) . 5. Conclusion : Since the line a x + b y = r 2 passes through the point ( a , b ) and is perpendicular to the radius at that point, it is tangent to the circle at ( a , b ) . Thus, we have shown that the line a x + b y = r 2 is tangent to the circle x 2 + y 2 = r 2 at the point ( a , b ) . Sripad Sambrani Knows Sanskrit · Author has 6.8K answers and 2.9M answer views · 4y Equation of tangent to the circle x²+y²-r²=0 at point A(x1,y1), in general, is given by xx1+yy1-r²=0. At (a,b) the equation of tangent to the circle is ax+by-r²=0. Trust this helps. Related questions Can anyone solve this maths question? The perpendicular bisector of a chord XY cuts XY at N and the circle at P. If XY=16 cm and NP=2cm, calculate he radius of the circle. The answer to this question is 17cm, but I can't solve it. A circle C1 is tangent to y axis, on which the centre of another circle C2 lies. What are the circles if their common tangent is √3y = x+2? In the xy'-plane below, the equation of the circle is x^2+ y^2 = 13. Line l is tangent to the circle at (-2,3). What is the slope of line l? When is a 2 + b 2 = x 2 + y 2 ? Can you answer this math problem? X²+y²=10x+6y-2=0. How would you identify the center and the radius of the circle? S ≡ x 2 + y 2 − r 2 = 0 is a circle. X ≡ ( x 1 , √ r 2 − x 2 ) is a point on circle. Q ≡ ( d , 0 ) is a point inside the circle. How can I find the intersection points of the circle and line XQ? How do you find the incenter of a triangle formed by the centers of three circles that touch each other, and why is this important in solving problems with common tangents? Which line is the tangent line to the circle given by x^2+y^2=25? Let T1, T2 be tangents from (-2,0) onto circle C: X^2+Y^2=1. How can I determine the circles touching C and having T1, T2 as their pair of tangents, and what are the equations of all possible common tangents to the circles when taken two at a time? The chords of contact of the pair of tangents drawn from each point on the line 2x+y=4 to the circle x^2+y^2=1 passes through the point (a,b) . Then find the value of a/b. I didn't understand the question, where does the point (a,b) lie? The figure below shows 13 circles of radius 1 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius 1? Could you help me solve this problem? There are two circles that go through two points (1,3) (2,4) and are tangents to the the y-axis, letting the radii of the circles be a,b implying that ab=? How do you find the length of the tangent from any point on the circle x 2 + y 2 + 2 f x + 2 g y + c 1 = 0 to the circle x 2 + y 2 + 2 f x + 2 g y + c 2 = 0 ? How do you evaluate circle s: $x^2 + y^2 = r^2$. Three congruent circles are externally tangent to each other and internally tangent, one touches s at a (0, r). (analytic geometry, circles, math)? Does (x×2×2×2) × (y÷2÷2÷2) = xy? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
6228	https://www.britannica.com/dictionary/eb/audio?file=egregi01&format=mp3&word=egregious&pron=%C9%AA%CB%88gri%CB%90%CA%A4%C9%99s	How to Pronounce egregious - (Audio) \| Britannica Dictionary "egregious" Listen to the audio pronunciation again /ɪˈgriːʤəs/ Having trouble hearing a pronunciation? Click here to listen with your default audio player.
6229	https://sco.wikipedia.org/wiki/Kilogramme	Kilogramme - Wikipedia Jump to content [x] Main menu Main menu move to sidebar Dern Navigation Main page Commonty Yett Mercat Cross Recent chynges Wale page allevolie Help Rake Rake [x] Appearance Appearance move to sidebar Dern Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Propines Mak accoont Log in [x] Personal tuils Propines Mak accoont Log in Kilogramme [x] 153 leids Аԥсшәа Afrikaans Alemannisch Aragonés Obolo العربية الدارجة مصرى অসমীয়া Asturianu Aymar aru Azərbaycanca تۆرکجه Башҡортса Basa Bali Žemaitėška Bikol Central Беларуская Беларуская (тарашкевіца) Български भोजपुरी বাংলা བོད་ཡིག Brezhoneg Bosanski Català 閩東語 / Mìng-dĕ̤ng-ngṳ̄ Нохчийн کوردی Čeština Словѣньскъ / ⰔⰎⰑⰂⰡⰐⰠⰔⰍⰟ Чӑвашла Cymraeg Dansk Deutsch Ελληνικά English Esperanto Español Eesti Euskara فارسی Suomi Français Nordfriisk Furlan Frysk Gaeilge 贛語 Galego Avañe'ẽ Hausa 客家語 / Hak-kâ-ngî עברית हिन्दी Fiji Hindi Hrvatski Kreyòl ayisyen Magyar Հայերեն Արեւմտահայերէն Interlingua Bahasa Indonesia Interlingue Ilokano Ido Íslenska Italiano 日本語 Patois Jawa ქართული Taqbaylit Tyap Қазақша ಕನ್ನಡ 한국어 Къарачай-малкъар Kurdî Kernowek Кыргызча Latina Lëtzebuergesch Limburgs Lombard Lingála Lietuvių Latviešu Македонски മലയാളം Монгол मराठी Bahasa Melayu Malti မြန်မာဘာသာ Plattdüütsch नेपाली नेपाल भाषा Nederlands Norsk nynorsk Norsk bokmål Occitan ଓଡ଼ିଆ ਪੰਜਾਬੀ Polski Piemontèis پنجابی Português Runa Simi Română Русский Русиньскый संस्कृतम् Саха тыла Sardu Sicilianu Srpskohrvatski / српскохрватски සිංහල Simple English Slovenčina Slovenščina Gagana Samoa Soomaaliga Shqip Српски / srpski Sunda Svenska Kiswahili Ślůnski தமிழ் తెలుగు Тоҷикӣ ไทย Tagalog Türkçe Xitsonga Українська اردو Oʻzbekcha / ўзбекча Vèneto Vepsän kel’ Tiếng Việt West-Vlams Walon Winaray 吴语 მარგალური ייִדיש Vahcuengh 中文文言閩南語 / Bân-lâm-gí 粵語 Eedit airtins Airticle Collogue [x] Scots Read Eedit Eedit soorce See history [x] Tuilkist Tuils move to sidebar Dern Actions Read Eedit Eedit soorce See history General Whit airts tae here Relatit chynges Uplaid file Permanent airtin Page information Cite this airticle Get shortened URL Download QR code Eedit interleid airtins Prent/export Create a beuk Doonload as PDF Prent version In ither projects Wikimedia Commons Wikidata item Frae Wikipedia, the free beuk o knawledge The kilogramme (cutty form kg) is a metric unit that descrives mass. The offeecial kilogramme equals the mass o a parteecular piece o platinum-iridium metal hauden in Paris. It is the anely SI unit left that is defined sic that it needs tae be evened wi some object. There nou attempts for tae define the kilogramme in a different wey, for ensaumple by specifyin a nummer o atoms o a certain substance (at a certain temperature). Ae kilogramme is a bit mair nor 2.2 poond. Ae tonne is ae thoosand kilogramme. Ae litre o watter weighs aboot ae kilogramme. Wikimedia Commons haes media relatit tae Kilogram. Taen frae " Categeries: SI base units Units o mass This page wis last eeditit on 27 August 2020, at 17:02. Text is available unner the Creative Commons Attribution-ShareAlike License; additional terms mey apply. See Terms o Uiss fur details. Preevacie policie Aboot Wikipedia Disclamation Code of Conduct Deveelopers Statistics Cookie statement Mobile view Edit preview settings Rake Rake Kilogramme 153 leidsEik topic
6230	https://blog.csdn.net/Always_away/article/details/145865682	26考研\|数学分析：微分中值定理及其应用_微分中值定理判断函数单调性-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作 26考研\|数学分析：微分中值定理及其应用最新推荐文章于 2025-09-26 22:58:42 发布原创于 2025-02-26 00:03:18 发布·692 阅读 · 7 · 2· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #考研#笔记#学习 2026考研\|数学分析专栏收录该内容 19 篇文章订阅专栏本章是整个一元函数微分学中最为精华和重点的地方，相对应的，本章的学习难度较大，在学习过程中，建议大家反复阅读教材，仔细揣摩课本习题的相关沿用知识。根据本章标题其实不难发现，本章节可分为两大部分：微分中值定理的定理部分与应用部分。本章节内容较多，学习过程中建议增加投入时间。定理部分其实包括费马定理、罗尔定理、拉格朗日中值定理、柯西中值定理以及泰勒定理（泰勒公式），应用部分，简而言之，就是通过导数来分析函数的相关性性质：单调性、凹凸性等，以及由其引出的极值点、拐点、单调区间、保凸区间等相关计算，同时还有利用微分中值定理进行极限的计算也是分析的重点。课本简单概括定理部分关于微分中值定理的具体内容，此处便不再赘述，大家可以参加课本。在学习过程中，一定要注意掌握各个定理之间的证明推出关系，比如：如何利用罗尔定理证明拉格朗日中值定理。除此之外，还应该掌握各个定理应用的具体条件，在适宜条件下才可以进行定理的运用。应用部分应用一：单调性、单调区间（#6.1节）掌握运用一阶导数判断函数单调性的方法，通过判断一阶导数的正负判断函数的单调性（单调递增、单调递减）。应用二：极值、最值（#6.4节）掌握极值判断的三个充分条件，并应用其对某一段定义域上的函数进行极值的求解。应用三：凹凸性、保凸区间（#6.5节）掌握凹（凸）函数的定义，通过判断二阶导数的正负判断函数的凹凸性，并且掌握对于保凸区间的求解与分析。应用四：极限求解（#6.2节、#6.3节）掌握运用经由柯西中值定理推导出的洛必达法则进行极限的求解（一定要注意使用的前提条件），掌握运用泰勒展开的方式进行极限的求解。应用五：函数的图象（#6.6节）对于函数的图象，应该综合运用单调性、凹凸性、渐近线、特殊点等关键信息进行图像的绘制。应用六：方程的近似解（#6.7节）课本经典习题确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 Always_away 关注关注 7点赞踩 2 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录考研高数专题4：微分中值定理及其应用（罗尔定理-拉格朗日中值定理-柯西中值定理）刘鑫磊 08-10 1419 一：方程的根（零点定理、罗尔定理）例题1：罗尔定理二：证明不等式（单调性、拉格朗日、柯西）例题1：拉格朗日中值定理三：求极限例题1：拉格朗日中值定理（中值相关的极限）例题2：拉格朗日中值定理四：证明存在一个中找到s使的一个式子成立（构造辅助函数用罗尔定理_分析法_ 微分方程法_规律）齐次方程公式例题1：分析法：（写在一边不要分母、一般思想f'和f化成f'/f其余的搬过来）微分方程... 考研数学第三章复习：微分中值定理以及导数的应用(1) lyy的博客 02-23 941 我一直坚信，学一个课程要把整个内容融会贯通起来，要用出题人的思维去解决题目，当一道题摆到你面前的时候你应该有一种或多种思考角度，然后逐一尝试来解决题目，而不是左右摇摆。上一章已经介绍了导数与微分，那么很明显，第三章就是要熟悉微分熟悉导数，并能够知道它们可以做些什么。首先第一节带领我们了解微分中值定理。什么是微分中值定理，每一个数学名词都有它之所以这么叫的原因，微分，中值？中间值？这些定... 参与评论您还未登录，请先登录后发表或查看评论高等数学第三章---微分中值定理与导数的应用(3.4~3.5)_大学微积分里导 ... 9-18 3. 函数单调性判别法定理 : 设函数 f(x)f(x)f(x)在开区间(a,b)(a, b)(a,b)内可导。若对任意x∈(a,b)x \in (a, b)x∈(a,b)都有f′(x)>0f'(x) > 0f′(x)>0,则f(x)f(x)f(x)在(a,b)(a, b)(a,b)内单调递增。若对任意x∈(a,b)x \in (a, b)x∈(a,b)都有f... 高数第三章微分中值定理与导数的应用 _高等数学matlab实验导数与微 ... 9-6 微分中值定理 : 费马引理,罗尔定理,拉格朗日中值定理,柯西中值定理洛必达法则 : 注意只适合0/0型泰勒公式 : 拉格朗日余项,佩亚诺余项; 麦克劳林公式。函数的单调性与曲线的凹凸性 : 驻点,拐点。拐点的定义(f''(x)=0且附近左右邻近点f''符号相反); 凹凸性结合f''(x)和f=xx来判断。函数的极值与最大... 数学分析(六)-微分中值定理及其应用 01：拉格朗日定理和函数的单调性 2401_82833469的博客 01-27 1665 § 1 拉格朗日定理和函数的单调性在这一章里, 我们要讨论怎样由导数 f′f^{\prime}f′ 的已知性质来推断函数 fff 所应具有的性质.微分中值定理（包括罗尔定理、拉格朗日定理、柯西定理、泰勒定理）正是进行这一讨论的有效工具. 本节首先介绍拉格朗日定理以及它的预备定理一一罗尔定理,并用此讨论函数的单调性. 一、罗尔定理与拉格朗日定理定理 6.1 (罗尔 (Rolle) 中值定理) 若函数 fff 满足如下条件 : (i) fff 在闭区间 [a,b][a, b][a,b] 上连续; (ii) f 高等数学：第三章微分中值定理与导数的应用（4）函数的单调性 gukedream的专栏 01-07 4329 §3.4 函数的单调性在上小于零，在上大于零。函数的单调性是否与导函数的符号有关呢?为此，我们进一步地作图，希望从中获得更多的感性认识。函数在上单调增加(减少)，则它的图形是一条沿轴正向上升(下降)的曲线，曲线上各点处的切线之斜率均为正的(负的)，即： () 这表明：函数的单调性确实与其导数的符号有关，因此，可以利用导数的符号来判定函数的单调性。二、函数单调性的判别... 微分中值定理及其在函数分析中的应用 8-28 如果函数 f(x)f(x)满足 : 1. 在闭区间[a,b][a,b]上连续; 2. 在开区间(a,b)(a,b)上可导; 3. 在区间端点处的函数值相等, 即f(a)=f(b)f(a)=f(b), 那么在(a,b)(a,b)内至少有一点ξ(a<ξ<b)ξ(a<ξ<b),使得f′(ξ)=0f′(ξ)=0 拉格朗日中值定理拉格朗日中值定理 : 如果函数... 高等数学 : 第三章微分中值定理与导数的应用(6)最小值与最大值问题_利用... 9-26 上的最值。二、非闭区间上定义的函数最值对于非闭区间上定义的函数,它有可能存在着最值,也有可能不存在着最值,这就给求函数最值带来了困难。探讨函数最值,可先求函数的可疑极值点(驻点,导数不存在的点),并讨论由这些点所形成的区间上函数的单调性,再利用函数的性态来判断函数在这些可疑点处是否有最值。第三章微分中值定理与导数的应用第四节函数的单调性与曲线的凹凸性 m0_49433560的博客 02-22 646 yfx)ab(ab)f′x≥0,yfx)abf′x≤0,yfx)[ab]fx)IIx1x2f2x1x22fx1fx2,fx)If2x1x22fx1fx2,fx)Ifx)ab(ab)f′′x0,fx)abf′′x。考研数学——高数：微分中值定理与导数应用（2）哈哈，被自己写的代码，蠢笑啦~ 02-28 1085 设函数 f(x)在区间[a，b]上连续，在（a，b）内可导，则1）如果（a，b）内 f'(x) ≥ 0，且等号只在有限个点上成立，则 f (x)在[a，b]上单调增加解释为什么是有限个点：因为如果是有限个点，那么这些导数为零的地方必然是零散的点，而零散的点可以将原区间分为若干个由f'(x)=0隔开的子区间，每个子区间可以由区间上导数恒大于（小于）0证出区间上单调递增（递减）。高等数学 : 第三章微分中值定理与导数的应用(3)曲线的凹凸拐点曲率_弯... 9-22 函数的一阶导数的符号可判断函数的单调性,二阶导数的符号又能确定函数的何种属性呢?一个最简单的例子,给我们以启迪。抛物线的二阶导数为 , 若 ,即 ,抛物线是开口向上的凹弧; 若 ,即 ,抛物线是开口向下的凸弧。三、凹凸性的判别法【定理】微分中值定理与导数的应用 _费马定理 9-29 这个定理即可以判断函数单调区间,又可以构造函数来验证不等式。曲线的凹凸性与拐点通过上述关系式可推出如下定理 : 拐点例题函数的极值与最值函数的极值与求法函数的极大值和极小值的概念是局部性的。极值定义 : 设函数 f(x)在点x0的某邻域中有定义,在邻域中任意x有 ... 26 考研 \| 数学分析：多元函数微分学 Always_away的博客 06-27 457 本章我们将进行多元函数微分学的学习，多元函数微分学与一元函数微分学相对应，涉及到可微性、中值定理、泰勒公式等诸多问题的探讨与研究，本章难度较大，在学习过程中需要进行深度思考与分析，才能真正掌握这一章的相关知识点。中值定理总结_ 数学分析 \| 第六章微分中值定理及其应用凸函数的定义与性质总结... weixin_32880357的博客 01-17 2191 当公式或文字展示不完全时，记得向左←滑动哦！摘要：本节给出了凸函数的三个等价条件，分别是从割线角度，导函数的角度，二阶导函数的角度来研究，并且给出了凸函数的三个推论，由此可知凸函数在定义域的每一点(除去端点)连续，且左右导数都存在。凸函数定义设为定义在区间上的函数，若对上的任意两点和任意实数总有则称为上的凸函数。注意：大家注意凸函数的定义是从函数值角度来研究的，这里面涉及到了三个点为和... ...微分中值定理与导数的应用——第四节函数的单调性与曲线的凹凸性... 8-19 1. 函数单调性的判定法 2. 曲线的凹凸性与拐点1. 函数单调性的判定法函数单调性的判定定理设函数 y = f ( x ) y = f(x) y=f(x) 在 [ a , b ] [a, b] [a,b] 上连续,在 ( a , b ) (a, b) (a,b) 内可导.(1)如果在 ( a , b ) (a, b) (a,b) 内 f ′ ( x )... 高数：Ch3.微分中值定理与导数应用 Edward2022的博客 09-12 4923 一元函数微分学：Ch1.函数、极限、连续 Ch2.导数与微分、Ch3.微分中值定理与导数应用考研数学笔记 - 微分中值定理 weixin_46637202的博客 05-06 730 考研数学笔记 - 微分中值定理文章目录考研数学笔记 - 微分中值定理 1.有界性定理2.最值性定理3.零点定理4.罗尔定理 1.有界性定理暂留 2.最值性定理 1.与积分配合使用（验证积分中值定理） 2. 3.零点定理证明唯一根 ps,端点不能取，因为f(a) f(b) < 0 ，所以原来两端就不能得0 4.罗尔定理（1）遇见证明的第一步就是先移动项，使一边为 0 （2）考点 1即使F（x）的寻找，二就是相等点的寻找（3）主要构造F（x）有三种形式f（x)f‘(x) ,f(x)e^x 高等数学基础篇（数二）之微分中值定理及导数应用常考题型心碎烤肠的博客 03-17 993 高等数学基础篇（数二）之微分中值定理及导数应用常考题型考研数学——高数：微分中值定理与导数应用（1）哈哈，被自己写的代码，蠢笑啦~ 02-27 1046 f 在 [a，b] 上连续f 在（a，b）内可导则一定有一点xo 在 (a，b)上，使得 f'(xo) = 0二、拉格朗日中值定理 f 在 [a，b] 上连续f 在（a，b）内可导则曲线上必有一点xo满足它的切线斜率，等于a、b两点连线斜率。高等数学微积分归纳总结：中值定理（大全包括函数、微分、积分） m0_65708726的博客 06-02 2701 本人总结的用于在考研中使用，力求全面和简洁的介绍《信息管理学基础（第三版）》考研知识点梳理总结 liweiweili126的博客 09-24 520 （★高频简答题） 20 26 考研时间，定了 2401_86286852的博客 09-25 188 网上报名结束后，各省级教育招生考试机构将组织网上确认工作，对考生报名信息进行审核。所有考生均须在规定时间内参加网上报名和网上确认，逾期不再补办。“中国研究生招生信息网”将于2025年9月25日至28日开展“20 26 年全国硕士研究生招生宣传咨询活动”，届时所有研究生招生单位将在线解答考生咨询。网上预报名时间为2025年10月10日至10月13日。网上报名时间为2025年10月16日至10月27日。初试时间为2025年12月20日至21日。20 26 考研时间，定了！【OpenGL】LearnOpenGL 学习笔记 28 - 延迟渲染 Deferred Rendering 最新发布 Duo1J的随笔 09-26 1244 LearnOpenGL 学习笔记 - 延迟渲染 Deferred Rendering MySQL 学习笔记 02：安装配置与程序结构深度解析 2301_76657443的博客 09-24 1414 在上一篇文章中，我分享了重新开始系统学习 MySQL的计划。今天开始第一课时的学习：MySQL安装与程序结构。虽然之前已经安装过MySQL，但这次我要更深入地理解MySQL的程序架构和配置细节。问题1：服务名称不匹配错误现象：sc query MySQL80 失败，提示"指定的服务未安装"解决方法：使用 sc query type= service \| find /i "mysql" 查找正确的服务名我的实际服务名：MySQL80_42问题2：端口占用冲突# 检查端口占用情况。 Mysql DBA 学习笔记（客户端常用工具） leo_yty的博客 09-24 1048 本文介绍了MySQL常用客户端工具的使用方法，包括mysql客户端连接本地/远程数据库、mysqladmin管理工具、mysqlbinlog日志查看工具、mysqlshow对象查找工具等。重点讲解了数据库备份工具mysqldump的语法和参数选项，以及数据恢复工具mysqlimport和source命令的使用方法。文章还提供了数据库备份与恢复的具体操作示例，如将数据库备份为SQL文件或文本文件，以及从备份文件恢复数据的过程。这些工具涵盖了MySQL数据库管理的主要操作场景，适合数据库管理员参考使用。 FPGA 学习笔记——图像滤波之均值滤波（线性） ZHP的博客 09-24 767 本文介绍了基于FPGA的均值滤波实现方法。均值滤波作为一种基础线性平滑滤波，通过计算像素3×3邻域的平均值来抑制噪声和模糊图像。文章详细阐述了实现思路：使用两个FIFO缓存1280像素的行数据，构建3×3卷积窗口，对9个像素数据求平均。提供了完整的Verilog代码实现，包括顶层模块、均值滤波模块和3×3模板模块，以及仿真测试代码。实验采用1280×720分辨率图像，通过读写BMP格式文件进行验证。文中还给出了FIFO IP核的调用方法，完整呈现了从理论到实现的均值滤波处理流程。关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 Always_away 博客等级码龄3年 47 原创735 点赞 404 收藏 272 粉丝关注私信 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告 TA的精选新数学分析\| 极限论\| 1.数列极限常用方法总结 518 阅读新 26考研\|高等代数：基础阶段总结 302 阅读热数据库系统概论\|第五章：数据库完整性—课程笔记 2411 阅读热 26考研\|数学分析：重难突破—间断点的类型 2188 阅读热数据库系统概论\|第二章：关系数据库—课程笔记3 1935 阅读查看更多 2025 08月 5篇 07月 2篇 06月 4篇 05月 7篇 04月 9篇 03月 8篇 02月 12篇大家在看基于SpringBoot的人力资源管理系统（源代码+文档+PPT+调试+讲解）基于单片机的智能照明控制系统（论文+源码）梦幻西游云手机网页版畅玩优势 VMware Workstation安装CentOS 7.9虚拟机计算机行业从业者 \| “长期主义” 的健康习惯 414 分类专栏 2026考研\|数学分析19篇数学分析强化专题1篇 2026考研\|高等代数10篇高等代数强化专题数据库系统概论15篇上一篇： 26考研\|数学分析：导数和微分下一篇： 26考研\|高等代数：矩阵目录课本简单概括定理部分应用部分应用一：单调性、单调区间（#6.1节）应用二：极值、最值（#6.4节）应用三：凹凸性、保凸区间（#6.5节）应用四：极限求解（#6.2节、#6.3节）应用五：函数的图象（#6.6节）应用六：方程的近似解（#6.7节）课本经典习题展开全部收起目录课本简单概括定理部分应用部分应用一：单调性、单调区间（#6.1节）应用二：极值、最值（#6.4节）应用三：凹凸性、保凸区间（#6.5节）应用四：极限求解（#6.2节、#6.3节）应用五：函数的图象（#6.6节）应用六：方程的近似解（#6.7节）课本经典习题展开全部收起 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告上一篇： 26考研\|数学分析：导数和微分下一篇： 26考研\|高等代数：矩阵分类专栏 2026考研\|数学分析19篇数学分析强化专题1篇 2026考研\|高等代数10篇高等代数强化专题数据库系统概论15篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
6231	https://es.wikipedia.org/wiki/Gratuidad	Gratuidad - Wikipedia, la enciclopedia libre Ir al contenido [x] Menú principal Menú principal mover a la barra lateral ocultar Navegación Portada Portal de la comunidad Actualidad Cambios recientes Páginas nuevas Página aleatoria Ayuda Notificar un error Páginas especiales Buscar Buscar [x] Apariencia Apariencia mover a la barra lateral ocultar Texto Pequeño Estándar Grande Esta página siempre usa texto pequeño Anchura Estándar Ancho Esta página siempre es ancha Color (beta) Automático Claro Oscuro Esta página está siempre en modo claro Donaciones Crear una cuenta Acceder [x] Herramientas personales Donaciones Crear una cuenta Acceder Páginas para editores desconectados más información Contribuciones Discusión [x] Cambiar a la tabla de contenidos Contenidos mover a la barra lateral ocultar Inicio 1 Etimología 2 EconomíaAlternar subsección Economía 2.1 TANSTAAFL 3 Gratuidad de los sacramentos 4 Cultura de la gratuidad 5 Gratis contra libre 6 Véase también 7 Referencias Gratuidad [x] 17 idiomas Afrikaans العربية Deutsch Euskara Français עברית हिन्दी Italiano 日本語 한국어 Bahasa Melayu Nederlands Svenska ייִדיש 中文閩南語 / Bân-lâm-gí 粵語 Editar enlaces Artículo Discusión [x] español Leer Editar Ver historial [x] Herramientas Herramientas mover a la barra lateral ocultar Acciones Leer Editar Ver historial General Lo que enlaza aquí Cambios en enlazadas Subir archivo Enlace permanente Información de la página Citar esta página Obtener URL acortado Descargar código QR Editar enlaces interlingüísticos Imprimir/exportar Crear un libro Descargar como PDF Versión para imprimir En otros proyectos Wikimedia Commons Elemento de Wikidata De Wikipedia, la enciclopedia libre Freebox (centro de distribución gratuita de material donado) en Berlín. Promoción de cerveza. Muestra gratuita de mantequilla de cacahuete. "Sopa, café y rosquillas gratis para los desempleados". La pagaba Al Capone (Chicago, 1931). Comida gratuita durante una festividad religiosa en la India. Fuente de vino para los peregrinos del Camino de Santiago. En el cartel se indica: "A beber sin abusar / le invitamos con agrado / para poderlo llevar / el vino ha de ser comprado". Gratuidad es un concepto económico: la dispensación de un bien o un servicio sin contraprestación o contrapartida aparente por parte del beneficiario, especialmente cuando no hay precio o este no se sustancia en un pago o cargo pecuniario que este haya de afrontar. Etimología [editar] Gratuidad, del francés gratuité, y este del latín medievalgratuitas, -atis «favor» - Cualidad de gratuito". Gratis (del latín gratis, contracción de gratiis "por gratitud, a cambio de nada", ablativo de gratiae, plural de gratia) comparte etimología con las distintas acepciones de los términos "gracias" y "gracia" ("favor, estima, cualidad agradable, buen deseo, gratitud", que en latín eclesiástico se usa para traducir el término griego χάρισμα -kharisma-, derivado de gratus -"grato, agradable, placentero", de la raíz preindoeuropea gwreto-, forma en sufijo de la raíz gwere- -"favorecer"-). La expresión latina gratis pro Deo ("gratuitamente por el amor de Dios"), utilizada desde la Edad Media, se emplea en la actualidad con sentido irónico. La primera acepción del DRAE para la palabra "gratuito" es "de balde o de gracia"; recogiéndose como segunda acepción "arbitrario, sin fundamento", proponiéndose como ejemplos las expresiones "acusación gratuita" y "suposición gratuita". Economía [editar] Es una noción ambigua, ya que en realidad ningún bien económico es gratuito socialmente, puesto que todos tienen algún coste de producción o alguna externalidad, y por tanto alguien tiene que asumirlos, aunque no los pague directamente el consumidor (en el caso de que un bien no tenga coste alguno para nadie, no es un bien económico sino uno de los llamados bienes libres). También ocurre que el que disfruta de un bien o servicio gratuito afronta al menos un coste de oportunidad (diferencia entre el valor del bien obtenido y el del bien al que se renuncia por el hecho de aceptarlo). Muchos conceptos, financiados de muy distintos modos, se confunden en el concepto de lo "gratuito": los artículos promocionales, lo comprendido en el precio (venta por lotes, tarifa plana, buffet libre), lo pagado por la molestia (por ejemplo, el acceso a ciertos contenidos a cambio de soportar publicidad -habitual en radiodifusión, televisión e internet-), lo pagado por los impuestos (servicios sociales, servicios públicos), las donaciones, los regalos (economía del don), las invitaciones o convites, los donativos, las limosnas, la ayuda humanitaria, algunos conceptos médicos (como los órganos para trasplante o algunos casos de medicina de urgencia), algunos seguros en caso de catástrofe (que se generalizan incluso a los que no han pagado previamente la cuota de un seguro), etc. Los bienes comunes y los bienes públicos pueden ser o no de acceso gratuito. TANSTAAFL [editar] Robert A. Heinlein popularizó en 1966 (en su novela La Luna es una cruel amante) el adagio preexistente (atestiguado al menos desde 1938) There Ain't No Such Thing As A Free Lunch ("no hay tal cosa como un almuerzo gratis" -equivalente a las expresiones castellanas "nadie da nada por nada" o "nadie da duros a peseta"-), reducida al acrónimo TANSTAAFL. Retomado por Milton Friedman, se utiliza en manuales de economía. Gratuidad de los sacramentos [editar] La cuestión de la gratuidad de los sacramentos en el cristianismo implica el pecado de la simonía cuando no se cumple. Cultura de la gratuidad [editar] Frente al contenido de pago, la cultura de la gratuidad (gratiskultur) es la que predomina en Internet. En palabras de Chris Anderson: "quien quiera estar presente en la Red, tiene que ofrecer su contenido de forma gratuita". Según una encuesta de 2001 por Media Metrix, el 69 por ciento de los consumidores no está dispuesto a pagar por el contenido en Internet. En muy distintos sectores económicos, la política de precios (pricing) puede incluir distintas formas que se aproximan a la gratuidad en distintos casos. Véanse también:Freemium, Paga lo que quieras, Paga lo que puedas, Pay what you want y Pay what you can. Gratis contra libre [editar] Artículo principal:Gratis contra libre Véase también [editar] Gratis (Ohio) Título gratuito Contrato gratuito Justicia gratuita Educación gratuita Sanidad gratuita Transporte gratuito Alimentación gratuita o comida gratuita Panem et circenses Comedor social Muestra gratuita Prensa gratuita Servicio de alojamiento gratuito (no debe confundirse con vivienda gratuita -véase derecho a la vivienda-) Software gratuito Número gratuito 900 Merced No venal (especialmente en el ámbito editorial: edición no venal) Sin ánimo de lucro Referencias [editar] ↑Real Academia Española. «gratuidad». Diccionario de la lengua española (23.ª edición). ↑"Gratis" en Online Etymology Dictionary - "Grace" en Online Etymology Dictionary. ↑Larousse ↑"Gratuitamente, sin coste alguno / En vano / Sin motivo, sin causa". Real Academia Española. «balde». Diccionario de la lengua española (23.ª edición). ↑"Gratuitamente, sin premio ni interés alguno". Real Academia Española. «gracia». Diccionario de la lengua española (23.ª edición). ↑Real Academia Española. «gratuito». Diccionario de la lengua española (23.ª edición). ↑"Donativo - Del latín donatīvum - Dádiva, regalo, cesión, especialmente con fines benéficos o humanitarios". Real Academia Española. «donativo». Diccionario de la lengua española (23.ª edición). "Dádiva - Del bajo latíndativa, plural neutro de dativum 'donativo', con influencia del latín debĭta 'deudas' - Acción de dar gratuitamente / Cosa que se da gratuitamente". Real Academia Española. «dádiva». Diccionario de la lengua española (23.ª edición). ↑ La gratuité totale des transports collectifs urbains: effets sur la fréquentation et intérêts, Bruno Cordier, Dans le cadre du PREDIT 3 (Programme de Recherche et d’Innovation dans les Transports Terrestres) Groupe opérationnel 11: Politique des transports. Texte gratuitement consultable sur internet De la gratuité, essais deJean-Louis Sagot-Duvauroux, éditions de l'Éclat. Texte gratuitement consultable sur internet ↑Fred Shapiro, Quotes Uncovered: The Punchline, Please, The New York Times, Freakonomics blog, 16 de julio de 2009. ↑Bruck, fuente citada en Paid content ↑Anarchie im Web: Die Gratis-Kultur. In Der Spiegel – Spiegel Special, 26. Juni 2007. Fuente citada en Gratiskultur ↑Medienkonzerne: Das Ende der Gratiskultur im Internet ist gekommen, Wirtschaftswoche vom 9. November 2011 ↑«La palabra ‘gratis’ se aleja de Internet». CPN Radio. Archivado desde el original el 3 de mayo de 2002. Consultado el 6 de enero de 2025. ↑William Poundstone, Priceless: The Myth of Fair Value (and How to Take Advantage of It), Hill and Wang, 2010. Fuente citada en en:Pricing. Conceptos económicos en torno al precio son también los de sistema de precios, estrategia de precios y otros. ↑Strom, fuente citada en en:Pay what you want - Pay what you can ↑Redirige a derecho a la educación. The Swedish School System, fuente citada en en:Free education ↑Perone, fuente citada en en:Free public transport ↑Véase también en:Free Food for Millionaires. No debe confundirse con en:Unionist Free Food League ↑De Bower, fuente citada en:Free sample. \| Control de autoridades \| Proyectos Wikimedia Datos:Q1543615 Multimedia:Zero prices / Q1543615 \| Datos:Q1543615 Multimedia:Zero prices / Q1543615 Obtenido de « Categoría: Economía Esta página se editó por última vez el 6 ene 2025 a las 00:21. El texto está disponible bajo la Licencia Creative Commons Atribución-CompartirIgual 4.0; pueden aplicarse cláusulas adicionales. Al usar este sitio aceptas nuestros términos de uso y nuestra política de privacidad. Wikipedia® es una marca registrada de la Fundación Wikimedia, una organización sin ánimo de lucro. Política de privacidad Acerca de Wikipedia Limitación de responsabilidad Código de conducta Desarrolladores Estadísticas Declaración de cookies Versión para móviles Edita configuración de previsualizaciones Buscar Buscar [x] Cambiar a la tabla de contenidos Gratuidad 17 idiomasAñadir tema
6232	https://pmc.ncbi.nlm.nih.gov/articles/PMC4206657/	Imaging in gout: A review of the recent developments - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Ther Adv Musculoskelet Dis . 2014 Aug;6(4):131–143. doi: 10.1177/1759720X14542960 Search in PMC Search in PubMed View in NLM Catalog Add to search Imaging in gout: A review of the recent developments Priya Varghese Chowalloor Priya Varghese Chowalloor 1 School of Medicine and Pharmacology, The University of Western Australia, 35 Stirling Highway, Crawley, Perth, Western Australia 6009, Australia Find articles by Priya Varghese Chowalloor 1,✉, Teck K Siew Teck K Siew 2 Diagnostic and Interventional Radiology, Royal Perth Hospital, Perth, WA, Australia Find articles by Teck K Siew 2, Helen Isobel Keen Helen Isobel Keen 3 School of Medicine and Pharmacology, University of Western Australia, Crawley, WA, Australia Find articles by Helen Isobel Keen 3 Author information Article notes Copyright and License information 1 School of Medicine and Pharmacology, The University of Western Australia, 35 Stirling Highway, Crawley, Perth, Western Australia 6009, Australia 2 Diagnostic and Interventional Radiology, Royal Perth Hospital, Perth, WA, Australia 3 School of Medicine and Pharmacology, University of Western Australia, Crawley, WA, Australia ✉ Email: priyachowalloor@outlook.com ✉ Corresponding author. Issue date 2014 Aug. © The Author(s), 2014 PMC Copyright notice PMCID: PMC4206657 PMID: 25342993 Abstract Gout is a common inflammatory arthritis and is caused by accumulation of monosodium urate crystals in joints and soft tissues. Apart from joint damage, untreated gout is associated with cardiovascular and renal morbidity. Gout, whilst in principle considered to be well understood and simple to treat, often presents diagnostic and management challenges, with evidence to suggest that it is often inadequately treated and poor compliance is a major issue. Imaging tools can aid clinicians in establishing the correct diagnosis, when histological crystal diagnosis is unable to be established, and also assess the burden of inflammatory and structural disease. Imaging can also be used to monitor treatment response. The imaging techniques that currently have a role in the imaging of gout include conventional radiography, ultrasound, computed tomography, dual energy computed tomography, magnetic resonance imaging and nuclear medicine. Despite the lack of major technological advances in imaging of gout in recent years, scientific studies of existing imaging modalities have improved our understanding of the disease, and how to best utilize imaging techniques in the clinical setting. Keywords: asymptomatic hyperuricemia, computed tomography, conventional radiography, dual energy computed tomography, erosions, magnetic resonance imaging, nuclear medicine, plane X-ray, tophus, ultrasound Introduction Gout is common, indeed the most common inflammatory arthritis in men [Richette and Bardin, 2010]. It is associated with mortality and significant morbidity. In addition to being a chronic deforming polyarthropathy, gout is associated with metabolic, cardiovascular and renal morbidity [Kim et al. 2003; Brook et al. 2006; Chen et al. 2007]. Gout is perhaps one of the best understood diseases in terms of aetiopathogenesis and effective treatments do exist. Despite this, the scientific evidence suggests that often patients and doctors do not understand the disease well [Pascual and Sivera, 2007]. Delays in establishing the diagnosis, commencing treatment, and targeting defined outcomes as well as poor compliance, contribute to suboptimal management of gout [Jackson et al. 2012; Riedel et al. 2004; Zhang et al. 2006b; Briesacher et al. 2008]. Gout results from the deposition of monosodium urate (MSU) crystals in joints and soft tissues when serum uric acid concentrations rise above the physiological saturation limit (approximately 380 µmol/liter or 6.4 mg/dl) [Choi et al. 2005]. The acute form of the arthritis is characterized by sudden onset, intense pain, swelling, warmth and erythema (the cardinal signs of inflammation). The great toe is characteristically affected, however almost all joints may be affected [Edwards, 2008]. The diagnosis is confirmed in acute gout by the presence of MSU crystals in the synovial fluid. Chronic tophaceous gout results from chronic hyperuricemia. Continued deposition of MSU crystals leads to increased frequency of acute attacks, progressive shortening of intercritical phase and development of tophi due to MSU deposition in soft tissues, bones and joints. The gold standard for diagnosis of gout is demonstration of negatively birefringent, needle-shaped MSU crystals in tissue or synovial fluid by polarized microscopy [Wallace et al. 1977]. Obtaining a histological diagnosis is not always feasible, and whilst application of international consensus definitions may assist in the diagnosis of gout in the absence of a crystal diagnosis, at times a definitive clinical diagnosis can be difficult [Wallace et al. 1977; Zhang et al. 2006b]. The management of gout should be holistic, incorporating patient education, lifestyle advice, pharmacotherapy of acute gout, prophylaxis to prevent chronic tophaceous gout, identification and management of comorbidities such as metabolic syndrome and renal disease [Zhang et al. 2006a; Khanna et al. 2012]. International guidelines recommend that prior to treating gout, the diagnosis must be accurately established, and the burden of disease should also be assessed [Zhang et al. 2006a; Khanna et al. 2012]. Imaging may play a useful role in this, particularly when uric acid crystals are unable to be identified to confirm the diagnosis. Imaging can assist with assessing disease burden and structural damage. Imaging can also be useful to monitor disease progression or treatment response and to assess efficacy of treatment in clinical trials. Imaging modalities that have clinical relevance in gout include conventional radiography (CR), ultrasonography (US), computed tomography (CT), dual energy computed tomography (DECT), magnetic resonance imaging (MRI) and nuclear medicine. Typically, imaging findings associated with joint inflammation are seen, as well as findings that are more specific to and even pathognomonic of gout. This manuscript aims to review imaging findings seen in gout, with a focus on recent developments. While there have been few recent technological imaging advances apart from DECT, application and understanding of the clinical utility of these imaging techniques in gout have been better understood, as has our understanding of the pathogenesis of gout. Conventional radiography CR has been the traditional imaging tool in the management of rheumatic disorders. During an acute attack of gout, soft tissue swelling and effusions may be seen by CR [Gentili, 2006], however these findings are nonspecific. The typical CR findings in chronic tophaceous gout, which differentiate it from other inflammatory arthritides, include well defined, ‘punched-out’ erosions with overhanging edges, soft tissue nodules (tophi), calcification of tophi and asymmetric involvement [Gentili, 2006] (Figure 1). The erosions are typically extra-articular, but may be intra-articular or para-articular [Gentili, 2006]. Tophi may be intraosseous or calcified. The joint space is usually preserved until late in the disease and there is lack of periarticular osteopenia. The most common site affected is the first metatarsophalangeal (MTP) joint, followed by the fifth MTP joint, midfoot and hand and wrist [Barthelemy et al. 1984]. These CR changes are relatively specific, with a diagnostic specificity of 93% using clinical diagnosis as the gold standard [Rettenbacher et al. 2008]. The diagnostic sensitivity is lower; 31% using clinical diagnosis as the gold standard [Rettenbacher et al. 2008]. This is particularly an issue in early disease, as radiographic changes may be delayed 10–15 years after the onset of gout [Barthelemy et al. 1984]. In the clinical setting, the low sensitivity of CR and the lag to developing radiographic changes means that CR has a limited role in the diagnosis or monitoring of this disease [Gentili, 2006]. Figure 1. Open in a new tab Conventional radiography of both hands showing multiple punched out erosions (arrowheads) and soft tissue densities (tophi; arrows) typical of gout. In the trial setting, CR has been utilized as an outcome tool. A radiographic scoring system developed for rheumatoid arthritis (RA) clinical trials (Sharp/van der Heijde system) has been modified to apply to gout [Dalbeth et al. 2007b]. This measures joint space narrowing and erosions in the small joints of the hands and feet. In contrast to the system developed for RA, this tool scores the distal interphalangeal joints. Initial validation work has demonstrated this scoring system to have excellent reproducibility and feasibility [Dalbeth et al. 2007b]. Responsiveness of this scoring system has not been tested extensively, however a recent small exploratory study showed a significant improvement in the erosion subscore using the modified Sharp/van der Heijde system in people with tophaceous gout with intensive urate-lowering therapy using pegloticase over a 12-month period [Dalbeth et al. 2013a]. The study, although small, is pivotal as it demonstrates healing of erosions in response to aggressive urate lowering, and also supports the as yet scientifically unproven hypothesis that suppression of uric acid will prevent structural joint damage [Dalbeth et al. 2013a]. CR may be a useful outcome tool in clinical trials as it is widely accessible and inexpensive, and while this is an old and simple imaging technique, this recent study showing erosion healing demonstrates that CR modality can still add to our understanding of this disease and assessment of outcomes in gout. Ultrasonography US is being used increasingly in gout and can assist in both the diagnosis and monitoring of disease. Advantages of US over other imaging modalities include the ease of access, lack of ionizing radiation and relatively low cost. US however has limitations, being reliant upon a good acoustic window to visualize a joint, and is generally less sensitive than MRI in detecting joint inflammation and structural changes [Chowalloor and Keen, 2013b]. The major limitation is its operator-dependent nature. Generic signs of joint inflammation and damage identifiable by US include synovitis and erosions. Synovitis on US is identified as synovial hypertrophy and effusion, with or without Doppler signal (suggestive of inflammation) [Wakefield et al. 2005]. Erosions are cortical breaks seen in two planes (as defined by Outcome Measures in Rheumatology Clinical Trials, OMERACT) [Wakefield et al. 2005]. More specific features of gout are also seen, such as the double contour sign (DC) and tophi [Thiele and Schlesinger, 2007; Filippucci et al. 2010a]. While recent decades have seen the validity and clinical utility of US in the setting of RA extensively explored [Terslev et al. 2012], there has been much less work specific to gout. Interpreting publications relating to US descriptions of gout are difficult due to a lack of internationally recognized descriptions and definitions of pathology seen in gout on US [Chowalloor and Keen, 2013b]. This is illustrated in the discussion of pathology in the paragraphs below. The OMERCT US group are working towards standardized definitions of pathology in gout, which should facilitate the development of this imaging modality as a clinical and outcome tool in gout in the future. In RA, US has been demonstrated to be sensitive and specific to the presence of these generic features and able to detect subclinical disease [Naredo et al. 2013b]. In the setting of gout, the synovium may have an ultrasound appearance thought to be more suggestive of gout than other inflammatory arthritis. Reported descriptions include ‘bright stippled foci’ and ‘hyper-echoic spots’, ‘hyper-echoic cloudy areas’ and a ‘snow storm’ appearance, which is thought to be a result of MSU crystals in synovial fluid or tissue producing small bright echoes [Wright et al. 2007; Rettenbacher et al. 2008; De Miguel et al. 2012] (Figure 2). While these findings have generally been considered as specific to people with a diagnosis of gout and asymptomatic hyperuricemia (AH) [Rettenbacher et al. 2008], ‘very small intra-articular hyper-echoic’ spots and ‘intra-articular or intra-bursal hyper-echoic aggregates’ have been reported to be seen in other types of arthritis [Wright et al. 2007]. More recently, a case-controlled multicenter study found that these were more commonly seen in gout, but are not specific to gout [Naredo et al. 2013a]. Additionally the “snowstorm” of synovial fluid, thought to be mobile crystals within the fluid described in acute gout [Grassi et al. 2006] may not be specific to gout [Wakefield, 2007]. Recent exploratory studies in gout reports that synovitis is most commonly seen in the first MTP joint, knee, ankle, wrist and second metacarpophalangeal (MCP) joint [Chowalloor and Keen, 2013a]. Other recent, small studies on gout established the presence of subclinical synovitis in gout in both the acute and intercritical phase [Schueller-Weidekamm et al. 2007; Chowalloor and Keen, 2012]. Of interest, in the single US study assessing subclinical synovitis longitudinally, the number of clinically active joints decreased in the intercritical phase, but subclinical inflammation did not differ between the acute and intercritical phases [Chowalloor and Keen, 2013a]. This has potential implications in the assessment of disease activity and burden of disease in gout. It is worth noting that in this small study, the serum uric acid level was not adequately suppressed, and that the long-term relevance of subclinical inflammation with regards to structural joint damage or comorbidities is uncertain. Figure 2. Open in a new tab Ultrasound of ring finger, dorsal distal interphalangeal joint longitudinal view, showing joint effusion (indicated with a cross) and hyper-echoic spots (arrow) suggestive of monosodium urate crystal deposition, the ‘snow-storm’ appearance. Ultrasound-detected erosions in the setting of gout are found most commonly in the first MTP joint (especially medial surface) and MCP joints [Thiele and Schlesinger, 2007; Wright et al. 2007; Filippucci et al. 2010a]. First MTP erosions are more characteristic of gout than other inflammatory arthritides [Wright et al. 2007]. Erosions may be found in previously asymptomatic MTP joints [Chowalloor and Keen, 2013b], they are commonly seen adjacent to tophi, and they may display Doppler signal [Wright et al. 2007] (Figure 3). In gout, US is more sensitive to erosions than CR, particularly smaller erosions [Wright et al. 2007]. Figure 3. Open in a new tab Ultrasound of left first metatarsophalangeal joint, longitudinal view, demonstrating intra-articular tophaceous material (arrows) and erosions (arrowheads). The DC sign is a hyper-echoic irregular band over the articular cartilage due to the deposition of MSU crystals, best seen on the dorsal side of the MTP joints [Filippucci et al. 2010b] and femoral condyles [Naredo et al. 2013a] (Figure 4). In most studies comparing subjects with gout or AH with controls (either patients with other inflammatory joint diseases or normouricemic individuals) the DC sign is reported to be specific (but not sensitive) to AH and gout [Filippucci et al. 2009; Pineda et al. 2011; De Miguel et al. 2012; Ottaviani et al. 2012]. In contrast, a recent study examining subjects with gout and controls found the DC sign was found in controls [Ottaviani et al. 2012]. The reported serum urate in the group ranged from 2.3 to 7.6 mg/dl, so some of the control cohort presumably had AH. Other explanations include differences in methodology, joints examined, the demographics of the population, or that US is a user-dependent technique, and that the cartilage interface may produce an artefact that may be mistaken for a DC. The validity of the DC sign in gout has not been compared with other imaging modalities [Chowalloor and Keen, 2013b], however in AH, the majority of joints with US-detected DC sign or hyper-echoic cloudy area demonstrated MSU crystals on joint aspirate [De Miguel et al. 2012]. Figure 4. Open in a new tab Ultrasound of ankle joint, longitudinal view, demonstrating the double contour sign, formed by monosodium urate crystal deposition across the top of the talar cartilage (arrows). Tophi are typical clinical features of gout and have been variably described in US studies [Wright et al. 2007; Ottaviani et al. 2010; De Ávila Fernandes et al. 2011; Howard et al. 2011; Pineda et al. 2011; De Miguel et al. 2012]. Tophi can be seen in various locations, such as within the joint, burse, in relation to the tendons, ligaments and in other soft tissues (Figure 3). Various descriptions of tophi are referred to in the literature, including ‘hyper-echoic heterogeneous soft tissue deposit with or without post echoic shadowing’, ‘iso-echoic/hyper-echoic nodular deposits’, ‘bright spots’ and ‘hyper-echoic areas’ [Chowalloor and Keen, 2013b]. A recent study reported that common intra-articular sites for tophaceous deposits include the first MTP joint, radiocarpal joint, midcarpal joint and knees [Naredo et al. 2013a]. The most common tendinous locations for tophi were the patellar and triceps tendon, followed by the quadriceps and Achilles tendons [Naredo et al. 2013a]. Naredo and colleagues tested the diagnostic value of US by extensively systematically examining a large number of joints, including cartilage, bursae, ligaments and tendons in subjects with gout and controls [Naredo et al. 2013a]. The study focused on US signs relatively specific for gout, such as the DC sign and hyper-echoic aggregates, and aimed to determine the optimal combination of pathologies and locations to assist in the diagnosis of gout. The results demonstrate that imaging the radiocarpal joint and the patella and triceps tendon for evidence of hyper-echoic aggregates, and the first metatarsal, talar and second metacarpal or femoral cartilage for the DC signs produced a sensitivity of 84.6% and specificity of 83.3%. The specificity of individual lesions, such as the DC sign, was not as high in this study as reported in other studies; however, the presence of hyper-echoic aggregates in both the patella and triceps tendon was highly specific for gout in this study. In clinical practice, US does add to the diagnostic certainty, but should be considered an adjuvant to the history, examination and biochemical findings. Figure 5. Open in a new tab Computed tomography scan showing tophaceous calcification in the gluteal tendon and trochanteric bursa (arrows). Ultrasound can be used to monitor response to therapy. The resolution of tophi and the DC sign have been demonstrated in response to urate-lowering therapy [Perez-Ruiz et al. 2007; Thiele and Schlesinger, 2010]. Computed tomography CT is not commonly utilized in the clinical management of gout. It is associated with ionizing radiation, and until recently, conferred little added benefit over CR, US and MRI. There has been some investigation into the utility of imaging tophi in gout, particularly with regards to differentiating tophus from other subcutaneous nodules noninvasively [Gerster et al. 1998; Gentili, 2006]. Tophus appears on CT as hyperdense lesions, with their specific density allowing differentiation from other hyperdense lesions [Gerster et al. 1996]. Soft tissue lesions have lower attenuation and calcifications have higher attenuation than that of tophus [Gentili, 2006]. CT can assess deep intra-articular and intraosseous tophi [Gerster et al. 1996], which may be undetectable by clinical or US examination. CT can also measure tophus volumes with excellent reproducibility [Dalbeth et al. 2007a], which has been used in clinical trials as an endpoint. However, the evidence suggests that if a tophus is able to be measured physically in the clinical setting with calipers, then this is an equivalent tool [Dalbeth et al. 2007a]. CT can identify gouty erosions. Dalbeth and colleagues recently demonstrated that CT erosions in gout are closely related to tophi, suggesting that tophus infiltration may have a pathogenic role in the development of erosions in gout [Dalbeth et al. 2009a]. The same group demonstrated a strong association between erosions and new bone formation (sclerosis, osteophytes and spurs), suggesting a relationship between bone resorption and new bone formation in gout [Dalbeth et al. 2012b]. A CT scoring method for quantifying bone erosions at the feet in gout has been developed and validated [Dalbeth et al. 2011]. Preliminary investigation has demonstrated sensitivity and reliability, and while responsiveness is yet to be established, this tool may assist in monitoring structural progression in gout in clinical studies. Arguably, the most exciting imaging advance in gout in recent years is the advent of DECT, which is able to detect MSU burden noninvasively. In addition to identifying crystals, advantages include shorter scanning times, the ability to scan multiple joints simultaneously and have excellent reproducible visualization compared with US. This technology was developed to determine the chemical composition of renal stones and atherosclerotic coronary plaques, but has been recognized to have utility in the clinical setting of diagnosing and managing gout. This technique relies on acquiring two datasets simultaneously through the use of two separate X-ray tubes producing differing energy levels [Nicolaou et al. 2010]. Differences in attenuation are able to be identified and color coded, allowing material rich in calcium (high attenuation) to be differentiated from material rich in MSU crystals (low attenuation) [Nicolaou et al. 2010]. DECT imaging is useful in identifying subclinical urate burden. MSU deposition can also be seen in joints, tendons, ligaments and soft tissues (Figure 6). When tophi identified by DECT imaging was compared with that of physical examination [Choi et al. 2009], DECT was able to pick up four times the tophi picked up by physical examination. The most common sites involved on DECT are MTP joints (85%), knees (85%) and ankles (70%), followed by wrists (50%), MCP joints and elbows (40%) [Choi et al. 2009]. The most commonly involved tendon/ligament site is the Achilles, followed by peroneal tendons [Dalbeth et al. 2013b]. Figure 6. Open in a new tab Dual energy computed tomography images of ankle and foot (a, b) showing monosodium urate crystal deposition in green color (arrows) and corresponding computed tomography (c, d) shows erosions (arrowheads). DECT has been shown to be highly sensitive and specific to the presence of MSU crystals, using aspirate of MSU crystals as the gold standard [Glazebrook et al. 2011]. Additionally, findings of MSU by DECT are able to differentiate between patients with well established gout and controls, with excellent sensitivity (78–84%) and specificity (93%) [Choi et al. 2012]. In studies with significant false negatives, the cohorts of patients are generally on urate-lowering therapies with serum urate concentrations below the physiological saturation threshold, which may affect the burden of MSU deposits in these patients. In addition to urate-lowering therapies, disease duration may also affect the sensitivity of DECT to detect gout, presumably related to the burden of deposited MSU crystals (the minimal detectable deposit size is 2 mm) [Glazebrook et al. 2012]. In addition to limitations in detecting small deposits, a recent case report illustrated that even large tophaceous deposits may be missed if they are not particularly dense [Melzer et al. 2014]. In assessing the utility of DECT compared with US, detection of MSU by DECT has been shown to have a similar sensitivity to the US detected DC sign [Gruber et al. 2014]. However, the sensitivity and specificity of DECT are dependent on the imaging protocol utilized. A recent study in assessing tophi compared DECT imaging with MRI scanning and found that results can vary depending on the software setting for DECT [McQueen et al. 2013]. When images were reconstructed with two different parameter ratios for DECT imaging, MRI only correlated with one DECT image and the other DECT image did not identify any tophi. Therefore it is important to standardize software settings to minimize these errors. The diagnostic sensitivity of DECT is much less in the absence of chronic gout, that is, in new presentations. For example, in one small study of 14 subjects with acute, nontophaceous gout, DECT identified MSU deposits in 78% of subjects, but only 50% of subjects presenting with an inaugural attack [Manger et al. 2012]. Thus, a negative DECT in the setting of an acute initial presentation of presumed gout is unhelpful to the clinician, as this is the situation when a clinical adjuvant tool is perhaps most needed. The reproducibility of DECT is very good [Glazebrook et al. 2011]. Excellent inter- and intra-observer reproducibility has been demonstrated [Choi et al. 2012] for tophus volume. DECT is more reproducible than physical methods (calipers or tape measurements) in assessing tophus size [Dalbeth et al. 2012a]. DECT may be responsive to change, with a reduction in the burden of MSU in response to treatment in multiple case reports [Desai et al. 2011; Bacani et al. 2012]. However, in the setting of longstanding stable gout treated with urate-lowering therapy, the measurable urate burden has been shown to be below the smallest detectable change, limiting the utility of DECT in monitoring therapeutic response [Rajan et al. 2013b]. Magnetic resonance imaging This is an excellent imaging modality to image synovium, cartilage, soft tissue and bone, as it lacks radiation and has excellent contrast and resolution. However, limitations include high cost, availability, long scanning time, use of contrast, patient acceptability, and exclusion of those patients with aneurysm clips or pacemaker. MRI can demonstrate generic features of inflammatory arthritis, such as synovial thickening, effusion, bone erosions, and bone marrow edema (BME) in gout [Popp et al. 1996; Cimmino et al. 2011] (Figures 7 and 8). MRI can demonstrate subclinical inflammation in asymptomatic joints in gout [Carter et al. 2009]. MRI is better than US and CR in detecting erosions [Carter et al. 2009]. A clue to the diagnosis of gout in the setting of generic inflammatory changes is that BME is uncommon and if present is often mild [McQueen et al. 2013], the presence of extensive BME on MRI should raise the question of infection. A retrospective review of patients with gout showed that severe BME on the MRI was much more common in gout plus osteomyelitis than in uncomplicated gout [Poh et al. 2011]. Figure 7. Open in a new tab Magnetic resonance imaging scans of the wrist at the dorsal ulnar aspect: (a) T2 fat-saturated transverse; (b) T1 transverse; and (c) T1 fat-saturated with gadolinium. The scans show a well circumscribed mixed soft tissue calcified mass (open arrow) which represents calcification of tophus with some gadolinium enhancement of soft tissue surrounding the lesion (filled arrow) and periarticular erosions (arrowheads) suggestive of gouty tophus. Figure 8. Open in a new tab Magnetic resonance imaging scans of the cervical spine: (a) T1 weighted; (b) T2 weighted. The scans demonstrate calcification and erosion of the odontoid peg (arrowhead). Tophi can have various appearances on MRI [Dhanda et al. 2011]. T1-weighted images characteristically show homogeneous, low to intermediate signal intensity, and variable signal intensity is shown on T2-weighted images depending on the degree of hydration of the tophus and its calcium concentration. Heterogeneous, intermediate to low signal is the most common pattern on T2-weighted images. Tophi show intense gadolinium enhancement due to hypervascular soft tissue and granulation tissue that surround tophi (Figure 7). MRI has been compared with DECT scanning for tophus detection [McQueen et al. 2013]. Compared with DECT, MRI had high specificity and moderate sensitivity for detecting tophi [Schumacher et al. 2006]. A recent study assessing the reproducibility of MRI scoring for assessment of erosion, tophus size, synovitis and bone edema [McQueen et al. 2013] found inter-observer reproducibility was high for scoring erosions and tophus size, and was moderate for assessment of synovitis and bone edema. Intra-observer reliability was very high for bone erosion, bone edema, synovitis and tophus size. The sensitivity to change with treatment has not been published for any of the features on MRI in gout. Assessment of tophus size with MRI correlates well with that of US [Perez-Ruiz et al. 2007]. Therefore one can assume that like US, MRI will be able to detect change in tophus size. Overall, MRI has a limited role in disease monitoring due to the high expense and limited availability. While MRI is not a new imaging technique, recent investigations into MRI in gout have altered our understanding of the disease and how the pathogenesis differs from other inflammatory arthritides. For example, a recent study demonstrated that, in contrast to RA, gout erosion was predicted by the presence of tophi, but not synovitis or BME [McQueen et al. 2014]. Nuclear medicine Few systematic publications exist with regards to nuclear imaging in gout. While bone scan has high sensitivity in detecting osseous abnormality, the scintigraphic findings in gout are often nonspecific (Figure 9). The current scientific literature has not shown white cell imaging to be useful in differentiating between infection and the acute inflammatory phase of gout [Palestro et al. 1990; Appelboom et al. 2003]. Case reports of positron emission tomography (PET) combined with CT (PET/CT) in gout showed articular and periarticular fluorodeoxyglucose (FDG) uptake [Steiner and Vijayakumar, 2009; Ito et al. 2012]. Soft tissue FDG uptake corresponding to tophi has also been reported [Popovich et al. 2006]. These findings are not specific for gout. Like with MRI, this may be useful when gout presents in unusual body locations, such as axial skeleton. Figure 9. Open in a new tab Bone scan: (a) blood pool; and (b) delayed bone images. Delayed phase imaging shows increased activity in both the first MTP region and in the mid feet. Conclusion In summary, different imaging modalities are available to assist clinicians with making an accurate diagnosis. Some of the imaging features can aid diagnostically in differentiating gout from other inflammatory arthritis conditions, but generally, these are less useful in early disease when the need is usually greater. CR, US and DECT have the ability to assist in monitoring response to treatment. While others, like CT and MRI, have been demonstrated to aid our understanding of this disease recently, and are likely to continue to be useful in proof-of-concept studies. Recent studies of imaging techniques have improved our understanding of gout and their clinical application. Acknowledgments We are grateful to Dr Brendan Adler, Envision Medical Imaging for providing dual energy computed tomography images. Footnotes Funding: This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors. Conflict of interest statement: The authors declare no conflict of interest in preparing this article. 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Ann Rheum Dis 65: 1301–1311 [DOI] [PMC free article] [PubMed] [Google Scholar] Articles from Therapeutic Advances in Musculoskeletal Disease are provided here courtesy of SAGE Publications ACTIONS View on publisher site PDF (1.3 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Conventional radiography Ultrasonography Computed tomography Magnetic resonance imaging Nuclear medicine Conclusion Acknowledgments Footnotes Contributor Information References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
6233	https://www.tutorchase.com/answers/ib/physics/what-forces-act-on-a-car-turning-on-a-flat-road	What forces act on a car turning on a flat road? \| TutorChase Revision PlatformHire a tutor All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More Revision PlatformHire a tutor Answers IB Physics Article What forces act on a car turning on a flat road? The forces acting on a car turning on a flat road are friction, gravity, and the centripetal force. When a car is turning on a flat road, there are several forces at play. The first is the force of gravity, which pulls the car downwards towards the earth. This force is constant and acts vertically downwards. It is balanced by the normal force, which is the force exerted by the road surface that supports the weight of the car. These two forces ensure the car does not move vertically as it turns. The second force is friction, specifically static friction, between the tyres and the road surface. This frictional force prevents the car from sliding outwards as it turns. The frictional force provides the necessary centripetal force required for circular motion. It acts towards the centre of the circle in which the car is turning. The amount of friction depends on the type of tyres, the condition of the road, and the speed of the car. If the car is moving too fast, the frictional force may not be sufficient to provide the necessary centripetal force, and the car may skid. Understanding the types of energy involved can help explain how these forces interact. The third force is the centripetal force, which is the net force causing the car to move in a circular path. This force is always directed towards the centre of the circle. In the case of a car turning on a flat road, the frictional force provides the centripetal force. The centripetal force is given by the equation F = mv^2/r, where m is the mass of the car, v is the velocity (speed) of the car, and r is the radius of the turn. For a deeper understanding of this force, explore the concept of centripetal force in more detail. IB Physics Tutor Summary: When a car turns on a flat road, it experiences gravity pulling it down, balanced by the road pushing it up. Friction between the tyres and the road stops it from sliding, acting towards the turn's centre, creating the needed centripetal force for circular motion. If the car goes too fast, it might skid due to insufficient friction. These dynamics are governed by Newton's first law of motion, which explains the need for a force to change the motion's state. Additionally, understanding banking and centrifugal force can shed light on how cars manage to turn without losing control at higher speeds. Answered byAaron - Qualified Tutor \| B.A. in Maths IB Physics tutor Study and Practice for Free Trusted by 100,000+ Students Worldwide Achieve Top Grades in your Exams with our Free Resources. Practice Questions, Study Notes, and Past Exam Papers for all Subjects! IB ResourcesA-Level ResourcesGCSE ResourcesIGCSE Resources Need help from an expert? 4.93/5 based on733 reviews in The world’s top online tutoring provider trusted by students, parents, and schools globally. #### Vera PGCE Qualified Teacher \| IB Examiner \| BA World Literature #### Stefan PGCE Qualified Teacher \| Cambridge University \| BA Mathematics #### Flora PGCE Qualified Teacher \| Cambridge University MSci Natural Sciences Hire a tutor Related Physics ib Answers Read All Answers Earn £ as a Student Ambassador! Loading... This website uses cookies to improve your experience, by continuing to browse we’ll assume you’re okay with this. Accept Loading...
6234	https://www.khanacademy.org/math/engageny-geo/geo-1	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
6235	https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts/section/1.10/primary/lesson/domain-and-range-of-a-function-alg-i/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 1.10 Domain and Range of a Function Written by:Andrew Gloag \| Eve Rawley Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Functions A function is a rule for relating two or more variables. For example, the price you pay for phone service may depend on the number of minutes you talk on the phone. We would say that the cost of phone service is a function of the number of minutes you talk. Consider the following situation: Josh goes to an amusement park where he pays $2 per ride. There is a relationship between the number of rides Josh goes on and the total amount he spends that day: To figure out the amount he spends, we multiply the number of rides by two. This rule is an example of a function. Functions usually—but not always—are rules based on mathematical operations. You can think of a function as a box or a machine that contains a mathematical operation. Whatever number we feed into the function box is changed by the given operation, and a new number comes out the other side of the box. When we input different values for the number of rides Josh goes on, we get different values for the amount of money he spends. The input is called the independent variable because its value can be any number. The output is called the dependent variable because its value depends on the input value. Functions usually contain more than one mathematical operation. Here is a situation that is slightly more complicated than the example above: Jason goes to an amusement park where he pays $8 admission and $2 per ride. The following function represents the total amount Jason pays. The rule for this function is "multiply the number of rides by 2 and add 8." When we input different values for the number of rides, we arrive at different outputs (costs). These flow diagrams are useful in visualizing what a function is. However, they are cumbersome to use in practice. In algebra, we use the following short-hand notation instead: input↓ f(x)⏟=y←outputfunctionbox First, we define the variables: x= the number of rides Jason goes on y= the total amount of money Jason spends at the amusement park. So, x represents the input and y represents the output. The notation f() represents the function or the mathematical operations we use on the input to get the output. In the last example, the cost is 2 times the number of rides plus 8. This can be written as a function: f(x)=2x+8 In algebra, the notations y and f(x) are typically used interchangeably. Technically, though, f(x) represents the function itself and y represents the output of the function. Identify the Domain and Range of a Function In the last example, we saw that we can input the number of rides into the function to give us the total cost for going to the amusement park. The set of all values that we can use for the input is called the domain of the function, and the set of all values that the output could turn out to be is called the range of the function. In many situations the domain and range of a function are both simply the set of all real numbers, but this isn’t always the case. Let's look at our amusement park example. Finding the Domain and Range of a Function Find the domain and range of the function that describes the situation: Jason goes to an amusement park where he pays $8 admission and $2 per ride. Here is the function that describes this situation: f(x)=2x+8=y In this function, x is the number of rides and y is the total cost. To find the domain of the function, we need to determine which numbers make sense to use as the input (x). The values have to be zero or positive, because Jason can't go on a negative number of rides. The values have to be integers because, for example, Jason could not go on 2.25 rides. Realistically, there must be a maximum number of rides that Jason can go on because the park closes, he runs out of money, etc. However, since we aren’t given any information about what that maximum might be, we must consider that all non-negative integers are possible values regardless of how big they are. For this function, the domain is the set of all non-negative integers. To find the range of the function we must determine what the values of y will be when we apply the function to the input values. The domain is the set of all non-negative integers: {0, 1, 2, 3, 4, 5, 6, ...}. Next we plug these values into the function for x. If we plug in 0, we get 8; if we plug in 1, we get 10; if we plug in 2, we get 12, and so on, counting by 2s each time. Possible values of y are therefore 8, 10, 12, 14, 16, 18, 20... or in other words all even integers greater than or equal to 8. The range of this function is the set of all even integers greater than or equal to 8. 2. Find the domain and range of the following functions. a) A ball is dropped from a height and it bounces up to 75% of its original height. Let’s define the variables: x= original height y= bounce height A function that describes the situation is y=f(x)=0.75x. x can represent any real value greater than zero, since you can drop a ball from any height greater than zero. A little thought tells us that y can also represent any real value greater than zero. The domain is the set of all real numbers greater than zero. The range is also the set of all real numbers greater than zero. b) y=x2 Since there is no word problem attached to this equation, we can assume that we can use any real number as a value of x. When we square a real number, we always get a non-negative answer, so y can be any non-negative real number. The domain of this function is all real numbers. The range of this function is all non-negative real numbers. In the functions we’ve looked at so far, x is called the independent variable because it can be any of the values from the domain, and y is called the dependent variable because its value depends on x. However, any letters or symbols can be used to represent the dependent and independent variables. Here are three different examples: y=f(x)=3xR=f(w)=3wv=f(t)=3t These expressions all represent the same function: a function where the dependent variable is three times the independent variable. Only the symbols are different. In practice, we usually pick symbols for the dependent and independent variables based on what they represent in the real world—like t for time, d for distance, v for velocity, and so on. But when the variables don’t represent anything in the real world—or even sometimes when they do—we traditionally use y for the dependent variable and x for the independent variable. Make a Table For a Function A table is a very useful way of arranging the data represented by a function. We can match the input and output values and arrange them as a table. For example, the values from Example 1 above can be arranged in a table as follows: x:01 2 3 4 5 6y:8101214161820 A table lets us organize our data in a compact manner. It also provides an easy reference for looking up data, and it gives us a set of coordinate points that we can plot to create a graph of the function. Constructing a Table of Values Make a table of values for the function f(x)=1x. Use the following numbers for input values: -1, -0.5, -0.2, -0.1, -0.01, 0.01, 0.1, 0.2, 0.5, 1. Make a table of values by filling the first row with the input values and the next row with the output values calculated using the given function. \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| x \| -1 \| -0.5 \| -0.2 \| -0.1 \| -0.01 \| 0.01 \| 0.1 \| 0.2 \| 0.5 \| 1 \| \| f(x)=1x \| 1-1 \| 1-0.5 \| 1-0.2 \| 1-0.1 \| 1-0.01 \| 10.01 \| 10.1 \| 10.2 \| 10.5 \| 11 \| \| y \| -1 \| -2 \| -5 \| -10 \| -100 \| 100 \| 10 \| 5 \| 2 \| 1 \| When you’re given a function, you won’t usually be told what input values to use; you’ll need to decide for yourself what values to pick based on what kind of function you’re dealing with. We will discuss how to pick input values throughout these lessons. Example Example 1 Identify the domain and then make a table of values for the function f(x)=1√x. Use the following numbers for input values: 0.01, 0.16, 0.25, 1, 4. Since you cannot compute the square root of negative numbers, these cannot be in the domain. Since we cannot have 0 in the denominator, 0 is also not in the domain. This means that the domain is all real numbers greater than zero. Make a table of values by filling the first row with the input values and the next row with the output values calculated using the given function. \| \| \| \| \| \| \| --- --- --- \| \| x \| 0.01 \| 0.16 \| 0.25 \| 1 \| 4 \| \| f(x)=1x \| 1√0.01 \| 1√0.16 \| 1√0.25 \| 1√1 \| 1√4 \| \| y \| 10 \| 2.5 \| 2 \| 1 \| 0.5 \| Review For 1-6, identify the domain and range of the following functions. Dustin charges $10 per hour for mowing lawns. Maria charges $25 per hour for tutoring math, with a minimum charge of $15. f(x)=15x−12 f(x)=2x2+5 f(x)=1x f(x)=3√x What is the range of the function y=x2−5 when the domain is -2, -1, 0, 1, 2? What is the range of the function y=2x−34 when the domain is -2.5, -1.5, 5? What is the domain of the function y=3x when the range is 9, 12, 15? What is the range of the function y=3x when the domain is 9, 12, 15? Angie makes $6.50 per hour working as a cashier at the grocery store. Make a table that shows how much she earns if she works 5, 10, 15, 20, 25, or 30 hours. The area of a triangle is given by the formula A=12bh. If the base of the triangle measures 8 centimeters, make a table that shows the area of the triangle for heights 1, 2, 3, 4, 5, and 6 centimeters. Make a table of values for the function f(x)=√2x+3 for input values -1, 0, 1, 2, 3, 4, 5. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? 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6236	https://core.ac.uk/download/pdf/222831788.pdf	Communication Complexity of Quasirandom Rumor Spreading ∗ Petra Berenbrink Robert Els¨ asser Thomas Sauerwald May 23, 2013 Abstract We consider rumor spreading on random graphs and hypercubes in the quasirandom phone call model. In this model, every node has a list of neighbors whose order is speciﬁed by an adversary. In step i every node opens a channel to its ith neighbor (modulo degree) on that list, beginning from a randomly chosen starting position. Then, the channels can be used for bi-directional communication in that step. The goal is to spread a message eﬃciently to all nodes of the graph. For random graphs (with suﬃciently many edges) we present an address-oblivious algorithm with runtime O(log n) that uses at most O(n log log n) message transmissions. For hypercubes of dimension log n we present an address-oblivious algorithm with runtime O(log n) that uses at most O(n(log log n)2) message transmissions. Together with a result of , our results imply that for random graphs the communication complexity of the quasirandom phone call model is signiﬁcantly smaller than that of the standard phone call model. 1 Introduction In this paper we consider rumor spreading (a.k.a. randomized broadcasting) in random graphs and hypercubes. This problem is motivated by overlay topologies in peer-to-peer (P2P) systems, in which each node possesses a list of neighboring peers. Our goal is to develop time-eﬃcient ru-mor spreading algorithms which produce a minimal number of message transmissions and use a small amount of randomness. Since P2P networks are decentralized platforms for sharing data and computing resources, it is very important to provide eﬃcient, simple, and robust rumor spreading algorithms for P2P overlays. Minimization of the number of transmission (communication com-plexity) is important for applications such as the maintenance of replicated databases in which often huge amounts of broadcasts are necessary to deal with frequent updates in the system. We consider the quasirandom phone call model, a variant of the standard phone call model. Let us ﬁrst introduce the standard phone call model (also known as random phone call model, see ). In this model, each node v may perform the following actions in every step: 1) create a new rumor to be spread, 2) establish a communication channel between itself and one randomly chosen neighbor, 3) transmit a message over incident channels (opened by v or by some neighbor of v) ∗An extended abstract of this paper appeared in the 18th Annual European Symposium (ESA’10) . 1 1 INTRODUCTION 2 and 4) close the channel opened in the current step. Note that open channels can be used for bi-directional communications. Calling nodes (i.e., the nodes that opened the channels) can send their messages to their neighbors. These are called push transmissions. Called nodes can also perform so called pull transmissions, i.e., they send the message to the calling nodes. These transmissions are simply called pull transmissions. If there are many rumors to be spread among the nodes, then it is assumed that nodes can combine several rumors to one larger message. Following , we therefore focus on the running time and message complexity produced by our algorithms w.r.t. one single message. Nodes can send messages over all their open channels in one time step. The major challenge for rumor spreading algorithms in the phone call model is to decide whether or not a node should forward the rumor over an open communication channel. An algorithm is called address-oblivious (see ) if the decision of node v to send a rumor over an open channel (v, w) or (w, v) does not depend on w. However, this decision can depend on the communication partners chosen in earlier rounds or on decisions made so far. Hence, according to such an algorithm a node has to decide whether to use a channel without knowing if the rumor is already known by the neighbor in question. If there are only very few rumors in the network, then many communication channels may be established without ever being used for transmissions. Thus, the phone call model is especially of interest in situations where rumors are frequently generated. Then, the cost of establishing communication channels amortizes over all message transmissions. In the case of the quasirandom phone call model it is assumed that every node has a cyclic list of all its neighbors, numbered from 1 to d, whose order is speciﬁed by an adversary. At the beginning, each node v chooses a random position in the list, independently of the other nodes. Assume that 1 ⩽ℓ⩽d is the random choice of node v, where d is the degree of v. Then v communicates in step i with the neighbor ((i + ℓ−2) mod d) + 1 from the list. To create the list we assume that the adversary has total knowledge about the topology of the network, but cannot foresee any node’s random choice such as the position selected at the beginning (cf. ). 1.1 Related Work Due to space constraints, we mention only results which focus on the theoretical study of push and push & pull algorithms. Runtime. Most rumor spreading studies analyze the runtime of the push algorithm in the stan-dard phone call model for diﬀerent graph classes. For complete graphs of size n, Pittel shows that (with probability 1 −o(1)) it is possible to spread a rumor in time log2(n) + ln(n) + f(n), where f(n) is a slowly growing function, improving a result of Frieze and Grimmett . In , Feige et al. determine asymptotically optimal upper bounds for the runtime on G(n, p) graphs (i.e., traditional Erd¨ os-R´ enyi random graphs ), bounded degree graphs, and hypercubes, which all hold w.h.p.1. Recently, Fountoulakis et al. prove a tighter bound for the runtime on suﬃciently dense G(n, p) graphs, similar to the result of for complete graphs. Also recently, Chierichetti et al. show that the runtime of the combined push & pull model is O(Φ−1 ·log n·log2(Φ−1)) w.h.p. for any graph G, where Φ denotes the conductance of G. This runtime bound has been recently tightened in who proved a runtime bound of O(Φ−1 · log n) for any graph G with conductance Φ. 1W.h.p. or “with high probability” means with probability at least 1 −n−c for some constant c > 0. 1 INTRODUCTION 3 In , Doerr et al. analyze the so called quasirandom rumor spreading. They show that for hypercubes and G(n, p) graphs O(log n) steps suﬃce to inform every node, w.h.p. These bounds are similar to the ones in the standard phone call model (push model). The results of are extended to further graph classes with good expansion properties in . Observe that in [7, 8] the authors mainly focus on the runtime eﬃciency, and the algorithms therein require Θ(n log n) message transmissions for hypercubes and G(n, p) graphs. Number of Message Transmissions. Karp et al. observe that in complete networks the pull approach is inferior to the push approach until roughly n/2 nodes receive the rumor. Then the pull approach becomes superior. They present a push & pull algorithm, together with a termination mechanism, which bounds the number of total transmissions to O(n log log n) (w.h.p.), and show that this result is asymptotically optimal. For sparser graphs and the standard phone call model it is not possible to get an oblivious algorithm that uses O(n log log n) message transmissions, together with a runtime of O(log n). In , the second author considers random G(n, p) graphs and shows a lower bound of Ω(n log n/ log(pn)) message transmissions for oblivious rumor spreading algorithms with a runtime of O(log n). For p > log2 n/n he develops an oblivious algorithm that spreads a rumor in time O(log n) using O(n · (log log n + log n/ log(pn))) transmissions, w.h.p. In the authors consider a simple modiﬁcation of the standard phone call model, called Random, where every node is allowed to open a channel to four diﬀerent randomly chosen neighbors in every time step. For G(n, p) graphs with p > log2 n/n, they show that this modiﬁcation results in a reduction of the number of message transmissions down to O(n log log n). Similar results are shown for random d-regular graphs in . The authors of present an extension of Random which they call Rr model. In their model each node has a randomly ordered cyclic list with all its neighbors. In step i, the node opens a communication channel to the ith neighbor in its list. The Rr model is the same as the quasirandom model except that the adversarial order is replaced by the random order. The authors present an oblivious algorithm for graphs with very good edge and node expansion properties which has a runtime O(log n) and which uses O(n√log n) message transmissions, w.h.p. The authors establish a lower bound of Ω(n p log n/ log d) on the number of message transmissions for oblivious rumor spreading algorithms (assuming a runtime of O(log n)), showing that their upper bound is tight up to a √log log n factor if d is polylogarithmic in n. The algorithms of [2, 9, 10, 17] spread the rumor using push transmissions until a constant fraction of the nodes receives the rumor (we call these nodes informed in the following). Then the algorithms spread the rumor via pull transmissions until every node is informed. To save on communications, the algorithms of [1, 2, 9, 10] only allow each node v a certain number of transmissions which depends on the age the rumor had at the time v received it for the ﬁrst time. 1.2 Model In this paper we consider random graphs G(n, p) = (V, E) and hypercubes Hd of dimension d. A random graph G(n, p) consists of n nodes. The probability that any pair of nodes is connected is p, we assume that (log2 n)/n ⩽p ⩽2o(√log n)/n. The expected number of edges for G(n, p) is pn · (n −1)/2. Let d(v) be the degree of node v and N(v) be the set of neighbors in V . For S ⊂V , let N(S) be the set of neighbors of nodes in S. Let α be the node expansion value of G(n, p). Then 1 INTRODUCTION 4 α = minS∈V,\|S\|⩽n/4 N(S)/\|S\|. It is known that for our choice of p, α is a constant close to 1 w.h.p. (). The d-dimensional hypercube Hd consists of n = 2d many nodes. A binary string of length 2 is assigned to every node and two nodes are connected if their binary strings diﬀer in exactly one bit. Hence, the degree of any node of Hd is d. Note that hypercubes have much smaller expansion than random graphs. We assume that every node has an estimation of n which is accurate to within a constant factor. We also assume that all nodes have access to a global clock, and that they work synchronously. As communication model we assume a variant of the phone call model. In the standard phone call model (see ) in each step t every node can create an arbitrary amount of rumors to be spread. To measure the communication cost we only count the number of message transmissions, i.e., opening a channel is not counted. Following [1, 2, 9, 17], we assume here that new pieces of information are generated frequently in the network, and then the cost of establishing communication channels amortizes over all message transmissions. However, we only concentrate on the distribution and lifetime of a single rumor. The quasirandom variant of the phone call model considered in this paper was introduced in . In the quasirandom phone call model every node v has a list ˜ Lv = ˜ Lv, ˜ Lv, . . . ˜ Lv[d(v) −1] of length d(v) with all its neighbors. The order of that list is arbitrary, i.e., it may be determined by an adversary. We assume that the rumor is initiated on a node at step 0. For spreading the rumor, every node v chooses a random position iv in the list, independently of the other nodes. For its j-th communication v will open a channel to node Lv[(iv + j −1) mod d(v)]. We deﬁne Lv = Lv, Lv, . . . , Lv[d(v)] as the list beginning at neighbor iv. Nodes that received the rumor will be called informed. By It (Ht) we denote the set of informed (uninformed) nodes in step t. Furthermore, let I+ t be the set of nodes that reveive the rumor for the ﬁrst time in step t. These nodes will also be called newly informed nodes. Note that we omit rounding of non-integers in our proofs but not in the statement of our algorithms. Furthermore, we assume that log n is the logarithm with base 2. 1.3 Our Contribution In this paper we show the following results. For random graphs with (log2 n)/n ⩽p ⩽2o(√log n)/n we present an oblivious algorithm (in the quasirandom model) that spreads a rumor in time O(log n) using O(n log log n) message transmissions, w.h.p. Compared to , we reduce the number of message transmissions by a factor of log n/ log log n. Moreover, our upper bound in the quasirandom model is signiﬁcantly smaller than the lower bound for the standard phone call model (cf. ). For the hypercube we show a result that is slightly weaker than our result for random graphs. We present an oblivious algorithm (which is similar to the algorithm for random graphs) that spreads a rumor in time O(log n) using O(n · (log log n)2) message transmissions, w.h.p. The communication complexity on the hypercube has not been analyzed before, neither in the standard nor in the quasirandom phone call model. Therefore the best known algorithms require O(log n) time, but produce Ω(n log n) message transmissions. In comparison to that, we reduce the number of message transmissions by a factor of log n/(log log n)2. Our results demonstrate that on two important networks rumor spreading can be done much more eﬃciently in the quasirandom phone model than in the standard phone call model. More-over, the results provide evidence that avoiding previously chosen communication partners is more 2 RANDOM GRAPHS 5 important than choosing all communication partners independently and uniformly at random. Note that the conference version of this paper () contained a lower bound on the message complexity for hypercubes. Unfortunately there is a mistake in the proof on the slower bound which we were not able to repair. 2 Random Graphs In this section we present an algorithm with runtime O(log n) and communication complexity O(n log log n) for random graphs. 2.1 Our Algorithm We assume that the rumor we want to spread is generated at time 0, i.e., at time t the age of the rumor equals t. The algorithm describes the behavior of the nodes w.r.t. one speciﬁc rumor. Depending on the age of the rumor, each node is in one of the following phases (in our algorithm, ρ is a suﬃciently large constant): Phase 0: [age ⩽⌈ρ log n⌉] The node which generates the rumor performs push in each step of this phase. No other node transmits the rumor in this phase. Phase 1: [⌈ρ log n⌉+ 1 ⩽age ⩽2 · ⌈ρ log n⌉+ 320] Nodes that received the rumor in Phase 0 use the ﬁrst 320 steps of this phase to perform push in each of these steps. If a node receives the rumor for the ﬁrst time in some step t ∈{⌈ρ log n⌉+ 1, . . . , 2 · ⌈ρ log n⌉}, then the node performs push in the steps t + 1, . . . , t + 320. Phase 2: [2⌈ρ log n⌉+321 ⩽age ⩽2·⌈ρ log n+ρ log log n⌉] Every informed node performs push in every step of this phase. Phase 3: [2⌈ρ log n + ρ log log n⌉+ 1 ⩽age ⩽3 · ⌈ρ log n⌉] Every node that becomes informed in some step of this phase performs pull transmissions for the rest of the phase, i.e., after receiving the message it transmits over all incoming channels in every step of the phase. All other informed nodes perform pull over all incoming channels with probability 1/ log n. Phase 4: [3⌈ρ log n⌉+ 1 ⩽age ⩽3 · ⌈ρ log n + ρ log log n⌉] All informed nodes perform pull transmissions. It is easy to see that at the end of Phase 0, exactly ρ log n + 1 nodes are informed (Observation 2.2). In Phase 1 we inform half of the nodes (see Lemma 2.3). At the end of Phase 2 we have n · (1 −2 log log n/ log n) informed nodes, w.h.p. (Lemma 2.5). Phase 3 and Phase 4 are analyzed in Lemma 2.6. There we show that w.h.p. at the end of Phase 4 all nodes are informed. 2 RANDOM GRAPHS 6 2.2 Analysis of the Algorithm For a graph G(n, p) and our choice of p the degree of each node is in the range [np·(1−1/ log n), np· (1 + 1/ log n)], with probability at least 1 −n−3. In the following, we condition on this event, and for simplicity we assume in our analysis that d = pn. Theorem 2.1. Consider G = G(n, p) with (log2 n)/n ⩽p ⩽2o(√log n)/n. Then, the algorithm above spreads a rumor in G in time O(log n) using O(n log log n) message transmissions, w.h.p. In the rest of this section we will prove the above theorem. The proof is split into several lemmas. It is easy to see that in Phase 0 the node that generated the rumor informs ρ log n diﬀerent neighbors, which results in the following observation. Observation 2.2. At the end of Phase 0 there are ρ log n informed nodes. Now we concentrate on Phase 1 and show the following lemma. Lemma 2.3. With probability 1 −n−2, at least n/2 nodes are informed at the end of Phase 1. Proof. Assuming that the nodes all have degree d we show that 1. After the ﬁrst ρ · (log n)/2 steps at least 6n/d nodes are informed, where ρ > 8. 2. After ρ · ((log n)/2 −320) additional steps we have at least n/40 informed nodes. 3. After the last 320 · ρ steps we have n/2 informed nodes for ρ large enough. Part 1). The statement follows from the claim below we adapted from Claim A.1 of with expansion factor α > 0.6. This (node-)expansion holds for the random graphs considered here, if the size of the set is bounded by n/d . Claim 2.4. Let τ1, τ2, . . . be consecutive time intervals in this phase, each of them consisting of 160 time steps. Furthermore, let ti be the beginning of interval τi, and let I+ τi be the set of nodes, which are informed in time interval τi for the ﬁrst time. Assume that \|Iti\| ⩽n d and \|Iti\| ⩾8 α2 · \|Iti−1\| Then, with probability at least 1 −n−3 we have \|I+ τi\| ⩾8 α2 · \|Iti\|. In the algorithm considered in , we assume that each node transmits the message for 80/α2 steps; in our algorithm we assume that the nodes transmit for 320 time steps, and therefore the claim applies here too. In order to have at 6n/d informed nodes at the end of Phase 1, we apply the claim to a time interval which starts right after n/d −1 nodes are informed. 2 RANDOM GRAPHS 7 Part 2). In this case the number of informed nodes lies in the range [6n/d, n/40]. We show inductively that with a very high probability the number of informed nodes grows by a factor of 2.1 every 160 steps. To do so we divide the time into ℓ= (ρ · ((log n)/2 −320)/160 subphases. For 0 ⩽i ⩽ℓ, subphase τi starts in step ρ·(log n)/2+160i+1 and ends in step ρ·(log n)/2+160(i+1). Let I+ τi be the newly informed nodes in Subphase τi, and Iτi are the informed nodes at the beginning of Subphase τi. Note that all nodes in I+ τi perform a push transmissions in Subphase τi+1. We show by induction that for 0 ⩽i ⩽ℓwe have \|I+ τi\| ⩾2.1 · \|Iτi\|, which then implies that \|I+ τi\| ⩾\|Iτi+1\|/2. Fix a subphase τi+1. From Lemma 4.5 (Part 1) we know that there are n/6 uninformed nodes at the beginning of the subphase so that, with probability at least 1 −εn, all of these nodes have at least \|I+ τi\| · d/(2n) neighbors in the set of nodes I+ τi. Hence, such an uninformed node remains uninformed in the time interval τi+1 with probability at most (1 −160/d)\|I+ τi\|·d/(2n). This holds since the ﬁrst positions are chosen independently and uniformly at random, and a neighbor misses a speciﬁc node in 160 steps with probability 1 −160/d. Thus, E h \|I+ τi+1\| i ⩾ 1 − 1 −160 d \|I+ τi\|·d/(2n)! · n 6 ⩾ 1 − 1 e 80\|I+ τi\|/n! · n 6 ⩾ 1 − 1 e 40·\|Iτi+1\|/n! · n 6 ⩾ 1 − 1 − 1 n/(40 · \|Iτi+1\|) + 1 · n 6 > 2.2 · \|Iτi+1\| Here, the third equation uses the induction hyphothesis. Now, we can construct a Martingale sequence Y0, Y1, . . . , Yn/6 on these n/6 uninformed nodes, where Yj is the expected value on the number of newly informed nodes in time interval τi+1 after the ﬁrst j uniformed nodes are exposed. Since this Martingale sequence satisﬁes the 160-Lipschitz condition, applying the Azuma-Hoeﬀding bound (see e.g. ) Pr h \|Yn/6 −Y0\| ⩾0.1 · E h \|I+ τi+1\| i i ⩽2 exp    0.01 · E h \|I+ τi+1\| i2 (160)2 · 2n/6    (1) we obtain with probability 1 −o(n−3) that \|I+ τi+1\| ⩾2.1 · \|Iτi+1\|. Part 3). Now the number of informed nodes lies in the range [n/40, n/2]. We divide the time into ℓ= 2ρ subphases. For 0 ⩽i ⩽ℓ, subphase τi starts in step ρ · (log n −320) + 160i + 1 and ends in step ρ · (log n −320) + 160(i + 1). Our goal is to show inductively that for all but the last phase \|I+ τi\| ⩾2.1 · \|Iτi\|. In the last phase we inform enough nodes so that half of the nodes are informed at the end of this phase. Similar to Part 2) we ﬁx a subphase τi+1 and deﬁne Hτi+1 as the number of uninformed nodes at the beginning of Subphase τi+1. From Lemma 4.5 (Part 2) it follows that \|Hτi+1\|/2 of the uninformed nodes have at least \|I+ τi\|d/(2n) neighbors in the set of nodes I+ τi, with probability 1 −εn. Such an 2 RANDOM GRAPHS 8 uninformed node remains uninformed in τi+1 with probability at most (1 −160/d)\|I+ τi\|d/(2n). Thus, E h \|I+ τi+1\| i ⩾ 1 − 1 −160 d \|I+ τi\|d/(2n)! · \|Hτi+1\| 2 ⩾ 1 − 1 e 80\|I+ τi\|/n! · \|Hτi+1\| 2 ⩾ 1 − 1 e 40\|Iτi+1\|/n! · \|Hτi+1\| 2 . The remainder of the proof is a case analysis depending on \|Iτi+1\|. If n/40 ⩽\|Iτi+1\| ⩽n/10, then 1 − 1 e 40\|Iτi+1\|/n! · \|Hτi+1\| 2 ⩾ 1 − 1 e · 9n 20 ⩾2.2 · n 10 . Using the method of bounded independent diﬀerences as in inequality (1) (with the adapted length of the Martingale to \|Hτi+1\|/2) one can show that with probability 1 −o(n−3) we obtain \|I+ τi+1\| ⩾2.1 · \|Iτi+1\|. For n/10 < \|Iτi+1\| ⩽n/6 1 − 1 e 40\|Iτi+1\|/n! · \|Hτi+1\| 2 ⩾ 1 − 1 e 4! · 5n 12 ⩾2.2 · n 6 . Then, with probability 1 −o(n−3) we have \|I+ τi+1\| ⩾2.1 · \|Iτi+1\| . For \|Iτi+1\| ⩾n/6 we get \|Iτi+1\| + 1 − 1 e 40\|Iτi+1\|/n! · \|Hτi+1\| 2 ⩾\|Iτi+1\| + 1 − 1 e 40/6! · n −\|Iτi+1\| 2 ⩾41n 80 . (2) Again, we obtain with probability 1 −o(n−3) that \|Iτi+2\| > n/2. Lemma 2.5. Assume ρ ⩾30. With probability at least 1−n−2, there are at most (n·2 log log n/ log n) uninformed nodes at the end of Phase 2. Proof. Note that in this phase every informed node performs a push transmission in every step. Let T be a random variable deﬁned as the ﬁrst time step directly after the subphase τ of the previous phase, for which \|Iτ ∪I+ τ \| ⩾n/2 for the ﬁrst time (this happens w.h.p. in Phase 1). For the sake of this proof we assume that only the nodes of I+ τ perform push transmissions in this phase. Due to the second term in the left hand side of inequality (2), we have \|I+ τ \| > n/5 with probability at least 1 −n−3. According to Lemma 4.6, with probability 1 −εn (ε > 0 is a constant) there are at most n · log log n/ log n nodes in HT which have fewer than d/10 neighbors in I+ τ . After ρ log log n additional steps each of the other (uninformed) nodes remains uninformed with probability at least 1 −ρ log log n d d/10 ⩽e−ρ log log n/10 ⩽log−3 n, 2 RANDOM GRAPHS 9 for ρ ⩾30. Thus, if there are at most n · log log n/ log n nodes in HT which have fewer than d/10 neighbors in I+ τ , the expected number of newly informed nodes in Phase 2 is at least \|HT \| −n log log n log n · (1 −log−3 n). Then, using one can show that with probability at least 1−n−2, the number of newly informed nodes in this phase is at least \|HT \| −2n log log n log n . Hence with probability at least 1−n−2, the number of uninformed nodes after this phase is at most n · 2 log log n/ log n. Finally, we concentrate on Phases 3 and 4. Lemma 2.6. Assume ρ ⩾30. With probability at least 1 −n−2 all nodes are informed at the end of Phase 4. Proof. For a node u and time interval τ = [t, t′], let Lu(τ) be the set of nodes chosen by u in steps τ = t, t + 1, . . . , t′. Deﬁne t2 = 3ρ · (log n + log log n) as the end of Phase 4, t1 = 3 · ρ log n as the beginning of Phase 4, and t0 = 2ρ(log n + log log n) as the beginning of Phase 3. First we consider Phase 4 and divide the time interval [t1 + 1, t2] into k′ = (t2 −t1)/320 subintervals of length 320. For any 0 ⩽i ⩽k′ −1 we deﬁne ˜ τi = [t2 −320i, t2 −320 · (i + 1) + 1]. For a node v, let U0(v) = Lv( ˜ τ0) and Ui(v) = ∪w∈Ui−1Lw(˜ τi). We can visualize ∪i⩽k′−1Ui(v) as tree of depth k′ −1 rooted in v (cf. Lemma 4.7). The level i nodes are the nodes in Ui(v). Then, according to Lemma 4.7 \|Uk′−1(v)\| = Ω(log3 n) with probability 1 −o(n−3). In the following we consider two cases. In the ﬁrst case, we assume that ∪i⩽k′−1Ui(v) ∩It1 ̸= ∅ for some node v. Then v is informed in Phase 4 since all informed nodes perform pull transmissions in that phase. In the second case, let Uk′−1(v) ∩It1 = ∅. For this case we know from Lemma 4.7 that in Phase 4 v will be the root of a communication tree consisting of nodes which are still all uninformed in step t1. Then we will show that w.h.p. at least one of the leaves of the tree will be informed in Phase 3. The rumor will be propagated to v via the path between v and the informed leaf. Now we need some additional deﬁnitions. We divide the time interval [t0 + 1, t1] into k′′ = (t1 −t0)/160 rounds of length 160. For any 0 ⩽i ⩽k′′ −1 ˜ τ ′ i = [t1 −160i, t1 −160 · (i + 1) + 1]. For 0 ⩽i ⩽ρ log n, let ˜ U H −1(v) = Uk′−1(v) ˜ U H i (v) = ∪w∈˜ UH i−1(v)Lw(˜ τ ′ i) ∩Ht0 ˜ U I i (v) = ∪w∈˜ UH i−1(v)Lw(˜ τ ′ i) ∩It0. 3 HYPERCUBES 10 A node ˜ wi ∈˜ U I i (v) is connected to a node ˜ w−1 ∈˜ U H −1(v) by a path P = ( ˜ wi, . . . , ˜ w0, ˜ w−1), where ˜ wi−1, . . . , ˜ w0, ˜ w−1 ∈Ht0, and for j = −1, . . . , i −1 we have ˜ wj+1 ∈L ˜ wj(˜ τ ′ j+1). Now deﬁne ˜ U H 0→i(v) = ∪i j=0 ˜ U H j (v) and ˜ U I 0→i(v) = ∪i j=0 ˜ U I j (v). Since \| ˜ U H −1(v)\| = Ω(log3 n), we can apply the same techniques as in Lemma 2.3 and obtain that \| ˜ U H i (v) ∪˜ U I i (v)\| ⩽2.1 · \| ˜ U H i−1(v)\| for any i ⩾1 as long as \| ˜ U I i−1(v)\| = O(log2 n) and \| ˜ U H i (v)\| < n/40. However, since \|Ht0\| ⩽ 2n log log n/ log n, there exists some i < k′′ such that \| ˜ U I 0→i(v)\| > ρ log2 n. Then, we can argue that every node u ∈˜ U I 0→i(v) performs pull transmissions with probability 1/ log n. Since for every u there is a s < k′′ such that a path (u, ˜ ws, . . . , ˜ w0, . . . , v) exists, that consists of nodes of Ht0 that perform pull transmissions in the corresponding rounds. Hence, there is a node u ∈L ˜ ws( ˜ τ ′s+1) which transmits the rumor at the right time, with probability at least 1 − 1 − 1 log n ρ log2 n = 1 −o(n−3), if ρ is large enough. Let us now prove Theorem 2.1. The correctness (every node gets informed w.h.p.) follows from the lemmas above. It remains to analyze the total number of message transmissions. In Phase 0, the algorithm uses O(log n) message transmissions. In Phases 1, 2 and 4, the algorithm uses O(n log log n) message transmissions. By Lemma 2.5, we know that after Phase 2 at most O(n log log n/(log n)) uninformed nodes remain. These nodes generate at most O(n log log n) mes-sage transmissions in Phase 3. Using a Chernoﬀbound, we can show that the nodes that are informed at the end of Phase 2 use at most O(n) message transmissions. Hence the total number of message transmissions is O(n log log n). 3 Hypercubes In this section we present an algorithm with runtime O(log n) and communication complexity O(n(log log n)2 for hypercubes. 3.1 Our Algorithm In the algorithm below, the total number of message transmissions is O(n(log log n)2), which can be shown as in the proof of Theorem 2.1 above (ρ > 0 is a suﬃciently large constant). Phase 1: [1 ⩽age ⩽⌈ρ log n⌉] If a node receives a rumor for the ﬁrst time in step t ∈ {1, . . . , ⌈ρ log n⌉}, then the node performs push for the next C log log n consecutive steps, where C > 0 is a suﬃciently large constant. Phase 2: [⌈ρ log n⌉+ 1 ⩽age ⩽2 · ⌈ρ log n⌉] Every node which becomes informed in this phase performs pull over each incoming channel. All other informed nodes perform pull with probability 1/ log n over each incoming channel. 3 HYPERCUBES 11 Phase 3: [2⌈ρ log n⌉+ 1 ⩽age ⩽2 · ⌈ρ log n + ρ(log log n)2⌉] All informed nodes perform pull transmissions in every step of this phase. 3.2 Analysis of the Algorithm Theorem 3.1. Assume that Hd is a hypercube of dimension d = log n. The algorithm above spreads a rumor in Hd in time O(log n) using O(n(log log n)2) message transmissions, w.h.p. The analysis of the above theorem is similar to the one for random graphs. However, the lack of strong expansion properties makes it more diﬃcult and one has to resort to the special structure and the symmetries of hypercubes. For any integer 0 ⩽k ⩽d, let Nk(0) be the set of nodes with distance k to the node 0d, i.e. the set of nodes with k ones. Sometimes we also simply write Nk for Nk(0) if the reference to 0d is clear from the context. In addition, we deﬁne N⩾k := ∪d j=kNj. For any node v and any subset S ⊆V , we denote by dS(v) the number of neighbors of v within the set S. 3.2.1 Analysis of Phase 1 For the analysis we ﬁrst subdivide this phase into two intervals [1, t1), [t1, t2), where t1 := (C2/2)(log log n)2, t2 := t1 + K · (log n/2 −(C/2) log log n), where K > 0 is a suﬃciently large constant. We further deﬁne ℓ1 := (C/2) log log n. Lemma 3.2. For all possible lists ∪v∈V Lv we have \|I+ t1 ∩Nℓ1\| ⩾(log n)C/2. Proof. To prove this lemma we consider a delayed version of our algorithm. We assume that a node in level Ni does not transmit before timestep iC log log n. Observe that any node in Ni has exactly d −i neighbors in Ni+1 and i neighbors in Ni−1. Let ℓ:= (C/2) log log n. Any informed node in Ni informs at least C log log n −i nodes in Ni+1 and any node in Ni+1 is informed by at most i + 1 nodes in Ni. Hence we obtain that the number of informed nodes in Nℓwithin ℓ·C log log n rounds is at least ℓ−1 Y i=0 C log log n −i i + 1 ⩾((C/2) log log n)(C/2) log log n ((C/2) log log n)! ⩾2(C/2) log log n = (log n)C/2, where we have used the fact that n! ⩽(n/2)n for every integer n. Lemma 3.3. Let ε > 0 be a constant. Assume that an adversary (who knows all random choices of the protocol) is allowed to choose a set of nodes of size log3 n so that these nodes never transmit a rumor within the time-interval [t1, t2]. Then there exists a constant δ = δ(ε) > 0 and ℓ2 := (1/2 −δ) · d, so that with probability 1 −n−ω(1), \|I+ t2 ∩Nℓ2\| ⩾2(1−ε)·d, and It2 ∩N>ℓ2 = ∅. 3 HYPERCUBES 12 Proof. The proof is similar to the one of [7, Theorem 2]. From Lemma 3.2 we get \|I+ t1 ∩Nℓ1\| ⩾(log n)C/2. In the following we consider a slowed-down version of our algorithm. The algorithm works in phases of K steps each, where K is a suﬃciently large constant to be determined later. In phase i, only nodes communicate that are in level Ni+ℓ1. Fix an arbitrary phase i with 0 ⩽i ⩽log n −t1 and a timestep t = t1 + K · i (ﬁrst round of phase i). Let us denote by ˜ It+K the nodes that would get informed if there was no adversary. Consider Nj with j = i + ℓ1 and 1 ⩽j ⩽ℓ2 ⩽d/2. Our goal is to show that a large subset of the nodes in Nj+1 will be informed in phase i. The probability that a node v ∈Nj+1 is still uninformed at the end of phase i is Pr h v ̸∈˜ It+K i ⩽ Y u∈Γ(v)∩It∩Nj 1 −K d = 1 −K d d˜ It∩Nj (v) . By linearity of expectation we get E h \|˜ It+K ∩Nj+1\| i = X v∈Nj+1 Pr h v ∈˜ It+K i ⩾ X v∈Nj+1 1 − 1 −K d dIt∩Nj (v) ⩾ X v∈Nj+1∩N(It) 1 −exp −K d · dIt∩Nj(v) = \|Nj+1 ∩N(It)\| − X v∈Nj+1∩N(It) exp −K d · dIt∩Nj(v) . Applying Lemma 4.3 with \|E(It ∩Nj, Nj+1)\| = P v∈Nj+1 dIt∩Nj(v) = \|It ∩Nj\| · (d −j) and for any v ∈Nj+1, dIt∩Nj(v) ∈[0, j + 1], we get E h \|˜ It+K ∩Nj+1\| i ⩾\|Nj+1 ∩N(It)\| −\|E(It ∩Nj, Nj+1)\| j + 1 · exp −K d · (j + 1) − \|Nj+1 ∩N(It)\| −\|E(It ∩Nj, Nj+1)\| j + 1 · 1 = \|E(It ∩Nj, Nj+1)\| j + 1 · 1 −exp −K d · (j + 1) ⩾\|It ∩Nj\| · d −j j + 1 · 1 − 1 K(j+1) d + 1 ! , where we have used in the last inequality the fact that exp(−x) ⩽ 1 x+1 for any x ∈R. Let the nodes in It ∩Nj be u1, u2, . . . , u\|It∩Nj\|. Consider the random variable \|˜ It+K ∩Nj+1\| as a function of iu1, iu2, . . . , iu\|It∩Nj\|, where iuj is the randomly chosen starting point of the list of node 3 HYPERCUBES 13 uj. Since the iuj are independent random variables and changing one iuj can change ˜ It+K ∩Nj+1 by at most K, Lemma 4.8 gives Pr \|˜ It+K ∩Nj+1\| ⩽ 1 − 1 log n · E h \|˜ It+K ∩Nj+1\| i ⩽exp   − 1 log n · E [ \|It+K ∩Nj+1\| ] 2 K · \|It ∩Nj\|    ⩽exp     − 1 log2 n \|It ∩Nj\| · 1 − 1 K(j+1) d +1 2 K · \|It ∩Nj\|      ⩽exp −Ω 1 log4 n \|It ∩Nj\| = exp −Ω(log2 n) , provided that \|It ∩Nj\| = Ω(log6 n) (which holds initially for timestep t1 by assumption). Recall now that the adversary is allowed to choose log3 n nodes that never transmit the rumor within the time-interval [t1, t2]. Since \|It+K ∩Nj+1\| ⩾\|˜ It+K ∩Nj+1\| −log3 n, Pr \|It+K ∩Nj+1\| ⩽ 1 − 2 log n · E [ \|It+K ∩Nj+1\| ] = exp(−Ω(log2 n)). Let us now compute how many nodes are ﬁnally informed in Nℓ2, provided that \|It+K ∩ Nj+1\| is always close to its expectation. Taking the union bound, it holds with probability 1 −exp −Ω(log2 n) that, for suﬃciently large constant δ = δ(ε) > 0, \|It0+K·(ℓ2−(C/2) log log n)\| ⩾ ℓ2 Y j=(C/2) log log n 1 − 2 log n · d −j j + 1 · 1 − 1 K(j+1) d + 1 ! · \|It1 ∩Nℓ1\| ⩾ 1 − 1 log n log n · Qℓ2 j=1 d−j j+1 Q(C/2) log log n j=1 d−j j+1 · d/2 Y j=1 1 − 1 K(j+1) d + 1 ! · 1 ⩾1 16 · d d/2 Qd/2−ℓ2−1 r=0 d/2−r d/2+(r+1) d (C/2) log log n · d/2 Y j=1 1 − 1 K(j+1) d + 1 ! ⩾1 16 · 1 2 n √ πd/2 · 2−(ε/2)n dO(log log n) · d/2 Y j=1 K(j+1) d 1 + K(j+1) d ! = 1 32 · n √ πd/2 · 2−(ε/2)n dO(log log n) · 1 Qd/2 j=1 1 + d K(j+1) ⩾1 32 · n √ πd/2 · 2−(ε/2)n dO(log log n) · 1 2(ε/3)d ⩾2(1−ε)d, 3 HYPERCUBES 14 where the penultimate inequality holds by Lemma 4.2. Lemma 3.4. Let ε > 0 and δ(ε) > 0 be the constant from Lemma 3.3. Let w be an arbitrary node, ℓ2 := (1/2 −δ) · d and ℓ3 := (1/2 −δ/2) · d > ℓ2. Suppose that there is a time step t2 and a subset U ⊆V satisfying the following three conditions: • \|I+ t2 ∩Nℓ2\| ⩾n1−ε, • It2 ∩N>ℓ2 = ∅, • \|U ∩N⩾ℓ3\| ⩾n1−ε. Assume further that ε is chosen small enough so that 3/2 −2ε −log(e) > 0. Then in step t2 + d, there exists an informed node in U with probability at least 1 −2n−3. Proof. We deﬁne a relation between random walks starting randomly chosen vertices and the spread of the information. The idea is that, if a random walk does not collide with another random walk (or with itself) and it starts from an informed node, then the random walk follows a path by which the message is spread. Note that the assumptions that the random walks start at random nodes is for technical reasons only; it allows us to conveniently bound the number of collisions while there will be still enough random walks that start from an uniformed node. We continue to deﬁne this relation precisely. Set x := n1/2−γ (γ > 0 is a suﬃciently small constant) and choose x vertices Xt2 1 , Xt2 2 , . . . , Xt2 x independently and uniformly at random from V . From each Xj with Xt2 j ∈It2 we consider a walk of maximum length log n that is deﬁned recursively for t ⩾t2 as follows. If Xt j transmits to a node v ∈Ht ∩N>ℓ2 then we set Xt+1 j := v. Otherwise the walk Xj terminates at step t. Our goal is to prove that with high probability there is at least one walk Xj that reaches U before step t2 + d, implying that U contains at least one informed node at step t2 + d. Note that it follows that with probability 1 −n−Ω(1), at least x 2 · n1−ε n = 1 2n1/2−γ−ε random walks start from I+ t2 ∩Nℓ2. Let us ﬁrst consider the number of collisions between the random walks. Here, a collision occurs if two diﬀerent random walks visit the same node. Let us expose the x random walks one after the other to estimate their collision probability. We say that a failure occurs if the random walk collides with a previously exposed one. Note that in order to have more than C2 failures (C > 0 is a constant to be speciﬁed later), there must be at least C diﬀerent random walks which are involved in a (not necessarily the same) collision. Therefore, we can upper bound the probability for having more than C2 failures by n1/2−γ C · (C −1) · log n · n1/2−γ n !C . This bound holds since (i) we have n1/2−γ C possibilities to choose the C random walks which should be involved in a collision, (ii) a random walk involved in a collision has to collide with one of the at 3 HYPERCUBES 15 most (C −1) log n · n1/2−γ nodes of the previous random walks and (iii) each random walks starts from a randomly chosen vertex. Let us deﬁne the constant C := ⌈2/γ⌉and let A be the event that there are at most C2 collisions among the n1/2−γ random walks. It follows that Pr [ ¬A ] ⩽ n1/2−γ ⌈2 γ ⌉ · (⌈2 γ ⌉−1) · log n n1/2+γ !⌈2 γ ⌉ ⩽ e · n1/2−γ ⌈2 γ ⌉ !⌈2 γ ⌉ ·  (⌈2 γ ⌉−1)⌈2 γ ⌉log⌈2 γ ⌉n n(1/2+γ)·⌈2 γ ⌉   = O n⌈2 γ ⌉·(−2γ) · log⌈2 γ ⌉n = O n−3.5 . Next we compute the probability that a particular walk Xj with starting node v := Xt2 j ∈It2 reaches a node in U. Note that there is a subset U ′ ⊆U with \|U ′\| ⩾\|U\|/d such that for all u′ ∈U ′, dist(v, u′) is ﬁxed (independent of u′). Let us set D := dist(v, u′). Let pv,u′ be the probability that the walk Xt2 j reaches the node u′ at step t2 + D and let pv,U′ be the probability that the walk Xt2 j reaches any node in U ′ at step t2 + D. By disjointness of events, pv,U′ = P u′∈U′ pv,u′, and by symmetry of the hypercube, pv,u′ is the same for each u′ ∈U ′. Let us now lower bound pv,u′ for a ﬁxed u′ ∈U ′. Certainly, the path Xj reaches u′ from v if it follows a shortest path from v to u′ that does not return to N⩽ℓ2. Note that since by assumption \|It2 ∩N>ℓ2\| = 0, a shortest path from v to u′ will not contain any node in It2 (except the starting node v) provided that the path always uses nodes in N>ℓ2. Let us now ignore all other paths (which could possibly result in a premature termination of Xj) and focus on the event that the path never visits a node in N⩽ℓ2 (except for the starting node v). Let ℓbe the number of ones in u′ and recall that D = dist(v, u′). Hence, v has exactly (1/2−δ)d ones and u′ has at least ℓ3 = (1/2 −δ/2)d ones. It follows that any shortest path from v to u′ has to change α ones into a zero, and has to change β = α + ℓ−(1/2 −δ)d zeros into a one. Since ℓ⩾(1/2 −δ/2)d, it follows that β ⩾α + (1/2)δd. Clearly, the number of shortest paths between v and u′ equals D!. Using Bertrand’s ballot theorem ([19, pages 299–300]), it follows that the number of shortest paths between v and u′ for which only the starting vertex v lies in N⩽ℓ2 equals D! · β −α β + α ⩾D! · (1/2)δd d = D! · (1/2)δ. Hence if we were to ignore all other paths (which could result in the termination of Xj), then we would get a lower bound on the probability that the path Xj reaches u′ of pv,u′ ⩾D! · (1/2)δ · d−D ⩾d! · (1/2)δ · d−d, since D ⩽d. For suﬃciently large integer n ∈N, Stirling’s formula gives n! ⩾1 2 √ 2πn · (n/e)n ⩾ (n/e)n and thus pv,u′ ⩾(1/2)δ · e−d. Consequently, X u′∈U′ pv,u′ ⩾\|U ′\| · (1/2)δe−d ⩾\|U\| d · (1/2)δe−d ⩾(1/2)δn1−ε d · e−d. 3 HYPERCUBES 16 Deﬁne now a random variable Z that counts the number of paths from I+ t2 ∩Nℓ2 to U, if we were to ignore that paths could terminate due to a collision. Linearity of expectation yields E [ Z ] ⩾1 2n1/2−ε−γ · (1/2)δn1−ε d · e−d = δ 4d · n3/2−2ε−log(e)−γ. Let B := {Z ⩾(1/2) · E [ Z ]}. Using a Chernoﬀbound, we obtain that Pr [ ¬B ] ⩽exp −1 8 · E [ Z ] = exp (−Ω(poly(n))) , as 3/2 −2ε −log(e) > 0 by assumption and γ > 0 can be made arbitrarily small. Note that the event A ∧B implies that at least one random walk starting from a node in I+ t2 ∩Nℓ2 will reach a node in U ′ ⊆U before terminating. Taking a union bound, we obtain Pr [ A ∧B ] ⩾1 −Pr [ ¬A ] −Pr [ ¬B ] ⩾1 −n−3 −exp (−Ω(poly(n))) ⩾1 −2n−3. This completes the proof. 3.2.2 Analysis of Phase 2 and Phase 3 Lemma 3.5. With probability at least 1 −n−2 all nodes are informed at the end of Phase 3. Proof. The proof is similar to the proof of Lemma 2.6. Again, deﬁne t5 := 2ρ·(log n+(log log n)2) as the end of Phase 3, t4 := 2·ρ log n as the beginning of Phase 3, and t3 := ρ log n ⩾t2 as the beginning of Phase 2 (we ignore the +1 at the beginning of Phases 2 and 3). We set σ := ρ(log log n)2. First we concentrate on Phase 3 and divide the time interval [t4 + 1, t5] into k′ = (t5 −t4)/√σ subintervals of length √σ. For any 0 ⩽i ⩽k′ −1 we deﬁne ˜ τi = [t5 −√σi, t5 −√σ · (i + 1) + 1]. We assume w.l.o.g. that node v = 0d is uninformed at the end of Phase 3, and show that then there are log3 n uninformed nodes in N√σ/2 at the beginning of Phase 3. For this, for 1 ⩽i ⩽√σ/2 we deﬁne U0(v) = Lv[˜ τ0] and Ui(v) = ∪w∈Ui−1(v)Lw[˜ τi]. Then, according to Lemma 3.2, we have \|U√σ/2(v)\| ⩾logC/2 n, if ρ is large enough. Now, all nodes of U√σ/2(v) must be uninformed at the beginning of Phase 3, since otherwise v becomes informed in the time-interval [t4 + 1, t5]. Now we consider the channels opened by the nodes of U√σ/2(v) in Phase 2. Recall that in this phase, the nodes informed in Phase 1 perform pull with probability 1/ log n. The nodes which become informed in this phase, perform pull with probability 1 in each step of this phase. For the analysis of Phase 2, we divide the time interval [t3 + 1, t4] into k′′ = (t4 −t3)/φ rounds of length φ, where φ is a large constant. For any 0 ⩽i ⩽k′′ −1 ˜ τ ′ i = [t4 −φi, t4 −φ · (i + 1) + 1]. For 0 ⩽i ⩽ρ log n, let ˜ U H −1(v) = U√σ/2(v) ˜ U H i (v) = ∪w∈˜ UH i−1(v)Lw[˜ τ ′ i] ∩Ht3 ˜ U I i (v) = ∪w∈˜ UH i−1(v)Lw[˜ τ ′ i] ∩It3. 4 APPLIED RESULTS 17 Note that a node ˜ wi ∈˜ U I i (v) is connected to a node ˜ w−1 ∈˜ U H −1(v) by a path P = ( ˜ wi, . . . , ˜ w0, ˜ w−1), where ˜ wi−1, . . . , ˜ w0, ˜ w−1 ∈Ht3, and ˜ wj+1 ∈L ˜ wj[˜ τ ′ j+1]. Assume that there is some i ⩽ρ log n/φ for which \| ˜ U I i (v)\| > ρ log2 n. Then, for every u ∈˜ U I i (v) there is a path (u, ˜ wi, . . . , ˜ w0, . . . , v), where all nodes ˜ wi, . . . , ˜ w0, . . . , v ∈Ht3. If u performs pull when ˜ wi opens a channel to u, then v becomes informed at the end of Phase 3. However, there are more than ρ log2 n such nodes u, and at least one of them performs pull at the right time with probability 1 − 1 − 1 log n ρ log2 n = 1 −o(n−3), if ρ is large enough. If there is no i ⩽ρ log n/φ for which \| ˜ U I i (v)\| > ρ log2 n, then we can apply Lemma 3.3, and obtain that \| ˜ U H ρ log n(v)\| > n1−ε′, where ε′ can be made an arbitrarily small constant by choosing ρ large enough. Since the hypercube is vertex-transitive, the same result holds for any v that is uninformed at the end of Phase 3. Our goal is now to apply Lemma 3.4. In that notation, we let U = ˜ U H ρ log n(v). Using Lemma 3.2 and Lemma 3.3, it follows that \|I+ t2 ∩Nℓ2\| ⩾n1−ε and It2 ∩N>ℓ2 = ∅, so that the ﬁrst two preconditions of Lemma 3.4 are satisﬁed. We now choose ε′ := min{ε/2, δ/2}, where ε > 0 and δ = δ(ε) > 0 are the constants from Lemma 3.3. By Lemma 4.4, \|U ∩N⩾(1/2−δ/2)d\| ⩾1 2n1−ε′ ⩾n1−ε, so that the third precondition of Lemma 3.4 also holds. Applying now Lemma 3.4 we obtain that, with probability 1 −n−3, at least one node in U = ˜ U H ρ log n(v) becomes informed in Phase 1, and thus v cannot be uninformed by the end of Phase 3. 4 Applied Results In this section we ﬁrst state some technical results (Section 4.1) that were used for the analysis of the algorithm for hypercubes. In Section 4.2 we show some properties of random graphs that were used for the analysis of the random graph algorithm. Finally, in Section 4.3 we state two tail estimates which we applied in this paper. 4.1 Technical Claims Lemma 4.1. There is a suﬃciently large constant K, such that for any d ∈N and for any 1 ⩽j ⩽d/ √ K, d + Kj Kj ⩽d −j K2/3j . Proof. The claim is equivalent to dK2/3 + K5/3j ⩽Kd −jK, and further rearranging gives j ⩽d · K −K2/3 K5/3 + K ⩽ d √ K , if K is suﬃciently large. 4 APPLIED RESULTS 18 Lemma 4.2. Let ε > 0 be any small constant. Then there is a suﬃciently large constant K > 0, so that it holds for any d ∈N that d/2 Y j=1 1 + d Kj ⩽2(ε/3)d. Proof. For suﬃciently large K we have, d/2 Y j=1 1 + d Kj ⩽ d/ √ K Y j=1 1 + d Kj · d/2 Y j=d/ √ K+1 1 + d √ Kd ⩽ d/ √ K Y j=1 d + Kj Kj · 1 + 1 √ K d/2−d/ √ K ⩽ d/ √ K Y j=1 d −j K2/3j · 1 + 1 √ K d/2−d/ √ K (by Lemma 4.1) ⩽ d d/ √ K · K−(2/3)·(d/ √ K) · 1 + 1 √ K d/2 ⩽ √ Ke d/ √ K · K−(2/3)·(d/ √ K) · 1 + 1 √ K d/2 ⩽2(1/100)·d. Lemma 4.3. Let x1, x2, . . . , xn ∈[0, M] and X := Pn i=1 xi. Then it holds for any λ > 0, n X i=1 λ−xi ⩽X M · λ−M + n −X M · λ0. Proof. As the function x 7→λ−x is convex, we have that λ−xi = λ−(xi/M)·M−(1−xi/M)·0 ⩽xi M · λ−M + 1 −xi M · λ−0. This implies n X i=1 λ−xi ⩽ n X i=1 xi M · λ−M + n X i=1 1 −xi M · λ−0 = X M · λ−M + n −X M · λ0, as needed. Lemma 4.4. Let U ⊆{0, 1}d be any subset of the hypercube. For any 0 ⩽i ⩽d, let N⩾i := {v ∈ {0, 1}d : \|v\|1 ⩾i}. If \|U\| ⩾n(1−ε) for a constant ε > 0, then it follows that \|U ∩N⩾(1/2−ε)d+1\| ⩾ (1/2)n1−ε for suﬃciently large d. 4 APPLIED RESULTS 19 Proof. We ﬁrst upper bound the size N⩽(1/2−ε)d. In order to do that, we use the probabilistic method. More speciﬁcally, let X1, . . . , Xd be independent 0/1-random variables with Pr [ Xi = 1 ] = Pr [ Xi = 0 ] = 1/2. Let X := Pd i=1 Xi and µ := E [ X ] = d/2. Then by the Hoeﬀding bound, N⩽(1/2−ε)d ⩽2d · Pr [ X ⩾E [ X ] −εd ] ⩽2d · e−(εd)2 d ⩽1 2 · 2(1−ε)d. Hence, \|U ∩N⩾(1/2−ε)d+1\| ⩾\|U\| − N⩽(1/2−ε)d ⩾1 2n1−ε. 4.2 Random Graph Properties In this section we show some combinatorial properties of random graphs. Some of these properties (Lemmas 4.5 and 4.6) are also derived (in a modiﬁed form) in for (almost) Ramanujan graphs. Recall that we consider G = G(n, p) with (log2 n)/n ⩽p ⩽2o(√log n)/n. Lemma 4.5. Let ε < 1 be a suitably chosen constant, and ﬁx x ∈[6n/d, n/2]. Let X = {(X, Y )\| Y ⊂X ⊂V, \|X\| = x, and \|Y \| = x/4}, and let N(u, S) := {v ∈S \| (u, v) ∈E} for some u ∈V and S ⊆V . 1. For x ⩽n/40 let A be the event that for all (X, Y ) ∈X there exists Y ′ ⊂V \ X such that 1) \|Y ′\| = n/6, and 2) for all y′ ∈Y ′ : \|N(y′, Y )\| ⩾\|Y \| · d/(2n). Then Pr [ A ] ⩾1 −εn. 2. For n/40 < x ⩽n/2 let B be the event that for all (X, Y ) ∈X there exists Y ′ ⊂V \ X such that 1) \|Y ′\| = (n −x)/2, and 2) for all y′ ∈Y ′ : \|N(y′, Y )\| ⩾\|Y \| · d/(2n). Then Pr [ B ] ⩾1 −εn. Proof. We deﬁne y = \|Y \|. Now ﬁx two subsets Y1 of size y and X1 of size x with Y1 ⊂X1. For 1 ⩽i ⩽n −x, let Zi = 1 if the ith node of V \ X1 has at most yd/(2n) neighbors in Y1 and 0 otherwise. Deﬁne Z = Z1 + Z2 · · · + Zn−x. Every node of V \ X1 is connected to a ﬁxed node of Y1 with probability p, independently form all other nodes. With d = pn we get for 1 ⩽i ⩽n −x Pr [ Zi = 1 ] ⩽ y X i=y−yd/(2n) y i · (1 −p)i · py−i ⩽ 1 −p 1 −d/(2n) y(1−d/(2n)) · p d/(2n) yd/2n ⩽ 1 −d 2n y(1−d/(2n)) · 2yd/2n ⩽ 1 e y· d 2n ·(1−d/(2n)) · 2yd/2n ⩽ 2 e · (1 −o(1)) yd/(2n) ⩽ 2 · (1 + o(1)) e yd/(2n) Now we consider two cases, depending on x. 4 APPLIED RESULTS 20 Case 1: x ⩽n/40. In this case we have Pr [ Zi = 1 ] ⩽ 2 · (1 + o(1)) e 3 < 1/2. This gives us E[Z] ⩽(n −x)/2. Now we can use Chernoﬀbounds (Equation (12) in ) and show that with x ⩽n/40 we get Pr [ Z ⩾5n/6 −x ] ⩽ (n −x)/2 5n/6 −x 5n/6−x · (n −x)/2 n −x −(5n/6 −x) n/6 ⩽ (n −n/40)/2 5n/6 −n/40 5n/6−n/40 · (n)/2 n −(5n/6) n/6 ⩽ 117 194 5n/6−n/40 · 3n/6 ⩽ 4 5 n Now we can conclude that, with probability 1 −(4/5)−n we have at least n/6 nodes in V \ X with yd/(2n) (or more) neighbors in Y1. There are n x diﬀerent ways to choose the nodes for the set X1. Furthermore, there are x x/4 possible ways to choose Y1 as subsets of size x/4 from the nodes of X. Thus, for every subset Y of size x/4 of an arbitrary set X there exists a subset Y ′ ⊂V \ X of size n/6 such that each node of Y ′ has at least yd/2n neighbors in Y with probability at least 1 − n x · x x/4 · 4 5 n ⩾ 1 − n · e x x · x · e x/4 x/4 · 4 5 n ⩾ 1 − n · e n/40 n/40 · n/40 · e n/160 n/160 · 4 5 n = 1 −(40e)n/40 · (40e)n/160 · 4 5 n ⩽1 −εn. Case 2: n/40 ⩽\|X\| ⩽n/2. In this case we have Pr [ Zi = 1 ] ⩽ 2 e · (1 −o(1)) yd/(2n) ⩽ 2 · (1 + o(1)) e yd/(2n) ⩽ 2 · (1 + o(1)) e d/(320) ⩽1 n2 and Pr [ Z ⩾(n −x)/2 ] ⩽ n −x (n −x)/2 · 1 n2 (n−x)/2 ⩽ 2e n2 (n−x)/2 . With the same probability there are at least (n −\|X\|)/2 nodes in V \ X with yd/(2n) or fewer neighbors in Y1. Again, for every subset Y of size x/4 of an arbitrary set X there exists a subset 4 APPLIED RESULTS 21 Y ′ ⊂V \X of size n/6 such that each node of Y ′ has at least yd/2n neighbors in Y with probability at least 1 − n x · x x/4 · 2e n2 (n−x)/2 ⩾1 −2n · (4e)x/4 · 2e n2 (n−x)/2 ⩾1 −2n · (4e)n/8 · 2e n2 n/4 ⩾1 −εn Next, we consider the connectivity between two very large sets. Lemma 4.6. Let ε < 1 be a suitably chosen constant, and let X = {(X, Y )\| Y ⊂X ⊂V, \|X\| = n/2, and \|Y \| = n/5}. Let C be the event that for all (X, Y ) ∈X there exists Y ′ ⊂V \ X such that 1) \|Y ′\| = n/2 − n log log n/ log n, and 2) for all y′ ∈Y ′ : \|N(y′, Y )\| ⩾d/10. Then Pr [ C ] ⩾1 −εn. Proof. The proof is similar to the proof of Lemma 4.5. Let X1 and Y1 ⊆X1 be two sets of size n/2 and n/5, respectively. For 1 ⩽i ⩽n −x, let Zi = 1 if the ith node of V \ X1 has at most d/10 neighbors in Y1 and 0 otherwise. Deﬁne Z = Z1 + Z2 · · · + Zn−x. Similar to Lemma 4.5 we get for 1 ⩽i ⩽n −x Pr [ Zi = 1 ] ⩽ 2(1 + o(1)) e d/10 . Thus, there are more than n log log n/ log n nodes in V \ X1 with fewer than d/10 neighbors in Y1 with probability n/2 X i= n log log n log n n/2 i p′d/10(1 −p′)n/2−i ⩽ p′ log log n/ log n n log log n log n 1 −p′ 1 −log log n/ log n n 2 1−log log n log n , which equals e−Ω(n log log n). Now there are n n/2 diﬀerent subdivisions of V into two subsets of size n/2 and there are n/2 n/5 diﬀerent subsets of size n/5 in a subset of size n/2. Thus, the statement of the lemma holds with probability 1 − n n/2 · n/2 n/5 · e−Ω(n log log n) ⩾1 −(2e)n/2 · 5e 2 n/5 e−Ω(n log log n) ⩾1 −e−Ω(n log log n). 4 APPLIED RESULTS 22 The next lemma deals with the local neighborhood-structure around a node in a G(n, p) graph. Lemma 4.7. Let v ∈V be an arbitrary node in G(n, p) and let T(v) be the graph induced by the nodes at distance at most ρ log log n from v. Then, with probability 1 −o(n−3), the graph T(v) is either a tree, or there are at most 4 edges which violate the tree property in T(v). Proof. In the following we denote v as the root of T(v). The nodes at distance ℓfrom v are called nodes on level ℓin the following. Then v is on level 0. For w, w′ ∈T(v), w is called ancestor of node w′ iﬀdist(v, w) < dist(v, w′). For ⩽i ⩽ρ log log n, Ni will be the set of nodes on level i. We can assume that G(n, p) is constructed by the following procedure. In the ﬁrst step v = v0 1 draws an edge to every other node with probability p. This gives our level 1 nodes N1 = {v1 1, v1 2, . . . }. In the second step the edges between v1 1 and V \ {N0, N1} are chosen in the same way. Then we choose the edges between v1 2 and V \ {N0, N1}, v1 3 and V \ {N0, N1}, . . ., until all nodes in N1 are considered. The nodes that are connected to nodes in N1 in this way are the nodes of N2 = v2 1, v2 2, . . .. We do the same for the nodes nodes in N2 (considering only nodes in V \ {N0, N1, N2}) , which gives us the set N3 = v3 1, v3 2, . . ., and so on. For 1 ⩽i ⩽ρ log log n −1 we use the nodes in Ni to create Ni+1 = vi+1 1 , vi+1 2 , . . .. For vi j ∈Ni we consider only the nodes in V \ {N0, N1, . . . Ni}. So far we created a tree with some additional edges. Two diﬀerent nodes on level j can be connected to the same node on level j +1. So far we did not evaluate all events for the nodes in the tree, since a node on level j can also be connected to other nodes on the same level. We evaluate these events at the end of the process. In the following we call these edges cycle edges and our goal is to upper bound the number of cycle edges in the tree. First we calculate the maximum number of nodes of T(v). Recall that p ⩽2o(√log n)/n. We know that with probability 1 −o(n−5) the maximum degree of G is 2 · pn . Hence, w.h.p. T(v) has at most (2 · pn)ρ log log n ⩽(2 · 2o(√log n))ρ log log n ⩽logO(1) n. many nodes. Hence, the probability that there are more than 4 cycle edges is at most (logO(1) n)2 5 · p5 ⩽(log n)O(1) · 2o(√log n) n !5 ⩽1 n4 4.3 Probabilistic and Combinatorial Tools Lemma 4.8 (Method of Bounded Independent Diﬀerences, ). Let Xi : Ωi →R, 1 ⩽i ⩽n, be mutually independent random variables. Let f : Qn i=1 Ωi →R satisfy the Lipschitz condition \|f(x) −f(x′)\| ⩽ci where x and x′ diﬀer only in the i-th coordinate, 1 ⩽i ⩽n. Let Y be the random variable f(X1, . . . , Xn). Then for any t ⩾0, Pr [ Y > E [ Y ] + t ] ⩽exp −2t2. n X i=1 c2 i ! , 5 CONCLUSIONS 23 and Pr [ Y < E [ Y ] −t ] ⩽exp −2t2. n X i=1 c2 i ! , We need the following standard Chernoﬀbound. Lemma 4.9. Consider some ﬁxed 0 < δ < 1. Suppose that X1, . . . , Xn are independent geometric random variables on N with Pr [ Xi = k ] = (1 −δ)k−1δ for every k ∈N. Let X = Pn i=1 Xi, µ = E [ X ]. Then it holds for all ε > 0 that Pr [ X ⩾(1 + ε)n/δ ] ⩽e−ε2n/2(1+ε) 5 Conclusions In this paper we consider rumor spreading on random graphs and hypercubes in the quasiran-dom phone call model. We show two results. For random graphs we present an address-oblivious algorithm with runtime O(log n) that uses at most O(n log log n) message transmissions. For hy-percubes of dimension log n we present an address-oblivious algorithm with runtime O(log n) that uses at most O(n(log log n)2) message transmissions. Together with a result of , our results imply that for random graphs the communication complexity of the quasi random phone call model is signiﬁcantly smaller than that of the standard phone call model. Open problems include a generalisation of our results for general random graphs where the nodes can have very diﬀerent degrees. Also, it might be interesting to show results for additional deterministic graphs like star graphs or grids. And, of course, another open problem is to show (this time a correct proof) of the lower bound of that is stated in . Acknowledgements. We would like to thank the reviewers of this journal version for their helpful comments. REFERENCES 24 References P. Berenbrink, R. Els¨ asser, T. Friedetzky. Eﬃcient Randomised Broadcasting in Random Regular Networks with Applications in Peer-to-Peer Systems. In Proc. of PODC’08, pages 155–164, 2008. P. Berenbrink, R. Els¨ asser and T. Sauerwald. Randomised Broadcasting: Memory vs. Randomness. In Proc. of LATIN’10, pages 306–319, 2010. P. Berenbrink, R. Els¨ asser and T. Sauerwald. Communication Complexity of Quasirandom Rumor Spreading. In Proc. of ESA’10, pages 134–145, 2010. B. Bollob´ as. Random Graphs. Academic Press, 1985. F. Chierichetti, S. Lattanzi, and A. Panconesi. Almost Tight Bounds for Rumour Spreading with Conductance. In Proc. of STOC’10, pages 399–408, 2010. A. Demers, D. Greene, C. Hauser, W. Irish, J. Larson, S. Shenker, H. Sturgis, D. Swinehart, and D. Terry. Epidemic algorithms for replicated database maintenance. In Proc. of PODC’87, pages 1–12, 1987. B. Doerr, T. Friedrich, T. Sauerwald. Quasirandom Rumor Spreading In Proc. of SODA’08, pages 773–781, 2008. B. Doerr, T. Friedrich, T. Sauerwald. Quasirandom rumor spreading: expanders, push vs. pull, and robustness In Proc. of ICALP’09, pages 366-377, 2009. R. Els¨ asser. On the communication complexity of randomized broadcasting in random-like graphs. In Proc. of SPAA’06, pages 148–157, 2006. R. Els¨ asser and T. Sauerwald. The power of memory in randomized broadcasting. In Proc. of SODA’08, pages 218–227, 2008. P. Erd˝ os and A. R´ enyi. On random graphs I. Publ. Math. Debrecen, 6:290–297, 1959. U. Feige, D. Peleg, P. Raghavan, and E. Upfal. Randomized broadcast in networks. Random Structures and Algorithms, 1(4):447–460, 1990. A.M. Frieze and G.R. Grimmett. The shortest-path problem for graphs with random arc-lengths. Discrete Applied Mathematics, 10:57–77, 1985. N. Fountoulakis, A. Huber and K. Panagiotou. Reliable broadcasting in random networks and the eﬀect of density. In Proc. of INFOCOM’10, pages 2552–2560, 2010. G. Giakkoupis. Tight bounds for rumor spreading in graphs of a given conductance. In Proc. of STACS’11, pages 57-68, 2011. T. Hagerup and C. R¨ ub. A guided tour of Chernoﬀbounds. Information Processing Letters, 36(6):305– 308, 1990. R. Karp, C. Schindelhauer, S. Shenker, and B. V¨ ocking. Randomized rumor spreading. In Proc. of FOCS’00, pages 565–574, 2000. C. McDiarmid. On the method of bounded diﬀerences. Surveys in Combinatorics, 1989, J. Siemons ed., London Mathematical Society Lectures Note Series 141, Cambridge University Press, 1989, 148-188. M. Mitzenmacher and E. Upfal. Probability and Computing: Randomized Algorithms and Probabilistic Analysis. Cambridge University Press, 2005. B. Pittel. On spreading a rumor. SIAM Journal on Applied Mathematics, 47(1):213–223, 1987.
6237	https://www.youtube.com/watch?v=E34CftP455k	Dot vs. cross product \| Physics \| Khan Academy Khan Academy 9090000 subscribers 6113 likes Description 793825 views Posted: 9 Aug 2008 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Understanding the differences between the dot and cross products. Created by Sal Khan. Watch the next lesson: Missed the previous lesson? Physics on Khan Academy: Physics is the study of the basic principles that govern the physical world around us. We'll start by looking at motion itself. Then, we'll learn about forces, momentum, energy, and other concepts in lots of different physical situations. To get the most out of physics, you'll need a solid understanding of algebra and a basic understanding of trigonometry. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Physics channel: Subscribe to Khan Academy: 204 comments Transcript: Let's do a little compare and contrast between the dot product and the cross product. Let me just make two vectors-- just visually draw them. And maybe if we have time, we'll, actually figure out some dot and cross products with real vectors. Let's call the first one-- That's the angle between them. OK. So let's just go over the definitions and then we'll work on the intuition. And hopefully, you have a little bit of both already. So what is a dot b? Well first of all, that's the exact same thing as b dot a. Order does not matter when you take the dot product because you end up with just a number. And that is equal to the magnitude of a times the magnitude of b times cosine of the angle between them. Let's look at the definition of the cross product. What is a cross b? Well first of all, that does not equal b cross a. It actually equals the opposite direction, or you could view it as the negative of b cross a. Because the vector that you end up with ends up flipped, whichever order you do it in. But a cross b, that is equal to the magnitude of vector a times the magnitude of vector b-- so far, it looks a lot like the dot product, but this is where the diverge is-- times the sine of the angle between them. The sine of the angle between them. And this is where it really diverges. When we took the dot product, we just ended up with a number. This is just a number. There's no direction here. This is just a scalar quantity. But the cross product, we take the magnitude of a times the magnitude of b, times the sine of the angle between them, and that provides a magnitude, but it also has a direction. And that direction is provided by this normal vector. It's a unit vector. A unit vector gets that little hat on it. It's a unit vector, and what direction is it? Well, that's defined by the right hand rule. This is a vector. It's perpendicular to both a and b. And then you might say, a and b, the way I drew them, they're both sitting in the plane of this video screen, or your video screen. So in order for something to be perpendicular to both of these, it either has to pop out of the screen or pop into the screen, right? And when you learned about the cross product, I said there's two ways of showing a vector popping out of the screen. It looks like that because that's the tip of an arrow. And to show a vector going into the screen, it's like that because that is the back of an arrow. The rear end of an arrow. So how do you know which of these two it is? Because both of these vectors are perpendicular to a and b. That's where you take your right hand and you use the right hand rule. So you take your index finger in the direction of a, your middle finger in the direction of b, and then your thumb points in the direction of n. So let's do that. I'm looking at my hand. It's not an easy thing to do with your right hand, but your right hand is going to look something like this. Your index finger will go in the direction of a. Your middle finger goes in the direction of b. So that's my middle finger. And then my other two fingers just do what they need to do. I like to just bend them out of the way. So they just curl around my hand. And then what direction is my thumb in? My thumb-- well, actually I drew it at the wrong angle. My thumb is actually going in this direction, right? Into the page. This is the top of my hand. These are like my veins. Or, if I actually drew it correctly, where you would see your hand from side-- so it would look like this. You would see your pinky. Your palm and your pinky would be like that. And your other finger like this. Your middle finger would go in the direction of b. Your index finger goes in the direction of a, and you wouldn't even see your thumb, because your thumb is pointing straight down. But I think you get the point. a cross b, this n vector is pointing straight down. It's a unit vector. And this provides the magnitude. Unit vector just means it has a magnitude of one. So the magnitudes of the cross and the dot products seem pretty close. They both have the magnitude of both vectors there. Dot product, cosine theta. Cross product sine of theta. But then, the huge difference is that sine of theta has a direction. It is a different vector that is perpendicular to both of these. Now, let's get the intuition. And if you've watched the videos on the dot and the cross product, hopefully you have a little intuition. But I review it because I think it all fits together when you see them with each other. First, let's study a, b cosine of theta. If you watched the dot product video, cosine of theta, if you took, let's say, b cosine of theta. What is b cosine of theta? b cosine of theta-- and you could work it out on your own time-- if you say cosine is adjacent over hypotenuse, the magnitude of b cosine theta is actually going to be the magnitude of, if you dropped a perpendicular-- I'll use a different color here-- if you dropped a perpendicular here, this length right here-- that's b cosine theta. Let me draw it separately. I don't want to mess up this picture too much. So if that's b. If that's a-- And that's b. That's a. This is theta. b cosine theta, if you drop a line perpendicular to a, this is a right angle. b cosine theta, adjacent over hypotenuse is equal to cosine theta. So it would be the projection of b going in the same direction as a. So it would be this magnitude. That is b cosine theta. So the magnitude of that vector right there is the magnitude of b cosine of theta. So when you're taking the dot product, at least the example I just did, if you view it as the magnitude of a times the magnitude of b cosine theta, you're saying what part of b goes in the same direction as a? And whatever that magnitude is, let me just multiply that times the magnitude of a. And I have the dot product. Let's take the pieces that go the same direction and multiply them. So how much do they move together? Or do they point together? Or you could view it the other way. You could view the dot product as-- and I did this in the dot product video-- you could view it as a cosine of theta, b. Because it doesn't matter. These are all scalar quantities, so it doesn't matter what order you take the multiplication in. And a cosine theta is the same thing. It's the magnitude of the a vector that's going in the same direction of b. Or the projection of a onto b. So this vector right here is a cosine theta; the magnitude of a cosine theta. And they're actually the same number. If you take how much of b goes in the direction of a, and multiply that with the magnitude of a, that gives you the same number as how much of a goes in the direction of b, and then multiply the two magnitudes. Now, what is a, b sine theta? a, b, sine theta. Well if this vector right here is a cosine theta-- and you learned this when you learned how to take the components of vectors. This vector right here is the magnitude of a sine theta. You could rewrite this as the magnitude of a sine theta times the magnitude of b in that normal vector direction. So if you take a sine theta times b, you're saying what part of a doesn't go the same direction as b. What part of a is completely perpendicular to b-- has nothing to do is b. They share nothing in common. It goes in a completely different direction. That's a sine theta. And so you take the product of this with b and then you get a third vector. And it almost says, how different are these two vectors? And it points in a different direction. It gives you this-- sometimes it's called a pseudo vector, because it applies to some concepts that are pseudo vectors. But the most important of these concepts is torque, when we talk about the magnetic field; the force of a magnetic field on electric charge. These are all forces, or these are all physical phenomena, where what matters isn't the direction of the force with another vector, it's the direction of the force perpendicular to another vector. And so that's where the cross product comes in useful. Anyway, hopefully, that gave you a little intuition. And you could have done it the other way. You could have written this as b sine theta. And then you would have said that's the component of b that is perpendicular to a. So b sine theta actually would have been this vector. Or let me draw it here. That would make more sense. This would be b sine theta. So you could switch orders. You could visualize it either way. You could say this is the magnitude of b that is completely perpendicular to a, multiply the two, and use the right hand rule to get that normal vector. And we just decided that we're going to use the right hand rule to have a common convention. But people could have used the left hand rule, or they might have used it a different way. It's just a way that we have a consistent framework, so that when we take the cross product we all know what direction that normal vector is pointing in. Anyway. In the next video I'll show you how to actually compute dot and cross products when you're given them in their component notation. See you in the next video.
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6239	https://math.stackexchange.com/questions/2995071/possibly-wrong-question-in-s-l-loney-coordinate-geometry	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Possibly wrong question in S L Loney Coordinate Geometry Ask Question Asked Modified 6 years, 10 months ago Viewed 151 times 0 $\begingroup$ Given question: $P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$ What I did: I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome $$RM = a(t_1^2-t_3^2)$$ $$RN = a(t_2^2-t_3^2)$$ $$RV = -a(t_1-t_3)(t_2-t_3)$$ Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong? analytic-geometry Share asked Nov 12, 2018 at 9:04 Shubhraneel PalShubhraneel Pal 1,14077 silver badges1717 bronze badges $\endgroup$ 3 $\begingroup$ Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve. $\endgroup$ Saradamani – Saradamani 2018-11-12 09:23:09 +00:00 Commented Nov 12, 2018 at 9:23 $\begingroup$ "Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse. $\endgroup$ Intelligenti pauca – Intelligenti pauca 2018-11-12 19:09:48 +00:00 Commented Nov 12, 2018 at 19:09 $\begingroup$ @Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right $\endgroup$ Shubhraneel Pal – Shubhraneel Pal 2018-11-13 08:56:51 +00:00 Commented Nov 13, 2018 at 8:56 Add a comment \| 1 Answer 1 Reset to default 3 $\begingroup$ In a parabola the abscissa is proportional to the square or the related ordinate, that is: $$ RM=kPM^2,\quad RN=kQN^2. $$ On the other hand, by similar triangles we have: $$ \begin{align} VM/VN &= PM/QN \ (VM+VN)/VN &= (PM+QN)/QN \ (RN-RM)/VN &= (PM+QN)/QN \ k(QN^2-PM^2)/VN &= (PM+QN)/QN \ k(QN-PM)QN &= VN \ kQN^2-kPM\cdot QN &= VN \ RN-kPM\cdot QN &= VN \ RN-VN &= kPM\cdot QN \ RV &= kPM\cdot QN. \end{align} $$ Hence: $$ RV^2 = kPM^2\cdot kQN^2 =RM\cdot RN. $$ Share edited Nov 13, 2018 at 7:08 answered Nov 12, 2018 at 20:43 Intelligenti paucaIntelligenti pauca 56.2k44 gold badges5050 silver badges9191 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. 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6240	https://www.youtube.com/watch?v=z06Hj4vO9FM&pp=0gcJCdgAo7VqN5tD	HOW TO FIND ARC LENGTH AND SECTOR AREA OF CIRCLES \| GEOMETRY your math tutor 14200 subscribers 2601 likes Description 183162 views Posted: 20 Apr 2021 Learn how to find the arc length and sector area of circles using proportions. We will go over the formula for arc length and sector area, but this video also walks through how to set up the equation without memorizing it. There are 3 arc length example problems and 3 sector area example problems. Timestamps: 00:00 - Intro 00:40 - Arc Length Problem 1 03:22 - Arc Length Problem 2 04:34 - Arc Length Problem 3 06:53 - Sector Area Problem 1 08:49 - Sector Area Problem 2 09:45 - Sector Area Problem 3 Be sure to subscribe for more remote learning and math videos: Tags: your math tutor, geometry, arc length geometry, sector area geometry, how to find arc length, how to find sector area, how to find arc length and sector area, arc length of circle, sector area of circle, length of an arc, arc length tutorial ,finding arc length, finding sector area, arc length examples 84 comments Transcript: Intro hey guys it's your math tutor today we're talking about how to find the arc length and sector area of circles in geometry let's go we should first talk about what we mean when we say arc length here we have a circle and you all know what a circumference is right it's the length of this outside perimeter of the circle and you can calculate the full perimeter by using the formula 2 pi r where r is the radius now when we're talking about arc length an arc is defined as any piece of the circumference so arc length is just how long that piece is Arc Length Problem 1 looking at our first arc length problem we're asked to find the arc length of a b our first step is just going to be to identify the arc because we're told it's arc a b it's going to be the distance between this point a to point b in our problem we're also given that our radius which is the distance from the center to the perimeter is 6 meters and this angle is 80 degrees our second step is to come up with an equation to solve for arc length and there is a formula for arc length which is this one here but instead of just memorizing it let's think through it because let's be honest there are too many formulas out there for us to memorize them all when we look at arc length we already talked about how it's just a piece of the circumference right well the length of the arc is proportional to how big or small this angle is the bigger the angle the bigger the arc length and the smaller the angle the smaller the arc length and this is exactly how the formula works it's just a proportion to put it into an equation we're trying to find the length of the arc i'm just going to call it l and this arc length is just a fraction of the total circumference which is 2 pi r where r is 6 meters now like we talked about these lengths we just wrote down are proportional to the angles the angle for this arc is 80 degrees and those 80 degrees are just a fraction of the total number of degrees in a circle which is 360. if we look at the equation it makes sense because if the angle of an arc is 360 degrees then the arc length is just the full perimeter of the circle which is 2 pi r we're just using that proportion to find if the angle is only 80 degrees what is the arc length and yeah we just came up with the arc length formula in class you might also see it written in terms of arc length but it's the same formula just rearranged okay we have all of the numbers we need to solve for our arc length so solving for l we'll get the length is about 8.4 meters which is our answer Arc Length Problem 2 for our second problem we're actually using the same circle but now it's asking us to find the arc length of atb which is this longer arc length here we can use the same idea and use proportions to come up with an equation we're trying to find arc length l which is just a piece of the total circumference 2 pi r and the lengths are proportional to the angle measures the angle of the arc we're looking at is 280 degrees i got that by taking the total number of degrees in a circle which is 360 and subtracting 80. let's put that over the total degrees in a circle which is 360. solving for l will get about 29.3 meters as our arc length Arc Length Problem 3 let's do a final challenge problem for arc links before we get into sector areas this problem is asking us to find the arc length of a b and c d which are these two arcs here we're given that the diameter is three inches and we're not directly given the angle measures that we want but we are given that this angle is 130 degrees we can still use proportions to figure out the equation but let's just do one arc at a time to make it easier i'm going to choose to do arc a b first if we try to set up the proportions we're trying to figure out our arc length l and let's put that over the entire circumference which is 2 pi r where r is 1.5 inches the lengths are proportional to the angle measures but we actually don't know the angle measure of our arc or do we if we look at this angle we know for sure that it's 130 and we know it's on the same straight line as the angle of our arc if you remember because it's on the same line we know that these two angles add up to 180 180 minus 130 is 50 degrees which is the measure of our arc let's put that over the total number of degrees in a circle and we can solve our equation we'll get the arc length of a b is 1.3 inches however we're not done yet because the question asked us to find the arc length of a b and c d but lucky for us these arc lengths are actually the same because if you look at these two angles they're directly across from each other which means they're vertical angles and if you remember vertical angles always have the same angle measures so that will also mean that their arc lengths are the same too if you don't believe me you can plug everything into the arc length formula again but i'm just going to double what we got for a b and the arc length of the two arcs added together is 2.6 inches that's it Sector Area Problem 1 all right moving on to sector areas we still have the same circle we're working with but now it's asking us to find the area of the shaded region this region this pie slice is called the sector area we can think about sector area the same way we thought about arc length where arc length is just a piece of the total circumference sector area is just a piece of the total area looking at this problem it's already done the first step for us which is identifying the sector area it's this slice shaded in bright orange now let's set up our equation we can also set up the equation using proportions just like how we did it for arc length because in the same way the bigger the angle the bigger the slice or sector area and the smaller the angle the smaller the slice we're trying to find a the area of the sector and we can put it over the total area of the circle which is defined as pi r squared where r is 6 meters this area is proportional to angle measure and we know the angle we're working with is 80 degrees let's put that over the total number of degrees in a circle which is 360. this makes sense if you think about it because if the angle of the sector area was a full 360 degrees the sector area would be the entire area of the circle which is pi r squared now if the angle were just 80 degrees we want to figure out what the area of the slice would be let's solve for a and we'll get 25.1 meters squared as our area that's our answer Sector Area Problem 2 for our next problem again we have the same circle but we're asked to find the area of this larger sector here again let's do the same thing and set up our equation we're trying to find the sector area a let's put it over the total area pi r squared and this is proportional to the angle measures our angle measure is 280 because 360 minus 80 equals 280 and let's put that over the total number of degrees 360. solving for area we'll get about 88 meters squared for our answer Sector Area Problem 3 all right last question and this doesn't look as challenging now right we're trying to figure out the area of these two orange sectors and again let's just do one sector at a time and set up the equation using proportions i'm gonna go for this one first we're trying to find the area a and let's put that over the total area of the circle pi r squared and in this case r is 1.5 the area is proportional to the angle measure and this problem doesn't directly tell us the angle measure of the sector but because it's on the same line as this 130 degree angle and we know two angles on the same line add up to 180 we know that this angle is equal to 50. let's put that over the angle measure of a full circle which is 360 and we'll solve for the area we'll get the area of this sector is one inch squared this is only one of the sectors but it's pretty easy to figure out the rest the area of the other sector is actually the same as this one because these two angles are vertical angles that means they're the same size and again if you don't believe me you can set up the equation for this sector but i'm just going to double the area that we got earlier and this is our final answer 2 inches squared all right those were all of the problems i wanted to go over today let me know in the comments below if you have any questions or other topics you'd want me to cover please consider giving this video a thumbs up and subscribe for more math tutorials like this i'll see you guys in the next one
6241	https://math.stackexchange.com/questions/999319/how-solutions-of-distinct-non-negative-solutions-are-there-to-k-1-cdotsk-n-k	combinatorics - How solutions of distinct non-negative solutions are there to $k_1+\cdots+k_n=k$? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How solutions of distinct non-negative solutions are there to k 1+⋯+k n=k k 1+⋯+k n=k? Ask Question Asked 10 years, 11 months ago Modified4 years, 2 months ago Viewed 2k times This question shows research effort; it is useful and clear 14 Save this question. Show activity on this post. How many distinct n n-tuples with distinct non-negative integer elements are there that add to k k. For example there are 6 6 triples that add to 4 4. Namely (0,1,3)(0,1,3) and its 6 6 permutations. Is there a formula for this amount? I have tried very hard to do it but with no luck. This question can also be rephrased as: How many sets of non-negative solutions are there to k 1+⋯+k n=k k 1+⋯+k n=k where k i≠k j k i≠k j. It is obvious that the smallest k k would be n(n−1)2 n(n−1)2. Another example would be How many pairs of distinct non-negative integers are there that add to 6 6? Clearly this is the number of compositions of length 2 2 with distinct terms and 2!2! times the number of compositions of length 1 1 with distinct terms (how to count the zeros). We get 0+6 0+6, 1+5 1+5, 2+3 2+3, 3+2 3+2, 5+1 5+1, 6+0 6+0 So there are 6 6 such pairs. I should note that the answer may be given in terms of the partition function. Which gives how many ways can an integer be written as a sum of positive .integers. combinatorics integer-partitions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 12, 2015 at 15:53 Ali CaglayanAli Caglayan asked Oct 31, 2014 at 2:45 Ali CaglayanAli Caglayan 5,833 10 10 gold badges 50 50 silver badges 74 74 bronze badges 1 That's an extremely difficult question... Look at this Wikipedia article to get an idea.Najib Idrissi –Najib Idrissi 2014-11-02 10:54:11 +00:00 Commented Nov 2, 2014 at 10:54 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 11 Save this answer. +100 This answer has been awarded bounties worth 100 reputation by Kaj Hansen Show activity on this post. By way of enrichment I would like to point out that using the Polya Enumeration Theorem the closed form is also given by n![z k]Z(P n)(1 1−z)n![z k]Z(P n)(1 1−z) where Z(P n)=Z(A n)−Z(S n)Z(P n)=Z(A n)−Z(S n) is the difference between the cycle index of the alternating group and the cycle index of the symmetric group. This cycle index is known in species theory as the set operator P=n P=n and the species equation here is P=n(E+Z+Z 2+Z 3+⋯).P=n(E+Z+Z 2+Z 3+⋯). Recall the recurrence by Lovasz for the cycle index Z(P n)Z(P n) of the set operator P=n P=n on n n slots, which is Z(P n)=1 n∑l=1 n(−1)l−1 a l Z(P n−l)where Z(P 0)=1.Z(P n)=1 n∑l=1 n(−1)l−1 a l Z(P n−l)where Z(P 0)=1. This recurrence lets us calculate the cycle index Z(P n)Z(P n) very easily. For example when n=3 n=3 as in the introduction to the problem the cycle index is Z(P 3)=1/6 a 1 3−1/2 a 2 a 1+1/3 a 3 Z(P 3)=1/6 a 1 3−1/2 a 2 a 1+1/3 a 3 and the generating function becomes 1/6(1−z)−3−1/2 1(−z 2+1)(1−z)+1/3(−z 3+1)−1 1/6(1−z)−3−1/2 1(−z 2+1)(1−z)+1/3(−z 3+1)−1 which gives the sequence 0,0,6,6,12,18,24,30,42,48,60,72,84,96,…0,0,6,6,12,18,24,30,42,48,60,72,84,96,… which is six times OEIS A069905. Similarly when n=5 n=5 we get the cycle index Z(P 5)=a 1 5 120−1/12 a 2 a 1 3+1/6 a 3 a 1 2+1/8 a 1 a 2 2−1/4 a 4 a 1−1/6 a 2 a 3+1/5 a 5 Z(P 5)=a 1 5 120−1/12 a 2 a 1 3+1/6 a 3 a 1 2+1/8 a 1 a 2 2−1/4 a 4 a 1−1/6 a 2 a 3+1/5 a 5 and the generating function becomes 1 120(1−z)5−1/12 1(−z 2+1)(1−z)3+1/6 1(−z 3+1)(1−z)2+1/8 1(−z 2+1)2(1−z)−1/4 1(−z 4+1)(1−z)−1/6 1(−z 2+1)(−z 3+1)+1/5(−z 5+1)−1 1 120(1−z)5−1/12 1(−z 2+1)(1−z)3+1/6 1(−z 3+1)(1−z)2+1/8 1(−z 2+1)2(1−z)−1/4 1(−z 4+1)(1−z)−1/6 1(−z 2+1)(−z 3+1)+1/5(−z 5+1)−1 which gives the sequence 0,0,0,0,0,0,0,0,0,120,120,240,360,600,…0,0,0,0,0,0,0,0,0,120,120,240,360,600,… which is 120 120 times OEIS A001401. The prefix of zeroes (these two examples start at one) compared to the two OEIS entries represents the fact that the minimum value attainable with n n distinct summands in [z k]Z(P n)(1 1−z)[z k]Z(P n)(1 1−z) is 0+1+2+⋯+n−1=1/2×n×(n−1).0+1+2+⋯+n−1=1/2×n×(n−1). These sequences match the formula by @bof, which is n!p n(k−1 2 n(n−3)).n!p n(k−1 2 n(n−3)). There are many more related links at MSE Meta on Burnside/Polya. The Maple code for these was as follows. p := proc(n, k) option remember; if k=1 then return 1 fi; if n<k then return 0 fi; p(n-1, k-1)+p(n-k,k) end; pet_cycleind_symm := proc(n) local p, s; option remember; if n=0 then return 1; fi; expand(1/nadd(a[l]pet_cycleind_symm(n-l), l=1..n)); end; pet_cycleind_set := proc(n) local p, s; option remember; if n=0 then return 1; fi; expand(1/nadd((-1)^(l-1) a[l]pet_cycleind_set(n-l), l=1..n)); end; pet_varinto_cind := proc(poly, ind) local subs1, subs2, polyvars, indvars, v, pot, res; res := ind; polyvars := indets(poly); indvars := indets(ind); for v in indvars do pot := op(1, v); subs1 := [seq(polyvars[k]=polyvars[k]^pot, k=1..nops(polyvars))]; subs2 := [v=subs(subs1, poly)]; res := subs(subs2, res); od; res; end; q1 := proc(n, k) option remember; local gf; gf := pet_varinto_cind(1/(1-z), pet_cycleind_set(n)); n!coeftayl(gf, z=0, k); end; q2 := proc(n, k) option remember; n!p(k-n(n-3)/2, n); end; Addendum. As per request we now give a mixed (combinatorial, algebraic) proof of the identity [z n]Z(P k)(1 1−z)=p k(n−1 2 k(k−3)).[z n]Z(P k)(1 1−z)=p k(n−1 2 k(k−3)). Observe that we have reverted to the standard convention of using n n for the sum of the partition and k k for the number of parts. By the same construction as before (PET) we have p k(n−1 2 k(k−3))=[z n−1 2 k(k−3)]Z(S k)(z 1−z)p k(n−1 2 k(k−3))=[z n−1 2 k(k−3)]Z(S k)(z 1−z) with Z(S k)Z(S k) being the cycle index of the symmetric group (unlabelled multisets with operator M=k M=k.) Using basic algebra this becomes [z n−1 2 k(k−3)]z k Z(S k)(1 1−z)=[z n−1 2 k(k−3)−1 2 2 k]Z(S k)(1 1−z)=[z n−1 2 k(k−1)]Z(S k)(1 1−z).[z n−1 2 k(k−3)]z k Z(S k)(1 1−z)=[z n−1 2 k(k−3)−1 2 2 k]Z(S k)(1 1−z)=[z n−1 2 k(k−1)]Z(S k)(1 1−z). But this is the species M=k(E+Z+Z 2+Z 3+⋯),M=k(E+Z+Z 2+Z 3+⋯), i.e. partitions with empty constitutents being permitted and constituents not necessarily distinct. There is however a straighforward bijection between these and partitions with potentially empty but distinct constituents. To go from the former to the latter add q q circles on the left of every row in the Ferrers diagram with row indices q q starting at zero. Now if we had two adjacent rows with the first above the second with length b 1 b 1 and a 1 a 1 where b 1≥a 1 b 1≥a 1 then the resulting pair is b 1+q b 1+q and a 1+q−1.a 1+q−1. The difference between these is b 1−a 1+1≥1 b 1−a 1+1≥1 so the new pair is distinct and in non-decreasing order seen from below. To go from the latter to the former remove q q circles from every row (index is q q), turning b 2 b 2 and a 2 a 2 where b 2>a 2 b 2>a 2 into b 2−q b 2−q and a 2−(q−1).a 2−(q−1). The difference is b 2−a 2−1≥0 b 2−a 2−1≥0 and the pair is non-decreasing order but not necessarily distinct. The number of circles being added/removed is ∑q=0 k−1 q=1 2 k(k−1).∑q=0 k−1 q=1 2 k(k−1). We have shown that [z n−1 2 k(k−1)]Z(S k)(1 1−z)=[z n]Z(P k)(1 1−z).[z n−1 2 k(k−1)]Z(S k)(1 1−z)=[z n]Z(P k)(1 1−z). Here we have n≥1 2 k(k−1),n≥1 2 k(k−1), both sides are zero otherwise. This concludes the argument. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 7, 2014 at 2:02 answered Nov 4, 2014 at 21:46 Marko RiedelMarko Riedel 64.9k 5 5 gold badges 67 67 silver badges 147 147 bronze badges 7 Hi! Your answer is an enrichment. Thanks for the nice link into MSE Meta. +1 Markus Scheuer –Markus Scheuer 2014-11-05 09:17:13 +00:00 Commented Nov 5, 2014 at 9:17 The sequences match but does there exist a derivation?Ali Caglayan –Ali Caglayan 2014-11-06 22:14:47 +00:00 Commented Nov 6, 2014 at 22:14 @Alizter Please find the requested proof in the addendum above.Marko Riedel –Marko Riedel 2014-11-07 02:02:52 +00:00 Commented Nov 7, 2014 at 2:02 Marko thank you for your answer. I have left the question unaccepted in order not to discourage any further contribution. Enjoy your bounty.Ali Caglayan –Ali Caglayan 2014-11-08 20:47:11 +00:00 Commented Nov 8, 2014 at 20:47 Another request. Could you add the generating function for the closed form as well?Ali Caglayan –Ali Caglayan 2014-11-12 20:27:25 +00:00 Commented Nov 12, 2014 at 20:27 \|Show 2 more comments This answer is useful 9 Save this answer. Show activity on this post. This is really just a commentary on yashg's (now deleted) answer; I just want to provide a bit of context and a reference. To save repetition: all variables in this post are restricted to integers. The title of the question is somewhat confusing: the vector (0,1,3)(0,1,3) is a solution, not a "set of solutions", of the multivariable equation k 1+k 2+k 3=4 k 1+k 2+k 3=4. You are asking for the number of solutions of the equation k 1+⋯+k n=k k 1+⋯+k n=k where k 1,…,k n k 1,…,k n are distinct nonnegative integers. By the way, I believe it's usual in the literature (at least, in the Wikipedia reference I'm about to give) to interchange the roles of k k and n n in such problems. The answer is in terms of the much-studied partition functionp k(n)p k(n) which is defined as the number of solutions of the equation x 1+⋯+x k=n x 1+⋯+x k=n where x 1≥x 2≥⋯≥x k≥1 x 1≥x 2≥⋯≥x k≥1; those solutions are called the partitions of n n into (exactly) k k parts. For fixed k k, the generating function is ∑k=0∞p k(n)x n=x k(1−x)(1−x 2)⋯(1−x k).∑k=0∞p k(n)x n=x k(1−x)(1−x 2)⋯(1−x k). For n≥k>1 n≥k>1 we have the recurrence equation p k(n)=p k−1(n−1)+p k(n−k),p k(n)=p k−1(n−1)+p k(n−k), since p k−1(n−1)p k−1(n−1) is the number of partitions of n n into k k parts with smallest part equal to 1 1, while p k(n−k)p k(n−k) is the number of partitions of n n into k k parts ≥2≥2. Using the recurrence equation and the obvious boundary conditions p k(n)=0 p k(n)=0 for n0 n>0, we can calculate values of p(n,k)p(n,k), and derive closed formulas for fixed k k, e.g., p 2(n)=⌊n 2⌋p 2(n)=⌊n 2⌋, p 3(n)=⌊n 2+3 12⌋p 3(n)=⌊n 2+3 12⌋, etc. Next, the transformation y j=x j+1−k+j y j=x j+1−k+j shows that the number of solutions of the equation x 1+⋯+x k=n x 1+⋯+x k=n with x 1>x 2>⋯>x k≥0 x 1>x 2>⋯>x k≥0 is the same as the number of solutions of y 1+⋯+y k=n−k(k−3)2 y 1+⋯+y k=n−k(k−3)2 with y 1≥y 2≥⋯≥y k≥1 y 1≥y 2≥⋯≥y k≥1, that is, p k(n−k(k−3)2)p k(n−k(k−3)2). Since your question allows the summands to be arranged in any order, and since they are all distinct, the number of solutions of x 1+⋯+x k=n x 1+⋯+x k=n where x 1,…,x k x 1,…,x k are distinct nonnegative integers is k!p k(n−k(k−3)2)k!p k(n−k(k−3)2); or in your notation: The number of solutions of k 1+⋯+k n=k k 1+⋯+k n=k where k 1,…,k n k 1,…,k n are distinct nonnegative integers is n!p n(k−n(n−3)2).n!p n(k−n(n−3)2). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 7, 2014 at 2:58 answered Nov 3, 2014 at 9:02 bofbof 82.8k 7 7 gold badges 106 106 silver badges 184 184 bronze badges 1 (+1) Please find an independent verification of your formula by Polya counting below.Marko Riedel –Marko Riedel 2014-11-04 21:47:54 +00:00 Commented Nov 4, 2014 at 21:47 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. I believe I've found a solution, but it's up to you all to check its correctness. Here it goes: We have to find the number of non-negative integral solutions to : k 1+k 2+....+k n=P k 1+k 2+....+k n=P where k i∈{0,1,2,3,.....}k i∈{0,1,2,3,.....} , P∈{1,2,3,4,....}P∈{1,2,3,4,....} and k i≠k j k i≠k j Since all the numbers in LHS are distinct, we can assume for simplicity that k i<k i+1 k i<k i+1 With this assumption, let: k 1=k 1+0 k 1=k 1+0 k 2=k 1+a 1 k 2=k 1+a 1 .. .. .. k n=k n−1+a n−1 k n=k n−1+a n−1 where a i∈{1,2,3,.....}a i∈{1,2,3,.....} Then, k 1+k 2+...+k n=k 1+(k 1+a 1)+(k 1+a 1+a 2)+...+(k 1+a 1+...a n−1)k 1+k 2+...+k n=k 1+(k 1+a 1)+(k 1+a 1+a 2)+...+(k 1+a 1+...a n−1) k 1+k 2+...+k n=n k 1+(n−1)a 1+(n−2)a 2+...+2 a n−2+a n−1 k 1+k 2+...+k n=n k 1+(n−1)a 1+(n−2)a 2+...+2 a n−2+a n−1 So we got to find the solutions to n k 1+(n−1)a 1+(n−2)a 2+...+2 a n−2+a n−1=P n k 1+(n−1)a 1+(n−2)a 2+...+2 a n−2+a n−1=P Since I don't know how to deal with natural numbers, I rewrite the equation as n k 1+(n−1)b 1+(n−2)b 2+...+2 b n−2+b n−1=P−(n 2)n k 1+(n−1)b 1+(n−2)b 2+...+2 b n−2+b n−1=P−(n 2) where b i∈{0,1,2,3,.....}b i∈{0,1,2,3,.....} Now very easily we can write the multinomial in which we will have the coefficient of x P−(n 2)x P−(n 2) as the number of solution sets. Which we can multiply by n!n! to get the desired number of tuples. I give the final answer as C C multiplied by n!n! , where C C is the coefficient of x P−(n 2)x P−(n 2) in the expansion of (1+x n+x 2 n+...+x n⌊P−(n 2)n⌋)(1+x n−1+x 2 n−2+...+x(n−1)⌊P−(n 2)n−1⌋).....(1+x+x 2+....+x P−(n 2))(1+x n+x 2 n+...+x n⌊P−(n 2)n⌋)(1+x n−1+x 2 n−2+...+x(n−1)⌊P−(n 2)n−1⌋).....(1+x+x 2+....+x P−(n 2)) NOTE: If it's not visible, the power of x x in the last terms of the brackets has a floor function. If it's still not clear, it's n⌊P−(n 2)n⌋n⌊P−(n 2)n⌋ in the first bracket, and then you just keep on decreasing n by 1 in the subsequent brackets. CHECKING THE FORMULA First, let's take your first case where n=3 n=3 and P=4 P=4. The expression will be (x 0)(x 0)(x 0)(x 0)(x 0)(x 0) in which the coefficient of x 0 x 0 is 1. Which we multiply by 3! to get the answer 6. Now let's take your second case where n=2 n=2 and P=6 P=6. The expression will be (1+x 2+x 4)(1+x+x 2+x 3+x 4+x 5)(1+x 2+x 4)(1+x+x 2+x 3+x 4+x 5) in which the coefficient of x 5 x 5 is 3. Which we multiply by 2! to get 6. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 25, 2014 at 15:22 answered Nov 2, 2014 at 10:49 najayaznajayaz 5,621 3 3 gold badges 23 23 silver badges 46 46 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. It is not my area of expertise, but I believe that what you are after are called in the literature compositions of k k into n n distinct parts. You can probably find the answer in the following 1995 paper by B.Richmond and A.Knopfmacher "Compositions with distinct parts" (link), to which I unfortunately have no access. The generating function for the number of composition of k k into n n distinct parts is also given at the very end of the following preprint and a close formula could possibly be derived from it. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 2, 2014 at 8:41 alezokalezok 1,424 1 1 gold badge 14 14 silver badges 26 26 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. If k i k i are not distinct then the total ways are: ∑i n k i=k,k i≥0⟶(k+n−1 n−1)∑i n k i=k,k i≥0⟶(k+n−1 n−1) Now we need to use inclusion-exclusion to remove cases where any two or more k i k i's are equal, note that you need to be more careful while using inclusion-exclusion because when taking for example k 1 k 1 and k 2 k 2 equal there may be the case that k 3=k 4,k 1≠k 3 k 3=k 4,k 1≠k 3, not even that for it may be that at the same time groups of 2,3 and 4 elements are equal and this may require much work which I skip here. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 12, 2015 at 16:13 user940 answered Nov 2, 2014 at 9:00 RE60KRE60K 18.1k 2 2 gold badges 37 37 silver badges 83 83 bronze badges 1 The inclusion-exclusion leads to the cycle index and is the actual meat of this problem, which was skipped in this answer qwr –qwr 2021-07-11 12:00:39 +00:00 Commented Jul 11, 2021 at 12:00 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. This isn't a closed-form solution, but here is a straightforward recurrence that admits a slow O(n k 2)O(n k 2) dynamic programming solution in O(k 2)O(k 2) space. Let f n(k,m)f n(k,m) be the number of sorted solutions to k 1+⋯+k n=k k 1+⋯+k n=k with k 1≤⋯≤k n≤m k 1≤⋯≤k n≤m, then f n(k,m)=∑k′=0 min(m,k)f n−1(k−k′,k′−1)f n(k,m)=∑k′=0 min(m,k)f n−1(k−k′,k′−1) Base case is f 1(k,m)=[k≤m]f 1(k,m)=[k≤m]. Since each integer is distinct, the number of solutions with any ordering is n!f n(k,k)n!f n(k,k). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 11, 2021 at 12:39 qwrqwr 11.4k 4 4 gold badges 49 49 silver badges 84 84 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics integer-partitions See similar questions with these tags. 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6242	https://artofproblemsolving.com/wiki/index.php/Inequality?srsltid=AfmBOop8_uk3EisD04EwUHD3osgbbKwEomOreKiykQQOpY0MT1lD3AZ_	Art of Problem Solving Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Inequality The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads. Contents [hide] 1 Overview 2 Solving Inequalities 2.1 Linear Inequalities 2.2 Polynomial Inequalities 2.3 Rational Inequalities 3 Complete Inequalities 4 List of Theorems 4.1 Introductory 4.2 Advanced 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Resources 6.1 Books 6.1.1 Intermediate 6.1.2 Olympiad 6.2 Articles 6.2.1 Olympiad 6.3 Classes 6.3.1 Olympiad 7 See also Overview Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: . For two numbers and : if is greater than , that is, is positive. if is smaller than , that is, is negative. if is greater than or equal to , that is, is nonnegative. if is less than or equal to , that is, is nonpositive. Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa. Some properties of inequalities are: If , then , where . If , then , where . If , then , where . Solving Inequalities In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable. We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict. Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means . Linear Inequalities Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign. Polynomial Inequalities The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots. Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality. Rational Inequalities A more complex example is . Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either. A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$. For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set. One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by . Complete Inequalities A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications. List of Theorems Here are some of the more useful inequality theorems, as well as general inequality topics. Introductory Arithmetic Mean-Geometric Mean Inequality Cauchy-Schwarz Inequality Titu's Lemma Chebyshev's Inequality Geometric inequalities Jensen's Inequality Nesbitt's Inequality Rearrangement Inequality Power mean inequality Triangle Inequality Trivial inequality Schur's Inequality Advanced Aczel's Inequality Callebaut's Inequality Carleman's Inequality Hölder's inequality Radon's Inequality Homogenization Isoperimetric inequalities Maclaurin's Inequality Muirhead's Inequality Minkowski Inequality Newton's Inequality Ptolemy's Inequality Can someone fix that Ptolemy's is in Advanced? Problems Introductory Practice Problems on Alcumus Inequalities (Prealgebra) Solving Linear Inequalities (Algebra) Quadratic Inequalities (Algebra) Basic Rational Function Equations and Inequalities (Intermediate Algebra) A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3) Intermediate Practice Problems on Alcumus Quadratic Inequalities (Algebra) Advanced Rational Function Equations and Inequalities (Intermediate Algebra) General Inequality Skills (Intermediate Algebra) Advanced Inequalities (Intermediate Algebra) Given that , and show that . (weblog_entry.php?t=172070 Source) Olympiad See also Category:Olympiad Inequality Problems Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2) Resources Books Intermediate Introduction to Inequalities Geometric Inequalities Olympiad Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess. The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya. Articles Olympiad Inequalities by MIT Professor Kiran Kedlaya. Inequalities by IMO gold medalist Thomas Mildorf. Classes Olympiad The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities. 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6243	https://www.cs.cornell.edu/courses/cs280/2002sp/homework/hw01sol.pdf	CS280 HW1 Solution Set Spring2002 Solutions by: Shaddin Doghmi 1-) a-) 2 2 1 4 1 = − = a 4 4 2 4 2 ) 1 ( 4 1 + = + − = − + = + n n a n n a b-) n n n n n a a a a − = + − − = + − − − + = − + = = − + = + + 2 1 ) 1 ( 1 1 ) 1 ( ) 1 ) ) 1 ( 1 (( ) 1 ( 1 0 ) 1 ( 1 1 1 1 1 c-) ) 1 ( 2 ) 1 ( 2 ) 1 ( ) 2 )( 1 ( 1 + + = + + + = + + = + n a n n n n n a n n now, we need to get rid of the n term. We know that: ) 1 ( 2 2 2 2 2 ) 1 ( ) 1 ( ) 1 ( 1 1 1 + = + = + − = − − + = − − = − − − n n a a n n n n n a a n n a n n n n n substitute into the recursive expression: 2 2 ) 2 ( 1 1 1 1 + − = + − + = − + − + n n n n n n n a a a a a a a so the solution is: 2 2 6 3 2 2 2 1 1 1 2 1 + − = = = = = − + n n n a a a a a d-) Two possible solutions I can think of: solution 1: 1 2 1 2 ) 1 ( 1 2 2 1 1 + + = + + = + = = + n n n a a n n n a a solution 2: by similar reasoning(actually its exactly the same!!) to c-), we get: 2 2 4 1 1 1 2 1 + − = = = − + n n n a a a a a 2-) We want to prove inductively that: n ∀ (The plane is divided by n lines) ⇒ (the regions can be colored using blue and red such that no two adjacent regions have the same color) Note that two regions are adjacent if and only if they share an edge (I will use the word edge instead of side here to avoid ambiguity) Basis: The plane is divided by 0 lines =>we can color the only region blue=> we can color it using blue and red such that no two adjacent regions have the same color Inductive Hypothesis: (The plane is divided by n lines) ⇒ (the regions can be colored using blue and red such that no two adjacent regions have the same color) Induction: Given any n+1 lines dividing the plane, we can remove a line in the plane to make a total of n lines in the plane. Therefore the arrangement of n+1 lines can be formed as such from an arrangement of n lines by adding the desired line to make an arrangement of n+1 lines in the plane. Let us start with our initial n lines. By the Induction Hypothesis, we can color the regions such that no two adjacent regions have the same color. Color them to achieve this. No two adjacent regions have the same color if and only if all edges obey predicate P: P(s): At any point on edge s, the color to that point’s right is different from the color to that point’s left Now, add the desired line, let’s call it line λ. Invert all the colors on the right of that line (blue becomes red and red becomes blue). For all edges not on λ, we know that prior to the inversion all such edges obeyed predicate P. After the inversion, each edge to the right of λ had the colors on both its right and its left inverted, which implies that the colors remained different! (Blue-Red became Red-Blue). Edges to the left of λ were not affected, and still obey predicate P. Therefore: Even after adding line λ and inverting colors to λ’s right, all edges NOT on λ still obey predicate P. Now, what about edges ON line λ? When we added line λ, and before inverting colors, each edge on λ passed through what was a single region before λ came into being (by the definition of an edge). Therefore each edge on λ had the same colors to both it’s left and it’s right. Then came our inversion, which inverted all colors to the right of λ, causing each edge on λ to have different colors to the left and right. Therefore, after adding λ and inverting the right side of λ: All edges on λ have different colors to the left and right Thus we conclude that after adding λ and inverting colors toλ’s right: All edges, be they on λ or not, have different colors to their right and left Hence, given the induction hypothesis, we have proved that an arbitrary arrangement of n+1 lines can be colored such that no two adjacent regions have the same color: (The plane is divided by n+1 lines) ⇒ (the regions can be colored using blue and red such that no two adjacent regions have the same color) Conclusion: Given the basis case and the induction, we conclude that: n ∀ (The plane is divided by n lines) ⇒ (the regions can be colored using blue and red such that no two adjacent regions have the same color) 3-) Basis: 0 lines in the plane ⇒ plane is divided into 1 region. ⇒ plane is divided into 1=1+0(0+1)/2 regions Inductive Hypothesis: n lines in the plane ⇒ plane is divided into 1+n(n+1)/2 regions Induction: In an arrangement of n+1 lines, we can remove a line to get an arrangement of n lines. Therefore, the arrangement of n+1 lines can be formed from an arrangement of n lines by adding the desired line. According to the inductive hypothesis, before adding the extra line the plane has 1+n(n+1)/2 regions. Now, we add the desired line λ. Since the line is infinite (not a segment) and is not parallel to any other line (according to the question specification), it must intersect each of the n lines already in the plane. Since the question also specifies that no 3 lines can intersect at a point, we know that our new line λ will intersect each of the n lines in the plane separately. Since λ will intersect n lines separately, these n lines will chop λ up into n+1 segments (this is easy to see intuitively, but can be proved inductively). Each of the n+1 segments passes through what was one region before λ was added. Hence, n+1 regions are divided into 2 each, generating n+1 more regions. So after adding λ we end up with this many regions: 1+n(n+1)/2 +n+1 =1+(n+1)(n+2)/2 Therefore, if our inductive hypothesis is true then: n+1 lines in the plane ⇒ plane is divided into 1+(n+1)(n+2)/2 regions Or, more precisely: IH (n) ⇒IH (n+1) Where IH is the inductive hypothesis predicate applied to variable n. Conclusion: n ∀ (n lines in the plane ⇒ plane is divided into 1+n(n+1)/2 regions) 4-) The statement is false. Proof by counter-example: Take n = 41 43 41 ) 1 1 41 ( 41 41 41 41 41 2 2 = + + = + + = + + n n 4143 is obviously divisible by 41, which is not itself nor 1. This means that it is not prime. Therefore it is not true that 41 2 + + n n is prime for all natural numbers n. 5-) We want to prove inductively that:         ≤ ∀ ∀ ∑ ∑ = = n k k n k k n x x x x x n 1 1 2 1 ... , Basis: For n=0, the absolute value of the sum of 0 numbers =0. The sum of the absolute values of 0 numbers also=0. They are equal. Therefore the inductive hypothesis below is true for n=0. Inductive Hypothesis:         ≤ ∀ ∑ ∑ = = n k k n k k n x x x x x 1 1 2 1 ... , Induction: For any 1 2 1 ... , + n x x x : 1 1 1 1 + = + = + = ∑ ∑ n n k k n k k x x x now, if ∑ = n k k x 1 and 1 + n x have the same sign(+ve or -ve) then the above expression is equal to: 1 1 + = + ∑ n n k k x x otherwise, if they have different signs, then the expression is equal to the difference between their absolute values: 1 1 + = − ∑ n n k k x x Therefore, the greatest value ∑ + = 1 1 n k k x can have is: 1 1 + = + ∑ n n k k x x and it follows that: IEQ1: 1 1 1 1 + = + = + ≤∑ ∑ n n k k n k k x x x Using our induction hypothesis: IH: ∑ ∑ = = ≤ n k k n k k x x 1 1 Combine IH and IEQ1 to get: 1 1 1 1 1 1 + = + = + = + ≤ + ≤ ∑ ∑ ∑ n n k k n n k k n k k x x x x x ∑ ∑ + = + = ≤ 1 1 1 1 n k k n k k x x So, given our IH (n) predicate, we proved that:         ≤ ∀ ∑ ∑ + = + = + 1 1 1 1 1 2 1 ... , n k k n k k n x x x x x Which is IH (n+1) Conclusion:         ≤ ∀ ∀ ∑ ∑ = = n k k n k k n x x x x x n 1 1 2 1 ... , 6-) The inductive by itself step is flawless. The basis step by itself is also flawless. However, it is not correct to combine them to form an inductive proof. The reason is that the inductive step uses n a and 1 − n a to prove the inductive hypothesis for n+1. Therefore, any valid Basis step MUST define the first two elements of the sequence. Only by defining the first two elements ( 0 a and 1 a ), and proving the inductive hypothesis for these elements can we begin to construct a valid inductive proof. If we modify the Basis step and define 0 a and 1 a , we would realize that the Inductive Hypothesis actually DOES NOT hold for the Basis case, hence destroying the inductive argument before it even begins.
6244	https://www.youtube.com/watch?v=dNUIJ0vCM8Y	What is Rate of Change of a Function? Intro to Derivatives in Calculus - [8-6-17] Math and Science 1610000 subscribers 220 likes Description 9534 views Posted: 9 Jun 2022 More Lessons: Twitter: In this lesson, you will learn what the rate of change is of a function and how to calculate it. We will examine many functions to determine its rate of change, also called its derivative or its instantaneous slope at a point. This has important uses in calculus where we use the rate of change (derivative) to solve many practical problems. The derivative of a function and its opposite, called the integral, is used in many branches of math, science, and engineering. 24 comments Transcript: hello welcome back the title of this lesson is called interpreting the rate of change of a function this is part one so this is actually a graphical lesson i'm going to show you various graphs and we're going to talk about what the rate of change of the function is doing at different points in the graph and we'll do that with a couple of kind of word problems that are attached to the graph so there's really not much math here but there's a whole lot of important understanding because as i've mentioned several times we talk about the rate of change how steep a line is or a line segment is what is how fast is whatever it is you're talking about changing we mentioned many times that that forms the basis later on of what we learn in calculus so even though these things seem like a simple thing really the basis of what we're learning here has far-reaching applications we're going to get to a lot more later on so it's actually really important so let's take a look at our first little example here we have a graph and we have the month of the year along the x-axis and on the y-axis we have the number of visitors now what we're talking about here is a public library so that's that's one thing i want to make sure you understand this is a library so you can see in january and i don't even have any numbers here okay so it's kind of a relative it's kind of a relative indication of traffic in a library in january we start out february and then march or constant and then april may june and july or over the summer are very low in the library because kids are out of school august so this is june july august a little bit higher september there's a peak in traffic to the library probably because kids are going back to school and then throughout the school year the traffic to the library dwindles down to get to the end of the year so it's january february march april may june july august september october november and then december all right now i have several little questions here but i'm not gonna sometimes i'll write the answer down and sometimes i won't i mostly want to just talk and i want you to make sure you agree with what i'm saying also so i want you to be thinking about these as well if i don't write anything down it doesn't mean it's not important i just don't want to write gigantic sentences down we will write some of these answers down as we go so question number one between what two months did the largest increase occur and the largest increase of what of the number of visitors to this library so we see for instance between january and february we had an increase of again i don't know the number but an increase of this many people and then between february and march we actually had a flat line the same number of people throughout that period and the decrease in increase big decrease decrease this is a pretty big increase this looks to be the biggest increase the biggest month over month increase in this graph between what two months did the largest increase occur this is june july this is august and then this peak here is september so between august and september we had a large jump of people again i don't have a scale here so we don't have to write numbers down but we want to say that between i'll put between august and september sorry that looks like something else but it's supposed to be august a-u-g like this in september and the reason probably is because again students going back to school they're starting to study they're going to the library because of that that makes sense from the data all right part b explain the slope of the function during the months september through december so let's go here this is june july august this is september so september is right here and then october november and then finally we have december at the end of the year and the question says explain the slope so if we go up here to this point in september and compare all the way through december what can you say is true about this slope now i'm not asking you to calculate the slope i'm just asking you in general what does it look like the slope is doing there because the slope is the rate of change of the visitors going to that library well first of all the slope here is slanted like this so it's a negative slope because these are positive slopes going up these are negative slopes so we can say it's a negative slope and we also can see that the the you can hold a straight edge to it you can see it doesn't change the slope doesn't change at all so it's a constant slope over those months and it's also a negative slope so we can just put constant negative slope what do these things mean the when i say a constant slope i mean that yes the number of visitors are decreasing each and every month but the rate of decrease is really the same month over month so in other words from september to october i lose the same number of people you can see it goes down from here to here but from october to november i lose again the same amount of people and then from november to december again i lose the same amount of people so when you look at one month then the next month and the next month i'm losing the same amount each month so the rate of change is not is not changing at all i'm losing a fixed amount of people every single month and of course it's negative that makes sense because negative slope means something is decreasing all right so see there's no math here it's just interpreting what the rate of change is telling you here so the largest rate of change as far as month over month is from here to here because we gain this many people over a one month period and then of course here is a constant negative slope there as well all right let's take a look at problem number two so here we have the height in meters of a football as we throw it in the air and this is the number of seconds that it stays into the air so you grab a football you throw it up it goes up for a ways and then it comes down now this is not a physics class okay so i'm not drawing uh very accurate representations right really when you throw things the exact shape of the graph of whatever it is you're talking about is going to obey the laws of physics here we're not drawing accurate accurate graphs okay but you can still see that as time goes on the altitude of this ball goes up up up up up up up up and then it reaches some maximum and then it begins falling back down and it goes down down down down down again in real life it's not going to follow a curve like this because the ball is going to be slowing down as soon as you release it from your hand but let's just pretend that this is you know reality here and answer a couple of questions about it so let's take a look at a little question number one for how many seconds is the ball's height increasing so here we start at zero we reach some maximum height here and all throughout this part of the journey the altitude or the height is increasing so we say that uh from zero seconds all the way to the peak here at five seconds maybe it's a little bit fuzzy because of the peak of this thing is not exactly right at five but it's intended to be at five right here then the question how many seconds is the ball height increasing it's going to be from from i'll put from zero to five seconds so for basically the first five seconds of flight the altitude or the height of this ball is increasing next question what is the rate of change of the ball's height between five and seven seconds now here we actually have numbers on each graph when you see somebody or someone or an exam or whatever asking you to find the rate of change what they're asking you for is the slope we've connected the slope and the rate of change and over many many lessons now that that is what we're trying to figure out so when someone says what is the rate of change of something what they're asking you is what is the slope in this region here all right now if you look really carefully if you hold a straight edge here you can see that this is not a totally straight line it kind of it's kind of bending over at the top and then it's kind of getting more straight kind of toward the bottom there but that's okay we can still look at the entire region as if it's a straight line and ask ourselves how would we find the slope here in that region well the slope is always rise over run right so the rise or the this direction here means it goes from a starting position of five to an ending position of zero there and then uh then it happens over uh over one second period over a two second period right because it starts from five and it goes to seven seconds so you can put your coordinates in all this stuff but you could basically just look at this and say all right we can subtract uh and say okay it goes from 0 there and then subtract 5 meters there so this is the subtraction of the height here to here and then if we go that way we have to go 7 minus 5 as well 7 minus 5. so we're doing rise over run we're subtracting the y value which is the altitude 0 and then subtract 5 and then we're subtracting the x value 7 minus 5. we're subtracting the same direction and the slope that you get is negative 5 over 2 which is negative 5 halves right now fractions are great but sometimes it's nice to put things in decimal that's negative 2.5 now what are the units here the units are in what the rise was in meters that's we subtracted the run was in seconds so it's meters per second so when they say what is the rate of change of the ball's height between 5 and 7 seconds you say it's negative 2.5 meters per second and this makes a lot of physical sense to us right meters per second is a is a velocity or a speed the negative sign just means that the ball is going down and not up that's what that means it's going toward the ground in other words and it's going at a rate of two and a half meters per second now again this curve is a little bit bendy at the top it's not a total straight line so really this is the average speed maybe in the beginning when it's just coming down it's traveling a little slower and at the end near the ground it's traveling a little faster because gravity is pulling it down but over the entire interval the average speed is based on the two end points even though it's changing very slightly between the end points we can still say more accurately that the average rate of change or the average velocity is two and a half meters per second negative just means that it's falling and it's going towards the ground all right let's take a look at problem number three here we have a wacky uh graph here this graph represents the temperature at various times of the day so this is 12 a.m 12 noon and 11 59 p.m about to roll over to 12 a.m all right so every little tick mark here on the x-axis represents a two-hour interval so this is 2 a.m and then 4 a.m and then 6 a.m and then 8 a.m and then 10 a.m and then 12 p.m that's noon and then again 2 p.m 4 p.m 6 p.m 8 p.m 10 p.m and then again midnight there so just keep that in mind and of course you can see the temperature is going down initially and then it's going up in the afternoon and it's coming back down in the evening that's what's happening now problem number one what name a time period during which the temperature was dropping now there are a couple of different spots on this when the temperature is dropping and i only asked you to name one but we can definitely see that the temperature is starting from a high and it hits a low here so it's dropping in this interval from here all the way to here it's increasing this is flat and then all throughout here it's decreasing and then it's flat again here so this is 2 a.m 4 a.m so from 12 a.m to 4 a.m it's dropping so 12 a.m to 4 a.m it's dropping and then again from here this is uh 2 p.m 4 p.m and then it ends here 6 8 uh 4 6 8 10. so 4 p.m to 10 p.m 4 pm 10 10 p.m it's also dropping as well these areas where the thing is flat it's not dropping there so we don't count that now it says if janice likes to take a walk when the temperature is warm and steady what is the best time for her to do so so when we say warm and steady steady is another way of saying things are not changing the temperature is not changing so what you want to do is find the parts of the graph where the temperature is flat across time here is one and here is one as well this of course is warmer than this so we're going to say this is the best block of time from 2 pm to 4 p.m so from 2 p.m to 4 p.m and the the better mathematical way to say it is that from 2 p.m to 4 p.m the slope there or the rate of change of the temperature is zero when we have a flat line remember i told you the slope is always zero because there is no slope it's not slanted at all so the rate of change in those regions is zero so here's just a a quick little lesson basically interpreting charts and really trying to pick out parts of the charts where the slope is high and the slope is low and tying that to the rate of change of real physical quantities because if you don't do something like this then you end up graphing lines and but you don't really know like how it could be used in real life i mean this is a real life thing you can measure the temperature here in different little segments and treat them almost as little line segments and then you can compare the slope which is the rate of change in different um in different regions of time and you can use that in real life for real you know applications whatever it is if you're studying the weather or whatever and this is what we do in calculus also when we study things like a circle we just chop it up into little line segments and we start to study how to handle things like that so we can handle everyday real life situations in math and science by taking a complex shape treating it as little line segments which is what we do in calculus and then we can study complex situations and all the things that you've learned about slope of a line apply in those situations so go over these again make sure you understand follow me on the part two will continue to build your skills
6245	https://elsevier-elibrary.com/contents/fullcontent/58913/epubcontent_v2/OEBPS/B9781416054511000256.htm	Pocket Companion to Guyton and Hall Textbook of Medical Physiology Page 181 CHAPTER 25 The Body Fluid Compartments Extracellular and Intracellular Fluids; Edema The total amount and composition of the body fluids are maintained relatively constant under most physiological conditions, as required for homeostasis. Some of the most important problems in clinical medicine, however, arise because of abnormalities in the control systems that maintain this constancy. In this section, we discuss the overall regulation of body fluid volume, control of the constituents of the extracellular fluid, regulation of the fluid exchange between the extracellular and intracellular compartments, and regulation of the acid-base balance. Fluid Intake and Output Are Balanced During Steady-State Conditions (p. 285) The total intakes of water and electrolytes must be carefully matched by equal outputs from the body to prevent fluid volumes and electrolyte concentrations from increasing or decreasing. Table 25–1 shows the routes of daily water intake and output from the body. Under most conditions, the primary means of regulating output is by altering renal excretion. Urine volume can be as low as 0.5 L/day in a dehydrated person or as high as 20 L/day in a person who has been drinking large amounts of fluids. This ability of the kidneys to adjust the output to such an extreme to match intake also occurs for the electrolytes of the body such as sodium, chloride, and potassium. Table 25–1Daily Intake and Output of Water \| Parameter \| Normal (mL/day) \| With Prolonged Heavy Exercise (mL/day) \| --- \| Intake \| \| Fluids ingested \| 2100 \| ? \| \| From metabolism \| 200 \| 200 \| \| Total intake \| 2300 \| ? \| \| Output \| \| Insensible skin \| 350 \| 350 \| \| Insensible lungs \| 350 \| 650 \| \| Sweat \| 100 \| 5000 \| \| Feces \| 100 \| 100 \| \| Urine \| 1400 \| 500 \| \| Total output \| 2300 \| 6600 \| Total Body Fluid Is Distributed between the Extracellular Fluid and the Intracellular Fluid (p. 286) The total amount of body water averages about 60% of the body weight, or about 42 L in a 70-kg adult man. Because women normally have more body fat than men, their total body water averages about 50% of the body weight. In premature and newborn babies, the total body water ranges from 70% to 75% of body weight. Therefore, when discussing the “average” body fluid compartments, we should realize that variations exist, depending on age, gender, and percentage of body fat. Page 182 Total body fluid is distributed into two main compartments: (1) the intracellular fluid, which is about 40% of body weight or 28 L, and (2) the extracellular fluid, which is about 20% of body weight, or 14 L in a 70-kg person. The two main compartments of the extracellular fluid are the interstitial fluid, which makes up about three fourths of the extracellular fluid, and the plasma, which makes up about one fourth of the extracellular fluid, or about 3 L. The plasma is the noncellular portion of the blood that mixes continuously with interstitial fluid through the pores of the capillary membranes. Blood Contains Extracellular and Intracellular Fluids The average blood volume in a normal adult human is 8% of the body weight, or about 5 L. About 60% of the blood is plasma, and about 40% is red blood cells. The hematocrit, the fraction of blood that is composed of red blood cells, is normally about 0.42 in men and about 0.38 in women. With severe anemia, the hematocrit may fall to as low as 0.10, which is barely sufficient to sustain life. When there is excessive production of red blood cells, resulting in polycythemia, the hematocrit can rise to as high as 0.65. Page 183 The Constituents of Extracellular and Intracellular Fluids Differ Table 25–2 compares the compositions of the intracellular and extracellular fluids. Table 25–2Chemical Compositions of Extracellular and Intracellular Fluids \| Chemical \| Intracellular Fluid \| Extracellular Fluid \| --- \| Na+ (mmol/L) \| 10 \| 142 \| \| K+ (mmol/L) \| 140 \| 4 \| \| Cl− (mmol/L) \| 4 \| 108 \| \| HCO 3− (mmol/L) \| 10 \| 24 \| \| Ca 2+ (mmol/L) \| 0.0001 \| 2.4 \| \| Mg 2+ (mmol/L) \| 58 \| 1.2 \| \| SO 4 2− (mmol/L) \| 2 \| 1 \| \| Phosphates (mmol/L) \| 75 \| 4 \| \| Glucose (mg/dL) \| 0–20 \| 90 \| \| Amino acids (mg/dL) \| 200? \| 30 \| \| Protein (g/dL) \| 16 \| 2 \| The plasma and interstitial fluid of the extracellular compartment are separated by highly permeable capillary membranes, so their ionic compositions are similar. The most important difference between these two compartments is that the plasma has a higher protein concentration. The capillaries have low permeability to proteins and therefore leak only small amounts of protein into the interstitial spaces in most tissues. The intracellular fluid is separated from the extracellular fluid by a highly selective cell membrane that is permeable to water but not to most electrolytes found in the body. For this reason, the concentration of water and the osmolarity of intracellular and extracellular fluids are approximately equal under steady-state conditions, although the concentrations of various solutes are markedly different in these fluid compartments. The Indicator-Dilution Principle Can Be Used to Measure Volumes of Body Fluid Compartments (p. 287) The volume of a fluid in a compartment in the body can be estimated by injecting a substance into the compartment, allowing it to disperse evenly, and then analyzing the extent to which the substance has become diluted. This method is based on the assumption that the total amount of substance remaining in the fluid compartment after dispersion is the same as the total amount that was injected into the compartment. Thus when a small amount of substance contained in syringe A is injected into compartment B and the substance is allowed to disperse throughout the compartment until it becomes mixed in equal concentrations in all areas, the following relation can be expressed: Page 184 This method can be used to measure the volume of virtually any compartment in the body if (1) the amount of indicator injected into the compartment (the numerator of the equation) is known, (2) the concentration of the indicator in the compartment is known, (3) the indicator disperses evenly throughout the compartment, and (4) the indicator disperses only in the compartment that is being measured. Table 25–3 shows some of the indicators that can be used to measure the fluid volumes of the body compartments. The volumes of two of the compartments, the intracellular and extracellular interstitial fluids, cannot be measured directly but, instead, are calculated from the values for other body fluid volumes. Table 25–3Measurement of Body Fluid Volume \| Volume \| Indicators \| --- \| \| Total body water \| 3 H 2 O, 2 H 2 O, antipyrine \| \| Extracellular fluid \| 22 Na, 125 I-iothalamate, inulin \| \| Intracellular fluid \| Calculated as: Total Body Water – Extracellular Fluid Volume \| \| Plasma volume \| 125 I-albumin, Evans blue dye (T-1824) \| \| Blood volume \| 51 Cr-labeled red blood cells; calculated as: Blood Volume = Plasma Volume/(1 − Hematocrit) \| \| Interstitial fluid \| Calculated as: Extracellular Fluid Volume – Plasma Volume \| Page 185 Intracellular and Extracellular Fluid Distribution Is Determined Mainly by the Osmotic Effect of Electrolytes Acting across the Cell Membrane (p. 290) Because the cell membrane is highly permeable to water but relatively impermeable to even small ions, such as sodium and chloride, the distribution of fluid between the intracellular and extracellular compartments is determined mainly by the osmotic effects of these ions. The basic principles of osmosis and osmotic pressure are presented in Chapter 4. Therefore, only the most important principles as they apply to volume regulation are discussed in this section. Osmosis Is the Net Diffusion of Water across a Selectively Permeable Membrane from a Region of High Water Concentration to One of Lower Water Concentration The addition of a solute to pure water reduces the water concentration and causes water to move toward the region of high solute concentration. The concentration term used to measure the total number of solute particles in solution is the osmole: 1 osmole is equal to 1 mole (6.02 × 10 23) of solute particles. For biological solutions, the term milliosmole (mOsm), which equals 1/1000 osmole, is commonly used. The osmolar concentration of a solution is called its osmolality when the concentration is expressed as osmoles per kilogram of water and osmolarity when it is expressed as osmoles per liter of solution. The amount of pressure required to prevent osmosis of water through a semipermeable membrane is called the osmotic pressure. Expressed mathematically, the osmotic pressure (π) is directly proportional to the concentration of osmotically active particles in that solution. where C is the concentration of solutes in osmoles per liter, R is the ideal gas constant, and T is the absolute temperature in degrees Kelvin. If π is expressed in millimeters of mercury (the unit of pressure commonly used for biologic fluids), π calculates to be about 19.3 mm Hg for a solution with an osmolarity of 1 mOsm/L. Thus, for each milliosmole concentration gradient across the cell membrane, 19.3 mm Hg of force is required to prevent water diffusion across the membrane. Very small differences in solute concentration across the cell membrane can therefore cause rapid osmosis of water. Page 186 Isotonic, Hypotonic, and Hypertonic Fluids A solution is said to be isotonic if no osmotic force develops across the cell membrane when a normal cell is placed in the solution. An isotonic solution has the same osmolarity as the cell, and the cells do not shrink or swell if placed in the solution. Examples of isotonic solutions include a 0.9% sodium chloride solution and a 5% glucose solution. A solution is said to be hypertonic when it contains a higher concentration of osmotic substances than does the cell. In this case, an osmotic force develops that causes water to flow out of the cell into the solution, thereby reducing the intracellular fluid volume and increasing the intracellular fluid concentration. A solution is said to be hypotonic if the osmotic concentration of substances in the solution is less than the concentration of the cell. The osmotic force develops immediately when the cell is exposed to the solution, causing water to flow by osmosis into the cell until the intracellular fluid has about the same concentration as the extracellular fluid or until the cell bursts as a result of excessive swelling. Volumes and Osmolarities of Extracellular and Intracellular Fluids in Abnormal States (p. 292) Some of the factors that can cause extracellular and intracellular volumes to change markedly are ingestion of large amounts of water, dehydration, intravenous infusion of various solutions, loss of large amounts of fluid from the gastrointestinal tract, and loss of abnormal amounts of fluid via sweating or from the kidneys. One can approximate the changes in intracellular and extracellular fluid volumes and the therapy that must be instituted if the following basic principles are kept in mind: • Water moves rapidly across cell membranes; therefore, the osmolarities of intracellular and extracellular fluids remain almost exactly equal to each other except for a few minutes after a change in one of the compartments. • Cell membranes are almost completely impermeable to most solutes; therefore, the number of osmoles in the extracellular and intracellular fluids remains relatively constant unless solutes are added to or lost from the extracellular compartment. Page 187 Effect of Adding Isotonic, Hypertonic, and Hypotonic Saline Solutions to Extracellular Fluid If an isotonic solution is added to the extracellular fluid compartment, the osmolarity of the extracellular fluid does not change, and there is no osmosis through the cell membranes. The only effect is an increase in the extracellular fluid volume (Fig. 25–1). Sodium and chloride mainly remain in the extracellular fluid because the cell membrane behaves as though it were virtually impermeable to sodium chloride. Figure 25–1Effect of adding isotonic, hypertonic, and hypotonic solutions to extracellular fluid after osmotic equilibrium. The normal state is indicated by the solid lines, and the shifts from normal are shown by the dashed lines. The volumes of intracellular and extracellular fluid compartments are shown on the abscissa of each diagram, and the osmolarities of these compartments are shown on the ordinates. If a hypertonic solution is added to the extracellular fluid, the extracellular fluid osmolarity increases and causes osmosis of water out of the cells into the extracellular compartment. The net effect is an increase in extracellular volume (greater than the volume of fluid that was added), a decrease in intracellular fluid volume, and an increase in the osmolarity of both compartments. If a hypotonic solution is added to the extracellular fluid, the osmolarity of the extracellular fluid decreases, and some of the extracellular water diffuses into the cells until the intracellular and extracellular compartments have the same osmolarity. Both the intracellular and extracellular volumes are increased by addition of hypotonic fluid, although the intracellular volume is increased to a greater extent. Edema: Excess Fluid in the Tissues (p. 296) Intracellular Edema: Increased Intracellular Fluid Three conditions especially likely to cause intracellular swelling are (1) hyponatremia, (2) depression of the metabolic systems of the tissues, and (3) lack of adequate nutrition to the cells. When the cell’s metabolic systems are depressed or they receive inadequate nutrition, sodium ions that normally leak into the interior of the cells can no longer be effectively pumped out of the cells, and the excess sodium ions cause osmosis of water into the cells. Intracellular edema can also occur in inflamed tissues. Inflammation usually has a direct effect on the cell membranes to increase their permeability, allowing sodium and other ions to diffuse into the interior of the cells with subsequent osmosis of water into the cells. Page 188 Extracellular Edema: Increased Fluid in Interstitial Spaces The two general causes of extracellular edema are (1) abnormal leakage of fluid from the plasma to the interstitial spaces across the capillaries and (2) failure of the lymphatics to return fluid from the interstitium to the blood, often called lymphedema. Factors Can Increase Capillary Filtration and Cause Interstitial Fluid Edema To understand the causes of excessive capillary filtration, it is useful to review the determinants of capillary filtration discussed in Chapter 16 as shown in the following equation: where K f is the capillary filtration coefficient (the product of the permeability and surface area of the capillaries), P if is the interstitial fluid hydrostatic pressure, π c is the capillary plasma colloid osmotic pressure, and π if is the interstitial fluid colloid osmotic pressure. Thus, any of the following changes can increase the capillary filtration rate: Page 189 1. Increased capillary filtration coefficient, which allows increased leakage of fluids and plasma proteins through the capillary membranes. This can occur as a result of allergic reactions, bacterial infections, and toxic substances that injure the capillary membranes and increase their permeability to plasma proteins. 2. Increased capillary hydrostatic pressure, which can result from obstruction of veins, excessive flow of blood from the arteries into the capillaries, or failure of the heart to pump blood rapidly out of the veins (heart failure). 3. Decreased plasma colloid osmotic pressure, which may result from failure of the liver to produce sufficient quantities of plasma proteins (cirrhosis), loss of large amounts of protein in the urine with certain kidney diseases (nephrotic syndrome), or loss of large quantities of protein through burned areas of the skin or other denuding lesions. 4. Increased interstitial fluid colloid osmotic pressure, which draws fluid out of the plasma into the tissues spaces. This situation occurs most often as a result of lymphatic blockage, which prevents the return of protein from interstitial spaces to the blood (discussed in the following sections). Lymphatic Blockage Causes Edema When lymphatic blockage occurs, edema can become especially severe because plasma proteins that leak into the interstitium have no other way to be returned to the plasma. The rise in protein concentration increases the colloid osmotic pressure of the interstitial fluid, which draws even more fluid out of the capillaries. Blockage of lymph flow can be especially severe with infections of the lymph nodes, such as occurs with infection by filarial nematodes. Lymph vessels may also be blocked with certain types of cancer or after surgery in which the lymph vessels are removed or obstructed. Safety Factors That Normally Prevent Edema Although many abnormalities can cause fluid accumulation in interstitial spaces, the disturbances must be substantial before clinically significant edema develops. Three major safety factors normally prevent fluid accumulation in the interstitial spaces: Page 190 1. The compliance of the tissues is low so long as interstitial fluid hydrostatic pressure is in the negative range. Low compliance (defined as the change in volume per millimeter of mercury pressure change) means that small increases in interstitial fluid volume are associated with relatively large increases in interstitial fluid hydrostatic pressure. When the interstitial fluid volume increases, the interstitial fluid hydrostatic pressure increases markedly, which opposes further excessive capillary filtration. The safety factor that protects against edema for this effect is about 3 mm Hg in many tissues such as skin. 2. Lymph flow can increase as much as 10- to 50-fold. Lymph vessels carry away large amounts of fluid and proteins in response to increased capillary filtration. The safety factor for this effect has been calculated to be about 7 mm Hg. 3. There is a “wash-down” of interstitial fluid protein as lymph flow increases. As increased amounts of fluid are filtered into the interstitium the interstitial fluid pressure increases, causing greater lymph flow. This decreases the protein concentration of the interstitium because more protein is carried away than can be filtered by the capillaries. A decrease in tissue fluid protein concentration lowers the net filtration force across the capillaries and tends to prevent further fluid accumulation. The safety factor for this effect has been calculated to be about 7 mm Hg in most tissues. Combining all of the safety factors, the total safety factor that protects against edema is about 17 mm Hg. Capillary pressure in peripheral tissues could therefore theoretically rise 17 mm Hg before significant interstitial edema would occur.
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Tools & Reference>Gastroenterology Hirschsprung Disease Workup Updated: Aug 02, 2021 Author: Justin P Wagner, MD; Chief Editor: BS Anand, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Hirschsprung Disease Sections Hirschsprung Disease Overview Background Pathophysiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Laboratory Studies Imaging Studies Other Tests Procedures Histologic Findings Show All Treatment Approach Considerations Medical Care Surgical Care Outpatient Monitoring Show All Guidelines Medication Medication Summary Antibiotics Toxins Show All Media Gallery;) References;) Workup Laboratory Studies Chemistry panel For most patients, electrolyte and renal panel findings are within the reference ranges. Children presenting with diarrhea may have findings consistent with dehydration. Test results may aid in directing fluid and electrolyte management. Complete blood cell (CBC) count This test is obtained to ensure that the preoperative hematocrit and platelet count are suitable for surgery. In most cases, values are within the reference ranges. Coagulation studies These studies are obtained to ensure that clotting disorders are corrected before surgery. Coagulation parameters are expected to be within the reference ranges. Next: Imaging Studies Plain abdominal radiographs may show distended bowel loops with a paucity of air in the rectum. With regard to barium enema, avoid washing out the distal colon with enemas before obtaining the contrast enema because this may distort a low transition zone. The catheter is placed just inside the anus without inflation of the balloon to avoid distortion of a low transition zone and perforation. Radiographs are taken immediately after hand injection of the contrast and again 24 hours later. The classic finding of Hirschsprung disease is a narrowed distal colon with proximal dilation; however, the findings are difficult to interpret in neonates (age < 1 mo) and do not demonstrate this transition zone in approximately 25% of the time. Retention of rectal contrast for longer than 24 hours after the barium enema also suggests a diagnosis of Hirschsprung disease. Previous Next: Other Tests Anorectal manometry detects the relaxation reflex of the internal sphincter after distention of the rectal lumen. This normal inhibitory reflex is presumed absent in patients with Hirschsprung disease. Swenson initially used this test. In the 1960s, it was refined but has fallen out of favor because of its many limitations. Sedation is usually necessary. Although some authors find this test useful, false-positive results have been reported in up to 62% and false-negative results have been reported in up to 24% of cases. Because of these limitations, anorectal manometry is not commonly performed in the United States. Echocardiography and chromosomal analyses may be warranted to evaluate for any associated congenital conditions. Previous Next: Procedures Full-thickness rectal biopsy The definitive diagnosis of Hirschsprung disease is confirmed by a full-thickness rectal biopsy demonstrating the absence of ganglion cells. The specimen must be obtained at least 1.5 cm above the dentate line because aganglionosis may normally be present below this level. Disadvantages of full-thickness rectal biopsy include the necessity of general anesthesia and the risks of bleeding and scarring. Suction rectal biopsy Simple suction rectal biopsy has been used to obtain tissue for histologic examination. Rectal mucosa and submucosa are sucked into the suction device, and a self-contained cylindrical blade excises the tissue. The distinct advantage of the suction biopsy is that it can be easily performed at the bedside. The diagnostic yield of the full-thickness rectal biopsy is significantly better than that of the suction biopsy. Previous Next: Histologic Findings Acetylcholinesterase staining reveals hypertrophied nerve trunks throughout the lamina propria and muscularis propria layers of the bowel wall. Although acetylcholinesterase histochemistry can be a useful ancillary technique to help in the diagnosis and preoperative planning, studies have suggested that immunohistochemical (IHC) staining for calretinin might be more accurate than acetylcholinesterase staining in diagnosing congenital aganglionosis in suction biopsy specimens. [49, 50] Guinard-Samuel and colleagues evaluated the diagnostic value of calretinin IHC for Hirschsprung disease in 131 pediatric rectal biopsies. Of the 131 biopsies, 130 were accurately diagnosed based on calretinin staining. When an additional 12 cases were considered doubtful based on the standard evaluation method, they were accurately diagnosed with calretinin IHC. The investigators found calretinin superior to acetylcholinesterase to complete histology. In another study that evaluated the diagnostic value of calretinin IHC staining compared with standard hematology and eosin (H&E) staining of rectal suction biopsies over 1 year in 188 children with Hirschsprung disease, Tran et al confirmed the disease in 80 children (42.6%), with 1 false positive, no false negatives, and no serious complications associated with the procedure. Calretinin and H&E staining both had a 100% sensitivity, but whereas the specificity was 99.1% for calretinin staining, it was 85.2% for H&E. Yang and colleagues identified the presence or absence of ganglion cells via IHC staining for calretinin and microtubule associated protein-2 (MAP-2) in 52 samples of normal and aganglionic bowel. 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J Pediatr Surg. 2000 Jun. 35(6):820-2. [QxMD MEDLINE Link]. Gosemann JH, Friedmacher F, Ure B, Lacher M. Open versus transanal pull-through for Hirschsprung disease: a systematic review of long-term outcome. Eur J Pediatr Surg. 2013 Apr. 23(2):94-102. [QxMD MEDLINE Link]. Chen Y, Nah SA, Laksmi NK, et al. Transanal endorectal pull-through versus transabdominal approach for Hirschsprung's disease: a systematic review and meta-analysis. J Pediatr Surg. 2013 Mar. 48(3):642-51. [QxMD MEDLINE Link]. Onishi S, Nakame K, Yamada K, et al. Long-term outcome of bowel function for 110 consecutive cases of Hirschsprung's disease: Comparison of the abdominal approach with transanal approach more than 30 years in a single institution - is the transanal approach truly beneficial for bowel function?. J Pediatr Surg. 2016 Dec. 51(12):2010-4. [QxMD MEDLINE Link]. Levitt MA, Hamrick MC, Eradi B, Bischoff A, Hall J, Pena A. Transanal, full-thickness, Swenson-like approach for Hirschsprung disease. J Pediatr Surg. 2013 Nov. 48(11):2289-95. [QxMD MEDLINE Link]. Tang ST, Yang Y, Li SW, et al. Single-incision laparoscopic versus conventional laparoscopic endorectal pull-through for Hirschsprung's disease: a comparison of short-term surgical results. J Pediatr Surg. 2013 Sep. 48(9):1919-23. [QxMD MEDLINE Link]. Li N, Zhang W, Yu D, et al. NOTES for surgical treatment of long-segment Hirschsprung's disease: report of three cases. J Laparoendosc Adv Surg Tech A. 2013 Dec. 23(12):1020-3. [QxMD MEDLINE Link]. Vahdad MR, Foroutan A, Najafi SM, et al. Totally transanal LESS pull-through colectomy: a novel approach for avoiding abdominal wall incision in children with long-segment intestinal aganglionosis. J Laparoendosc Adv Surg Tech A. 2013 Mar. 23(3):276-80. [QxMD MEDLINE Link]. Hotta R, Stamp LA, Foong JP, et al. Transplanted progenitors generate functional enteric neurons in the postnatal colon. J Clin Invest. 2013 Mar 1. 123(3):1182-91. [QxMD MEDLINE Link].[Full Text]. Frith TJR, Gogolou A, Hackland JOS, et al. Retinoic acid accelerates the specification of enteric neural progenitors from in-vitro-derived neural crest. Stem Cell Reports. 2020 Aug 11. [QxMD MEDLINE Link].[Full Text]. Levitt MA, Martin CA, Olesevich M, Bauer CL, Jackson LE, Pena A. Hirschsprung disease and fecal incontinence: diagnostic and management strategies. J Pediatr Surg. 2009 Jan. 44(1):271-7; discussion 277. [QxMD MEDLINE Link]. [Guideline] Kyrklund K, Sloots CEJ, de Blaauw I, et al. ERNICA guidelines for the management of rectosigmoid Hirschsprung's disease. Orphanet J Rare Dis. 2020 Jun 25. 15(1):164. [QxMD MEDLINE Link].[Full Text]. Thapar N. New frontiers in the treatment of Hirschsprung disease. J Pediatr Gastroenterol Nutr. 2009 Apr. 48 suppl 2:S92-4. [QxMD MEDLINE Link]. Versteegh HP, Johal NS, de Blaauw I, Stanton MP. Urological and sexual outcome in patients with Hirschsprung disease: A systematic review. J Pediatr Urol. 2016 Dec. 12(6):352-60. [QxMD MEDLINE Link]. Kapur RP. Histology of the transition zone in Hirschsprung disease. Am J Surg Pathol. 2016 Dec. 40(12):1637-46. [QxMD MEDLINE Link]. Witvliet MJ, van Gasteren S, van den Hondel D, Hartman E, van Heurn L, van der Steeg A. Predicting sexual problems in young adults with an anorectal malformation or Hirschsprung disease. J Pediatr Surg. 2018 Aug. 53(8):1555-9. [QxMD MEDLINE Link]. Media Gallery A: Plain abdominal radiograph showing a transition zone (PARTZ) at the rectosigmoid. B: Plain abdominal radiograph showing a PARTZ at the midsigmoid. C: Plain abdominal radiograph showing a PARTZ at the descending colon. D: Contrast enema showing a contrast enema transition zone (CETZ) at the rectosigmoid. E: Contrast enema showing a CETZ at the midsigmoid. F: Contrast enema showing a CETZ at descending colon. Images courtesy of Pratap A, Gupta DK, Tiwari A, et al. BMC Pediatr. 2007 Jan 27;7:5. [Open access.] PMID: 17257439, PMCID: PMC1790893. Hirschsprung disease. Contrast enema demonstrating transition zone in the rectosigmoid region. of 2 Tables Back to List Contributor Information and Disclosures Author Justin P Wagner, MD Resident Physician, Department of Surgery, University of California, Los Angeles, David Geffen School of MedicineJustin P Wagner, MD is a member of the following medical societies: American College of Surgeons, American Pediatric Surgical Association, Association for Academic Surgery, Association for Surgical Education, Association of Program Directors in SurgeryDisclosure: Nothing to disclose. Coauthor(s) Shant Shekherdimian, MD, MPH Resident Physician, Department of Pediatric Surgery, Hospital for Sick Children; Toronto, Ontario, CanadaDisclosure: Nothing to disclose. Steven L Lee, MD Chief of Pediatric Surgery, Harbor-UCLA Medical Center; Associate Clinical Professor of Surgery and Pediatrics; University of California, Los Angeles, David Geffen School of MedicineSteven L Lee, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Pediatrics, American College of Surgeons, American Pediatric Surgical Association, Association for Academic Surgery, Society of Laparoscopic and Robotic Surgeons, International Pediatric Endosurgery Group, Pacific Association of Pediatric Surgery, Society of American Gastrointestinal and Endoscopic SurgeonsDisclosure: Nothing to disclose. Specialty Editor Board Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment. Chief Editor BS Anand, MD Professor, Department of Internal Medicine, Division of Gastroenterology, Baylor College of MedicineBS Anand, MD is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, American Gastroenterological Association, American Society for Gastrointestinal EndoscopyDisclosure: Nothing to disclose. Additional Contributors Vivek V Gumaste, MD Associate Professor of Medicine, Mount Sinai School of Medicine of New York University; Adjunct Clinical Assistant, Mount Sinai Hospital; Director, Division of Gastroenterology, City Hospital Center at Elmhurst; Program Director of GI Fellowship (Independent Program); Regional Director of Gastroenterology, Queens Health NetworkVivek V Gumaste, MD is a member of the following medical societies: American College of Gastroenterology, American Gastroenterological AssociationDisclosure: Nothing to disclose. Acknowledgements Jeffrey J DuBois, MD Chief of Children's Surgical Services, Division of Pediatric Surgery, Kaiser Permanente, Women and Children's Center, Roseville Medical Center Jeffrey J DuBois, MD, is a member of the following medical societies: Alpha Omega Alpha, American Academy of Pediatrics, American College of Surgeons, American Pediatric Surgical Association, and California Medical Association Disclosure: Nothing to disclose. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Hirschsprung Disease Overview Background Pathophysiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Laboratory Studies Imaging Studies Other Tests Procedures Histologic Findings Show All Treatment Approach Considerations Medical Care Surgical Care Outpatient Monitoring Show All Guidelines Medication Medication Summary Antibiotics Toxins Show All Media Gallery;) References;) encoded search term (Hirschsprung Disease) and Hirschsprung Disease What to Read Next on Medscape Related Conditions and Diseases Severe Pediatric Constipation Constipation Pediatric Constipation Gastrointestinal Disease and Pregnancy Fast Five Quiz: Opioid-Induced Constipation Laxatives, Stool Softeners, and Prokinetic Agents Rapid Review Quiz: Constipation News & Perspective Ultraprocessed Food Linked to Constipation Review: Does Caffeine Cut Constipation? 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6247	https://www.youtube.com/watch?v=Mz3f8pTznEE	Chemistry Lesson: Significant Digits & Rounding GetChemistryHelp 22300 subscribers 270 likes Description 37058 views Posted: 24 Jan 2013 All numbers in measurements are significant, but calculations often generate nonsignificant digits. This lesson explains how to round the nonsignificant digits, including the concept of placeholder zeros. 27 comments Transcript: Introduction welcome to gemistry health.com my name is Dr Ken and today's lesson is on significant digits and rounding now as we saw in our previous lesson on significant digits and measurements all numbers from a measurement are significant however we will often generate nonsignificant digits when it comes to performing a calculation so we need to get rid of those non-significant digits in a very systemized way and the way we do that is through rounding so there are three rules when it comes to Rounding Rules rounding rule number one says if the first nonsignificant digit is less than five then we'll drop all of the non-significant digits so if the first non-significant one is less than five then we'll just drop all of the non-significant ones rule two says if the first non-significant digit is greater than or equal to five so now if it's greater than or equal to five then we will increase the last significant digit by one but still drop all of the non-significant digits and the third rule says that if you have a calculation with two or more operations well you want to keep at least one non-significant digit until you get down to the final operation and then round off the answer so what that means is there's going to be times when you have to do multiple operations so you might have to multiply a number and then add another number and then divide a number and then subtract a number so you don't want to round off after every every operation you want to wait and round off at the very end okay so let's Examples just look at a couple of examples and see how these rules play out okay so here's a number 17.84% well it's this four right here and according to rounding rule number one less than five tells me just to drop all the non-significant ones so that would round off to become 17.8 okay how about at the calculator displayed 17.85% of five so rounding rule number two said five are greater add one onto the last significant digit which is this eight and then still drop all of the non-significant ones so if I add one onto that last nonsignificant place 17.8 would become 17.9 okay well how about 151 if I want Rounding to round that off to two significant digits okay so the first two significant digits are this one and this five the first non-significant one one is this one so that's less than five so I have to round down and just get rid of it but I can't just turn 151 into 15 I have to put something in this one's place so I put in a zero as you might recall in our previous lesson these are just called placeholder zeros and placeholder zeros are never significant so this is still just one two sigfigs how about 2788 well how would I round that off to show three sigfigs okay well we want to keep the first three so one 2 3 again I look here at the first non-significant digit and that's an eight so five or greater tells me to round up so I'm going to add one onto this last significant digit so I would make that 278 will become 279 but again I can't just leave that 279 I have to put something in this place so I'm going to put a zero in there so it becomes 2790 okay so let's do a few last examples and we'll wrap this up how about if I was going to round 41278 off to three significant digits okay so again I'm going to keep the first three significant digits so one 2 3 the first non-significant digit is a seven that's five or greater so I round up on the last significant one so 4.12 would become 4.1 three how about 41278 but off to two significant digits okay so I keep the first two significant digits the first non-significant digit is this two so that's less than five so I round down and just drop all of the non-significant digits so just become 4.1 how about 63,5 to three significant digits okay so I look at the first three significant digits so this six this three and four are significant the first non-significant digit is this zero so that tells me to round down and drop all the non- significant digits so I'm going to keep the six the three and the four but again I can't just leave this 10's place and this one's place empty because 634 is very different from 63,5 so I have to put zeros in the 10 place and the ones place so that would round off to 63,500 which still is just 3 significant figures because we learned before that trailing zeros are never significant anyway how about 0.05 62 off to two significant digits okay again leading zeros as we saw in our previous lesson are never significant so these zeros aren't significant so the five and the six are the first two significant digits the first non-significant digit is this two that's less than five so I just round It Off and drop those so I'm going to keep just the five and the six here so that would become 0.056 okay great well I hope you enjoyed this lesson uh for more video lessons and chemistry practice problems please come and visit me at getchemistryhelp ccom and we'll see you next time thank you
6248	https://math.stackexchange.com/questions/78406/how-can-i-prove-the-inequality-frac1x-frac1y-frac1z-geq-f	algebra precalculus - How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How can I prove the inequality 1 x+1 y+1 z≥9 x+y+z 1 x+1 y+1 z≥9 x+y+z? Ask Question Asked 13 years, 11 months ago Modified12 years, 5 months ago Viewed 7k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. For x>0 x>0, y>0 y>0, z>0 z>0, prove: 1 x+1 y+1 z≥9 x+y+z.1 x+1 y+1 z≥9 x+y+z. I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it? (You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.) algebra-precalculus inequality Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 3, 2011 at 3:14 Srivatsan 26.8k 7 7 gold badges 95 95 silver badges 148 148 bronze badges asked Nov 3, 2011 at 0:59 Lenar HoytLenar Hoyt 1,072 11 11 silver badges 20 20 bronze badges 3 Presumably, you want >0>0 rather than ≥0≥0, otherwise undefined Thomas Andrews –Thomas Andrews 2011-11-03 01:02:19 +00:00 Commented Nov 3, 2011 at 1:02 You are right, I fixed it.Lenar Hoyt –Lenar Hoyt 2011-11-03 01:03:37 +00:00 Commented Nov 3, 2011 at 1:03 2 In view of the large number of very nice answers, perhaps the question should have been, is there any way to avoid proving the inequality?Gerry Myerson –Gerry Myerson 2011-11-03 05:23:52 +00:00 Commented Nov 3, 2011 at 5:23 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 31 Save this answer. Show activity on this post. Since w+1 w≥2 w+1 w≥2 for all w>0 w>0, we have (x+y+z)(1 x+1 y+1 z)=3+x y+y x+x z+z x+y z+z y≥3+2+2+2=9(x+y+z)(1 x+1 y+1 z)=3+x y+y x+x z+z x+y z+z y≥3+2+2+2=9 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 3, 2011 at 1:12 HeikeHeike 1,176 8 8 silver badges 8 8 bronze badges 0 Add a comment\| This answer is useful 11 Save this answer. Show activity on this post. This also follows from the Cauchy-Schwarz inequality: Let u=(x,y,z)u=(x,y,z) and v=(1 x,1 y,1 z)v=(1 x,1 y,1 z). Then u⋅v≤∥u∥∥v∥u⋅v≤‖u‖‖v‖ is exactly your inequality. P.S. I personally find the other proofs mentioned above much simpler... And I would use the C-S inequality for numbers not vectors. But from my experience, many undergrad students know the C-S vector inequality and yet they don't know the HM-GM-AM inequality, this is the only reason I mentioned this proof :) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 3, 2011 at 2:42 user940 answered Nov 3, 2011 at 2:01 N. S.N. S. 135k 12 12 gold badges 150 150 silver badges 271 271 bronze badges 1 Wow , thank you! I didn't know about this form of CS.A Googler –A Googler 2015-04-20 12:38:47 +00:00 Commented Apr 20, 2015 at 12:38 Add a comment\| This answer is useful 8 Save this answer. Show activity on this post. 1 x+1 y+1 z≥9 x+y+z 1 x+1 y+1 z≥9 x+y+z ↔(x+y+z)(1 x+1 y+1 z)≥9↔(x+y+z)(1 x+1 y+1 z)≥9. Using Cauchy Inequality, we have x+y+z≥3 x y z−−−√3 x+y+z≥3 x y z 3 and 1 x+1 y+1 z≥3 1 x y z−−−√3 1 x+1 y+1 z≥3 1 x y z 3 ⇒(x+y+z)(1 x+1 y+1 z)≥3 x y z−−−√3⋅3 1 x y z−−−−√3=9⇒(x+y+z)(1 x+1 y+1 z)≥3 x y z 3⋅3 1 x y z 3=9 . Equality occurs when x=y=z x=y=z. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 3, 2011 at 9:16 CommunityBot 1 answered Nov 3, 2011 at 4:33 qwerty89qwerty89 746 2 2 gold badges 7 7 silver badges 16 16 bronze badges Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. We don't need all above. We can use simple inequality arithmetic mean ≥ geometric ≥ harmonic mean x+y+z 3≥3 1 x+1 y+1 z⟺(x+y+z)(1 x+1 y+1 z)≥9 x+y+z 3≥3 1 x+1 y+1 z⟺(x+y+z)(1 x+1 y+1 z)≥9 hence proved.. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 10, 2013 at 12:01 answered Apr 10, 2013 at 6:52 chilchinchilchin 149 1 1 silver badge 7 7 bronze badges 3 I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.A.P. –A.P. 2013-04-10 07:19:02 +00:00 Commented Apr 10, 2013 at 7:19 let us continue this discussion in chatA.P. –A.P. 2013-04-10 07:40:19 +00:00 Commented Apr 10, 2013 at 7:40 The right hand side of the ⟺⟺ should be a product, not a ratio. Also, in Heike's answer, the statement w+1/w≥2 w+1/w≥2 is the AM-GM inequality.Willie Wong –Willie Wong 2013-04-10 08:26:46 +00:00 Commented Apr 10, 2013 at 8:26 Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. You can also brute-force it along with some careful re-grouping of terms: 1 x+1 y+1 z≥9 x+y+z 1 x+1 y+1 z≥9 x+y+z ⇔x y+x z+y z x y z≥9 x+y+z⇔x y+x z+y z x y z≥9 x+y+z ⇔x 2 y+x y z+x 2 z+x y 2+x y z+y 2 z+x y z+x z 2+y z 2≥9 x y z⇔x 2 y+x y z+x 2 z+x y 2+x y z+y 2 z+x y z+x z 2+y z 2≥9 x y z ⇔x 2 y−2 x y z+y z 2+x 2 z−2 x y z+y z 2+x z 2−2 x y z+x y 2≥0⇔x 2 y−2 x y z+y z 2+x 2 z−2 x y z+y z 2+x z 2−2 x y z+x y 2≥0 ⇔y(x−z)2+z(x−y)2+x(y−z)2≥0⇔y(x−z)2+z(x−y)2+x(y−z)2≥0. The last inequality holds (since x,y,z>0 x,y,z>0), therefore the first inequality holds. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 3, 2011 at 5:23 user5137 user5137 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. In general, the harmonic mean is at most the arithmetic mean, ((1/n)(x−1 1+⋯+x−1 n))−1≤(1/n)(x 1+⋯+x n)((1/n)(x 1−1+⋯+x n−1))−1≤(1/n)(x 1+⋯+x n) (assuming all quantities positive). Your question is the case n=3 n=3. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 3, 2011 at 1:24 Gerry MyersonGerry Myerson 187k 13 13 gold badges 235 235 silver badges 408 408 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus inequality See similar questions with these tags. 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6249	https://www.rutlandherald.com/news/firefighters-catch-up-on-burning-issues/article_ca503ee6-09aa-5d91-afa4-4b40877f546d.html	Rutland, VT (05701) Today Clear skies. Low 47F. Winds light and variable.. Tonight Clear skies. Low 47F. Winds light and variable. Updated: September 28, 2025 @ 8:36 pm Firefighters catch up on burning issues By SETH HARKNESS Staff Writer NORTH CLARENDON — Hundreds of firefighters from around the state studied Saturday for the real-life tests that come with the ring of the fire bell. The parking lot at Mill River Union High School was crowded with pickup trucks carrying emergency lights and fire department placards as 341 firefighters converged at the Southern Vermont Regional Fire School. With classes such as “Advanced Pumps,” “Foam and Appliances,” and “Extrication and New Vehicle Techniques,” the two-day school helps firefighters stay abreast of the latest techniques and innovations, according to Robert Schlachter, chief of the Rutland City Fire Department and the school’s director. In “Extrication and New Vehicle Technology,” for instance, students looked at the special considerations involved in removing someone from one of the new generation of hybrid gas/electric cars. With their large batteries and high-voltage electrical systems, removing people from these vehicles poses a number of dangers not encountered with gas and diesel cars, Schlachter said. The school continues today, as students trade their books and pencils for fire suits. Firefighters will gather at the Rutland State Airport, a local junk yard, the Pittsford Fire Department’s training site and other locations for practical training in different emergency situations. In past years this has meant staging real fires, but this year instructors will use theatrical smoke as they practice carrying their colleagues out of a simulated blaze. “This year we’re not burning much,” Schlachter said. Of Saturday’s classes, the most heavily subscribed course was on how firefighters can rescue colleagues who become trapped while fighting a fire. “Saving Your Own With Rapid Intervention” attracted about 60 students and was taught by Lt. Rick Kolomay of the Schaumburg, Ill., Fire Department. “We’re the crazy people who run into the buildings when everybody else is leaving,” Schlachter said. “It carries risk and we’re trying to reduce that risk.” In his class, Kolomay shared several case studies involving rescue operations where the rescuers suddenly became the victims. In one case, he showed a video made by a Chicago fire chief who was in the habit of wearing a video camera when responding to a call, so he could review the tape for training purposes. The shaky video depicted a fire that began in a basement of a three-story building. In the midst of the effort to extinguish the fire, a floor collapsed and three firefighters fell into the basement. The scene quickly became one of pandemonium and failed communications and discipline. A dead radio, firefighters who did not leave the building as ordered, and the chief’s sudden focus on his men rather than fighting the fire all contributed to the mayhem. In this case, the three firefighters were not harmed, but Kolomay said lessons still could be learned from the near-disaster. He said it underscored his recommendation that fire crews place a five-member team on standby for emergency rescues of their own members. While most of the audience seemed to agree with his plan in principle, a Weston firefighter asked if Kolomay was aware of how difficult it is for small, rural fire departments to muster these sort of numbers for standby crews. “Where do we get five, where do we get four, where do we get two?” asked Lee Phillips. “In rural settings today, finding the basic volunteers is hard enough,” he said during a lunch break. “Manpower is a constant problem.” The fire school is expected to swell to nearly 400 firefighters from 83 departments for today’s demonstrations. The Southern Vermont Regional Fire School, which started in 1972, costs $35 to $40 per person. If the school turns a profit, Schlachter said, it is split between two organizations that benefit all fire departments in Rutland and Bennington counties. Contact Seth Harkness at seth.harkness@rutlandherald.com. Browser Compatibility Your browser is out of date and potentially vulnerable to security risks. We recommend switching to one of the following browsers:
6250	https://www.comberprimary.com/maths-1/	Maths \| Comber Primary School Log in Sitemap Cookie Information Home Page Home Children Class Pages P5 P4/5 Mrs Drysdale Maths Maths Multiplication Word Problems Today we are going to look at word problems using RUCSAC. Hopefully you remember from class what each letter stands for. To help you recap over this have a look at the PowerPoint below before you start your work today. This morning the calculation you choose will be multiplication. Example - John and his friend Sam had 12 sweets each. How many sweets did they have altogether? 12x2 = 24 sweets. Word Problems rucsac-powerpoint.ppt As an extension of your work there are lots of word problems in each of the multiplication pods I have set up on Studyladder. Subtraction Word Problems On Tuesday you are going to continue to use the RUCSAC method looking at word problems. This time the calculation you will choose is subtraction. Remember to highlight the key words and numbers to help you. Set your sum out clearly with TH H T U columns if needed. Don't forget to exchange if required. Lastly, label your answer and check your sum. Use the second slide in the PDF below to revise your terms of subtraction before you begin your task today. Terms of subtraction math-vocabulary-posters_.pdf 2D Shape We will continue our 2D shape booklet today. Use the co-ordinates given to plot a mark and build a 2d shape. Then try to identify the shape you have made. To help you identify the shapes have a look at the PowerPoint below. If you are not ready for this stage yet use today to catch up with work in your booklet. 2D shape quiz 2D-Shape Quiz.ppt When you have finished go onto Studyladder and have a go at the 2d shape pod. Simple Grid References If you are ready to move onto the co-ordinates sheet on Thursday you will need to have a look at the PowerPoint below -This will recap over grid references with some examples to help refresh your memory. Coordinates coordinates-powerpoint.ppt If you are continuing to work on grid references in your shape booklet today, please have a look at the PowerPoint below which will give you some similar examples using shapes. Plot the Shape position-and-direction-powerpoint.pptx Finish off Friday For your maths lesson on Friday make sure you have completed your mental maths booklet. About Us Welcome Prospectus New and Prospective Parents Who's Who Contact Details Key Information Curriculum Numeracy Literacy ICT Safer Internet Day - Tuesday 9th February 2021 The World Around Us Geography History Science Fun experiments STEM Outdoor learning Summer Autumn Winter Spring Active Travel Science Week Forest schools STEM - Engineering Nursery P1/2 P1 P2 P3/4 P3 P4 P5 P6 P7 The Arts P.E. 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6251	https://www.superteacherworksheets.com/brain-teasers.html	Log In Become a Member Membership Info Addition (Basic) Addition (Multi-Digit) Algebra & Pre-Algebra Comparing Numbers Daily Math Review (Math Buzz) Division (Basic) Division (Long Division) Hundreds Charts Multiplication (Basic) Multiplication (Multi-Digit) Order of Operations Place Value Skip Counting Telling Time Word Problems (Daily) More Math Worksheets Reading Comprehension Reading Comprehension Pre-K/K Reading Comprehension Grade 1 Reading Comprehension Grade 2 Reading Comprehension Grade 3 Reading Comprehension Grade 4 Reading Comprehension Grade 5 Reading Comprehension Grade 6 Reading Comprehension Reading & Writing Reading Worksheets Cause & Effect Daily ELA Review (ELA Buzz) Fact & Opinion Fix the Sentences Graphic Organizers Synonyms & Antonyms Writing Prompts Writing Story Pictures Writing Worksheets More ELA Worksheets Phonics Consonant Sounds Vowel Sounds Consonant Blends Consonant Digraphs Word Families More Phonics Worksheets Early Literacy Alphabet Build Sentences Sight Word Units Sight Words (Individual) More Early Literacy Grammar Nouns Verbs Adjectives Adverbs Pronouns Punctuation Syllables Subjects and Predicates More Grammar Worksheets Spelling Lists Spelling Grade 1 Spelling Grade 2 Spelling Grade 3 Spelling Grade 4 Spelling Grade 5 Spelling Grade 6 More Spelling Worksheets Chapter Books Bunnicula Charlotte's Web Magic Tree House #1 Boxcar Children More Literacy Units Science Animal (Vertebrate) Groups Butterfly Life Cycle Electricity Human Body Matter (Solid, Liquid, Gas) Simple Machines Space - Solar System Weather More Science Worksheets Social Studies 50 States Explorers Landforms Maps (Geography) Maps (Map Skills) More Social Studies Art & Music Colors Worksheets Coloring Pages Learn to Draw Music Worksheets More Art & Music Holidays Autumn Back to School Johnny Appleseed Halloween More Holiday Worksheets Puzzles & Brain Teasers Brain Teasers Logic: Addition Squares Mystery Graph Pictures Number Detective Lost in the USA More Thinking Puzzles Teacher Helpers Teaching Tools Award Certificates More Teacher Helpers Pre-K and Kindergarten Alphabet (ABCs) Numbers and Counting Shapes (Basic) More Kindergarten Worksheet Generator Word Search Generator Multiple Choice Generator Fill-in-the-Blanks Generator More Generator Tools S.T.W. Full Website Index Help Files Contact Brain Teaser & Riddle Worksheets for Kids These printable brain teasers will help build vocabulary, creative thinking and logic skills. The riddles and puzzles are great for kids who need an extra brain-charged challenge. Word Challenges What's the Word? #1 (Mixed Levels) Figure out the mystery word that matches each clue. 2nd through 4th Grades View PDF What's the Word? #2 (Mixed Levels) Figure out the mystery word that matches each clue on this kids' riddle printable. 2nd through 4th Grades View PDF Monkey Message (Easy) Can you decode Marvin the Monkey's secret message to you? Kindergarten and 1st Grade View PDF Thinking Riddles FREE Can your students figure out the answers to these tricky riddle puzzles? 2nd through 4th Grades View PDF Word Puzzle - Soccer (Easy) FREE Figure out the mystery word that matches each clue on this soccer themed fun riddle worksheet. 2nd through 4th Grades View PDF Diltoid Puzzles: Number-Letter Teasers Students solve the number-letter teasers, also known as diltoids. (example: 26 L in the A = 26 letters in the alphabet) 4th through 8th Grades View PDF 5-Letter Word Whiz (Medium Difficulty) Can you unscramble all twenty-six of these 5-letter words? Includes a fun scoring key for kids. 2nd through 4th Grades View PDF Code Riddles (Medium) Each letter in the code represents the letter that comes before it in the alphabet. (example: IFMMP = HELLO) Kindergarten to 2nd Grade View PDF What Letter Comes Next? (Very Hard) Find the pattern of letters and tell which comes next. Kindergarten and 1st Grade View PDF Weird Words Learn about oddities in the English language. Find the weird words in the puzzle, then match them to the given clues. 2nd through 4th Grades View PDF Complete the Proverbs This worksheet contains a list of partially-completed proverbs. Students have to fill in the blank lines to complete each proverb. If they're unsure of the answer, have them make their best guess. 2nd through 4th Grades View PDF Acrostic Squares (Easy) Use the clues to complete the 4 by 4 acrostic crossword squares. 2nd through 4th Grades View PDF More Acrostic Squares (Easy) More 4 by 4 acrostic crossword square puzzles for kids. 2nd through 4th Grades View PDF Rebus Puzzles Rebus Puzzles Riddle (Easy) FREE Can you tell what these picture puzzles are trying to say? 2nd through 5th Grades View PDF Rebus Puzzle Activity: A-1 Can you figure out the word and phrases illustrated in the pictures? 4th through 6th Grades View PDF Rebus Puzzle Activity A-2 What common words and phrases are drawn in the pictures on this rebus riddle activity for kids? 4th through 6th Grades View PDF Rebus Puzzle Activity A-3 Challenge your students to solve these mind-bending rebus puzzles. 4th through 6th Grades View PDF Rebus Puzzle Activity B-1 Here's another set of nine brain teaser rebus puzzles. 4th through 8th Grades View PDF Rebus Puzzle Activity B-2 How many of these picture-word rebus puzzle challenges will your students be able to solve? 4th through 8th Grades View PDF Rebus Puzzle Activity B-3 Will your class be able to figure out the answers to these challenging brain teaser rebus puzzles? 4th through 8th Grades View PDF Hink Pinks Hink Pinks: Level A-1 FREE A hink pink is a pair of rhyming words that can be used to define a silly clue. What do you call a lengthy tune? A long song! 4th through 6th Grades View PDF Hink Pinks Riddles: Level A-2 Here are some more easy hink pink riddles (rhyming pairs). What might you call a rip in a teddy? A bear tear! 4th through 6th Grades View PDF Hink Pinks: Level B-1 These hink pinks are a little bit harder, for kids with a more complex vocabulary. What phrase describes a lawful bird? A legal eagle! 4th through 6th Grades View PDF Hink Pinks: Level C-1 These tough hink pinks will really twist your mind and drive you crazy. What might you call an unhappy friend? A glum chum! 4th through 6th Grades View PDF Hink Pinks Riddles: Level C-2 Even more super-tough hink pink clues to solve. A baseball player who just got a raise is a... richer pitcher. 4th through 6th Grades View PDF Hink Pinks Riddles: Level C-3 What do you call a large rock that's colder than other rocks? A colder boulder! How about a plastic glass for a young dog? A pup cup! 4th through 6th Grades View PDF Hink Pinks Riddles: Level C-4 What do you call a tired, yellow flower? A lazy daisy! What do you call a magical creature who milks cows? A dairy fairy! 4th through 6th Grades View PDF Visual-Spatial Challenges Jigsaw Pieces (Easy) Which jigsaw puzzle piece fits into each puzzle? 2nd through 4th Grades View PDF Stroop Test Say the colors you see printed on the page, not the color word that is written. This game is tricky for fluent readers. (Requires a color printer.) 2nd through 8th Grades View PDF Number Challenges Calculator Words Type numbers into a calculator, turn the calculator upside down and read the words. View PDF Circle the 10s (Easy) Find the pairs of numbers in each row that have a sum of 10. Kindergarten and 1st Grade View PDF Circle the 25s (Medium) Figure out which pairs of numbers add up to 25. 1st and 2nd Grades View PDF Circle the 100s (Hard) Find and circle the pairs of numbers that add up to 100. 1st through 3rd Grades View PDF How Old Am I? (Grade 5 and Up) Calculate how many weeks, months, and days old you are. Problem-solving worksheet requires higher-level multiplication skills. 3rd and 4th Grades View PDF See Also: Addition Squares Addition square mathematical logic puzzles What Am I? Challenges Use the clues to determine the mystery object. Number Detective Read the clues and figure out the secret number. Secret Code Math On these printable math worksheets, students will use a cypher key to decode the numbers on math problems. Penelope Peabody - 50 States What state is Penelope Peabody visiting? Use the clues to determine her location. 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6252	https://www.ck12.org/flexi/physical-science/Temperature-in-Physics/what-is-100-degrees-celsius-in-kelvin-(k-celsius-plus-273)/	Flexi answers - What is 100 degrees Celsius in Kelvin? (K = Celsius + 273) \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Physical Science Temperature Question What is 100 degrees Celsius in Kelvin? (K = Celsius + 273) Flexi Says: To convert from Celsius to Kelvin, we add 273 to the Celsius temperature. So, 100∘C+273=373 K So, 100∘C=373 K. Analogy / Example Try Asking: What is 67 degrees Fahrenheit in Celsius?What is 73 degrees Fahrenheit in Celsius?How do you convert 21 degrees Celsius to Fahrenheit? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
6253	https://books.google.com/books/about/Esau_s_Plant_Anatomy.html?id=0DhEBA5xgbkC	Esau's Plant Anatomy: Meristems, Cells, and Tissues of the Plant Body: Their ... - Ray F. Evert - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Good for: Web Tablet / iPad eReader Smartphone#### Features: Flowing text Scanned pages Help with devices & formats Learn more about books on Google Play Buy eBook - $205.00 Get this book in print▼ Wiley.com Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» My library My History Esau's Plant Anatomy: Meristems, Cells, and Tissues of the Plant Body: Their Structure, Function, and Development ================================================================================================================= Ray F. Evert John Wiley & Sons, Aug 28, 2006 - Science - 624 pages This revision of the now classic Plant Anatomy offers a completely updated review of the structure, function, and development of meristems, cells, and tissues of the plant body. The text follows a logical structure-based organization. Beginning with a general overview, chapters then cover the protoplast, cell wall, and meristems, through to phloem, periderm, and secretory structures. "There are few more iconic texts in botany than Esau’s Plant Anatomy... this 3rd edition is a very worthy successor to previous editions..." ANNALS OF BOTANY, June 2007 More » Preview this book » Selected pages Page 4 Title Page Table of Contents Index References Contents Chapter 1 Structure and Development of the Plant BodyAn Overview1 Chapter 2 The Protoplast Plasma Membrane Nucleus and Cytoplasmic Organelles 15 Chapter 3 The Protoplast Endomembrane System Secretory Pathways Cytoskeleton and Stored Compounds 45 Chapter 4 Cell Wall 65 Chapter 5 Meristems and Differentiation 103 Chapter 6 Apical Meristems 133 Chapter 7 Parenchyma and Collenchyma 175 Chapter 8 Sclerenchyma 191 Chapter 12 Vascular Cambium 323 Chapter 13 Phloem Cell Types and Developmental Aspects 357 Chapter 14 Phloem Secondary Phloem and Variations in Its Structure 407 Chapter 15 Periderm 427 Chapter 16 External Secretory Structures 447 Chapter 17 Internal Secretory Structures 473 Addendum Other Pertinent References Not Cited in the Text 503 Glossary 521 More Chapter 9 Epidermis 211 Chapter 10 Xylem Cell Types and Developmental Aspects 255 Chapter 11 Xylem Secondary Xylem and Variations in Wood Structure 291 Author Index541 Subject Index567 Copyright Less Other editions - View all Esau's Plant Anatomy: Meristems, Cells, and Tissues of the Plant Body: Their ... Ray F. Evert No preview available - 2006 Esau's Plant Anatomy: Meristems, Cells, and Tissues of the Plant Body: Their ... Ray F. Evert No preview available - 2006 Common terms and phrases actinAnatomyangiospermsapical meristemArabidopsisauxinbarkbundlescallosecambial activitycell divisioncell wallcellulosechloroplastscollenchymacompanion cellsconiferscortexcorticalcuticlecytoplasmdifferentiationductselongationendoplasmic reticulumepidermal cellsepidermisEsaueudicotsfibersformationfunctionfusiform initialsgeneGolgiguard cellsIAWAintercellular spaceslaticiferslayerleafleavesmaturemeristematicmesophyllmicrofibrilsmicrotubulesmother cellnectariesoccurorganizationparenchymaparenchyma cellspatternpericlinalperidermphellogenPinusPlant Biolplant bodyPlant CellPlant PhysiolPlant Sciplasma membraneplasmodesmataplastidsporesprimary wallproteinprotoplastradialroleroot hairsrootcapsclereidssecondary phloemsecondary wallsecondary xylemsecretoryshoot apexshoot apicalsieve cellssieve elementssieve platessieve tubessieve-tube elementsspeciesstemstomatalstructuresurfacetangentialthickeningtiontracheary elementstracheidstransporttransverse sectiontreestrichomesUltrastructurevacuolevascular cambiumvascular tissuesvessel elementswoodzone References to this book Cell Biology of Plant Nematode Parasitism R. Howard Berg,Chris Taylor Limited preview - 2008 About the author(2006) Ray F. Evert, Ph.D. is Professor of Botany and Plant Pathology at the University of Wisconsin, Madison. Bibliographic information Title Esau's Plant Anatomy: Meristems, Cells, and Tissues of the Plant Body: Their Structure, Function, and Development AuthorRay F. Evert Contributor Susan E. Eichhorn Edition 3, illustrated Publisher John Wiley & Sons, 2006 ISBN 0470047372, 9780470047378 Length 624 pages SubjectsScience › Life Sciences › Botany Science / General Science / Life Sciences / Botany Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
6254	https://journals.asm.org/doi/10.1128/jvi.68.6.3512-3526.1994	Preexisting nuclear architecture defines the intranuclear location of herpesvirus DNA replication structures \| Journal of Virology Request Username Can't sign in? Forgot your username? Enter your email address below and we will send you your username Email Close If the address matches an existing account you will receive an email with instructions to retrieve your username Create a new account Email Returning user Change Password Old Password New Password Too Short Weak Medium Strong Very Strong Too Long Your password must have 8 characters or more and contain 3 of the following: a lower case character, an upper case character, a special character or a digit Too Short Password Changed Successfully Your password has been changed close Login Register Email Password Forgot password? Reset it here - [x] Keep me logged in Fields with are mandatory Don't have an account? 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Your Phone has been verified close CLOSE This Journal This Journal Anywhere Enter words / phrases / DOI / ISBN / keywords / authors / etc Search Search Enter words / phrases / DOI / ISBN / keywords / authors / etc Search Search Advanced search Suggested Terms: open access policy peer review at ASM what it costs to publish with ASM calls for editors asm journals collections Skip to main content BACK Antimicrobial Agents and Chemotherapy Applied and Environmental Microbiology ASM Animal Microbiology ASM Case Reports Clinical Microbiology Reviews EcoSal Plus Infection and Immunity Journal of Bacteriology Journal of Clinical Microbiology Journal of Microbiology & Biology Education Journal of Virology mBio Microbiology and Molecular Biology Reviews Microbiology Resource Announcements Microbiology Spectrum mSphere mSystems 0 Login / Register Skip main navigation JOURNAL HOME FOR AUTHORS WRITING YOUR PAPER SUBMITTING PEER REVIEW PROCESS TRANSFERRING ACCEPTED PAPERS OPEN ACCESS EDITORIAL POLICIES PUBLISHING ETHICS ARTICLES CURRENT ISSUE LATEST ARTICLES COLLECTIONS ARCHIVE ABOUT THE JOURNAL ABOUT JVI SUBMIT SCOPE EDITORIAL BOARD FORMATTING SUBMISSION AND REVIEW PROCESS ACTIVE CONTRIBUTOR TRACK FOR REVIEWERS SEMINAR SERIES FAQ CONTACT US FOR SUBSCRIBERS MEMBERS INSTITUTIONS ALERTS RSS JOURNAL HOME FOR AUTHORS WRITING YOUR PAPER SUBMITTING PEER REVIEW PROCESS TRANSFERRING ACCEPTED PAPERS OPEN ACCESS EDITORIAL POLICIES PUBLISHING ETHICS ARTICLES CURRENT ISSUE LATEST ARTICLES COLLECTIONS ARCHIVE ABOUT THE JOURNAL ABOUT JVI SUBMIT SCOPE EDITORIAL BOARD FORMATTING SUBMISSION AND REVIEW PROCESS ACTIVE CONTRIBUTOR TRACK FOR REVIEWERS SEMINAR SERIES FAQ CONTACT US FOR SUBSCRIBERS MEMBERS INSTITUTIONS ALERTS RSS JOURNALS 0 Login / Register ASM Journals Journal of Virology Vol. 68, No. 6 Preexisting nuclear architecture defines the intranuclear location of herpesvirus DNA replication structures Advertisement Research Article 1 June 1994 Share on Preexisting nuclear architecture defines the intranuclear location of herpesvirus DNA replication structures Authors: A de Bruyn Kops, D M KnipeAuthors Info & Affiliations 72 696 Metrics Total Citations 72 Total Downloads 696 View all metrics Cite PDF/EPUB Contents JVI Volume 68, Number 6 June 1994 Abstract Information & Contributors Metrics & Citations References Figures Tables Media Share Abstract Herpes simplex virus DNA replication proteins localize in characteristic patterns corresponding to viral DNA replication structures in the infected cell nucleus. The intranuclear spatial organization of the HSV DNA replication structures and the factors regulating their nuclear location remain to be defined. We have used the HSV ICP8 DNA-binding protein and bromodeoxyuridine labeling as markers for sites of herpesviral DNA synthesis to examine the spatial organization of these structures within the cell nucleus. Confocal microscopy and three-dimensional computer graphics reconstruction of optical series through infected cells indicated that viral DNA replication structures extend through the interior of the cell nucleus and appear to be spatially separate from the nuclear lamina. Examination of viral DNA replication structures in infected, binucleate cells showed similar or virtually identical patterns of DNA replication structures oriented along a twofold axis of symmetry between many of the sister nuclei. These results demonstrate that HSV DNA replication structures are organized in the interior of the nucleus and that their location is defined by preexisting host cell nuclear architecture, probably the internal nuclear matrix. Formats available You can view the full content in the following formats: PDF/ePub Information & Contributors Information Contributors Information Published In Journal of Virology Volume 68 • Number 6 • June 1994 Pages: 3512 - 3526 PubMed: 8189490 History Published online: 1 June 1994 Permissions Request permissions for this article. Request permissions Download PDF Contributors Expand All Authors A de Bruyn Kops Department of Microbiology and Molecular Genetics, Harvard Medical School, Boston, Massachusetts 02115. View all articles by this author D M Knipe Department of Microbiology and Molecular Genetics, Harvard Medical School, Boston, Massachusetts 02115. View all articles by this author Metrics & Citations Metrics Citations Metrics Article Metrics View all metrics Downloads Citations No data available. 696 72 Total 6 Months 12 Months Total number of downloads Note: For recently published articles, the TOTAL download count will appear as zero until a new month starts. There is a 3- to 4-day delay in article usage, so article usage will not appear immediately after publication. Citation counts come from the Crossref Cited by service. 83 6 64 0 Smart Citations 83 6 64 0 Citing Publications Supporting Mentioning Contrasting View Citations See how this article has been cited at scite.ai scite shows how a scientific paper has been cited by providing the context of the citation, a classification describing whether it supports, mentions, or contrasts the cited claim, and a label indicating in which section the citation was made. Citations Citation text copied Copy de Bruyn Kops A, Knipe D M. 1994. 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6255	https://support.microsoft.com/en-us/office/var-p-function-73d1285c-108c-4843-ba5d-a51f90656f3a	Microsoft Support Sign in Sign in with Microsoft Sign in or create an account. Select a different account. You have multiple accounts Choose the account you want to sign in with. VAR.P function Applies To Excel for Microsoft 365 Excel for Microsoft 365 for Mac Excel for the web Excel 2024 Excel 2024 for Mac Excel 2021 Excel 2021 for Mac Excel 2019 Excel 2016 Calculates variance based on the entire population (ignores logical values and text in the population). Syntax VAR.P(number1,[number2],...) The VAR.P function syntax has the following arguments: Number1 Required. The first number argument corresponding to a population. Number2, ... Optional. Number arguments 2 to 254 corresponding to a population. Remarks VAR.P assumes that its arguments are the entire population. If your data represents a sample of the population, then compute the variance by using VAR.S. Arguments can either be numbers or names, arrays, or references that contain numbers. Logical values, and text representations of numbers that you type directly into the list of arguments are counted. If an argument is an array or reference, only numbers in that array or reference are counted. Empty cells, logical values, text, or error values in the array or reference are ignored. Arguments that are error values or text that cannot be translated into numbers cause errors. If you want to include logical values and text representations of numbers in a reference as part of the calculation, use the VARPA function. The equation for VAR.P is: where x is the sample mean AVERAGE(number1,number2,…) and n is the sample size. Example Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data. \| \| \| Strength \| \| 1,345 \| \| 1,301 \| \| 1,368 \| \| 1,322 \| \| 1,310 \| \| 1,370 \| \| 1,318 \| \| 1,350 \| \| 1,303 \| \| 1,299 \| \| Formula \| Description \| Result \| \| =VAR.P(A2:A11) \| Variance of breaking strengths for all the tools, assuming that only 10 tools are produced (the entire population is used). \| 678.84 \| \| =VAR.S(A2:A11) \| The variance, using the VAR.S function, which assumes only a sample of the population is tested. The result is different from VAR.P. \| 754.27 \| Need more help? Want more options? Explore subscription benefits, browse training courses, learn how to secure your device, and more. Microsoft 365 subscription benefits Microsoft 365 training Microsoft security Accessibility center Communities help you ask and answer questions, give feedback, and hear from experts with rich knowledge. Ask the Microsoft Community Microsoft Tech Community Windows Insiders Microsoft 365 Insiders Thank you for your feedback! ×
6256	https://www.youtube.com/watch?v=l5VCku4Eo_o	Prove that cos(sin^(-1)) = sqrt(1-x^2). Inverse Trig Functions Ms Shaws Math Class 51100 subscribers 210 likes Description 18922 views Posted: 7 Mar 2021 18 comments Transcript: Intro hi everyone we're going to prove that cosine of inverse sine of x equals the square root of 1 minus x squared Inner function so i'm going to look at this inner function first and let y equal this so we have y equals inverse sine of x and that means our y has to be in between quadrants four and quadrants 1 there so that's our restrictions and therefore our cosine of y is going to be positive or greater than or equal to 0. so now we have that cosine of inverse sine of x i'm going to use substitution equals y all right now i know that cosine of y Proof remember your identities cosine squared of let's just write y plus sine squared of y equals 1. so i can write this as 1 minus sine squared of y if i put the square root so i'm just subtracting this from both sides and square rooting both sides and remember has to be positive all right so now that i have that i know that my um this equals square root of 1 minus and sine squared is x so let's look at this if i rewrite this i can write this as sine of y equals x so now since this is sine squared i can write sine squared of y equals x squared so basically this is going to be 1 minus x squared and we just finished the proof put that all in red 1 minus all right thank you have a nice day bye bye you
6257	https://www.studocu.com/en-us/messages/question/3311988/from-the-measured-data-determine-the-early-voltage-va-for-the-transistor	[Solved] from the measured data determine the early voltage va for the - Engineering Technology (Electroni) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent from the measured data determine the early voltage va for the transistorEngineering Technology (Electroni) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist St. Petersburg College Engineering Technology Question from the measured data determine the early voltage va for the St. Petersburg College Engineering Technology Question ### Nicole 2 years ago from the measured data determine the early voltage va for the transistor Like 0 Answer Created with AI 2 years ago Determining the Early Voltage (Va) for a Transistor The Early voltage, also known as Va, is a parameter used in the model of transistors, particularly in the Ebers-Moll model for a bipolar junction transistor (BJT). It is named after its discoverer, James M. Early. To determine the Early voltage from measured data, you need to follow these steps: Plot the Output Characteristics: First, you need to plot the output characteristics of the transistor. This is a graph of the collector current (Ic) against the collector-emitter voltage (Vce) for different base currents (Ib). Draw the Load Line: The load line is a straight line drawn from the saturation point to the cut-off point on the output characteristics graph. Find the Intersection Points: The points where the load line intersects the output characteristics curves are the operating points of the transistor. Draw the Early Voltage Line: The Early voltage line is a line drawn from the origin through the operating points. The slope of this line is inversely proportional to the Early voltage. Calculate the Early Voltage: The Early voltage can be calculated from the slope of the Early voltage line. The formula is: js Va = -1 / slope Here is an example of how the output characteristics graph might look: \| Vce (V) \| Ic (mA) for Ib = 20uA \| Ic (mA) for Ib = 40uA \| Ic (mA) for Ib = 60uA \| --- --- \| \| 0.2 \| 1.0 \| 2.0 \| 3.0 \| \| 0.4 \| 2.0 \| 4.0 \| 6.0 \| \| 0.6 \| 3.0 \| 6.0 \| 9.0 \| \| 0.8 \| 4.0 \| 8.0 \| 12.0 \| \| 1.0 \| 5.0 \| 10.0 \| 15.0 \| Remember, the Early voltage is a measure of the variation of the base-emitter junction area with the collector-emitter voltage. It is an important parameter in the design and analysis of transistor circuits. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Engineering Technology Electroni St. Petersburg College 2 Documents Go to course 3 Chapter 6 - Engineering Technology Engineering Technology None 1 Shmaba Gabapentin booty Engineering Technology None Discover more from: Engineering Technology ElectroniSt. Petersburg College2 Documents Go to course 3 Chapter 6 - Engineering Technology Engineering Technology None 1 Shmaba Gabapentin booty Engineering Technology None Related Answered Questions 2 years ago Given voltage = 240 V, current = 10 A, power factor = 0.1. Compute for the electrical power for a three phase AC motor. Engineering Technology (Electroni) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. Copyright © 2025 StudeerSnel B.V., Keizersgracht 424-sous, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01 Cookies give you a personalised experience We’re not talking about the crunchy, tasty kind. These cookies help us keep our website safe, give you a better experience and show more relevant ads. We won’t turn them on unless you accept. Want to know more or adjust your preferences? Reject all Accept all cookies Manage cookies
6258	https://www.kristakingmath.com/blog/tangent-lines-of-circles	Tangent lines of circles — Krista King Math \| Online math help Tangent lines of circles What is a tangent line to a circle? A tangent line to a circle intersects the circle at exactly one point on its circumference. Line XY is tangent to circle A at point X. Point X is called the point of tangency. The radius drawn from the center of the circle to the point of tangency is always perpendicular to the tangent line. In the figure below, Radius AX is perpendicular to line XY. How to use tangent lines of circles Take the course Want to learn more about Geometry? I have a step-by-step course for that. :) Learn More Determining whether or not a line is tangent to a circle Example The radius of circle C is 5 units. AB=6 units and DA=3 units. Determine whether line AB is tangent to circle C. If line AB is tangent to circle C, then the radius will be perpendicular to line AB and angle ∠CBA will be a right angle. If the triangle formed in the diagram is a right triangle, then the Pythagorean theorem will be satisfied for the triangle, so we want to verify the following equation. CB2+AB2=CA2 CB is a radius so we know CB=5 units, and we know that AB=6 units. We know that CA=CD+DA, and that CD is also a radius, so CD=5 units. Since DA=3 units, CA=5+3=8. Now we can check the Pythagorean theorem. CB2+AB2=CA2 52+62=82 25+36=64 61≠64 This means that angle ∠CBA is not a right angle, and line AB is therefore not tangent to circle C. Let’s do one more. Example Find the length of the radius of circle C, given that line AB is a tangent line. A radius drawn to point A will be perpendicular to line AB and form right triangle BAC. Let’s call the length of the radius x, then AC=x and CD=x. Now we can use the Pythagorean Theorem to set up an equation and solve for x. x2+15.22=(x+7.6)2 x2+231.04=x2+15.2x+57.76 Subtract x2 and 57.76 from both sides and rearrange. 15.2x=173.28 x=11.4 The radius of circle C is 11.4 units. Get access to the complete Geometry course Get started Learn mathKrista Kingmath, learn online, online course, online math, geometry, circles, geometry of circles, tangent lines of circles, circle tangent lines, tangent lines, circle tangent line problems Facebook0TwitterLinkedIn0RedditTumblrPinterest00 Likes
6259	https://physics.stackexchange.com/questions/227942/why-are-significant-figure-rules-in-multiplication-division-different-than-in-ad	units - Why are significant figure rules in Multiplication/Division different than in Addition/Subtraction? - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why are significant figure rules in Multiplication/Division different than in Addition/Subtraction? Ask Question Asked 9 years, 8 months ago Modified9 years, 5 months ago Viewed 8k times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. I've never understood specifically why this is. Here's what I mean. In Addition/Subtraction, what matters are the digits after the decimal point. So for example: 1.689 + 4.3 = ``` 1.629 + 4.3XX 5.929 5.9 ``` This makes sense to me. I filled in uncertain values with X, and it makes sense why I can't use the 0.029 in the answer - because I added it to an uncertain value. However, I don't understand the rules when it comes to Multiplication/Division. The same little trick with X's doesn't help me here. I know that what matters in Multiplication/Division are the significant figures. So for example: 12.3 4.6 = ``` 12.3 4.6 738 492X 56.58 57 ``` The answer is 57 according to significant figure rules of Multiplication/Division, but I just can't make sense of those rules like the way I did with Addition/Subtraction. Does anyone have an intuitive explanation for the significant figure rules of Multiplication/Division? units error-analysis Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Jan 6, 2016 at 21:07 Qmechanic♦ 222k 52 52 gold badges 636 636 silver badges 2.6k 2.6k bronze badges asked Jan 6, 2016 at 20:59 b0yfriendb0yfriend 213 2 2 silver badges 5 5 bronze badges 8 2 I am reasonably certain that this has been asked before.Kyle Kanos –Kyle Kanos 2016-01-06 21:04:35 +00:00 Commented Jan 6, 2016 at 21:04 More on significant figures.Qmechanic –Qmechanic♦ 2016-01-06 21:09:03 +00:00 Commented Jan 6, 2016 at 21:09 I just looked at the posts you linked. None of them address my question.b0yfriend –b0yfriend 2016-01-06 21:12:59 +00:00 Commented Jan 6, 2016 at 21:12 2 You're not going to get a better answer, because significant figures have no theoretical basis. They're a cheap approximation to actual error analysis.knzhou –knzhou 2016-01-06 22:01:26 +00:00 Commented Jan 6, 2016 at 22:01 1 @knzhou That (" They're a cheap approximation to actual error analysis.") is the basis of a good answer, if you'd like to expand it.dmckee --- ex-moderator kitten –dmckee --- ex-moderator kitten 2016-01-06 23:23:33 +00:00 Commented Jan 6, 2016 at 23:23 \|Show 3 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. The number of significant figures is a representation of the uncertainty of a number. 123.4 123.4 has an uncertainty of 0.1 0.1 since the first uncertain digit is usually included. So, your multiplication of 12.3×4.6 12.3×4.6 is better represented as (12.3±0.1)(4.6±0.1).(12.3±0.1)(4.6±0.1). I'm going to use @barrycarter's trick of using scientific notation to better represent the values (1.23±0.01)⋅10 1×(4.6±0.1)⋅10 0.(1.23±0.01)⋅10 1×(4.6±0.1)⋅10 0. (1.23±0.01)(4.6±0.1)×10 1.(1.23±0.01)(4.6±0.1)×10 1. From this we can see that the number with more significant digits has less relative uncertainty. Relative uncertainty is approximately the measured uncertainty (also called absolute uncertainty) divided by the value. Multiplying out: ((1.23)(4.6)+(1.23)(±0.1)+(4.6)(±0.01)+(±0.01)(±0.1))×10 1.((1.23)(4.6)+(1.23)(±0.1)+(4.6)(±0.01)+(±0.01)(±0.1))×10 1. (5.658+(±0.123)+(±0.046)+(±0.001))×10 1.(5.658+(±0.123)+(±0.046)+(±0.001))×10 1. The number of digits we can write down is determined by the total uncertainty, which here is dominated by ±0.123±0.123. So, our result is (5.658±0.123)×10 1.(5.658±0.123)×10 1. (5.7±0.1)×10 1.(5.7±0.1)×10 1. 57±1 57±1 which results in 57 57. To summarize, in multiplication, the largest relative uncertainty dominates the uncertainty of the final answer. This translates into the the number with fewer significant digits determining the significant digits of the final answer. This contrasts with addition, where the largest absolute uncertainty determines the uncertainty of the answer. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 7, 2016 at 6:39 Mark HMark H 25.6k 3 3 gold badges 68 68 silver badges 88 88 bronze badges 4 This is great. Thanks for the explanation. One question: If we were to be more precise about the uncertainty, wouldn't we (±0.123)+(±0.046)+(±0.001)=±0.17(±0.123)+(±0.046)+(±0.001)=±0.17 To get the total uncertainty? But we don't do that because significance arithmetic "is just a cheap approximation of actual error analysis?"b0yfriend –b0yfriend 2016-01-07 17:15:36 +00:00 Commented Jan 7, 2016 at 17:15 Uncertainties don't add that way because errors in measurements aren't always in the same direction. The actual errors in a given measurement could be +0.123+0.123, −0.046−0.046, and 0 0, yielding 0.077 0.077. Because of this, errors add in quadrature e r r 2 1+e r r 2 2+⋯−−−−−−−−−−−−−√e r r 1 2+e r r 2 2+⋯. In this case, the total error is closer to 0.13 0.13, which is close to 0.123 0.123. See @DavidZ's answer below for a much more precise explanation.Mark H –Mark H 2016-01-07 19:17:56 +00:00 Commented Jan 7, 2016 at 19:17 What if we try with a different example like (1.0±0.1)(1.0±0.1)(1.0±0.1)(1.0±0.1)? In this case the computation would give an error ±0.2±0.2 and if we keep multiplying to obtain (1.0±0.1)n(1.0±0.1)n we get an error n⋅0.1 n⋅0.1 that grows eventually reducing the number of significant figures, this seems that would make the product rule false. Wouldn't it?Marco Disce –Marco Disce 2016-12-23 20:48:46 +00:00 Commented Dec 23, 2016 at 20:48 @MarcoDisce That's true, which is why the rule is just that: a heuristic rule to estimate the total error. In equations with lots of moving parts, you need more sophisticated error analysis, like in DavidZ's answer.Mark H –Mark H 2016-12-23 22:53:47 +00:00 Commented Dec 23, 2016 at 22:53 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Actually, I have two explanations, depending on what you consider more intuitive. Interval arithmetic This one is best explained by example. Remember that a value with a certain number of significant figures is supposed to represent that exact value ±5±5 in the first insignificant digit. For example, 12.3 12.3 with three significant figures represents anything in the range [12.25,12.35][12.25,12.35], and 4.6 4.6 with two significant figures represents [4.55,4.65][4.55,4.65]. If you multiply these two ranges, you can get anything between a minimum of 4.55×12.25=55.7375 4.55×12.25=55.7375 and a maximum of 4.65×12.35=57.4275 4.65×12.35=57.4275. So strictly speaking, the result is [55.7375,57.4275][55.7375,57.4275]. However, it's a lot more convenient to represent this range as a single number, using the same significant figure convention as before, where the uncertainty is ±5±5 in the first insignificant digit. We can't actually do this, though; the size of the range is 1.69 1.69, which is not 5×10 n 5×10 n for any n n. The sensible options are either to write 60 60, which represents the range [55,65][55,65], or 57 57, which represents the range [56.5,57.5][56.5,57.5]. But there isn't a straightforward rule that justifies the first choice, without doing interval arithmetic every time. The second choice, on the other hand, can be justified by keeping the lower number of significant figures from either of the original operands. So that's what we do. General error propagation For a better justification, significant figure rules are chosen as they are because they're a decent match to the general formula for propagation of uncertainties. That general formula is as follows: if you have some quantity y y which is a function of several other variables, y=f(x 1,x 2,…)y=f(x 1,x 2,…), then the uncertainty in y y is given by σ y=(∂y∂x 1)2 σ 2 x 1+(∂y∂x 2)2 σ 2 x 2+⋯−−−−−−−−−−−−−−−−−−−−−−−−−−√σ y=(∂y∂x 1)2 σ x 1 2+(∂y∂x 2)2 σ x 2 2+⋯ Actually, this formula only applies under certain conditions, but they're very common conditions, and it would take a whole separate answer (or even more) to explain all the subtleties. (See here and here for examples.) So let's just take that formula as a starting point. If a value x x has n n significant figures, that means its relative uncertainty r=σ x\|x\|r=σ x\|x\| is between (5×10−n)(5×10−n), if x x is a power of 10 10, and (5×10−(n+1))(5×10−(n+1)), if x x is just less than a power of 10 10 (like 99.99999 99.99999). Or within an order of magnitude, anyway. You could argue that maybe the 5 5 should be a 3 3 or something like that, but that gets into the conditions I already decided to skip discussing. Multiplication and division I do this case first because it's easier. If the function is f(x 1,x 2)=x 1 x 2 f(x 1,x 2)=x 1 x 2, the general error propagation formula gives σ y=x 2 2 σ 2 x 1+x 2 1 σ 2 x 2−−−−−−−−−−−√σ y=x 2 2 σ x 1 2+x 1 2 σ x 2 2 Or substituting in r x r x for the uncertainty, all the x 1 x 1 s and x 2 x 2 s cancel out and we get r y=r 2 1+r 2 2−−−−−−√r y=r 1 2+r 2 2 Going through the argument for division, the algebra is a little different but you still get to this same formula for relative uncertainty. There are three cases to consider here: r 1≪r 2 r 1≪r 2: r 1 r 1 is basically insignificant and you can write r y≈r 2 r y≈r 2 r 1≫r 2 r 1≫r 2: r 2 r 2 is basically insignificant and you can write r y≈r 1 r y≈r 1 r 1≈r 2 r 1≈r 2: in this case you can approximate them as roughly equal and write r y≈2–√r 1 r y≈2 r 1. Depending on the exact relative magnitudes, this factor might be a little different, but both r 1 r 1 and 2 r 1 2 r 1 are likely to be around the same order of magnitude, which is all that matters for an uncertainty. One can make this argument a little more precise, but the point is, the resulting uncertainty r y r y is more or less the same as the larger of r 1 r 1 and r 2 r 2. But remember, r r is related to the number of significant figures: r≈10−n r≈10−n, meaning that a larger relative uncertainty r r corresponds to a smaller number of significant figures. So the final quantity y y will typically have the same number of significant figures as the less precise (fewer significant figures) of the two operands x 1 x 1 and x 2 x 2. Addition and subtraction It's also worth checking that this works for addition and subtraction. If the function is f(x 1,x 2)=x 1±x 2 f(x 1,x 2)=x 1±x 2, then the partial derivatives are both equal to 1 1. So the error propagation formula gives σ y=σ 2 x 1+σ 2 x 2−−−−−−−−√σ y=σ x 1 2+σ x 2 2 Here substituting in the relative uncertainty gives r y(x 1±x 2)=r 2 1 x 2 1+r 2 2 x 2 2−−−−−−−−−√r y(x 1±x 2)=r 1 2 x 1 2+r 2 2 x 2 2 but this isn't so useful because things don't cancel out. So let's go back to the previous formula, with the σ σ s. This time, the three cases break down as follows: σ 1≪σ 2 σ 1≪σ 2: σ 1 σ 1 is negligible and you can set σ y≈σ 2 σ y≈σ 2 σ 1≫σ 2 σ 1≫σ 2: σ 2 σ 2 is negligible, so σ y≈σ 1 σ y≈σ 1 σ 1≈σ 2 σ 1≈σ 2: following more or less the same reasoning as before, σ y≈σ 1 σ y≈σ 1 to within an order of magnitude or so Now, the actual uncertainty σ σ is related, not to the number of significant figures, but to their position. The larger uncertainty corresponds to the leftmost insignificant figure among the two operands, which is exactly what you use. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 30, 2016 at 8:01 answered Jan 7, 2016 at 8:27 David ZDavid Z 77.9k 27 27 gold badges 186 186 silver badges 295 295 bronze badges 4 Could you please explain as to how "depending on the exact relative magnitudes, you might have as much as r y≈2 r 1 r y≈2 r 1". Thanks.Procyon –Procyon 2016-04-30 00:50:45 +00:00 Commented Apr 30, 2016 at 0:50 @Procyon Actually, I don't remember exactly what I meant by that. I think I'll take it out.David Z –David Z 2016-04-30 07:56:40 +00:00 Commented Apr 30, 2016 at 7:56 If r 1≈r 2 r 1≈r 2 then you can actually have an error that is greater than both r 1 r 1 and r 2 r 2 and this could result in a smaller number of significant figures. Doesn't it make the general rule of the product false?Marco Disce –Marco Disce 2016-12-23 20:46:58 +00:00 Commented Dec 23, 2016 at 20:46 @MarcoDisce The significant figure rules are only approximate anyway.David Z –David Z 2016-12-23 21:12:24 +00:00 Commented Dec 23, 2016 at 21:12 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. It makes more sense if you convert to scientific notation first, so 12.3 4.6 becomes: (1.2310^1 4.610^1). In this case, you can't expect an answer more accurate than one place after the decimal point, since you don't know what comes after the 6 in 4.6. In general, in scientific notation, you will essentially be multiplying numbers between 1.000 and 9.999... (and adding the powers of 10), so the accuracy you'd expect is the last decimal position of the least precise number. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 7, 2016 at 4:51 user854 user854 0 Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. 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6260	https://math.stackexchange.com/questions/3626559/lines-formed-mid-point-of-sides-and-cevians-from-opposite-point-are-concurrent	Skip to main content Lines formed mid point of sides and cevians from opposite point are concurrent Ask Question Asked Modified 5 years, 4 months ago Viewed 324 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. AD,BE,CF are three concurrent lines in △ABC meeting opposite sides in D,E,F respectively. Show that the joins of the midpoints of BC,CA,AB to the midpoints of AD,BE,CF are concurrent. (Should be done by Ceva's theorem, Menelaus theorem, Stewart's theorem) I tried by using trig form of Ceva's theorem and tried to do something similar to Cevian nests proof by connecting A'B'C' triangle but I failed. So please consider giving a hint or something and post the answer later on if I need it. Source:CTPCM geometry contest-math euclidean-geometry triangles Share CC BY-SA 4.0 Follow this question to receive notifications edited Apr 15, 2020 at 22:37 peter.petrov 13k22 gold badges2323 silver badges4040 bronze badges asked Apr 15, 2020 at 11:32 user765842user765842 4 There should be an elementary solution here using vectors and these theorems which you mention. I used to do contest math long time ago, I think I've seen this problem. – peter.petrov Commented Apr 15, 2020 at 11:58 I don't know vectors more than what is used in high school physics. – user765842 Commented Apr 15, 2020 at 12:00 1 You don't need much, just some classic properties like AB−→−+BC−→−+CA−→−=0 and AB−→−=k.AX−→− iff X lies on AB and similar. – peter.petrov Commented Apr 15, 2020 at 12:01 I abandoned the vectors after calculating for 2 hours or so and realizing that this path is really really very difficult. Then suddenly solved it with a much simpler idea. – peter.petrov Commented Apr 15, 2020 at 22:42 Add a comment \| 3 Answers 3 Reset to default This answer is useful 0 Save this answer. Show activity on this post. Look at the drawing here. What do we have? AD,BE,CF - they intersect in a single/common point - point O A′ - midpoint of BC B′ - midpoint of CA C′ - midpoint of AB D′ - midpoint of AD E′ - midpoint of BE F′ - midpoint of CF From Ceva's theorem for triangle ABC we get: AFFBBDDCCEEA=1(1) Now the trick is to realize that: B′F′F′A′=AFFB(2) C′D′D′B′=BDDC(3) A′E′E′C′=CEEA(4) Why is this so? Because B′C′\|\|BC , C′A′\|\|CA and A′B′\|\|AB so these relations follow from the Intercept theorem. Multiplying the last 3 equations and using (1) we get: B′F′F′A′C′D′D′B′A′E′E′C′=AFFBBDDCCEEA=1 Thus: B′F′F′A′A′E′E′C′C′D′D′B′=1(5) Now using the inverse Ceva's theorem (for the triangle A′B′C′ and for the points D′,E′,F′), we can conclude from (5) that the three lines A′D′,B′E′,C′F′ intersect at a single/common point. This is what we had to prove hence the problem is solved. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Apr 16, 2020 at 12:35 answered Apr 15, 2020 at 22:24 peter.petrovpeter.petrov 13k22 gold badges2323 silver badges4040 bronze badges 2 Well the answer is looking awesome to me but how come we know that D', E', F', lie on the medial triangle. Is it due to the reason that B'A'\|\|AB and Parallel lines cut transervals in a fixed ratio and here that ratio will be CB'/B'A=1? – user765842 Commented Apr 16, 2020 at 3:00 1 @AryanRaina Well... Interceptor theorem... B′C′ must divide AD in a 1:1 ratio. Agree? Hence D′ which is the midpoint of AD must lie on B′C′ – peter.petrov Commented Apr 16, 2020 at 9:57 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Hints: (1) Pick 2 vectors e.g. AB−→− and AC−→− as a basis in the plane, and express all other vectors AX−→− as linear combination of them (where X is any point on this drawing or whichever of the points you need to involve in your solution). This means, for any point X you should be able to find 2 numbers k1,k2 such that AX−→−=k1AB−→−+k2AC−→− Then it's all a matter of picking good start (number) parameters and playing with the equations. Good parameters maybe are: AF:FB = a -> number AE:EC = b -> number Then BD:DC you can express in terms of a,b using Ceva. (2) Denote: point S1=A′D′∩B′E′ point S2=A′D′∩C′F′ Prove that AS1−→−=AS2−→− If you do this, it would mean the points S1 and S2 coincide. My high school math is rusty but that (or similar) should be the general idea. (3) Also, for arbitrary points in the plane ABCD, prove and use this lemma: FE−→−=1/2(CA−→−+DB−→−) (on the drawing below F and E are midpoints). Seems it might be useful for this problem. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Apr 15, 2020 at 12:31 answered Apr 15, 2020 at 12:08 peter.petrovpeter.petrov 13k22 gold badges2323 silver badges4040 bronze badges 8 How would I know k1, k2 wouldn't assuming too many variables be a problem – user765842 Commented Apr 15, 2020 at 12:19 Well, these are hints/ideas. You need to try, probe, suffer a bit, and see what happens. But at the end some of the attempts will work :) Of course if you want a direct solution (just to read it), you will miss all the fun because the solutions usually don't say what insights took them there. – peter.petrov Commented Apr 15, 2020 at 12:27 And this is a math contest preparation problem. It's not supposed to be of the form " 2+x=5 Find x ". See especially the lemma, it might be useful here. – peter.petrov Commented Apr 15, 2020 at 12:29 Ok thanks dude I will try it for an hour or so if I still fail then I will ask you the solution. – user765842 Commented Apr 15, 2020 at 12:44 I am still unable to think how to get k1, k2 and now I want to see the answer, my brain freezes from vectors I haven't started it yet. By the way I proved that (3) theorem. – user765842 Commented Apr 15, 2020 at 13:17 \| Show 3 more comments This answer is useful -1 Save this answer. Show activity on this post. I would do this using Cartesian coordinates. Given any triangle, we can set up a coordinate system with origin at one vertex and x-axis along one side. In that coordinate system, the vertices of the triangle are at (0, 0), (a, 0), and (b, c) for some numbers, a, b, and c. The midpoints of the three sides are (a/2, 0), (b/2, c/2), and ((a+b)/2, c/2). The medians are given by y= ((a+b)/c)x, y= (2c/(2b-a))x+ 2c/(a-2b), and y= (ac/(4b-2a))x+ ac/(2a-4b). Show that the solutions to any two of those equation are the same. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 15, 2020 at 11:51 user247327user247327 19k22 gold badges1515 silver badges2121 bronze badges 4 1 1) this is not quite an elementary solution (for contest math) 2) this proves only that the 3 medians intersect in a single point ... no? – peter.petrov Commented Apr 15, 2020 at 11:52 @user247327 What does medians have to do here? By the way I haven't started coordinate geometry I just know basics(straight lines and circles) and my books is asking me solve this via the theorems mentioned above. – user765842 Commented Apr 15, 2020 at 11:56 YOU said "the joins of the midpoint of BC,CA,AB to the midpoints of AD,BE,CF". Those are "medians"! – user247327 Commented Apr 15, 2020 at 12:09 How are they medians the don't even touch the points A, B, C – user765842 Commented Apr 15, 2020 at 12:21 Add a comment \| You must log in to answer this question. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 1 If three cevians are concurrent at a point and form triangles of equal area, the point is the centroid 1 Prove that lines passing through the midpoints of sides of a triangle and the midpoints of cevians are also concurrent 7 Cevians AD, BE, CF are concurrent, as are cevians DP, EQ, FR; show that AP, BQ, CR are concurrent 0 Ceva's Theorem- Homothetic Triangles 1 AD, BE, CF are concurrent in △ABC. Show that lines through midpoints of BC, CA, AB parallel to AD, BE, CF are concurrent. 2 Prove that external bisectors of the angles of a triangle meet the opposite sides in three collinear points. 0 Concurrency of lines made with end points of concurrent lines of a triangle made by end point of concurrent lines and points of given triangle. 0 Prove that the lines joining the midpoints of opposite sides of a quadrilateral...... 3 Prove that perpendiculars dropped from midpoints of orthic triangle to opposite sides are concurrent. 0 Ceva's and Menelaus in Olympiad problems - Possible generalization of Monge's Theorem? Hot Network Questions tcolorbox/pgfkeys: making an argument the title if it contains no pgfkeys; otherwise, pass the argument as pgfkeys to tcolorbox Dovecot login rejected with password having illegal chars Ethical considerations of line editing How are the crosswords made for mass-produced puzzle books? 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6261	https://dlmf.nist.gov/10	DLMF: Chapter 10 Bessel Functions DLMF Index Notations Search Help? Citing Customize Annotate UnAnnotate About the Project Chapter 10 Bessel Functions F.W.J.OlverInstitute for Physical Science and Technology and Department of Mathematics, University of Maryland, College Park, Maryland. L.C.MaximonCenter for Nuclear Studies, Department of Physics, The George Washington University, Washington, D.C. ⓘ Acknowledgements:This chapter is based in part on Abramowitz and Stegun (1964, Chapters 9, 10, and 11) by F. W. J. Olver, H. A. Antosiewicz, and Y. L. Luke, respectively. The authors are pleased to acknowledge assistance of Martin E. Muldoon with §§10.21 and 10.42, Adri Olde Daalhuis with the verification of Eqs.(10.15.6)–(10.15.9), (10.38.6), (10.38.7), (10.60.7)–(10.60.9), and (10.61.9)–(10.61.12), Peter Paule and Frédéric Chyzak for the verification of Eqs. (10.15.6)–(10.15.9), (10.38.6), (10.38.7), (10.56.1)–(10.56.5), (10.60.4), (10.60.6), (10.60.10), and (10.60.11) by application of computer algebra, Nico Temme with the verification of Eqs. (10.15.6)–(10.15.9), (10.38.6), and (10.38.7), and Roderick Wong with the verification of §§10.22(v) and 10.43(v).Notes:The main references used in writing this chapter are Watson (1944) and Olver (1997b).Referenced by:§3.10(ii), New Associate Editors for the DLMF, Profile Diego Dominici, Profile Frank W.J.Olver, Profile Gergő Nemes, Profile Leonard C.Maximon, § ‣ Chapter Authors, § ‣ Chapter Authors, § ‣ Associate Editors, § ‣ Associate Editors, Erratum (V1.1.6) for Chapters 10 Bessel Functions, 18 Orthogonal Polynomials, 34 3 j, 6 j, 9 j Symbols, Erratum (V1.1.6) for Chapters 1 Algebraic and Analytic Methods, 10 Bessel Functions, 14 Legendre and Related Functions, 18 Orthogonal Polynomials, 29 Lamé Functions, Version 1.2.0 (March 27, 2024), § ‣ Software Cross IndexPermalink: Notation 10.1 Special Notation Bessel and Hankel Functions 10.2 Definitions 10.3 Graphics 10.4 Connection Formulas 10.5 Wronskians and Cross-Products 10.6 Recurrence Relations and Derivatives 10.7 Limiting Forms 10.8 Power Series 10.9 Integral Representations 10.10 Continued Fractions 10.11 Analytic Continuation 10.12 Generating Function and Associated Series 10.13 Other Differential Equations 10.14 Inequalities; Monotonicity 10.15 Derivatives with Respect to Order 10.16 Relations to Other Functions 10.17 Asymptotic Expansions for Large Argument 10.18 Modulus and Phase Functions 10.19 Asymptotic Expansions for Large Order 10.20 Uniform Asymptotic Expansions for Large Order 10.21 Zeros 10.22 Integrals 10.23 Sums 10.24 Functions of Imaginary Order Modified Bessel Functions 10.25 Definitions 10.26 Graphics 10.27 Connection Formulas 10.28 Wronskians and Cross-Products 10.29 Recurrence Relations and Derivatives 10.30 Limiting Forms 10.31 Power Series 10.32 Integral Representations 10.33 Continued Fractions 10.34 Analytic Continuation 10.35 Generating Function and Associated Series 10.36 Other Differential Equations 10.37 Inequalities; Monotonicity 10.38 Derivatives with Respect to Order 10.39 Relations to Other Functions 10.40 Asymptotic Expansions for Large Argument 10.41 Asymptotic Expansions for Large Order 10.42 Zeros 10.43 Integrals 10.44 Sums 10.45 Functions of Imaginary Order 10.46 Generalized and Incomplete Bessel Functions; Mittag-Leffler Function Spherical Bessel Functions 10.47 Definitions and Basic Properties 10.48 Graphs 10.49 Explicit Formulas 10.50 Wronskians and Cross-Products 10.51 Recurrence Relations and Derivatives 10.52 Limiting Forms 10.53 Power Series 10.54 Integral Representations 10.55 Continued Fractions 10.56 Generating Functions 10.57 Uniform Asymptotic Expansions for Large Order 10.58 Zeros 10.59 Integrals 10.60 Sums Kelvin Functions 10.61 Definitions and Basic Properties 10.62 Graphs 10.63 Recurrence Relations and Derivatives 10.64 Integral Representations 10.65 Power Series 10.66 Expansions in Series of Bessel Functions 10.67 Asymptotic Expansions for Large Argument 10.68 Modulus and Phase Functions 10.69 Uniform Asymptotic Expansions for Large Order 10.70 Zeros 10.71 Integrals Applications 10.72 Mathematical Applications 10.73 Physical Applications Computation 10.74 Methods of Computation 10.75 Tables 10.76 Approximations 10.77 Software 9.20 Software10.1 Special Notation © 2010–2025 NIST / Disclaimer / Feedback; Version 1.2.4; Release date 2025-03-15. Site PrivacyAccessibilityPrivacy ProgramCopyrightsVulnerability DisclosureNo Fear Act PolicyFOIAEnvironmental PolicyScientific IntegrityInformation Quality StandardsCommerce.govScience.govUSA.gov
6262	https://artofproblemsolving.com/wiki/index.php/Difference_of_squares?srsltid=AfmBOoqv1lSdLTmElQ7R5oSBMBLwXQuVKtILtccuh2sxiGUsk7asKCg5	Art of Problem Solving Difference of squares - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Difference of squares Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Difference of squares Recognizing a difference of squares is a commonly used factoring technique in algebra. It refers to the identity . Note that this identity depends only on the (right and left) distributive property and the commutative property of multiplication and so holds not only for real or complex numbers but also for polynomials, in arithmetic modulo for any positive integer, or more generally in any commutative ring. However, due to matrices not being commutative under multiplication, this identity doesn't hold for matrices. See Also Sum and difference of powers This article is a stub. Help us out by expanding it. Retrieved from " Categories: Stubs Algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6263	https://www.sciencedirect.com/topics/veterinary-science-and-veterinary-medicine/parthenogenesis	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Comparative Reproduction Parthenogenesis Parthenogenesis is the development of a new offspring from an unfertilized egg. Parthenogenetic lineages occur in many insect species, but are widespread among other invertebrate taxa (Table 1). It entails modification or absence of meiosis so that the eggs remain diploid and do not have to fuse with sperm to give rise to a diploid zygote. This kind of parthenogenesis is termed as apomictic, which produces genetically identical modules of the same genet (typical asexual reproduction). In contrast, automixis (meiotic parthenogenesis) restores diploidy by the fusion of the egg with the second polar body (e.g., free-living nematode, Rhabditis). Obviously, the resultant modules may not be the exact genetic replicates of the mother. In insects, parthenogenesis may be thelytokous (female producing), arrhenotokous (male producing) or amphitokous (producing either sex). Haploid parthenogenesis is a special case in which the oocytes undergo regular meiotic division. If the eggs are fertilized, the offspring is a female, and if the eggs remain unfertilized, then parthenogenetic development results in a male offspring, which is haploid in its somatic tissues (e.g., the haplodiploid Hymenopteran insects). In social insects like the subterranean termite, Reticuliter messperatus, the queens produce new queens asexually by thelytokous parthenogenesis, but produce other colony members (workers and soldiers) by sexual reproduction. The parthenogenetic production of these new queens is achieved by the closing of the egg’s micropyle (sperm gates) to prevent sperm entry. Yet another type, namely obligate parthenogenesis, occurs in bdelloid rotifers, in which sexual reproduction never takes place due to the lack of males in the population. In the cladoceran rotifers and aphid insects, parthenogenesis occurs cyclically together with bouts of sexual reproduction. This is called cyclical parthenogenesis. In Daphnia, parthenogenetic reproduction takes place for one to several generations during favourable conditions, followed by sexual reproduction under unfavourable environments. The sexually produced long-lived dormant eggs hatch once favourable conditions return. By this alternation of generations, favourable environmental conditions can be exploited to increase the number of offspring by parthenogenetic reproduction, whereas the periodical appearance of one or more sexual generations will ensure genetic advantages such as increased heterosis, and re-assortment of genetic characters. View chapterExplore book Read full chapter URL: Reference work2018, Encyclopedia of Reproduction (Second Edition)Thanumalaya Subramoniam Chapter Parthenogenesis 2013, Brenner's Encyclopedia of Genetics (Second Edition)B.B. Normark Introduction Parthenogenesis (partheno-, ‘virgin’ + -genesis, ‘birth’) is the development of offspring from unfertilized eggs. In humans and other mammals, parthenogenetically produced individuals usually die as early embryos, but in many other groups of organisms, parthenogenesis can result in fully viable reproductive individuals. Some sexual life cycles include parthenogenesis as an essential component. For instance, in haplodiploid organisms, fertilized eggs develop as females, but unfertilized eggs develop as males. Haplodiploidy is found in the whole insect order Hymenoptera (ants, bees, wasps, and sawflies), as well as all thrips and whiteflies and some beetles, scale insects, mites, rotifers and roundworms. Another sexual life cycle with a parthenogenetic component is cyclical parthenogenesis, in which one sexual generation alternates with one or more parthenogenetic generations. There may be one sexual generation per year, as in gall wasps and aphids, or the timing may be more flexible, as in digenean trematode flatworms, which are parthenogenetic in their intermediate hosts and sexual in their primary hosts. Often the switch from parthenogenetic to sexual generations is cued by deteriorating local conditions, as in some insects that feed in rotting wood (cecidomyiid flies, micromalthid beetles) and some aquatic invertebrates (monogonont rotifers, Daphnia water fleas). Other sexual life cycles include parthenogenesis more sporadically, as an occasional occurrence. This is known as facultative parthenogenesis, or sometimes – if only a small percentage of unfertilized eggs develop parthenogenetically – tychoparthenogenesis. Facultative parthenogenesis is seen in Komodo dragons, turkeys, some shark species, and several orders of insects (especially mayflies, grasshoppers, and stick insects). But many lineages of animals have given up sex altogether and become obligately parthenogenetic – there are probably at least a few thousand such separately originated obligately parthenogenetic lineages alive today. Related Sperm-Dependent Systems A peculiar variation on parthenogenesis is seen in many organisms in which sperm are required to initiate the development of an egg, but the sperm’s genome is not incorporated into the developing offspring. Although not parthenogenesis (‘virgin birth’) in the literal sense, this form of reproduction – called gynogenesis or pseudogamy – is genetically equivalent to parthenogenesis. Gynogenetic development of individuals can often be induced, especially in fishes, by heat or cold shocks. Obligate gynogenesis is the natural reproductive mode of some fishes, salamanders, planthoppers, and beetles. Some plants, such as dandelions, have an unusual variant form of gynogenesis in which the sperm fertilizes only the endosperm of the seed, but makes no genetic contribution to the embryo. The mirror image of gynogensis is androgenesis, in which the egg’s nuclear genome is replaced by that of the sperm, and the resulting offspring are essentially asexual clones of the father with respect to their nuclear DNA. Androgenesis can also be induced artificially in some groups, especially plants and fishes, by cold shocks or chemical treatments. Just a few obligately androgenetic lineages are found in nature, and these have all been recently discovered: in clams, stick insects, ants, and cypresses. Both gynogenetic and androgenetic lineages depend upon a sexual ‘host’ species from which to obtain sperm (for gynogens) or eggs (for androgens). Apomixis and Automixis There are a number of genetically distinct forms of asexual reproduction that characterize parthenogenetic lineages (as well as gynogenetic and androgenetic ones). The simplest is apomictic parthenogensis, or apomixis, in which the egg is produced by mitosis, and is thus essentially a faithful genetic copy of the mother. More complex forms of parthenogenesis involve meiosis and are known collectively as automictic parthenogenesis, or automixis. In a type of automixis called premeiotic doubling, the genome is duplicated prior to meiosis, such that meiosis simply restores the mother’s diploid genotype. Thus, although this is a type of automixis, it is genetically equivalent to apomixis. Premeiotic doubling is found in some lizards, fish, tardigrades, earthworms, flatworms, mites and beetles, and in gynogenetic salamanders. A radically different type of automixis is gamete duplication: duplication of the genome of the haploid egg nucleus after meiosis is complete. Gamete duplication causes complete homozygosity at all loci and is found in some mites, brine shrimp, and several groups of insects. The other forms of automixis are intermediate in their genetic effects between premeiotic doubling (which preserves all of the heterozygosity found in the mother’s diploid genome) and gamete duplication (which eliminates all heterozygosity). Three of these are diagrammed in Figure 1. Terminal fusion (Figure 1(a)) is found in some roundworms, annelids, crustaceans, and insects. It eliminates heterozygosity near the centromeres. Fusion of the egg with a derivative of the first polar body (Figure 1(b)) is found in some moths, and central fusion (Figure 1(c)) is found in some moths, bees, and flies. These latter two systems retain heterozygosity in regions near the centromeres. In all three of these systems, the probability of loss of heterozygosity approaches 50% toward the telomeres, depending upon crossing over. Origins Origins of parthenogenesis are not well understood in mechanistic terms. One recurring pattern is that obligately parthenogenetic lineages are often of hybrid origin. Possibly this is related to the disruption of meiosis and to the sexual sterility that frequently results from hybridization. A related pattern is that parthenogenetic lineages are often polyploid: polyploidy might disrupt meiosis but is fully compatible with apomictic parthenogenesis. In recent decades, it has become clear that many origins of obligate parthenogenesis in invertebrates have been induced by Wolbachia and other intracellular bacteria. Because Wolbachia are transmitted vertically from mother to offspring through eggs – but not transmitted through sperm – the bacteria clearly benefit from distorting the sex ratios of their hosts toward production of more female offspring, and parthenogenesis induction is one of several means the bacteria have evolved to accomplish this, though the mechanisms remain largely unclear. Some groups clearly have a much lower rate of origin of parthenogenetic lineages than other groups, with a stark example being the total lack of successful parthenogenesis in mammals. It has now been shown that this is due to the genomic imprinting system of mammals – experiments show that parthenogenetic mice are viable if imprinted loci are suitably manipulated. Evolutionary Significance Parthenogenesis has played a large role in shaping biologists’ understanding of the adaptive significance of sexual reproduction. The mere existence of parthenogenesis demonstrates that sex is not essential to reproduction and raises the question of why sex exists at all. In the 1880s, August Weismann proposed a hypothesis that is still widely accepted today: sexual lineages persist longer than obligately parthenogenetic lineages because traits in sexual populations evolve more readily by natural selection in response to environmental change. Almost all organismal traits are thought to have evolved due to their individual-level selective advantage, but obligate parthenogenesis looks like an exception to this rule: it is a population-level character with population-level evolutionary advantages or disadvantages. Because parthenogens do not need to mate, this may give them an advantage in colonizing new areas, which may help to explain the relatively high incidence of parthenogenesis in invasive pests. Indeed, parthenogenesis would seem to convey a huge demographic advantage over sexuality: an all-female population should produce about twice as many babies per capita, on an average, than a 50% female population. This is sometimes called the ‘twofold cost of sex’ and it sharpens the paradox that parthenogenesis is rare (perhaps 0.1% of animal species are obligately parthenogenetic or contain obligately parthenogenetic lineages). Although some parthenogenetic populations may reach high levels of abundance, with few exceptions they are of recent origin in evolutionary terms. This observation tends to support Weismann’s hypothesis that – although parthenogenetic lineages may enjoy some short-term advantages – they are at a long-term evolutionary disadvantage. Ecology and Red Queen Hypothesis Although there is a consensus that Weismann’s hypothesis is correct, the twofold cost of sex presents a paradox. For sex to confer an adaptive advantage, given its enormous cost, it must confer a benefit that is even more enormous. In mathematical models, it is challenging to find conditions in which sexual lineages can outcompete parthenogenetic ones. For instance, sex is favored when selection is very strong, and when different alleles are favored in each successive generation. One feature of the environment that might exert this kind of strong, rapidly varying selection is natural enemies – especially parasites and pathogens that can evolve rapidly in comparison to their hosts. The idea that sex is an adaptation for keeping up with rapidly evolving natural enemies is often called the Red Queen hypothesis (after the Red Queen’s remark to Alice, in Lewis Carroll’s Through the Looking-Glass, “Now, here, you see, it takes all the running you can do, to keep in the same place”). The ecological distribution of parthenogenesis provides a measure of support for the Red Queen hypothesis, as parthenogenetic lineages are more frequent in environments with low biotic diversity (high latitudes, high altitudes, disturbed habitats, arid environments, islands), than in high-diversity environments. The problem of the evolutionary significance of sex versus parthenogenesis is far from solved, and remains an active area of research. View chapterExplore book Read full chapter URL: Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)B.B. Normark Chapter Physiology of Elasmobranch Fishes: Structure and Interaction with Environment 2015, Fish PhysiologyCynthia A. Awruch 3.2 Parthenogenesis True parthenogenesis (Greek for “virgin birth”) is a form of asexual reproduction in which the production of offspring occurs in the absence of any male genetic contribution. Females produce unfertilized eggs that will develop into viable embryos (Neaves and Baumann, 2011). Parthenogenesis has been documented in bony fishes, reptiles, amphibians, birds (Neaves and Baumann, 2011), and most recently in elasmobranchs (Chapman et al., 2007; Feldheim et al., 2010; Portnoy et al., 2014). There are two forms of parthenogenesis: in apomictic parthenogenesis the germ line cells (gametocytes) bypass meiosis to undergo only regular cell divisions (mitosis) to produce an egg that is genetically identical to its mother (i.e., the maternal genome is transmitted to the embryo intact), while in automictic parthenogenesis (automixis) there is a fusion of post-meiotic cells in the mother producing offspring with elevated homozygosity compared to its mother (i.e., genetic diversity is lost in transmission) (Lampert et al., 2007; Watts et al., 2006). Parthenogenesis is defined as “obligate” when organisms exclusively reproduce through asexual means, while it is “facultative” when species that ordinarily rely on sexual reproduction can resort to facultative parthenogenesis under extenuating circumstances that isolate females from males (Booth et al., 2012; Neaves and Baumann, 2011). Parthenogenesis has been confirmed, using genetic data, in five species of captive elasmobranchs: S. tiburo (Chapman et al., 2007); C. limbatus (Chapman et al., 2008); the whitespotted bamboo shark, Chiloscyllium plagiosum (Feldheim et al., 2010; Voss et al., 2001); S. fasciatum (Robinson et al., 2011); and the whitetip reef shark, Triaenodon obesus (Portnoy et al., 2014). The studies provided evidence that all these species utilized an automixis form of parthenogenesis as a reproductive strategy and viable offspring were produced. In some of these cases females have produced several healthy offspring over multiple reproductive cycles, and although in many cases the mortality rate across the litters was very high (more than 80%), some other pups were able to live for at least 5 years (Feldheim et al., 2010; Robinson et al., 2011; Voss et al., 2001). Increasing cases of parthenogenesis have been reported in elasmobranchs; however, if this reproductive strategy benefits and how it may benefit elasmobranch populations remains unknown. Although the ability for females to reproduce through facultative parthenogenesis would be selectively advantageous in situations where males are sparse (e.g., a possibility in the wild due to low population densities caused by overexploitation), very little is known about the long-term viability of automictic parthenogenesis in elasmobranchs and the occurrence of parthenogenesis in wild populations. The reduction in genomic diversity in automictic parthenogenesis could reduce female fitness by increasing expression of recessive genetic disorders (Neaves and Baumann, 2011; Watts et al., 2006). Thus, the genetic costs of parthenogenesis may offset the benefit of having a mechanism to avoid occasional reproductive failure in elasmobranch populations. However, if females have no other options available for passing on genes, then this strategy would require that at least some portion of the offspring produce viable gametes (Portnoy et al., 2014). View chapterExplore book Read full chapter URL: Book series2015, Fish PhysiologyCynthia A. Awruch Chapter Phylum Tardigrada 2015, Thorp and Covich's Freshwater Invertebrates (Fourth Edition)Diane R. Nelson, ... Lorena Rebecchi Parthenogenesis Parthenogenesis is unknown in marine species of tardigrades, but it is very frequent in less stable habitats, such as limnic and terrestrial environments. Many taxa are known only or mostly as parthenogenetic and thelytokous. Among them are many terrestrial echiniscid species, the partly limnic eutardigrade family Murrayidae, some limnic species of Pseudobiotus, Thulinius (Isohypsibiidae) and Hypsibius (Hypsibiidae), and several limnoterrestrial or terrestrial species of Isohypsibiidae, Hypsibiidae, and Macrobiotidae (Bertolani, 2001). Thelytoky represents a good strategy for colonization of new territories by animals subject to passive dispersal. Parthenogenesis appears to be continuous in any case. No heterogony has been found. Both automictic (e.g., in Murrayidae and H. dujardini) and apomictic (as in most other cases) parthenogenesis have been found. Automixis is always associated with diploid animals, whereas apomixis is often tied to polyploidy (most frequently triploidy). In terrestrial and limnic environments, there are several cases of morphospecies including amphimictic diploid populations and apomictic polyploid, or rarely automictic diploid populations. As in the case of self-fertilization, automixis repeated for several generations leads to a complete homozygosity and to loss of genetic variability except that due to mutations. Apomixis maintains the heterozygosity level, which can be increased only as a result of mutations. Since parthenogenesis may be advantageous for invading new habitats, the evolution of parthenogenesis may be correlated with the evolution of cryptobiosis and the invasion of the terrestrial habitat. View chapterExplore book Read full chapter URL: Book2015, Thorp and Covich's Freshwater Invertebrates (Fourth Edition)Diane R. Nelson, ... Lorena Rebecchi Chapter Uniparental Inheritance 2001, Encyclopedia of GeneticsA.C. Ferguson-Smith Parthenogenesis in Invertebrates Parthenogenesis can be defined as the production of an embryo from a female gamete without any genetic contribution from a male gamete, with or without the eventual development into an adult. It is distinct from asexual reproduction since it involves the production of egg cells. Parthenogenesis is a normal method of reproduction in many lower organisms, but does not lead to viable mammalian offspring. Parthenogenetic development can proceed by various routes depending on whether meiosis has occurred or has been supressed, in which case the egg develops as as result of mitotic divisions. Whenever sex is determined by chromosome constitution, parthenogenetic offspring, in the absence of effective meiosis, all will be, mostly female. In birds, however (see below), the offspring are male as in this case females are the heterogametic sex. In bees, males originate by haploid parthenogenesis while diploid females are produced by fertilization in the normal way. Other aphids, such as greenfly (Hemiptera) have generations which alternate between parthenogenesis and fertilization, so called cyclical parthenogenesis. The formation of female parthenogenetic offspring is widespread among many order of insects. For example in Drosophila parthenogenetica, a small proportion of eggs laid by virgin females develop to produce viable adults. Another example is the parthenogenetic grasshopper Warramaba virgo, a species which consists of females only. Parthenogenesis is also successful in some Crustaceae such as the brine shrimp, Artemia salina. View chapterExplore book Read full chapter URL: Reference work2001, Encyclopedia of GeneticsA.C. Ferguson-Smith Chapter Comparative Reproduction 2018, Encyclopedia of Reproduction (Second Edition)Marvalee H. Wake Parthenogenesis Parthenogenesis is reproduction without fertilization, an ovum developing into a new individual without fertilization by a sperm. In vertebrates, parthenogenetic “species” are the result of modification of ovum development, usually changes in meiosis, leading to eggs produced with multiple sets of chromosomes. Some species are facultatively parthenogenetic, induced to produce asexually as ova that have undergone “meiotic error” are laid and develop into adults, but these females usually are sterile and cannot produce a new generation of offspring. They typically are evolutionary/ecological modifications of species that have “standard” reproductive modes, but have hybridized or undergone other speciation modes. In some cases, courtship by a male of a closely related species is required to induce ovum development, but his sperm do not fertilize the ovum. Often, this mode of reproduction produces all-female species, given that only ova develop. Consequently the broods of young are effectively clones of the mother. Several species of teleost fishes, at least three sharks, several lizards, a snake, and frogs and salamanders are known to be strictly parthenogenetic, while yet others are considered facultative or incomplete parthenogens. There are no known cases of natural parthenogenesis in mammals, but it has been artificially induced in rabbits, mice, pigs, and monkeys, often with consequent abnormal development. The low genetic variation and modification of size and other components of development typically cause reduced fitness of offspring in parthenogenetic species. The patterns of evolution of various forms of natural parthenogenesis in several vertebrates have been studied, and are diverse and complex (summarized by Avise (2008)). The complexity, variation, and frequent convergences of the evolution of vertebrate reproductive modes is exciting, and not yet fully understood. View chapterExplore book Read full chapter URL: Reference work2018, Encyclopedia of Reproduction (Second Edition)Marvalee H. Wake Review article Special Edition dedicated to European COST Action FA0702 Maternal Interaction with Gametes and Embryos, GEMINI 2012, TheriogenologyT.A.L. Brevini, ... F. Gandolfi 1 Introduction Parthenogenesis is the process by which an oocyte can begin development without the involvement of the male gamete. While Mammals are not spontaneously capable of this form of reproduction, a variety of lower organisms, such as fish, ants, flies, honeybees, amphibians, lizards and snakes may routinely reproduce in this manner. However, it must be noted that mammalian oocytes can be successfully activated in vitro, inducing them to perform several cleavage divisions and thus initiating embryonic development. These events can be obtained by mimicking the intracellular calcium wave induced by the spermatozoon at fertilization, a basic mechanism which controls many cellular events. In particular, oscillations of this ion causes cortical granule exocytosis, decreasing MPF and MAP kinase activities with cell cycle resumption and the recruitment of maternal mRNAs which regulate embryonic development . In the past years, parthenogenetic development has been investigated in connection with reproductive cloning, since oocyte activation is an essential part of the procedure. As a result it is well known that activated mammalian oocytes can develop in vitro at rates comparable to IVF oocytes. However, parthenotes transferred in vivo can reach a variable level of development, depending on the species, but are inherently unable to develop to term because of the epigenetic modification of mammalian DNA, known as imprinting , which is responsible for silencing of genes that are normally expressed by the paternal allele. Because of their inability to develop to term, parthenotes may represent a promising tool for studies of the mechanisms driving early embryogenesis and for the preclinical testing of experimental protocols in human assisted reproduction (i.e., different oocyte cryopreservation procedures, oocyte in vitro maturation or polar body genetic screening) that would otherwise imply the destruction of a viable embryo. Moreover, research in mouse [5,6] and non-human primate models [7,8] it has recently demonstrated that, although unable to form a new individual, parthenotes develop enough to allow the successful derivation of stable pluripotent cell lines and therefore they can represent a stimulating research tool as well as an alternative, less controversial source of embryonic stem cell lines in the human [2–10]. Parthenogenetic stem cells have been derived in human [3,9–12] as well as non-human primates [13,14], pig [15,16], cow [17–19] and buffalo (Table 1). Table 1. Parthenogenetic cell lines derived in non-rodent species. \| Species \| References \| --- \| \| Human \| [2,3,9,10,11,12] \| \| Non-human primates \| [13,14] \| \| Pig \| [15,16] \| \| Cow \| [17,18,19] \| \| Buffalo \| \| These exciting perspectives and applications are, however, hampered by the fact that many aspects related to the biology of parthenogenetic embryos and parthenogenetic derived cell lines are largely unknown and still need to be elucidated. In an attempt to better understand some of these aspects, parthenogenetic cell lines, recently derived from different species in our laboratory [3,15,16,21], were studied and characterized for their pluripotency and differentiation plasticity, both in vitro and in vivo. Here we describe some features of these cell lines that are common to biparental embryonic stem cells. By contrast, we also report the presence of an intrinsic deregulation of molecules controlling cell proliferation and cell adhesion and suggest that the uniparental origin of these cells may represent one of the possible cause. View article Read full article URL: Journal2012, TheriogenologyT.A.L. Brevini, ... F. Gandolfi Chapter Regulation of apomixis 2012, Plant Biotechnology and AgriculturePeggy Ozias-Akins, Joann A. Conner Parthenogenesis Parthenogenesis is fertilization independent development of the embryo during seed formation. Parthenogenesis is rare, but well documented, in sexual plants and usually leads to the formation of haploids (Kimber and Riley, 1963; Dunwell, 2010). Induction of gynogenic haploids is one breeding technology used for crops recalcitrant to induction of androgenic haploids (Forster et al., 2007). Spontaneous formation of gynogenic haploids may be accompanied by a diploid embryo in the same seed. The haploid embryo in a twin-embryo seed probably arises most frequently from a synergid cell that assumes egg cell fate (Kimber and Riley, 1963). In the maize ig1 mutant described earlier, the micropylar-oriented embryo sac cells are competent to form embryos without fertilization. Embryogenesis most likely is triggered post-pollination even though fertilization is avoided. The strict sense of parthenogenesis as observed among apomicts would entail egg activation in the absence of either pollination or fertilization. Evidence that eggs from the “Salmon” system of wheat are activated comes from their in vitro response upon isolation from the surrounding tissues. The Salmon system is characterized by a 1BS/1RS wheat/rye chromosomal translocation that is essential to confer the capacity for parthenogenesis in specific heterocytoplasmic backgrounds. Chromosome 1BS is postulated to carry a parthenogenesis-suppressing gene while 1RS has a parthenogenesis-inducing gene (Matzk et al., 1995). The mechanisms underlying this complex nuclear–cytoplasmic interaction are not understood, but the isogenic lines developed with different cytoplasms provide valuable genetic tools for comparing gene expression differences between parthenogenetic and non-parthenogenetic eggs (Kumlehn et al., 2001). Of particular importance was the demonstration that egg cells isolated from the parthenogenetic (cS and kS) lines initiated embryos in up to 25% of the cases, whereas unfertilized egg cells from the non-parthenogenetic (aS) isogenic line showed no egg cell division in vitro. Metabolic differences were indicated by ultrastructural comparison of aS and kS lines three days prior to anthesis where nuclear size, number of nucleoli, and number of ribosomes were greater in the kS than aS eggs (Naumova and Matzk, 1998). The precocious metabolic activity did not lead to egg cell division prior to anthesis, as has been demonstrated for some apomicts. The factors responsible for egg activation in plants are unknown. In fact, the egg of Arabidopsis assumes a quiescent state with an abundance of stored transcripts whose translation predominates for approximately three days after zygote formation and prior to the maternal to zygotic transition (Vielle-Calzada et al., 2000; Pillot et al., 2010). Enhancer trap lines with β-glucuronidase (GUS) as the reporter gene and representing 19 embryo expressed genes showed GUS expression in seeds only when maternally transmitted. Furthermore, the GUS evidence for maternal allele expression of some genes was validated using allele-specific RT-PCR. Paternal alleles of these genes could not be detected in seeds, either embryo or endosperm, until more than three days after pollination. Data generated to address the timing of maternal to zygotic transition in maize are contrasting, however, perhaps due to different analytical methods and crosses tested (Grimanelli et al., 2005; Meyer and Scholten, 2007). Allele-specific RT-PCR showed no evidence for paternal allele expression of 16 genes in maize seeds three days after pollination (Grimanelli et al., 2005), whereas a single nucleotide polymorphism (SNP)-detection assay identified paternal alleles for 24 active genes in microdissected zygotes 1 day after pollination (Meyer and Scholten, 2007). Paternal allele expression was consistent with observation of heterosis early in embryo development, manifested as a higher rate of cell division. In Arabidopsis as well as maize, there is evidence for repression of paternal allele expression in endosperm and embryo, although the two fertilization products were shown to have different transcriptional states. The requirement for transcription during early seed development was tested by knockdown of RNA POLYMERASE II through RNA interference (Pillot et al., 2010). Transcription was shown to be essential for functional megaspore mitosis and endosperm development, but dispensable for embryo initiation. Microarray analysis of seeds with developing embryos but no endosperm (apomictic maize-Tripsacum hybrid with fertilization-independent and precocious embryo initiation), and seeds with embryo and endosperm three days after pollination (sexual maize) showed no differential gene expression for the embryo-only sample compared to unfertilized maize ovules. In contrast, more than 2% of the genes were differentially expressed once endosperm began to develop (Grimanelli et al., 2005). These collective data support that zygotic transcription is not required for embryo initiation, nor is there zygote-specific gene expression. Yet there must be similar signals generated during the formation of a zygote that are precociously generated and perceived by unreduced eggs in apomictic plants, or repressive signals that are absent. In apomicts, egg activation and assumption of a zygotic fate does not require a signal from pollination or pollen tube penetration of a synergid, such as generation of a calcium wave. Furthermore, synergids undergo accelerated degeneration compared with those in the egg apparatus of a sexual plant (Vielle et al., 1995). Perhaps a quiescent state is never imposed on the egg of apomicts. The egg of sexual buffelgrass has a large chalazal vacuole, whereas many small vacuoles are more centrally positioned in the egg of apomictic buffelgrass. A similar pattern of vacuolation was observed in aposporous P. maximum (Naumova and Willemse, 1995). How is fertilization of an unreduced egg prevented? Mechanisms could be physical or physiological, although there is little direct evidence for either. The cell wall of the egg in a sexual individual is incomplete, particularly at the chalazal end and where it is in contact with a synergid. The exposed membrane presumably would be necessary for gamete recognition via membrane localized receptor molecules (Peng and Sun, 2008). In apomictic buffelgrass, the cell wall around the egg is completed shortly after pollination, imposing a physical barrier to syngamy (Vielle et al., 1995), but this is not a universal phenomenon in apomicts as demonstrated for P. maximum (Naumova and Willemse, 1995). Early pollination of Pennisetum and Paspalum species has been shown to increase the frequency of fertilization events leading to generation of 2n + n (BIII) hybrids with elevated ploidy levels (Martinez et al., 1994; Burson et al., 2002), implying that the fertilization barrier intensifies shortly before anthesis. Lack of cell cycle synchronization also could influence fertilization if the egg is receptive only at G1, yet early egg activation in an apomict presumably would yield eggs at multiple phases of the cell cycle. What genes might therefore be responsible for initiating embryogenesis in the absence of fertilization? Positive regulators of embryogenesis have been studied more thoroughly in somatic rather than zygotic embryogenesis. Over-expression of numerous genes provides evidence for a role in embryo morphogenesis. Expression profiling of mutants of some of these genes also has provided evidence for their direct and indirect targets. The embryogenesis gene SOMATIC EMBRYOGENESIS RECEPTOR-LIKE KINASE (SERK) is a leucine-rich repeat receptor-like kinase that is a marker for embryogenically competent cells, both somatic and zygotic (Schmidt et al., 1997). In Arabidopsis, AtSERK1 is expressed in the ovule primordium, continuing through megasporogenesis in the distal region of the ovule, including the MMC, in all cells of the megagametophyte, and post-fertilization in both early embryo and endosperm (Hecht et al., 2001). Mutants of AtSERK1, however, do not show attenuation of embryogenesis suggesting that other SERK genes may have overlapping functions. Over-expression of AtSERK1 in Arabidopsis led to enhanced competence for somatic embryo induction, although somatic embryos did not form in the absence of auxin (Hecht et al., 2001). Among natural apomicts, a SERK gene differentially expressed in apomictic versus sexual ovules was identified in Poa pratensis (Albertini et al., 2005), suggesting that its expression could play a role in acquisition of embryogenic competence by either the MMC or aposporous initials. The pattern of expression of AtSERK:GUS in Hieracium ovules, however, did not precisely parallel that of Arabidopsis; for example, no expression was observed during megagametogenesis or in the egg cell, but globular stage embryos were clearly marked by GUS signal. Furthermore, no differences in AtSERK:GUS expression were observed between sexual and apomictic reproduction (Tucker et al., 2003). Several transcriptional regulators that play a role in embryogenesis have been identified including LEAFY COTYLEDON 1 (LEC1) and LEC2, FUSCA3 (FUS3), AGAMOUS-LIKE 15 (AGL15), WUSCHEL (WUS), and BABY BOOM (BBM). LEC1 encodes a HAP3 subunit of the CCAAT-box binding factor whereas LEC2 and FUS3 encode B3-domain proteins. Over-expression of LEC1 or LEC2 in vegetative tissues causes ectopic embryo formation indicating that either is sufficient for embryo formation under such permissive conditions, and their loss-of-function mutants have impaired embryo development indicating their requirement for embryo morphogenesis. LEC1 and LEC2 both act to target FUS3 as well as each other. Other targets of LEC2 are auxin biosynthesis genes, YUC2 and YUC4 (Stone et al., 2008), a finding that suggests LEC2 may indirectly influence levels or gradients of auxin, a growth regulator known to stimulate somatic embryogenesis and establish polarity in zygotic embryos (Weijers and Jurgens, 2005). LEC2 also upregulates AGL15, a MADS domain transcription factor that activates expression of GA2ox6, an inhibitor of gibberellin biosynthesis. Furthermore, ChIP-chip experiments showed that LEC2 and FUS3 were direct targets of AGL15 (Zheng et al., 2009), suggesting positive feedback regulation. While a role for LEC1 and LEC2 in early embryogenesis is implicated by these experiments, LEC2 has been shown to bind to RY elements found in the promoter of many seed-specific genes, particularly those involved in seed maturation such as seed storage proteins (Braybrook et al., 2006). Similarly, LEC1 promotes expression of fatty acid biosynthetic genes during seed maturation albeit indirectly (Mu et al., 2008). Evidence therefore supports that LEC1 and LEC2 have functions in both early (Lotan et al., 1998) and late (Braybrook and Harada, 2008) embryogenesis. Another transcriptional regulator, WUS, is necessary for shoot apical meristem development, including the shoot meristem of embryos; however, ectopic expression of WUS is sufficient to induce somatic embryo formation (Zuo et al., 2002). WUS is a homeobox gene whose product confers stem cell identity. In addition to being expressed in shoot meristems, it is expressed in the nucellus (thought to have evolved from a shoot meristem; Kenrick and Crane, 1997), but acts non-cell-autonomously through a signaling pathway to stimulate integument initiation (Gross-Hardt et al., 2002). WUS expression is positively regulated by NOZZLE/SPOROCYTELESS (Sieber et al., 2004), a putative transcription factor that affects ovule ontogeny and specification of the MMC (Yang et al., 1999; Schiefthaler et al., 2004). Since WUS has multiple signaling functions, its expression in embryogenically competent cells and response to auxin gradients is probably related to its regulation of stem cell function (Su et al., 2009). BABY BOOM is an AP2-domain transcription factor, a member of the AP2-ERF gene family (Weigel, 1995; Ohme-Takagi and Shinshi, 1995). BABY BOOM falls in the AINTEGUMENTA (ANT) clade, members of which are involved in root, ovule, and embryo development (Kim et al., 2006). The discovery of BBM was as a gene expressed during Brassica microspore embryogenesis. Over-expression of BBM leads to ectopic embryo formation in seedlings of Brassica and Arabidopsis (Boutilier et al., 2002). It is likely that BBM functions through a different pathway than LEC, given the gene targets that have been identified (Passarinho et al., 2008). Interestingly, a novel homolog of BBM has been identified in two apomicts, Pennisetum squamulatum and C. ciliaris, that is tightly linked with apomixis (Conner et al., 2008). The gene is expressed one to two days prior to anthesis and during embryo development, whereas transcript from the most closely related homolog from a sexual genotype is barely detectable on the day of anthesis (Huo, 2008; Zeng, 2009). Non-viable parthenogenetic haploid embryos of up to 20 cells have been associated with the loss-of-function of the Arabidopsis gene MULTICOPY SUPPRESSOR IRA 1 (MSI1; Guitton and Berger, 2005). Rodrigues et al. (2010) found similar expression of the Hieracium homolog HMSI1 in both apomictic and sexual Hieracium ovules, suggesting that autonomous egg cell development is not caused by a lack of HMS1 expression in apomicts. Furthermore, HMS1 did not map to the loss of parthenogenesis (LOP) locus, which controls both autonomous embryo and endosperm development in Hieracium. Whether mis-expression or expression of novel forms of any of the previously mentioned transcriptional regulators or signaling pathway components plays a role in parthenogenesis is still open to speculation and hypothesis testing. Regulated expression of several of these transcription factors in the ovule and egg cell has failed to induce parthenogenesis (reported without supporting data in a review by Curtis and Grossniklaus, 2008), although these experiments were likely to have been carried out in sexual, diploid Arabidopsis, with genes from sexual diploids, and would not have tested the effect of a novel gene or its regulatory elements from an apomict. View chapterExplore book Read full chapter URL: Book2012, Plant Biotechnology and AgriculturePeggy Ozias-Akins, Joann A. Conner Chapter Parthenogenesis 2013, Brenner's Encyclopedia of Genetics (Second Edition)B.B. Normark Abstract Parthenogenesis is the development of offspring from unfertilized eggs. Parthenogenesis forms a regular part of some sexual life cycles, but there are also many lineages of animals that have given up sexual reproduction and become obligately parthenogenetic. There are a number of genetically different types of parthenogenesis, ranging from gamete duplication, which immediately eliminates all heterozygosity, to apomixis and prezygotic doubling, which both preserve all the heterozygosity that was present in the mother. The existence of successful parthenogenetic lineages demonstrates that sex is not necessary for reproduction and raises the question of why sex exists. Parthenogenetic lineages are often successful in the short term but almost never persist in the evolutionary long term. An influential hypothesis holds that sexual populations persist longer than parthenogenetic populations because they evolve more rapidly in response to environmental challenges such as rapidly evolving parasites and pathogens. But the adaptive significance of sexual reproduction remains an open question, and parthenogenetic lineages play an important role in ongoing testing of this and other hypotheses. View chapterExplore book Read full chapter URL: Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)B.B. Normark Chapter Tardigrada (Water Bears) 2009, Encyclopedia of Inland WatersR. Bertolani, ... D.R. Nelson Parthenogenesis Parthenogenesis, unknown in marine species of tardigrades, is very frequent in limnic and terrestrial environments. Many taxa are known only or mostly as thelytokous. Among them are many terrestrial echiniscid species, the partly limnic eutardigrade family Murrayidae, some limnic species of Pseudobiotus, Thulinius, and Hypsibius (Hypsibiidae), and several limnoterrestrial or terrestrial species of Hypsibiidae and Macrobiotidae. Thelytoky represents a good strategy for colonization of new territories by animals subject to passive dispersal. Parthenogenesis appears to be continuous in any case; no heterogony has been found. Both automictic (e.g., in Murrayidae and in Hypsibius dujardini) and apomictic (as in most other cases) parthenogenesis have been found. View chapterExplore book Read full chapter URL: Reference work2009, Encyclopedia of Inland WatersR. Bertolani, ... D.R. Nelson Related terms: P53 Stem Cell Protocerebrum Mouse Bee Transcription Factor Leukocyte Zygote Adipocyte Escherichia coli View all Topics We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy
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Koenigshofer" ] [ "article:topic", "transcluded:yes", "license:ccbync", "showtoc:yes", "program:oeri", "autonumheader:yes1", "licenseversion:40", "authorname:oeri", "source-socialsci-113223", "stage:draftoeri", "author@Kenneth A. Koenigshofer" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Irvine Valley College 4. Physiological Psychology (IVC PSYCH003) 5. 6: Answer Key 6. 6.20: Supplemental Content 7. 6.20.12: Chapter 14- Problem Solving, Categories and Concepts Expand/collapse global location Physiological Psychology (IVC PSYCH003) Front Matter 1: Part I- Foundations of Physiological Psychology 2: Part II- Sensory and Motor Systems 3: Part III- Rhythms, Drives, and Motivation 4: Part IV- Cognition and Emotion 5: Part V- Drugs, Disorders, and Brain Health 6: Answer Key 7: Appendix 8: About This Book Back Matter 6.20.12: Chapter 14- Problem Solving, Categories and Concepts Last updated Jul 9, 2025 Save as PDF 6.20.11: Chapter 14- Traditional Models of Human Intelligence 6.20.13: Chapter 14- The Brain and Evolution of Cognition Page ID 274881 This page is a draft and under active development. Please forward any questions, comments, and/or feedback to the ASCCC OERI (oeri@asccc.org). Kenneth A. Koenigshofer ASCCC Open Educational Resources Initiative (OERI) ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Overview 3. Defining Problems 1. Well-defined Problems 2. Ill-defined Problems 3. How is a problem represented in the mind? 4. Insight 5. Fixation 1. Functional fixedness 2. Mental fixedness Problem-Solving Strategies Problem Solving as a Search Problem Means-End Analysis Analogies Restructuring by Using Analogies Schemas How Do Experts Solve Problems? Creative Cognition Divergent Thinking Convergent Thinking Brain Mechanisms in Problem Solving Moral Reasoning Summary References Brain Mechanisms in Problem Solving Attributions Categories and Concepts Learning Objectives Introduction Nature of Categories Fuzzy Categories Borderline Items Typicality Source of Typicality Category Hierarchies Theories of Concept Representation Knowledge Summary Discussion Questions Vocabulary References Attributions Authors Creative Commons License Adapted by Kenneth Koenigshofer, PhD, from Categories and Concepts by Gregory Murphy, licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Permissions beyond the scope of this license may be available in our Licensing Agreement. How to cite this Noba module using APA Style Learning Objectives Define problem types Describe problem solving strategies Define algorithm and heuristic Describe the role of insight in problem solving Explain some common roadblocks to effective problem solving What is meant by a search problem Describe means-ends analysis How do analogies and restructuring contribute to problem solving Explain how experts solve problems and what gives them an advantage over non-experts Describe the brain mechanisms in problem solving Overview In this section we examine problem-solving strategies. People face problems every day—usually, multiple problems throughout the day. Sometimes these problems are straightforward: To double a recipe for pizza dough, for example, all that is required is that each ingredient in the recipe be doubled. Sometimes, however, the problems we encounter are more complex. For example, say you have a work deadline, and you must mail a printed copy of a report to your supervisor by the end of the business day. The report is time-sensitive and must be sent overnight. You finished the report last night, but your printer will not work today. What should you do? First, you need to identify the problem and then apply a strategy, usually a set of steps, for solving the problem. Defining Problems We begin this module on Problem Solving by giving a short description of what psychologists regard as a problem. Afterwards we are going to present different approaches towards problem solving, starting with gestalt psychologists and ending with modern search strategies connected to artificial intelligence. In addition we will also consider how experts do solve problems and finally we will have a closer look at two topics: The neurophysiological background on the one hand and the question what kind of role can be assigned to evolution regarding problem solving on the other. The most basic definition is “A problem is any given situation that differs from a desired goal”. This definition is very useful for discussing problem solving in terms of evolutionary adaptation, as it allows to understand every aspect of (human or animal) life as a problem. This includes issues like finding food in harsh winters, remembering where you left your provisions, making decisions about which way to go, repeating and varying all kinds of complex movements by learning, and so on. Though all these problems were of crucial importance during the evolutionary process that created us the way we are, they are by no means solved exclusively by humans. We find a most amazing variety of different solutions for these problems of adaptation in animals as well (just consider, e.g., by which means a bat hunts its prey, compared to a spider). However, for this module, we will mainly focus on abstract problems that humans may encounter (e.g. playing chess or doing an assignment in college). Furthermore, we will not consider those situations as abstract problems that have an obvious solution: Imagine a college student, let's call him Knut. Knut decides to take a sip of coffee from the mug next to his right hand. He does not even have to think about how to do this. This is not because the situation itself is trivial (a robot capable of recognizing the mug, deciding whether it is full, then grabbing it and moving it to Knut’s mouth would be a highly complex machine) but because in the context of all possible situations it is so trivial that it no longer is a problem our consciousness needs to be bothered with. The problems we will discuss in the following all need some conscious effort, though some seem to be solved without us being able to say how exactly we got to the solution. Still we will find that often the strategies we use to solve these problems are applicable to more basic problems, as well as the more abstract ones such as completing a reading or writing assignment for a college class. Non-trivial, abstract problems can be divided into two groups: Well-defined Problems For many abstract problems it is possible to find an algorithmic solution. We call all those problems well-defined that can be properly formalized, which comes along with the following properties: The problem has a clearly defined given state. This might be the line-up of a chess game, a given formula you have to solve, or the set-up of the towers of Hanoi game (which we will discuss later). There is a finite set of operators, that is, of rules you may apply to the given state. For the chess game, e.g., these would be the rules that tell you which piece you may move to which position. Finally, the problem has a clear goal state: The equations is resolved to x, all discs are moved to the right stack, or the other player is in checkmate. Not surprisingly, a problem that fulfills these requirements can be implemented algorithmically (also see convergent thinking). Therefore many well-defined problems can be very effectively solved by computers, like playing chess. Ill-defined Problems Though many problems can be properly formalized (sometimes only if we accept an enormous complexity) there are still others where this is not the case. Good examples for this are all kinds of tasks that involve creativity, and, generally speaking, all problems for which it is not possible to clearly define a given state and a goal state: Formalizing a problem of the kind “Please paint a beautiful picture” may be impossible. Still this is a problem most people would be able to access in one way or the other, even if the result may be totally different from person to person. And while Knut might judge that picture X is gorgeous, you might completely disagree. Nevertheless ill-defined problems often involve sub-problems that can be totally well-defined. On the other hand, many every-day problems that seem to be completely well-defined involve a great deal of creativity and many ambiguities. For example, suppose Knut has to read some technical material and then write an essay about it. If we think of Knut's fairly ill-defined task of writing an essay, he will not be able to complete this task without first understanding the text he has to write about. This step is the first sub-goal Knut has to solve. Interestingly, ill-defined problems often involve subproblems that are well-defined. Knut’s situation could be explained as a classical example of problem solving: He needs to get from his present state – an unfinished assignment – to a goal state - a completed assignment - and has certain operators to achieve that goal. Both Knut’s short and long term memory are active. He needs his short term memory to integrate what he is reading with the information from earlier passages of the paper. His long term memory helps him remember what he learned in the lectures he took and what he read in other books. And of course Knut’s ability to comprehend language enables him to make sense of the letters printed on the paper and to relate the sentences in a proper way. Same place, different day. Knut is sitting at his desk again, staring at a blank paper in front of him, while nervously playing with a pen in his right hand. Just a few hours left to hand in his essay and he has not written a word. All of a sudden he smashes his fist on the table and cries out: "I need a plan! How is a problem represented in the mind? Generally speaking, problem representations are models of the situation as experienced by the agent. Representing a problem means to analyze it and split it into separate components: objects, predicates state space operators selection criteria Therefore the efficiency of Problem Solving depends on the underlying representations in a person’s mind. Analyzing the problem domain according to different dimensions, i.e., changing from one representation to another, results in arriving at a new understanding of a problem. This is basically what is described as restructuring. Insight There are two very different ways of approaching a goal-oriented situation. In one case an organism readily reproduces the response to the given problem from past experience. This is called reproductive thinking. The second way requires something new and different to achieve the goal, prior learning is of little help here. Such productive thinking is (sometimes) argued to involve insight. Gestalt psychologists even state that insight problems are a separate category of problems in their own right. Tasks that might involve insight usually have certain features – they require something new and non-obvious to be done and in most cases they are difficult enough to predict that the initial solution attempt will be unsuccessful. When you solve a problem of this kind you often have a so called "AHA-experience" – the solution pops up all of a sudden. At one time you do not have any ideas of the answer to the problem, you do not even feel to make any progress trying out different ideas, but in the next second the problem is solved. Fixation Sometimes, previous experience or familiarity can even make problem solving more difficult. This is the case whenever habitual directions get in the way of finding new directions – an effect called fixation. Functional fixedness Functional fixedness concerns the solution of object-use problems. The basic idea is that when the usual way of using an object is emphasised, it will be far more difficult for a person to use that object in a novel manner. An example is the two-string problem: Knut is left in a room with a chair and a pair of pliers given the task to bind two strings together that are hanging from the ceiling. The problem he faces is that he can never reach both strings at a time because they are just too far away from each other. What can Knut do? Figure 6.20.12.1: Put the two strings together by tying the pliers to one of the strings and then swing it toward the other one. Mental fixedness Functional fixedness as involved in the examples above illustrates a mental set – a person’s tendency to respond to a given task in a manner based on past experience. Because Knut maps an object to a particular function he has difficulties to vary the way of use (pliers as pendulum's weight). Problem-Solving Strategies When you are presented with a problem—whether it is a complex mathematical problem or a broken printer, how do you solve it? Before finding a solution to the problem, the problem must first be clearly identified. After that, one of many problem solving strategies can be applied, hopefully resulting in a solution. Regardless of strategy, you will likely be guided, consciously or unconsciously, by your knowledge of cause-effect relations among the elements of the problem and the similarity of the problem to previous problems you have solved before. As discussed in earlier sections of this chapter, innate dispositions of the brain to look for and represent causal and similarity relations are key components of general intelligence (Koenigshofer, 2017). A problem-solving strategy is a plan of action used to find a solution. Different strategies have different action plans associated with them. For example, a well-known strategy is trial and error. The old adage, “If at first you don’t succeed, try, try again” describes trial and error. In terms of your broken printer, you could try checking the ink levels, and if that doesn’t work, you could check to make sure the paper tray isn’t jammed. Or maybe the printer isn’t actually connected to your laptop. When using trial and error, you would continue to try different solutions until you solved your problem. Although trial and error is not typically one of the most time-efficient strategies, it is a commonly used one. Table 1: Problem-Solving Strategies\| Method \| Description \| Example \| --- \| Trial and error \| Continue trying different solutions until problem is solved \| Restarting phone, turning off WiFi, turning off bluetooth in order to determine why your phone is malfunctioning \| \| Algorithm \| Step-by-step problem-solving formula \| Instruction manual for installing new software on your computer \| \| Heuristic \| General problem-solving framework \| Working backwards; breaking a task into steps \| Another type of strategy is an algorithm. An algorithm is a problem-solving formula that provides you with step-by-step instructions used to achieve a desired outcome (Kahneman, 2011). You can think of an algorithm as a recipe with highly detailed instructions that produce the same result every time they are performed. Algorithms are used frequently in our everyday lives, especially in computer science. When you run a search on the Internet, search engines like Google use algorithms to decide which entries will appear first in your list of results. Facebook also uses algorithms to decide which posts to display on your newsfeed. Can you identify other situations in which algorithms are used? A heuristic is another type of problem solving strategy. While an algorithm must be followed exactly to produce a correct result, a heuristic is a general problem-solving framework (Tversky & Kahneman, 1974). You can think of these as mental shortcuts that are used to solve problems. A “rule of thumb” is an example of a heuristic. Such a rule saves the person time and energy when making a decision, but despite its time-saving characteristics, it is not always the best method for making a rational decision. Different types of heuristics are used in different types of situations, but the impulse to use a heuristic occurs when one of five conditions is met (Pratkanis, 1989): When one is faced with too much information When the time to make a decision is limited When the decision to be made is unimportant When there is access to very little information to use in making the decision When an appropriate heuristic happens to come to mind in the same moment Working backwards is a useful heuristic in which you begin solving the problem by focusing on the end result. Consider this example: You live in Washington, D.C. and have been invited to a wedding at 4 PM on Saturday in Philadelphia. Knowing that Interstate 95 tends to back up any day of the week, you need to plan your route and time your departure accordingly. If you want to be at the wedding service by 3:30 PM, and it takes 2.5 hours to get to Philadelphia without traffic, what time should you leave your house? You use the working backwards heuristic to plan the events of your day on a regular basis, probably without even thinking about it. Another useful heuristic is the practice of accomplishing a large goal or task by breaking it into a series of smaller steps. Students often use this common method to complete a large research project or long essay for school. For example, students typically brainstorm, develop a thesis or main topic, research the chosen topic, organize their information into an outline, write a rough draft, revise and edit the rough draft, develop a final draft, organize the references list, and proofread their work before turning in the project. The large task becomes less overwhelming when it is broken down into a series of small steps. Problem Solving as a Search Problem The idea of regarding problem solving as a search problem originated from Alan Newell and Herbert Simon while trying to design computer programs which could solve certain problems. This led them to develop a program called General Problem Solver which was able to solve any well-defined problem by creating heuristics on the basis of the user's input. This input consisted of objects and operations that could be done on them. As we already know, every problem is composed of an initial state, intermediate states and a goal state (also: desired or final state), while the initial and goal states characterise the situations before and after solving the problem. The intermediate states describe any possible situation between initial and goal state. The set of operators builds up the transitions between the states. A solution is defined as the sequence of operators which leads from the initial state across intermediate states to the goal state. The simplest method to solve a problem, defined in these terms, is to search for a solution by just trying one possibility after another (also called trial and error). As already mentioned above, an organised search, following a specific strategy, might not be helpful for finding a solution to some ill-defined problem, since it is impossible to formalise such problems in a way that a search algorithm can find a solution. As an example we could just take Knut and his essay: he has to find out about his own opinion and formulate it and he has to make sure he understands the sources texts. But there are no predefined operators he can use, there is no panacea how to get to an opinion and even not how to write it down. Means-End Analysis In Means-End Analysis you try to reduce the difference between initial state and goal state by creating sub-goals until a sub-goal can be reached directly (in computer science, what is called recursion works on this basis). An example of a problem that can be solved by Means-End Analysis is the "Towers of Hanoi" Figure 6.20.12.2: Towers of Hanoi with 8 discs – A well defined problem (image from Wikimedia Commons; by User:Evanherk.licensed under the Creative CommonsAttribution-Share Alike 3.0 Unported license). The initial state of this problem is described by the different sized discs being stacked in order of size on the first of three pegs (the “start-peg“). The goal state is described by these discs being stacked on the third pegs (the “end-peg“) in exactly the same order. Figure 6.20.12.3: This animation shows the solution of the game "Tower of Hanoi" with four discs. (image from Wikimedia Commons; by André Karwath aka Aka; licensed under the Creative CommonsAttribution-Share Alike 2.5 Generic license). There are three operators: You are allowed to move one single disc from one peg to another one You are only able to move a disc if it is on top of one stack A disc cannot be put onto a smaller one. In order to use Means-End Analysis we have to create sub-goals. One possible way of doing this is described in the picture: Moving the discs lying on the biggest one onto the second peg. Shifting the biggest disc to the third peg. Moving the other ones onto the third peg, too You can apply this strategy again and again in order to reduce the problem to the case where you only have to move a single disc – which is then something you are allowed to do. Strategies of this kind can easily be formulated for a computer; the respective algorithm for the Towers of Hanoi would look like this: move n-1 discs from A to B move disc #n from A to C move n-1 discs from B to C where n is the total number of discs, A is the first peg, B the second, C the third one. Now the problem is reduced by one with each recursive loop. Means-end analysis is important to solve everyday-problems – like getting the right train connection: You have to figure out where you catch the first train and where you want to arrive, first of all. Then you have to look for possible changes just in case you do not get a direct connection. Third, you have to figure out what are the best times of departure and arrival, on which platforms you leave and arrive and make it all fit together. Analogies Analogies describe similar structures and interconnect them to clarify and explain certain relations. In a recent study, for example, a song that got stuck in your head is compared to an itching of the brain that can only be scratched by repeating the song over and over again. Useful analogies appears to be based on a psychological mapping of relations between two very disparate types of problems that have abstract relations in common. Applied to STEM problems, Gray and Holyoak (2021) state: "Analogy is a powerful tool for fostering conceptual understanding and transfer in STEM and other fields. Well-constructed analogical comparisons focus attention on the causal-relational structure of STEM concepts, and provide a powerful capability to draw inferences based on a well-understood source domain that can be applied to a novel target domain." Note that similarity between problems of different types in their abstract relations, such as causation, is a key feature of reasoning, problem-solving and inference when forming and using analogies. Recall the discussion of general intelligence in module 14.2. There, similarity relations, causal relations, and predictive relations between events were identified as key components of general intelligence, along with ability to visualize in imagination possible future actions and their probable outcomes prior to commiting to actual behavior in the physical world (Koenigshofer, 2017). Restructuring by Using Analogies One special kind of restructuring, the way already mentioned during the discussion of the Gestalt approach, is analogical problem solving. Here, to find a solution to one problem – the so called target problem, an analogous solution to another problem – the source problem, is presented. An example for this kind of strategy is the radiation problem posed by K. Duncker in 1945: As a doctor you have to treat a patient with a malignant, inoperable tumour, buried deep inside the body. There exists a special kind of ray, which is perfectly harmless at a low intensity, but at the sufficient high intensity is able to destroy the tumour – as well as the healthy tissue on his way to it. What can be done to avoid the latter? When this question was asked to participants in an experiment, most of them couldn't come up with the appropriate answer to the problem. Then they were told a story that went something like this: A General wanted to capture his enemy's fortress. He gathered a large army to launch a full-scale direct attack, but then learned, that all the roads leading directly towards the fortress were blocked by mines. These roadblocks were designed in such a way, that it was possible for small groups of the fortress-owner's men to pass them safely, but every large group of men would initially set them off. Now the General figured out the following plan: He divided his troops into several smaller groups and made each of them march down a different road, timed in such a way, that the entire army would reunite exactly when reaching the fortress and could hit with full strength. Here, the story about the General is the source problem, and the radiation problem is the target problem. The fortress is analogous to the tumour and the big army corresponds to the highly intensive ray. Consequently a small group of soldiers represents a ray at low intensity. The solution to the problem is to split the ray up, as the general did with his army, and send the now harmless rays towards the tumour from different angles in such a way that they all meet when reaching it. No healthy tissue is damaged but the tumour itself gets destroyed by the ray at its full intensity. M. Gick and K. Holyoak presented Duncker's radiation problem to a group of participants in 1980 and 1983. Only 10 percent of them were able to solve the problem right away, 30 percent could solve it when they read the story of the general before. After given an additional hint – to use the story as help – 75 percent of them solved the problem. With this results, Gick and Holyoak concluded, that analogical problem solving depends on three steps: Noticing that an analogical connection exists between the source and the target problem. Mapping corresponding parts of the two problems onto each other (fortress → tumour, army → ray, etc.) Applying the mapping to generate a parallel solution to the target problem (using little groups of soldiers approaching from different directions → sending several weaker rays from different directions) Schemas The concept that links the target problem with the analogy (the “source problem“) is called problem schema. Gick and Holyoak obtained the activation of a schema on their participants by giving them two stories and asking them to compare and summarize them. This activation of problem schemata is called “schema induction“. The two presented texts were picked out of six stories which describe analogical problems and their solution. One of these stories was "The General." After solving the task the participants were asked to solve the radiation problem. The experiment showed that in order to solve the target problem reading of two stories with analogical problems is more helpful than reading only one story: After reading two stories 52% of the participants were able to solve the radiation problem (only 30% were able to solve it after reading only one story, namely: “The General“). The process of using a schema or analogy, i.e. applying it to a novel situation, is called transduction. One can use a common strategy to solve problems of a new kind. To create a good schema and finally get to a solution using the schema is a problem-solving skill that requires practice and some background knowledge. How Do Experts Solve Problems? With the term expert we describe someone who devotes large amounts of his or her time and energy to one specific field of interest in which he, subsequently, reaches a certain level of mastery. It should not be of surprise that experts tend to be better in solving problems in their field than novices (people who are beginners or not as well trained in a field as experts) are. They are faster in coming up with solutions and have a higher success rate of right solutions. But what is the difference between the way experts and non-experts solve problems? Research on the nature of expertise has come up with the following conclusions: When it comes to problems that are situated outside the experts' field, their performance often does not differ from that of novices. Knowledge: An experiment by Chase and Simon (1973a, b) dealt with the question how well experts and novices are able to reproduce positions of chess pieces on chessboards when these are presented to them only briefly. The results showed that experts were far better in reproducing actual game positions, but that their performance was comparable with that of novices when the chess pieces were arranged randomly on the board. Chase and Simon concluded that the superior performance on actual game positions was due to the ability to recognize familiar patterns: A chess expert has up to 50,000 patterns stored in his memory. In comparison, a good player might know about 1,000 patterns by heart and a novice only few to none at all. This very detailed knowledge is of crucial help when an expert is confronted with a new problem in his field. Still, it is not pure size of knowledge that makes an expert more successful. Experts also organise their knowledge quite differently from novices. Organization: In 1982 M. Chi and her co-workers took a set of 24 physics problems and presented them to a group of physics professors as well as to a group of students with only one semester of physics. The task was to group the problems based on their similarities. As it turned out the students tended to group the problems based on their surface structure (similarities of objects used in the problem, e.g. on sketches illustrating the problem), whereas the professors used their deep structure (the general physical principles that underlay the problems) as criteria. By recognizing the actual structure of a problem experts are able to connect the given task to the relevant knowledge they already have (e.g. another problem they solved earlier which required the same strategy). Analysis: Experts often spend more time analyzing a problem before actually trying to solve it. This way of approaching a problem may often result in what appears to be a slow start, but in the long run this strategy is much more effective. A novice, on the other hand, might start working on the problem right away, but often has to realise that he reaches dead ends as he chose a wrong path in the very beginning. Creative Cognition Divergent Thinking The term divergent thinking describes a way of thinking that does not lead to one goal, but is open-ended. Problems that are solved this way can have a large number of potential 'solutions' of which none is exactly 'right' or 'wrong', though some might be more suitable than others. Solving a problem like this involves indirect and productive thinking and is mostly very helpful when somebody faces an ill-defined problem, i.e. when either initial state or goal state cannot be stated clearly and operators are either insufficient or not given at all. The process of divergent thinking is often associated with creativity, and it undoubtedly leads to many creative ideas. Nevertheless, researches have shown that there is only modest correlation between performance on divergent thinking tasks and other measures of creativity. Additionally it was found that in processes resulting in original and practical inventions things like searching for solutions, being aware of structures and looking for analogies are heavily involved, too. Figure 6.20.12.4: functional MRI images of the brains of musicians playing improvised jazz revealed that a large brain region involved in monitoring one's performance shuts down during creative improvisation, while a small region involved in organizing self-initiated thoughts and behaviors is highly activated (Image and modified caption from Wikimedia Commons. File:Creative Improvisation (24130148711).jpg; by NIH Image Gallery; As a work of the U.S. federal government, the image is in the public domain. Convergent Thinking Convergent thinking patterns are problem solving techniques that unite different ideas or fields to find a solution. The focus of this mindset is speed, logic and accuracy, also identification of facts, reapplying existing techniques, gathering information. The most important factor of this mindset is: there is only one correct answer. You only think of two answers, namely right or wrong. This type of thinking is associated with certain science or standard procedures. People with this type of thinking have logical thinking, are able to memorize patterns, solve problems and work on scientific tests. Most school subjects sharpen this type of thinking ability. Research shows that the creative process involves both types of thought processes. Brain Mechanisms in Problem Solving Presenting Neurophysiology in its entirety would be enough to fill several books. Instead, let's focus only on the aspects that are especially relevant to problem solving. Still, this topic is quite complex and problem solving cannot be attributed to one single brain area. Rather there are systems or networks of several brain areas working together to perform a specific problem solving task. This is best shown by an example, playing chess: \| Task \| Location of Brain activity \| --- \| \| Identifying chess pieces determining location of pieces Thinking about making a move Remembering a pieces move Planning and executing strategies \| Pathway from Occipital to Temporal Lobe (also called the "what"-pathway of visual processing) Pathway from Occipital to parietal Lobe (also called the "where"-pathway of visual processing) Premotor area Hippocampus (forming new memories) Prefrontal cortex \| Table 2: Brain areas involved in a complex cognitive task. One of the key tasks, namely planning and executing strategies, is performed by the prefrontal cortex (PFC), which also plays an important role for several other tasks correlated with problem solving. This can be made clear from the effects of damage to the PFC on ability to solve problems. Patients with a lesion in this brain area have difficulty switching from one behavioral pattern to another. A well known example is the wisconsin card-sorting task. A patient with a PFC lesion who is told to separate all blue cards from a deck, would continue sorting out the blue ones, even if the experimenter next told him to sort out all brown cards. Transferred to a more complex problem, this person would most likely fail, because he is not flexible enough to change his strategy after running into a dead end or when the problem changes. Another example comes from a young homemaker, who had a tumour in the frontal lobe. Even though she was able to cook individual dishes, preparing a whole family meal was an impossible task for her. Mushiake et al. (2009) note that to achieve a goal in a complex environment, such as problem‐solving situations like those above, we must plan multiple steps of action. When planning a series of actions, we have to anticipate future outcomes that will occur as a result of each action, and, in addition, we must mentally organize the temporal sequence of events in order to achieve the goal. These researchers investigated the role of lateral prefrontal cortex (PFC) in problem solving in monkeys. They found that "PFC neurons reflected final goals and immediate goals during the preparatory period. [They] also found some PFC neurons reflected each of all the forthcoming steps of actions during the preparatory period and they increased their [neural] activity step by step during the execution period. [Furthermore, they] found that the transient increase in synchronous activity of PFC neurons was involved in goal subgoal transformations. [They concluded] that the PFC is involved primarily in the dynamic representation of multiple future events that occur as a consequence of behavioral actions in problem‐solving situations" (Mushiake et al., 2009, p. 1). In other words, the prefrontal cortex represents in our imagination the sequence of events following each step that we take in solving a particular problem, guiding us step by step to the solution. As the examples above illustrate, the structure of our brain is of great importance regarding problem solving, i.e. cognitive life. But how was our cognitive apparatus designed? How did perception-action integration as a central species-specific property of humans come about? The answer, as argued extensively in earlier sections of this book, is, of course, natural selection and other forces of genetic evolution. Clearly, animals and humans with genes facilitating brain organization that led to good problem solving skills would be favored by natural selection over genes responsible for brain organization less adept at solving problems. We became equipped with brains organized for effective problem solving because flexible abilities to solve a wide range of problems presented by the environment enhanced ability to survive, to compete for resources, to escape predators, and to reproduce (see chapter on Evolution and Genetics in this text). In short, good problem solving mechanisms in brains designed for the real world gave a competitive advantage and increased biological fitness. Consequently, humans (and many other animals to a lesser degree) have "innate ability to problem-solve in the real world. Solving real world problems in real time given constraints posed by one's environment is crucial for survival . . . Real world problem solving (RWPS) is different from those that occur in a classroom or in a laboratory during an experiment. They are often dynamic and discontinuous, accompanied by many starts and stops . . . Real world problems are typically ill-defined, and even when they are well-defined, often have open-ended solutions . . . RWPS is quite messy and involves a tight interplay between problem solving, creativity, and insight . . . In psychology and neuroscience, problem-solving broadly refers to the inferential steps taken by an agent [human, animal, or computer] that leads from a given state of affairs to a desired goal state" (Sarathy, 2018, p. 261-2). According to Sarathy (2018), the initial stage of RWPS requires defining the problem and generating a representation of it in working memory. This stage involves activation of parts of the "prefrontal cortex (PFC), default mode network (DMN), and the dorsal anterior cingulate cortex(dACC)." TheDMN includes the medial prefrontal cortex, posterior cingulate cortex, and the inferior parietal lobule. Other structures sometimes considered part of the network are the lateral temporal cortex, hippocampal formation, and the precuneus.This network of structures is called "default mode" because these structures show increased activity when one is not engaged in focused, attentive, goal-directed actions, but rather a "resting state" (a baseline default state) and show decreased neural activity when one is focused and attentive to a particular goal-directed behavior (Raichle, et al., 2001). Moral Reasoning Jeurissen, et al., (2014) examined a special type of reasoning, moral reasoning, using TMS (Transcranial Magnetic Stimulation). The dorsolateral prefrontal cortex (DLPFC) and temporal-parietal junction (TPJ)have both been shown to be involved in moral judgments, but this study by Jeurissen, et al., (2014) uses TMS to tease out the different roles these brain areas play in different scenarios involving moral dilemmas. Moral dilemmas have been categorized by researchers as moral-impersonal(e.g., trolley or switch dilemma--save the lives of five workmen at the expense of the life of one by switching train to another track) and moral-personaldilemmas (e.g., footbridge dilemma--push a stranger in front of a train to save the lives of fiveothers). In the first scenario, the person just pulls a switch resulting in death of one person to save five, but in the second, the person pushes the victim to their death to save five others. Dual-process theory proposes that moral decision-making involves two components: an automatic emotional response and a voluntary application of a utilitarian decision-rule (in this case, one death to save five people is worth it). The thought of being responsible for the death of another person elicits an aversive emotional response, but at the same time, cognitive reasoning favors the utilitarian option. Decision making and social cognitionare oftenassociated withthe DLPFC.Neuronsin theprefrontal cortexhave been found to be involved in cost-benefit analysis and categorize stimuli based on the predicted consequences. Theory-of-mind (TOM) is a cognitive mechanism which is used when one tries to understand and explain the knowledge, beliefs, and intention of others. TOMandempathyare often associated withTPJ functioning. In the article by Jeurissen, et al., (2014), brain activity is measured by BOLD. BOLD refers to Blood-oxygen-level-dependent imaging, or BOLD-contrast imaging,which is a way to measure neural activity in different brain areas in MRI images. Greene et al., 2001 (Links to an external site.), 2004 (Links to an external site.) reported that activity in the prefrontal cortex is thought to be important for the cognitive reasoning process, which can counteract the automatic emotional response that occurs in moral dilemmas like the one in Jeurissen, et al., (2014). Greene et al. (2001) (Links to an external site.) found that the medial portionsof themedial frontal gyrus,theposterior cingulate gyrus,and thebilateral angular gyrusshowed ahigher BOLDresponse in themoral-personal conditionthan the moral-impersonal condition. The right middle frontal gyrusand thebilateral parietal lobes showed a lower BOLDresponse in themoral-personal condition than in the moral impersonal. Furthermore, Greene et al. (2004) (Links to an external site.) showed anincreasedBOLD response for the bilateral amygdale in personal compared to the impersonal dilemmas. Given the role of the prefrontal cortex in moral decision-making, Jeurissen, et al., (2014)hypothesized that when magnetically stimulatingprefrontal cortex, they will selectively influencethe decision process of the moral personal dilemmas because the cognitive reasoningfor which theDLPFCis important is disrupted, thereby releasingthe emotional component making it more influential in the resolution of the dilemma. Because the activityin theTPJis related toemotional processingand theory of mind (Saxe and Kanwisher, 2003 (Links to an external site.); Young et al., 2010 (Links to an external site.)), Jeurissen, et al., (2014)hypothesized that when magnetically stimulating this area, the TPJ, during a moral decision, this will selectively influence the decision process of moral-impersonaldilemmas. Results of this study by Jeurissen, et al., (2014) showed an important roleof theTPJinmoral judgment. Experiments using fMRI (Greene et al., 2004 (Links to an external site.)), have found the cingulate cortex to be involved in moral judgment. In earlier studies, the cingulate cortex was found to be involved in the emotionalresponse. Since the moral-personal dilemmas are more emotionally salient, the higher activity observed for TPJin the moral-personalcondition (more emotional)is consistent with this view. Another area that is hypothesized to be associated with the emotional response is the temporal cortex. In this study by Jeurissen, et al., (2014), magnetic stimulationof the right DLPFCand right TPJ shows roles for right DLPFC (reasoning and utilitarian) and right TPJ (emotion) in moral impersonal and moral personal dilemmas respectively. TMS over the right DLPFC (disrupting neural activity here) leads to behavior changes consistent with less cognitive control over emotion. After right DLPFC stimulation, participants show less feelings of regret than after magnetic stimulation of the right TPJ. This last finding indicates that the right DLPFC is involved in evaluatingthe outcomeof the decision process. In summary, this experiment by Jeurissen, et al., (2014) adds to evidence of a critical role of right DLPFC and right TPJinmoral decision-makingand supports that hypothesis that theformeris involved in judgments based oncognitive reasoningand anticipation of outcomes,whereas thelatteris involved inemotional processing related tomoral dilemmas. Summary Many different strategies exist for solving problems. Typical strategies include trial and error, applying algorithms, and using heuristics. To solve a large, complicated problem, it often helps to break the problem into smaller steps that can be accomplished individually, leading to an overall solution. The brain mechanisms involved in problem solving vary to some degree depending upon the sensory modalities involved in the problem and its solution, however, the prefrontal cortex is one brain region that appears to be centrally involved in all problem-solving. The prefrontal cortex is required for flexible shifts in attention, for representing the problem in working memory, and for holding steps in problem solving in working memory along with representations of future consequences of those actions permitting planning and execution of plans. Also implicated is the Default Mode Network (DMN) including medial prefrontal cortex, posterior cingulate cortex, and the inferior parietal module, and sometimes the lateral temporal cortex, hippocampus, and the precuneus. Moral reasoning involves a different set of brain areas, primarily the dorsolateral prefrontal cortex (DLPFC) and temporal-parietal junction (TPJ). Review Questions A specific formula for solving a problem is called __. an algorithm a heuristic a mental set trial and error A mental shortcut in the form of a general problem-solving framework is called __. an algorithm a heuristic a mental set trial and error References Gray, M. E., & Holyoak, K. J. (2021). Teaching by analogy: From theory to practice. Mind, Brain, and Education, 15(3), 250-263. Hunt, L. T., Behrens, T. E., Hosokawa, T., Wallis, J. D., & Kennerley, S. W. (2015). Capturing the temporal evolution of choice across prefrontal cortex. Elife, 4, e11945. Mushiake, H., Sakamoto, K., Saito, N., Inui, T., Aihara, K., & Tanji, J. (2009). Involvement of the prefrontal cortex in problem solving. International review of neurobiology, 85, 1-11. Jeurissen, D., Sack, A. T., Roebroeck, A., Russ, B. E., & Pascual-Leone, A. (2014). TMS affects moral judgment, showing the role of DLPFC and TPJ in cognitive and emotional processing. Frontiers in neuroscience, 8, 18. Kahneman, D. (2011). Thinking, fast and slow. New York: Farrar, Straus, and Giroux. Koenigshofer, K. A. (2017). General Intelligence: Adaptation to Evolutionarily Familiar Abstract Relational Invariants, Not to Environmental or Evolutionary Novelty. The Journal of Mind and Behavior, 119-153. Pratkanis, A. (1989). The cognitive representation of attitudes. In A. R. Pratkanis, S. J. Breckler, & A. G. Greenwald (Eds.), Attitude structure and function (pp. 71–98). Hillsdale, NJ: Erlbaum. Raichle, M. E., MacLeod, A. M., Snyder, A. Z., Powers, W. J., Gusnard, D. A., & Shulman, G. L. (2001). A default mode of brain function. Proceedings of the National Academy of Sciences, 98(2), 676-682. Sawyer, K. (2011). The cognitive neuroscience of creativity: a critical review. Creat. Res. J. 23, 137–154. doi: 10.1080/10400419.2011.571191 Tversky, A., & Kahneman, D. (1974). Judgment under uncertainty: Heuristics and biases. Science, 185(4157), 1124–1131. Brain Mechanisms in Problem Solving Hunt, L. T., Behrens, T. E., Hosokawa, T., Wallis, J. D., & Kennerley, S. W. (2015). Capturing the temporal evolution of choice across prefrontal cortex. Elife, 4, e11945. Mushiake, H., Sakamoto, K., Saito, N., Inui, T., Aihara, K., & Tanji, J. (2009). Involvement of the prefrontal cortex in problem solving. International review of neurobiology, 85, 1-11. Sawyer, K. (2011). The cognitive neuroscience of creativity: a critical review. Creat. Res. J. 23, 137–154. doi: 10.1080/10400419.2011.571191 Attributions "Overview," "Problem Solving Strategies," adapted from Problem Solving by OpenStax Colleg licensedCC BY-NC 4.0 via OER Commons "Defining Problems," "Problem Solving as a Search Problem," "Creative Cognition," "Brain Mechanisms in Problem-Solving" adapted by Kenneth A. Koenigshofer, Ph.D., from 2.1, 2.2, 2.3, 2.4, 2.5, 2.6 in Cognitive Psychology and Cognitive Neuroscience (Wikibooks) unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Legal; the LibreTexts libraries are Powered by MindTouch Moral Reasoning was written by Kenneth A. Koenigshofer, Ph.D, Chaffey College. Categories and Concepts By Gregory Murphy, New York University People form mental concepts of categories of objects, which permit them to respond appropriately to new objects they encounter. Most concepts cannot be strictly defined but are organized around the “best” examples or prototypes, which have the properties most common in the category. Objects fall into many different categories, but there is usually a most salient one, called the basic-level category, which is at an intermediate level of specificity (e.g., chairs, rather than furniture or desk chairs). Concepts are closely related to our knowledge of the world, and people can more easily learn concepts that are consistent with their knowledge. Theories of concepts argue either that people learn a summary description of a whole category or else that they learn exemplars of the category. Recent research suggests that there are different ways to learn and represent concepts and that they are accomplished by different neural systems. Learning Objectives Understand the problems with attempting to define categories. Understand typicality and fuzzy category boundaries. Learn about theories of the mental representation of concepts. Learn how knowledge may influence concept learning. Introduction Figure 6.20.12.2: Although you’ve (probably) never seen this particular truck before, you know a lot about it because of the knowledge you’ve accumulated in the past about the features in the category of trucks. Image: CC0 Public Domain, [ Consider the following set of objects: some dust, papers, a computer monitor, two pens, a cup, and an orange. What do these things have in common? Only that they all happen to be on my desk as I write this. This set of things can be considered a category, a set of objects that can be treated as equivalent in some way. But, most of our categories seem much more informative—they share many properties. For example, consider the following categories: trucks, wireless devices, weddings, psychopaths, and trout. Although the objects in a given category are different from one another, they have many commonalities. When you know something is a truck, you know quite a bit about it. The psychology of categories concerns how people learn, remember, and use informative categories such as trucks or psychopaths. The mental representations we form of categories are called concepts. There is a category of trucks in the actual physical world, and I also have a concept of trucks in my head. We assume that people’s concepts correspond more or less closely to the actual category, but it can be useful to distinguish the two, as when someone’s concept is not really correct. Concepts are at the core of intelligent behavior. We expect people to be able to know what to do in new situations and when confronting new objects. If you go into a new classroom and see chairs, a blackboard, a projector, and a screen, you know what these things are and how they will be used. You’ll sit on one of the chairs and expect the instructor to write on the blackboard or project something onto the screen. You do this even if you have never seen any of these particular objects before, because you have concepts of classrooms, chairs, projectors, and so forth, that tell you what they are and what you’re supposed to do with them. Furthermore, if someone tells you a new fact about the projector—for example, that it has a halogen bulb—you are likely to extend this fact to other projectors you encounter. In short, concepts allow you to extend what you have learned about a limited number of objects to a potentially infinite set of entities (i.e. generalization). Notice how categories and concepts arise from similarity, one of the abstract features of the world that has been genetically internalized into the brain during evolution, creating an innate disposition of brains to search for and to represent groupings of similar things, forming one component of general intelligence. One property of the human brain that distinguishes us from other animals is the high degrees of abstraction in similarity relations that the human brain is capable of encoding compared to the brains of non-human animals (Koenigshofer, 2017). Simpler organisms, such as animals and human infants, also have concepts (Mareschal, Quinn, & Lea, 2010). Squirrels may have a concept of predators, for example, that is specific to their own lives and experiences. However, animals likely have many fewer concepts and cannot understand complex concepts such as mortgages or musical instruments. You know thousands of categories, most of which you have learned without careful study or instruction. Although this accomplishment may seem simple, we know that it isn’t, because it is difficult to program computers to solve such intellectual tasks. If you teach a learning program that a robin, a swallow, and a duck are all birds, it may not recognize a cardinal or peacock as a bird. However, this shortcoming in computers may be at least partially overcome when the type of processing used is parallel distributed processing as employed in artificial neural networks (Koenigshofer, 2017), discussed in this chapter. As we’ll shortly see, the problem for computers is that objects in categories are often surprisingly diverse. Nature of Categories Figure 6.20.12.3: Here is a very good dog, but one that does not fit perfectly into a well-defined category where all dogs have four legs. Image: State Farm, CC BY 2.0, [ Traditionally, it has been assumed that categories are well-defined. This means that you can give a definition that specifies what is in and out of the category. Such a definition has two parts. First, it provides the necessary features for category membership: What must objects have in order to be in it? Second, those features must be jointly sufficient for membership: If an object has those features, then it is in the category. For example, if I defined a dog as a four-legged animal that barks, this would mean that every dog is four-legged, an animal, and barks, and also that anything that has all those properties is a dog. Unfortunately, it has not been possible to find definitions for many familiar categories. Definitions are neat and clear-cut; the world is messy and often unclear. For example, consider our definition of dogs. In reality, not all dogs have four legs; not all dogs bark. I knew a dog that lost her bark with age (this was an improvement); no one doubted that she was still a dog. It is often possible to find some necessary features (e.g., all dogs have blood and breathe), but these features are generally not sufficient to determine category membership (you also have blood and breathe but are not a dog). Even in domains where one might expect to find clear-cut definitions, such as science and law, there are often problems. For example, many people were upset when Pluto was downgraded from its status as a planet to a dwarf planet in 2006. Upset turned to outrage when they discovered that there was no hard-and-fast definition of planethood: “Aren’t these astronomers scientists? Can’t they make a simple definition?” In fact, they couldn’t. After an astronomical organization tried to make a definition for planets, a number of astronomers complained that it might not include accepted planets such as Neptune and refused to use it. If everything looked like our Earth, our moon, and our sun, it would be easy to give definitions of planets, moons, and stars, but the universe has not conformed to this ideal. Fuzzy Categories Borderline Items Experiments also showed that the psychological assumptions of well-defined categories were not correct. Hampton (1979) asked subjects to judge whether a number of items were in different categories. He did not find that items were either clear members or clear nonmembers. Instead, he found many items that were just barely considered category members and others that were just barely not members, with much disagreement among subjects. Sinks were barely considered as members of the kitchen utensil category, and sponges were barely excluded. People just included seaweed as a vegetable and just barely excluded tomatoes and gourds. Hampton found that members and nonmembers formed a continuum, with no obvious break in people’s membership judgments. If categories were well defined, such examples should be very rare. Many studies since then have found such borderline members that are not clearly in or clearly out of the category. Table 1. Examples of two categories, with members ordered by typicality (from Rosch & Mervis, 1975) McCloskey and Glucksberg (1978) found further evidence for borderline membership by asking people to judge category membership twice, separated by two weeks. They found that when people made repeated category judgments such as “Is an olive a fruit?” or “Is a sponge a kitchen utensil?” they changed their minds about borderline items—up to 22 percent of the time. So, not only do people disagree with one another about borderline items, they disagree with themselves! As a result, researchers often say that categories are fuzzy, that is, they have unclear boundaries that can shift over time. Typicality A related finding that turns out to be most important is that even among items that clearly are in a category, some seem to be “better” members than others (Rosch, 1973). Among birds, for example, robins and sparrows are very typical. In contrast, ostriches and penguins are very atypical (meaning not typical). If someone says, “There’s a bird in my yard,” the image you have will be of a smallish passerine bird such as a robin, not an eagle or hummingbird or turkey. You can find out which category members are typical merely by asking people. Table 1 shows a list of category members in order of their rated typicality. Typicality is perhaps the most important variable in predicting how people interact with categories. The following text box is a partial list of what typicality influences. We can understand the two phenomena of borderline members and typicality as two sides of the same coin. Think of the most typical category member: This is often called the category prototype. Items that are less and less similar to the prototype become less and less typical. At some point, these less typical items become so atypical that you start to doubt whether they are in the category at all. Is a rug really an example of furniture? It’s in the home like chairs and tables, but it’s also different from most furniture in its structure and use. From day to day, you might change your mind as to whether this atypical example is in or out of the category. So, changes in typicality ultimately lead to borderline members. Table 2: Text Box 1 Source of Typicality Intuitively, it is not surprising that robins are better examples of birds than penguins are, or that a table is a more typical kind of furniture than is a rug. But given that robins and penguins are known to be birds, why should one be more typical than the other? One possible answer is the frequency with which we encounter the object: We see a lot more robins than penguins, so they must be more typical. Frequency does have some effect, but it is actually not the most important variable (Rosch, Simpson, & Miller, 1976). For example, I see both rugs and tables every single day, but one of them is much more typical as furniture than the other. The best account of what makes something typical comes from Rosch and Mervis’s (1975) family resemblance theory. They proposed that items are likely to be typical if they (a) have the features that are frequent in the category and (b) do not have features frequent in other categories. Let’s compare two extremes, robins and penguins. Robins are small flying birds that sing, live in nests in trees, migrate in winter, hop around on your lawn, and so on. Most of these properties are found in many other birds. In contrast, penguins do not fly, do not sing, do not live in nests or in trees, do not hop around on your lawn. Furthermore, they have properties that are common in other categories, such as swimming expertly and having wings that look and act like fins. These properties are more often found in fish than in birds. Figure 6.20.12.4: When you think of “bird,” how closely does the robin resemble your general figure? Image: CC0 Public Domain, [ According to Rosch and Mervis, then, it is not because a robin is a very common bird that makes it typical. Rather, it is because the robin has the shape, size, body parts, and behaviors that are very common (i.e. most similar) among birds—and not common among fish, mammals, bugs, and so forth. In a classic experiment, Rosch and Mervis (1975) made up two new categories, with arbitrary features. Subjects viewed example after example and had to learn which example was in which category. Rosch and Mervis constructed some items that had features that were common in the category and other items that had features less common in the category. The subjects learned the first type of item before they learned the second type. Furthermore, they then rated the items with common features as more typical. In another experiment, Rosch and Mervis constructed items that differed in how many features were shared with a different category. The more features were shared, the longer it took subjects to learn which category the item was in. These experiments, and many later studies, support both parts of the family resemblance theory. Category Hierarchies Many important categories fall into hierarchies, in which more concrete categories are nested inside larger, abstract categories. For example, consider the categories: brown bear, bear, mammal, vertebrate, animal, entity. Clearly, all brown bears are bears; all bears are mammals; all mammals are vertebrates; and so on. Any given object typically does not fall into just one category—it could be in a dozen different categories, some of which are structured in this hierarchical manner. Examples of biological categories come to mind most easily, but within the realm of human artifacts, hierarchical structures can readily be found: desk chair, chair, furniture, artifact, object. Brown (1958), a child language researcher, was perhaps the first to note that there seems to be a preference for which category we use to label things. If your office desk chair is in the way, you’ll probably say, “Move that chair,” rather than “Move that desk chair” or “piece of furniture.” Brown thought that the use of a single, consistent name probably helped children to learn the name for things. And, indeed, children’s first labels for categories tend to be exactly those names that adults prefer to use (Anglin, 1977). Figure 6.20.12.5: This is a highly simplified illustration of hierarchically organized categories, with the superordinate, basic, and subordinate levels labeled. Keep in mind that there may be even more specific subordinates (e.g., wire-haired terriers) and more general superordinates (e.g., living thing) This preference is referred to as a preference for the basic level of categorization, and it was first studied in detail by Eleanor Rosch and her students (Rosch, Mervis, Gray, Johnson, & Boyes-Braem, 1976). The basic level represents a kind of Goldilocks effect, in which the category used for something is not too small (northern brown bear) and not too big (animal), but is just right (bear). The simplest way to identify an object’s basic-level category is to discover how it would be labeled in a neutral situation. Rosch et al. (1976) showed subjects pictures and asked them to provide the first name that came to mind. They found that 1,595 names were at the basic level, with 14 more specific names (subordinates) used. Only once did anyone use a more general name (superordinate). Furthermore, in printed text, basic-level labels are much more frequent than most subordinate or superordinate labels (e.g., Wisniewski & Murphy, 1989). The preference for the basic level is not merely a matter of labeling. Basic-level categories are usually easier to learn. As Brown noted, children use these categories first in language learning, and superordinates are especially difficult for children to fully acquire. People are faster at identifying objects as members of basic-level categories (Rosch et al., 1976). Rosch et al. (1976) initially proposed that basic-level categories cut the world at its joints, that is, merely reflect the big differences between categories like chairs and tables or between cats and mice that exist in the world. However, it turns out that which level is basic is not universal. North Americans are likely to use names like tree, fish, and bird to label natural objects. But people in less industrialized societies seldom use these labels and instead use more specific words, equivalent to elm, trout, and finch (Berlin, 1992). Because Americans and many other people living in industrialized societies know so much less than our ancestors did about the natural world, our basic level has “moved up” to what would have been the superordinate level a century ago. Furthermore, experts in a domain often have a preferred level that is more specific than that of non-experts. Birdwatchers see sparrows rather than just birds, and carpenters see roofing hammers rather than just hammers (Tanaka & Taylor, 1991). This all suggests that the preferred level is not (only) based on how different categories are in the world, but that people’s knowledge and interest in the categories has an important effect. One explanation of the basic-level preference is that basic-level categories are more differentiated: The category members are similar to one another, but they are different from members of other categories (Murphy & Brownell, 1985; Rosch et al., 1976). (The alert reader will note a similarity to the explanation of typicality I gave above. However, here we’re talking about the entire category and not individual members.) Chairs are pretty similar to one another, sharing a lot of features (legs, a seat, a back, similar size and shape); they also don’t share that many features with other furniture. Superordinate categories are not as useful because their members are not very similar to one another. What features are common to most furniture? There are very few. Subordinate categories are not as useful, because they’re very similar to other categories: Desk chairs are quite similar to dining room chairs and easy chairs. As a result, it can be difficult to decide which subordinate category an object is in (Murphy & Brownell, 1985). Experts can differ from novices in which categories are the most differentiated, because they know different things about the categories, therefore changing how similar the categories are. This is a controversial claim, as some say that infants learn superordinates before anything else (Mandler, 2004). However, if true, then it is very puzzling that older children have great difficulty learning the correct meaning of words for superordinates, as well as in learning artificial superordinate categories (Horton &Markman, 1980; Mervis, 1987). However, it seems fair to say that the answer to this question is not yet fully known. Theories of Concept Representation Now that we know these facts about the psychology of concepts, the question arises of how concepts are mentally represented. There have been two main answers. The first, somewhat confusingly called the prototype theory suggests that people have a summary representation of the category, a mental description that is meant to apply to the category as a whole. (The significance of summary will become apparent when the next theory is described.) This description can be represented as a set of weighted features (Smith & Medin, 1981). The features are weighted by their frequency in the category. For the category of birds, having wings and feathers would have a very high weight; eating worms would have a lower weight; living in Antarctica would have a lower weight still, but not zero, as some birds do live there. Figure 6.20.12.6: If you were asked, “What kind of animal is this?” according to prototype theory, you would consult your summary representations of different categories and then select the one that is most similar to this image—probably a lizard! Image: Adhi Rachdian, CC BY 2.0, [ The idea behind prototype theory is that when you learn a category, you learn a general description that applies to the category as a whole: Birds have wings and usually fly; some eat worms; some swim underwater to catch fish. People can state these generalizations, and sometimes we learn about categories by reading or hearing such statements (“The kimodo dragon can grow to be 10 feet long”). When you try to classify an item, you see how well it matches that weighted list of features. For example, if you saw something with wings and feathers fly onto your front lawn and eat a worm, you could (unconsciously) consult your concepts and see which ones contained the features you observed. This example possesses many of the highly weighted bird features, and so it should be easy to identify as a bird. This theory readily explains the phenomena we discussed earlier. Typical category members have more, higher-weighted features. Therefore, it is easier to match them to your conceptual representation. Less typical items have fewer or lower-weighted features (and they may have features of other concepts). Therefore, they don’t match your representation as well (less similarity). This makes people less certain in classifying such items. Borderline items may have features in common with multiple categories or not be very close to any of them. For example, edible seaweed does not have many of the common features of vegetables but also is not close to any other food concept (meat, fish, fruit, etc.), making it hard to know what kind of food it is. A very different account of concept representation is the exemplar theory (exemplar being a fancy name for an example; Medin & Schaffer, 1978). This theory denies that there is a summary representation. Instead, the theory claims that your concept of vegetables is remembered examples of vegetables you have seen. This could of course be hundreds or thousands of exemplars over the course of your life, though we don’t know for sure how many exemplars you actually remember. How does this theory explain classification? When you see an object, you (unconsciously) compare it to the exemplars in your memory, and you judge how similarit is to exemplars in different categories. For example, if you see some object on your plate and want to identify it, it will probably activate memories of vegetables, meats, fruit, and so on. In order to categorize this object, you calculate how similarit is to each exemplar in your memory. These similarity scores are added up for each category. Perhaps the object is very similar to a large number of vegetable exemplars, moderately similar to a few fruit, and only minimally similar to some exemplars of meat you remember. These similarity scores are compared, and the category with the highest score is chosen. Why would someone propose such a theory of concepts? One answer is that in many experiments studying concepts, people learn concepts by seeing exemplars over and over again until they learn to classify them correctly. Under such conditions, it seems likely that people eventually memorize the exemplars (Smith & Minda, 1998). There is also evidence that close similarity to well-remembered objects has a large effect on classification. Allen and Brooks (1991) taught people to classify items by following a rule. However, they also had their subjects study the items, which were richly detailed. In a later test, the experimenters gave people new items that were very similar to one of the old items but were in a different category. That is, they changed one property so that the item no longer followed the rule. They discovered that people were often fooled by such items. Rather than following the category rule they had been taught, theyseemed to recognize the new item as being very similar to an old one and so put it, incorrectly, into the same category. Many experiments have been done to compare the prototype and exemplar theories. Overall, the exemplar theory seems to have won most of these comparisons. However, the experiments are somewhat limited in that they usually involve a small number of exemplars that people view over and over again. It is not so clear that exemplar theory can explain real-world classification in which people do not spend much time learning individual items (how much time do you spend studying squirrels? or chairs?). Also, given that some part of our knowledge of categories is learned through general statements we read or hear, it seems that there must be room for a summary description separate from exemplar memory. Many researchers would now acknowledge that concepts are represented through multiple cognitive systems. For example, your knowledge of dogs may be in part through general descriptions such as “dogs have four legs.” But you probably also have strong memories of some exemplars (your family dog, Lassie) that influence your categorization. Furthermore, some categories also involve rules (e.g., a strike in baseball). How these systems work together is the subject of current study. Actually, the decision of which category is chosen is more complex than this, but the details are beyond this discussion. Knowledge The final topic has to do with how concepts fit with our broader knowledge of the world. We have been talking very generally about people learning the features of concepts. For example, they see a number of birds and then learn that birds generally have wings, or perhaps they remember bird exemplars. From this perspective, it makes no difference what those exemplars or features are—people just learn them. But consider two possible concepts of buildings and their features in Table 2. Table 3. Examples of two fictional concepts Imagine you had to learn these two concepts by seeing exemplars of them, each exemplar having some of the features listed for the concept (as well as some idiosyncratic features). Learning the donker concept would be pretty easy. It seems to be a kind of underwater building, perhaps for deep-sea explorers. Its features seem to go together. In contrast, the blegdav doesn’t really make sense. If it’s in the desert, how can you get there by submarine, and why do they have polar bears as pets? Why would farmers live in the desert or use submarines? What good would steel windows do in such a building? This concept seems peculiar. In fact, if people are asked to learn new concepts that make sense, such as donkers, they learn them quite a bit faster than concepts such as blegdavs that don’t make sense (Murphy & Allopenna, 1994). Furthermore, the features that seem connected to one another (such as being underwater and getting there by submarine) are learned better than features that don’t seem related to the others (such as being red). Such effects demonstrate that when we learn new concepts, we try to connect them to the knowledge we already have about the world. If you were to learn about a new animal that doesn’t seem to eat or reproduce, you would be very puzzled and think that you must have gotten something wrong. By themselves, the prototype and exemplar theories don’t predict this. They simply say that you learn descriptions or exemplars, and they don’t put any constraints on what those descriptions or exemplars are. However, the knowledge approach to concepts emphasizes that concepts are meant to tell us about real things in the world, and so our knowledge of the world is used in learning and thinking about concepts. We can see this effect of knowledge when we learn about new pieces of technology. For example, most people could easily learn about tablet computers (such as iPads) when they were first introduced by drawing on their knowledge of laptops, cell phones, and related technology. Of course, this reliance on past knowledge can also lead to errors, as when people don’t learn about features of their new tablet that weren’t present in their cell phone or expect the tablet to be able to do something it can’t. One important aspect of people’s knowledge about categories is called psychological essentialism(Gelman, 2003;Medin & Ortony, 1989). People tend to believe that some categories—most notably natural kinds such as animals, plants, or minerals—have an underlying property that is found only in that category and that causes its other features. Most categories don’t actually have essences, but this is sometimes a firmly held belief. For example, many people will state that there is something about dogs, perhaps some specific gene or set of genes, that all dogs have and that makes them bark, have fur, and look the way they do. Therefore, decisions about whether something is a dog do not depend only on features that you can easily see but also on the assumed presence of this cause. Figure 6.20.12.7: Although it may seem natural that different species have an unchangeable “essence,” consider evolution and everything’s development from common ancestors. Image: Marc Dragiewicz, CC BY-NC-SA 2.0, [ Belief in an essence can be revealed through experiments describing fictional objects. Keil (1989) described to adults and children a fiendish operation in which someone took a raccoon, dyed its hair black with a white stripe down the middle, and implanted a “sac of super-smelly yucky stuff” under its tail. The subjects were shown a picture of a skunk and told that this is now what the animal looks like. What is it? Adults and children over the age of 4 all agreed that the animal is still a raccoon. It may look and even act like a skunk, but a raccoon cannot change its stripes (or whatever!)—it will always be a raccoon. Importantly, the same effect was not found when Keil described a coffeepot that was operated on to look like and function as a bird feeder. Subjects agreed that it was now a bird feeder. Artifacts don’t have an essence. Signs of essentialism include (a) objects are believed to be either in or out of the category, with no in-between; (b) resistance to change of category membership or of properties connected to the essence; and (c) for living things, the essence is passed on to progeny. Essentialism is probably helpful in dealing with much of the natural world, but it may be less helpful when it is applied to humans. Considerable evidence suggests that people think of gender, racial, and ethnic groups as having essences, which serves to emphasize the difference between groups and even justify discrimination (Hirschfeld, 1996). Historically, group differences were described by inheriting the blood of one’s family or group. “Bad blood” was not just an expression but a belief that negative properties were inherited and could not be changed. After all, if it is in the nature of “those people” to be dishonest (or clannish or athletic ...), then that could hardly be changed, any more than a raccoon can change into a skunk. Research on categories of people is an exciting ongoing enterprise, and we still do not know as much as we would like to about how concepts of different kinds of people are learned in childhood and how they may (or may not) change in adulthood. Essentialism doesn’t apply only to person categories, but it is one important factor in how we think of groups. Summary Concepts are central to our everyday thought. When we are planning for the future or thinking about our past, we think about specific events and objects in terms of their categories. If you’re visiting a friend with a new baby, you have some expectations about what the baby will do, what gifts would be appropriate, how you should behave toward it, and so on. Knowing about the category of babies helps you to effectively plan and behave when you encounter this child you’ve never seen before. Such inferences from knowledge about a category are highly adaptive and an important component of thinking and intelligence. Learning about those categories is a complex process that involves seeing exemplars (babies), hearing or reading general descriptions (“Babies like black-and-white pictures”), general knowledge (babies have kidneys), and learning the occasional rule (all babies have a rooting reflex). Current research is focusing on how these different processes take place in the brain. It seems likely that these different aspects of concepts are accomplished by different neural structures (Maddox & Ashby, 2004). However, it is clear that the brain is genetically predisposed to seek out similarities in the environment and to represent groupings of things forming categories that can be used to make inferences about new instances of the category which have never been encountered before. In this way knowledge is organized and expectations from this knowledge allow improved adaptation to newly encountered environmental objects and situations by virtue of their similarity to a known category previously formed (Koenigshofer, 2017). Another interesting topic is how concepts differ across cultures. As different cultures have different interests and different kinds of interactions with the world, it seems clear that their concepts will somehow reflect those differences. On the other hand, the structure of the physical world also imposes a strong constraint on what kinds of categories are actually useful. The interplay of culture, the environment, and basic cognitive processes in establishing concepts has yet to be fully investigated. Discussion Questions Pick a couple of familiar categories and try to come up with definitions for them. When you evaluate each proposal (a) is it in fact accurate as a definition, and (b) is it a definition that people might actually use in identifying category members? For the same categories, can you identify members that seem to be “better” and “worse” members? What about these items makes them typical and atypical? Going around the room, point to some common objects (including things people are wearing or brought with them) and identify what the basic-level category is for that item. What are superordinate and subordinate categories for the same items? List some features of a common category such as tables. The knowledge view suggests that you know reasons for why these particular features occur together. Can you articulate some of those reasons? Do the same thing for an animal category. Choose three common categories: a natural kind, a human artifact, and a social event. Discuss with class members from other countries or cultures whether the corresponding categories in their cultures differ. Can you make a hypothesis about when such categories are likely to differ and when they are not? Vocabulary Basic-level categoryThe neutral, preferred category for a given object, at an intermediate level of specificity.CategoryA set of entities that are equivalent in some way. Usually the items are similar to one another.ConceptThe mental representation of a category.ExemplarAn example in memory that is labeled as being in a particular category.Psychological essentialismThe belief that members of a category have an unseen property that causes them to be in the category and to have the properties associated with it.TypicalityThe difference in “goodness” of category members, ranging from the most typical (the prototype) to borderline members. References Allen, S. W., & Brooks, L. R. (1991). Specializing the operation of an explicit rule.Journal of Experimental Psychology: General, 120, 3–19. Anglin, J. M. (1977). Word, object, and conceptual developmen t. New York, NY: W. W. Norton. Berlin, B. (1992). Ethnobiological classification: Principles of categorization of plants and animals in traditional societies. Princeton, NJ: Princeton University Press. Brown, R. (1958). How shall a thing be called? Psychological Review, 65, 14–21. Gelman, S. A. (2003). The essential child: Origins of essentialism in everyday thought. Oxford, UK: Oxford University Press. Hampton, J. A. (1979). Polymorphous concepts in semantic memory.Journal of Verbal Learning and Verbal Behavior, 18, 441–461. Hirschfeld, L. A. (1996). Race in the making: Cognition, culture, and the child's construction of human kinds. Cambridge, MA: MIT Press. Horton, M. S., & Markman, E. M. (1980). Developmental differences in the acquisition of basic and superordinate categories. Child Development, 51, 708–719. Keil, F. C. (1989). Concepts, kinds, and cognitive development. Cambridge, MA: MIT Press. Koenigshofer, K. A. (2017). General Intelligence: Adaptation to Evolutionarily Familiar Abstract Relational Invariants, Not to Environmental or Evolutionary Novelty. The Journal of Mind and Behavior, 38(2):119-153. Maddox, W. T., & Ashby, F. G. (2004). Dissociating explicit and procedural-based systems of perceptual category learning. Behavioural Processes, 66, 309–332. Mandler, J. M. (2004). The foundations of mind: Origins of conceptual thought. Oxford, UK: Oxford University Press. Mareschal, D., Quinn, P. C., & Lea, S. E. G. (Eds.) (2010). The making of human concepts. Oxford, UK: Oxford University Press. McCloskey, M. E., & Glucksberg, S. (1978). Natural categories: Well defined or fuzzy sets? Memory & Cognition, 6, 462–472. Medin, D. L., & Ortony, A. (1989). Psychological essentialism. In S. Vosniadou & A. Ortony (Eds.), Similarity and analogical reasoning (pp. 179–195). Cambridge, UK: Cambridge University Press. Medin, D. L., & Schaffer, M. M. (1978). Context theory of classification learning. Psychological Review, 85, 207–238. Mervis, C. B. (1987). Child-basic object categories and early lexical development. In U. Neisser (Ed.), Concepts and conceptual development: Ecological and intellectual factors in categorization (pp. 201–233). Cambridge, UK: Cambridge University Press. Murphy, G. L., & Allopenna, P. D. (1994). The locus of knowledge effects in concept learning. Journal of Experimental Psychology: Learning, Memory, and Cognition, 20, 904–919. Murphy, G. L., & Brownell, H. H. (1985). Category differentiation in object recognition: Typicality constraints on the basic category advantage. Journal of Experimental Psychology: Learning, Memory, and Cognition, 11, 70–84. Norenzayan, A., Smith, E. E., Kim, B. J., & Nisbett, R. E. (2002). Cultural preferences for formal versus intuitive reasoning. Cognitive Science, 26, 653–684. Rosch, E., & Mervis, C. B. (1975). Family resemblance: Studies in the internal structure of categories. Cognitive Psychology, 7, 573–605. Rosch, E., Mervis, C. B., Gray, W., Johnson, D., & Boyes-Braem, P. (1976). Basic objects in natural categories. Cognitive Psychology, 8, 382–439. Rosch, E., Simpson, C., & Miller, R. S. (1976). Structural bases of typicality effects. Journal of Experimental Psychology: Human Perception and Performance, 2, 491–502. Rosch, E. H. (1973). On the internal structure of perceptual and semantic categories. In T. E. Moore (Ed.), Cognitive development and the acquisition of language (pp. 111–144). New York, NY: Academic Press. Smith, E. E., & Medin, D. L. (1981). Categories and concepts. Cambridge, MA: Harvard University Press. Smith, J. D., & Minda, J. P. (1998). Prototypes in the mist: The early epochs of category learning. Journal of Experimental Psychology: Learning, Memory, and Cognition, 24, 1411–1436. Tanaka, J. W., & Taylor, M. E. (1991). Object categories and expertise: Is the basic level in the eye of the beholder? Cognitive Psychology, 15, 121–149. Wisniewski, E. J., & Murphy, G. L. (1989). Superordinate and basic category names in discourse: A textual analysis. Discourse Processes, 12, 245–261. Attributions Authors Gregory Murphy is Professor of Psychology at New York University. He previously taught at the University of Illinois and Brown University. His research focuses on concepts and reasoning, and he is the author of The Big Book of Concepts (MIT Press, 2002). Creative Commons License Adapted by Kenneth Koenigshofer, PhD, from Categories and Concepts by Gregory Murphy, licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Permissions beyond the scope of this license may be available in our Licensing Agreement. How to cite this Noba module using APA Style Murphy, G. (2021). Categories and concepts. In R. Biswas-Diener & E. Diener (Eds), Noba textbook series: Psychology. Champaign, IL: DEF publishers. Retrieved from This page titled 6.20.12: Chapter 14- Problem Solving, Categories and Concepts is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Kenneth A. Koenigshofer (ASCCC Open Educational Resources Initiative (OERI)) . 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6265	https://online.stat.psu.edu/stat200/	STAT 200 Elementary Statistics STAT 200 Welcome to STAT 200! User Preferences Content Preview Arcu felis bibendum ut tristique et egestas quis: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris Duis aute irure dolor in reprehenderit in voluptate Excepteur sint occaecat cupidatat non proident Lorem ipsum dolor sit amet, consectetur adipisicing elit. Odit molestiae mollitia laudantium assumenda nam eaque, excepturi, soluta, perspiciatis cupiditate sapiente, adipisci quaerat odio voluptates consectetur nulla eveniet iure vitae quibusdam? Excepturi aliquam in iure, repellat, fugiat illum voluptate repellendus blanditiis veritatis ducimus ad ipsa quisquam, commodi vel necessitatibus, harum quos a dignissimos. Keyboard Shortcuts Help : F1 or ? Previous Page : ← + CTRL (Windows) : ← + ⌘ (Mac) Next Page : → + CTRL (Windows) : → + ⌘ (Mac) Search Site : CTRL + SHIFT + F (Windows) : ⌘ + ⇧ + F (Mac) Close Message : ESC Welcome to STAT 200! About this Course Welcome to the course notes for STAT 200: Elementary Statistics. These notes are designed and developed by Penn State's Department of Statistics and offered as open educational resources. These notes are free to use under Creative Commons license CC BY-NC 4.0. Currently enrolled? If you are a current student in this course, please see Canvas for your syllabus, assignments, lesson videos and communication from your instructor. How to enroll? If you would like to enroll and experience the entire course for credit please see 'How to enroll in a course' on the World Campus website. Course Overview Section As a time expectation, most students require 8-16 hours per week to be successful in the course. This is comparable to the resident version of the course where students meet with an instructor 4 hours per week in a classroom plus time outside of class completing assignments and preparing for exams. We find that the most successful students in this course are those who do not wait until the last minute to complete assignments and those who post questions on the discussion boards. Students are encouraged to make a weekly schedule for themselves. Below is an example from a student whose assignments are all due Tuesday night: Weekly Schedule Example \| Weekday \| Activity \| --- \| \| Wednesday \| Work through the assigned readings Start the WileyPLUS assignment \| \| Thursday \| Continue to work through the readings and the WileyPLUS assignment \| \| Friday \| Finish the WileyPLUS assignment \| \| Saturday \| Relax Catch up / get ahead \| \| Sunday \| Start the lab assignment \| \| Monday \| Finish the lab assignment Take the quiz \| \| Tuesday \| Confirm that all assignments have been submitted Relax \| This course uses the textbook Statistics: Unlocking the Power of Data by Lock, Frazier Lock, Lock Morgan, Lock, & Lock. (All images used in this website are from the public domain unless otherwise noted.)
6266	https://en.wikipedia.org/wiki/Power_series	Jump to content Search Contents (Top) 1 Examples 1.1 Polynomial 1.2 Geometric series, exponential function and sine 1.3 On the set of exponents 2 Radius of convergence 3 Operations on power series 3.1 Addition and subtraction 3.2 Multiplication and division 3.3 Differentiation and integration 4 Analytic functions 4.1 Behavior near the boundary 5 Formal power series 6 Power series in several variables 7 Order of a power series 8 Notes 9 References 10 External links Power series العربية Azərbaycanca Беларуская Беларуская (тарашкевіца) Български Bosanski Català Чӑвашла Čeština Dansk Deutsch Eesti Ελληνικά Español Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Bahasa Indonesia Íslenska Italiano עברית ಕನ್ನಡ Magyar Nederlands 日本語 Norsk bokmål Norsk nynorsk Oʻzbekcha / ўзбекча Polski Português Romnă Русский Simple English Slovenščina Српски / srpski Suomi Svenska Tagalog தமிழ் Türkçe Українська اردو 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia For other uses, see Power series (disambiguation). Infinite sum of monomials In mathematics, a power series (in one variable) is an infinite series of the form where represents the coefficient of the nth term and c is a constant called the center of the series. Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. In many situations, the center c is equal to zero, for instance for Maclaurin series. In such cases, the power series takes the simpler form The partial sums of a power series are polynomials, the partial sums of the Taylor series of an analytic function are a sequence of converging polynomial approximations to the function at the center, and a converging power series can be seen as a kind of generalized polynomial with infinitely many terms. Conversely, every polynomial is a power series with only finitely many non-zero terms. Beyond their role in mathematical analysis, power series also occur in combinatorics as generating functions (a kind of formal power series) and in electronic engineering (under the name of the Z-transform). The familiar decimal notation for real numbers can also be viewed as an example of a power series, with integer coefficients, but with the argument x fixed at 1⁄10. In number theory, the concept of p-adic numbers is also closely related to that of a power series. Examples [edit] Polynomial [edit] Every polynomial of degree d can be expressed as a power series around any center c, where all terms of degree higher than d have a coefficient of zero. For instance, the polynomial can be written as a power series around the center as or around the center as One can view power series as being like "polynomials of infinite degree", although power series are not polynomials in the strict sense. Geometric series, exponential function and sine [edit] The geometric series formula which is valid for , is one of the most important examples of a power series, as are the exponential function formula and the sine formula valid for all real x. These power series are examples of Taylor series (or, more specifically, of Maclaurin series). On the set of exponents [edit] Negative powers are not permitted in an ordinary power series; for instance, is not considered a power series (although it is a Laurent series). Similarly, fractional powers such as are not permitted; fractional powers arise in Puiseux series. The coefficients must not depend on , thus for instance is not a power series. Radius of convergence [edit] A power series is convergent for some values of the variable x, which will always include x = c since and the sum of the series is thus for x = c. The series may diverge for other values of x, possibly all of them. If c is not the only point of convergence, then there is always a number r with 0 < r ≤ ∞ such that the series converges whenever \|x – c\| < r and diverges whenever \|x – c\| > r. The number r is called the radius of convergence of the power series; in general it is given as or, equivalently, This is the Cauchy–Hadamard theorem; see limit superior and limit inferior for an explanation of the notation. The relation is also satisfied, if this limit exists. The set of the complex numbers such that \|x – c\| < r is called the disc of convergence of the series. The series converges absolutely inside its disc of convergence and it converges uniformly on every compact subset of the disc of convergence. For \|x – c\| = r, there is no general statement on the convergence of the series. However, Abel's theorem states that if the series is convergent for some value z such that \|z – c\| = r, then the sum of the series for x = z is the limit of the sum of the series for x = c + t (z – c) where t is a real variable less than 1 that tends to 1. Operations on power series [edit] Addition and subtraction [edit] When two functions f and g are decomposed into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if and then The sum of two power series will have a radius of convergence of at least the smaller of the two radii of convergence of the two series, but possibly larger than either of the two. For instance it is not true that if two power series and have the same radius of convergence, then also has this radius of convergence: if and , for instance, then both series have the same radius of convergence of 1, but the series has a radius of convergence of 3. Multiplication and division [edit] With the same definitions for and , the power series of the product and quotient of the functions can be obtained as follows: The sequence is known as the Cauchy product of the sequences and . For division, if one defines the sequence by then and one can solve recursively for the terms by comparing coefficients. Solving the corresponding equations yields the formulae based on determinants of certain matrices of the coefficients of and Differentiation and integration [edit] Once a function is given as a power series as above, it is differentiable on the interior of the domain of convergence. It can be differentiated and integrated by treating every term separately since both differentiation and integration are linear transformations of functions: Both of these series have the same radius of convergence as the original series. Analytic functions [edit] Main article: Analytic function A function f defined on some open subset U of R or C is called analytic if it is locally given by a convergent power series. This means that every a ∈ U has an open neighborhood V ⊆ U, such that there exists a power series with center a that converges to f(x) for every x ∈ V. Every power series with a positive radius of convergence is analytic on the interior of its region of convergence. All holomorphic functions are complex-analytic. Sums and products of analytic functions are analytic, as are quotients as long as the denominator is non-zero. If a function is analytic, then it is infinitely differentiable, but in the real case the converse is not generally true. For an analytic function, the coefficients an can be computed as where denotes the nth derivative of f at c, and . This means that every analytic function is locally represented by its Taylor series. The global form of an analytic function is completely determined by its local behavior in the following sense: if f and g are two analytic functions defined on the same connected open set U, and if there exists an element c ∈ U such that f(n)(c) = g(n)(c) for all n ≥ 0, then f(x) = g(x) for all x ∈ U. If a power series with radius of convergence r is given, one can consider analytic continuations of the series, that is, analytic functions f which are defined on larger sets than { x \| \|x − c\| < r } and agree with the given power series on this set. The number r is maximal in the following sense: there always exists a complex number x with \|x − c\| = r such that no analytic continuation of the series can be defined at x. The power series expansion of the inverse function of an analytic function can be determined using the Lagrange inversion theorem. Behavior near the boundary [edit] The sum of a power series with a positive radius of convergence is an analytic function at every point in the interior of the disc of convergence. However, different behavior can occur at points on the boundary of that disc. For example: Divergence while the sum extends to an analytic function: has radius of convergence equal to and diverges at every point of . Nevertheless, the sum in is , which is analytic at every point of the plane except for . Convergent at some points divergent at others: has radius of convergence . It converges for , while it diverges for . Absolute convergence at every point of the boundary: has radius of convergence , while it converges absolutely, and uniformly, at every point of due to Weierstrass M-test applied with the hyper-harmonic convergent series . Convergent on the closure of the disc of convergence but not continuous sum: Sierpiński gave an example of a power series with radius of convergence , convergent at all points with , but the sum is an unbounded function and, in particular, discontinuous. A sufficient condition for one-sided continuity at a boundary point is given by Abel's theorem. Formal power series [edit] Main article: Formal power series In abstract algebra, one attempts to capture the essence of power series without being restricted to the fields of real and complex numbers, and without the need to talk about convergence. This leads to the concept of formal power series, a concept of great utility in algebraic combinatorics. Power series in several variables [edit] An extension of the theory is necessary for the purposes of multivariable calculus. A power series is here defined to be an infinite series of the form where j = (j1, …, jn) is a vector of natural numbers, the coefficients a(j1, …, jn) are usually real or complex numbers, and the center c = (c1, …, cn) and argument x = (x1, …, xn) are usually real or complex vectors. The symbol is the product symbol, denoting multiplication. In the more convenient multi-index notation this can be written where is the set of natural numbers, and so is the set of ordered n-tuples of natural numbers. The theory of such series is trickier than for single-variable series, with more complicated regions of convergence. For instance, the power series is absolutely convergent in the set between two hyperbolas. (This is an example of a log-convex set, in the sense that the set of points , where lies in the above region, is a convex set. More generally, one can show that when c=0, the interior of the region of absolute convergence is always a log-convex set in this sense.) On the other hand, in the interior of this region of convergence one may differentiate and integrate under the series sign, just as one may with ordinary power series. Order of a power series [edit] Let α be a multi-index for a power series f(x1, x2, …, xn). The order of the power series f is defined to be the least value such that there is aα ≠ 0 with , or if f ≡ 0. In particular, for a power series f(x) in a single variable x, the order of f is the smallest power of x with a nonzero coefficient. This definition readily extends to Laurent series. Notes [edit] ^ Howard Levi (1967). Polynomials, Power Series, and Calculus. Van Nostrand. p. 24. ^ Erwin Kreyszig, Advanced Engineering Mathematics, 8th ed, page 747 ^ Wacław Sierpiński (1916). "Sur une série potentielle qui, étant convergente en tout point de son cercle de convergence, représente sur ce cercle une fonction discontinue. (French)". Rendiconti del Circolo Matematico di Palermo. 41. Palermo Rend.: 187–190. doi:10.1007/BF03018294. JFM 46.1466.03. S2CID 121218640. ^ Beckenbach, E. F. (1948). "Convex functions". Bulletin of the American Mathematical Society. 54 (5): 439–460. doi:10.1090/S0002-9904-1948-08994-7. References [edit] Solomentsev, E.D. (2001) , "Power series", Encyclopedia of Mathematics, EMS Press External links [edit] Weisstein, Eric W. "Formal Power Series". MathWorld. Weisstein, Eric W. "Power Series". MathWorld. Powers of Complex Numbers by Michael Schreiber, Wolfram Demonstrations Project. \| v t e Sequences and series \| \| List of mathematical series \| \| Integer sequences \| \| \| \| --- \| \| Basic \| Arithmetic progression Geometric progression Harmonic progression Square number Cubic number Factorial Powers of two Powers of three Powers of 10 \| \| Advanced (list) \| Complete sequence Fibonacci sequence Figurate number Heptagonal number Hexagonal number Lucas number Pell number Pentagonal number Polygonal number Triangular number + array \| \| \| Properties of sequences \| Cauchy sequence Monotonic function Periodic sequence \| \| Properties of series \| \| \| \| --- \| \| Series \| Alternating Convergent Divergent Telescoping \| \| Convergence \| Absolute Conditional Uniform \| \| \| Explicit series \| \| \| \| --- \| \| Convergent \| 1/2 − 1/4 + 1/8 − 1/16 + ⋯ 1/2 + 1/4 + 1/8 + 1/16 + ⋯ 1/4 + 1/16 + 1/64 + 1/256 + ⋯ 1 + 1/2s + 1/3s + ... (Riemann zeta function) \| \| Divergent \| 1 + 1 + 1 + 1 + ⋯ 1 − 1 + 1 − 1 + ⋯ (Grandi's series) 1 + 2 + 3 + 4 + ⋯ 1 − 2 + 3 − 4 + ⋯ 1 + 2 + 4 + 8 + ⋯ 1 − 2 + 4 − 8 + ⋯ Infinite arithmetic series 1 − 1 + 2 − 6 + 24 − 120 + ⋯ (alternating factorials) 1 + 1/2 + 1/3 + 1/4 + ⋯ (harmonic series) 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ⋯ (inverses of primes) \| \| \| Kinds of series \| Taylor series Power series Formal power series Laurent series Puiseux series Dirichlet series Trigonometric series Fourier series Generating series \| \| Hypergeometric series \| Generalized hypergeometric series Hypergeometric function of a matrix argument Lauricella hypergeometric series Modular hypergeometric series Riemann's differential equation Theta hypergeometric series \| \| \| Retrieved from " Categories: Real analysis Complex analysis Multivariable calculus Series (mathematics) Hidden categories: Articles with short description Short description matches Wikidata Power series Add topic
6267	https://structurepoint.org/pdfs/manuals/spslab-manual.pdf	Revision: March-17-2025 Version 10.00 This Computer program (including software design, programming structure, graphics, manual, and on-line help) was created and published by STRUCTUREPOINT, formerly the Engineering Software Group of the Portland Cement Association (PCA) for the engineering analysis and design of reinforced concrete beams and slab floor systems. While STRUCTUREPOINT has taken every precaution to utilize the existing state-of-the-art and to assure the correctness of the analytical solution techniques used in this program, the responsibilities for modeling the structure, inputting data, applying engineering judgment to evaluate the output, and implementing engineering drawings remain with the structural engineer of record. Accordingly, STRUCTUREPOINT does and must disclaim any and all responsibility for defects or failures of structures in connection with which this program is used. Neither this manual nor any part of it may be reproduced, edited, transmitted or altered by any means electronic or mechanical or by any information storage and retrieval system, without the written permission of STRUCTUREPOINT, LLC. All products, corporate names, trademarks, service marks, and trade names referenced in this material are the property of their respective owners and are used only for identification and explanation without intent to infringe. spSlab® and spBeam® are registered trademarks of STRUCTUREPOINT, LLC. Copyright © 1988 – 2024, STRUCTUREPOINT, LLC. All Rights Reserved. \| 1 \| Chapter 1: INTRODUCTION 1.1. Program Features ............................................................................................................. 14 1.2. Program Capacity ............................................................................................................ 16 1.3. System Installation Requirements ................................................................................... 17 1.4. Terms & Conventions ..................................................................................................... 18 Chapter 2: SOLUTION METHODS 2.1. Introduction ..................................................................................................................... 20 2.1.1. Slab Systems ............................................................................................................ 21 2.1.1.1. Two-Way System ............................................................................. ...21 2.1.1.2. One-Way System ........................................................................................... 24 2.1.2. Coordinate Systems ................................................................................................. 27 2.2. Codes and Standards Provisions ...................................................................................... 30 2.2.1. Code Checks ............................................................................................................ 30 2.2.1.1. Geometry Considerations .............................................................................. 30 Minimum Thickness – One-Way Construction ................................................. 30 Minimum Slab Thickness – Two-Way Construction ........................... ...31 Drop Panel Dimensions ........................................................................ ...36 Beam Dimensions .............................................................................................. 38 2.2.1.2. Material Considerations ................................................................................ 39 2.2.1.3. Loading Considerations ................................................................................. 42 Load Cases and Load Combinations .................................................................. 42 Load Patterns ..................................................................................................... 42 Off-Centered Loads for Two-Way ........................................................ ...43 Loads Due to Openings ......................................................................... ...44 2.2.1.4. Special Considerations for Joist Systems ...................................................... 45 Rib Dimensions .................................................................................................. 45 Minimum Thickness for Joist Systems .............................................................. 45 Joist System Analysis and Design ..................................................................... 45 \| 2 \| 2.2.2. Geometry Checks ..................................................................................................... 49 2.2.2.1. Slabs .............................................................................................................. 49 2.2.2.2. Drop Panels ...................................................................................... ...49 2.2.2.3. Column Capitals ............................................................................................ 49 2.2.3. Definitions and Assumptions ................................................................................... 51 2.3. Analysis Methods ............................................................................................................ 52 2.3.1. Overview of Equivalent Frame Method ..................................................... ...52 2.3.1.1. Stiffness Characteristics ................................................................... ...53 2.3.1.2. Slab Beams ....................................................................................... ...54 2.3.1.3. Columns ............................................................................................ ...54 2.3.2. Modeling of Supports .............................................................................................. 64 2.3.3. Modeling of Loads ................................................................................................... 65 2.3.3.1. Self-Weight ................................................................................................... 65 2.3.3.2. Superimposed Loading .................................................................................. 65 2.3.3.3. Lateral Loading ............................................................................................. 66 2.3.3.4. Load Patterns ................................................................................................. 67 2.3.4. Column and Middle Strip Widths ............................................................... ...70 2.3.5. Strip Design Moments ................................................................................ ...74 2.3.5.1. ACI 318 and CSA A23.3-94 ......................................................................... 75 2.3.5.2. CSA A23.3-14/04 .......................................................................................... 79 2.3.6. Moment Redistribution ............................................................................................ 81 2.3.7. Shear Analysis of Slabs .............................................................................. ...83 2.3.7.1. Critical Section ................................................................................. ...87 For Interior Supports of Interior Frames ............................................... ...87 For Exterior Supports of Interior Frames.............................................. ...88 For Interior Supports of Exterior Frames.............................................. ...90 For Exterior Supports of Exterior Frames ............................................ ...91 2.3.7.2. Computation of Allowable Shear Stress at Critical Section ............. ...92 2.3.7.3. Computation of Factored Shear Force at Critical Section ................ ...94 \| 3 \| 2.3.7.4. Computation of Unbalanced Moment at Critical Section ................ ...95 2.3.7.5. Computation of Shear Stresses at Critical Section ........................... ...96 2.3.7.6. Shear Resistance at Corner Columns ............................................... ...98 2.3.7.7. Shear Resistance in Slab Bands ........................................................ ...99 2.3.8. One-Way Shear Analysis of Longitudinal Beams and Slabs................................. 100 2.3.8.1. Shear Calculations for ACI 318 and CSA A23.3-94 ................................... 101 2.3.8.2. Shear Calculation for CSA A23.3-14/04 ..................................................... 105 2.3.8.3. Shear Distribution ........................................................................................ 108 2.3.8.4. One-Way Shear in Slab Bands (CSA A23.3-14/04) ................................... 109 2.3.8.5. Shear in Drop Panels ................................................................................... 109 2.3.9. Torsion and Shear .................................................................................................. 110 2.3.9.1. Additional Longitudinal Reinforcement for ACI 318 and CSA A23.3-94 . 115 2.3.9.2. Additional Longitudinal Reinforcement for CSA A23.3-14/04 .................. 116 2.3.9.3. Investigation Mode ...................................................................................... 118 2.3.10. Deflections ............................................................................................................. 119 2.3.10.1. Instantaneous Deflections ............................................................................ 119 2.3.10.2. Cracking ...................................................................................................... 120 2.3.10.3. Long-Term Deflections ............................................................................... 122 2.3.10.4. Deflections of Two-Way Systems .................................................. ...124 2.4. Design Methods ............................................................................................................. 128 2.4.1. Area of Reinforcement ........................................................................................... 128 2.4.1.1. Design for Combined Flexure, Shear, and Torsion ..................................... 133 2.4.2. Concentration and Additional Reinforcement .......................................... ...135 2.5. Detailing Provisions ...................................................................................................... 138 2.5.1. Reinforcement Detailing ........................................................................................ 138 2.5.1.1. Minimum Clear Spacing ............................................................................. 138 2.5.1.2. Development Length Computation ............................................................. 142 2.5.2. Structural Integrity Reinforcement ........................................................................ 145 2.5.3. Corner Reinforcement ............................................................................... ...146 2.6. Special Topics ............................................................................................................... 147 \| 4 \| 2.6.1. Lateral Load Effects & Restraint ........................................................................... 147 2.6.2. Loads Along the Span Imposed by Lateral Loading ............................................. 149 2.6.3. Gravity Frame Analysis under Sidesway ............................................................... 150 2.6.4. Openings in Concrete Slabs ...................................................................... ...151 2.6.5. Support Conditions ................................................................................................ 152 2.6.6. Moment Redistribution .......................................................................................... 153 2.6.7. Default Load Assignment Impact on Deflections .................................................. 155 2.6.8. Reinforcing Bar Arrangement Impact on Deflections ........................................... 156 2.6.9. Design of Doubly Reinforced Beam Sections ....................................................... 157 2.6.10. Waffle Slabs Analysis & Design Approach........................................................... 158 2.6.11. Material Quantities................................................................................................. 159 2.7. References ..................................................................................................................... 160 Chapter 3: PROGRAM INTERFACE 3.1. Start Screen .................................................................................................................... 163 3.2. Main Program Window ................................................................................................. 165 3.2.1. Quick Access Toolbar ............................................................................................ 165 3.2.2. Title Bar ................................................................................................................. 165 3.2.3. Ribbon .................................................................................................................... 166 3.2.4. Left Panel ............................................................................................................... 167 3.2.5. Left Panel Toolbar ................................................................................................. 167 3.2.6. Viewport ................................................................................................................ 168 3.2.7. View Controls ........................................................................................................ 169 3.2.8. Status Bar ............................................................................................................... 169 3.3. Tables Window .............................................................................................................. 170 3.3.1. Toolbar ................................................................................................................... 171 3.3.2. Explorer Panel ........................................................................................................ 173 3.4. Reporter Window .......................................................................................................... 174 3.4.1. Toolbar ................................................................................................................... 175 3.4.2. Export / Print Panel ................................................................................................ 177 3.4.3. Explorer Panel ........................................................................................................ 179 \| 5 \| 3.5. Print/Export Window .................................................................................................... 180 3.5.1. Toolbar ................................................................................................................... 181 3.5.2. Export / Print Panel ................................................................................................ 182 Chapter 4: MODELING METHODS 4.1. Model Creation Concepts .............................................................................................. 184 4.1.1. Physical Modeling Terminology............................................................................ 185 4.1.2. Structural Elements ................................................................................................ 186 4.1.3. Properties ............................................................................................................... 187 4.1.4. Input Preparation .................................................................................................... 188 4.1.5. Modeling Considerations ....................................................................................... 190 4.1.5.1. Sway and Non-Sway Considerations .......................................................... 190 4.1.5.2. Unbalanced Moments in Column Design ................................................... 191 4.1.5.3. One-Way Slabs on Transverse Beams ........................................................ 193 4.1.5.4. Boundary Condition Effects on Continuous Beam Deflections .................. 195 4.1.5.5. Design of New Buildings vs. Investigation of Existing Buildings .............. 197 4.1.5.6. Modeling and Design of T-Beams .............................................................. 198 4.1.5.7. One-Way Slab System Considerations ........................................................ 200 4.2. Model Editing Concepts ................................................................................................ 201 4.2.1. Editing Elements .................................................................................................... 201 4.2.2. Span Length and Grid Spacing .............................................................................. 203 4.2.3. Changing Floor System Location .......................................................................... 205 4.2.4. Editing Loads ......................................................................................................... 206 4.2.5. Managing Load Locations ..................................................................................... 207 Chapter 5: MODEL DEVELOPMENT 5.1. Opening Existing Models .............................................................................................. 212 5.2. Creating New Models .................................................................................................... 213 5.2.1. Project Information ................................................................................................ 213 5.2.2. Structural Grids ...................................................................................................... 214 \| 6 \| 5.2.2.1. Working with Grids ..................................................................................... 214 5.2.2.2. Generating Spans ......................................................................................... 215 5.2.2.3. Using the Span Table .................................................................................. 216 5.2.2.4. Working with Frame Location .................................................................... 217 5.2.2.5. Grids Display Options ................................................................................. 218 5.2.3. Generating Definitions ........................................................................................... 219 5.2.3.1. Materials ...................................................................................................... 220 Concrete ........................................................................................................... 220 Reinforcing Steel ............................................................................................. 222 5.2.3.2. Reinforcement Criteria ................................................................................ 223 Slabs & Ribs .................................................................................................... 223 Beams ............................................................................................................... 225 Beam Stirrups................................................................................................... 227 Bar Set .............................................................................................................. 229 5.2.3.3. Options ........................................................................................................ 230 Design & Modeling.......................................................................................... 230 5.2.3.4. Load Case / Combo. .................................................................................... 231 Load Cases ....................................................................................................... 231 Load Combinations .......................................................................................... 233 5.2.4. Creating Model Elements ...................................................................................... 234 5.2.4.1. Spans Elements ............................................................................................ 234 Slab .................................................................................................................. 235 Longitudinal Beam........................................................................................... 235 Rib .................................................................................................................... 236 Longitudinal Slab Bands ..................................................................... ...237 Strip Moment Distribution Factors ..................................................... ...238 5.2.4.2. Supports Elements ....................................................................................... 239 Column ............................................................................................................. 240 Drop Panel .......................................................................................... ...241 Column Capital ................................................................................................ 244 Transverse Beams ............................................................................................ 245 \| 7 \| Transverse Slab Bands ........................................................................ ...246 Restraint ........................................................................................................... 247 Moment Redistribution .................................................................................... 248 5.2.4.3. Loads ........................................................................................................... 249 Assigning Area Loads ...................................................................................... 250 Assigning Line Loads ...................................................................................... 251 Assigning Point Loads ..................................................................................... 253 Assigning Support Loads and Displacements.................................................. 255 Assigning Lateral Load Effects ....................................................................... 257 5.2.4.4. Rebars .......................................................................................................... 258 Column Strip Rebars for Two-Way Slab Systems ............................. ...258 Middle Strip Rebars for Two-Way Slab Systems ............................... ...260 Beam Rebars for Two-Way Slab Systems .......................................... ...261 Beam Stirrups for Two-Way Slab Systems ........................................ ...262 Flexure Rebars for Beams and One-Way Slab Systems .................................. 264 Stirrups for Beams and One-Way Slab Systems.............................................. 265 Torsional Longitudinal Rebars for Beams ....................................................... 267 5.2.5. Editing Model Elements ........................................................................................ 268 5.2.5.1. Using the Left Panel .................................................................................... 268 Spans ................................................................................................................ 268 Supports ........................................................................................................... 270 5.2.5.2. Using the Left Panel Toolbar ...................................................................... 271 Delete ............................................................................................................... 271 Move ................................................................................................................ 271 Duplicate .......................................................................................................... 271 Advance Copy .................................................................................................. 271 5.2.5.3. Using the Right Click Menu at Viewport .................................................... 273 5.3. Modeling with Templates .............................................................................................. 274 5.3.1. Utilizing Templates ................................................................................................ 274 5.3.2. Template Ribbon .................................................................................................... 276 \| 8 \| 5.3.2.1. New Pattern ................................................................................................. 276 5.3.2.2. Discard & Exit ............................................................................................. 276 5.3.2.3. Save & Exit ................................................................................................. 276 5.3.3. Template Left Panel ............................................................................................... 277 5.3.4. Types of Templates ................................................................................................ 279 5.3.4.1. Two-Way Systems ......................................................................... ...279 5.3.4.2. One-Way Systems ....................................................................................... 284 5.4. Utilizing Predefined Examples ...................................................................................... 288 Chapter 6: MODEL SOLUTION 6.1. Design Options .............................................................................................................. 290 6.1.1. Reinforcement ........................................................................................................ 290 6.1.2. Shear Design .......................................................................................................... 291 6.1.3. Torsion Analysis and Design ................................................................................. 291 6.1.4. Beam Design .......................................................................................................... 291 6.1.5. Live Loads ............................................................................................................. 292 6.2. Deflection Options ........................................................................................................ 293 6.2.1. Section Used .......................................................................................................... 293 6.2.2. Ig & Mcr Calculation ............................................................................................... 293 6.2.3. Long-Term Deflection ........................................................................................... 294 6.3. Running the Model ........................................................................................................ 295 Chapter 7: MODEL OUTPUT 7.1. Tabular Output .............................................................................................................. 298 7.1.1. Input Echo .............................................................................................................. 299 7.1.1.1. General Information .................................................................................... 299 7.1.1.2. Solve Options .............................................................................................. 299 7.1.1.3. Material Properties ...................................................................................... 299 7.1.1.4. Reinforcement Database .............................................................................. 299 7.1.1.5. Span Data .................................................................................................... 299 \| 9 \| 7.1.1.6. Support Data ................................................................................................ 300 7.1.1.7. Load Data .................................................................................................... 300 7.1.1.8. Reinforcement Criteria ................................................................................ 300 7.1.1.9. Reinforcing Bars .......................................................................................... 300 7.1.2. Design Results ....................................................................................................... 301 7.1.2.1. Solver Messages .......................................................................................... 301 7.1.2.2. Moment Redistribution Factors ................................................................... 301 7.1.2.3. Strip Widths and Distribution Factors ............................................ ...301 7.1.2.4. Top Reinforcement ...................................................................................... 301 7.1.2.5. Top Bar Details and Development Lengths ................................................ 302 7.1.2.6. Band Reinforcement at Supports .................................................... ...302 7.1.2.7. Bottom Reinforcement ................................................................................ 303 7.1.2.8. Bottom Bar Details and Development Lengths ........................................... 303 7.1.2.9. Flexural Capacity ........................................................................................ 304 7.1.2.10. Longitudinal Beam Combined M-V-T Capacity ......................................... 304 7.1.2.11. Longitudinal Beam Transverse Reinforcement Capacity ........................... 304 7.1.2.12. Longitudinal Beam Transverse Reinforcement Demand and Capacity ...... 304 7.1.2.13. Beam Shear (and Torsion) Capacity ........................................................... 304 7.1.2.14. Longitudinal Beam Shear and Torsion Reinforcement Required ............... 305 7.1.2.15. Longitudinal Slab Band Shear Capacity ........................................ ...305 7.1.2.16. Slab Shear Capacity ........................................................................ ...305 7.1.2.17. Flexural Transfer of Negative Unbalanced Moments .................... ...305 7.1.2.18. Flexural Transfer of Positive Moments .......................................... ...306 7.1.2.19. Punching Shear Around Columns .................................................. ...306 7.1.2.20. Punching Shear Around Drops ....................................................... ...307 7.1.2.21. Integrity Reinforcement at Supports .............................................. ...307 7.1.2.22. Corner Reinforcement .................................................................... ...307 7.1.2.23. Shear Resistance at Corner Columns ............................................. ...307 7.1.2.24. Material Takeoff .......................................................................................... 308 7.1.3. Deflection Results: Summary ................................................................................ 309 \| 10 \| 7.1.3.1. Section Properties ........................................................................................ 309 7.1.3.2. Instantaneous Deflections ............................................................................ 309 7.1.3.3. Long-Term Deflections ............................................................................... 309 7.1.4. Detailed Results ..................................................................................................... 310 7.1.4.1. Column Forces and Redistributed Column Forces ...................................... 310 7.1.4.2. Non – Redistributed and Redistributed Internal Forces: M – V .................. 310 7.1.4.3. Internal Forces: T ........................................................................................ 311 7.1.4.4. Deflections – Load Cases ............................................................................ 312 7.1.4.5. Required Reinforcement .............................................................................. 312 7.2. Graphical Output ........................................................................................................... 313 7.2.1. Internal Forces ....................................................................................................... 313 7.2.2. Moment Capacity ................................................................................................... 313 7.2.3. Shear Capacity ....................................................................................................... 313 7.2.4. Deflection ............................................................................................................... 313 7.2.5. Reinforcement ........................................................................................................ 313 7.2.6. Diagrams Display Options ..................................................................................... 314 7.2.7. Viewing Aids ......................................................................................................... 314 7.2.7.1. Multiple Viewports ...................................................................................... 315 7.2.7.2. View Controls .............................................................................................. 316 7.2.7.3. Display Options ........................................................................................... 317 Chapter 8: EXAMPLES 8.1. Example 1 – Spandrel Beam with Moment Redistribution ........................................... 321 8.1.1. Problem Formulation ............................................................................................. 321 8.1.2. Preparing the Input ................................................................................................. 323 8.1.3. Assigning Spans ..................................................................................................... 334 8.1.4. Assigning Supports ................................................................................................ 335 8.1.5. Assigning Loads..................................................................................................... 337 8.1.6. Solving ................................................................................................................... 339 8.1.7. Viewing and Printing Results ................................................................................ 342 8.2. Example 2 – Spandrel Beam with Torsion .................................................................... 349 \| 11 \| 8.2.1. Problem Formulation ............................................................................................. 349 8.2.2. Preparing the Input ................................................................................................. 351 8.2.3. Assigning Spans ..................................................................................................... 363 8.2.4. Assigning Supports ................................................................................................ 365 8.2.5. Assigning Loads..................................................................................................... 367 8.2.6. Solving ................................................................................................................... 375 8.2.7. Viewing and Printing Results ................................................................................ 378 8.3. Example 3 – Design of a Continuous Beam .................................................................. 389 8.3.1. Problem Formulation ............................................................................................. 389 8.3.2. Preparing the Input ................................................................................................. 391 8.3.3. Assigning Spans ..................................................................................................... 402 8.3.4. Assigning Supports ................................................................................................ 403 8.3.5. Assigning Loads..................................................................................................... 404 8.3.6. Solving ................................................................................................................... 406 8.3.7. Viewing and Printing Results ................................................................................ 409 8.4. Example 4 – Flat Plate Floor System ............................................................... ...416 8.4.1. Problem Formulation ............................................................................................. 416 8.4.2. Preparing the Input ................................................................................................. 418 8.4.3. Assigning Spans ..................................................................................................... 427 8.4.4. Assigning Supports ................................................................................................ 428 8.4.5. Assigning Loads..................................................................................................... 430 8.4.6. Solving ................................................................................................................... 433 8.4.7. Viewing and Printing Results ................................................................................ 436 8.5. Example 5 – Two-way Slab System ................................................................. ...443 8.5.1. Problem Formulation ............................................................................................. 443 8.5.2. Preparing the Input ................................................................................................. 445 8.5.3. Assigning Spans ..................................................................................................... 456 8.5.4. Assigning Supports ................................................................................................ 458 8.5.5. Assigning Loads..................................................................................................... 462 8.5.6. Solving ................................................................................................................... 465 8.5.7. Viewing and Printing Results ................................................................................ 468 \| 12 \| Chapter: APPENDIX A.1. Default Load Case and Combination Factors ................................................................ 476 A.1.1. For ACI 318-14/11 ................................................................................................. 477 A.1.2. For ACI 318-08/05/02............................................................................................ 478 A.1.3. For ACI 318-99 ...................................................................................................... 479 A.1.4. For CSA A23.3-14 ................................................................................................. 480 A.1.5. For CSA A23.3-04 ................................................................................................. 481 A.1.6. For CSA A23.3-94 ................................................................................................. 482 A.2. Conversion Factors – English to SI ............................................................................... 483 A.3. Conversion Factors – SI to English ............................................................................... 484 A.4. Technical Resources ...................................................................................................... 485 A.5. Contact Information ...................................................................................................... 486 \| 13 \| 1. INTRODUCTION Formerly pcaSlab and ADOSS, spSlab is a computer software program for the analysis and design of reinforced concrete beams and slab floor systems. Two-way slab systems are analyzed using the Equivalent Frame Method. Beams and frames of up to 22 spans can be analyzed and designed. In addition to the design option, spBeam and spSlab have the capability of investigating existing beams and slab systems. spSlab includes provisions for slab band systems as well as punching shear check and deflection calculations using cracked or gross sections. For beams, moment redistribution, as well as combined shear and torsion design, are available. In addition to the required area of reinforcing steel at the critical sections, spSlab provides a complete reinforcing bar schedule that includes the number of bars and bar sizes and lengths. spSlab checks the applicable provisions of the selected code. Formerly pcaBeam, spBeam is a limited version of spSlab. It includes all elements that apply to beams and one-way slab systems. Two-way slab systems are available in spSlab only and topics related to two-way slab systems are denoted with icon. CHAPTER 1 INTRODUCTION \| 14 \| 1.1. Program Features • Code support for ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, ACI 318-02, and ACI 318-99 • Code support for CSA A23.3-14, CSA A23.3-04, and CSA A23.3-94 • English (U.S.) and Metric (SI) unit systems • Design and investigation of beams, one and two-way slabs including one-way joist systems (standard and wide module) and two-way joist systems (waffle slabs) • Slab band system design and investigation for CSA A23.3-14/04 • Flexure and shear design and investigation with live load reduction and patterning • Torsion design and investigation for beams/one-way slab systems • Longitudinal reinforcement for combined flexure, shear, and torsion per CSA A23.3-14/04 • Automatic or manual moment distribution factors and strip widths • Moment redistribution for beams/one-way slab systems • Calculation of instantaneous deflections at three load levels; dead load, dead load plus sustained live load, and dead load plus live load • Calculation of incremental long-term deflections • Instantaneous and long-term design strip deflections for two-way systems • Analytical modeling of variable support stiffness in systems with rectangular, and circular supports • One and two-way (punching) shear investigation considering the effects of drop panels, column capitals, longitudinal beams, transverse beams, and slab bands. • Boundary conditions including vertical and rotational springs • Top and bottom bar details including development lengths and material quantities • Specialty design requirements including crack control, integrity reinforcing, and corner column checking • Mixed span types within one-way or two-way systems • Import input data from PCA-Beam and pcaSlab • Object-based modeling of two-way and one-way systems with a full featured graphical interface • Templates of predefined and loaded models allowing the user to select and generate quick INTRODUCTION \| 15 \| models for two-way systems (flat plate, flat slab, slab on beams, and two-way joist) and one-way systems (one-way slab, one-way joist, rectangular beams, and flanged beams) • Structural grids to facilitate the placement of structural members in plan view • Point, uniform line, variable line, and uniform area load types to model typical slab and beam loading conditions • Automatic computation of element self-weights with the option to include or exclude them in the analysis • Isometric (3D) view of the modeled system with ability to view grids, loads and other model features in typical CAD environment in multi viewports with up to 6 concurrent views • Data validation during input • Frame solution results including tabulated column axial forces and moments • Diagrams to visualize the analysis and design results • Print/Export module for viewing, customizing, and exporting screenshots • Tables module for viewing and exporting input and output data • Reporter module for customizing, generating, viewing, exporting, and printing results • User-controlled screen display settings including a full color palette • Ability to save defaults and settings for future input sessions • Detailed manual and online resources INTRODUCTION \| 16 \| 1.2. Program Capacity • 21 supports (22 spans including left and right cantilevers) • 6 load cases • 50 load combinations • 999 partial dead loads per case • 999 partial live loads per case • 2 top bar layers (Design mode) • 2 bottom bar layers (Design mode) • 15 bar sets per span INTRODUCTION \| 17 \| 1.3. System Installation Requirements Any computer running Microsoft Windows 10 or Windows 11 operating system is sufficient to run the spSlab and spBeam programs provided that .NET 7.0 is installed. If it is not detected by the installation program, then it will be installed automatically. The actual program capacity depends on system resources available on the computer on which spSlab and spBeam are running. To solve models with the maximum number of supports and load combinations, a 64-bit operating system with at least 8GB of RAM is required. It is recommended to run the model on the local computer hard drive for fastest response. For instructions on how to purchase, download, install, license, and troubleshoot issues, please refer to support pages on the StructurePoint website at Structurepoint.org/slg.asp. INTRODUCTION \| 18 \| 1.4. Terms & Conventions The following terms are used throughout this manual. A brief explanation is given to help familiarize you with them. Windows refers to the Microsoft Windows environment as listed in System Requirements. [ ] indicates equivalent value expressed in metric unit or CSA code requirement corresponding ACI code requirement. Click on means to position the cursor on top of a designated item or location and press and release the left-mouse button (unless instructed to use the right-mouse button). Double-click on means to position the cursor on top of a designated item or location and press and release the left-mouse button twice in quick succession. Marquee select means to depress the mouse button and continue to hold it down while moving the mouse. As you drag the mouse, a rectangle (known as a marquee) follows the cursor. Release the mouse button and the area inside the marquee is selected. INTRODUCTION \| 19 \| Various styles of text and layout have been used in this manual to help differentiate between different kinds of information. The styles and layout are explained below placed in a topic header means that the topic applies to spSlab only Italic indicates a glossary item, or emphasizes a given word or phrase. Bold All bold typeface makes reference to either a menu or a menu item command such as File or Save, or a tab such as General Information or Columns. Mono-space indicates something you should enter with the keyboard. For example type “c:\filename.txt”. KEY + KEY indicates a key combination. The plus sign indicates that you should press and hold the first key while pressing the second key, then release both keys. For example, “ALT + F” indicates that you should press the “ALT” key and hold it while you press the “F” key. Then release both keys. SMALL CAPS Indicates the name of an object such as a dialog box or a dialog box component. For example, the OPEN dialog box or the CANCEL or MODIFY buttons. \| 20 \| 2. SOLUTION METHODS 2.1. Introduction spSlab and spBeam are advanced software tools used worldwide for the modeling, analysis, and design of reinforced concrete floor slab and beam systems. They are designed to handle two-way and one-way slab systems, including flat plates, flat slabs, slabs on beams, slab bands, two-way waffle slabs, one-way solid slabs, one-way ribbed slabs, rectangular beams, and flanged beams. Equipped with the American (ACI 318) and Canadian (CSA A23.3) concrete codes, spSlab and spBeam provide robust solutions for analyzing and designing conventional concrete floor systems under various loading conditions. The programs leverage sophisticated methods, such as the Equivalent Frame Method and the Matrix Stiffness Method, to perform analyses that comply with code provisions while offering flexibility in modeling and adaptability to diverse design and investigation scenarios. Their comprehensive approach includes rigorous geometry and code checks to ensure accurate and code-compliant design outcomes. Additionally, the programs flag potential issues, such as inadequate reinforcement, excessive deflections, or geometry conflicts, requiring further attention or adjustment. As industry-leading tools, spSlab and spBeam simplify complex design challenges, enabling engineers to confidently address the most demanding structural scenarios encountered in reinforced concrete buildings and structures. CHAPTER 2 SOLUTION METHODS \| 21 \| 2.1.1. Slab Systems spSlab and spBeam can be used to model, analyze, and design two-way and one-way systems such as flat plate, flat slab, slab on beams, slab bands, two-way joist slab (waffle slab), one-way slab (solid slab), one-way joist slab (ribbed slab), rectangular and flanged beams. Samples of such systems are illustrated below: 2.1.1.1. Two-Way System Flat Plate Flat Plate with Column Capitals with Spandrel Beams with Spandrel Beams & Column Capitals SOLUTION METHODS \| 22 \| Flat Slab Flat Slab with Column Capitals with Spandrel Beams with Spandrel Beams & Column Capitals Slab on Beams Two-Way Beam-Supported Slab SOLUTION METHODS \| 23 \| Slab Bands Longitudinal Bands Transverse Bands Longitudinal Bands with Column Capitals Transverse Bands with Column Capitals Two-Way Joist (Waffle) Waffle Slab Waffle Slab with Column Capitals SOLUTION METHODS \| 24 \| 2.1.1.2. One-Way System One-Way Slab (Solid) Simply Supported Fixed Ends Simple Cantilever Propped Cantilever Continuous – Pinned Ends Continuous – Fixed Ends Continuous – Column Supports Continuous – Beam Supports Column Supports – End Walls Beam Supports – End Walls SOLUTION METHODS \| 25 \| One-Way Joist (Ribbed) Joist-standard Module Joist-wide Module Rectangular Beams Simply Supported Fixed Ends Simple Cantilever Propped Cantilever Continuous – Pinned Ends Continuous – Fixed Ends Continuous – Column Supports SOLUTION METHODS \| 26 \| Flanged Beams Simply Supported Fixed Ends Simple Cantilever Propped Cantilever Continuous – Pinned Ends Continuous – Fixed Ends Continuous – Column Supports SOLUTION METHODS \| 27 \| 2.1.2. Coordinate Systems The top surface of the slab/beam lies in the XY plane of the right-handed XYZ rectangular coordinate system shown in Figure 2.1. The slab thickness (and/or beam depth) is measured in the direction of the Z-axis. When looking at the screen, the positive X-axis points horizontally to the right on the screen, the positive Y-axis points directly out of the screen towards the user, and positive Z-axis points vertically downwards on the screen. Thus, the XY plane is defined as being in the plane of the screen. Note that the loads shown in the figure are all positive and may not match the typical sign conventions. More details related to Span Loads can be found in Section 5.2.4.3. Figure 2.1 – Coordinate System SOLUTION METHODS \| 28 \| Consistent with this sign convention, the tabular results in the program output such as reinforcement values are presented for each span, beginning with the zero value of the X coordinate referred to as the Left zone and working towards the full span length in the X direction also designated as the Right zone. between the Right and Left zones of a span is the Midspan zone represents commonly important values from the structural analysis and design. Figure 2.2 – Span Zones SOLUTION METHODS \| 29 \| As a result of the coordinate system described above, results output is presented as follows for shear force diagram, bending moment diagram, and torsion force diagram. Figure 2.3 – Result Output Sign Convention SOLUTION METHODS \| 30 \| 2.2. Codes and Standards Provisions 2.2.1. Code Checks 2.2.1.1. Geometry Considerations Minimum Thickness – One-Way Construction The program checks beam or one-way slab thickness based on minimum requirement for ACI-3181 code as specified in Table 9.5(a) or for CSA code2 according to Table 9.2. For lightweight concrete with density 90 lb/ft3 ≤ wc ≤ 115 lb/ft3 [1440 kg/m3 ≤ wc ≤ 1840 kg/m3] for ACI 318-14, ACI 318-11, and ACI 318-08 90 lb/ft3 ≤ wc ≤ 120 lb/ft3 [1440 kg/m3 ≤ wc ≤ 1920 kg/m3] for ACI 318-05 90 lb/ft3 ≤ wc ≤ 120 lb/ft3 [1500 kg/m3 ≤ wc ≤ 2000 kg/m3] for ACI 318-02, and ACI 318-99, the minimum slab thickness is additionally increased by adjustment factor (1.65 – 0.005wc), but not less than 1.09. For CSA standards, the adjustment is calculated as (1.65 – 0.0003wc), but not less than 1.0, for structural low density (wc ≤ 1850 kg/m3) and structural semi-low density (1850 kg/m3 ≤ wc ≤ 2150 kg/m3) concrete3. 1 ACI 318-14,7.3.1.1.,9.3.1.1; ACI 318-11, 9.5.2.1; ACI 318-08, 9.5.2.1; ACI 318-05, 9.5.2.1; ACI 318-02, 9.5.2.1; ACI 318-99, 9.5.2.1 2 CSA A23.3-14, 9.8.2.1; CSA A23.3-04, 9.8.2.1 (Ref. ) 3 CSA A23.3-04, 2.2; CSA A23.3-94 (Ref. ), 2.1 SOLUTION METHODS \| 31 \| Minimum Slab Thickness – Two-Way Construction The program checks slab thickness against minimum slab thickness defined by design standards for two-way systems with long to short span ratio not greater than 2.04. Slabs with thickness below the minimum value will be flagged by the program, however, they are allowed provided that calculated deflections do not exceed maximum permissible computed deflections5. For two-way system with a long to short span ratio greater than 2.0, the program will calculate minimum thickness requirements based on the provisions of one-way construction including any cantilevered spans. Minimum thickness of slabs with beams spanning between supports on all sides is calculated for ACI 318 codes in US customary units from6: ( ) 0.8 200,000 5 in. if 0.2 2.0 36 5 0.2 0.8 200,000 3.5 in. if 2.0 36 9 y n fm fm y n fm f l h f l        +           + −  =     +         +   Eq. 2-1 4 ACI 318-14, 8.3.1.1; ACI 318-11, 9.5.3.1, 13.6.1.2; ACI 318-08, 9.5.3.1, 13.6.1.2; ACI 318-05, 9.5.3.1, 13.6.1.2; ACI 318-02, 9.5.3.1, 13.6.1.2; ACI 318-99, 9.5.3.1, 13.6.1.2; CSA A23.3-14, 13.2.2, 2.2; CSA A23.3-04, 13.2.2, 2.2; CSA A23.3-94, 13.3.2, 13.1 5 ACI 318-14, 8.3.2.1; ACI 318-11, 9.5.3.4; ACI 318-08, 9.5.3.4; ACI 318-05, 9.5.3.4; ACI 318-02, 9.5.3.4; ACI 318-99, 9.5.3.4; CSA A23.3-14, 13.2.7; CSA A23.3-04, 13.2.7; CSA A23.3-94, 13.3.6 6 ACI 318-14, 8.3.1.2; ACI 318-11, 9.5.3.3; ACI 318-08, 9.5.3.3; ACI 318-05, 9.5.3.3; ACI 318-02, 9.5.3.3; ACI 318-99, 9.5.3.3 SOLUTION METHODS \| 32 \| in metric unit system for ACI 318-14, ACI 318-11, ACI 318-08, and ACI 318-05 from7: ( ) 0.8 1,400 125 mm if 0.2 2.0 36 5 0.2 0.8 1,400 90 mm if 2.0 36 9 y n fm fm y n fm f l h f l        +           + −  =     +         +   Eq. 2-2 and in metric unit system for ACI 318-02 and ACI 318-99 from8: ( ) 0.8 1,500 120 mm if 0.2 2.0 36 5 0.2 0.8 1,500 90 mm if 2.0 36 9 y n fm fm y n fm f l h f l        +           + −  =     +         +   Eq. 2-3 where: ln = longer clear span measured face-to-face of beams, β = ratio of the clear spans in long to short direction, fy = yield stress of reinforcing steel, αfm = average value of αf, the ratio of flexural stiffness of a beam section to the flexural stiffness of a width of slab bounded laterally by centerlines of adjacent panels on either side of the beam, for all beams supporting the edges of a slab panel. 7 ACI 318M-11, 9.5.3.3; ACI 318M-08, 9.5.3.3; ACI 318M-05, 9.5.3.3 8 ACI 318M-02, 9.5.3.3; ACI 318M-99, 9.5.3.3 SOLUTION METHODS \| 33 \| The program assumes that beams are present on all sides of a panel if the span under consideration includes a longitudinal beam and there are transverse beams defined at both ends of the span. If this assumption is satisfied but in reality beams are not present on all sides (e.g. design strip next to the one under consideration has no longitudinal beam) then the user is advised to check deflections even if slab thickness is larger than the minimum slab thickness reported by the program. For the design of ACI slabs without beams (αfm ≤ 0.2) spanning between interior supports the minimum thickness shall conform to ACI 318 Table 9.5(c) and will not be less than 5.0 in. [125 mm for ACI 318M-11/08/05 or 120 mm for ACI 318M-02/99] for flat plates (slabs without drop panel) and not less than 4.0 in. [100 mm] for two-way flat slab systems (slab with drop panels)9. For flat slabs that contain valid drop panels (see Figure 2.4), Table 9.5(c) reduces the minimum thickness by approximately 10%. For values of fy between the ones given in the table, minimum thickness is determined by linear interpolation. For design strips that have neither beams between all supports nor beams between interior supports (e.g. exterior strips with beams on the outside edges only), the program reports maximum value of minimum slab thickness resulting from both Table 9.5(c) and Equations. However, since this case is not explicitly covered by the ACI code, the user is advised to check deflections even if slab thickness is larger than the minimum slab thickness reported by the program. 9 ACI 318-14, 8.3.1.1; ACI 318-11, 9.5.3.2; ACI 318-08, 9.5.3.2; ACI 318-05, 9.5.3.2; ACI 318-02, 9.5.3.2; ACI 318-99, 9.5.3.2; ACI 318M-08, 9.5.3.2; ACI 318M-05, 9.5.3.2; ACI 318M-02, 9.5.3.2; ACI 318M-99, 9.5.3.2 SOLUTION METHODS \| 34 \| For CSA A23.3 standard10, the minimum thickness of slab with beams spanning between all supports is: 0.6 1,000 taken 2.0 30 4 y n s m m f l h     +       + Eq. 2-4 with the value of αm evaluated for CSA A23.3-04 using the following beam moment of inertia: 3 2.5 1 12 w s b b h h I h   = −     Eq. 2-5 For flat plates and slabs with column capitals11, the minimum slab thickness is: 0.6 1,000 30 y n s f l h   +      Eq. 2-6 For slabs with drop panels12, the minimum slab thickness satisfies the conditions: ( ) 0.6 1,000 CSA A23.3-94 2 30 1 y n s d d s n s f l h x h h l h   +           − +           Eq. 2-7 ( ) 0.6 1,000 2 CSA A23.3-14/04 30 y n d s h n f l x h l   +      −  Eq. 2-8 10 CSA A23.3-14, 13.2.5; CSA A23.3-04, 13.2.5; CSA A23.3-94, 13.3.5 11 CSA A23.3-14, 13.2.3; CSA A23.3-04, 13.2.3; CSA A23.3-94, 13.3.3 12 CSA A23.3-14, 13.2.4; CSA A23.3-04, 13.2.4; CSA A23.3-94, 13.3.4 SOLUTION METHODS \| 35 \| where (hd – hs) shall not be greater than hs and xd = dimension from face of column to edge of drop panel, but not more than ln / 4, 2xd /ln = the smaller of the values determined in the two directions, Δh = additional thickness of the drop panel below the soffit of the slab and shall not be taken more than hs. The minimum thickness in a span that contains a discontinuous edge will be increased by 10%, if the edge beam provided has a stiffness ratio, αf, of less than 0.8013. The first and last spans are considered to contain a discontinuous edge as well as a span that contains an exterior edge. The minimum thickness of slab bands follows the requirements for the beams14. 13 ACI 318-14, 8.3.1.2; ACI 318-11, 9.5.3.3 (d); ACI 318-08, 9.5.3.3 (d); ACI 318-05, 9.5.3.3 (d); ACI 318-02, 9.5.3.3 (d); ACI 318-99, 9.5.3.3 (d), CSA A23.3-14, 13.2.3, 13.2.4; CSA A23.3-04, 13.2.3, 13.2.4; CSA A23.3-94, 13.3.3 14 CSA A23.3-04, 13.2.6 SOLUTION METHODS \| 36 \| Drop Panel Dimensions Per ACI15, a valid drop must extend in each direction at least one-sixth the center-to-center span length in that direction (Figure 2.4). The depth of an invalid drop will not be used in the calculation of the depth used to reduce the amount of negative reinforcement required over a column16. If the valid drop depth is greater than one-quarter the distance from the edge of the drop panel to the face of the column (x) the excess depth exceeding ¼ x will not be considered in the calculation of the effective depth used to reduce the amount of negative reinforcement required at a column (Figure 2.5)17. Slabs that contain valid drops are allowed a 10% decrease in minimum slab depth18. 15 ACI 318-14, 8.2.4; ACI 318-11, 13.2.5; ACI 318-08, 13.2.5; ACI 318-05, 13.2.5; ACI 318-02, 13.3.7.1, 13.3.7.2; ACI 318-99, 13.3.7.1, 13.3.7.2 16 ACI 318-14, 8.2.4; ACI 318-11, 13.2.5; ACI 318-08, 13.2.5; ACI 318-05, 13.2.5, 13.3.7; ACI 318-02, 13.3.7; ACI 318-99, 13.3.7 17 ACI 318-14, 8.5.2.2; ACI 318-11, 13.3.7; ACI 318-08, 13.3.7; ACI 318-05, 13.3.7; ACI 318-02, 13.3.7.3; ACI 318-99, 13.3.7.3; CSA A23.3-14, 13.10.7; CSA A23.3-04, 13.10.7; CSA A23.3-94, 13.11.6 18 ACI 318-14, 8.3.1.1; ACI 318-11, 9.5.3.2; ACI 318-08, 9.5.3.2; ACI 318-05, 9.5.3.2; ACI 318-02, 9.5.3.2, ACI 318-99, 9.5.3.2 SOLUTION METHODS \| 37 \| Figure 2.4 – Valid Drop Dimensions The input drop dimensions will be used for self-weight computations, when computing slab stiffness to determine deflections, moments, shears, and when computing punching shear around a column19. Figure 2.5 – Excess Drop Depth 19 ACI 318-14, 8.2.4; ACI 318-11, R13.2.5; ACI 318-08, R13.2.5; ACI 318-05, R13.2.5 SOLUTION METHODS \| 38 \| Beam Dimensions The program follows linear distribution of strain (plain section) assumption20 for flexure design which is applicable to shallow flexural members. In case of deep beams21, design standards recommend using non-linear distribution of strain or strut-and-tie method. The program checks beam dimensions and if a beam with the following clear span, ln, to overall depth, h, ratio is found: 4 ACI 318-11, ACI 318-08, ACI 318-05 and ACI 318-02 2.5 ACI 318-99, continous spans 1.25 ACI 318-99, simple spans 2 CSA A23.3-14, CSA A23.3-04 and CSA A23.3-94 n n n n l h l h l h l h     Eq. 2-9 a warning is issued alerting the user that additional deep beam design and detailing is required. For cantilevers, the warning is issued only if their clear span is larger than overall depth. 20 ACI 318-14, 22.2.1.2; ACI 318-11, 10.2.2; ACI 318-08, 10.2.2; ACI 318-05, 10.2.2; ACI 318-02, 10.2.2; ACI 318-99, 10.2.2; CSA A23.3-14, 10.1.2; CSA A23.3-04, 10.1.2; CSA A23.3-94, 10.1.2 21 ACI 318-14, 9.9.1.2; ACI 318-11, 10.7.1; ACI 318-08, 10.7.1; ACI 318-05, 10.7.1; ACI 318-02, 10.7.1; ACI 318-99, 10.7.1; CSA A23.3-14; 10.7.1; CSA A23.3-04; 10.7.1; CSA A23.3-94, 10.7.1 SOLUTION METHODS \| 39 \| 2.2.1.2. Material Considerations By entering the concrete density and compressive strength of the members, default values for the other concrete properties are determined. The slabs/beams and columns may have different concrete properties. The density of concrete is used to determine the type of concrete, modulus of elasticity, and self-weight. The concrete type is determined in accordance with Table 2.1. Type ACI 318-14 ACI 318-11 ACI 318-08 ACI 318-05 ACI 318-02 ACI 318-99 wc wc pcf kg/m3 pcf kg/m3 Normal 135 ≤ wc 2,155 ≤ wc 130 ≤ wc 2,000 ≤ wc Sand-Lightweight 115 < wc < 135 1,840 < wc < 2,155 105 < wc < 130 1,700 < wc < 2,000 All-Lightweight wc ≤ 115 wc ≤ 1,840 wc ≤ 105 wc ≤ 1,700 Type CSA A23.3-14 CSA A23.3-04 CSA A23.3-94 γc γc kg/m3 pcf kg/m3 pcf Normal 2,150 ≤ γc 134.2 ≤ γc 2,000 ≤ γc 124.8 ≤ γc Low Density 1,850 < γc < 2,150 115.5 < γc < 134.2 1,700 < γc < 2,000 106.1 < γc < 124.8 Semi-low Density γc ≤ 1,850 γc ≤ 115.5 γc ≤ 1,700 γc ≤ 106.1 Table 2.1 – Concrete Weight Classification SOLUTION METHODS \| 40 \| Once the compressive strength of concrete fc′ is input, various parameters are set to their default values. The modulus of elasticity is computed as22: 1.5 33 c c c E w f  = Eq. 2-15 where: wc = the unit weight of concrete. For CSA A23.3 standard23 ( ) 1.5 3,300 6,900 2,300 c c c E f     = +     Eq. 2-16 where: γc = the density of concrete. The square root of fc′ is limited to 100 psi for the computation of shear strength provided by concrete, Vc, and development lengths.24 For CSA A23.3-14/04 standard the value of square root of fc′ used to calculate factored shear resistance vr shall not exceed 8 MPa.25 22 ACI 318-14,19.2.2.1 (a); ACI 318-11, 8.5.1; ACI 318-08, 8.5.1; ACI 318-05, 8.5.1; ACI 318-02, 8.5.1; ACI 318-99, 8.5.1; CSA A23.3-14, 8.6.2.2; CSA A23.3-04, 8.6.2.2; CSA A23.3-94, 8.6.2.2 23 CSA A23.3-14, 8.6.2.2; CSA A23.3-04, 8.6.2.2; CSA A23.3-94, 8.6.2.2 24 ACI 318-14, 22.5.3.1, 22.6.3.1; ACI 318-11, 11.1.2; ACI 318-08, 11.1.2; ACI 318-05, 11.1.2; ACI 318-05, 11.1.2; ACI 318-99, 11.1.2 25 CSA A23.3-14, 13.3.4.2; CSA A23.3-04, 13.3.4.2 SOLUTION METHODS \| 41 \| The modulus of rupture is used to determine the cracking moment when computing the effective moment of inertia in deflection calculations. For ACI 318 code, the default value of modulus of rupture, fr, is set equal to:26 7.5 r c f f   = Eq. 2-17 and for the CSA A23.3 standard, the default value of modulus of rupture fr is:27 0.6 r c f f   = Eq. 2-18 For two-way slabs analyzed in accordance with CSA A23.3-94 as well as for beams, one-way, and two-way slabs analyzed in accordance with CSA A23.3-14/04, the default value is reduced to its half value28, i.e. 0.6 2 c r f f   = Eq. 2-19 Factor λ reflecting the reduced mechanical properties of lightweight concrete is equal to29: 1.00 for normal density concrete 0.85 for sand-lightweight (structural semi-low-density) concrete 0.75 for all-lightweight (structural low-density) concrete    =    Refer to Table 2.1 for determination of concrete type. There is no limit imposed on fr. Entering a large value of fr will produce deflections based on gross properties (i.e. uncracked sections). The default values for the longitudinal reinforcement yield strength, fy, and shear reinforcement yield strength, fyv, if applicable, are set equal to 60 ksi [413 MPa] for ACI and 400 MPa for CSA. 26 ACI 318-14, 19.2.3.1; ACI 318-11, 9.5.2.3; ACI 318-08, 9.5.2.3; ACI 318-05, 9.5.2.3; ACI 318-02, 9.5.2.3; ACI 318-99, 9.5.2.3 27 CSA A23.3-14, 8.6.4; CSA A23.3-04, 8.6.4; CSA A23.3-94, 8.6.4 28 CSA A23.3-14, 9.8.2.3; CSA A23.3-04, 9.8.2.3; CSA A23.3-94, 13.3.6 29 ACI 318-14, 19.2.4; ACI 318-11, 8.6; ACI 318-08, 8.6; ACI 318-05, 11.2.1.2; ACI 318-02, 11.2.1.2; ACI 318-99, 11.2.1.2; CSA A23.3-14, 8.6.5; CSA A23.3-04, 8.6.5; CSA A23.3-94, 8.6.5 SOLUTION METHODS \| 42 \| 2.2.1.3. Loading Considerations Load Cases and Load Combinations Each load is applied to the span element under one of the 6 (A through F) load cases. The span element is analyzed and designed under load combinations. A load combination is the algebraic sum of each of the load cases multiplied by a user specified load factor. The program allows defining up to 50 load combinations. The user has full control over the combinations used. The program contains predefined (built into the program) default primary load combinations for the supported codes. These default combinations are created when starting a new project. For the design of span elements, the required area of steel is calculated due to the element internal forces from the ultimate level combinations. On the other hand, displacement envelopes are determined using the displacement from the service level combinations. Basic load cases and the corresponding load factors for load combinations are suggested as defaults to facilitate user’s input. The default load cases and load combination factors should be modified as necessary at the discretion of the user to satisfy project and code requirements. Given the numerous applications of spSlab/spBeam to structural floor systems, the load combinations must be examined in detail to consider the myriad of possible conditions that can exist. A detailed discussion of load case and combination factors is provided in Appendix A.1. Load Patterns The evaluation of floor systems requires consideration of various live load configurations to capture the critical effects on structural responses. To address these scenarios, the program automatically generates live load cases using predefined patterns. The program also incorporates a live load pattern reduction ratio, defaulting to 75% for two-way systems per code requirements, with user customization available between 0–100%. A detailed discussion of load patterns is provided in Section 2.3.3.4. SOLUTION METHODS \| 43 \| Off-Centered Loads for Two-Way For two-way slab systems, the distribution of loads along the analysis direction becomes simplified while utilizing the Equivalent Frame Method (EFM), eliminating the need to consider additional torsion and moments for “Off-Centered” loads. The following figure highlights how different load types - uniform distributed loads (A), concentrated loads (B), and line loads (C) - are translated into equivalent load input values for spSlab. Each load is converted into the corresponding program input format, ensuring accurate representation of the structural behavior. This approach streamlines the modeling process while maintaining the integrity of the applied loads in the analysis. It is critical that the engineer and user exercise maximum judgment in this area based on his thorough understanding of the EFM and how its application in two directions should be managed. The simplification of taking a two-way concrete floor system and breaking it into two individual one-way systems should not overlook how the overall system will ultimately behave. As such, loads handled in one direction will have to also be reconsidered in the orthogonal direction where they may present a more governing condition. Figure 2.6 – Entering “Off-Centered” loads into spSlab model as equivalent load input SOLUTION METHODS \| 44 \| Loads Due to Openings When a floor system contains an opening, usually it is due to an important Mechanical, Electrical or Piping (MEP) equipment that are passing through such opening. As such, the presence of the opening complicates the analysis using the EFM. The engineer has to consider the absence of some or an entire part of the column strip or the middle strip. In addition, the additional loads from attachments borne by the MEP equipment or stairs or any other accessories have to also be added to an area that is already compromised by the opening. Such loads are commonly concentrated point loads that are close or in direct contact with the support creating concentrated shear forces that have to be handled carefully depending on their proximity to the support. The impact of such floor openings and concentrated forces on the shear force and bending moment diagrams and ultimately the reinforcement required will depend greatly on a number of variables that are difficult to quantify, nor are they clearly explained in the current codes and standards. StructurePoint's experience indicates that occasionally engineers have to forego the intended two-way behavior and resort to one-way action in certain spans or in an entire design strip. In some instances, additional beams and framing should be added to accommodate such loss of section and the concentration of forces. The engineering end-user is advised to e-mail and contact StructurePoint's experts to discuss such conditions and find the proper application of the EFM to model their floor system in spSlab to that specific condition. SOLUTION METHODS \| 45 \| 2.2.1.4. Special Considerations for Joist Systems The following considerations are applicable to one-way (standard and wide module) and two-way (waffle) joist systems unless noted. More details about this topic can be found in “Two-Way Joist Concrete Slab Floor (Waffle Slab) System Analysis and Design” and “One-Way Wide Module (Skip Joist) Concrete Floor System Design” Design Examples from StructurePoint. Rib Dimensions Rib dimensions will be considered valid if the rib width is at least 4 in. [100 mm], the depth is no more than 3-1/2 times the rib width, and the clear spacing between ribs does not exceed 30 in. [800 mm]30. If rib dimensions do not meet these requirements (e.g. wide module joist systems) the code requires such ribs to be designed as beams31. The program treats the design of wide-spaced joists the same way as for valid slabs, regardless of code limitation. If the code limits are exceeded, the condition is flagged and the 10% increase of rib shear capacity is not used. The user is then responsible to validate the resulting design and reconcile the code requirements. Minimum Thickness for Joist Systems The minimum slab thickness allowed for joist slabs is one-twelfth the clear rib spacing, or 1.5 in. [40 mm] for ACI code32 and 50 mm for CSA code33. Joist System Analysis and Design For the purposes of analysis and design, the program replaces the ribbed slab with solid slabs of equivalent moment of inertia, weight, punching shear capacity, and one-way shear capacity. 30 ACI 318-14, 8.8.1.2, 8.8.1.4, 9.8.1.2, 9.8.1.4; ACI 318-11, 8.13.2, 8.13.3; ACI 318-08, 8.13.2, 8.13.3; ACI 318-05, 8.11.2, 8.11.3; ACI 318-02, 8.11.2, 8.11.3; ACI 318-99, 8.11.2, 8.11.3; CSA A23.3-14, 10.4.1; CSA A23.3-04, 10.4.1; CSA A23.3-94, 10.4.1 31 ACI 318-14, 8.8.1.8, 9.8.1.8; ACI 318-11, 8.13.4; ACI 318-08, 8.13.4; ACI 318-05, 8.11.4; ACI 318-02, 8.11.4; ACI 318-99, 8.11.4; CSA A23.3-14, 10.4.2; CSA A23.3-04, 10.4.2; CSA A23.3-94, 10.4.2 32 ACI 318-14, 8.8.2.1.1, 9.8.2.1.1; ACI 318-11, 8.13.5.2; ACI 318-08, 8.13.5.2; ACI 318-05, 8.11.5.2; ACI 318-02, 8.11.5.2; ACI 318-99, 8.11.5.2 33 CSA A23.3-14, 10.4.1; CSA A23.3-04, 10.4.1; CSA A23.3-94, 10.4.1 SOLUTION METHODS \| 46 \| The equivalent thickness based on system weight is used to compute the system self-weight. This thickness, hw, is given by: mod mod w V h A = Eq. 2-10 where: Vmod = the volume of one joist module, Amod = the plan area of one joist module. Figure 2.7 – Valid Rib Dimensions The equivalent thickness based on moment of inertia is used to compute slab stiffness. The ribs spanning in the transverse direction are not considered in the stiffness computations. This thickness, hMI, is given by: 1 3 12 rib MI rib I h b   =     Eq. 2-11 where: Irib = moment of inertia of one joist section between centerlines of ribs, brib = the center-to-center distance of two ribs (clear rib spacing plus rib width). SOLUTION METHODS \| 47 \| The drop panel depth for two-way joist (waffle) slab systems is set equal to the rib depth. The equivalent drop depth based on moment of inertia, dMI, is given by: dMI = hMI + hrib Eq. 2-12 where: hrib = rib depth below slab, hMI = equivalent slab thickness based on moment of inertia. A drop depth entered for a waffle slab system other than 0 will be added to dMI, thus extending below the ribs. One-way shear capacity, Vc (increased by 10% for ACI code34), is calculated assuming the shear cross-section area consisting of ribs and the portion of slab above, decreased by concrete cover. For such section the equivalent shear width of single rib is calculated from the formula: 12 v d b b = + Eq. 2-13 where: b = rib width, d = distance from extreme compression fiber to tension reinforcement centroid. 34 ACI 318-14, 8.8.1.5, 9.8.1.5; ACI 318-11, 8.13.8; ACI 318-08, 8.13.8; ACI 318-05, 8.11.8; ACI 318-02, 8.11.8; ACI 318-99, 8.11.8 SOLUTION METHODS \| 48 \| The equivalent thickness based on shear area is used to compute the area of concrete section resisting punching shear transfer, Ac around drop panels in two-way joist (waffle) systems. The equivalent slab thickness, hV, used to compute Ac, is given by: rib v reinf rib A h d b = + Eq. 2-14 where: Arib = the entire rib area below the slab plus the slab thickness minus the distance to the reinforcement centroid, dreinf, within the rib width, i.e., the slab depth between the ribs is not considered as contributing to shear capacity, brib = the center-to-center distance of two ribs (clear rib spacing plus rib width), dreinf = the distance to reinforcement centroid from the slab top at the support. When calculating flexural capacity for negative bending moments, the distance between center of top reinforcement to the soffit of the rib is used as an effective depth, d, while assuming the width of compression zone as equal to the center-to-center distance of two ribs (brib). This assumption results in higher estimates of negative moment capacity since the space between ribs is void. The user may switch to a single rib design or investigation as a beam in order to consider rib width only in the compression zone. SOLUTION METHODS \| 49 \| 2.2.2. Geometry Checks spSlab and spBeam provide various geometric checks to avoid an analysis with an inconsistent system. Dimensions of slabs, beams, drops, bands and column capitals are checked and modified to produce a code compliant system. 2.2.2.1. Slabs If a slab cantilever length is less than one-half the column dimension in the direction of analysis, c1, or less than the lateral extension of the transverse beam into the cantilever, the cantilever length will be increased to the larger of these two lengths. If the slab width is less than one-half the column dimension transverse to the direction of analysis, c2, or less than one-half the longitudinal beam width, the slab width will be increased to the larger of these two widths. 2.2.2.2. Drop Panels If a drop panel extends beyond the end of a slab cantilever, the drop panel dimensions will be reduced so that it extends only to the cantilever tip. Guidance is absent from all standards and reference documents regarding continuous extension of drop panels between supports. If a slab band is discontinuous, to model this condition, it must be completed with a user-defined drop panel of corresponding width and thickness on a discontinuous end, as a minimum, in order to complete the analysis. 2.2.2.3. Column Capitals When a column capital is defined, the program checks if capital side slope (depth/extension ratio) is more than 1, i.e. the angle between capital side and column axis is no greater than 45 degrees35. The upper limit for the side slope is 50. If a column with capital frames into a drop panel (or a beam), extension of the capital will be automatically adjusted – if necessary – so that projected sides of the capital do not fall outside of the drop panel (or the beam) edges before reaching slab 35 ACI 318-14 (Ref. ), 8.4.1.4; ACI 318-11 (Ref. ), 13.1.2; ACI 318-08 (Ref. ),13.1.2; ACI 318-05 (Ref. ), 13.1.2; ACI 318-02 (Ref. ), 13.1.2; ACI 318-99 (Ref. ), 13.1.2; CSA A23.3-04, 2.2; CSA A23.3-94, 13.1 SOLUTION METHODS \| 50 \| soffit (see Figure 2.8). The modified column capital extension will be used when computing column stiffness and in punching shear calculations. Figure 2.8 – Maximum Capital Width SOLUTION METHODS \| 51 \| 2.2.3. Definitions and Assumptions The analysis of the reinforced concrete members performed by spSlab/spBeam conforms to the provisions of the Strength Design Method and Unified Design Provisions and is based on the following basic assumptions: • All conditions of strength satisfy the applicable conditions of equilibrium and strain compatibility • Strain in the concrete and in the reinforcement is directly proportional to the distance from the neutral axis. In other words, plane sections normal to the axis of bending are assumed to remain plane after bending. • An equivalent uniform rectangular concrete stress block is used with a maximum usable ultimate strain at the extreme concrete compression fiber equal to 0.003 for ACI codes and 0.0035 for CSA codes • Tensile strength of concrete in flexural calculations is neglected • For reinforcing steel, the elastic-plastic stress-strain distribution is used • Assumptions related to the analysis method, along with detailed provisions and equations for the design codes and unit systems supported by spSlab/spBeam, are outlined in Sections 2.3, 2.4, and 2.5. SOLUTION METHODS \| 52 \| 2.3. Analysis Methods 2.3.1. Overview of Equivalent Frame Method The equivalent frame method, as described in the code36, is used by spSlab for both analysis and design. The code specifies procedures for the analysis and design of slab systems reinforced for flexure in more than one direction, with or without beams between the supports. A two-way slab37 system, including the slab and its supporting beams, columns, and walls may be designed by either of the following procedures • The Direct Design Method • The Equivalent Frame Method spSlab uses the Equivalent Frame Method of analysis which is based on extensive analytical and experimental studies conducted at the University of Illinois. Note also that there are no restrictions on the number of slab spans or on dead-to-live load ratios in this method of analysis. The first step in the frame analysis is to divide the three-dimensional building into a series of two-dimensional frames extending to the full height of the building. Horizontal members for each frame are formed by slab strips as shown in Figure 2.9. For vertical loads, each story (floor and/or roof) may be analyzed separately with the supporting columns being considered fixed at their remote ends (Figure 2.10). The required reinforcing and resulting deflections for an interior or exterior panel in a floor system shall be combined from the analysis of two equivalent frames in orthogonal directions in order to arrive at the final design. 36 ACI 318-14, 8.11; ACI 318-11, 13.7; ACI 318-08, 13.7; ACI 318-05, 13.7; ACI 318-02, 13.7; ACI 318-99, 13.7; CSA A23.3-14, 13.8; CSA A23.3-04, 13.8; CSA A23.3-94, 13.9 37 Implies a slab supported by isolated supports which permits the slab to bend in two orthogonal directions. SOLUTION METHODS \| 53 \| 2.3.1.1. Stiffness Characteristics The stiffness factors for the horizontal members (the slab beams) and the vertical members (the equivalent columns) are determined using segmental approach. Figure 2.9 – Design Strips SOLUTION METHODS \| 54 \| 2.3.1.2. Slab Beams The moment of inertia of the slab beam elements between the faces of the columns (or column capitals) is based on the uncracked section of the concrete including beams or drop panels. The moment of inertia from the face of the column (or capital) to the centerline of the column (or capital) is considered finite and is dependent on the transverse dimensions of the panel and support. This reduced stiffness (as compared to the infinite stiffness assumed in previous codes) is intended to soften the slab at the joint to account for the flexibility of the slab away from the support. This is consistent with provisions of the code.38 Figure 2.11 shows the changes in stiffness between a slab, and a drop panel, and a column (or capital). Figure 2.10 – Analytical Model for Vertical Loads for a Typical Story 2.3.1.3. Columns The computation of the column stiffness is more complicated as it utilizes the concept of an equivalent column. Theoretical slab studies have shown that the positive moment in a slab may increase under pattern loads, even if rigid columns are used, because of the flexibility of the slab away from the column. However, if a two-dimensional frame analysis is applied to a structure with rigid columns, pattern loads will have little effect. 38 ACI 318-14, 8.11.3; ACI 318-11, 13.7.3; ACI 318-08, 13.7.3; ACI 318-05, 13.7.3; ACI 318-02, 13.7.3; ACI 318-99, 13.7.3; CSA A23.3-14, 13.8.2.3; CSA A23.3-04, 13.8.2.3; CSA A23.3-94, 13.9.2.3 SOLUTION METHODS \| 55 \| Figure 2.11 – Sections for Calculating Slab-Beam Stiffness, Ksb SOLUTION METHODS \| 56 \| To account for this difference in behavior between slab structures and frames, the equivalent column torsional member, as shown in Figure 2.13, runs transverse to the direction in which the moments are being determined. The transverse slab beam can rotate even though the column may be infinitely stiff, thus permitting moment distribution between adjacent panels. It is seen that the stiffness of the equivalent column is affected by both the flexural stiffness of the columns and the torsional stiffness of the slabs or beams framing into the columns. Note that the method of computation of column stiffness is in accordance with the requirements of the code39. Figure 2.15 shows a schematic representation of the stiffness of typical columns. 39 ACI 318-14, 8.11.4; ACI 318-11, 13.7.4; ACI 318-08, 13.7.4; ACI 318-05, 13.7.4; ACI 318-02, 13.7.4; ACI 318-99, 13.7.4; CSA A23.3-14, 13.8; CSA A23.3-04, 13.8; CSA A23.3-94, 13.9 SOLUTION METHODS \| 57 \| Figure 2.12 – Continued SOLUTION METHODS \| 58 \| The column stiffness is based on the column height, lc, measured from mid-depth of the slab above, to the mid-depth of the slab below. spSlab calculates the stiffness of the column below the design slab, taking into account the design slab system at its top end. spSlab calculates the stiffness of the column above the design slab taking only the slab depth into account at its bottom end; column capitals, beams, or drops are ignored. The computation of the torsional stiffness of the member requires several simplifying assumptions. The first step is to assume dimensions of the transverse torsional slab-beam members. Assumptions for dimensions of typical torsional members are shown in Figure 2.14. The stiffness, Kt, of the torsional member is given by the following expression40: 3 2 9 1 cs t t t E C K c l l =   −      Eq. 2-20 where: Σ = denotes summation over left and right side torsional member, Ecs = modulus of elasticity for slab concrete, C = cross-sectional constant defined in Eq. 2-21, c2 = size of rectangular column or capital measured transverse to the direction in which moments are being determined, lt = l2L and l2R, lengths of span transverse to l1, measured on each side of the column for ACI 318; for CSA A23.3 value of lt is taken as the smaller of l1a and l2a where l1a is the average l1 and l2a is the average l2 on each side of an interior column. In case of an exterior columns, l1a and l2a are taken respectively as l1 (if the column is exterior with respect to the direction of analysis) and l2 (if column is exterior in the transverse direction) of the adjacent span, i.e. cantilevers, if any, are neglected. 40 ACI 318-14, 8.11.5; ACI 318-11, 13.7.5; ACI 318-08, 13.7.5; ACI 318-05, 13.7.5; ACI 318-02, 13.7.5; ACI 318-99, 13.7.5; CSA A23.3-14, 13.8.2.8; CSA A23.3-04, 13.8.2.8; CSA A23.3-94, 13.9.2.8 SOLUTION METHODS \| 59 \| Figure 2.13 – Equivalent Column The constant C is evaluated for the cross section by dividing it into separate rectangular parts and by carrying out the following summation41: 3 1 0.63 3 x x y C y   = −      Eq. 2-21 where: x = short overall dimension of the rectangular part of a cross section, y = long overall dimension of the rectangular part of a cross section. 41 ACI 318-14, 8.10.5.2; ACI 318-11, 13.6.4.2; ACI 318-08, 13.6.4.2; ACI 318-05, 13.6.4.2; ACI 318-02, 13.0; ACI 318-99, 13.0; CSA A23.3-14, 13.8.2.9; CSA A23.3-04, 13.8.2.9; CSA A23.3-94, 13.9.2.9 SOLUTION METHODS \| 60 \| The program divides the section into rectangles in such a way that the value of constant C is maximum (Figure 2.14). Walls perpendicular to the direction analysis can be modeled as wide columns. If a column/wall runs full length of the total design strip42, the program modifies moment distribution factors to achieve uniform distribution of moments along the column and middle strips. If the width of the wall is less than 75% of the total design strip then no modification of distribution factors is applied. For column/wall widths between 75% and 100% of total strip widths, moment distribution factors are linearly interpolated between regular values and uniformly distributed values. When beams frame into the column in the direction of analysis, the value of Kt as computed in Eq. 2-20 is multiplied by the ratio of the moment of inertia of the slab with the beam (Isb) to the moment of inertia of the slab without the beam (Is), as shown: sb ta t s I K K I = Eq. 2-22 With reference to Figure 2.13, Is is computed from part A (slab without beam), whereas Isb is computed from both parts A and B (slab with beam). 42 For walls running full width of the slab (c2 = l2), the program slightly adjusts the width of the wall to avoid singularity in the denominator of Eq. 2-20. SOLUTION METHODS \| 61 \| Figure 2.14 – Section of the attached Torsional Members SOLUTION METHODS \| 62 \| Figure 2.15 – Section for Calculating the Stiffness (Kc) of the Column Below the Design Floor (lc-input, lc-computed) SOLUTION METHODS \| 63 \| Having the column stiffness, Kc, and the stiffness of the attached torsional member, Kt, the stiffness of the equivalent column, Kec, is computed from the equation: 1 ct cb ec ct cb l r ta ta K K K K K K K + = + + + Eq. 2-23 where: Kct = top column stiffness, Kcb = bottom column stiffness, Kta = torsional stiffness of the left ( l ta K ) and the right ( r ta K ) member. SOLUTION METHODS \| 64 \| 2.3.2. Modeling of Supports By default spSlab assumes that column-slab/beam joints can only rotate and that they do not undergo any translational displacements. Rotation of a joint is affected by the stiffness of elements it connects i.e. slabs/beams, transverse beams, and columns. Columns are assumed by default to be fixed at their far ends as shown in Figure 2.10. These default assumptions can be altered using the Column and Restraint commands. By specifying vertical spring support constant with Kz value other than 0, you can allow the joint to displace vertically. This movement is then controlled by the stiffness of the spring Kz in addition to the stiffness of the column below. The column above is assumed not to constrain the vertical movement of the joint. Additional rotational spring support can be applied to the joint by specifying the value of Kry. Also, the far end column conditions can be selected as either fixed or pinned as shown in the Figure 2.16. All elements controlling the displacements of a joint are shown in the Figure 2.16. More information about creating support elements in spSlab/spBeam can be found in Section 5.2.4.2. Figure 2.16 – (a) Elements controlling joint displacements (b) Far End Column Boundary Conditions SOLUTION METHODS \| 65 \| 2.3.3. Modeling of Loads All applied loads are input as unfactored loads. There are no limitations imposed on the ratio of dead to live loads in the Equivalent Frame Method. Results of gravity load and lateral load analyses may be combined, however, the effects of cracking and reinforcement on stiffness must be accounted for in the lateral load analysis. 2.3.3.1. Self-Weight The self-weight of the system is automatically calculated and assigned to the reserved load self-weight load case, SELF, which is by default defined in all new data files. The weights of the slabs, drops, and longitudinal and transverse beams are considered in the self-weight computations. Only the concrete weight is considered, the reinforcement weight is ignored. The weight of longitudinal beams is ignored starting at the column centerline, for a length equal to one-half c1, the column dimension in the direction of analysis. This will produce slightly less self-weight than actually present for beams wider than c2, the column’s transverse dimension. If load case SELF is removed then the program will ignore self-weight in all ultimate load combinations as well as in internally defined service load combination used to calculate displacements. 2.3.3.2. Superimposed Loading All superimposed vertical loading is considered to act over the entire transverse width of the slab. For slab systems with beams, loads supported directly by the beam (such as the weight of the beam stem or a wall supported directly by the beams) are also assumed to be distributed over the entire transverse width of the strip. An additional analysis may be required, with the beam section designed to carry these loads in addition to the portion of the slab moments assigned to the beam. SOLUTION METHODS \| 66 \| 2.3.3.3. Lateral Loading For lateral loads, each frame should be analyzed as a unit for the entire height of the building (Figure 2.17). Computer programs, such as spFrame, are available for performing such analyses. It should be realized that, for lateral load analysis, slab-beam elements may have a reduced stiffness due to cracking as well as other assumptions made for the effective slab width used for the lateral analysis. The moments obtained from such an analysis may then be input into the equivalent frame model using the program to determine the appropriate design moments under combined vertical and lateral loads. Figure 2.17 – Analytical Model for Lateral Loads The program distributes the effect of the superimposed lateral load moments to the column strip and middle strip according to the moment distribution factors computed for gravity loads (see Table 2.2 through Table 2.4 later in this chapter). SOLUTION METHODS \| 67 \| 2.3.3.4. Load Patterns The analysis of floor systems requires the consideration of several loading configurations. For example, the two adjacent spans loaded may produce the maximum shear stress around a column, while the alternate spans loaded may produce the maximum flexural moments. To cover different loading scenarios the program generates live load case based on the following load patterns (Figure 2.18): • Pattern No. l (All): All spans loaded with live load, • Pattern No. 2 (Odd): Starting at span 1, alternate spans loaded with live load, • Pattern No. 3 (Even): Starting at span 2, alternate spans loaded with live load, • Pattern No. 3+N (SN): Two spans adjacent to support No. N loaded with live load. SOLUTION METHODS \| 68 \| Figure 2.18 – Live Load Patterns SOLUTION METHODS \| 69 \| The program reduces the magnitude of live load patterns No. 2 through No. (3+n) by a predefined ratio. For two-way systems, the default live load pattern ratio selected by the program equals 75% as permitted by the code43. The user has the ability to select different value for the pattern ratio within the range 0-100%. If 0% is selected then load patterning effects will be neglected. However, the pattern No. 1 with all spans loaded (as specified by the user) is always considered with full unreduced magnitude. 43 ACI 318-14, 6.4.3.3; ACI 318-11, 13.7.6.3; ACI 318-08, 13.7.6.3; ACI 318-05, 13.7.6.3; ACI 318-02, 13.7.6.3; ACI 318-99, 13.7.6.3; CSA A23.3-14, 13.8.4; CSA A23.3-04, 13.8.4; CSA A23.3-94, 13.9.4 SOLUTION METHODS \| 70 \| 2.3.4. Column and Middle Strip Widths The code44 defines the width of the column strip on each side of the column centerline as being one-fourth of the smaller of either the transverse or the longitudinal span. These widths are printed as part of the design results. The strip widths at a support are computed by (see Figure 2.19) • column strip 2, 2, 1 1 2, 2, 1 1 1 1 min , min , 4 4 4 4 min min , min , 4 4 4 4 l r i i cs l r i i l l l l w l l l l + +       +             =         +             Eq. 2-24 • middle strip   2, 2, 1 min , ms i i cs w l l w + = − Eq. 2-25 44 ACI 318-14, 8.4.1.5; ACI 318-11, 13.2.1; ACI 318-08, 13.2.1; ACI 318-05, 13.2.1; ACI 318-02, 13.2.1; ACI 318-99, 13.2.1; CSA A23.3-14, 2.2; CSA A23.3-04, 2.2; CSA A23.3-94, 13.1 SOLUTION METHODS \| 71 \| Figure 2.19 – Strips Widths at Support The strip widths in the span are defined as (see Figure 2.20): • column strip 2, 2, 1 1 min , min , 4 4 4 4 l r cs l l l l w     = +         Eq. 2-26 • middle strip 2 ms cs w l w = − Eq. 2-27 SOLUTION METHODS \| 72 \| where: l1 = span length in the direction of analysis, l2,l = the input transverse strip widths on the left of column centerline, l2,r = the input transverse strip widths on the right of column centerline, l2 = l2,l / 2 + l2,r / 2, the total input transverse strip width. Figure 2.20 – Strips Widths in Span If a longitudinal slab band is defined (CSA A23.3-14/04 standard only) then the column strip width is automatically adjusted to be equal to the band width: wcs = wband Eq. 2-28 If a longitudinal beam exists then the adjusted column strip width, cs w , is calculated by subtracting the beam width, wbeam, from the width of the column strip: cs cs beam w w w = − Eq. 2-29 SOLUTION METHODS \| 73 \| If the user selects the BEAM T-SECTION DESIGN option in Design Options tab under Solve command in the Ribbon, the beam width, wbeam, used by the program will include portion of the slab on each side of the beam equal to projection of the beam below the slab, but not greater than slab thickness45. Otherwise, only web width is used. If the beam width, wbeam, is greater than the column strip width, wcs, then the adjusted column strip width is set to zero and moment distribution factors are adjusted to apply all column strip moment to the beam. This may occur when modeling a slab band with a wide longitudinal beam for codes other than CSA A23.3-14/04. In case of CSA A23.3-14/04 standard, the dedicated LONGITUDINAL SLAB BAND option in the Project Left Panel is available to model slab band systems explicitly. By selecting USER SLAB STRIP WIDTH and USER STRIP DISTRIBUTION FACTORS options in the Design & Modeling panel under Define command in the Ribbon, the user has the ability to manually override strip widths and moment distribution factors calculated automatically by the program and requires engineering judgment. Note: For exterior frames, the edge width should be specified to the edge of the slab from the column centerline. Entering edge width greater than l1 / 4 involves engineering judgment regarding two-way behavior of the system and the applicability of the equivalent frame method. 45 ACI 318-14, 8.4.1.8; ACI 318-11, 13.2.4; ACI 318-08, 13.2.4; ACI 318-05, 13.2.4; ACI 318-02, 13.2.4; ACI 318-99, 13.2.4; CSA A23.3-04 Figure N13.1.2.(a) in Ref. , CSA A23.3-94, 13.1 SOLUTION METHODS \| 74 \| 2.3.5. Strip Design Moments For design purposes46, spSlab considers negative moments as those producing tension at the top of the slab and positive moments as those producing tension at the bottom of the slab. The negative design moment is taken at the face of the column below the slab, or at the face of the column capital, but in no case is it considered at a location greater than 0.175 of the longitudinal span length, l1, away from the center of the column.47 This imposes a limit on long narrow supports, in order to prevent undue reduction in the design moment. For slab systems with transverse beams, the face of a beam is not considered as the face of support. For end columns with capitals, the negative moments are taken at the midpoint of the capital extension.48 If a positive moment occurs at a support then its value at the face of the column above the slab is considered (or at the support centerline if there is no column above the slab). spSlab computes the amount of reinforcement for the moments on the left and the right side of the support. The negative design moment is the moment which requires the most area of reinforcement to be resisted. The location, left or right of the support, of the maximum moment may vary when systems differ on each side of the support (for example, a system with beams on one side only). spSlab automatically calculates the values of strip moment distribution factors for column strips and longitudinal beams (if present). Portion of the total factor moment not assigned to a column strip or a beam is then proportionally assigned to the remaining middle strip. Note: By checking USER STRIP DISTRIBUTION FACTORS option in Design & Modeling panel under Define command in the Ribbon, the user has the ability to manually adjust strip moment distribution factors calculated automatically by the program. 46 ACI and CSA provisions for the location of critical section for flexure referred to in this paragraph apply to two-way systems. Due to lack of similar provisions for one-way systems and beams in ACI and CSA standards, the program consistently applies the same rules (with the exception of 0.1751 limitation) to one-way systems and beams. 47 ACI 318-14, 8.11.6.1; ACI 318-11, 13.7.7.1; ACI 318-08, 13.7.7.1; ACI 318-05, 13.7.7.1; ACI 318-02, 13.7.7.1; ACI 318-99, 13.7.7.1; CSA A23.3-14, 13.8.5.1; CSA A23.3-04, 13.8.5.1; CSA A23.3-94, 13.9.5.1 48 ACI 318-14, 8.11.6.2, 8.11.6.3; ACI 318-11, 13.7.7.2; ACI 318-08, 13.7.7.2; ACI 318-05, 13.7.7.2; ACI 318-02, 13.7.7.2; ACI 318-99, 13.7.7.2; CSA A23.3-14, 13.8.5.2; CSA A23.3-04, 13.8.5.2; CSA A23.3-94, 13.9.5.2 SOLUTION METHODS \| 75 \| 2.3.5.1. ACI 318 and CSA A23.3-9449 The column strips are proportioned to resist the portions in percent of interior negative factored moments according to Table 2.2.50 l2/l1 0.5 1.0 2.0 (αf1 l2/l1) = 0 75 75 75 (αf1 l2/l1) ≥ 1.0 90 75 45 Table 2.2 – Column Strip Percent of Interior Negative Factored Moments at Supports The column strips are proportioned to resist the portions in percent of exterior negative factored moments according to Table 2.3.51 l2/l1 0.5 1.0 2.0 (αf1 l2/l1) = 0 βt = 0 100 100 100 βt ≥ 2.5 75 75 75 (αf1 l2/l1) ≥ 1.0 βt = 0 100 100 100 βt ≥ 2.5 90 75 45 Table 2.3 – Column Strip Percent of Exterior Negative Factored Moments at Supports 49 For CSA A23.3-94 standard, the program assumes by default values given by ACI code which fall within the ranges specified in CSA A23.3-94, 13.12.2 50 ACI 318-14, 8.10.5.1; ACI 318-11, 13.6.4.1; ACI 318-08, 13.6.4.1; ACI 318-05, 13.6.4.1; ACI 318-02, 13.6.4.1; ACI 318-99, 13.6.4.1; CSA A23.3-14, 13.9.5; CSA A23.3-04, 13.9.5; CSA A23.3-94, 13.9.5.1; CSA A23.3-94, 13.8.5.1 51 ACI 318-14, 8.10.5.2, 8.10.5.3; ACI 318-11, 13.6.4.2; ACI 318-08, 13.6.4.2; ACI 318-05, 13.6.4.2; ACI 318-02, 13.6.4.2; ACI 318-99, 13.6.4.2 SOLUTION METHODS \| 76 \| The values αf1 in Table 2.2 and Table 2.3 and βt in Table 2.3 are defined as: αf1 = ratio of flexural stiffness of the beam section to flexural stiffness of a width of slab bounded by centerlines of adjacent panels (if any) on each side of the beam in the direction of analysis. For flat plates, flat slabs, and waffle (αf1 l2 / l1) = 0, βt = ratio of torsional stiffness of an edge beam section to flexural stiffness of a width of slab equal to the span length of the beam, center-to-center of supports.52 When no transverse beams are present, βt = 0, otherwise 2 cb t cs s E C E I = Eq. 2-30 where: Ecb = modulus of elasticity of beam concrete, Ecs = modulus of elasticity of slab concrete, C = cross-sectional constant, see Eq. 2-21, Is = moment of inertia of the gross section of the slab about its centroidal axis. For intermediate values of (l2 / l1), (αf1 l2 / l1) and βt the values in Table 2.2 and Table 2.3 are interpolated using equations Eq. 2-31 and Eq. 2-32. Percentage of negative factored moment at interior support to be resisted by column strip: 1 2 2 1 1 75 30 1 f l l l l     + −       Eq. 2-31 Percentage of negative factored moment at exterior support to be resisted by column strip: 1 2 2 1 1 100 10 12 1 f t t l l l l      − + −       Eq. 2-32 52 ACI 318-14, 8.10.5.2, 8.10.5.3; ACI 318-11, 13.6.4.2; ACI 318-08, 13.6.4.2; ACI 318-05, 13.6.4.2; ACI 318-02, 13.0; ACI 318-99, 13.0; CSA A23.3-94, 13.0 SOLUTION METHODS \| 77 \| When a column width, c2, is equal to or greater than 75 percent of the tributary strip width l2, the distribution factor for negative column strip moment is linearly interpolated between the factor for regular support, and the factor equal 0.50 (moment uniformly distributed across l2). This extends the requirement of the design code53, by providing continuous linear transition between standard and uniform moment distributions, depending on the relative dimension of the support with respect to strip width. User may override software assumptions by selecting user defined distribution factors. When designing by the CSA A23.3-94 code, a portion of the total positive or interior negative moment equivalent to54: 1 2 2 1 1 f l l    +     Eq. 2-33 is resisted by the beam. For exterior supports, the beam is proportioned to resist 100% of the negative moment. That portion of the moment not resisted by the beam is resisted by the slab. The reinforcement required to resist this moment is distributed evenly across the slab. For ACI designs the longitudinal beams are proportioned to resist 85 percent of the column strip moments if αf1 l2 / l1 is equal to or greater than 1.0. For values of αf1 l2 / l1 between 0 and 1.0, the beam is designed to resist a proportionate percentage of the column strip moment between 0 and 85.55 The middle strips are proportioned to resist the portion of the total factored moments that is not resisted by the column strips. 53 ACI 318-14, 8.10.5.4; ACI 318-11, 13.6.4.3; ACI 318-08, 13.6.4.3; ACI 318-05, 13.6.4.3; ACI 318-02, 13.6.4.3; ACI 318-99, 13.6.4.3 54 CSA A23.3-94, 13.13.2.1 55 ACI 318-14, 8.10.5.7; ACI 318-11, 13.6.5; ACI 318-08, 13.6.5; ACI 318-05, 13.6.5; ACI 318-02, 13.6.5; ACI 318-99, 13.6.5 SOLUTION METHODS \| 78 \| The column strips are proportioned to resist the portions in percent of positive factored moments according to Table 2.4.56 l2/l1 0.5 1.0 2.0 (αf1 l2/l1) = 0 60 60 60 (αf1 l2/l1) ≥ 1.0 90 75 45 Table 2.4 – Column Strip Percent of Positive Factored Moments For intermediate values of (l2 / l1) and (αf1 l2 / l1) the values in Table 2.4 are interpolated using Eq. 2-34 as follows: 1 2 2 1 1 60 30 1.5 f l l l l     + −       Eq. 2-34 Note: For flat plates, flat slabs, and waffle slabs, αf1 l2 / l1 = 0. 56 ACI 318-14, 8.10.5.5; ACI 318-11, 13.6.4.4; ACI 318-08, 13.6.4.4; ACI 318-05, 13.6.4.4; ACI 318-02, 13.6.4.4; ACI 318-99, 13.6.4.4 SOLUTION METHODS \| 79 \| 2.3.5.2. CSA A23.3-14/04 For slabs without drop panels (with or without transverse beams) the following moment factors are used57: • Negative moment at interior column, factor = 0.80 • Negative moment at exterior column, factor = 1.00 • Positive moment at all spans, factor = 0.60 For slabs with drop panels (with or without transverse beams) the following moment factors are used58: • Negative moment at interior column, factor = 0.825 • Negative moment at exterior column, factor = 1.00 • Positive moment at all spans, factor = 0.60 For slabs with longitudinal slab bands59: • Negative moment at interior column, factor = 0.90 • Negative moment at exterior column, factor = 1.00 • Positive moment at all spans, factor = 0.90 57 CSA A23.3-14, 13.11.2.2; CSA A23.3-04, 13.11.2.2 58 CSA A23.3-14, 13.11.2.3; CSA A23.3-04, 13.11.2.3 59 CSA A23.3-14, 13.11.2.4; CSA A23.3-04, 13.11.2.4 SOLUTION METHODS \| 80 \| For slabs with transverse slab bands60: • Negative moment at interior column in width bb, factor from 0.05 to 0.15 range is selected so that the remaining moment is distributed evenly over the entire frame width (including bb width) and at least one-third of the total factored moment61 is applied to the band width bb. • Negative moment at exterior column, factor = 1.00 • Positive moment at all spans where (l1 / l2) ≥ 1.0, factor = 0.55 • Positive moment at all spans where (l1 / l2) < 1.0, factor = 0.55 (l1 / l2) For slabs with beams between all the supports62, the positive and interior negative factored moments are distributed as follows: 1 2 1 1 1 0.3 3 l l     −   +   Eq. 2-35 Factored negative moments at exterior supports are assigned in 100% proportion to beams. CSA A23.3-14/04 does not stipulate requirement for distributing moments in slab systems with beams between some (but not all) supports. For estimation of the moment resisted by the beams in this case, the program applies the ACI approach described in the previous section where longitudinal beams are proportioned to resist 85 percent of the column strip moments if αf1 l2 / l1 is equal to or greater than 1.0. For values of αf1 l2 / l1 between 0 and 1.0, the beam is designed to resist a proportionate percentage of the column strip moment between 0 and 85%. 60 CSA A23.3-14, 13.11.2.5; CSA A23.3-04, 13.11.2.5 61 CSA A23.3-14, 13.11.2.7; CSA A23.3-04, 13.11.2.7 62 CSA A23.3-14, 13.12.2; CSA A23.3-04, 13.12.2 SOLUTION METHODS \| 81 \| 2.3.6. Moment Redistribution Redistribution of negative moments applies to one-way and beam systems only. It can be engaged by checking the MOMENT REDISTRIBUTION option in the Design & Modeling panel under Define command in the Ribbon. The program allows for redistribution of negative moments at supports. Only reduction in negative moments is considered. Increase of negative moments at the support is not taken into account even though it is allowed by the code63. Static equilibrium is maintained meaning that bending moments and shear forces along the span are adjusted in accordance with the reduction of moments applied at the supports. The following procedure is followed to obtain moment redistribution factors at the supports. From elastic static analysis, the largest moments from all load combinations and load patterns are determined at support faces on both ends of each span except cantilevers. These moments are used to calculate the maximum percentage adjustment of moments, δ, allowed by the codes. For ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, and ACI 318-0264: 0 if 0.0075 1,000 if 0.0075 t t t       =    Eq. 2-36 where εt is net tensile strain in extreme tension steel at nominal strength. 63 ACI 318-14, 6.6.5.1; ACI 318-11, 8.4.1; ACI 318-08, 8.4.1; ACI 318-05, 8.4.1; ACI 318-02, 8.4.1; ACI 318-99, 8.4.1; CSA A23.3-14, 9.2.4; CSA A23.3-04, 9.2.4; CSA A23.3-94, 9.2.4 64 ACI 318-14, 6.6.5.3; ACI 318-11, 8.4.1 and 8.4.3; ACI 318-08, 8.4.1 and 8.4.3; ACI 318-05, 8.4.1 and 8.4.3; ACI 318-02, 8.4.1 and 8.4.3 SOLUTION METHODS \| 82 \| For ACI 318-9965: ( ) ( ) 0 if 0.5 20 1 if 0.5 b b b            −    =     −  − −        Eq. 2-37 where: ρ = tension reinforcement ratio. ρ' = compression reinforcement ratio. ρb = balanced reinforcement ratio. For CSA A23.366: 30 50 c d = − Eq. 2-38 where: c = distance from extreme compression fiber to neutral axis. d = distance from extreme compression fiber to centroid of tension reinforcement. In the investigation mode, program uses the area of provided reinforcement to obtain redistribution factors. In the design mode the required reinforcement area is used. Additionally, δ is limited to 20% and not to exceed the maximum values specified by the user. Negative moments at span ends are reduced by the amount of redistribution factors and new moment values are iteratively used to obtain new redistribution factors. This iterative procedure is repeated until the change in distribution factor is negligible (does not exceed 0.01%), but no more than 10 times. 65 ACI 318-99, 8.4.1 and 8.4.3 66 CSA A23.3-14, 9.2.4; CSA A23.3-04, 9.2.4; CSA A23.3-94, 9.2.4 SOLUTION METHODS \| 83 \| 2.3.7. Shear Analysis of Slabs Shear analysis in spSlab takes into account one way shear and two-way shear. For two-way shear, the program considers contributions of factored shear force67, Vu, and fraction of unbalanced moment transferred by shear68, γvMunb. spSlab does not consider torsional stresses in the slab. If in the engineer’s judgment this may control, it must be computed manually. spSlab checks one-way shear at a critical section located at a distance not less than the effective depth away from the face of the support69. If a concentrated load is applied closer than the effective depth away from the face of the support then critical section is located at the face of the support. Factored shear force at the critical section is obtained from the analysis of the equivalent frame70. 67 ACI 318-14, 8.4.4.2.3; ACI 318-11, 11.11.7.2; ACI 318-08, 11.11.7.2; ACI 318-05, 11.12.6.2; ACI 318-02, 11.12.6.2; ACI 318-99, 11.12.6.2; CSA A23.3-14, 13.3.5.2; CSA A23.3-04, 13.3.5.2; CSA A23.3-94, 13.4.5.2 68 ACI 318-14, 8.4.4.2.1; ACI 318-11, 11.11.7.1; ACI 318-08, 11.11.7.1; ACI 318-05, 11.12.6.1; ACI 318-02, 11.12.6.1; ACI 318-99, 11.12.6.1; CSA A23.3-14, 13.4.5.3; CSA A23.3-04, 13.4.5.3; CSA A23.3-94, 13.4.5.3 69 ACI 318-14, 7.4.3.2, 8.4.3.2, 9.4.3.2; ACI 318-11, 13.5.4, 11.1.3; ACI 318-08, 13.5.4, 11.1.3; ACI 318-05, 13.5.4, 11.1.3; ACI 318-02, 13.5.4, 11.1.3; ACI 318-99, 13.5.4, 11.1.3; CSA A23.3-14, 13.3.6.1, 11.3.2; CSA A23.3-04, 13.3.6.1, 11.3.2; CSA A23.3-94, 13.4.6.1; CSA A23.3-94, 13.4.6.1; 11.3.2 70 ACI 318-14, 8.11.1.1; ACI 318-11, 13.7.1; ACI 318-08, 13.7.1; ACI 318-05, 13.7.1; ACI 318-02, 13.7.1; ACI 318-99, 13.7.1; CSA A23.3-04, 13.8.1.1; CSA A23.3-94, 13.9.1.1 SOLUTION METHODS \| 84 \| Figure 2.21 – Critical Section for Two-Way Shear Figure 2.21 shows the general two-way shear area71 used by spSlab. Note that the shaded area represents the general case and is modified for special considerations as explained below. Beams are considered in the two-way shear as indicated in Figure 2.21 by areas B1, B2, B3, B4, B5, and B6. Ordinarily, transverse beams transfer unbalanced moment to the column through torsion along the beam and not through shear between the slab and column. However, the code leaves the transfer method to the engineer’s judgment concerning the point at which punching shear is no longer applicable and beam shear becomes the dominate element in shear transfer to the column. spSlab makes no such distinction and computes unbalanced moment transfer stress without regard to any beams framing into the column. When a beam is present, the depth of the beam increases the depth of the critical section where it intersects with the beam. The distances from the face of the support to the critical section will also be increased, i.e. effective depth of the beam will be used to calculate the distance instead of effective depth of the slab, if it results in a critical section that is still within the beam. Otherwise, distances to the critical section are not increased. 71 ACI 318-14, 22.6.4.1; ACI 318-11, 11.11.1.2; ACI 318-08, 11.11.1.2; ACI 318-05, 11.12.1.2; ACI 318-02, 11.12.1.2; ACI 318-99, 11.12.1.2; CSA A23.3-14, 13.3.3; CSA A23.3-04, 13.3.3; CSA A23.3-94, 13.4.3 SOLUTION METHODS \| 85 \| For circular supports (column or column capital), ACI code and CSA standard differ in their treatment and do not provide clear guidance towards the applicability of an equivalent rectangular section for checking punching shear around circular supports. Therefore, spSlab provides as a default option the calculation of properties of the critical section for punching shear based on circular critical shear perimeter. This option is possible given that both the soffit around the perimeter of circular support and the soffit around the perimeter of circular critical shear perimeter are flat & critical shear perimeter stays circular. If circular critical shear perimeter is not achievable or possible, then, the program calculates properties of the critical section for punching shear based on an equivalent rectangular support with the same centroid and equal perimeter length72. The equivalent rectangular support is a square with side length equal to π D/4 ≈ 0.785D where D is the diameter of the circular support as shown in Figure 2.22. Figure 2.22 – Critical Section for Circular Column While this approach is widely used, it produces an equivalent section but not an equivalent shear perimeter. It is, therefore, left to the end-user discretion to judge the use of the circular shear perimeter as it produces more conservative results compared with the traditional equivalent square option. 72 See Fig.13-38(b) and Fig. 13-57 in Ref. , and Fig. 10.5(f) in Ref. SOLUTION METHODS \| 86 \| The critical section is considered closed if the concrete slab around a column extends to a distance greater than or equal to the specified threshold value. In spSlab, the user may define the distance extended beyond the column face in order to consider the section closed. If the critical section does not meet the distance requirement, it is considered open. ACI 318-08 code introduced the definition of the shear cap73 which, alternatively to column capital, can be used to increase the critical section around the column. spSlab users can use the capital geometry to model a shear cap and calculate the punching shear through the thickness of the slab itself (shear cap acting as capital). Other failure modes, such as punching within the perimeter of the shear cap, need to be verified by the user manually. The dimensions of the substitute capital have to be selected such that the resulting critical section is equivalent to critical section for a column with a shear cap. ACI code74 requires shear caps to extend beyond the face of the column by at least the distance equal to cap depth, and so depth/extension ratio should not exceed 1.0. For column capitals depth/extension ratio should not be less than 1.0. Therefore to model shear cap acting as capital, the substituted capital should have depth/extension ratio equal to 1.0. 73 ACI 318-14, 2.3; ACI 318-11, 2.2; ACI 318-08, 2.2 74 ACI 318-14, 8.2.5; ACI 318-11, 13.2.6; ACI 318-08, 13.2.6 SOLUTION METHODS \| 87 \| 2.3.7.1. Critical Section For Interior Supports of Interior Frames The critical section (Figure 2.23) consists of four vertical surfaces through the slab, located at distances of d/2 beyond the support faces. The critical section for interior supports of interior frames is always closed. A closed section will have all its faces defined in Figure 2.21 resisting shear as indicated by Eq. 2-39: 8 1 c i i A A = = Eq. 2-39 If beams frame75 into the column, then the critical section includes the dimensions of the beams (B1 through B6 in Figure 2.21). Figure 2.23 – Interior Supports of Interior Frames 75 A beam is considered as framing into the column if the beam is within a face of the column. SOLUTION METHODS \| 88 \| For Exterior Supports of Interior Frames The critical section for exterior supports of interior frames (Figure 2.24) will be either closed (full A7 and A6 for the first column or A1 and A2 for the last column in Figure 2.21) or open, depending upon the length of the cantilever in relation to slab thickness. The critical section will be considered closed when the clear cantilever span, lc, is greater than or equal to the distance defined by the user beyond the column face. The default value of the distance is 4h when an ACI code is selected76 and 5d for the CSA standard77. The user can modify the default value to accommodate scenarios when larger distances are required, e.g. 10h for slabs with openings78. If beams frame into the column then the critical section includes the contributions from the beam dimensions (B1 through B6 in Figure 2.21). 76 Critical Sections near Holes and at Edges in Ref. , pp.672, Fig 13-59 (b) and (c) 77 CSA A23.3-04 Figure N13.3.3.4 (b) in Ref. ; CSA A23.3-94 Figure N13.4.3.4 (b) in Ref. 78 ACI 318-14, 22.6.4.3; ACI 318-11, 11.11.6; ACI 318-08, 11.11.6; ACI 318-05, 11.12.5; ACI 318-05, 11.12.5; ACI 318-05, 11.12.5; CSA A23.3-14, 13.3.3.4; CSA A23.3-04, 13.3.3.4; CSA A23.3-94, 13.4.3.4 SOLUTION METHODS \| 89 \| Figure 2.24 – Exterior Supports of Interior Frames SOLUTION METHODS \| 90 \| For Interior Supports of Exterior Frames Figure 2.25 shows the critical section for shear for an interior support of an exterior frame. Note that the section is considered as U-shaped (A5 = 0, A8 = 0, B3 = 0, B4 = 0 in Figure 2.21) and it extends up to the edge of the exterior face of the support. If beams frame into the column, then the critical section includes the contribution from the beam dimensions (B1 through B6 in Figure 2.21). If the exterior cantilever span, lc, is greater than or equal to the distance defined by the user beyond the column face (the default value is 4h when an ACI code is selected81 and 5d for the CSA standard82), the section is treated as closed, that is, the support is treated as an interior support of an interior frame. Figure 2.25 – Interior Supports of Exterior Frames SOLUTION METHODS \| 91 \| For Exterior Supports of Exterior Frames The critical section for an exterior support of an exterior frame will typically be L-shaped as shown in the Figure 2.26. If the cantilever span, lc, (in the direction of analysis) is greater than or equal to the distance defined by the user beyond the column face (the default value is 4h when an ACI code is selected81 and 5d for the CSA standard82), then the section is treated as a U-shaped interior support. If, in addition, the cantilever span in transverse direction is greater than or equal to the distance defined by the user beyond the column face, the section is treated as closed. If beams frame into the column, then the critical section includes the contributions from the beam dimensions. Figure 2.26 – Exterior Supports of Exterior Frames SOLUTION METHODS \| 92 \| 2.3.7.2. Computation of Allowable Shear Stress at Critical Section One-way shear strength of slabs is limited79 to 2 c f  . Two-way shear strength of slabs is affected by concrete strength, relationship between size of loaded area and slab thickness, loaded area aspect ratio, and shear-to-moment ratio at slab-column connections. These variables are taken into account in the allowable shear stress, vc, computed at distances of d/2 around the columns and drops (if applicable). For the ACI 318 code, vc is taken as the smallest of the 3 quantities80: 4 2 c c c v f      = +     Eq. 2-40 2 s c c o d v f b      = +     Eq. 2-41 4 c c v f   = Eq. 2-42 where: βc = the ratio of the long to the short side of the column, αs = a constant dependent on the column location, (40 for an interior 4-sided effective critical area, 30 for an exterior 3-sided critical area, 20 for a corner 2-sided effective critical area), d = distance from the slab bottom to centroid of the slab tension reinforcement at support (average value is used if d changes along critical section perimeter), bo = the perimeter of the critical section, λ = factor81 reflecting the reduced mechanical properties of lightweight concrete equal to 0.75 for all-lightweight concrete, 0.85 for sand-lightweight concrete and 1.0 for normal weight concrete. Refer to Table 2.1 for determination of concrete type. 79 ACI 318-14, 22.5.5.1; ACI 318-11, 11.2.1.1; ACI 318-08, 11.2.1.1; ACI 318-05, 11.3.1.1; ACI 318-02, 11.3.1.1; ACI 318-99, 11.3.1.1 80 ACI 318-14, 22.6.5.2, 22.6.5.3; ACI 318-11, 11.11.2.1; ACI 318-08, 11.11.2.1; ACI 318-05, 11.12.2.1; ACI 318-02, 11.12.2.1; ACI 318-99, 11.12.2.1 81 ACI 318-14, 22.6.5.2, 22.6.5.3; ACI 318-11, 11.11.2.1; ACI 318-08, 11.11.2.1; ACI 318-05, 11.2.1.2; ACI 318-02, 11.2.1.2; ACI 318-99, 11.2.1.2 SOLUTION METHODS \| 93 \| For the CSA A23.382, the allowable shear stresses are calculated as the minimum of the following metric equations: 2 1 c c c c v f      = +     Eq. 2-43 s c c c o d v f b      = +     Eq. 2-44 2 c c c v f   = Eq. 2-45 where: η = 0.19 for CSA A23.3-14/04 and 0.20 for CSA A23.3-94, αs = a constant dependent on the column location, (4 for an interior 4-sided effective critical area, 3 for an edge column, 2 for a corner column), d = distance from the slab bottom to centroid of the slab tension reinforcement at support (average value is used if d changes along critical section perimeter), ϕc = resistance factor for concrete83 equal to 0.60 for CSA A23.3-94 and for CSA A23.3-14/04 it is equal to 0.65 for regular and 0.70 for precast concrete, λ = factor84 reflecting the reduced mechanical properties of lightweight concrete equal to 0.75 for structural low-density concrete, 0.85 structural semi-low-density concrete and 1.0 for normal density concrete. Refer to Table 2.1 for determination of concrete type, c f  ≤ 8 MPa. When the value of d is greater than 300 mm, allowable stress vc obtained from the above three equations shall be multiplied by 1300 / (1000 + d) as required by CSA A23.3-14/04 code85. 82 CSA A23.3-14, 13.3.4; CSA A23.3-04, 13.3.4; CSA A23.3-94, 13.4.4 83 CSA A23.3-14 8.4.2, 16.1.3; CSA A23.3-04 8.4.2, 16.1.3; CSA A23.3-94 8.4.2 84 CSA A23.3-14, 8.6.5; CSA A23.3-04, 8.6.5; CSA A23.3-94, 8.6.5 85 CSA A23.3-14, 13.3.4.3; CSA A23.3-04, 13.3.4.3 SOLUTION METHODS \| 94 \| The allowable shear stress around drops when waffle slabs are used is computed as: 2 for ACI 0.20 for CSA A23.3-94 0.19 for CSA A23.3-04 c c c c c c f v f f         =      Eq. 2-46 For waffle slab systems with valid ribs defined earlier in this chapter, the allowable shear stress is increased by 10% for ACI designs86. 2.3.7.3. Computation of Factored Shear Force at Critical Section The factored shear force Vu in the critical section, is computed as the reaction at the centroid of the critical section (e.g., column centerline for interior columns) minus the self-weight and any superimposed surface dead and live load acting within the critical section. If the section is considered open, two 45° lines are drawn from the column corners to the nearest slab edge (lines AF and DE in Figure 2.25) and the self-weight and superimposed surface dead and live loads acting on the area ADEF are omitted from Vu. 86 ACI 318-14, 8.8.1.5, 9.8.1.5; ACI 318-11, 8.13.8; ACI 318-08, 8.13.8; ACI 318-05, 8.11.8; ACI 318-02, 8.11.8; ACI 318-99, 8.11.8 SOLUTION METHODS \| 95 \| 2.3.7.4. Computation of Unbalanced Moment at Critical Section The factored unbalanced moment used for shear transfer, Munb, is computed as the sum of the joint moments to the left and right. Moment of the vertical reaction with respect to the centroid of the critical section is also taken into account by: ( ) , , unb u left u right u g M M M V c = − − Eq. 2-47 where: Mu,left = factored bending moment at the joint on the left hand side of the joint, Mu,right = factored bending moment at the joint on the right hand side of the joint, Vu = factored shear force in the critical section described above, cg = location of the centroid of the critical section with respect to the column centerline (positive if the centroid is to the right in longitudinal direction with respect to the column centerline). SOLUTION METHODS \| 96 \| 2.3.7.5. Computation of Shear Stresses at Critical Section The punching shear stress computed by the program is based on the following87: u u c V v A = Eq. 2-48 where: Vu = factored shear force in the critical section described above, Ac = area of concrete, including beam if any, resisting shear transfer. Under conditions of combined shear, Vu, and unbalanced moment, Munb, γvMunb is assumed to be transferred by eccentricity of shear about the centroidal axis of the critical section. The shear stresses computed by the program for this condition correspond to88: u v unb AB AB c c V M c v A J  = + Eq. 2-49 u v unb CD CD c c V M c v A J  = − Eq. 2-50 where: Munb = factored unbalanced moment transferred directly from slab to column, as described above, γv = (1 – γf), Eq. 2-51 is a fraction of unbalanced moment considered transferred by the eccentricity of shear about the centroid of the assumed critical section89, c = distance from centroid of critical section to the face of section where stress is being 87 ACI 318-14, 8.4.4.2.3; ACI 318-11, 11.11.7.2; ACI 318-08; 11.11.7.2; ACI 318-05, 11.12.6.2; ACI 318-02, 11.12.6.2; ACI 318-99, 11.12.6.2; CSA A23.3-14, 13.3.5; CSA A23.3-04, 13.3.5; CSA A23.3-94, 13.4.5 88 ACI 318-14, R8.4.4.2.3; ACI 318-11, R11.11.7.2; ACI 318-08, R11.11.7.2; ACI 318-05, R11.12.6.2; ACI 318-02, R11.12.6.2; ACI 318-99, R11.12.6.2; CSA A23.3-14, 13.3.5.5; CSA A23.3-04, 13.3.5.5; CSA A23.3-94, 13.4.5.5; Ref. 89 ACI 318-14, 8.4.4.2.1, 8.4.4.2.2; ACI 318-11, 11.11.7.1; ACI 318-08, 11.11.7.1; ACI 318-05, 11.12.6.1; ACI 318-02, 11.12.6.1; ACI 318-99, 11.12.6.1; CSA A23.3-04, Eq. 13-8; CSA A23.3-94, Eq. 13-8 SOLUTION METHODS \| 97 \| computed, Jc = property of the assumed critical section analogous to polar moment of inertia. Factor γf in Eq. 2-51 is calculated as90: ( ) 1 2 1 1 2 / 3 / f b b  = + Eq. 2-52 where: b1 = width of critical section in the direction of analysis, b2 = width of the critical section in the transverse direction. If an ACI 318 standard is selected then the program provides an option to use an increased value91 of γf. For edge and corner columns with unbalanced moment about an axis parallel to the edge, the value can be increased to 1.0 if the factored shear force at the support doesn’t exceed 0.75ϕVc for edge columns and 0.5ϕVc for corner columns. For ACI 318-99, ACI 318-02, and ACI 318-05, condition that reinforcement ratio in the effective slab width doesn’t exceed 0.375ρb must also be satisfied to apply the increase. For interior columns and for edge columns with unbalanced moment perpendicular to the edge, γf can be increased 25% but the final value of γf cannot exceed 1.0. The increase can be applied if the shear doesn’t exceed 0.4ϕVc. Also, the net tensile strain in the effective slab has to exceed 0.010 for the ACI 318-08, ACI 318-11, and ACI 318-14. For earlier ACI 318 editions, the condition that reinforcement ratio does not exceed 0.375ρb applies. spSlab calculates vu as the absolute maximum of vAB and vCD. Local effects of concentrated loads are not computed by spSlab and must be calculated manually. 90 ACI 318-14, 8.4.2.3.2; ACI 318-11, 13.5.3.2; ACI 318-08, 13.5.3.2; ACI 318-05, 13.5.3.2; ACI 318-02, 13.5.3.2; ACI 318-99, 13.5.3.2; CSA A23.3-04, Eq. 13-8; CSA A23.3-94, Eq. 13-7 91 ACI 318-14, 8.4.2.3.4; ACI 318-11, 13.5.3.3; ACI 318-08, 13.5.3.3; ACI 318-05, 13.5.3.3; ACI 318-02, 13.5.3.3; ACI 318-99, 13.5.3.3 SOLUTION METHODS \| 98 \| 2.3.7.6. Shear Resistance at Corner Columns For the CSA A23.3 code in design mode, the program performs one-way shear resistance check in the vicinity of corner columns. A corner column is determined in spSlab as the exterior support along an exterior left or exterior right equivalent frame. For slabs with edge beams or drop panels a supplementary check including the contribution of these components should be performed manually. For the CSA A23.3-94 edition, a critical shear section is located d/2 from the column corner. The minimum length section is selected using an optimization algorithm which analyzes sections at different angles. The extension to the cantilevered portion is considered by a length not to exceed effective slab thickness d. For the CSA A23.3-14/04 edition, a critical shear section is located not further than d/2 from the edge of the column or column capital. The extension to the cantilevered portion is considered by a length not to exceed effective slab thickness d. The factored shear resistance is calculated as follows92: c c c v f   = Eq. 2-53 where: β = factor accounting for shear resistance of cracked concrete93. 92 CSA A23.3-14, 13.3.6.2; CSA A23.3-04, 13.3.6.2 93 CSA A23.3-14, 11.3.6.2 and 11.3.6.3; CSA A23.3-04, 11.3.6.2 and 11.3.6.3 SOLUTION METHODS \| 99 \| 2.3.7.7. Shear Resistance in Slab Bands When performing two-way shear analysis for models with non-continuous longitudinal slab bands a non-standard partial drop panel is anticipated to close the slab band and the calculations are performed as follows. Punching shear around the column is checked using effective depth of the slab band on one side of the column and the depth of the extension drop panel on the other side of the column. On each four sides of the column the critical section is located ½ of the respective depth from the face of the column. Punching shear calculation around the drop panel/slab band assumes that the plane of critical section, which cuts perpendicularly through slab band, is located ½ d of the slab band from the face of column. For three remaining column faces critical section is located ½ d of the slab measured from the respective edges of drop panel or slab band. SOLUTION METHODS \| 100 \| 2.3.8. One-Way Shear Analysis of Longitudinal Beams and Slabs When longitudinal beams are present in a span, the program computes the shear reinforcement requirements for the beams. Beam Transverse Reinforcement Capacity dialog box in the program output provides values of Vu, Vc, and Av/s for selected segment locations of each span. Segment lengths are chosen not to exceed the beam section depth. The beginning of first segment and the end of last segment correspond to the locations of critical sections on the left and right support respectively. The critical sections are located at a distance d, the effective beam depth, away from the column face at both the left and the right ends of the beam. However, if concentrated loads are present within distance d from the column face, critical section is selected at the column face. Vu is computed from the load acting over the entire width of the design strip. The program makes no distinction between shallow beams (αf1 l2 / l1 < 1) and deeper beams (αf1 l2 / l1 > 1). SOLUTION METHODS \| 101 \| 2.3.8.1. Shear Calculations for ACI 318 and CSA A23.3-94 Shear strength provided by concrete, Vc, is computed by94: 2 for ACI 318 0.17 for ACI 318M-11/08/05 6 for ACI 318M-02/99 0.20 for CSA A23.3-94 c w c w c c w c c w f b d f b d V f b d f b d           =        Eq. 2-54 where: ϕc = resistance factor for concrete95 equal to 0.60. In CSA A23.3-94 design, for beams without minimum stirrup reinforcement and greater than 300 mm deep, Vc is calculated from the following equation96: 260 0.10 1,000 c c c w c c w V f b d f b d d       =    +   Eq. 2-55 When Vu > ϕVc /2, the beam must be provided with at least a minimum shear reinforcement of97: ( ) ( ) ( ) ,min max 0.75 ,50 for ACI 318-11/08/05/02 50 for ACI 318-99 max 0.062 ,0.35 for ACI 318M-11/08/05 max 16,0.33 for ACI 318M-02 1/ 3 for ACI 318M-99 0.06 for CSA A23.3-94 c c v c c f f A f f         =          Eq. 2-56 94 ACI 318-14, 22.5.5.1; ACI 318-11, 11.2.1.1; ACI 318-08, 11.2.1.1; ACI 318-05, 11.3.1.1; ACI 318-02, 11.3.1.1; ACI 318-99, 11.3.1.1; ACI 318M-08, 11.2.1.1; ACI 318M-05, 11.3.1.1; ACI 318M-02, 11.3.1.1; ACI 318M-99, 11.3.1.1; CSA A23.3-94, 11.3.5.1 95 CSA A23.3-94, 8.4.2 96 CSA A23.3-94, 11.3.5.2 97 ACI 318-14, 9.6.3.3; ACI 318-11, 11.4.6.3; ACI 318-08, 11.4.6.3; ACI 318-05, 11.5.6.3; ACI 318-02, 11.5.5.2; ACI 318-99, 11.5.5.2; ACI 318M-08, 11.4.6.3; ACI 318M-05, 11.5.6.3; ACI 318M-02, 11.5.5.2; ACI 318M-99, 11.5.5.2; CSA A23.3-94, 11.2.8.4 SOLUTION METHODS \| 102 \| where: Av = area of all stirrup legs, s = stirrups spacing, bw = longitudinal beam width, fyt = yield strength of the shear reinforcement. In the investigation mode, if the ACI-318 spacing requirement for shear reinforcement98 or minimum shear reinforcement requirement are not met, the shear strength of the section is taken as one-half of the shear strength provided by concrete. When Vu > ϕVc, shear reinforcement must be provided so that: for ACI 318-11/08/05/02 for CSA A23.3-94 u c s yt yt v u c s s yt s yt V V V f d f d A V V V s f d f d      −  =   =  −  =   Eq. 2-57 where: Vu = factored shear force at the section being considered, Vs = shear strength provided by shear reinforcement, d = effective depth of the beam at the same location, ϕ = strength reduction factor for shear calculations99 equal to 0.85 for ACI 318-99 and equal to 0.75 for ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, and ACI 318-02, ϕs = resistance factor for reinforcement100 equal to 0.85. 98 ACI 318-14, 9.7.6.2.2, 9.7.6.2.3; ACI 318-11, 11.4.5; ACI 318-08, 11.4.5; ACI 318-05, 11.5.5; ACI 318-02, 11.5.4; ACI 318-99, 11.5.4 99 ACI 318-14, 21.2.1; ACI 318-11, 9.3.2.3; ACI 318-08, 9.3.2.3; ACI 318-05, 9.3.2.3; ACI 318-02, 9.3.2.3; ACI 318-99, 9.3.2.3 100 CSA A23.3-94, 8.4.3 SOLUTION METHODS \| 103 \| The capacity of shear reinforcement Vs is limited to , 8 s max c w V f b d  = ( , 0.8 s max c c w V f b d   = for CSA A23.3-94). When Vu exceeds ϕVc + ϕVs,max (Vc + Vs,max for CSA A23.3-94), the beam section dimensions must be increased or a higher concrete strength must be provided101. When 10 u c w V f b d    , the spacing is computed as: ( ) 1 sb v s n A A s = Eq. 2-58 where: Asb = stirrup bar area (one leg), n = number of stirrup legs. The maximum stirrup spacing for ACI codes102 must not exceed d/2 or 24 in when 4 s c w V f b d   0.33 s c w V f b d      . When 4 s c w V f b d   , the maximum stirrup spacing must be reduced by half, to d/4 or 12 in. For the CSA A23.3-94 standard103, maximum spacing must not exceed the smaller of 0.7d and 600 mm when 0.1 u c c w V f b d    or the smaller of 0.35d and 300 mm when 0.1 u c c w V f b d    . When 8 0.66 s c w s c w V f b d V f b d         for ACI codes and 0.8 s c c w V f b d    for CSA A23.3-94 code, the beam section dimensions must be increased or a higher concrete strength must be provided104. 101 ACI 318-14, 22.5.1.2; ACI 318-11, 11.4.7.9; ACI 318-08, 11.4.7.9; ACI 318-05, 11.5.7.9; ACI 318-02, 11.5.6.9; ACI 318-99, 11.5.6.9; CSA A23.3-94, 11.3.4 102 ACI 318-14, 9.7.6.2.2, 9.7.6.2.3; ACI 318-11, 11.4.5; ACI 318-08, 11.4.5; ACI 318-05, 11.5.5; ACI 318-02, 11.5.4; ACI 318-99, 11.5.4; CSA A23.3-94, 11.2.11 103 CSA A23-3-94, 11.2.11 104 ACI 318-14, 22.5.1.2; ACI 318-11, 11.4.7.9; ACI 318-08, 11.4.7.9; ACI 318-08, ACI 318-05, 11.5.7.9; ACI 318-02, 11.5.6.9; ACI 318-99, 11.5.6.9, CSA A23.3-94, 11.3.4 SOLUTION METHODS \| 104 \| The minimum shear reinforcement requirement is waived105 for joist construction and for beams satisfying the following criteria: For ACI 318-14, ACI 318-11 and ACI 318-08: • Beams with depth not exceeding 10 in. [250 mm]. • Beams integral with slabs (assumed by the program for all beams in two-way systems and beams within one-way slabs with overall width larger than effective beam flange width), with beam depth not exceeding 24 in. [600 mm] and not greater than the larger of 2.5 times flange thickness and 0.5 times web width. For ACI 318-05/02/99: • Beams with depth not exceeding the largest of 10 in. [250 mm], 2.5 times flange thickness, and half of web width (rectangular beams are assumed by the program to have flange thickness equal to zero and web width equal to beam width). For CSA A23.3-94: • Beams with depth not exceeding 250 mm. • Beams integral with slabs (assumed by the program for all beams in two-way systems and beams within one-way slabs with overall width larger than effective beam flange width), with beam depth not exceeding the larger of 600 mm and 0.5 times web width. 105 ACI 318-14, 7.6.3.1, 9.6.3.1; ACI 318-11, 11.4.6.1; ACI 318-08, 11.4.6.1; ACI 318-05, 11.5.6.1; ACI 318-02, 11.5.5.1; ACI 318-99, 11.5.5.1; ACI 318M-08, 11.4.6.1; ACI 318M-05, 11.5.6.1; ACI 318M-02, 11.5.5.1; ACI 318M-99, 11.5.5.1; CSA A23.3-94, 11.2.8.1 SOLUTION METHODS \| 105 \| 2.3.8.2. Shear Calculation for CSA A23.3-14/04 For CSA A23.3-04 code, the program calculates shear strength Vc provided by concrete from the following equation106: c c c w v V f b d   = Eq. 2-59 where: ϕc = resistance factor for concrete equal to 0.65 for regular and 0.70 for precast concrete, λ = factor to account for low-density concrete, bw = beam web width, dv = effective shear depth equal to greater of 0.9d or 0.72h, β = factor accounting for shear resistance of cracked concrete, c f  ≤ 8 MPa. When Vu > Vc, the beam must be provided with at least minimum shear reinforcement107. Additionally minimum shear reinforcement is required for beam sections with overall thickness exceeding 750 mm. Minimum area of shear reinforcement is calculated from the following formula108: 0.06 w v c y b s A f f  = Eq. 2-60 106 CSA A23.3-14, 11.3.4; CSA A23.3-04, 11.3.4 107 CSA A23.3-14, 11.2.8.1; CSA A23.3-04, 11.2.8.1 108 CSA A23.3-14, 11.2.8.2; CSA A23.3-04, 11.2.8.2 SOLUTION METHODS \| 106 \| Shear strength provided by shear reinforcement, Vs, is calculated from the following equation109: ( ) cot s v y v s A f d V s   = Eq. 2-61 where: ϕs = resistance factor for reinforcement steel110 equal to 0.85, Av = area of shear reinforcement within distance s, fy = yield strength of reinforcement, dv = effective shear depth equal to greater of 0.9d or 0.72h, s = spacing of transverse reinforcement, θ = the angle of inclination of diagonal compressive stresses. Spacing of transverse reinforcement, s, must not exceed the smaller111 of 0.7d and 600 mm when 0.125 u c c w V f b d    or the smaller112 of 0.35d and 300 mm when 0.125 u c c w V f b d    . When 0.25 u c w v V f b d    , the beam section dimensions must be increased or a higher concrete strength must be provided113. 109 CSA A23.3-14, 11.3.5.1; CSA A23.3-04, 11.3.5.1 110 CSA A23.3-14, 8.4.3(a); CSA A23.3-04, 8.4.3(a) 111 CSA A23.3-14, 11.3.8.1; CSA A23.3-04, 11.3.8.1 112 CSA A23.3-14, 11.3.8.3; CSA A23.3-04, 11.3.8.3 113 CSA A23.3-94, 11.3.3 SOLUTION METHODS \| 107 \| The program recognizes special member types and assumes values of β = 0.21 and θ = 42° in the following cases114: • Slabs (including slab bands for CSA code) having thickness not exceeding 350 mm. • Beams having thickness not exceeding 250 mm. • Concrete joist construction. • Beams cast monolithically with the slab and having the depth below the slab not exceeding one-half of the width or 350 mm. For other general cases the program utilizes the so called simplified method. The value of θ is assumed as 35°. For sections having or requiring at least minimum transverse reinforcement β = 0.18 is assumed. For sections with no transverse reinforcement the value of β is calculated as follows115: 230 1,000 v d = + Eq. 2-62 114 CSA A23.3-14, 11.3.6.2; CSA A23.3-04, 11.3.6.2 115 CSA A23.3-14, 11.3.6.3(b); CSA A23.3-04, 11.3.6.3(b) SOLUTION METHODS \| 108 \| 2.3.8.3. Shear Distribution When no ribs are present, one way shear is proportioned to the slab and beam according to the following ratios: 1 2 1 1 2 1 / , 1 / f f l l l l   − Eq. 2-63 When ribs are present (joist systems), one way shear is proportioned to the slab and beam according to the following ratios of cross-section areas: , ribs beam ribs beam ribs beam A A A A A A + + Eq. 2-64 Per requirement116 of CSA A23.3-14/04, the program allows distributing one-way shear in the slab between column and middle strips using the distribution factors which are proportional to the factors used for negative moment distribution. The fraction of the shear transferred to the beam remains unchanged irrespective of the use of this feature. This functionality is also provided for other design codes, to be selected at engineer’s discretion. 116 CSA A23.3-14, 13.3.6.1; CSA A23.3-04, 13.3.6.1 SOLUTION METHODS \| 109 \| 2.3.8.4. One-Way Shear in Slab Bands (CSA A23.3-14/04) One-way shear calculations in slab bands are done similar to shear in two-way slabs, except the column strip is substituted by the band strip. Shear forces are distributed between the band and the middle strip proportionally to negative moment distribution factors. Transverse reinforcement is not considered. 2.3.8.5. Shear in Drop Panels When calculating one-way shear capacity for two-way solid and waffle slabs, the contribution of the drop panel cross-section can be optionally selected. For such slabs, the shear capacity is calculated in three regions, with increased Vc values in support (drop panel) locations. In case shear is distributed into column and middle strips, drop panel contribution is divided according to the share of drop panel cross-section area in each strip. SOLUTION METHODS \| 110 \| 2.3.9. Torsion and Shear Torsion analysis can be engaged for beam and one-way systems by selecting YES for Consider Torsion run option located in the Project Left Panel under Project command in the Ribbon. As far as torsional analysis is concerned, it is assumed that columns provide perfectly rigid supports so there is no transfer of torsional moments between spans. Within a span, torsional moments are considered only if a longitudinal beam is present. Torsion can be induced by concentrated and redistributed torsional loads and also, in the case of a beam with unsymmetrical cross sections, by self weight and area loads. A T-section with different flange widths is an example of a cross section which is not symmetrical. It can be obtained if a beam and a slab with different left and right widths are combined in the same span. However, in order for a flange to be considered in the torsional analysis its thickness has to be greater than twice the cover. If a flange is wider than the effective width then only the effective width is taken into account. The design for torsion is based on a thin-walled tube, space truss analogy. For the Canadian code the simplified method is used. The program allows both equilibrium and compatibility torsion conditions. In the equilibrium mode, which is assumed by default, unreduced total value of the torsional design moment is used in the design. In the compatibility mode117, factored torsional moments that exceed cracking moment Tcr (0.67Tcr for CSA) are reduced to the value of Tcr (0.67Tcr for CSA). However, it is user’s responsibility to determine which mode is appropriate and the program does not perform any redistribution of internal forces if compatibility torsion is selected. If torsion analysis is engaged then both torsion and shear actions contribute to the amount of required transverse (stirrup) reinforcement. However, additional longitudinal bars distributed along the perimeter of a cross-section are also required to provide torsional capacity. 117 ACI 318-14, 22.7.5.1; ACI 318-11, 11.5.2.2; ACI 318-08, 11.6.2.2; ACI 318-05, 11.6.2.2; ACI 318-02, 11.6.2.2; ACI 318-99, 11.6.2.2; CSA A23.3-14, 11.2.9.2; CSA A23.3-04, 11.2.9.2; CSA A23.3-94, 11.2.9.2 SOLUTION METHODS \| 111 \| For torsion design a span is divided into segments in the same way as for shear design. Governing values within a segment are used to design the whole segment. For stirrups, the governing values of torsional moment and shear force (acting simultaneously) will be these that produce the highest intensity of required stirrup area. On the other hand, the required area of longitudinal bars depends only on the torsional moment so the highest absolute value of torsional moment will govern. Since stirrup area depends both on shear and torsion whereas longitudinal bar area depends only on torsion, the governing values for stirrups and longitudinal bars can occur at different locations within a segment and for different load combinations. Governing values along with their location and associated load combination are provided in the design results report. Effect of torsion within a segment will be neglected if the factored torsional moment, Tu, at every segment location is less than one fourth of the torsion cracking moment, Tcr, which equals: for ACI code118 2 4 cp cr c cp A T f p   = Eq. 2-65 for CSA A23.3-94 code119 2 0.4 cp cr c c cp A T f p   = Eq. 2-66 for CSA A23.3-14/04 code120 2 0.38 cp cr c c cp A T f p   = Eq. 2-67 Acp denotes the area enclosed by outside perimeter of concrete section and pcp is equal to the outside perimeter of concrete section. 118 ACI 318-08, R11.5.1; ACI 318-05, R11.6.1; ACI 318-02, R11.6.1; ACI 318-99, R11.6.1 119 CSA A23.3-94, 11.2.9.1 120 CSA A23.3-04, 11.2.9.1 SOLUTION METHODS \| 112 \| To be adequate for torsion design, a section has to be proportioned in such a way that combined shear stress due to shear and torsion does not exceed the limit value specified by the code. In ACI code this condition reads as121: 2 2 2 8 1.7 u u h c c w oh w V T p V f b d A b d         +  +             Eq. 2-68 The simplified method of CSA A23.3-94 standard defines this relation as122: 2 0.25 u u h c c w oh V T p f b d A   +  Eq. 2-69 Similar requirement for CSA A23.3-04 reads as follows123: 2 2 2 0.25 1.7 u u h c c w oh V T p f b d A       +          Eq. 2-70 In above relations, Aoh is the area enclosed by centerline of the outermost closed transverse reinforcement and ph is the perimeter of that area. By default, flanges do not contribute to Aoh and ph. For sections with flanges, flanges will only be taken into account for Aoh and ph if the option to include stirrups in flanges is engaged in the torsion design. In the program output, the combined stress (left hand side of the above inequalities) is denoted as vf and the limit value as ϕsvt. 121 ACI 318-14, 22.7.7.1; ACI 318-11, 11.5.3.1; ACI 318-08, 11.5.3.1; ACI 318-05, 11.6.3.1; ACI 318-02, 11.6.3.1; ACI 318-99, 11.6.3.1 122 CSA A23.3-94, 11.3.9.8 123 CSA A23.3-14, 11.3.10.4(b); CSA A23.3-04, 11.3.10.4(b) SOLUTION METHODS \| 113 \| The required intensity of stirrup area to provide required torsional capacity is calculated from the following formula124: ( ) for ACI 318 2 for CSA A23.3-94 2 for CSA A23.3-14/04 2 cot u o yt t u s o yt u s o yt T A f A T s A f T A f          =       Eq. 2-71 where the gross area enclosed by the shear path125, Ao, is taken as 0.85Aoh × At /s is the quantity per stirrup leg. Concrete shear and torsion strength reduction factor126, ϕ, for ACI-318 codes is equal to 0.75 for the 99 edition and 0.75 for later editions. The total requirement for stirrup intensity combining shear and torsion equals127: 2 2 v t v t A A A s s s + = + Eq. 2-72 124 ACI 318-14, 22.7.6.1; ACI 318-11, 11.5.3.6; ACI 318-08, 11.5.3.6; ACI 318-05, 11.6.3.6; ACI 318-02, 11.6.3.6; ACI 318-99, 11.6.3.6; CSA A23.3-14, 11.3.10.3; CSA A23.3-04, 11.3.10.3; CSA A23.3-94, 11.3.9.4 125 CSA A23.3-14, 11.3.10.3; CSA A23.3-04, 11.3.10.3; CSA A23.3-94, 11.3.9.7 126 ACI 318-14, 21.2.1; ACI 318-11, 9.3.2.3; ACI 318-08, 9.3.2.3; ACI 318-05, 9.3.2.3; ACI 318-02, 9.3.2.3; ACI 318-99, 9.3.2.3 127 ACI 318-14, R9.5.4.3; ACI 318-11, R11.5.2.8; ACI 318-08, R11.5.3.8; ACI 318-05, R11.6.3.8; ACI 318-02, R11.6.3.8; ACI 318-99, R11.6.3.8 SOLUTION METHODS \| 114 \| This value cannot be taken less than minimum stirrup area required by the codes. The minimum code requirements can be written in the following form128: ( ) ( ) ( ) 2 max 0.75 ,50 for ACI 318-11/08/05/02 50 for ACI 318-99 max 0.062 ,0.35 for ACI 318M-11/08/05 max 16,0.33 for ACI 318M-02 0.33 for ACI 318M-99 0.06 for CSA A23.3-04/94 c c w v t yt c c f f b s A f f f +         =          Eq. 2-73 In addition to stirrup spacing requirement defined for shear, program imposes one more torsion specific requirement for all ACI codes129 which limits the spacing to the smallest of ph/8, and 12 in [300 mm]. Based on the total required stirrup area intensity and spacing requirements, the program attempts to select stirrups taking also into account that if stirrups with more than two legs have to be used then the area of an outer leg must not be less than At. 128 ACI 318-14, 9.6.4.2; ACI 318-11, 11.5.5.2; ACI 318-08, 11.5.5.2; ACI 318-05, 11.6.5.2; ACI 318-02, 11.6.5.2; ACI 318-99, 11.6.5.2; ACI 318M-11, 11.5.5.2; ACI 318M-08, 11.5.5.2; ACI 318M-05, 11.6.5.2; ACI 318M-02, 11.6.5.2; ACI 318M-99, 11.6.5.2; CSA A23.3-14, 11.2.8.2; CSA A23.3-04, 11.2.8.2; CSA A23.3-94, 11.2.8.4 129 ACI 318-14, 9.7.6.3.3; ACI 318-11, 11.5.6.1; ACI 318-08, 11.5.6.1; ACI 318-05, 11.6.6.1; ACI 318-02, 11.6.6.1; ACI 318-99, 11.6.6.1 SOLUTION METHODS \| 115 \| 2.3.9.1. Additional Longitudinal Reinforcement for ACI 318 and CSA A23.3-94 The area of additional longitudinal reinforcement, Al, is calculated from130: 2 u h l o y T p A A f  = Eq. 2-74 For ACI code it is also checked against the following minimum value131: ,min 5 c cp yt t l h y y f A f A A p f s f    = −    Eq. 2-75 where At / s is calculated from Eq. 2-56 but is not taken less than 25bw / fyt. Longitudinal bars are selected in such a way that their area is not less than Al ≥ Al,min and that number of longitudinal bars in a section is enough to provide a bar in every corner of a stirrup and preserve spacing between bars not higher than 12 in [300 mm]. Also, bar sizes are selected not to have diameter less than No. 3 bar and not less than 1/24 of stirrup spacing for ACI codes132 and 1/16 for CSA standard133. 130 ACI 318-14, 22.7.6.1; ACI 318-11, 11.5.3.7; ACI 318-08, 11.5.3.7; ACI 318-05, 11.6.3.7; ACI 318-02, 11.6.3.7; ACI 318-99, 11.6.3.7; CSA A23.3-94, 11.3.9.5 131 ACI 318-14, 9.6.4.3, 9.7.5.1; ACI 318-11, 11.5.5.3; ACI 318-08, 11.5.5.3; ACI 318-05, 11.6.5.3; ACI 318-02, 11.6.5.3; ACI 318-99, 11.6.5.3 132 ACI 318-14, 9.7.5.1, 9.7.5.2; ACI 318-11, 11.5.6.2; ACI 318-08, 11.5.6.2; ACI 318-05, 11.6.6.2; ACI 318-02, 11.6.6.2; ACI 318-99, 11.6.6.2 133 CSA A23.3-14, 11.2.7; CSA A23.3-04, 11.2.7; CSA A23.3-94, 11.2.7 SOLUTION METHODS \| 116 \| 2.3.9.2. Additional Longitudinal Reinforcement for CSA A23.3-14/04 The additional longitudinal reinforcement, Al, will only be calculated for CSA A23.3-14/04 if COMBINED M-V-T REINF. DESIGN option is unchecked in the Design Options tab under Solve command in the Ribbon. If this option is checked (default setting) then no additional longitudinal reinforcement is calculated because the regular top and bottom reinforcement will automatically be proportioned to resist combined action of flexure, shear and torsion. Proportioning of longitudinal reinforcement for sections subjected to combined shear and torsion in flexural regions is based on the requirement that the resistance of the longitudinal reinforcement has to be greater or equal to the axial force that can be developed in this reinforcement. In sections with no axial action (Nf = 0 and Vp = 0) that force is equal to134: • flexural tension side ( ) 2 2 , , , , 0.45 cot 0.5 2 f h f lt f s lt flexure lt shear v o lt flexure lt shear M p T F V V F F d A F F    = + − + = +     Eq. 2-76 • flexural compression side ( ) 2 2 , , , , 0.45 cot 0.5 2 f h f lc f s lc flexure lc shear v o lc flexure lc shear M p T F V V F F d A F F    = − + − + = +     Eq. 2-77 134 CSA A23.3-14, 11.3.9.2, 11.3.9.3, 11.3.10.6; CSA A23.3-04, 11.3.9.2, 11.3.9.3, 11.3.10.6 SOLUTION METHODS \| 117 \| These forces can be decomposed135 into flexure and shear components. The flexure components, Flt,flexure and Flc,flexure, account for the action of the bending moment, Mf, whereas the shear components, Flt,shear and Flc,shear, account for the action of the shear force, Vf, and the torsional moment, Tf. The amounts of reinforcement needed to resist the flexure components are calculated separately in the flexure and axial design procedure. The total amount of the additional longitudinal reinforcement, Al, needed to resist shear and torsion will be determined as follows: ( ) 2 2 , , 0.45 2cot 0.5 2 h f f s o lt shear lc shear l s y s y p T V V A F F A f f      − +   +   = = Eq. 2-78 If only torsion is present (Vf = 0 and Vs = 0), then (assuming136 θ = 35º) Al would reduce to ( ) 0.45 2 2cot 35º 1.285 2 h f h f o l s y o s y p T p T A A f A f         = = Eq. 2-79 which is comparable (and conservative) to the additional amount of longitudinal reinforcement due to torsion required in accordance with the previous edition of the CSA A23.3 standard137. 135 See Eq. 7-42, pp 294 in Ref. 136 CSA A23.3-14, 11.3.6.3; CSA A23.3-04, 11.3.6.3 137 CSA A23.3-94, 11.3.9.5 SOLUTION METHODS \| 118 \| 2.3.9.3. Investigation Mode In the investigation mode when transverse and longitudinal reinforcement is input by the user, the program checks the combined shear and torsional capacity of the system in terms of required and provided reinforcement area. In other words, the provided area of reinforcement is compared to the area of reinforcement required to resist applied loads. This is a different approach than for flexure and shear actions without coupling where design forces are directly compared to capacity. In the case where torsion and shear stirrup requirements are combined, the approach of comparing total reinforcement area is more convenient since it does not require dividing stirrup area into a part that resists torsion only and a part that resists shear only. For consistency, additional longitudinal reinforcement required for torsion and shear is also checked in terms of provided and required area. Other requirements, e.g. bar or stirrup spacing, number of longitudinal bars, area of stirrup outer leg, and combined stresses in concrete due to shear and torsion are checked also. Exceeded capacity and other conditions are flagged in the Design Results section of the report. SOLUTION METHODS \| 119 \| 2.3.10. Deflections 2.3.10.1. Instantaneous Deflections Instantaneous deflections are obtained directly by the program from elastic analysis of the defined system for three load levels. The first corresponds to dead load only, the second corresponds to dead load plus sustained part of live load only, and the third corresponds to dead load plus live load on all spans (total deflection). The deflection occurring when the live load is applied can be computed as the total load deflection (due to the dead and the live load) minus the dead load only deflection138. Depending on the option selected by the user, the program will calculate flexural stiffness of the members based on either gross moment of inertia or the effective moment of inertia which takes cracking into account. Live Total Dead  =  − The program results section provides detailed summary of the frame section properties, frame effective section properties, column and middle strip properties at midspan, and a summary of extreme deflection values for each load level along the span. 138 Example 9-5 Calculation of Immediate Deflections in Ref. , pp. 443, Step 5 SOLUTION METHODS \| 120 \| 2.3.10.2. Cracking When calculating the deflections for effective (cracked) section properties, the frame solution is obtained for three load levels: dead load, dead load plus sustained part of live load, and dead load plus full live load on all spans. Flexural stiffness is assumed corresponding to the load level. A reduction in the flexural stiffness caused by cracking leads to an increase in deflections. Several methods of deflection analyses taking cracking into account are reviewed in Ref. . The program uses the approach based on the effective moment of inertia as permitted by the code139. The effective moment of inertia, Ie, developed by Branson (Ref. ) and incorporated into the code equals: 3 3 max max 1 cr cr e g cr M M I I I M M         = + −             Eq. 2-102 where: Ig = moment of inertia of the gross uncracked concrete section, Icr = moment of inertia of the cracked transformed concrete section140, Mcr = cracking moment, Mmax = maximum bending moment at the load level for which the deflection is computed. To calculate Ie for two-way slabs, the values of all terms for the full width of the equivalent frame are used in Eq. 2-102. This approach averages the effects of cracking in the column and middle strips. 139 ACI 318-14, 19.2.3.1, 24.2.3.5; ACI 318-11, 9.5.2.3; ACI 318-08, 9.5.2.3; ACI 318-05, 9.5.2.3; ACI 318-02, 9.5.2.3; ACI 318-99, 9.5.2.3; CSA A23.3-14, 9.8.2.3; CSA A23.3-04, 9.8.2.3; CSA A23.3-94, 9.8.2.3 140 See formulas for various cross sections in Table 10-2 in Ref. SOLUTION METHODS \| 121 \| The value of Ie at midspan for a simple span and at support for a cantilever is taken141 to calculate flexural stiffness of a member. For other conditions, an averaged effective moment of inertia, Ie,avg is used. For spans with both ends continuous, Iframe is given by142: ( ) , , , 0.70 0.15 e avg e e l e r I I I I + − − = + + Eq. 2-103 where: Ie+ = effective moment of inertia for the positive moment region, Ie,l– = effective moment of inertia for the negative moment region at the left support, Ie,r– = effective moment of inertia for the negative moment region at the right support. For spans with one end continuous the value of Iframe is given by143: , 0.85 0.15 e avg e e I I I + − = + Eq. 2-104 where: Ie+ = effective moment of inertia for the positive moment region, Ie– = effective moment of inertia for the negative moment region at the continuous end. 141 ACI 318-14, 24.2.3.6, 24.2.3.7; ACI 318-11, 9.5.2.4; ACI 318-08, 9.5.2.4; ACI 318-05, 9.5.2.4; ACI 318-02, 9.5.2.4; ACI 318-99, 9.5.2.4 142 ACI 435R-95 (Ref. ), 2.5.1, Eq. (2.15a); CSA A23.3-14, 9.8.2.4(a); CSA A23.3-04, 9.8.2.4(a); CSA A23.3-94, 9.8.2.4, Eq. 9.3 143 ACI 435R-95 (Ref. ), 2.5.1, Eq. (2.15b); CSA A23.3-14, 9.8.2.4(b); CSA A23.3-04, 9.8.2.4(b); CSA A23.3-94, 9.8.2.4, Eq. 9.4 SOLUTION METHODS \| 122 \| 2.3.10.3. Long-Term Deflections The program estimates additional long-term deflection resulting from creep and shrinkage, Δcs, by multiplying the immediate deflection due to sustained load, Δsust, by the factor, λΔ, equal to144: 1 50    =  + Eq. 2-105 where: ξ = time dependent factor with the maximum value of 2.0 (the actual value is interpolated from the values and the chart given in the code145 based on the load duration specified by the user in the input), ρ' = ratio of compressive reinforcement at midspan for simple and continuous spans and at support for cantilevers. Deflection due to the sustained load, Δsust, is the deflection induced by the dead load (including self weight), plus sustained portion of the live load. And long-term deflection resulting from creep and shrinkage equals: cs sust  =  Eq. 2-106 The program calculates incremental deflection which occurs after partitions are installed in two ways. In the first approach, it is assumed that the live load has been applied before installing the partitions and the incremental deflection equals146: ( ) cs lu cs total sust +  =  +  − Eq. 2-107 144 ACI 318-14, 24.2.4.1.1, 24.2.4.1.2, 24.2.4.1.3; ACI 318-11, 9.5.2.5; ACI 318-08, 9.5.2.5; ACI 318-05, 9.5.2.5; ACI 318-02, 9.5.2.5; ACI 318-99, 9.5.2.5; CSA A23.3-14, 9.8.2.5; CSA A23.3-04, 9.8.2.5; CSA A23.3-94, 9.8.2.5A23.3 145 Fig. R24.2.4.1; Fig. R9.5.2.5 in ACI 318-11, ACI 318-08; ACI 318-05; ACI 318-02, and ACI 318-99; Fig. N9.8.2.6 in CSA A23.3-04 and CSA A23.3-94 146 CSA A23.3-04 N9.8.2.5, CSA A23.3-94 N9.8.2.5 SOLUTION METHODS \| 123 \| In the second approach, the assumption is that the full live load, including the sustained portion of the live load, has been applied after the partitions are installed which results in the incremental deflection equal to147: cs l cs live +  =  +  Eq. 2-108 The total long-term deflection (Δtotal)lt is also calculated as148: ( ) ( ) ( ) 1 total sust total sust lt   =  + +  − Eq. 2-109 147 See Example 10.1 in Ref. 148 CSA A23.3-04 N9.8.2.5; CSA A23.3-94 N9.8.2.5 SOLUTION METHODS \| 124 \| 2.3.10.4. Deflections of Two-Way Systems Calculation of deflections of reinforced concrete two-way slabs is complicated by a large number of significant parameters such as: the aspect ratio of the panels, the vertical and torsional deflection of supporting beams, the stiffening effect of drop panels and column capitals, cracking, and the time-dependent nature of the material response. Based on studies (Ref. -), an approximate method consistent with the equivalent frame method was developed (Ref. ) to estimate the column and middle strip deflections. Under vertical loads, Reference 20 indicates that the midspan deflection of an equivalent frame can be considered as the sum of three parts: that of the panel assumed to be fixed at both ends of its span, Δf,ref and those due to the known rotation at the two support lines, Δθ,L and Δθ,r. Calculation of midspan deflection of the column strip or the middle strip under fixed-end conditions is based on M/EI ratio of the strip to that of the full-width panel: , , strip c frame f strip f ref frame c strip M E I M E I  =  Eq. 2-110 The ratio (Mstrip/Mframe) can be considered as a lateral distribution factor, LDF. For ACI and CSA A23.3-94 codes the lateral distribution factor, LDF, at an exterior negative moment region is: 2 2 , 1 1 1 100 10 12 1 neg ext t t f l l LDF l l       = − + −       Eq. 2-111 The LDF at an interior negative moment region is 2 2 , 1 1 1 75 30 1 neg int f l l LDF l l     = − −       Eq. 2-112 The LDF at a positive moment region is 2 2 1 1 1 60 30 1.5 pos f l l LDF l l     = + −       Eq. 2-113 SOLUTION METHODS \| 125 \| where: αf1 = the ratio of flexure stiffness of a beam section to the flexural stiffness of a width of slab bounded laterally by centerlines of adjacent panels on either side of the beam, βt = ratio of torsional stiffness of an edge beam section to the flexural stiffness of a width of slab equal to the span length of the beam, center-to-center of the supports (see Eq. 2-30). For CSA A23.3-14/04 code lateral distribution factors are based on tabulated values presented earlier in the chapter. When αf1 l2 / l1 is greater than 1.0, αf1 l2 / l1 will be set equal to 1.0. The column and middle strip LDF’s can be computed by: , , 2 2 neg l neg r pos c LDF LDF LDF LDF + + = Eq. 2-114 100 m c LDF LDF = − Eq. 2-115 where: LDFneg,l = LDF for the negative moment region at the left end of the span, LDFneg,r = LDF for the negative moment region at the right end of the span. The total midspan deflection for the column or middle strip is the sum of three parts: , , , strip f strip l r    =  +  +  Eq. 2-116 where: Δθ,l, Δθ,r = midspan deflection due to rotation of left and right supports, respectively. The above procedure was implemented starting in v5.00 to follow the reference recommendations exactly and eliminate overestimation of the column strip deflection and underestimation of the middle strip deflection especially for the exterior span. SOLUTION METHODS \| 126 \| The deflections should be used in conjunction with the deflections obtained from an analysis in the transverse direction. For square panels (l1 = l2), the mid-panel deflection is obtained from the following equation as shown in Figure 2.27. cy mx cx my =  +  =  +  Eq. 2-117 For rectangular panels, (l1 ≠ l2), the mid panel deflection is obtained from: ( ) ( ) 2 cy mx my cx  +  +  +  = Eq. 2-118 SOLUTION METHODS \| 127 \| Figure 2.27 – Deflection Computation for a Square Panel SOLUTION METHODS \| 128 \| 2.4. Design Methods 2.4.1. Area of Reinforcement The program calculates the required area of reinforcement (top and bottom) based on the values of bending moment envelope within the clear span. For rectangular sections with no compression reinforcement, the design flexural strength of the column strip, middle strip and beam must equal the factored design moment: ( ) 2 0.85 s y u y s c A f M f A d f b    = −        Eq. 2-80 The reinforcement can therefore be computed from: 2 0.85 2 0.85 c u s y c f b M A d d f f b     = − −        Eq. 2-81 For CSA A23.3: 1 2 s s y r s y s c c A f M f A d f b      = −      Eq. 2-82 The effective depth of the section is taken as the overall section depth minus the distance from the extreme tension fiber to the tension reinforcement centroid. The column strip depth may include all or part of the drop panel depth. The drop depth will not be included in the effective depth of the column strip when the drop does not extend at least one-sixth the center-to-center span length in all directions, or when the drop depth below the slab is less than one-quarter the slab depth. If the drop extends at least one-sixth the center-to-center span length and the drop depth is greater than one-quarter the distance from the edge of the drop panel to the face of the column or column capital, the excess depth will not be included in the column strip effective depth. If the drop width is less than the column strip width, the drop width will be used in the computation of the required reinforcement. SOLUTION METHODS \| 129 \| When computing negative slab reinforcement and additional reinforcement for negative unbalanced moments over the supports, the contribution of the depth of transverse beam can be optionally selected. The contribution of transverse beam will be considered, if it extends beyond the critical section and if its depth exceeds the depth of the drop panel. The increase of the slab thickness is limited to ¼ of the extent of the transverse beam beyond the face of support, identical to design depth limitations for drop panels. If transverse beam depth exceeds the limit, excess depth is disregarded in the reinforcement calculations. For two-way slabs with beams, an option exists when designing reinforcement for positive bending moments, to include a portion of slab as beam flanges149(T-Section). The width of the column strip is then decreased accordingly. The extent of the flanges on each side is limited to four times slab thickness and not more that the projection of the beam under the slab. When this option is not selected, beam geometry is treated as rectangular. When calculating required reinforcement for negative bending moments the geometry of the beam is treated as rectangular, having beam width equal to web width. However, when a T-Section is selected, reinforcing bar design is performed assuming that they are distributed across the beam width including the flanges. For the ACI 318-99 code the strength reduction factor for flexure calculations is specified as ϕ = 0.90150. For the ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, and ACI 318-02 codes the strength reduction factor for tension-controlled sections (εt ≥ 0.005) is equal ϕ = 0.90. For transition sections (fy / Es ≤ εt ≤ 0.005) the strength reduction factor can be linearly interpolated by the formula151: ( ) 0.90 0.65 0.65 / 0.005 / t y s y s f E f E   − = + − − Eq. 2-83 149 See footnote 50 150 ACI 318-99, 9.3.2 151 ACI 318-14, 21.2.1; ACI 318-11, 9.3.2; ACI 318-08, 9.3.2; ACI 318-05, 9.3.2; ACI 318-02, 9.3.2 SOLUTION METHODS \| 130 \| ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, and ACI 318-02 codes specify the strength reduction factor for compression controlled sections (εt < fy / Es) as equal ϕ = 0.65. The reduction factors for transition or compression controlled sections have application primarily in investigation mode of the program. In design mode the program performs the calculations assuming a tension controlled section (εt ≥ 0.005) or a section with compressive reinforcement (if enabled). The ACI 318-99 code152 requires keeping the steel ratio below the maximum value, ρmax, equal to 75% of steel ratio producing balanced strain condition, ρb, where153: 1 87 0.85 87 c d y y f f f    = + Eq. 2-84 with 1 0.85 for 4 ksi 0.65 for 8 ksi 1.05 0.05 for 4 ksi 8 ksi c c c c f f f f      =      −    For CSA code the value of ρmax equals ρb and is calculated as follows154: max 1 1 700 700 c c b s y y f f f      = = + Eq. 2-85 where: α1 = 0.85 0.0015 0.67 c f  −  , β1 = 0.97 0.0025 0.67 c f  −  . 152 ACI 318-99, 10.3.3 153 ACI 318-99, 8.4.3 154 CSA A23.3-14, 10.5.2; CSA A23.3-04, 10.5.2; CSA A23.3-94, 10.5.2; Eq. 4-24, pp 110 in Ref. SOLUTION METHODS \| 131 \| The ACI 318-14, ACI 318-11, ACI 318-08, ACI 318-05, and ACI 318-02 codes control the amount of reinforcement by limiting the value of net tensile strain (εt ≥ 0.004)155. The program satisfies this condition by assuming a tensioned controlled section with εt ≥ 0.005. From this assumption the equivalent maximum reinforcement ratio for rectangular section can be written as: 1 max 0.85 0.003 0.003 0.005 c y f f    = + Eq. 2-86 If the calculated reinforcement exceeds the maximum allowed, a message will appear in the output. In such cases, it is recommended that the engineer review the slab thickness to ensure a more satisfactory design. If compression reinforcement calculations are enabled, the program will attempt to add compression reinforcement to the section. The program is capable to design compressive reinforcement for any design strip (column, middle, and beam) including also unbalanced moment strip156. The amount of reinforcement provided will not be less than the code prescribed minimum. For the ACI 318 code, the minimum ratio of reinforcement area to the gross sectional area of the slab strip using Grade 60 reinforcement is taken as 0.0018. When reinforcement yield strength exceeds 60 ksi, the minimum ratio is set to 0.0018 × 60 / fy. For reinforcement with yield strength less than 60 ksi, the minimum ratio is set to 0.0020. In no case will this ratio be less than 0.0014 (See Table 2.5)157. The CSA Standard requires a minimum ratio of slab reinforcement area to gross sectional area of the slab strip equal to 0.002 for all grades of reinforcement158. 155 ACI 318-14, 7.3.3.1, 8.3.3.1, 9.3.3.1; ACI 318-11, 10.3.5; ACI 318-08, 10.3.5; ACI 318-05, 10.3.5; ACI 318-02, 10.3.5 156 ACI 318-14, 8.4.2.3.2, 8.4.2.3.3; ACI 318-11, 13.5.3.2; ACI 318-08, 13.5.3.2; ACI 318-05, 13.5.3.2; ACI 318-02, 13.5.3.2; ACI 318-99, 13.5.3.2; CSA A23.3-14, 13.3.5.3; CSA A23.3-04, 13.3.5.3; CSA A23.3-94, 13.11.2 157 ACI 318-14, 7.6.1.1, 8.6.1.1; ACI 318-11, 7.12.2.1; ACI 318-08, 7.12.2.1; ACI 318-05, 7.12.2.1; ACI 318-02, 7.12.2.1; ACI 318-99, 7.12.2.1 158 CSA A23.3-14, 7.8.1; CSA A23.3-04, 7.8.1; CSA A23.3-94, 7.8.1 SOLUTION METHODS \| 132 \| fy (ksi) As / Ag < 60 0.0020 ≥ 60 0.0018 60 0.0014 y f   Table 2.5 – Minimum Ratios of Reinforcement to Gross Concrete Area According to ACI code for beams and positive moment regions of joist slabs, minimum reinforcement provided will not be less than159: ,min 3 c s w y f A b d f  = Eq. 2-87 and not less than 200bwd / fy where bw is the web width of the section. For statically determinate sections with flange in tension, bw is replaced by the smaller of 2bw and the width of the flange. Similar equation prescribed by CSA A23.3 code has the form160: ,min 0.2 c s t y f A b h f  = Eq. 2-88 where bt is the width of the tension zone of the section. Additionally, for T-sections having flange in tension the CSA code limits value of bt to 1.5bw for single sided flanges and to 2.5bw for double sided flanges. When designing reinforcement for longitudinal slab bands according to CSA code, program assumes identical minimum steel requirements as for beams. 159 ACI 318-14, 9.6.1.1, 9.6.1.2; ACI 318-11, 10.5.1; ACI 318-08, 10.5.1; ACI 318-05, 10.5.1; ACI 318-02, 10.5.1; ACI 318-99, 10.5.1 160 CSA A23.3-14, 10.5.1.2(b); CSA A23.3-04, 10.5.1.2(b); CSA A23.3-94, 10.5.1.2(b) SOLUTION METHODS \| 133 \| 2.4.1.1. Design for Combined Flexure, Shear, and Torsion CSA A23.3-14/04 requires, in proportioning of longitudinal reinforcement, to include additional tension forces caused by shear and torsion161. To achieve this, the program calculates forces developed in the longitudinal reinforcement due to flexure, shear, and torsion. On the flexural tension side the force in longitudinal reinforcement is equal to162: ( ) 2 2 0.45 cot 0.5 2 f h f lt f s v o M p T F V V d A    = + − +    Eq. 2-89 On the flexural compression side the force in longitudinal reinforcement is equal to163: ( ) 2 2 0.45 cot 0.5 2 f h f lc f s o v M p T F V V A d    = − + −     Eq. 2-90 but not less than zero. For these forces, longitudinal reinforcement area is calculated from the following equations164: lt lt c y F A f  = Eq. 2-91 lc lc c y F A f  = Eq. 2-92 161 CSA A23.3-14, 11.3.9; CSA A23.3-04, 11.3.9 162 CSA A23.3-14, 11.3.9.2 and 11.3.10.6; CSA A23.3-04, 11.3.9.2 and 11.3.10.6 163 CSA A23.3-14, 11.3.9.3 and 11.3.10.6; CSA A23.3-04, 11.3.9.3 and 11.3.10.6 164 CSA A23.3-14, 11.3.9.1; CSA A23.3-04, 11.3.9.1 SOLUTION METHODS \| 134 \| Taking into account both positive and negative bending moments (resulting from all load combinations and load patterns) and checking against area of steel required for flexure only, the final areas of top and bottom reinforcement can be calculated from:     max , if 0 max , if 0 s lc f top s lt f A A M A A A M     =     Eq. 2-93     max , if 0 max , if 0 s lt f bot s lc f A A M A A A M    =      Eq. 2-94 SOLUTION METHODS \| 135 \| 2.4.2. Concentration and Additional Reinforcement spSlab computes the fraction of the unbalanced moment, γf Mu, that must be transferred by flexure within an effective slab width (a band) equal to the column width plus one and one-half the slab or drop panel depth (1.5h) on either side of the column where165 1 2 1 2 1 / 3 f b b  = + Eq. 2-100 The amount of reinforcement required to resist this moment is computed. The amount of reinforcement already provided for flexure is then computed from the bar schedule (i.e. the number of bars that fall within the effective slab width multiplied by the area of each bar). Depending on load conditions, additional negative or positive reinforcement may be required. If the reinforcement area provided for flexure is greater than or equal to the reinforcement requirements to resist moment transfer by flexure, no additional reinforcement is provided, and the number of additional bars will be set to 0. If the amount of reinforcement provided for flexure is less than that required for moment transfer by flexure, additional reinforcement is required. The additional reinforcement is the difference between that required for unbalanced moment transfer by flexure and that provided for design bending moment in the slab, and it is selected based on the bar size already provided at the support. For ACI codes the value of γf on selected supports can be automatically adjusted to the maximum permitted value. The corresponding value of γv = 1 – γf is adjusted accordingly. This option allows relaxing stress levels for two-way shear around the columns by transferring increased part of the unbalanced moment through flexure. The adjustment is performed independently for each load case and pattern. If for given load case the corresponding two-way shear Vu exceeds the appropriate limits 0.75ϕVc at an edge support, 0.5ϕVc at a corner support, or 0.4ϕVc at an interior support, adjustment of both factors is not performed. When the adjustment of γf and γv factors is selected, the reinforcement calculated within the transfer width should be limited according to the code to 165 ACI 318-14, 8.4.2.3.2, 8.4.2.3.3; ACI 318-11, 13.5.3.2; ACI 318-08, 13.5.3.2; ACI 318-05, 13.5.3.2; ACI 318-02, 13.5.3.2; ACI 318-99, 13.5.3.2; CSA A23.3-14, 13.3.5.3; CSA A23.3-04, 13.3.5.3; CSA A23.3-94, 13.11.2 and 13.4.5.3 SOLUTION METHODS \| 136 \| reinforcement ratio ρ < 0.375ρb, as stipulated in ACI 318-99/02/05166, or limitation of net tensile strain εt > 0.010, as required by ACI 318-14, ACI 318-11 and ACI 318-08167. Violation of this requirement is reported by the software as exceeding maximum allowable reinforcement indicating that the option to adjust the factor γf should be turned off by the user at the support where the violation occurs. It should be noted that the ACI code168 requires either concentration of reinforcement over the column by closer spacing, or additional reinforcement, to resist the transfer moment within the effective slab width. spSlab satisfies this requirement by providing additional reinforcement without concentrating existing reinforcement. When computing additional reinforcement for the transfer of negative and positive unbalanced moments over the supports through flexure in systems with longitudinal beams, the contribution of the longitudinal beam cross-section can be optionally selected. If selected, this contribution will be considered. For CSA designs this functionality extends also to design of banded reinforcement in bb strip. The CSA A23.3 code requires at least one-third of the total negative reinforcement for the entire design strip at interior supports to be concentrated in the band width, bb, extending 1.5hs from the sides of the columns169. The program fulfills this requirement by concentrating a portion of reinforcement assigned to the design strip that includes width bb. This strip will typically be the column strip. However, if longitudinal slab bands or slab-band-like beams wider than band width bb are present, then reinforcement assigned to these elements is concentrated. At exterior supports, the total negative reinforcement is placed in the bb band width170 or if a beam narrower than bb is 166 ACI 318-14, 8.4.2.3.4; ACI 318-11, 13.5.3.3; ACI 318-05, 13.5.3.3; ACI 318-02, 13.5.3.3; ACI 318-99, 13.5.3.3 167 ACI 318-14, 8.4.2.3.4; ACI 318-11, 13.5.3.3; ACI 318-08, 13.5.3.3 168 ACI 318-14, 8.4.2.3.5; ACI 318-11, 13.5.3.4; ACI 318-08, 13.5.3.4; ACI 318-05, 13.5.3.4; ACI 318-02, 13.5.3.4; ACI 318-99, 13.5.3.4 169 CSA A23.3-14, 13.11.2.7; CSA A23.3-04, 13.11.2.7; CSA A23.3-94, 13.12.2.1 170 CSA A23.3-14, 13.10.3; CSA A23.3-04, 13.10.3; CSA A23.3-94, 13.12.2.2, 13.13.4.2 SOLUTION METHODS \| 137 \| present, then the total reinforcement is placed within the beam width171. The reinforcement in the bb and the remaining portions of the design strip is also checked for compliance with spacing and minimum reinforcement requirements. 171 CSA A23.3-04, 13.12.2.2; CSA A23.3-04, 13.12.2.2; CSA A23.3-94, 13.13.2.2 SOLUTION METHODS \| 138 \| 2.5. Detailing Provisions Proper reinforcement detailing is a critical step in ensuring the performance and safety of structural systems designed using spSlab and spBeam. The detailing process involves selecting the size, spacing, and extension of reinforcement in accordance with the requirements of design codes, such as ACI 318 and CSA A23.3. Adhering to these guidelines ensures that the structural elements achieve the intended capacities while maintaining durability, ductility, and resistance to cracking and progressive collapse. This section outlines the methodology and criteria employed by spSlab and spBeam to assist users in determining optimal reinforcement layouts. It covers key aspects of reinforcement detailing, spacing limitations, development lengths, and structural integrity reinforcement. The iterative design approach used by the programs incorporates considerations such as bar sizes, clear spacing, and crack control requirements to meet both minimum and maximum limits required by the code. Additionally, provisions for structural integrity and corner reinforcement are discussed to enhance the redundancy and ductility of the structural system. 2.5.1. Reinforcement Detailing 2.5.1.1. Minimum Clear Spacing According to ACI-318 code172, the default minimum clear spacing of reinforcement for both slabs and beams is taken as the larger of the two prescribed minima of one bar diameter, db, or 1 in. According to CSA code173, the default minimum clear spacing of reinforcement for both slabs and beams is taken as the larger of the two prescribed minima of 1.4 times the bar diameter, db, or 1.2 in (30mm). The user may select a clear spacing greater than the default value to take into account 172 ACI 318-14, 25.2.1; ACI 318-11, 7.6.1; ACI 318-08, 7.6.1; ACI 318-05, 7.6.1; ACI 318-02, 7.6.1; ACI 318-99, 7.6.1 173 CSA A23.3-14, Annex A, 6.6.5.2; CSA A23.3-04, Annex A, 6.6.5.2; CSA A23.3-94, Annex A, A12.5.2 SOLUTION METHODS \| 139 \| tolerances for reinforcement placement174 and other project specific considerations. For two-way systems, the maximum spacing of reinforcement is kept at two times the slab thickness for the ACI code175 and three times the slab thickness for the CSA code176, but no more than 18 in. or 500 mm respectively. For joist systems the limit is increased to 5 times the slab thickness177. When calculating negative support reinforcement for the CSA code178, the program assumes that banded reinforcement over supports is spaced at a maximum of 1.5hs and no more than 250 mm. For one-way slabs, the maximum spacing is limited to179 the smaller of three times the slab thickness and 18 in. [500 mm]. Additionally, the maximum spacing of reinforcement, s, in beams and one-way slabs is selected so that the following crack control requirements of the ACI and the CSA codes180 are met: ( ) ( ) ( ) ( ) 1 3 max 900,000 480,000 min 2.5 , ACI 318-11/08/05 900 432 min 2.5 , ACI 318-02/99 0.6 CSA A23.3-14/04/94 c y y c y y y c s c f f s c f f f d A z    −          −        Eq. 2-95 where: 174 See ACI 317-06 (Ref. ) 175 ACI 318-14, 8.7.2.2; ACI 318-11, 13.3.2; ACI 318-08, 13.3.2; ACI 318-05, 13.3.2; ACI 318-02, 13.3.2; ACI 318-99, 13.3.2 176 CSA A23.3-14, 13.10.4; CSA A23.3-04, 13.10.4; CSA A23.3-94, 13.11.3(b) 177 ACI 318-14, 7.7.6.2.1, 8.7.2.2; ACI 318-11, 7.12.2.2; ACI 318-08, 7.12.2.2; ACI 318-05, 7.12.2.2; ACI 318-02, 7.12.2.2; ACI 318-99, 7.12.2.2; CSA A23.3-14, 7.8.3; CSA A23.3-04, 7.8.3; CSA A23.3-94, 7.8.3 178 CSA A23.3-14, 13.10.4; CSA A23.3-04, 13.10.4; CSA A23.3-94, 13.11.3(a) 179 ACI 318-14, 7.7.2.3, 8.7.2.2; ACI 318-11, 7.6.5; ACI 318-08, 7.6.5; ACI 318-05, 7.6.5; ACI 318-02, 7.6.5, ACI 318-99, 7.6.5; CSA A23.3-14, 7.4.1.2; CSA A23.3-04, 7.4.1.2; CSA A23.3-94, 7.4.1.2 180 ACI 318-14, 24.3.2, 24.3.3; ACI 318-11, 10.6.4; ACI 318-08, 10.6.4; ACI 318-05, 10.6.4; ACI 318-02, 10.6.4; ACI 318-99, 10.6.4; CSA A23.3-14, 10.6.1; CSA A23.3-04, 10.6.1; CSA A23.3-94, 10.6.1 SOLUTION METHODS \| 140 \| cc = least distance from the surface of bar to the tension face, dc = distance from extreme tension fiber to center of the closest longitudinal bar, A = effective tension area of concrete surrounding the flexural tension reinforcement and extending from the extreme tension fiber to the centroid of the flexural tension reinforcement and an equal distance past that centroid, divided by the number of bars or wires, zmax = 30,000 N/mm for interior exposure or 25,000 N/mm for exterior exposure, multiplied by a factor of 1.2 for epoxy-coated reinforcement. An iterative process is performed to determine the number of bars and bar size. The initial number of bars is determined by dividing the total reinforcement area required, As, by the area of one bar, Asb, of the input minimum bar size. Next, the spacing is determined. If the minimum spacing limitations are violated, the bar size is increased and the iterative process is repeated until all bars sizes have been checked. If the maximum spacing limitations are not met, the number of bars required to satisfy these limitations is computed and the iteration process terminates. Figure 2.28 – Width due to Stirrup Bend For beams, layered reinforcement is provided if sufficient beam width is not available. The clear distance between layers is assumed 1.0 in [30 mm] but the user can change this value. By default, the program assumes a 1.5 in [40 mm] side cover to stirrup for width calculations and this value SOLUTION METHODS \| 141 \| can also be changed by the user. The program also assumes that the longitudinal bar makes contact at the middle of the stirrup bend where the minimum inside diameter of the bend is four times stirrup diameter181. Therefore, an additional width is added to the cover for longitudinal bars less than size #14 (#45 for CAN/CSA-G30.18) (Figure 2.28 – Figure 2.29). This additional width due to the bend, wbend, is equal to: 2 1 2 2 b bend d w r    = − −         Eq. 2-96 where: db = diameter of the longitudinal bar, r = inside radius of bend for stirrup. Figure 2.29 – Detail Reinforcement in Longitudinal Beams 181 ACI 318-14, 25.3.2; ACI 318-11, 7.2.2; ACI 318-08, 7.2.2; ACI 318-05, 7.2.2; ACI 318-02, 7.2.2; ACI 318-99, 7.2.2; CSA A23.3-04, 7.1.1 and Table 16 in Annex A; CSA A23.3-94, 7.1.1 and Table 16 in Annex A SOLUTION METHODS \| 142 \| 2.5.1.2. Development Length Computation Bar-length computations are performed for two-way slabs and longitudinal beams. For top reinforcement at the supports, the length for long bars is given by: ( ) ( )   50% , , max max max max ,12 , /16 d long long pi b n fos cr long l l l l d d l l l +    = +   +   Eq. 2-97 and the length for short bars is given by: ( )     50% , , max max ,12 max max , b short fos d short cr short l d d l l l l +   =  +   Eq. 2-98 where: max (l50%) = maximum distance to the points of 50% demand, max (lpi) = maximum distance to the points of inflection (P.I.), ld = bar development length182, d = effective depth, db = bar diameter, ln = clear span length, lfos = distance to the face of support (column), lcr = minimum code prescribed extension. These bar lengths are then compared and adjusted if necessary to meet the minimum extension requirements for reinforcement specified by the code.183 Additionally the program may select 182 Chapter 25 in ACI 318-14; Chapter 12 in ACI 318-11, ACI 318-08, ACI 318-05, ACI 318-02, and ACI 318-99; CSA A23.3-14, Clause 12.2; CSA A23.3-04, Clause 12.2; CSA A23.3-94, Clause 12.2 183 Figure 8.7.4.3a in ACI 318-14; Figure 13.3.8 in ACI 318-11, ACI 318-08, ACI 318-05, ACI 318-02, and ACI 318-99; CSA A23.3-14, Figure 13.1; CSA A23.3-04, Figure 13.1; CSA A23.3-94, Figure 13.1 SOLUTION METHODS \| 143 \| continuous top bars in those spans where steel is required by calculation in mid-span at top. If the computed bar lengths overlap, it is recommended that such reinforcement be run continuously. The printed bar lengths do not include hooks or portions of bars bent down into spandrel beams or other bar-bend configurations. If a bar starts (or ends) at a column support the length of the bar is measured from (or to) the center line of the column. The selection of bar lengths for positive reinforcement for flat plates, flat slabs, and beam-supported slabs, is based strictly on the minimum values of the code. The development length depends on the following factors: concrete cover, minimum transverse reinforcement, special transverse reinforcement, layer location bar size and bar clear spacing. The development length is calculated from the general expression184 below, but not less185 than 12 in [300 mm]: 3 for ACI 318-11/08 40 1 for ACI 318M-11/08 1.1 3 for ACI 318-05/02/99 40 for ACI 318M-05/02/99 1.1 1.15 for CSA A23.3-14/04/94 4 y t e s d b c d tr b f l d f c K d              =     +            Eq. 2-99 where: Ψt = reinforcement location factor equal to 1.3 if more than 12 in [300 mm] of fresh concrete is cast in the member below the development length or splice, or equal to 1.0 otherwise, Ψe = coating factor equal to 1.0 for uncoated reinforcement; for epoxy coated reinforcement with covers less than 3db or clear spacing less than 6db the factor is equal to 1.5 and for all 184 ACI 318-14, 25.4.2.3; ACI 318-11, 12.2.3; ACI 318-08, 12.2.3; ACI 318-05, 12.2.3; ACI 318-02, 12.2.3; ACI 318-99, 12.2.3; ACI 318M-11,12.2.3; ACI 318M-08, 12.2.3; ACI 318M-05, 12.2.3; ACI 318M-02, 12.2.3; ACI 318M-99, 12.2.3; CSA A23.3-14, 12.2.2; CSA A23.3-04, 12.2.2; CSA A23.3-94, 12.2.2 185 ACI 318-14, 25.4.2.1; ACI 318-11, 12.2.1; ACI 318-08, 12.2.1; ACI 318-05, 12.2.1; ACI 318-02, 12.2.1; ACI 318-99, 12.2.1; CSA A23.3-14, 12.2.1; CSA A23.3-04, 12.2.1; CSA A23.3-94, 12.2.1 SOLUTION METHODS \| 144 \| other epoxy coated bars it equals 1.2, Ψs = reinforcement size factor equal to 1.0 for bars #7 and larger or equal to 0.8 for bars #6 and smaller if ACI 318 [ACI 318M] is selected; for CSA A23.3 the factor is equal to 1.0 for bars 25M and larger or equal to 0.8 for bars 20M and smaller, λ = lightweight aggregate concrete factor equal to 1.0 for normal concrete and: 0.75 for lightweight concrete per ACI 318-14, ACI 318-11 and ACI 318-08 1.3 for lightweight concrete per ACI 318-05/02/99 1.3 for low density concrete per CSA A23.3-14/04/94 1.2 for semi low density concrete per CSA A23.3-14/04/94 Ktr = transverse reinforcement index conservatively assumed zero, cb = smaller of the distance form bar surface to the closest concrete surface and one-half (two thirds for CSA186) center-to-center bar spacing187. Additionally, the product of ΨtΨe is not taken greater than 1.7 and the development length, ld, is reduced188 by the factor of As,req to As,prov where the provided area of flexural reinforcement, As,prov, exceeds the area required by analysis, As,req. The final calculated or minimum development length for each bar is tabulated in the design results section of the program results report. In two-way slab systems without beams, the development length presented is often controlled by the minimum development length. Where flexural reinforcement is terminated in a tension zone, spSlab and spBeam provide a warning to require an extension of the bar beyond what is required for flexure. For ACI code, the shear capacity at the cutoff point for each bar is evaluated for satisfying the shear demand does not 186 Denoted as dcs in CSA A23.3-14, 3.2; CSA A23.3-04, 2.3 and CSA A23.3-94, 12.0 187 ACI 318-05, 2.1; ACI 318-02, 12.2.4; ACI 318-99, 12.2.4 188 ACI 318-14, 25.4.10.1; ACI 318-11, 12.2.5; ACI 318-08, 12.2.5; ACI 318-05, 12.2.5; ACI 318-02, 12.2.5; ACI 318-99, 12.2.5; CSA A23.3-14, 12.2.5; CSA A23.3-04, 12.2.5; CSA A23.3-94, 12.2.5 SOLUTION METHODS \| 145 \| exceed permissible shear limit189. Final bar length shall be extended beyond the minimum reported to meet one of the three conditions outlined in ACI 318190. 2.5.2. Structural Integrity Reinforcement Enhancing redundancy and ductility is necessary in the event of damage to a major supporting element resulting from an abnormal shock or blast loading event. Minor changes in reinforcement detailing typically result in substantial enhancement in the overall integrity of a structure by confining the resulting damage to a small area and improving the resistance to progressive collapse. The ACI code requires all bottom bars in the column strip to extend continuously (or with splices) in the entire span and at least two of these bars to pass within the column core and to be anchored at exterior supports191. In continuous beams, including longitudinal beams in two-way slab systems, spSlab and spBeam produce, in design mode, reinforcement that satisfies ACI requirements for structural integrity. In perimeter (exterior) beams, at least one sixth of the negative tension reinforcement and not less than two bars are continuous192. Also, at least one fourth of the positive tension reinforcement and not less than two bars are continuous in all beams193. For the CSA code, the program performs calculation of the amount of integrity reinforcement at slab column connections in design mode. The integrity reinforcement is required for slabs without 189 ACI 318-14, 7.7.3.5(a); ACI 318-11, 12.10.5.1; ACI 318-08, 12.10.5.1; ACI 318-05, 12.10.5.1; ACI 318-02, 12.10.5.1; ACI 318-99, 12.10.5.1 190 ACI 318-14, 7.7.3.5, 7.7.3.5(a), 7.7.3.5(b), 7.7.3.5(c); ACI 318-11, 12.10.5, 12.10.5.1, 12.10.5.2, 12.10.5.3; ACI 318-08, 12.10.5, 12.10.5.1, 12.10.5.2, 12.10.5.3; ACI 318-05, 12.10.5, 12.10.5.1, 12.10.5.2, 12.10.5.3; ACI 318-02, 12.10.5, 12.10.5.1, 12.10.5.2, 12.10.5.3; ACI 318-99, 12.10.5, 12.10.5.1, 12.10.5.2, 12.10.5.3 191 ACI 318-14, 8.7.4.2.1, 8.7.4.2.2; ACI 318-11, 13.3.8.5; ACI 318-08, 13.3.8.5; ACI 318-05, 13.3.8.5; ACI 318-02, 13.3.8.5; ACI 318-99, 13.3.8.5 192 ACI 318-14, 9.7.7.1(a); ACI 318-11, 7.13.2.2(a); ACI 318-08, 7.13.2.2(a); ACI 318-05, 7.13.2.2(a); ACI 318-02, 7.13.2.2(a); ACI 318-99, 7.13.2.2 193 ACI 318-14, 9.7.7.1(b); ACI 318-11, 7.13.2.2(b) and 7.13.2.4; ACI 318-08, 7.13.2.2(b) and 7.13.2.4; ACI 318-05, 7.13.2.2(b) and 7.13.2.4; ACI 318-02, 7.13.2.2(b) and 7.13.2.4; ACI 318-99, 7.13.2.2 and 7.13.2.3 SOLUTION METHODS \| 146 \| beams. Integrity reinforcement is not required if there are beams containing shear reinforcement in all spans framing into the column. Otherwise, the sum of all bottom reinforcement connecting the slab to the column on all faces of the periphery should consist of at least two bars and meet the condition194: 2 se sb y V A f   Eq. 2-101 where Vse is the larger of shear force transmitted to column or column capital due to specified (unfactored) loads and shear force corresponding to twice the self-weight of the slab. Figure 2.30 – Integrity Reinforcement at Supports 2.5.3. Corner Reinforcement The program performs calculation of the amount of reinforcement in exterior corners of slabs with stiff edge beams (α greater than 1.0)195. This reinforcement is required within a region equal to 1/5 of the shorter span. The amount of corner reinforcement is calculated from the moment per unit width intensity corresponding to the maximum positive moment in span. The code allows the corner reinforcement to be placed at top and bottom of the slab in bands parallel to the sides of the slab edges. 194 CSA A23.3-14, 13.10.6.1 and 13.10.6.2; CSA A23.3-04, 13.10.6.1 and 13.10.6.2; CSA A23.3-94, 13.11.5.1 and 13.11.5.2 195 ACI 318-14, 8.7.3.1; ACI 318-11, 13.3.6; ACI 318-08, 13.3.6; ACI 318-05, 13.3.6; ACI 318-02, 13.3.6; ACI 318-99, 13.3.6; CSA A23.3-14, 13.12.5; CSA A23.3-04, 13.12.5; CSA A23.3-94, 13.13.5 SOLUTION METHODS \| 147 \| 2.6. Special Topics Conventionally reinforced concrete floor systems, including slabs and beams, contain a diverse variety of parameters and considerations that could be achieved by the flexibility of cast-in-place concrete forming systems. Notwithstanding prestressing and post-tensioning, engineers and design professionals will encounter numerous special conditions to handle and deal with in the design of a concrete floor system, ranging from the placement of concrete to optimizing of shapes and cross sections for gravity and lateral load effects. 2.6.1. Lateral Load Effects & Restraint When analyzing lateral loads, each frame may be evaluated as a single unit for the full height of the building. Structural analysis software, such as spFrame and ETABS, can be used to perform this type of analysis. It is important to recognize that for lateral load assessment, slab-beam elements may experience reduced stiffness due to cracking, along with modifications to the effective slab width used in the analysis. The bending moments generated at the two ends of a span due to lateral loads, such as wind or seismic forces, can be obtained from such an analysis. These moments can then be incorporated into the spSlab model as Lateral Load Effects input, allowing for the determination of the appropriate design moments when considering both gravity and lateral loading. The scope of lateral load consideration in spSlab is limited to this process, where externally computed lateral moments can be applied within the equivalent frame model to evaluate their effects on the structural design as shown in the following figure. More information about this topic can be found in Section 2.3.3.3 and Section 5.2.4.3. SOLUTION METHODS \| 148 \| Figure 2.31 – Incorporation of Bending Moments due to Lateral Loads from spFrame into spSlab Gravity Load Analysis SOLUTION METHODS \| 149 \| 2.6.2. Loads Along the Span Imposed by Lateral Loading In some conditions, lateral loads acting on an equipment supported on a span may impose point loads and/or moments in that given span. The Program does not have a lateral type span load option in order to accommodate such point forces or moments at a span due to lateral loads. Although it is neither ideal nor recommended, the user may consider utilizing Snow load case to enter point forces or moments imposed by such lateral loads along the span in order to obtain the internal forces and for the flexural and shear design purposes. In the Program, since Snow load case is a Dead load type, its implications to deflection calculations would require considerable engineering judgment. Figure 2.32 – Loads along the Span imposed by Lateral Loading StructurePoint anticipates the inclusion of additional load cases to accommodate lateral loads being applied to the span in future releases once adequate consideration and research has been done to the effects of live load patterning and deflection calculations for lateral forces. It is anticipated that the inclusion of lateral forces will complicate greatly the procedure of live load arrangement and patterning, in addition to the added complexity of calculating sustained live loads that are an essential part of the deflection calculations specifically long term. SOLUTION METHODS \| 150 \| 2.6.3. Gravity Frame Analysis under Sidesway The spBeam program is a powerful tool for modeling one-way, multi-story, two-dimensional concrete frames, allowing users to analyze individual stories as separate frames. Each story is modeled with slab beam elements and columns above and below, employing an equivalent column concept and restraining horizontal translational degrees of freedom at the story level. However, in scenarios where unsymmetrical vertical loading, variations in support stiffness, or boundary conditions induce horizontal sidesway, the spFrame program provides a more comprehensive analysis. Unlike spBeam, spFrame accommodates the entire height of a two-dimensional frame, accurately accounting for sidesway effects on internal force magnitudes. A comparative analysis, conducted in “Comparison of Gravity Loaded Concrete Frame Models in spBeam and spFrame under Sidesway” Technical Article from StructurePoint, illustrates that while spBeam assumes sidesway restraint, spFrame considers its effects, leading to more reliable results for design under such conditions. SOLUTION METHODS \| 151 \| 2.6.4. Openings in Concrete Slabs Openings in concrete slabs significantly influence their structural behavior, particularly in terms of shear strength and load distribution. The presence of openings alters the flow of internal forces, potentially reducing the slab's capacity to resist shear and flexure while introducing stress concentrations around the perimeters of the openings. Theoretical approaches outlined in design codes such as ACI 318 and CSA A23.3 provide guidance on analyzing and designing slabs with openings, emphasizing the importance of maintaining adequate shear transfer mechanisms. The scope of these provisions includes practical design scenarios for slabs with various types and sizes of openings, ensuring that safety and functionality are preserved. The purpose of these guidelines is to enable engineers to design reinforced concrete slabs with openings that meet structural performance criteria while accommodating architectural or functional requirements, such as ductwork, piping, and access points, without compromising the slab's integrity or serviceability. More information about this topic can be found in “Shear Strength of Concrete Slabs with Openings (ACI 318)” and “Shear Strength of Concrete Slabs with Openings (CSA A23.3)” Design Examples from StructurePoint. Figure 2.33 – Effect of Openings on Slabs Shear Strength SOLUTION METHODS \| 152 \| 2.6.5. Support Conditions In spSlab and spBeam, defining appropriate support conditions is essential for accurate analysis and design of structural elements in a concrete floor system. By default, column-slab/beam joints are assumed to rotate freely while restricting translational displacement. The rotational stiffness of a joint is influenced by connected elements (such as slabs, beams, transverse beams, and columns) and can be adjusted using the Restraint command. Columns are typically modeled with fixed far-end boundary conditions but can be adjusted using the Column command to meet specific design needs. Engineers can specify vertical spring constants (Kz) to allow vertical displacement of joints or adjust rotational stiffness using rotational spring constants (Kry). Far-end column conditions can also be set as either fixed or pinned to match design assumptions. These options enable modeling of support conditions to achieve a closer reflection of physical conditions into the analytical models and generate more accurate results. For more details, refer to Section 2.3.2, Section 5.2.4.2, and “Deflection Observations in Girder-Supported Beams and One-way Slabs” Technical Article from StructurePoint. SOLUTION METHODS \| 153 \| 2.6.6. Moment Redistribution Moment redistribution is a design approach applied to optimize reinforcement placement and reduce material usage. The technique involves leveraging the plastic behavior of structures, which allows for a redistribution of bending moments beyond the elastic analysis results. This redistribution reduces peak negative moments at supports, typically shifting them to positive moments in spans, enabling more efficient utilization of structural capacity. Governed by standards such as ACI 318 and CSA A23.3, moment redistribution requires sufficient ductility in plastic hinge regions to ensure stability and maintain static equilibrium. By reducing reinforcement congestion in critical areas like the support regions, this method as permissible by codes, offers advantages in material savings, labor efficiency, and improved constructability. Advanced software tools, such as spSlab and spBeam, streamline the iterative calculations involved, making the process more accessible and precise. In some instances, while investigating existing buildings and their concrete floor systems for added forces or loads and changing occupancy, this technique has proven to eliminate the need for very costly and expensive repairs that may be very time-consuming, deeming the building uninhabitable. Used properly and with caution, this technique can also result in the ability to repurpose an existing building for long-term, durable, and safe use with new higher loads. This alone can be the main difference between increasing the carbon footprint with new construction compared with the reuse of an existing durable structure towards an additional cycle of service life and contributing to the sustainable development of the built environment. For more details regarding moment redistribution, refer to Section 2.3.6, “Continuous Beam Design with Moment Redistribution” Design Example, and “spSlab/spBeam Moment Redistribution Applications” Technical Article from StructurePoint. SOLUTION METHODS \| 154 \| Figure 2.34 – Redistribution of Moments SOLUTION METHODS \| 155 \| 2.6.7. Default Load Assignment Impact on Deflections In spBeam and spSlab, the deflection calculations are performed by considering all assigned loading belonging to Load Cases of “Dead” and “Live” Load Type with a service load factor of 1.0 automatically. This process is followed by default by the Program regardless of the Load Cases utilized with non-zero load factors in the Load Combinations Menu for the ultimate-level design of the model. The models generated through the Template Module incorporate Dead and Live Load assignments by default unless the load magnitude is modified to zero (0) by the user. These Load Cases with default assigned load factors of 1.0 will be utilized in deflection calculations even if they are given a zero (0) load factor in the Load Combinations Menu for project-specific reasons. Therefore, to obtain accurate deflection results, the user is advised to review all the load assignments before running the model and eliminate any unnecessary default load assignments that may have been assigned in the Templates Module. While this condition is not frequently needed in practical analysis and design applications, the software approaches the deflection calculations systematically for the traditional deflection calculations. SOLUTION METHODS \| 156 \| 2.6.8. Reinforcing Bar Arrangement Impact on Deflections In spSlab and spBeam, reinforcing bar arrangement plays an important role in accurately calculating deflections in concrete structural elements. These programs offer flexible solve options, including the activation of the compression reinforcement, which influences how reinforcing bars are considered in the calculation of the cracked moment of inertia (Icr). This feature affects deflection results significantly, especially when both continuous and discontinuous reinforcement types are present. By understanding the impact of these solve options and the associated bar arrangements, engineers can optimize structural analysis and achieve design outcomes tailored to project requirements. More information about this topic can be found in the "spSlab/spBeam Reinforcing Bar Arrangement Impact on Deflections" Technical Article from StructurePoint. SOLUTION METHODS \| 157 \| 2.6.9. Design of Doubly Reinforced Beam Sections The design of doubly reinforced beam sections incorporates both tension and compression reinforcements, allowing for enhanced ductility and strength while maintaining the section within the tension-controlled region. This approach is particularly valuable in scenarios where architectural constraints limit beam dimensions or where reducing beam weight is a priority. By adding compression reinforcement, engineers can effectively utilize higher percentages of tension reinforcement without exceeding strain limits, thereby achieving the required moment capacity efficiently. This design methodology aligns with the principles outlined in ACI 318 and CSA A23.3 codes, ensuring structural performance and safety. The process includes selecting an appropriate tension reinforcement ratio, integrating compression reinforcement to enhance ductility, and verifying section strength against design requirements. For more details regarding doubly reinforced beam design, refer to “Doubly Reinforced Concrete Beam Design” Design Example from StructurePoint. SOLUTION METHODS \| 158 \| 2.6.10. Waffle Slabs Analysis & Design Approach A waffle slab, also known as a two-way ribbed slab, is a structural system designed for applications requiring long spans and the ability to carry heavy loads. Its unique geometry, consisting of ribs in two directions, provides an efficient and economical solution for floors and roofs. The Equivalent Frame Method (EFM) is commonly used to analyze and design waffle slabs, simplifying the complexities of their ribbed structure while ensuring compliance with design standards. For a detailed explanation of the analysis and design methodology, including considerations for rib dimensions, slab thickness, drop panels (drop heads), shear analysis, and deflection calculations, refer to Section 2.2.1.4 and “Two-Way Joist (Waffle) Slab Design Approach and Methodology” Technical Article from StructurePoint. Figure 2.35 – Immediate Deflection Considerations for Negative Moment Sections SOLUTION METHODS \| 159 \| 2.6.11. Material Quantities The program computes concrete and reinforcing steel quantities. The quantity of concrete is based on an average of the slab, drop, and beam sizes. The total quantity of reinforcing steel computed by the program corresponds to the actual bar sizes and lengths required by design. No allowance is made for bar hooks, anchorage embedment, and so forth. It should be noted that the quantity of reinforcement printed by the program pertains to bending in one direction only. In practice, the total amount of reinforcement for the structure should also include the quantities obtained for the appropriate transverse equivalent frames. Figure 2.36 – spSlab Material Takeoff Table SOLUTION METHODS \| 160 \| 2.7. References Building Code Requirements for Structural Concrete (ACI 318-14) and Commentary (ACI 318R-14), American Concrete Institute, 2014. Building Code Requirements for Structural Concrete (ACI 318-11) and Commentary (ACI 318R-11), American Concrete Institute, 2011. Building Code Requirements for Structural Concrete (ACI 318-08) and Commentary (ACI 318R-08), American Concrete Institute, 2008. Building Code Requirements for Structural Concrete (ACI 318-05) and Commentary (ACI 318R-05), American Concrete Institute, 2005. Building Code Requirements for Structural Concrete (ACI 318-02) and Commentary (ACI 318R-02), American Concrete Institute, 2002. Building Code Requirements for Structural Concrete (ACI 318-99) and Commentary (ACI 318R-99), American Concrete Institute, 1999. Specification for Tolerances for Concrete and Materials and Commentary, An ACI Standard (ACI 117-06), American Concrete Institute, 2006 National Building Code of Canada 2010, Volume 2, Canadian Commission on Buildings and Fire Codes, National Research Council of Canada, 2010 National Building Code of Canada 2005, Volume 1, Canadian Commission on Buildings and Fire Codes, National Research Council of Canada, 2005 CSA A23.3-14, Design of Concrete Structures, Canadian Standards Association, 2014. CSA A23.3-04, Design of Concrete Structures, Canadian Standards Association, 2004. Explanatory Notes on CSA Standard A23.3-04 in Concrete Design Handbook, Third Edition, Cement Association of Canada, 2006 SOLUTION METHODS \| 161 \| CSA A23.3-94, Design of Concrete Structures, Canadian Standards Association, 1994 (Reaffirmed 2000). Explanatory Notes on CSA Standard A23.3-94 in Concrete Design Handbook, Second Edition, Canadian Portland Cement Association, 1995 Wight J.K., MacGregor J.G., Reinforced Concrete, Mechanics and Design, Fifth Edition, Pearson Prentice Hall, 2009 MacGregor J.G., Bartlett F.M., Reinforced Concrete – Mechanics and Design, First Canadian Edition, Prentice Hall Canada Inc., 2000 Branson, D. E., Instantaneous and Time-Dependent Deflections of Simple and Continuous Reinforced Concrete Beams, HPR Report No. 7, Pt. I, Alabama Highway Department in Cooperation with U.S. Department of Commerce, Bureau of Public Roads, August 1965. Notes on ACI 318-05 Building Code Requirements for Structural Concrete with Design Applications, Edited by Mahmoud E. Kamara and Basile G. Rabbat, Portland Cement Association, 2005 Control of Deflection in Concrete Structures (ACI 435R-95), Reported by ACI Committee 435, American Concrete Institute, 1995 (Reapproved 2000). ACI Committee 435, Subcommittee 7, Deflections of Continuous Concrete Beams, Journal of the American Concrete Institute, Proceedings V. 70, No. 12, December 1973, pp. 781-787. Nilson, A. H., and Walters, D. B., Jr., Deflection of Two-Way Floor Systems by the Equivalent Frame Method, Journal of the American Concrete Institute, Proceedings, V. 72, No. 5, May, 1975, pp. 210-218. ACI Committee 435, State-of-the-Art Report, Deflection of Two-Way Floor Systems, Special Publication SP43-3, American Concrete Institute, 1974. SOLUTION METHODS \| 162 \| Kripanarayanan, K. M., and Branson, D. E., Short-Time Deflections of Flat Plates, Flat Slabs and Two-Way Slabs, Journal of the American Concrete Institute, Proceedings, V. 73, No. 12, December 1976, pp. 686-690. Wight, J. K., Falconer, D., Checking Punching Shear Strength by the ACI Code, Concrete International, November 2005, pp. 76. Park, W. and Gamble, W. L., Reinforced Concrete Slabs, Second Edition, John Wiley & Sons, Inc., 2000 \| 163 \| 3. PROGRAM INTERFACE 3.1. Start Screen When the Program is launched, a start screen appears as shown below. The Start Screen consists of options to start New Project, Open existing Project, open Examples folder, open Templates, links to available program Resources and a list of Recent files. The program name and copyright information are located in the bottom right of the start screen. CHAPTER 3 PROGRAM INTERFACE \| 164 \| When selecting New Project, a dialog box appears allowing the user to choose between starting a blank project or using a predefined template. PROGRAM INTERFACE \| 165 \| 3.2. Main Program Window The Main Program Window shown above consists of the following: 3.2.1. Quick Access Toolbar The Quick Access Toolbar includes New, Open, Save and Undo and Redo commands. 3.2.2. Title Bar The Title Bar displays the name of the program, along with the filename of the current data file in use. If the file is new and has not yet been saved, the word “Untitled” is displayed in the Title Bar. It also displays “(Modified)” if the file has been changed and not saved yet. PROGRAM INTERFACE \| 166 \| 3.2.3. Ribbon The Ribbon consists File and Home tabs. File Tab consists of commands to go Back to Home Tab, create New file, Open an existing file, Save a file, Save as, and Exit. In addition, the entire Start Screen is present under the File Tab. Home Tab gives quick access to commands which are needed to complete the task of creating a model, executing it and analyzing solutions. These commands are: Project: enables to enter GENERAL, MATERIALS, RUN OPTIONS, and PROJECT DESCRIPTION. Define: enables to define Concrete, and Reinforcing Steel materials; Slabs & Ribs, Beams, Beam Stirrups, and Bar Set reinforcement criteria; Design & Modeling options; Load Cases and Load Combinations. Grids: enables to add new or edit existing grids and spans. Select: enables to select various model items. Spans: enables to create slabs, longitudinal beams and ribs. Supports: enables to create columns, drop panels, column capitals, transverse beams and restraints. Loads: enables to assign area loads, line loads, point loads, support loads, and lateral load effects to the model. Rebars: For two-way floor systems, the Rebars command allows the user to specify longitudinal reinforcement details for the column strip, middle strip, and beams, as well as shear reinforcement for beams. In beam or one-way slab systems, the user can specify flexural bars, stirrups, and torsional longitudinal reinforcement. The Rebars command is disabled when the Design run mode is selected in the RUN OPTIONS from the Project left panel. To enable it, select the Investigation run mode. PROGRAM INTERFACE \| 167 \| Note: When switching from DESIGN Mode to INVESTIGATION Mode, spSlab automatically assumes the results of the DESIGN Mode as an input for INVESTIGATION Mode. Solve: enables to specify design options, deflection options, and solve the model. Please note that design options for two-way systems are different from beams/one-way slab systems. Results: after a successful run, enables to view graphical results such as internal forces, moment capacity, shear capacity, deflection and reinforcement. Tables: enables to open Tables module to view tabular input and output. Reporter: enables to open Reporter module to view the report. Display: enables to toggle on/off model items. Viewports: enables to select from a predefined viewport configuration. Settings: enables to modify various program settings. 3.2.4. Left Panel The properties of active commands under Home Tab or the properties of items selected in the Viewport are displayed in the Left Panel which can then be used to execute the commands or edit the selected items. After execution the Left Panel also displays various commands and options which can be used to investigate the solution diagrams in the Viewport. 3.2.5. Left Panel Toolbar The Left Panel Toolbar contains commands that can be used to edit various items in the Viewport. PROGRAM INTERFACE \| 168 \| 3.2.6. Viewport The Viewport covers the majority of the main program window. It is the space where models can be created and graphical results can be viewed. Up to 6 Viewports can be used at once. Viewports can be moved and docked in a number of predefined locations using the docking tool. A viewport may be split out to a separate screen entirely for added flexibility and to enlarge the model view work area providing more accurate drafting controls. PROGRAM INTERFACE \| 169 \| 3.2.7. View Controls The View Controls contains various commands which can be used to adjust the views of Viewport both during modeling or viewing the graphical results. 3.2.8. Status Bar The Status Bar displays key information, including the design code, run options, cursor position, and current units. PROGRAM INTERFACE \| 170 \| 3.3. Tables Window The Tables Module interface shown above enables the user to view program inputs and outputs in tables and export them in different formats. The Tables Module is accessed from within the Main Program Window by clicking the Tables button from the Ribbon. Alternatively, Tables Module can also be accessed by pressing the F6 key. If the model has not been executed yet, then the Tables Module will only contain a list of input data tables. When a model has been successfully executed, the Tables Module will also display the output data tables. PROGRAM INTERFACE \| 171 \| 3.3.1. Toolbar The Toolbar contains commands which can be used to navigate through various Tables Previous table Displays the previous table. Next table Displays the next table. Table number box Displays the table with the table number entered in the box. Auto fit column width to view area When toggled on always fits the width of table to the Preview Area width. Maintain maximum column width Restores all table columns to their default maximum width. Export current table Exports the table being viewed in the selected format. PROGRAM INTERFACE \| 172 \| Settings Contains settings for the Explorer Panel. • LOCATION: Displays Explorer Panel on the left or right side of screen depending on selection. • HIDE INACTIVE ITEMS: Hides unused tables from the explorer view. • KEEP EXPLORER CONFIGURATION: Saves the explorer configuration i.e., information about selected tables and opened/closed sections so that it is available the next time user opens Tables Module. Explorer Shows or hides the Explorer Panel. PROGRAM INTERFACE \| 173 \| 3.3.2. Explorer Panel The Explorer Panel consists of all the available items of the inputs and results classified into sections and arranged hierarchically. Any item in the Explorer Panel can be clicked on to display the corresponding table in the Preview Area. Expand all Expands item list. Collapse all Collapses item list. PROGRAM INTERFACE \| 174 \| 3.4. Reporter Window The Reporter Module interface shown above enables the user to view, customize, print and export reports in different formats. The Reporter Module is accessed from within the Main Program Window by clicking the Reporter button from the Ribbon. Alternatively, Reporter Module can also be accessed by pressing the F7 key. If the model has not been solved then the Reporter Module will only contain a list of input data reports. When a model has been successfully executed, the Reporter Module will also display the output data reports. Immediately after opening the Reporter Module, you can export and/or print the default report by pressing Export/Print button. Various options to customize the report before printing and/or exporting it are also provided. Once the work in Reporter Module is complete, click the close button in the top right corner to exit Reporter window. PROGRAM INTERFACE \| 175 \| 3.4.1. Toolbar Previous page Displays the previous page of the report. Next page Displays the next page of the report. Page number box Displays the page with the page number entered in the box. Zoom in Zooms in on the report (Ctrl + Mouse wheel up). Zoom out Zooms out on the report (Ctrl + Mouse wheel down). Zoom box Zooms on the report preview to the extent typed in the box or selected from the dropdown list. Fit to window width and enable scrolling Fits the width of report to the preview space width and enables scrolling. Fit one full page to window Fits one full page in the preview space. Pan When toggled on and report is bigger than preview window, enables panning the report. Text selection When toggled on enables selecting text in the report. PROGRAM INTERFACE \| 176 \| Settings Modifies settings for Report and Explorer Panel. Report settings • FONT SIZE: Provides the options to use small, medium or large font sizes in the report. • REGENERATE AUTOMATICALLY: Enables automatic regeneration of report when content selection is modified by the user. • SPLIT LONG TABLES: Displays table headings in all pages when tables are split along several pages. Explorer settings • LOCATION: Displays Explorer Panel on the left or right side of screen depending on selection. • HIDE INACTIVE ITEMS: Hides unused tables from the explorer view. PROGRAM INTERFACE \| 177 \| • KEEP EXPLORER CONFIGURATION: Saves the explorer configuration i.e., information about selected tables and opened/closed sections so that it is available the next time user opens Reporter. Explorer Shows or hides the Explorer Panel. 3.4.2. Export / Print Panel Export Exports the report in the selected format, with an option to automatically open the report or its file location. Print Prints the report in the selected format when the option is available. Type Provides 5 format options to print and/or export the reports • WORD: produces a Microsoft Word file with .docx extension. • PDF: produces an Adobe Acrobat file with .pdf extension. • TEXT: produces a Text file with .txt extension. • EXCEL: produces a Microsoft Excel file with .xlsx extension. • CSV: produces a Comma Separated file with .csv extension. Printer Provides the option to select available printers and change printer properties. PROGRAM INTERFACE \| 178 \| Settings Provides the options to modify print settings. • PAPER: Provides the options to select from available paper sizes. • ORIENTATION: Provides the options to select between landscape or portrait paper orientation. • MARGINS: Provides the options to use narrow, normal, wide or custom margins to the report • PRINT RANGE: Provides the options to select the pages to print and/or export. PROGRAM INTERFACE \| 179 \| 3.4.3. Explorer Panel The Explorer Panel consists of all the available report items classified into sections and arranged hierarchically. Each item listed in the Explorer Panel is preceded by a checkbox. The user can check/uncheck the checkbox to include or exclude from the report, the items or sections. Expand all Expands item list. Collapse all Collapses item list PROGRAM INTERFACE \| 180 \| 3.5. Print/Export Window Print/Export Module interface shown above enables the user to view, customize, print and export diagrams in different formats. The Print/Export Module is accessed from within the Main Program Window by using the Right Click Menu or from the Reporter Submenu in the Ribbon. PROGRAM INTERFACE \| 181 \| Alternatively pressing the “CTRL + P” also opens the Print/Export Module. Once the module is open the rest of the program is locked until the Print/Export Module is closed. Immediately after opening the Print/Export Module, you can export and/or print the generated diagram by pressing Export/Print button. Options to customize the diagram orientation, paper size and margins are provided. Once the work in Print/Export Module is complete, click the close button in the top right corner to exit the module. 3.5.1. Toolbar Zoom in Zooms in on the report (Ctrl + Mouse wheel up). Zoom out Zooms out on the report (Ctrl + Mouse wheel down). Zoom box Zooms on the report preview to the extent typed in the box or selected from the dropdown list. Fit one full page to window Fits one full page in the preview space. Pan When toggled on and report is bigger than preview window, enables panning the report. PROGRAM INTERFACE \| 182 \| 3.5.2. Export / Print Panel Export Exports the report in the selected format, with an option to automatically open the report or its file location. Print Prints the displayed diagram. Type Provides 4 format options to export the reports • EMF produces a file with .emf extension • BMP produces a file with .bmp extension • TO REPORT adds the diagram to the report • TO CLIPBOARD copies the diagram to clipboard to be pasted elsewhere Printer Provides the option to select available printers and change printer properties. PROGRAM INTERFACE \| 183 \| Settings Provides the options to modify print settings. • PAPER: Provides the options to select from available paper sizes. • ORIENTATION: Provides the options to select between landscape or portrait paper orientation. • MARGINS: Provides the options to use narrow, normal, wide or custom margins to the report. \| 184 \| 4. MODELING METHODS 4.1. Model Creation Concepts The key to effectively implementing spSlab/spBeam in a project is understanding the program’s robust approach to modeling, analyzing, designing, and evaluating reinforced concrete slabs and beams under a variety of loading conditions. Of utmost importance is the understanding of the methods utilized in the program for the analysis of reinforced concrete floor slab systems. This section provides insights into the methods, assumptions, and factors that the design professional must consider while modeling using spSlab/spBeam for analysis, design, and detailing. As a foundational guideline, the geometry of the analytical model should represent the physical structure as closely as possible to ensure accurate analysis results. Users must confirm that the project criteria align with applicable design codes and standards. This includes considerations for load types, load factors, load combinations, material properties, reinforcement requirements, deflection criteria, and detailing provisions, ensuring compliance with industry standards and best practices. CHAPTER 4 MODELING METHODS \| 185 \| 4.1.1. Physical Modeling Terminology In spSlab, the terminology around elements and members is critical for understanding how the program models reinforced concrete slabs and beams. spSlab utilizes elements to represent structural components in the Equivalent Frame Method (EFM). Elements are the primary modeling units within spSlab, corresponding closely to the physical structural members in the project. The EFM is a very well established analysis method used exclusively for two-way concrete floor systems. In this method, a great variety of floor systems are covered in its scope. As a result. Many structural elements contribute to the making of a concrete floor system. These elements allow users to capture the geometry and material properties of slabs, beams, slab bands, columns, drop panels, and column capitals within the analytical model. These elements are used at the engineer's discretion to combat design challenges pertaining to one-way shear, two-way shear, inadequate flexural strength, or unacceptable deflections. The EFM in spSlab simplifies modeling by focusing on frame strips representing the slab-beam-column system. Users only need to define the primary elements, which spSlab discretizes into equivalent frame members for analysis. This streamlined modeling method saves time and simplifies the design process. By using the EFM, spSlab enables both efficient modeling and accurate structural behavior prediction without requiring more elaborate and intricate analysis processes such as the Finite Element Method (FEM) of analysis. Instead, users can focus on defining element properties and ensuring that the geometry of the analytical model accurately reflects the physical structure. It is crucial to understand the concept of elements in spSlab, as it forms the foundation for creating effective and efficient models. Once familiar with this approach, users can appreciate the simplicity and power of the EFM in modeling complex slab and beam systems. MODELING METHODS \| 186 \| 4.1.2. Structural Elements The spSlab program uses elements to represent physical structural members. When creating a model, users begin by defining the geometry of slabs, beams, and other structural elements within the program using drawing area common Computer-Aided Design (CAD) tools and then assign properties and loading to these elements, fully specifying the structural model for analysis. To complete the model of a concrete floor system, two essential element types are required in spSlab: • Span Elements: Used to represent slabs, longitudinal beams, ribs, and longitudinal slab bands. • Support Elements: Used to represent columns, drop panels, column capitals, transverse beams, and restraints. As a general rule, the geometry of each element should represent that of the actual physical member as closely as possible. This approach enhances visualization of the model and reduces potential errors during input. However, engineers can omit small changes in shape and geometry where added model accuracy or complexity is not consequential to the analysis & design results. A great deal of engineering judgment is involved in the conversion of a physical structure into an analytical model. However, significant gains can be achieved by keeping model simple & practical to the extent possible. It is also essential that beginner users must establish smaller simpler models at first to gain a better understanding of the program and its features. This will help greatly in understanding the method of solution as well as figure out what to do when there is an issue to diagnose or verify in the output. It is always much easier to discover the source of an error when working with a simple model and a few loading conditions. Once a model becomes complicated with numerous loads and load combinations, it becomes increasingly difficult to discover the source of an issue or a concern in the output, graphical or tabular. MODELING METHODS \| 187 \| 4.1.3. Properties Material and geometric properties for model elements in spSlab and spBeam are categorized into two groups: Defined Properties and Unique Properties. This distinction provides users with the flexibility to manage elements properties consistently across the model while accommodating variations for specific spans and supports. Defined Properties refer to material properties, including concrete material properties (compressive strength fc′, unit density Wc, Young's modulus Ec, and rupture modulus fr), and reinforcing steel material properties (yield stress for flexural steel fy, yield stress for stirrups fyt, and Young's modulus Es). Concrete material properties are consistent across all slab and beam elements but can differ for column elements, allowing for tailored input to reflect different structural requirements. Reinforcing steel material properties, however, remain the same throughout the model, including slabs, beams, and columns, ensuring uniformity in reinforcing steel characteristics. The program enables users to define these properties globally, maintaining consistency and alignment with design standards such as ACI 318 and CSA A23.3, while accommodating necessary distinctions between slabs, beams, and columns where applicable. Unique Properties, on the other hand, refer to element geometric properties that are specific to individual spans and supports. These properties (such as slab/beam width, thickness, column width, depth, or height) are assigned on a per-element basis, enabling the program to account for variations in geometry that influence structural behavior. Each span and support can therefore have distinct geometric attributes, allowing for a more refined and realistic representation of the structure. MODELING METHODS \| 188 \| 4.1.4. Input Preparation If you do not find a suitable template to begin your project, you can start with a blank project. A blank project requires a lot more care in the construction of the model and much more effort to complete the details of the input. To begin, start by setting up the grids in the Grid command to define the layout. Next, use the Spans command to add slab, beam, or rib elements and adjust their dimensions as needed. Finally, use the Supports command to add support elements such as columns, beams, or restraints, and configure their properties, including height, size, and end conditions. The following guidelines can be used in this scenario: • To establish a comprehensive model for structural analysis, the user begins by defining grids and specifying span lengths to ensure an accurate layout representation. If the span lengths need to be adjusted later, the user can return to the Grid command and modify the span lengths using the Span table. • The user must define frame location and evaluate which location best aligns with the design intent. Detailed insights and guidance on selecting appropriate frame location can be found in Section 5.2.2.4 of the manual. • A critical decision involves addressing cantilever extensions - whether they function as true cantilevers (e.g., balcony or canopy projections) or are intended to encompass the column to contribute to two-way shear resistance. In the Span table under Grid command, users can select USER-DEFINED for cantilevers on the left and right to simulate a true cantilever or choose ADJUST TO SUPPORT FACE to encompass the column. • Building on the selection of cantilever behavior, choosing ADJUST TO SUPPORT FACE in the Span table allows the cantilever to align its geometry with the support face. When selected, the program assigns a placeholder cantilever length of 5 ft until an exterior support is defined. If the model is run without assigning the required support, the process will be interrupted, and the program will display an error message, indicating the requirement for a column or transverse beam at the cantilever edge joint. The figure below illustrates this process. MODELING METHODS \| 189 \| Figure 4.1 – Adjust to Support Face Option • When NONE is selected for the right and left cantilevers in the Span table, the model assumes that no cantilever extensions are required. For that case, if the user runs the program without assigning any additional supports, the program will automatically assign default pin supports at the slab joints. This allows the model to run without interruptions, ensuring the structural analysis proceeds under simplified support conditions. MODELING METHODS \| 190 \| 4.1.5. Modeling Considerations 4.1.5.1. Sway and Non-Sway Considerations When considering sway versus non-sway conditions in structural analysis, it is essential to account for lateral effects in the model to ensure accurate results. In non-sway conditions, programs like spSlab and spBeam assume horizontal translational restraint, excluding lateral effects directly from the analysis. This approach simplifies the modeling process and is suitable for frames where lateral displacements are minimal or restrained, such as single-story frames with balanced vertical loads and consistent boundary conditions. However, in sway conditions, where unsymmetrical vertical loading, support stiffness variations, or boundary conditions induce lateral movements, the lateral effects must be explicitly included. These can be added as member-end forces, such as moments acting at the ends of each span, using Lateral Load Effects, which can be assigned from the Lateral command in the Left Panel under Loads command. This allows users to input lateral loads like wind or seismic forces as moments at the ends of members, enabling the program to account for combined vertical and lateral effects accurately. For cases where significant lateral displacements or sidesway effects are expected, it is recommended to use spFrame, which models the frame as a two-dimensional unit for the entire building height. spFrame accurately captures the influence of sidesway on internal force magnitudes, ensuring reliable results for design. This dual approach - leveraging spSlab and spBeam for simplified non-sway analysis and spFrame for sway-sensitive scenarios - ensures flexibility and precision in addressing varying structural conditions. Additional details about this topic can be found in Section 2.6.10, Section 5.2.4.3, and “Comparison of Gravity Loaded Concrete Frame Models in spBeam and spFrame under Sidesway” Technical Article from StructurePoint. MODELING METHODS \| 191 \| 4.1.5.2. Unbalanced Moments in Column Design Unbalanced moments in a floor system play a critical role in the design of columns located above and below the slab-beam structure. These moments arise from the distribution of loads and are transferred from the slab to the support columns in proportion to their relative stiffness. This transfer creates forces in the columns, which must be accurately calculated and incorporated into their design to ensure structural integrity. Additional details about this topic can be found in Section 2.3.7.4. Column end moments and axial forces can be found in the results table. These forces for various load combinations and live load patterns can be then exported to spColumn for detailed design or investigation of a column cross-section. StructurePoint prepared several design examples for a complete floor system including the calculation of unbalanced moments in columns and the investigation of column section reinforcing to provide the required strength in the column. Detailed discussion of unbalanced moments in column design can be found in the following design examples from StructurePoint: “Two-Way Flat Plate Concrete Slab Floor Analysis and Design” Section 2.2.7. and Section 3. “Two-Way Flat Slab (Drop Panels) Concrete Floor Analysis and Design” Section 2.1.7. and Section 3. “Two-Way Concrete Floor Slab with Beams System Analysis and Design” Section 2.7. and Section 3. “Two-Way Joist Concrete Slab Floor (Waffle Slab) System Analysis and Design” Section 2.1.7. and Section 3. MODELING METHODS \| 192 \| Figure 4.2 – Column Moments (Unbalanced Moments from Slab-Beam) MODELING METHODS \| 193 \| 4.1.5.3. One-Way Slabs on Transverse Beams In modeling one-way slabs supported by transverse beams, several important and consequential considerations must be addressed to ensure accurate analysis and design. The rotational resistance of exterior supports plays a significant role in moment distribution. Two common conditions are recognized: • Unrestrained exterior supports: Treated as pin supports, they provide no rotational resistance at the slab end. This is the default assumption in spSlab/spBeam, with the rotational stiffness (Kry) set to zero. • Integral exterior supports: Provide rotational resistance and are modeled as semi-rigid connections. For example, spandrel beams integrated with the slab can be assigned rotational stiffness approximated iteratively using design codes provisions or other sources for moment factors, offering a more realistic representation. Additionally, modeling transverse beams or girders can incorporate their added stiffness, reducing the effective slab span and yielding a closer approximation to actual conditions. While this approach enhances the accuracy of moment distribution, it introduces additional modeling parameters, requiring careful engineering judgment. Additional details about this topic can be found in “One-Way Slab Analysis and Design” Design Example from StructurePoint. MODELING METHODS \| 194 \| Figure 4.3 – One-Way Slabs on Transverse Beams MODELING METHODS \| 195 \| 4.1.5.4. Boundary Condition Effects on Continuous Beam Deflections Boundary conditions significantly influence the deflection behavior of continuous beams in reinforced concrete structures. Properly modeling these conditions in spBeam ensures accurate analysis and design results. This section explores the effects of different support conditions: • Beam Supported by Columns: The stiffness of columns above and below the beam is modeled to calculate rotational stiffness at the joint. This accurately determines beam end moments. Support stiffness is adjustable, with values ranging from 0 (pinned) to 999 (fixed support), providing flexibility in simulating various boundary conditions. • Beam Supported by Transverse Beams: Transverse beams are modeled using rotational stiffness values. ACI and CSA codes approximate moments at transverse beam supports as two-thirds of those at column supports. For more precise design, dummy columns can be used to define critical sections for shear and moment calculations. • Beam Supported by Transverse Walls: Beams cast monolithically with shear walls are modeled as having walls defined as elongated columns to simulate integral behavior. • Beam Supported by Masonry Bearing Walls: Masonry walls are modeled as pinned supports with zero stiffness, effectively simulating their bearing behavior without rotational constraints. • Beam Supported by Longitudinal Walls: Beams are modeled up to the face of the wall with fixed supports at the wall face, ignoring wall width in the analysis. Deflection comparisons highlight the variations due to boundary conditions. For example, beams with transverse beam supports show higher deflections compared to those supported by walls, emphasizing the importance of modeling boundary conditions accurately. For further details, refer to “Reinforced Concrete Continuous Beam Analysis and Design” Design Example from StructurePoint. MODELING METHODS \| 196 \| Figure 4.4 – Plan View of Continuous Beams with Different Boundary Conditions Figure 4.5 – Comparison of Continuous Beams with Different Boundary Conditions (Deflections) MODELING METHODS \| 197 \| 4.1.5.5. Design of New Buildings vs. Investigation of Existing Buildings In spSlab/spBeam, the modeling approach depends on whether the project involves the design of a new building or the investigation of an existing building. The program offers two distinct run modes to address these scenarios: DESIGN mode and INVESTIGATION mode. In DESIGN mode, the program performs structural analysis and determines the required flexural, shear, and torsional reinforcement based on the selected design code. This mode is ideal for new structures, as it provides a baseline design that ensures compliance with applicable codes and standards. In INVESTIGATION mode, the user inputs the existing flexural, shear, and torsional reinforcement. The program then evaluates the adequacy of the provided reinforcement given the section shape and material properties used. This mode is particularly suited for assessing the safety and performance of existing structures or making modifications to them. Even for investigations, it is recommended to initially use DESIGN mode to establish a reinforcement baseline that aligns with current codes. This design output can then serve as a starting point for further analysis and refinement in INVESTIGATION mode, ensuring consistency and accuracy throughout the modeling process. For detailed guidance, refer to Sections 5.2.3.2 and 5.2.4.4. MODELING METHODS \| 198 \| 4.1.5.6. Modeling and Design of T-Beams In spSlab/spBeam, T-beams are modeled to account for the contribution of flanges, which can be formed either by monolithic casting of slabs and beams or as isolated T-beams, such as precast elements. The inclusion of flanges enhances structural performance by utilizing the flange as part of the flexural member, improving both strength and efficiency. One advantage of flanges in T-beams is their ability to increase the compression area, boosting the beam's flexural capacity without requiring significant additional reinforcement. This makes T-beams highly effective for handling larger moments compared to rectangular beams. Another advantage lies in material efficiency. In T-beams, integrating the slab as part of the beam section reduces the concrete and steel required for the same structural capacity. Shear design for T-beams can be more complex due to the nonuniform stress distribution in the web caused by the flange. This requires careful modeling and analysis to ensure accurate results. For precast T-beams, connection details between the beam and the surrounding structure must be considered to achieve the intended performance. Proper construction practices are crucial in both cases. For monolithic T-beams, ensuring the slab and beam are cast together effectively is essential for accurate structural behavior. For precast T-beams, the quality of the connections and proper alignment during installation are critical. In spSlab/spBeam, flange dimensions and configurations are defined during modeling using the Spans command. Additionally, users can engage relevant code provisions for T-beam analysis, design, and deflection calculations through the Design Options and Deflection Options available under the Solve command. These features provide flexibility and precision in modeling and designing T-beams in accordance with applicable standards. StructurePoint has prepared a number of T-beam case studies to assist in understanding the complexity of this important structural element. These cases illustrate the number of ways to handle the flexural design given single or doubly reinforce configurations as shown in the following table. MODELING METHODS \| 199 \| Case Study Description Notes Case One • Rectangular Section Behavior • Singly reinforced • The stress block depth "a" is less than the flange thickness • 2 layers of tension reinforcement Case Two • T-Section Behavior • Singly reinforced • The stress block depth "a" is more than the flange thickness • 2 layers of tension reinforcement Case Three • T-Section Behavior • Singly reinforced • The stress block depth "a" is more than the flange thickness • 3 layers of tension reinforcement Case Four • Rectangular Section Behavior • Doubly reinforced • The stress block depth "a" is less than the flange thickness • 3 layers of tension reinforcement • 1 layer of compression reinforcement Case Five • T-Section Behavior • Doubly reinforced • The stress block depth "a" is more than the flange thickness • 3 layers of tension reinforcement • 1 layer of compression reinforcement Table 4.1 – StructurePoint T-Beams Case Studies MODELING METHODS \| 200 \| 4.1.5.7. One-Way Slab System Considerations In spSlab/spBeam, users have the option to model span elements as either slabs or beams; however, it is crucial to exercise sound engineering judgment to select the appropriate span element type. While the program permits the use of beam elements to model a one-way slab system, users must understand the implications of this choice, particularly concerning design code requirements. Design codes, such as ACI 318 and CSA A23.3, differentiate between the minimum reinforcement ratios for slabs and beams. When modeling one-way systems, it is advisable to use slab elements for components behaving as slabs and reserve beam elements for structural members specifically designed as beams. Misusing these span elements may lead to incorrect reinforcement calculations as well as impact shear strength and deflection estimates. spSlab/spBeam empowers users with the tools to model various structural configurations but relies on the engineer’s expertise to ensure the modeling choices align with the intended design purpose and applicable standards. Figure 4.6 – Modeling Span Elements as Either Slabs or Beams MODELING METHODS \| 201 \| 4.2. Model Editing Concepts 4.2.1. Editing Elements During the course of creating a model, it may be necessary to edit the model. This can be done through the Select button in the Ribbon. Then, any element that is present on the active Viewport can be selected and edited by the tools available within the Left Panel Toolbar and Left Panel. The editing tools at the Left Panel Toolbar can also be invoked by right-clicking the mouse button in the active Viewport. The editing tools that are available at the Left Panel Toolbar per element are: • Span Element: Span elements can be duplicated or advanced copied. • Support Element: Support elements can be deleted, moved, duplicated, or advanced copied. Figure 4.7 – Editing Tools The Duplicate tool provides a straightforward way to copy all assigned elements and loads directly from the original to the destination span or support. In contrast, the Advanced Copy tool offers a more flexible approach by displaying a dialog box with selectable options. MODELING METHODS \| 202 \| Figure 4.8 – Advance Copy Dialog Box The Left Panel can also be utilized to edit an Element further: • Span Elements: In addition to the editing tools available in the Left Panel Toolbar, users can edit slab or flange dimensions (including thickness and width), longitudinal beam dimensions (including width, depth, and offset), rib dimensions (including bottom width, depth, and clear spacing at the bottom), as well as area loads, line loads, point loads, and lateral load effects. • Support Elements: In addition to the editing tools available in the Left Panel Toolbar, users can edit columns (including type, height, cross-sectional dimensions, far end condition, punching shear check and γf adjustment), drop panels (including type, thickness, and punching shear check), column capitals (including depth and side slope), transverse beams (including width, depth, and offset), as well as support restraints, support springs, support loads, supports displacements, and redistribution limits. MODELING METHODS \| 203 \| 4.2.2. Span Length and Grid Spacing In spSlab/spBeam, the grid spacing defined by the user determines the span lengths of the model. Any modifications to the grid spacing directly translate into changes in span lengths, which can significantly impact the placement and application of loads. For instance, if a point load is initially applied at 15ft relative to a specific span and the grid spacing is later modified (e.g., reducing the span from 30ft to 10ft), the load retains its global position within the model. This may result in the load appearing in a different span or outside its intended span entirely. Figure 4.9 – Effect of Grid Spacing on Span Length and Load Location MODELING METHODS \| 204 \| When such discrepancies occur and the model is executed, the program issues an error, such as “Span 2, case: Dead, load falls outside the span”, to alert the user of invalid load placement. Figure 4.10 – Warnings / Errors Tab in Solver Dialog To avoid such issues, it is strongly recommended to finalize the grid spacing - and therefore the span lengths - before assigning supports and loads. Any subsequent adjustments to grid spacing will require careful review and reassignment of loads to ensure accurate structural analysis and design. MODELING METHODS \| 205 \| 4.2.3. Changing Floor System Location In spSlab/spBeam, the Templates module allows users to define the floor level of the modeled system, providing flexibility to represent various configurations such as first floors, intermediate floors, or roofs. The selected floor level determines the presence and arrangement of supports, such as columns, above and below the slab. For first and intermediate floors, setting the floor level to STORY enables the program to include columns both above and below the slab, accurately reflecting the structural conditions. When the floor level is changed to ROOF, the program automatically removes columns above the slab, as roofs typically do not have supporting elements above them. Figure 4.11 – Floor Level Location While users can manually create similar scenarios by modifying supports, the templates module provides a faster and more efficient way to set up these common layouts. By automating much of the setup, templates save time and simplify the modeling process, making it easier to ensure consistency and accuracy in model generation. MODELING METHODS \| 206 \| 4.2.4. Editing Loads In spSlab/spBeam, editing applied loads can be accomplished using two options: REPLACE EXISTING LOAD or ADD TO EXISTING LOAD. These options provide users with the flexibility to update load conditions without needing to reset or delete previously applied loads. The REPLACE EXISTING LOAD option allows users to completely overwrite the current load on the element or span with a new value. This is particularly useful when the load conditions have changed entirely, and the previous load no longer applies to the design scenario. The ADD TO EXISTING LOAD option enables users to incrementally add new loads to the currently applied load. This approach is especially valuable for modeling additional load components, such as live loads or lateral load effects, while preserving the baseline loads already assigned. Figure 4.12 – Adding / Editing Loads These features apply to all load types supported by the program, including area loads, line loads, point loads, support loads, and lateral load effects. By providing these options, spSlab/spBeam streamlines the process of refining and updating load conditions to accommodate a wide range of design requirements efficiently. MODELING METHODS \| 207 \| 4.2.5. Managing Load Locations In spSlab/spBeam, the location of line loads and point loads can be specified using two methods: absolute distance or distance as a ratio of span. These options provide users with flexibility in controlling how loads are positioned on spans, particularly when span lengths are subject to modification during the modeling process. Using the absolute distance method, the load location is defined as a fixed value measured from the start of the span. This method is active when the user unchecks the DISTANCE LOCATION AS RATIO OF SPAN option. While this approach provides precise control in static models, any changes to the span length will leave the load at its original fixed position, which may result in the load being misplaced relative to the updated span geometry as shown in the previous section. Figure 4.13 – Managing Load Locations – Absolute Distance MODELING METHODS \| 208 \| To assist users in such cases, the program issues a warning: “Span length(s) has changed. Check the validity of span load locations”. This notification reminds users to reevaluate and adjust load positions after modifying span lengths. Figure 4.14 – Span Length Warning Dialog Box Alternatively, the distance as a ratio of span method allows the load location to be defined as a percentage of the span length. This method is enabled when the user checks the DISTANCE LOCATION AS RATIO OF SPAN option. It ensures that the load location adjusts proportionally when the span length changes, maintaining its relative position within the span. This approach is particularly advantageous for projects where span lengths may be updated, as it eliminates the need for manual load adjustments following changes to the span configuration. MODELING METHODS \| 209 \| Figure 4.15 – Managing Load Locations – Distance as a Ratio of Span \| 210 \| 5. MODEL DEVELOPMENT In spSlab/spBeam, models can be started by utilizing one of the four methods under Projects within Start Screen. These are namely; Open Project, New Project, Templates, and Examples. Each of these methods can be used to create a new model from scratch, edit a model developed previously for an earlier project, start with pre-defined template, or use an existing example file from the provided library. Each of the methods are described in detail in this chapter. CHAPTER 5 MODEL DEVELOPMENT \| 211 \| MODEL DEVELOPMENT \| 212 \| 5.1. Opening Existing Models In the Start Screen under Projects select the Open Project option and browse to the folder that contains an existing spSlab/spBeam input file. The input files created in spSlab/spBeam v5.50 (.slb) and in spSlab/spBeam v10 (.slbx) can be opened. The input files for the prior versions of the Program require to be saved in consecutively newer version until .slb file is obtained. Then, that file can be opened in v10. Input file created in the newer version of the spSlab/spBeam program cannot be opened by a previous version. MODEL DEVELOPMENT \| 213 \| 5.2. Creating New Models In the Start Screen under Projects, select the New Project option, then select Empty Project option from the New Project dialog box. The model development process may require general input regarding a specific project. Project Information is entered through the Project command button, and Structural Grids are entered through the Grid command button. 5.2.1. Project Information The project information regarding to DESIGN CODE, UNIT SYSTEM, BAR SET, CONCRETE STRENGTH, REINFORCING STEEL STRENGTH, RUN MODE, FLOOR SYSTEM, PROJECT, FRAME, and ENGINEER can be entered into the model through the Project Left Panel. The Program supports American (ACI 318) and Canadian (CSA A23.3) Design Codes, and English and Metric unit systems. MODEL DEVELOPMENT \| 214 \| 5.2.2. Structural Grids Structural grids must be defined to create a model in spSlab/spBeam. These grids establish the spans and support locations. The spans and supports created by grids then act as placeholders for model elements. Once the spans and supports are set, the user can populate the model with various span elements (slabs, ribs and longitudinal beams/bands) and support elements (columns, capitals, drop panels and transverse beams/bands) to complete the model. 5.2.2.1. Working with Grids You can select the Grid command button from the Ribbon. The corresponding Grids Left Panel provides various tools and options for effectively working with grids. MODEL DEVELOPMENT \| 215 \| 5.2.2.2. Generating Spans • Click the Generate icon to display the GENERATE SPANS dialog box • To create multiple spans at once enter: number of spans x span length in the SPANS LENGTH(S) input box • You can create spans at different lengths by separating the span lengths by space • To add cantilevers, select either USER-DEFINED or ADJUST TO SUPPORT FACE in the LEFT CANTILEVER and RIGHT CANTILEVER input boxes • You can select the frame location from the FRAME LOCATION drop-down list MODEL DEVELOPMENT \| 216 \| 5.2.2.3. Using the Span Table You can use the Span table to change the cantilever options and the LENGTH of the spans. The table can also be used to add or delete specific spans. • Click on the button to add a span. • Click on the button to delete the selected span in the table. • Click on the button to move up the selected span in the table. • Click on the button to move down the selected span in the table. • LENGTH of the span can also be changed by clicking on the respective field and typing in the desired value. • For cantilevers: • Select NONE if no cantilever is required. • Select ADJUST TO SUPPORT FACE to extend the span to the support face. • Select USER-DEFINED to manually specify the cantilever length. MODEL DEVELOPMENT \| 217 \| 5.2.2.4. Working with Frame Location You can select the span location from the FRAME LOCATION drop-down list. Three types of locations are available: INTERIOR, EXTERIOR LEFT and EXTERIOR RIGHT. If the frame being analyzed has more than one of the above-mentioned locations, then VARIABLE can be selected and the individual span locations provided accordingly. The "left" and "right" are defined as you look along the direction of analysis. If a span has design strips on both sides, it should be an "INTERIOR" span. If a span has only a left design strip, it should be an "EXTERIOR RIGHT" span. If a span has only a right design strip, it should be an "EXTERIOR LEFT" span. MODEL DEVELOPMENT \| 218 \| 5.2.2.5. Grids Display Options • Use the checkboxes to toggle the display of various grid items in the model. • You can use the slider to adjust the SIZE of the grid LABELS, UNITS and DIMENSIONS displayed. MODEL DEVELOPMENT \| 219 \| 5.2.3. Generating Definitions The DEFINITIONS dialog box contains four categories: Materials, Reinforcement Criteria, Design and Modeling Options, and Load Case/Combo. You can access this dialog box by selecting the Define command from the Ribbon. MODEL DEVELOPMENT \| 220 \| 5.2.3.1. Materials The Materials that can be defined are: Concrete and Reinforcing Steel. Concrete The following properties for slab, beam, and column concrete, are entered by the user: • COMPRESSIVE STRENGTH, f'c, and • UNIT DENSITY, Wc Other concrete properties: • YOUNG’S MODULUS, Ec, and • RUPTURE MODULUS, fr are automatically computed and displayed when STANDARD option is selected. The user can manually modify any of the values by deselecting the option. When STANDARD option is selected, Design Code mandated default values for the young’s modulus, Ec , and the rupture modulus, fr , for the slabs, beams, and columns are computed automatically and are based on f'c and λ where λ is function of Wc . The modulus of rupture is used to determine the cracking moment when computing the effective moment of inertia in deflection calculations. Refer to Section 2.2.1.2 for default Ec , and fr equations per ACI 318 and CSA A23.3 Design Codes. Unchecking the STANDARD option shall enable the User to input Ec and fr manually that deviates from Equations in Section 2.2.1.2. Therefore, it may lead to an output not in compliance with ACI 318 and CSA A23.3 if used without sound Engineering Judgment. If precast concrete is used, check PRECAST CONCRETE checkbox 1. 1 CSA A23.3-14, 8.4.2, 16.1.3; CSA A23.3-04, 8.4.2, 16.1.3 MODEL DEVELOPMENT \| 221 \| MODEL DEVELOPMENT \| 222 \| Reinforcing Steel The following properties are entered by the user. • YIELD STRESS OF FLEXURAL STEEL, fy • YIELD STRESS OF STIRRUPS, fyt • YOUNG’S MODULUS, Es • If reinforcement is epoxy-coated, check REINFORCING BARS ARE EPOXY-COATED checkbox. This selection affects development lengths. MODEL DEVELOPMENT \| 223 \| 5.2.3.2. Reinforcement Criteria The Reinforcement Criteria that can be defined are for: Slabs & Ribs, Beams, Beam Stirrups, and Bar Set. Slabs & Ribs To define Reinforcement Criteria for Slab and Ribs: • Enter the minimum bar size to start the iteration for determining flexural reinforcement. • Enter the maximum bar size. This number will be used as a stop in the iteration for determining flexural bars in beams. • Enter minimum bar spacing for slab and rib flexural reinforcement. This number should be based on aggregate size or detailing considerations. Default spacing is 1 in. [30 mm] for slabs and ribs. • Enter maximum bar spacing for slab and rib flexural reinforcement. Default spacing is 18 in. [450 mm] for slabs and ribs. • Enter minimum Reinforcement Ratio for slab and rib flexural reinforcement. Default ratio is 0.14% [0.14%] for slabs and ribs. If the user specified value is smaller than 0.14%, 0.14% is used by spSlab. If the user specified value is greater than 0.14%, the specified value is used by spSlab. • Enter maximum Reinforcement Ratio for slab and rib flexural reinforcement. Default ratio is 5% for slabs and ribs. • Enter the clear covers for top and bottom reinforcing bars. For the top reinforcement, this distance is from the top of the slab to the top of the top bars. For the bottom reinforcement, this distance is from the bottom of the slab to the bottom of the bottom bars (see Figure 5.1). The default value is 0.75 in. [20 mm] for both input items. • If the top bars have more than 12 in. [300 mm] of concrete below them, check the corresponding check box. MODEL DEVELOPMENT \| 224 \| Figure 5.1 – Clear Cover to Reinforcement MODEL DEVELOPMENT \| 225 \| Beams To define Reinforcement Criteria for Beams: • Enter the minimum bar size for top and bottom bars to start the iteration for determining flexural reinforcement. • Enter the maximum bar size for top and bottom bars. This number will be used as a stop in the iteration for determining flexural bars in beams. • Enter the minimum bar spacing for beam flexural reinforcement. This number should be based on aggregate size or detailing considerations. The default minimum reinforcement bar spacing is 1 in. [30 mm]. • Enter the maximum bar spacing for beam flexural reinforcement. The default maximum reinforcement spacing is 18 in. [450 mm]. • Enter the minimum Reinforcement Ratio for beam flexural reinforcement. Default ratio is 0.14% [0.14%] for beams. If the user specified value is smaller than 0.14%, 0.14% is used by spSlab. If the user specified value is greater than 0.14%, the specified value is used by spSlab. • Enter the maximum Reinforcement Ratio for beam flexural reinforcement. Default ratio is 5% for beams. • Enter the covers for top and bottom reinforcing bars for beams. For the top reinforcement, this distance is from the top of the beam to the top of the top bars; and for the bottom reinforcement, this distance is from the bottom of the beam to the bottom of the bottom bars (see Figure 5.1). The default value is 1.5 in. [30mm] for both input items. • Enter the clear distance between bar layers to use if the program needs to distribute flexural bars in multiple layers. Default distance is 1 in. [30 mm] for beams. • If the top bars have more than 12 in. [300 mm] of concrete below them, check the corresponding check box. MODEL DEVELOPMENT \| 226 \| MODEL DEVELOPMENT \| 227 \| Beam Stirrups To define Reinforcement Criteria for Beam Stirrups: • Enter the minimum bar size for stirrups to start the iteration for determining shear reinforcement. • Enter the maximum bar size for stirrups. This number will be used as a stop in the iteration for determining shear reinforcement in beams. • Enter the minimum spacing for stirrups. This number should be based on aggregate size or detailing considerations. The default stirrup spacing is 6 in. [150 mm]. • Enter the maximum spacing for stirrups. The default maximum stirrup spacing is 18 in. [450 mm]. • Enter the side cover which is measured from the side face of a beam to the face of the stirrup (see Figure 2.28). The default value is 1.5 in [30 mm]. • Enter the distance from face of support (FOS) to first stirrup. The default value is 3.0 in [75 mm]. MODEL DEVELOPMENT \| 228 \| MODEL DEVELOPMENT \| 229 \| Bar Set The Bar Set can be selected from pre-defined standard sets or can be USER-DEFINED. For a new USER-DEFINED set entry, the SIZE, DIAMETER, AREA and WEIGHT of the bar are required. USER-DEFINED set can also be imported or exported. MODEL DEVELOPMENT \| 230 \| 5.2.3.3. Options Design & Modeling To specify Design & Modeling Options for two-way systems: • Check USER SLAB STRIP WIDTHS to enable manual input of column strip width. The default values are calculated according to design code selected. The validity of the assumptions when entering user defined values are to be decided by the Designer. • Check USER STRIP DISTRIBUTION FACTORS to enable manual input of moment distribution factors. The default values are calculated according to design code selected. The validity of the assumptions when entering user defined values are to be decided by the Designer. To specify Design & Modeling Options for beams/one-way slab systems: • Check MOMENT REDISTRIBUTION checkbox if it is to be considered in the analysis. This option has to be checked for the Moment Redistribution tab to be available in the Supports command. MODEL DEVELOPMENT \| 231 \| 5.2.3.4. Load Case / Combo. The Load Case / Combo. that can be defined are: Load Cases and Load Combinations. Load Cases The Load Cases definition consists of the CASE, TYPE, LABEL, and whether the Load Cases is USED in the model or not. Up to 6 load cases of dead load, live load or lateral load can be defined in the Load Cases dialog box. The default five load case labels (types) are SELF (dead load), Dead (dead load), Live (live load), Wind (wind load), and EQ (seismic load). Once the maximum number is reached, the NEW button will be disabled. Note: Only one case of live load can be defined. Load case label must be unique for each of the load cases. To ignore self-weight in both strength and deflection calculations remove load case SELF from the list of load cases. MODEL DEVELOPMENT \| 232 \| MODEL DEVELOPMENT \| 233 \| Load Combinations The Load Combinations definition consists of the LOAD CASES, LOAD CASE TYPE, LOAD COMBINATION NUMBER, LOAD COMBINATION LABEL, and LOAD FACTORS. Up to fifty load combinations may be defined. All the combinations are indexed automatically from U1 to U50. MODEL DEVELOPMENT \| 234 \| 5.2.4. Creating Model Elements The Elements that can be created are: Slabs, Longitudinal Beams/Bands, Ribs, Columns, Drop Panels, Column Capitals and Transverse Beams/Bands. 5.2.4.1. Spans Elements You can create Spans Elements using the Spans command from the Ribbon. You can create Slabs, Longitudinal Beams and Ribs by using one of the following three tools in the Spans Left Panel. MODEL DEVELOPMENT \| 235 \| Slab You can enter the slab THICKNESS, WIDTH – LEFT, and WIDTH – RIGHT data in the Left Panel. Then click on the desired span (or drag over multiple spans) to assign the slab. Longitudinal Beam You can enter the beam WIDTH, DEPTH, and OFFSET data in the Left Panel. Then click on the desired span (or drag over multiple spans) to assign the beam. MODEL DEVELOPMENT \| 236 \| Rib For joist systems, you must define the rib geometry. The ribs are assumed to be the same throughout the strip. You can enter the rib WIDTH BOTTOM, DEPTH, and CLEAR SPACING BOTTOM data in the Left Panel. Then click on the desired span (or drag over multiple spans) to assign the ribs. MODEL DEVELOPMENT \| 237 \| Longitudinal Slab Bands Longitudinal slab bands are only available for two-way floor systems when LONGITUDINAL is selected for Slab Bands under Run Options in Project Left Panel (CSA A23.3-14/04 only). You can enter the slab band WIDTH, DEPTH, and OFFSET data in the Left Panel. Then click on the desired span (or drag over multiple spans) to assign the slab band. MODEL DEVELOPMENT \| 238 \| Strip Moment Distribution Factors User has the ability to manually adjust strip moment distribution factors for two-way floor systems when USER STRIP DISTRIBUTION FACTORS option is checked in the Design & Modeling Options under the Define command. However, in spSlab v10.00, strip distribution factor is a select only property. For details on how to assign or edit these factors, refer to Editing Model Elements Using the Left Panel section. In such case, the strip Moment Distribution table will become available under Spans Left Panel when the Select command button is toggled on. This table contains fields for inputting distribution factors in column and beam strips. The distribution factors for middle strip are recalculated internally. MODEL DEVELOPMENT \| 239 \| 5.2.4.2. Supports Elements You can create Supports Elements using the Supports command from the Ribbon. You can create Columns, Drop Panels, column Capitals, Transverse Beams, and Restraints by using one of the following tools in the Supports Left Panel. MODEL DEVELOPMENT \| 240 \| Column Column data is optional. If no column is specified at the supports, the support is assumed pinned. You can enter the COLUMN ABOVE, COLUMN BELOW, and COLUMN OPTIONS data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the column. More information about columns FAR END CONDITION can be found in the Restraint section. MODEL DEVELOPMENT \| 241 \| Drop Panel Drops are available for the flat slab or waffle slab systems and can be defined at all the support locations. You can enter the TYPE, THICKNESS, and CHECK PUNCHING SHEAR data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the drop panel. MODEL DEVELOPMENT \| 242 \| If spSlab is to compute the drop dimensions, the STANDARD option should be selected from the TYPE drop-down list and then only the drop depth will be available. When the STANDARD drop option is selected, spSlab will calculate drop panel dimensions in accordance with ACI 3182. Similar requirements contained in previous editions of the CSA A23.3 Standard have been removed from the 1994 edition. As a result, the ACI minimum specifications for drop panels are also used in CSA A23.3 runs when the STANDARD drop option is selected. If you would like to specify drop dimensions other than those computed by spSlab, you must select USER-DEFINED from the TYPE drop-down list. Enter the dimension in the direction of analysis from the column centerline to the edge of the drop left of the column (see Figure 5.2). If this is a STANDARD drop, this dimension will not be available and the length left is set equal to the slab span length left/6 for interior columns or the left cantilever length for the first column. Enter the width dimension in the transverse direction (see Figure 5.2). If this is a STANDARD drop, this dimension will not be available and the width is set equal to slab width/3. In order for spSlab to recognize drops, drop thickness are required for the flat slab systems even if STANDARD drop is selected. Enter the thickness of the drop from the span with the smaller slab depth (see Figure 5.2). For waffle slab systems, the thickness is automatically assumed to be equal to the rib depth below the slab. A value entered will be considered to exist below the rib depth during calculations. Select whether spSlab should compute punching shear around drop panel. 2 ACI 318-14, 8.5.2.2, 8.2.4; ACI 318-11, 13.3.7, 13.2.5; ACI 318-08, 13.3.7, 13.2.5; ACI 318-05, 13.3.7, 13.2.5 MODEL DEVELOPMENT \| 243 \| Figure 5.2 – Required Drop Panel Dimensions MODEL DEVELOPMENT \| 244 \| Column Capital You can enter the DEPTH and SIDE SLOPE data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the column capital. Enter the capital DEPTH which is the distance from the bottom of the soffit (slab, drop, or beam), to the bottom of the capital. Enter the capital SIDE SLOPE which is the rate of depth to extension of the capital and it must be greater than 1 and smaller than 50. For circular column capitals, in Design Options under Solve command, ensure the preferred option for punching shear perimeter is selected. MODEL DEVELOPMENT \| 245 \| Transverse Beams You can enter the WIDTH, DEPTH, and OFFSET data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the beam. MODEL DEVELOPMENT \| 246 \| Transverse Slab Bands Transverse slab bands are only available for two-way floor systems when TRANSVERSE is selected for Slab Bands under Run Options in Project Left Panel (CSA A23.3-14/04 only). It is not required to input bands for every support. Supports where slab bands are not defined are modeled similar to regular two-way systems. You can enter the WIDTH, DEPTH, and OFFSET data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the slab band. MODEL DEVELOPMENT \| 247 \| Restraint You can enter the SUPPORT RESTRAINTS, and SUPPORT SPRINGS data in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the restraint/spring. MODEL DEVELOPMENT \| 248 \| Moment Redistribution This command is only available for one-way/beam floor systems when MOMENT REDISTRIBUTION option is checked in the Design & Modeling Options under the DEFINITIONS dialog box. You can enter the redistribution factor you want to allow on the LEFT and RIGHT side of the support in the Left Panel. Then click on the desired support (or drag over multiple supports) to assign the redistribution limits. MODEL DEVELOPMENT \| 249 \| 5.2.4.3. Loads spSlab computes the self-weight of the floor system. Other loads applied to the structure have to be specified by the user. You can create Loads using the Loads command from the Ribbon. You can assign Area loads, Uniform and Variable Line loads, Point loads, Support loads and Displacements, and Lateral Load Effects loads by using one of the five options that are presented in the Loads Left Panel. MODEL DEVELOPMENT \| 250 \| Assigning Area Loads Area loads can only be assigned to Spans with a Slab and/or a Longitudinal Beam. To assign an Area load, make sure that the Area command in the Left Panel is selected. • Select the LOAD CASE you want the Area load to belong to. You can always define LOAD CASES in the DEFINITIONS dialog. • In the W box, type in the required load value. Note that the thumbnail in the Left Panel shows the sign convention for the loads. • From the OPTIONS select if you want to ADD TO EXISTING LOAD on the span or REPLACE EXISTING LOAD completely. • Next, click on the span you want the load to be assigned to. You can also marquee select a group of spans to assign all of them the same area load. MODEL DEVELOPMENT \| 251 \| Assigning Line Loads Line loads can only be assigned to spans with a Slab and/or a Longitudinal Beam. MODEL DEVELOPMENT \| 252 \| To assign a Line load, make sure that the Line command in the Left Panel is selected. • Select the LOAD CASE you want the Line load to belong to. You can always define LOAD CASES in the DEFINITIONS dialog. • Select the line load TYPE (UNIFORM LINE LOADS, UNIFORM LINE TORQUES, VARIABLE LINE LOADS, and VARIABLE LINE TORQUES). Note that UNIFORM LINE TORQUES and VARIABLE LINE TORQUES are only available for one-way floor systems when YES is selected for Consider Torsion under Run Options in Project Left Panel. • In the boxes, type in the required location and load values. Note that the thumbnail in the Left Panel shows the sign convention for the loads. • From the OPTIONS: • Select if you want to ADD TO EXISTING LOAD on the span or REPLACE EXISTING LOAD completely. • If you want to enter the required location as a ratio of the span length, check DISTANCE LOCATION AS RATIO OF SPAN checkbox. • Next, click on the span you want the load to be assigned to. You can also marquee select a group of spans to assign all of them the same line loads. MODEL DEVELOPMENT \| 253 \| Assigning Point Loads Point loads can only be assigned to spans with a Slab and/or a Longitudinal Beam. To assign a Point load, make sure that the Point command in the Left Panel is selected. The Left Panel should be displaying various Point loads options. • Select the LOAD CASE you want the Point load to belong to. You can always define LOAD CASES in the DEFINITIONS dialog. • In the boxes, type in the required load values and their locations. Note that the thumbnail in the Left Panel shows the sign convention for the loads, moments, and lengths. • From the OPTIONS: • Select if you want to ADD TO EXISTING LOAD on the span or REPLACE EXISTING LOAD completely. MODEL DEVELOPMENT \| 254 \| • If you want to enter the required location as a ratio of the span length, check DISTANCE LOCATION AS RATIO OF SPAN checkbox. • Next, click on the span you want the load to be assigned to. You can also marquee select a group of spans to assign all of them the same point loads. MODEL DEVELOPMENT \| 255 \| Assigning Support Loads and Displacements Support loads and displacements can only be assigned at support locations. To assign a Support load and displacement, make sure that the Support command in the Left Panel is selected. The Left Panel should be displaying various Support loads and displacements options. • Select the LOAD CASE you want the Support load and displacement to belong to. You can always define LOAD CASES in the DEFINITIONS dialog. • In the SUPPORT LOADS boxes, type in the required FORCE and/or MOMENT values. Note that the thumbnail in the Left Panel shows the sign convention for the support loads. MODEL DEVELOPMENT \| 256 \| • In the SUPPORT DISPLACEMENTS boxes, type in the required DISPLACEMENT and/or ROTATION values. Note that the thumbnail in the Left Panel shows the sign convention for the support displacements. • From the OPTIONS select if you want to ADD TO EXISTING LOAD on the node or REPLACE EXISTING LOAD completely. • Next, click on the support location you want the support load and displacement to be assigned to. You can also marquee select a group of supports to assign all of them the same support load and displacement. MODEL DEVELOPMENT \| 257 \| Assigning Lateral Load Effects Lateral Load Effects can only be assigned to Spans with a Slab and/or a Longitudinal Beam. To assign a Lateral Load Effects, make sure that the Lateral command in the Left Panel is selected. The Left Panel should be displaying various Lateral loads options. • Select the LOAD CASE you want the Lateral Load Effects to belong to. You can always define LOAD CASES in the DEFINITIONS dialog. Please note that at least one lateral load case is required to assign lateral load effects. • In the boxes, type the moment values. Note that the thumbnail in the Left Panel show the sign convention for the lateral load effects. • From the OPTIONS select if you want to ADD TO EXISTING LOAD on the span end or REPLACE EXISTING LOAD completely. • Next, click on the span you want the lateral load effect to be assigned to. You can also marquee select a group of spans to assign all of them the same lateral load effect. MODEL DEVELOPMENT \| 258 \| 5.2.4.4. Rebars Rebars command is only available if the RUN MODE of INVESTIGATION is selected in the Project Left Panel (This command is not available if RUN MODE of DESIGN is selected). Note: When switching from DESIGN mode to INVESTIGATION mode, spSlab automatically assumes the results of the DESIGN mode as an input for INVESTIGATION mode. You can create Rebars using the Rebars command from the Ribbon. Column Strip Rebars for Two-Way Slab Systems The column strip rebars can be defined by users if the RUN MODE of INVESTIGATION and TWO-WAY FLOOR SYSTEM are selected in the Project Left Panel. To define column strip rebars, make sure that the Column Strip command in the Left Panel is selected. The Left Panel should be displaying various Column Strip rebars types. Five types are available: Top Right, Top Left, Top Continuous, Bottom Continuous and Bottom MODEL DEVELOPMENT \| 259 \| Discontinuous. • Select the rebar type for which Column Strip rebars will be defined. • In the boxes, type in the NO. OF BARS, BAR SIZE, and CLEAR COVER (also bar LENGTH and START location, if discontinuous rebars types are selected). • Next, click on the span (or spans) you want the Column Strip rebars to be assigned to. You can also marquee select a group of spans to assign all of them the same Column Strip rebars. MODEL DEVELOPMENT \| 260 \| Middle Strip Rebars for Two-Way Slab Systems The middle strip rebars can be defined by users if the RUN MODE of INVESTIGATION and TWO-WAY FLOOR SYSTEM are selected in the Project Left Panel. To define middle strip rebars, make sure that the Middle Strip command in the Left Panel is selected. The Left Panel should be displaying various Middle Strip rebars types. Five types are available: Top Right, Top Left, Top Continuous, Bottom Continuous and Bottom Discontinuous. • Select the rebar type for which Middle Strip rebars will be defined. • In the boxes, type in the NO. OF BARS, BAR SIZE, and CLEAR COVER (also bar LENGTH and START location, if discontinuous rebars types are selected). • Next, click on the span (or spans) you want the Middle Strip rebars to be assigned to. You can also marquee select a group of spans to assign all of them the same Middle Strip rebars. MODEL DEVELOPMENT \| 261 \| Beam Rebars for Two-Way Slab Systems The beam rebars can be defined by users if the RUN MODE of INVESTIGATION and TWO-WAY FLOOR SYSTEM are selected in the Project Left Panel. To define beam rebars, make sure that the Beam command in the Left Panel is selected. The Left Panel should be displaying various Beam rebars types. Five types are available: Top Right, Top Left, Top Continuous, Bottom Continuous and Bottom Discontinuous. • Select the rebar type for which Beam rebars will be defined. • In the boxes, type in the NO. OF BARS, BAR SIZE, and CLEAR COVER (also bar LENGTH and START location, if discontinuous rebars types are selected). • Next, click on the span (or spans) you want the Beam rebars to be assigned to. You can also marquee select a group of spans to assign all of them the same Beam rebars. MODEL DEVELOPMENT \| 262 \| Beam Stirrups for Two-Way Slab Systems The beam stirrups can be defined by users if the RUN MODE of INVESTIGATION and TWO-WAY FLOOR SYSTEM are selected in the Project Left Panel. To define beam stirrups, make sure that the Stirrups command in the Left Panel is selected. The Left Panel should be displaying two Stirrups configurations: Stirrups Entire Span and Stirrups General. • Select the stirrups configuration for which Stirrups will be defined. • In the boxes, type in the NO. OF STIRRUPS, BAR SIZE, SPACING and NO. OF LEGS (also FIRST STIRRUP LOCATION when applicable). • Next, click on the span (or spans) you want the Stirrups to be assigned to. You can also marquee select a group of spans to assign all of them the same Stirrups. • In order to mirror stirrup set or sets at the end of the span, use APPEND button. MODEL DEVELOPMENT \| 263 \| MODEL DEVELOPMENT \| 264 \| Flexure Rebars for Beams and One-Way Slab Systems The flexure rebars can be defined by users if the RUN MODE of INVESTIGATION and ONE-WAY/BEAM FLOOR SYSTEM are selected in the Project Left Panel. To define flexure rebars, make sure that the Flexure command in the Left Panel is selected. The Left Panel should be displaying various Flexure rebars types. Five types are available: Top Right, Top Left, Top Continuous, Bottom Continuous and Bottom Discontinuous. • Select the rebar type for which Flexure rebars will be defined. • In the boxes, type in the NO. OF BARS, BAR SIZE, and CLEAR COVER (also bar LENGTH and START location, if discontinuous rebars types are selected). • Next, click on the span (or spans) you want the Flexure rebars to be assigned to. You can also marquee select a group of spans to assign all of them the same Flexure rebars. MODEL DEVELOPMENT \| 265 \| Stirrups for Beams and One-Way Slab Systems The stirrups can be defined by users if the RUN MODE of INVESTIGATION and ONE-WAY/BEAM FLOOR SYSTEM are selected in the Project Left Panel. To define stirrups, make sure that the Stirrups command in the Left Panel is selected. The Left Panel should be displaying two Stirrups configurations: Stirrups Entire Span and Stirrups General. • Select the stirrups configuration for which Stirrups will be defined. • In the boxes, type in the NO. OF STIRRUPS, BAR SIZE, SPACING and NO. OF LEGS (also FIRST STIRRUP LOCATION when applicable). • Next, click on the span (or spans) you want the Stirrups to be assigned to. You can also marquee select a group of spans to assign all of them the same Stirrups. • In order to mirror stirrup set or sets at the end of the span, use APPEND button. MODEL DEVELOPMENT \| 266 \| MODEL DEVELOPMENT \| 267 \| Torsional Longitudinal Rebars for Beams The torsional longitudinal rebars can be defined by users if the RUN MODE of INVESTIGATION and ONE-WAY/BEAM FLOOR SYSTEM are selected in the Project Left Panel. To define torsional longitudinal rebars, make sure that the Torsion command in the Left Panel is selected. The Left Panel should be displaying various Torsion rebars types. Two types are available: Torsion Continuous and Torsion Discontinuous. • Select the rebar type for which Torsion rebars will be defined. • In the boxes, type in the NO. OF BARS, BAR SIZE, and CLEAR COVER (also bar LENGTH and START location, if discontinuous rebars type is selected). • Next, click on the span (or spans) you want the Torsion rebars to be assigned to. You can also marquee select a group of spans to assign all of them the same Torsion rebars. MODEL DEVELOPMENT \| 268 \| 5.2.5. Editing Model Elements The model can be edited by using the Left Panel, Left Panel Toolbar or by using right-click at Viewport. To edit a model element, its corresponding placeholder (Span or Support) must be selected. 5.2.5.1. Using the Left Panel The Elements that can be edited are: Slabs, Longitudinal Beams/Bands, Ribs, Columns, Drop Panels, Column Capitals and Transverse Beams/Bands. Spans The corresponding Spans Left Panel provides various tools and options for effectively working with Span Elements. You must have the Select command button toggled on to edit the Span Elements. Editing Span Elements To edit span elements: • Click on the span (or spans) you want to edit to display its properties in the Left Panel. • Then in the Elements tab in the Span Left Panel, simply change the desired parameter. • Additionally, you can add new elements or remove existing elements to the selected span (or spans) by clicking on or next to the desired element, respectively. Editing Span Loads To edit span loads: • Click on the span (or spans) you want to edit to display its properties in the Left Panel. • Click on the Loads tab in the Span Left Panel, then simply change the load value as desired. MODEL DEVELOPMENT \| 269 \| • Additionally, you can add new loads or remove existing loads to the selected span by clicking on or next to the desired load, respectively. Editing Strip Moment Distribution Factors To edit strip Moment Distribution Factors: • Enable check box USER STRIP DISTRIBUTION FACTORS under Design & Modeling Options under the Define command. • You must have the Select command button toggled on to edit the Spans. • Click on the span (or spans) you want to edit to display its properties in the Left Panel. • Click on next to the strips Moment Distribution table. • Then in Moment Distribution table in the Span Left Panel, simply change the desired factors. MODEL DEVELOPMENT \| 270 \| Supports The corresponding Support Left Panel provides various options for effectively working with Support. You must have the Select command button toggled on to edit the Support. Editing Support Elements To edit support elements: • Click on the support (or supports) you want to edit to display its properties in the Left Panel. • Then in the Elements tab in the Span Left Panel, simply change the desired parameter. • Additionally, you can add new elements or remove existing elements to the selected support (or supports) by clicking on or next to the desired element, respectively. Editing Support Loads & Restraints To edit support loads & restraints: • Click on the support (or supports) you want to edit to display its properties in the Left Panel. • Click on the Loads & Restraints tab in the Span Left Panel, then simply change the load or restraint value as desired. • Additionally, you can add new loads (or restraints) or remove existing loads (or restraints) to the selected support by clicking on or next to the desired load (or restraint), respectively. MODEL DEVELOPMENT \| 271 \| 5.2.5.2. Using the Left Panel Toolbar You must have the Select command button toggled on in order to use the tools available in the Left Panel Toolbar. You can use the tools in the Left Panel Toolbar to edit various model items. Delete The Delete command is active only when one or more supports are selected. • Select the support or supports you want to remove from the model and click Delete to remove. Move The Move command is active only when one support is selected. • Select the support you want to move and click the Move command. • Specify the destination support to complete moving. Duplicate The Duplicate command is active only when one span or support is selected. • Select the span or support you want to duplicate of and click the Duplicate command. • Specify the destination span or support to complete duplication. Advance Copy The Advanced Copy command is active only when one span or support is selected. MODEL DEVELOPMENT \| 272 \| • Select the span or support you want to copy of and click the Advanced Copy command. • Select the properties, load types and load cases you want to copy and click the Ok button. • Specify the destination span or support to complete advanced copy. MODEL DEVELOPMENT \| 273 \| 5.2.5.3. Using the Right Click Menu at Viewport All the tools in the Left Panel Toolbar are also available in the Right Click Menu at Viewport when the Select command button is toggled on. MODEL DEVELOPMENT \| 274 \| 5.3. Modeling with Templates 5.3.1. Utilizing Templates Templates are an option for creating new models in the spSlab program. It enables the user to select from a set of pre-defined templates and edit them for quick model generation. To begin, go to the Start Screen under Projects and select the Templates option. This will take you to the TEMPLATE selection Dialog Box. Here you can select the desired template along with the DESIGN CODE and UNIT SYSTEM. Clicking on a template image will open the Template Module and load the selected Template. MODEL DEVELOPMENT \| 275 \| Once you are done editing the template to reflect your project criteria, you can click the Save & Exit button to take it to spSlab for further modification or execution. MODEL DEVELOPMENT \| 276 \| 5.3.2. Template Ribbon The Template Ribbon provides the following options: 5.3.2.1. New Pattern Opens the TEMPLATE Selection Dialog Box. Selecting a New template will discard the old template and load the new one. 5.3.2.2. Discard & Exit Discards the current template and exits to spSlab home screen. 5.3.2.3. Save & Exit Imports the current template to spSlab. MODEL DEVELOPMENT \| 277 \| 5.3.3. Template Left Panel The Left Panel lists various template properties that can be modified by the user. The bottom part of the Left Panel consists of Display Options. You can use these to toggle on/off several Viewport items and also switch between displaying Load Cases. MODEL DEVELOPMENT \| 278 \| MODEL DEVELOPMENT \| 279 \| 5.3.4. Types of Templates 5.3.4.1. Two-Way Systems Flat Plate Systems MODEL DEVELOPMENT \| 280 \| Flat Slab Systems MODEL DEVELOPMENT \| 281 \| Slab on Beams System MODEL DEVELOPMENT \| 282 \| Slab Bands Systems These systems are only available when CSA A23.3-14 or CSA A23.3-04 are selected under Design Code. MODEL DEVELOPMENT \| 283 \| Two-Way Joist (Waffle) Systems MODEL DEVELOPMENT \| 284 \| 5.3.4.2. One-Way Systems One-Way Slab (Solid) Systems MODEL DEVELOPMENT \| 285 \| One-Way Joist (Ribbed) Systems MODEL DEVELOPMENT \| 286 \| Rectangular Beams MODEL DEVELOPMENT \| 287 \| Flanged Beams MODEL DEVELOPMENT \| 288 \| 5.4. Utilizing Predefined Examples In the Start Screen under Projects select the Examples option. This will take you to the Examples folder under the spSlab installation folder. The Examples folder contains predefined slab models that can be further utilized by the user. \| 289 \| 6. MODEL SOLUTION Once the model creation and development are completed, the analysis can begin using the spSlab Equivalent Frame Method (EFM) Solver by clicking on the Solve command. Solve Menu containing Design Options and Deflection Options will appear on the Left Panel. Any solver warnings or errors encountered during the solution are issued in the Solver Dialog and recorded as solver messages in the results table. CHAPTER 6 MODEL SOLUTION \| 290 \| 6.1. Design Options The Design Options tab allows the user to select options and specify parameters that affect the analysis and design results. Changing these settings involves engineering judgment and so has to be done cautiously. The set of parameters is different for two-way and one-way systems. 6.1.1. Reinforcement • Check COMPRESSION REINFORCEMENT checkbox if it is to be considered when needed. • Check DECREMENTAL REINFORCEMENT DESIGN to use alternative reinforcement design algorithm. • Check COMBINED M-V-T REINF. DESIGN to proportion longitudinal reinforcement for combined action of flexure, shear, and torsion. This option is available only when CSA A23.3-14 or CSA A23.3-04 are selected. MODEL SOLUTION \| 291 \| 6.1.2. Shear Design These options are available only when TWO-WAY FLOOR SYSTEM Run Option is selected. • Check ONE-WAY SHEAR IN DROP PANELS to include drop panel cross-section in slab one-way shear capacity calculations in support locations. • Check DISTRIBUTE SHEAR TO SLAB STRIPS to distribute slab one-way shear between column and middle strips in proportion to moment distribution factors. • Select whether circular critical section around circular supports is to be used or traditional equivalent rectangular critical section. • Enter the multiplier that defines the distance between a column face and a free edge of a slab, within which a segment of punching shear critical section is to be ignored. 6.1.3. Torsion Analysis and Design These options are available only when: ONE-WAY/BEAM FLOOR SYSTEM Run Option is selected, and YES is selected for CONSIDER TORSION Run Option. • Select whether the TORSION TYPE is EQUILIBRIUM or COMPATIBILITY. • Select YES or NO for STIRRUPS IN FLANGES option. 6.1.4. Beam Design The following Beam Design options are available only when TWO-WAY FLOOR SYSTEM Run Option is selected. • Check BEAM T-SECTION DESIGN to include portions of slab as flanges in beam cross-section for reinforcement design. • Check LONG. BEAM SUPPORT DESIGN to include cross-section of longitudinal beam in reinforcement design for unbalanced moments over supports. This feature can be useful for slabs having wide longitudinal beams. When used together with USER STRIP DISTRIBUTION FACTORS, it can produce solutions consistent with the solutions for models MODEL SOLUTION \| 292 \| with longitudinal slab bands for CSA A23.3-14/04 code. • Check TRANSVERSE BEAM SUPPORT DESIGN to include cross-section of transverse beam in reinforcement design for negative moments and unbalanced moments over supports. This feature is useful for slabs having wide transverse beams. When used together with USER STRIP DISTRIBUTION FACTORS, it can produce solutions consistent with the solutions for systems with transverse slab bands for CSA A23.3-14/04 code. The following Beam Design options are available only when ONE-WAY/BEAM FLOOR SYSTEM Run Option is selected. • Check EFFECTIVE FLANGE WIDTH if instead of the full flange width only the effective flange width is to be considered in the flexural design. • Check RIGID BEAM-COLUMN JOINT to consider beam-column joint as rigid. 6.1.5. Live Loads • Enter the live load pattern ratio. The default value is 75% for TWO-WAY FLOOR SYSTEM and 100% for ONE-WAY/BEAM FLOOR SYSTEM. MODEL SOLUTION \| 293 \| 6.2. Deflection Options The Deflection Options tab allows the user to select options and specify parameters that affect the analysis and design results. Changing these settings involves engineering judgment and so has to be done cautiously. The set of parameters is the same for two-way and one-way systems. 6.2.1. Section Used • Choose if GROSS (UNCRACKED) or EFFECTIVE (CRACKED) sections are to be considered in the deflection calculations. 6.2.2. Ig & Mcr Calculation • Choose if in the case of a section with flanges in the negative moment region, only the web (RECTANGULAR SECTION) or the whole section (T-SECTION) is to be used to calculate the gross moment of inertia (Ig) and the cracking moment (Mcr). MODEL SOLUTION \| 294 \| 6.2.3. Long-Term Deflection • Select YES for CALCULATE LONG-TERM DEFLECTIONS option if you want the program to calculate long-term deflections. • Enter the LOAD DURATION in months. The default value is 60 months. • Enter the percentage of the live load which is considered as sustained load (SUSTAINED PART OF LIVE-LOAD). The default percentage is 0%. MODEL SOLUTION \| 295 \| 6.3. Running the Model After inputting the model, the solver portion of the program can be invoked using the Run button in the Solve panel. After you click the Run button, the program then switches control to the Solver Module. A dialog box reporting the progress and status of the solution is displayed under the Messages tab. When the solution is successfully completed, the program automatically switches to the Results scope. Warnings and/or errors are displayed under the Warnings/Errors tab. Detailed information on the solution can be found in the Solver Messages Table in the Tables Window. MODEL SOLUTION \| 296 \| \| 297 \| 7. MODEL OUTPUT The results of the analysis and following design and code calculations are presented by spSlab model output in two key categories. 1. Tabular Output: Text results organized in tables including all relevant exact numerical results formatted in columnar tables. 2. Graphical Output: Visual representations including diagrams of internal forces, moment capacity, shear capacity, deflections, and reinforcement details. A detailed description of all the features of both output types is given below. CHAPTER 7 MODEL OUTPUT \| 298 \| 7.1. Tabular Output The Tabular Output can be accessed through both the Tables Module and the Reporter Module. The Tables Module allows users to view and export output data at any stage of model development, while the Reporter Module is used for creating, exporting, and printing customized reports once the design is complete. Both modules share the same output sections, with a few distinctions: the Reporter Module includes additional sections for cover, contents, and screenshots, while the Tables Module includes a Solver Messages section to display warnings or error messages generated during model execution. Diagram views are also available for selected output results to facilitate graphical analysis; however, final modeling decisions should be based on the comprehensive information provided in the tabular reports. The Tabular output contains the following common input and results sections: MODEL OUTPUT \| 299 \| 7.1.1. Input Echo Input Echo section reports the data used in the analysis. Carefully check the contents of the section and compare it with the intended design model. This section contains the following input data blocks: 7.1.1.1. General Information This block contains the information regarding the Project such as FILE NAME, PROJECT NAME, FRAME, ENGINEER, DESIGN CODE, UNITS, REINFORCEMENT DATABASE, MODE, NUMBER OF SUPPORTS, and FLOOR SYSTEM. 7.1.1.2. Solve Options This block contains the information regarding to the Solve Options (Design and Deflection Options) input entered by the user. 7.1.1.3. Material Properties This subsection has data blocks for CONCRETE: SLABS / BEAMS, CONCRETE: COLUMNS, and REINFORCING STEEL input data. 7.1.1.4. Reinforcement Database This block contains the Bar Set input data such as SIZE, BAR DIAMETER, CROSS-SECTION AREA and UNIT WEIGHT for each bar. 7.1.1.5. Span Data This subsection contains the information regarding to the Spans input data for spans utilized in the model. This subsection has data blocks for SLABS, LONGITUDINAL SLAB BANDS, and RIBS AND LONGITUDINAL BEAMS input data. This subsection provides information about the span input data used in the model. MODEL OUTPUT \| 300 \| 7.1.1.6. Support Data This subsection contains the information regarding to the Supports input data for supports utilized in the model. This subsection has data blocks for COLUMNS, DROP PANELS, COLUMN CAPITALS, TRANSVERSE BEAMS, MOMENT REDISTRIBUTION LIMITS, and BOUNDARY CONDITIONS input data. Notes: For circular column the transverse dimension c2 is reported as zero. If dimensions of a drop panel are invalid it will be marked. Invalid or excessive drop panel geometry is not used in the analysis. 7.1.1.7. Load Data This subsection contains the information regarding to the Loads input data for loads utilized in the model. This subsection has data blocks for LOAD CASES AND COMBINATIONS, AREA LOADS, LINE LOADS, POINT FORCES, POINT MOMENTS, LINE TORQUE, POINT TORQUE, SUPPORT LOADS, SUPPORT DISPLACEMENTS, and LATERAL LOAD EFFECTS input data. 7.1.1.8. Reinforcement Criteria This subsection contains the information regarding to the Reinforcement Criteria input data for slabs, ribs, and beams reinforcement utilized in the model. This subsection has data blocks for SLABS & RIBS, LONGITUDINAL SLAB BANDS, and BEAMS input data. 7.1.1.9. Reinforcing Bars This block is available only when Investigation Mode is selected. This subsection contains the information regarding to the Rebars input data for reinforcement utilized in the model. This subsection has data blocks for TOP BARS, BOTTOM BARS, TORSION BARS, and TRANSVERSE REINFORCEMENT input data. Note: When switching from Design Mode to Investigation Mode, spSlab automatically assumes the results of the Design Mode as an input for Investigation Mode. MODEL OUTPUT \| 301 \| 7.1.2. Design Results This section contains the summary of the design results of the slab/beam system. The following paragraphs describe the blocks included in the section. 7.1.2.1. Solver Messages This block displays the progress and status of the solution. It also displays warnings or error messages generated during model execution. 7.1.2.2. Moment Redistribution Factors This block is only available for ONE-WAY/BEAM floor systems when MOMENT REDISTRIBUTION checkbox is selected. The calculated moment redistribution factor (including undistributed ultimate moment, number of iterations, and net tensile strain in extreme tension steel), user limit, and applied moment redistribution factor for both sides of each supports are displayed. 7.1.2.3. Strip Widths and Distribution Factors This block is available only for TWO-WAY floor systems. It contains the information on design strip widths, moment distribution factors, and shear distribution factors (for CSA A23.3-14/04 standard and optionally for other standards if DISTRIBUTE SHEAR TO SLAB STRIPS is selected). 7.1.2.4. Top Reinforcement This block is available only when DESIGN run mode is selected. It reports the negative reinforcement requirements. The block contains the values of corresponding design strip widths (column, middle, and beam), maximum factored design moments per strip and critical location, minimum and maximum steel areas, spacing for bars selected based on required reinforcement area, steel areas required by ultimate condition, selected bar sizes and numbers. The quantities are given for left, center and right location of each span. For a detailed discussion, see Chapter 2, Section 2.4.1. MODEL OUTPUT \| 302 \| Note: This block does not include reinforcement quantities necessary to transfer unbalanced moments at supports. In case of CSA standard, the reported spacing is averaged between reinforcement placed in the bb band and in the remaining portion of the column strip outside of the bb band. 7.1.2.5. Top Bar Details and Development Lengths The block contains a span-by-span listing of the longitudinal bars selected in column, middle and beam strips. This reinforcement schedule is intended as a guide for bar placement. In more complex cases the bar schedule selected by the program may have to be adjusted by the user for constructability reasons. The selected bar sizes are limited by user specified minimum and maximum sizes. Bar sizes and numbers are selected to satisfy the minimum and required steel areas in conjunction with the bar spacing requirements of the code. The program calculates the bar lengths based on the computed inflection points and the recommended minima of the code. The bar lengths are adjusted by appropriate development lengths. Hooks and bends are not included in bar length tables and figures. For beams, bars are placed in a single layer (see Figure 2.29), provided there is sufficient beam width. For a detailed discussion, see Chapter 2, Section 2.5.1. Note: This block does not include additional reinforcement bars necessary to transfer negative unbalanced moment at supports. 7.1.2.6. Band Reinforcement at Supports This block is available only when the CSA code is selected. This section describes how the negative reinforcement in column strips should be concentrated over supports. It reports the total width of the strip from which reinforcement is concentrated, the width of the bb band, and the remaining width. The total width of the subdivided strip will be equal to the column strip width when no wide beams (width greater than bb) and no slab bands are present. If either one is present, then its width will be used. If a beam narrower than bb frames into an exterior support, then both the total strip width and the bb width will be reduced to the beam width. The section also gives the area of reinforcement and the number of bars required in each strip. The sum of number of bars in the band strip and in the remaining strip should be equal to the total number of bars over each support in the strip from which the bars were concentrated. The total MODEL OUTPUT \| 303 \| number of bars in this strip should also be consistent with the number of bars listed in the Top Bar Details dialog box. Note: This output block does not include additional reinforcement bars necessary to transfer negative unbalanced moment at supports. 7.1.2.7. Bottom Reinforcement This block is available only when DESIGN run mode is selected. It reports the positive reinforcement requirements. The block contains the values of corresponding design strip widths (column, middle, and beam), maximum factored design moments per strip and critical location, minimum and maximum steel areas, spacing for bars selected based on required reinforcement area, steel areas required by ultimate condition, selected bar sizes and numbers. The quantities are given for mid-span regions of each span. For a detailed discussion, see Chapter 2, Section 2.4.1. 7.1.2.8. Bottom Bar Details and Development Lengths This block contains a span-by-span listing of the longitudinal bars selected in column, middle and beam strips. The reinforcement schedule is intended as a guide for bar placement. In more complex cases the bar schedule selected by the program may have to be adjusted by the user for constructability reasons. The selected bar sizes are limited by user specified minimum and maximum sizes. Bar sizes and numbers are selected to satisfy the minimum and required steel areas in conjunction with the bar spacing requirements of the code. The program calculates the bar lengths based on the computed inflection points and the recommended minima of the code. The bar lengths are adjusted by appropriate development lengths. Bottom bar development lengths are tabulated directly below the bottom bar details block. Hooks and bends are not included in bar length tables and figures. For beams, bars are placed in a single layer (see Figure 2.29), provided there is sufficient beam width. For a detailed discussion, see Chapter 2, Section 2.5.1. MODEL OUTPUT \| 304 \| 7.1.2.9. Flexural Capacity This block lists the selected top and bottom steel areas and corresponding negative and positive moment capacity values in each span. The data is subdivided between column, middle and beam strips. Each span is subdivided into segments reflecting the changes in geometry and bar placement. 7.1.2.10. Longitudinal Beam Combined M-V-T Capacity This block is available if Combined M-V-T reinforcement design option is checked under CSA A23.3-04/14. It reports the requirement of top and bottom reinforcement for a longitudinal beam based on the required tension force due to the combined effects of moment, shear and torsion forces at top and bottom respectively. 7.1.2.11. Longitudinal Beam Transverse Reinforcement Capacity This block is available only when Investigation Mode is selected and model contains longitudinal beam. It is applicable if Design Code is ACI 318, CSA A23.3-94, and CSA A23.3-04/14 given that Combined M-V-T reinforcement design option is unchecked. This block contains two subblocks namely Section Properties, and Beam Transverse Reinforcement Capacity. 7.1.2.12. Longitudinal Beam Transverse Reinforcement Demand and Capacity This block is available only when Design Mode is selected and model contains longitudinal beam. It is applicable if Design Code is ACI 318, CSA A23.3-94, and CSA A23.3-04/14 given that Combined M-V-T reinforcement design option is unchecked. This block contains four subblocks namely Section Properties, Beam Transverse Reinforcement Demand, Beam Transverse Reinforcement Details, and Beam Transverse Reinforcement Capacity. 7.1.2.13. Beam Shear (and Torsion) Capacity If torsion is not considered then this block lists the concrete section shear capacity ϕVc, selected stirrup intensities and spacing, and corresponding beam shear capacity ϕVn values in each span. For CSA A23.3-14/04 code, the program additionally reports value of factor β. The maximum factored shear forces Vu in beam strip along the span is also reported. MODEL OUTPUT \| 305 \| In the case of combined shear and torsion analysis (beams/one-way slab systems only), this block lists section properties, shear and torsion transverse reinforcement capacity, and longitudinal torsional reinforcement capacity. The provided and required capacities are expressed in terms of the provided and required areas of reinforcement. 7.1.2.14. Longitudinal Beam Shear and Torsion Reinforcement Required This block is available only when Design Mode is selected, torsion is considered and model contains longitudinal beam. This block contains nine subblocks namely Section Geometrical Properties, Section Strength Properties, Transverse Reinforcement Demand, Required Longitudinal Reinforcement, Beam Transverse Reinforcement Details, Longitudinal Torsional Reinforcement Details, Beam Transverse Reinforcement Capacity (Required Area), Beam Transverse Reinforcement Capacity (Provided Area), and Beam Torsion Reinforcement Capacity. 7.1.2.15. Longitudinal Slab Band Shear Capacity This block is available only for two-way slab systems containing longitudinal slab bands in CSA A23.3-04/14. This block lists longitudinal slab band shear capacity ϕVc in each span. 7.1.2.16. Slab Shear Capacity This block lists the values of one-way slab shear capacity ϕVc in each span. For CSA A23.3-04 code, the program additionally reports value of factor β. The maximum factored shear force Vu and the location of the critical section Xu are also reported. 7.1.2.17. Flexural Transfer of Negative Unbalanced Moments This block contains the results for critical (effective) section width as per the code, width of the effective section on the tension side for negative unbalanced moments, distance to the centroid of reinforcement, the maximum negative unbalanced moment, the corresponding load combination and governing load pattern, the reinforcing steel area provided and additional steel required. The provided reinforcement area (main longitudinal bars) is reduced by the ratio of critical (effective) strip width to total strip width and does not include the required area due to unbalanced moments. The additional reinforcement is the difference between that required by unbalanced moment MODEL OUTPUT \| 306 \| transfer by flexure and that provided for design bending moment. When additional reinforcement is required, it is selected based on the bar sizes already provided at the support. For a detailed discussion, see Chapter 2, Section 2.4.1, and Section 2.4.2. 7.1.2.18. Flexural Transfer of Positive Moments This block contains the results for critical (effective) section width as per the code, width of the effective section on the compression side for positive unbalanced moments, distance to the centroid of reinforcement, the maximum positive unbalanced moment, the corresponding load combination and governing load pattern, the reinforcing steel area provided and additional steel required. The provided reinforcement area (main longitudinal bars) is reduced by the ratio of critical (effective) strip width to total strip width and does not include the required area due to unbalanced moments. The additional reinforcement is the difference between that required by unbalanced moment transfer by flexure and that provided for design bending moment. When additional reinforcement is required, it is selected based on the bar sizes already provided at the support. For a detailed discussion, see Chapter 2, Section 2.4.1, and Section 2.4.2. 7.1.2.19. Punching Shear Around Columns The block contains two tables with values pertaining to punching shear check in critical sections around the columns. The first table lists geometrical properties of punching shear critical section. The reported properties of the critical section are overall dimensions in the direction of analysis, b1, and in the perpendicular direction, b2, perimeter, b0, location of centroid with respect to column center line, CG, average distance from the slab bottom to centroid of the slab tension reinforcement, davg, distance from centroid to the left, cleft, and right, cright, edge of critical section, area of concrete resisting shear transfer, Ac, and moment of inertia of critical section, Jc. The second table lists two sets of punching shear calculations – direct shear alone and direct shear with moment transfer. The output contains the values of the allowable shear stress ϕvc, reactions Vu, unbalanced moments Munb, governing load pattern, fraction of unbalanced moment ϕV, punching shear stress vu. The calculation for moment transfer adjusts the unbalanced moment to the centroid of the critical section. The "shear transfer" is the unbalanced moment multiplied by γv. When calculated shear stress vu exceeds the allowable value ϕvc, the program prints a warning flags for this support. For a detailed discussion, see Chapter 2, Section 2.3.7. MODEL OUTPUT \| 307 \| 7.1.2.20. Punching Shear Around Drops The block contains two tables with values pertaining to punching shear check in a critical section around the drop panels. The first table lists geometrical properties of punching shear critical section (see description in the Punching shear around Columns block). The second table displays the reactions Vu, governing load pattern, the punching shear stress around the drop vu, and the allowable shear stress ϕvc. When calculated shear stress vu exceeds the allowable value ϕvc, the program prints a warning flags for this drop panel. For a detailed discussion, see Chapter 2, Section 2.3.7. 7.1.2.21. Integrity Reinforcement at Supports This section is available only when the CSA code is selected. It lists the shear transferred to the column and the minimum area of bottom reinforcement crossing one face of the periphery of a column and connecting slab to the column to provide structural integrity. For a detailed discussion, see Chapter 2, Section 2.5.2. 7.1.2.22. Corner Reinforcement This block refers to the reinforcement required in the exterior corners of a slab with beams between columns. The ratio of flexural stiffness of beam section to flexural stiffness of slab is listed as well as the area of reinforcement and the distance over which the reinforcement is required. The area applies to each layer of reinforcement in each direction. For a detailed discussion, see Chapter 2, Section 2.5.3. 7.1.2.23. Shear Resistance at Corner Columns This section is available only when the CSA code is selected in design mode. It reports results of one-way shear check at corner columns. The results include the factored shear resistance and the factored shear force at the column. Also, the minimum length of the critical shear section and, for the 1994 edition, the angle at which the minimum length is obtained are listed. For a detailed discussion, see Chapter 2, Section 2.3.7.6 in Section 2.3.7. MODEL OUTPUT \| 308 \| 7.1.2.24. Material Takeoff This block lists the approximate total and unit quantities of concrete, and reinforcement. Note that the reinforcement estimate is for one direction only and ignores items such as hooks, bends, and waste. For a detailed discussion, see Chapter 2, Section 2.6.11. MODEL OUTPUT \| 309 \| 7.1.3. Deflection Results: Summary Deflection Results: Summary section reports the summary of the deflection results of the slab system. This section contains the following output blocks: 7.1.3.1. Section Properties This subsection has output summary blocks of FRAME SECTION PROPERTIES, for positive and negative moments, FRAME EFFECTIVE SECTION PROPERTIES for load levels, and STRIP SECTION PROPERTIES AT MIDSPAN (two-way systems only). 7.1.3.2. Instantaneous Deflections This subsection has output summary blocks of EXTREME INSTANTANEOUS FRAME DEFLECTIONS (COLUMN STRIP DEFLECTIONS and MIDDLE STRIP DEFLECTIONS only for two-way systems). 7.1.3.3. Long-Term Deflections This subsection, if selected in Deflection Options, has output summary blocks of LONG-TERM DEFLECTION FACTORS (COLUMN STRIP DEFLECTION FACTORS and MIDDLE STRIP DEFLECTION FACTORS only for two-way systems) and EXTREME LONG-TERM FRAME DEFLECTIONS (COLUMN STRIP DEFLECTIONS and MIDDLE STRIP DEFLECTIONS only for two-way systems). If the deflection option GROSS (UNCRACKED) section is selected, only gross moment of inertia, Ig, is reported in the Section Properties table and the values of all deflections reported are based on gross section properties. If deflection option EFFECTIVE (CRACKED) section is used, the values of deflections reported are based on averaged effective moments of inertia, Ie,avg, which are then reported in the Section Properties table together with other properties of cracked sections. For a detailed discussion, see Chapter 2, Section 2.3.10.3. MODEL OUTPUT \| 310 \| 7.1.4. Detailed Results 7.1.4.1. Column Forces and Redistributed Column Forces Sections Column Forces and Redistributed Column Forces present the summary of axial forces (reactions) and bending moments in bottom and top columns and in springs attached to the column-slab joints. Also, moments at far ends of columns are reported. All reported values represent forces and moments at column ends, i.e. at joint level (not at slab or drop panel surface level). If moment redistribution is selected (beams/one-way slab systems only) both redistributed and un-redistributed values can be included. The values reported represent the loading of a single floor only. Any actions on the columns from the floors above must be added to this story’s actions to properly analyze/design the columns. The output contains column axial forces and moments due to all load cases, including all live load patterns, and all load combinations. Positive axial forces mean compression and positive moments mean that fibers on the left hand side are in tension for top columns and for bottom column fibers on the right hand side are in tension. Also, reactions of additional translational and rotational springs applied at joints are reported in this block. Positive values mean upward translational spring reaction and clockwise rotational spring reaction. 7.1.4.2. Non – Redistributed and Redistributed Internal Forces: M – V Load Cases This section presents the summary of unfactored bending moments and shear forces for individual load cases including selfweight, dead load, live load and lateral cases. The reported values are presented using span-by-span segmental approach. If moment redistribution is selected (beams/one-way slab systems only) both redistributed and un-redistributed values can be included. Load Combinations This section presents the summary of bending moments and shear forces for each load combination. The reported values for each load combination are presented using span-by-span segmental approach. The negative and positive values of bending moments and shear forces are presented in separate columns in order to provide consistent format with enveloped output. If moment redistribution is selected (beams/one-way slab systems only) both redistributed and un- MODEL OUTPUT \| 311 \| redistributed values can be included. Envelopes This section presents the summary of bending moments and shear forces for envelope of all load combinations. The reported values are presented using span-by-span segmental approach. The negative and positive values of bending moments and shear forces are presented in separate columns for user convenience. The factored values presented in this section are used for design purposes (longitudinal and transverse reinforcement). If moment redistribution is selected (beams/one-way slab systems only) both redistributed and un-redistributed values can be included. 7.1.4.3. Internal Forces: T Load cases This section presents the summary of unfactored beam torsion forces (beams/one-way slab systems with torsion analysis and design only) for individual load cases including selfweight, dead load, live load and lateral cases. The reported values are presented using span-by-span segmental approach. Load Combinations This section presents the summary of beam torsion forces (beams/one-way slab systems with torsion analysis and design only) for each load combination. The reported values for each load combination are presented using span-by-span segmental approach. The negative and positive values of torsion forces are presented in separate columns in order to provide consistent format with enveloped output. Envelopes This section presents the summary of beam torsion forces (beams/one-way slab systems with torsion analysis and design only) for envelope of all load combinations. The reported values are presented using span-by-span segmental approach. The negative and positive values of torsion forces are presented in separate columns for user convenience. The factored values presented in this section are used for design purposes (longitudinal and transverse reinforcement). MODEL OUTPUT \| 312 \| 7.1.4.4. Deflections – Load Cases This section presents the summary of instantaneous deflections for unfactored (service) load cases including selfweight and dead load (DL), live load (LL), sustained load (DL + LLsustained), and total load (DL + LL) cases, and summary of long-term deflections for unfactored incremental and total deflections for one-way systems. For two-way systems, instantaneous frame deflections for fixed-end, end-rotation, and total (fixed-end and end rotation combined), instantaneous strip deflections, and long-term strip deflections. The reported values are presented using span-by-span segmental approach. For a detailed discussion, see Chapter 2, Section 2.3.10. 7.1.4.5. Required Reinforcement This section presents the summary of enveloped design moments and the required areas of longitudinal reinforcement required for flexure. If combined M-V-T option (available only for beam design/investigation per CSA A23.3-04) is selected in the Solve Option window then longitudinal reinforcement required for combined flexure, shear, and torsion (M-V-T) is also reported with the corresponding values of bending moment, shear force, and torsional moment. The values are tabulated for every design strip at every design segment. MODEL OUTPUT \| 313 \| 7.2. Graphical Output The graphical output is organized into diagrams that may be viewed, printed, exported or added to Report. spSlab provides the following diagrams: 7.2.1. Internal Forces This graphical view displays the internal shear force, the internal bending moment and the torsional moment diagrams for ultimate load envelope or any individual ultimate load combinations. 7.2.2. Moment Capacity This graphical view displays the design strips moment capacity diagrams for any span. For beams designed/investigated per CSA A23.3-14/04, longitudinal reinforcement capacity due to combined M-V-T action is also displayed. 7.2.3. Shear Capacity This graphical view displays the design strips one-way shear (and torsion) capacity diagrams for any span. 7.2.4. Deflection This graphical view displays instantaneous deflection diagrams for any span. 7.2.5. Reinforcement This graphical view displays the design strips flexural and shear reinforcement diagrams for any span. MODEL OUTPUT \| 314 \| 7.2.6. Diagrams Display Options The diagrams Display Options allow the user to control the view of DIAGRAM GRIDS, LEGEND, FILL DIAGRAMS, VALUES, and SCALE. 7.2.7. Viewing Aids Viewing aids are those features in the program that facilitate viewing the graphical output results produced by the program. MODEL OUTPUT \| 315 \| 7.2.7.1. Multiple Viewports Multiple viewports can be used to view different diagrams and model views at the same time. The Viewports Command in the Ribbon can be used to select from a set of pre-defined viewport configurations or create a new viewport window. A maximum of 6 viewports can be used at one time. MODEL OUTPUT \| 316 \| 7.2.7.2. View Controls When a viewport is active it has a set of View Controls located in the top right corner. These commands can be used to aid in viewing the model and diagrams. View Controls offer commands like ZOOM TO MODEL (ZOOM TO WORKSPACE), ZOOM TO WINDOW, ZOOM IN, ZOOM OUT, and PAN. Note that controls such as VIEW HOME, FRONT VIEW, VIEW 3D, ROTATE, and EXTRUDE are not available for diagrams views. Users can also: • Rotate Section in 3D: Enables rotating the model in three dimensions (shift + middle mouse button). This control is not available for diagrams views. • Zoom in and zoom out using the mouse wheel and panned by holding the middle mouse button and moving the mouse around. MODEL OUTPUT \| 317 \| 7.2.7.3. Display Options The Display Command in the Ribbon can be used to open the DISPLAY OPTIONS dialog. This dialog facilitates toggling on/off the different Objects, Loads, Restraints, and Grids. \| 318 \| 8. EXAMPLES 8.1. Example 1 – Spandrel Beam with Moment Redistribution ............................................ 321 8.1.1. Problem Formulation .............................................................................................. 321 8.1.2. Preparing Input........................................................................................................ 323 8.1.3 Assigning Spans ....................................................................................................... 334 8.1.4 Assigning Supports .................................................................................................. 335 8.1.5 Assigning Loads....................................................................................................... 337 8.1.6. Solving .................................................................................................................... 339 8.1.7. Viewing and Printing Results ................................................................................. 342 8.2. Example 2 – Spandrel Beam with Torsion ..................................................................... 349 8.2.1. Problem Information ............................................................................................... 349 8.2.2. Preparing Input........................................................................................................ 351 8.2.3 Assigning Spans ....................................................................................................... 363 8.2.4 Assigning Supports .................................................................................................. 365 8.2.5 Assigning Loads....................................................................................................... 367 8.2.6. Solving .................................................................................................................... 375 8.2.7. Viewing and Printing Results ................................................................................. 378 CHAPTER 8 EXAMPLES \| 319 \| 8.3. Example 3 – Design of a Continuous Beam ................................................................... 389 8.3.1. Problem Information ............................................................................................... 389 8.3.2. Preparing the Input .................................................................................................. 391 8.3.3 Assigning Spans ....................................................................................................... 402 8.3.4 Assigning Supports .................................................................................................. 403 8.3.5 Assigning Loads....................................................................................................... 404 8.3.6. Solving .................................................................................................................... 406 8.3.7. Viewing and Printing Results ................................................................................. 409 8.4. Example 4 – Flat Plate Floor System ................................................................ ...416 8.4.1. Problem Information ............................................................................................... 416 8.4.2. Preparing the Input .................................................................................................. 418 8.4.3 Assigning Spans ....................................................................................................... 427 8.4.4 Assigning Supports .................................................................................................. 428 8.4.5 Assigning Loads....................................................................................................... 430 8.4.6. Solving .................................................................................................................... 433 8.4.7. Viewing and Printing Results ................................................................................. 436 8.5. Example 5 – Two-way Slab System .................................................................. ...443 8.5.1. Problem Information ............................................................................................... 443 8.5.2. Preparing the Input .................................................................................................. 445 8.5.3 Assigning Spans ....................................................................................................... 456 8.5.4 Assigning Supports .................................................................................................. 458 8.5.5 Assigning Loads....................................................................................................... 462 8.5.6. Solving .................................................................................................................... 465 8.5.7. Viewing and Printing Results ................................................................................. 468 EXAMPLES \| 320 \| In this chapter, several examples are presented to demonstrate some of the many features and capabilities of the program. Generally, program results match closely the results found in the referenced textbooks. When discrepancies are observed, they result from variations in assumptions and solutions methods, and numerical accuracy. Both beams/one-way slab systems as well as two-way slab systems are presented in the examples. The output of beams/one-way slab examples shows that spBeam program was used to solve them. This is to illustrate that spBeam program is available as a limited version of spSlab that includes only beams/one-way slab capabilities. EXAMPLES \| 321 \| 8.1. Example 1 – Spandrel Beam with Moment Redistribution 8.1.1. Problem Formulation Determine the required reinforcement for the spandrel beam at an intermediate floor level as shown below, using moment redistribution to reduce total reinforcement required. (Note: the self-weight is already included in the specified dead load below.)1 Figure 8.1 – Example Spandrel Beam Problem 1 Notes on ACI 318-11 Building Code Requirements for Structural Concrete, Twelfth Edition, 2013 Portland Cement Association, Example 8-2 EXAMPLES \| 322 \| Design data DL = 1,167 lb/ft LL = 450 lb/ft fc’ = 4,000 psi fy = 60,000 psi Columns: 16 × 16 in. Story height: 10 ft Spandrel beam: 12 × 16 in. EXAMPLES \| 323 \| 8.1.2. Preparing the Input 1. From the Start screen, select New Project. 2. From the Main Program Window, select Project from the Ribbon. • In the General section, select the DESIGN CODE, UNIT SYSTEM, and BAR SET. • In the Materials section, input the following: f'c (SLABS & BEAMS): 4.00 ksi f'c (COLUMNS): 4.00 ksi fy: 60.00 ksi Alternatively, detailed material properties for Concrete and Reinforcement Steel can be entered using Definitions dialog box (see Steps 4 and 5). • In the Run Options section, select the following: RUN MODE: Design FLOOR SYSTEM: One-Way/Beam CONSIDER TORSION: No • In the Description section, enter the PROJECT, FRAME, and ENGINEER. EXAMPLES \| 324 \| EXAMPLES \| 325 \| 3. From the Ribbon, select Grids. • Click on the Generate in the left panel to have the program surface the following: • Enter the following values in the corresponding text boxes: SPANS LENGTH(S): 25 15 20 LEFT CANTILEVER: None RIGHT CANTILEVER: None FRAME LOCATION: Interior • Click on the GENERATE button to return to the main window. Notice how the grid lines now appear in the VIEWPORT. EXAMPLES \| 326 \| EXAMPLES \| 327 \| 4. From the Ribbon, select Define, then choose Concrete from Materials to display the Concrete dialog box. • Check STANDARD for SLABS AND BEAMS and COLUMNS. • Enter the following for SLABS AND BEAMS and COLUMNS: COMPRESSIVE STRENGTH, f’c: 4.00 ksi UNIT DENSITY, Wc: 150.00 pcf EXAMPLES \| 328 \| 5. Click on Reinforcing Steel from Materials to display the Reinforcing Steel dialog box. • Enter the following: YIELD STRESS OF FLEXURAL STEEL, fy: 60.00 ksi YIELD STRESS OF STIRRUP, fyt: 60.00 ksi YOUNG’S MODULUS, Es: 29000.00 ksi EXAMPLES \| 329 \| 6. Click on Beams from Reinforcement Criteria to display the Beams dialog box. • Enter the following for TOP BARS and BOTTOM BARS: Min. Max. BAR SIZE: #8 #8 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in CLEAR DISTANCE BETWEEN BAR LAYERS (SL): 1.00 in EXAMPLES \| 330 \| 7. Click on Beam Stirrups from Reinforcement Criteria to display the Beam Stirrups dialog box. • Enter the following: Min. Max. BAR SIZE: #3 #5 BAR SPACING (S): 6.00 in 18.00 in NUMBER OF LEGS: 2 6 SIDE COVER – CLEAR (CS): 1.50 in FIRST STIRRUP FROM FOS (S1): 3.00 in EXAMPLES \| 331 \| 8. Click on Design & Modeling from Options to display the Design & Modeling dialog box. • Check MOMENT REDISTRIBUTION. EXAMPLES \| 332 \| 9. Click on Load Cases from Load Case/Combo. to display the Load Cases dialog box. • Add SELF-WEIGHT for CASE A. • Enter the following: CASE B: Dead CASE C: Live EXAMPLES \| 333 \| 10. Click on Load Combinations from Load Case/Combo. to display the Load Combinations dialog box. • Enter the following load combination shown in the figure below: EXAMPLES \| 334 \| 8.1.3. Assigning Spans 11. From the Ribbon, select Spans command. • In the left panel, select Beam and enter the following: WIDTH (W): 12.00 in DEPTH (D): 16.00 in • Apply to all spans as shown in the figure below. EXAMPLES \| 335 \| 8.1.4. Assigning Supports 12. From the Ribbon, select Supports command. • In the left panel, select Column and enter the following for COLUMN – ABOVE and COLUMN – BELOW: TYPE: Rectangular HEIGHT: 10.00 ft c1: 16.00 in c2: 16.00 in FAR END CONDITION: Fixed • Apply to all supports as shown in the figure below. EXAMPLES \| 336 \| • In the left panel, select Moment Redistribution and enter the following: LEFT: 20.00 % RIGHT: 20.00 % • Apply to Support 2 and Support 3 as shown in the figure below. EXAMPLES \| 337 \| 8.1.5. Assigning Loads 13. From the Ribbon, select Loads command. • In the left panel, select Uniform Line Loads then select B-DEAD from LOAD CASE and enter the following: W1: 1167.00 plf • Apply to all spans as shown in the figure below. EXAMPLES \| 338 \| • Select C-LIVE from LOAD CASE and enter the following: W1: 450.00 plf • Apply to all spans as shown in the figure below. EXAMPLES \| 339 \| 8.1.6. Solving 14. From the Ribbon, select Solve command. For Design Options: • Leave all Design Options to their default settings. EXAMPLES \| 340 \| For Deflection Options: • Leave all Design Options to their default settings. EXAMPLES \| 341 \| • Click on the Run button. • The spBeam Solver window is displayed and the solver messages are listed. After the solution is done, the design will be performed and then the focus will immediately be passed to the Results scope. EXAMPLES \| 342 \| 8.1.7. Viewing and Printing Results 15. After a successful run, results can be viewed by selecting Internal Forces, Moment Capacity, Shear Capacity, Deflection, or Reinforcement from the left panel. EXAMPLES \| 343 \| EXAMPLES \| 344 \| EXAMPLES \| 345 \| EXAMPLES \| 346 \| EXAMPLES \| 347 \| 16. Results can be also viewed in table format by selecting the Tables command from the Ribbon. EXAMPLES \| 348 \| 17. Results can be printed or exported in different formats by selecting the Reporter command from the Ribbon. EXAMPLES \| 349 \| 8.2. Example 2 – Spandrel Beam with Torsion 8.2.1. Problem Formulation Design a precast, nonprestressed concrete spandrel beam for combined shear and torsion. Roof members are simply supported on spandrel ledge. Spandrel beams are connected to columns to transfer torsion. Continuity between spandrel beams is not provided.2 2 Notes on ACI 318-11 Building Code Requirements for Structural Concrete, Twelfth Edition, 2013 Portland Cement Association, Example 13-1 EXAMPLES \| 350 \| Design data Dead load = 90 lb/ft2 (double tee + topping + insulation + roofing) Live load = 30 lb/ft2 fc’ = 5,000 psi (wc = 150 pcf) fy = 60,000 psi Roof members are 10 ft wide double tee units, 30 in. deep with 2 in. topping. Design of these units is not included in this design example. For lateral support, alternate ends of roof members are fixed to supporting beams. EXAMPLES \| 351 \| 8.2.2. Preparing the Input 1. From the Start screen, select New Project. 2. From the Main Program Window, select Project from the Ribbon. • In the General section, select the DESIGN CODE, UNIT SYSTEM, and BAR SET. • In the Materials section, input the following: f'c (SLABS & BEAMS): 5.00 ksi f'c (COLUMNS): 5.00 ksi fy: 60.00 ksi Alternatively, detailed material properties for Concrete and Reinforcement Steel can be entered using Definitions dialog box (see Steps 4 and 5). • In the Run Options section, select the following: RUN MODE: Design FLOOR SYSTEM: One-Way/Beam CONSIDER TORSION: Yes • In the Description section, enter the PROJECT, FRAME, and ENGINEER. EXAMPLES \| 352 \| EXAMPLES \| 353 \| 3. From the Ribbon, select Grids. • Click on the Generate in the left panel to have the program surface the following: • Enter the following values in the corresponding text boxes: SPANS LENGTH(S): 40 LEFT CANTILEVER: None RIGHT CANTILEVER: None FRAME LOCATION: Interior • Click on the GENERATE button to return to the main window. Notice how the grid lines now appear in the VIEWPORT. EXAMPLES \| 354 \| EXAMPLES \| 355 \| 4. From the Ribbon, select Define, then choose Concrete from Materials to display the Concrete dialog box. • Check STANDARD for SLABS AND BEAMS and COLUMNS. • Enter the following for SLABS AND BEAMS and COLUMNS: COMPRESSIVE STRENGTH, f’c: 5.00 ksi UNIT DENSITY, Wc: 150.00 pcf EXAMPLES \| 356 \| 5. Click on Reinforcing Steel from Materials to display the Reinforcing Steel dialog box. • Enter the following: YIELD STRESS OF FLEXURAL STEEL, fy: 60.00 ksi YIELD STRESS OF STIRRUP, fyt: 60.00 ksi YOUNG’S MODULUS, Es: 29000.00 ksi EXAMPLES \| 357 \| 6. Click on Slabs & Ribs from Reinforcement Criteria to display the Slabs & Ribs dialog box. • Enter the following for TOP BARS and BOTTOM BARS: Min. Max. BAR SIZE: #5 #8 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.75 in EXAMPLES \| 358 \| 7. Click on Beams from Reinforcement Criteria to display the Beams dialog box. • Enter the following for TOP BARS: Min. Max. BAR SIZE: #5 #5 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.75 in • Enter the following for BOTTOM BARS: Min. Max. BAR SIZE: #11 #11 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.75 in CLEAR DISTANCE BETWEEN BAR LAYERS (SL): 1.00 in EXAMPLES \| 359 \| EXAMPLES \| 360 \| 8. Click on Beam Stirrups from Reinforcement Criteria to display the Beam Stirrups dialog box. • Enter the following: Min. Max. BAR SIZE: #4 #4 BAR SPACING (S): 6.00 in 18.00 in NUMBER OF LEGS: 2 6 SIDE COVER – CLEAR (CS): 1.25 in FIRST STIRRUP FROM FOS (S1): 3.00 in EXAMPLES \| 361 \| 9. Click on Load Cases from Load Case/Combo. to display the Load Cases dialog box. • Add SELF-WEIGHT for CASE A. • Enter the following: CASE B: Dead CASE C: Live EXAMPLES \| 362 \| 10. Click on Load Combinations from Load Case/Combo. to display the Load Combinations dialog box. • Enter the following load combination shown in the figure below: EXAMPLES \| 363 \| 8.2.3. Assigning Spans 11. From the Ribbon, select Spans command. • In the left panel, select Slab and enter the following: THICKNESS (T): 16.00 in WIDTH – LEFT (L): 0.667 ft WIDTH – RIGHT (R): 1.333 ft • Apply to the span as shown in the figure below. EXAMPLES \| 364 \| • In the left panel, select Beam and enter the following: WIDTH (W): 16.00 in DEPTH (D): 48.00 in • Apply to the span as shown in the figure below. EXAMPLES \| 365 \| 8.2.4. Assigning Supports 12. From the Ribbon, select Supports command. • In the left panel, select Column and enter the following for COLUMN – ABOVE and COLUMN – BELOW: TYPE: Rectangular HEIGHT: 10.00 ft c1: 16.00 in c2: 16.00 in FAR END CONDITION: Fixed • Apply to all supports as shown in the figure below. EXAMPLES \| 366 \| • In the left panel, select Restraint and select the following for SUPPORT RESTRAINTS. TYPE: Pinned • Apply to all supports as shown in the figure below. EXAMPLES \| 367 \| 8.2.5. Assigning Loads 13. From the Ribbon, select Loads command. • In the left panel, select Uniform Line Loads then select B-DEAD from LOAD CASE and enter the following: W1: 4080.00 plf (Note: this value was obtained by converting the area loads on the roof and the beam’s self weight into line loads.) Dead Load = Superimposed Load + Self Weight of Spandrel Beam = ( ) ( ) 70 ft 90 psf 1.33 ft 4.00 ft 1.33 ft 0.67 ft 150 pcf 4.080 kip/ft 2    +  +   =         • Apply to the span as shown in the figure below. EXAMPLES \| 368 \| EXAMPLES \| 369 \| • Select C-LIVE from LOAD CASE and enter the following: W1: 1050.00 plf (Note: this value was obtained by converting the area loads on the roof to line loads on the beam.) Live Load = 70 ft 30 psf 1,050 lb/ft 2    =     • Apply to the span as shown in the figure below. EXAMPLES \| 370 \| • In the left panel, select Uniform Line Torques then select B-DEAD from LOAD CASE and enter the following: Mx1: 3.28 kip-ft/ft (Note: this value was obtained by multiplying the superimposed line load by the moment arm of 12 in.) Torsion Line Load (Dead) = 2 70 ft 16 8 12 in. 90 psf ft 150 pcf 3.28 kip-ft/ft 2 12 12 in./ft     +   =     • Apply to the span as shown in the figure below. EXAMPLES \| 371 \| EXAMPLES \| 372 \| • Select C-LIVE from LOAD CASE and enter the following: Mx1: 1.05 kip-ft/ft (Note: this value was obtained by multiplying the live line load by the moment arm of 12 in.) 70 ft 12 in. 30 psf 1.05 kip-ft/ft 2 12 in./ft       =         • Apply to the span as shown in the figure below. EXAMPLES \| 373 \| • In the left panel, select Point Loads then select B-DEAD from LOAD CASE and enter the following: Fz: 0.001 kips L: 0.667 ft • Apply to the span as shown in the figure below. • Select ADD TO EXISTING LOAD from OPTIONS and enter the following: Fz: 0.001 kips L: 39.333 ft • Apply to the span as shown in the figure below. • Critical section for torsion is at the face of the support because of concentrated torques applied by the double tee stems at a distance less than d from the face of the support. The critical section for shear is also at the face of support because the load on the spandrel beam is not applied close to the top of the member and because the concentrated forces transferred by the double tee stems are at a distance less than d from the face of support. A small dummy load of 0.001 kips at the face of support is therefore introduced in order to move the critical section for shear from the default location of d away from the support to the face of the support. EXAMPLES \| 374 \| EXAMPLES \| 375 \| 8.2.6. Solving 14. From the Ribbon, select Solve command. For Design Options: • Select EQUILIBRIUM for TORSION TYPE, and YES for STIRRUPS IN FLANGES in TORSION ANALYSIS AND DESIGN section. • Uncheck EFFECTIVE FLANGE WIDTH for BEAM DESIGN section. EXAMPLES \| 376 \| For Deflection Options: • Leave all Design Options to their default settings. EXAMPLES \| 377 \| • Click on the Run button. • The spBeam Solver window is displayed and the solver messages are listed. After the solution is done, the design will be performed and then the focus will immediately be passed to the Results scope. EXAMPLES \| 378 \| 8.2.7. Viewing and Printing Results 15. After a successful run, results can be viewed by selecting Internal Forces, Moment Capacity, Shear Capacity, Deflection, or Reinforcement from the left panel. EXAMPLES \| 379 \| 16. Results can be also viewed in table format by selecting the Tables command from the Ribbon. EXAMPLES \| 380 \| EXAMPLES \| 381 \| EXAMPLES \| 382 \| EXAMPLES \| 383 \| EXAMPLES \| 384 \| EXAMPLES \| 385 \| EXAMPLES \| 386 \| EXAMPLES \| 387 \| EXAMPLES \| 388 \| 17. Results can be printed or exported in different formats by selecting the Reporter command from the Ribbon. EXAMPLES \| 389 \| 8.3. Example 3 – Design of a Continuous Beam 8.3.1. Problem Formulation The system shown in the following figure consists of five spans symmetric about the centerline. We will be designing beam ABCD assuming that the other half of the beam will be loaded and designed the same way. All beams have a width of 12 in. and a depth of 22 in. – including the 5 in. thick deck. Span length and widths are shown in the figure. Columns have a 12 in. × 12 in. cross-section and a length equal to a typical story height of 13 ft. The system will be analyzed and designed under a uniform live load of 130 psf and a dead load that consists of the slab system’s own weight plus 80 psf. Use f’c = 4 ksi, fy = 60 ksi, and γconcrete = 150 pcf.3 3 M. N. Hassoun and A. Al-Manaseer, Structural Concrete: Theory and Design, John Wiley & Sons, Inc., Sixth Edition, 2015, Example 16.1 EXAMPLES \| 390 \| Design data DL = 80 psf LL = 130 lb/ft fc’ = 4,000 psi fy = 60,000 psi γconcrete = 150 pcf Beams: 12 × 22 in. Columns: 12 × 12 in. Story height: 13 ft EXAMPLES \| 391 \| 8.3.2. Preparing the Input 1. From the Start screen, select New Project. 2. From the Main Program Window, select Project from the Ribbon. • In the General section, select the DESIGN CODE, UNIT SYSTEM, and BAR SET. • In the Materials section, input the following: f'c (SLABS & BEAMS): 4.00 ksi f'c (COLUMNS): 4.00 ksi fy: 60.00 ksi Alternatively, detailed material properties for Concrete and Reinforcement Steel can be entered using Definitions dialog box (see Steps 4 and 5). • In the Run Options section, select the following: RUN MODE: Design FLOOR SYSTEM: One-Way/Beam CONSIDER TORSION: No • In the Description section, enter the PROJECT, FRAME, and ENGINEER. EXAMPLES \| 392 \| EXAMPLES \| 393 \| 3. From the Ribbon, select Grids. • Click on the Generate in the left panel to have the program surface the following: • Enter the following values in the corresponding text boxes: SPANS LENGTH(S): 24 3x26 24 LEFT CANTILEVER: None RIGHT CANTILEVER: None FRAME LOCATION: Interior • Click on the GENERATE button to return to the main window. Notice how the grid lines now appear in the VIEWPORT. EXAMPLES \| 394 \| EXAMPLES \| 395 \| 4. From the Ribbon, select Define, then choose Concrete from Materials to display the Concrete dialog box. • Check STANDARD for SLABS AND BEAMS and COLUMNS. • Enter the following for SLABS AND BEAMS and COLUMNS: COMPRESSIVE STRENGTH, f’c: 4.00 ksi UNIT DENSITY, Wc: 150.00 pcf EXAMPLES \| 396 \| 5. Click on Reinforcing Steel from Materials to display the Reinforcing Steel dialog box. • Enter the following: YIELD STRESS OF FLEXURAL STEEL, fy: 60.00 ksi YIELD STRESS OF STIRRUP, fyt: 60.00 ksi YOUNG’S MODULUS, Es: 29000.00 ksi EXAMPLES \| 397 \| 6. Click on Beams from Reinforcement Criteria to display the Beams dialog box. • Enter the following for TOP BARS: Min. Max. BAR SIZE: #9 #9 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in • Enter the following for BOTTOM BARS: Min. Max. BAR SIZE: #8 #8 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in CLEAR DISTANCE BETWEEN BAR LAYERS (SL): 1.00 in EXAMPLES \| 398 \| EXAMPLES \| 399 \| 7. Click on Beam Stirrups from Reinforcement Criteria to display the Beam Stirrups dialog box. • Enter the following: Min. Max. BAR SIZE: #3 #5 BAR SPACING (S): 6.00 in 18.00 in NUMBER OF LEGS: 2 6 SIDE COVER – CLEAR (CS): 1.50 in FIRST STIRRUP FROM FOS (S1): 3.00 in EXAMPLES \| 400 \| 8. Click on Load Cases from Load Case/Combo. to display the Load Cases dialog box. • Add SELF-WEIGHT for CASE A. • Enter the following: CASE B: Dead CASE C: Live EXAMPLES \| 401 \| 9. Click on Load Combinations from Load Case/Combo. to display the Load Combinations dialog box. • Enter the following load combination shown in the figure below: EXAMPLES \| 402 \| 8.3.3. Assigning Spans 10. From the Ribbon, select Spans command. • In the left panel, select Beam and enter the following: WIDTH (W): 12.00 in DEPTH (D): 22.00 in (Note: Since there will be no slab assigned, we must convert the area loads to line loads along the beam and also add the self-weight of the slab to the dead load. This calculation will be shown in Step 12.) • Apply to all spans as shown in the figure below. EXAMPLES \| 403 \| 8.3.4. Assigning Supports 11. From the Ribbon, select Supports command. • In the left panel, select Column and enter the following for COLUMN – ABOVE and COLUMN – BELOW: TYPE: Rectangular HEIGHT: 13.00 ft c1: 12.00 in c2: 12.00 in FAR END CONDITION: Fixed • Apply to all supports as shown in the figure below. EXAMPLES \| 404 \| 8.3.5. Assigning Loads 12. From the Ribbon, select Loads command. • In the left panel, select Uniform Line Loads then select B-DEAD from LOAD CASE and enter the following: W1: 1647.50 plf (Note: This value was obtained by converting the area loads of the slab’s self weight (without the beam) and superimposed dead load into a line load) Dead Load = ( ) 2 2 5 5 in. 12 in. ft 150 pcf 12 ft 80 psf 12 ft 150 pcf 12 144 in. /ft        +  −          = 1647.50 lb/ft • Apply to all spans as shown in the figure below. EXAMPLES \| 405 \| • Select C-LIVE from LOAD CASE and enter the following: W1: 1560.00 plf Live Load = (130 psf × 12 ft) = 1560 lb/ft • Apply to all spans as shown in the figure below. EXAMPLES \| 406 \| 8.3.6. Solving 13. From the Ribbon, select Solve command. For Design Options: • Leave all Design Options to their default settings. EXAMPLES \| 407 \| For Deflection Options: • Leave all Design Options to their default settings. EXAMPLES \| 408 \| • Click on the Run button. • The spBeam Solver window is displayed and the solver messages are listed. After the solution is done, the design will be performed and then the focus will immediately be passed to the Results scope. EXAMPLES \| 409 \| 8.3.7. Viewing and Printing Results 14. After a successful run, results can be viewed by selecting Internal Forces, Moment Capacity, Shear Capacity, Deflection, or Reinforcement from the left panel. EXAMPLES \| 410 \| EXAMPLES \| 411 \| EXAMPLES \| 412 \| EXAMPLES \| 413 \| EXAMPLES \| 414 \| 15. Results can be also viewed in table format by selecting the Tables command from the Ribbon. EXAMPLES \| 415 \| 16. Results can be printed or exported in different formats by selecting the Reporter command from the Ribbon. EXAMPLES \| 416 \| 8.4. Example 4 – Flat Plate Floor System 8.4.1. Problem Formulation An office building is planned using a flat plate floor system with the column layout as shown in figure below. No beams, drop panels, or column capitals are permitted. Specified live load is 100 psf and dead load will include the weight of the slab plus an allowance of 20 psf for finish floor plus suspended loads. The columns will be 18 in. square, and the floor-to-floor height of the structure will be 12 ft. The slab thickness will be 8.50 in. according to ACI Code. Design the interior panel C, using material strengths fy = 60,000 psi and f’c = 4,000 psi. Straight-bar reinforcement will be used.1 1 A. H. Nilson, D. Darwin, and C. W. Dolan, Design of Concrete Structures, Fifteenth Edition, 2016 McGraw-Hill Education, Example 13-2 EXAMPLES \| 417 \| Design data Dead load = 20 psf Live load = 100 psf fc’ = 4,000 psi (wc = 150 pcf) fy = 60,000 psi Columns: 18 × 18 in. Story height: 12 ft Slab thickness: 8.50 in. EXAMPLES \| 418 \| 8.4.2. Preparing the Input 1. From the Start screen, select New Project. 2. From the Main Program Window, select Project from the Ribbon. • In the General section, select the DESIGN CODE, UNIT SYSTEM, and BAR SET. • In the Materials section, input the following: f'c (SLABS & BEAMS): 4.00 ksi f'c (COLUMNS): 4.00 ksi fy: 60.00 ksi Alternatively, detailed material properties for Concrete and Reinforcement Steel can be entered using Definitions dialog box (see Steps 4 and 5). • In the Run Options section, select the following: RUN MODE: Design FLOOR SYSTEM: Two-Way • In the Description section, enter the PROJECT, FRAME, and ENGINEER. EXAMPLES \| 419 \| EXAMPLES \| 420 \| 3. From the Ribbon, select Grids. • Click on the Generate in the left panel to have the program surface the following: • Enter the following values in the corresponding text boxes: SPANS LENGTH(S): 3x22 LEFT CANTILEVER: Adjusted to support face RIGHT CANTILEVER: Adjusted to support face FRAME LOCATION: Interior • Click on the GENERATE button to return to the main window. Notice how the grid lines now appear in the VIEWPORT. EXAMPLES \| 421 \| EXAMPLES \| 422 \| 4. From the Ribbon, select Define, then choose Concrete from Materials to display the Concrete dialog box. • Check STANDARD for SLABS AND BEAMS and COLUMNS. • Enter the following for SLABS AND BEAMS and COLUMNS: COMPRESSIVE STRENGTH, f’c: 4.00 ksi UNIT DENSITY, Wc: 150.00 pcf EXAMPLES \| 423 \| 5. Click on Reinforcing Steel from Materials to display the Reinforcing Steel dialog box. • Enter the following: YIELD STRESS OF FLEXURAL STEEL, fy: 60.00 ksi YIELD STRESS OF STIRRUP, fyt: 60.00 ksi YOUNG’S MODULUS, Es: 29000.00 ksi EXAMPLES \| 424 \| 6. Click on Slabs & Ribs from Reinforcement Criteria to display the Slabs & Ribs dialog box. • Enter the following for TOP BARS and BOTTOM BARS: Min. Max. BAR SIZE: #5 #6 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in EXAMPLES \| 425 \| 7. Click on Load Cases from Load Case/Combo. to display the Load Cases dialog box. • Add SELF-WEIGHT for CASE A. • Enter the following: CASE B: Dead CASE C: Live EXAMPLES \| 426 \| 8. Click on Load Combinations from Load Case/Combo. to display the Load Combinations dialog box. • Enter the following load combination shown in the figure below: EXAMPLES \| 427 \| 8.4.3. Assigning Spans 9. From the Ribbon, select Spans command. • In the left panel, select Slab and enter the following: THICKNESS (T): 8.50 in WIDTH – LEFT (L): 11.00 ft WIDTH – RIGHT (R): 11.00 ft • Apply to all spans as shown in the figure below. EXAMPLES \| 428 \| 8.4.4. Assigning Supports 10. From the Ribbon, select Supports command. • In the left panel, select Column and enter the following for COLUMN – ABOVE and COLUMN – BELOW: TYPE: Rectangular HEIGHT: 12.00 ft c1: 18.00 in c2: 18.00 in FAR END CONDITION: Fixed CHECK PUNCHING SHEAR: Yes INCREASE GAMMAF: No • Apply to all supports as shown in the figure below. • Notice how the cantilevers adjusted to the column faces when the exterior columns are assigned. EXAMPLES \| 429 \| EXAMPLES \| 430 \| 8.4.5. Assigning Loads 11. From the Ribbon, select Loads command. • In the left panel, select Area Load then select B-DEAD from LOAD CASE and enter the following: W: 20.00 psf • Apply to all spans as shown in the figure below. EXAMPLES \| 431 \| • Select C-LIVE from LOAD CASE and enter the following: W: 100.00 psf • Apply to all spans as shown in the figure below. EXAMPLES \| 432 \| • Also, you can click on the VIEW 3D icon from View Controls (top right of the active viewport) to get a better view of the applied loads. EXAMPLES \| 433 \| 8.4.6. Solving 12. From the Ribbon, select Solve command. For Design Options: • Uncheck DISTRIBUTE SHEAR TO SLAB STRIPS for SHEAR DESIGN section. • Leave all the other Design Options to their default settings. EXAMPLES \| 434 \| For Deflection Options: • Leave all Design Options to their default settings. EXAMPLES \| 435 \| • Click on the Run button. • The spSlab Solver window is displayed and the solver messages are listed. After the solution is done, the design will be performed and then the focus will immediately be passed to the Results scope. EXAMPLES \| 436 \| 8.4.7. Viewing and Printing Results 13. After a successful run, results can be viewed by selecting Internal Forces, Moment Capacity, Shear Capacity, Deflection, or Reinforcement from the left panel. EXAMPLES \| 437 \| EXAMPLES \| 438 \| EXAMPLES \| 439 \| EXAMPLES \| 440 \| EXAMPLES \| 441 \| 14. Results can be also viewed in table format by selecting the Tables command from the Ribbon. EXAMPLES \| 442 \| 15. Results can be printed or exported in different formats by selecting the Reporter command from the Ribbon. EXAMPLES \| 443 \| 8.5. Example 5 – Two-way Slab System 8.5.1. Problem Formulation Using the Equivalent Frame Method, determine design moments for the slab system in the direction shown, for an intermediate floor.2 2 Notes on ACI 318-11 Building Code Requirements for Structural Concrete, Twelfth Edition, 2013 Portland Cement Association, Example 20-2 EXAMPLES \| 444 \| Design data Dead load = Self-weight Service live load = 100 psf fc’ = 4,000 psi (for all members), normal weight concrete fy = 60,000 psi Column dimensions = 18 × 18 in. Story height = 12 ft Edge beam dimensions = 14 × 27 in. Interior beam dimensions = 14 × 20 in. EXAMPLES \| 445 \| 8.5.2. Preparing the Input 1. From the Start screen, select New Project. 2. From the Main Program Window, select Project from the Ribbon. • In the General section, select the DESIGN CODE, UNIT SYSTEM, and BAR SET. • In the Materials section, input the following: f'c (SLABS & BEAMS): 4.00 ksi f'c (COLUMNS): 4.00 ksi fy: 60.00 ksi Alternatively, detailed material properties for Concrete and Reinforcement Steel can be entered using Definitions dialog box (see Steps 4 and 5). • In the Run Options section, select the following: RUN MODE: Design FLOOR SYSTEM: Two-Way • In the Description section, enter the PROJECT, FRAME, and ENGINEER. EXAMPLES \| 446 \| EXAMPLES \| 447 \| 3. From the Ribbon, select Grids. • Click on the Generate in the left panel to have the program surface the following: • Enter the following values in the corresponding text boxes: SPANS LENGTH(S): 3x17.5 LEFT CANTILEVER: Adjusted to support face RIGHT CANTILEVER: Adjusted to support face FRAME LOCATION: Interior • Click on the GENERATE button to return to the main window. Notice how the grid lines now appear in the VIEWPORT. EXAMPLES \| 448 \| EXAMPLES \| 449 \| 4. From the Ribbon, select Define, then choose Concrete from Materials to display the Concrete dialog box. • Check STANDARD for SLABS AND BEAMS and COLUMNS. • Enter the following for SLABS AND BEAMS and COLUMNS: COMPRESSIVE STRENGTH, f’c: 4.00 ksi UNIT DENSITY, Wc: 150.00 pcf EXAMPLES \| 450 \| 5. Click on Reinforcing Steel from Materials to display the Reinforcing Steel dialog box. • Enter the following: YIELD STRESS OF FLEXURAL STEEL, fy: 60.00 ksi YIELD STRESS OF STIRRUP, fyt: 60.00 ksi YOUNG’S MODULUS, Es: 29000.00 ksi EXAMPLES \| 451 \| 6. Click on Slabs & Ribs from Reinforcement Criteria to display the Slabs & Ribs dialog box. • Enter the following for TOP BARS and BOTTOM BARS: Min. Max. BAR SIZE: #5 #8 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in EXAMPLES \| 452 \| 7. Click on Beams from Reinforcement Criteria to display the Beams dialog box. • Enter the following for TOP BARS and BOTTOM BARS: Min. Max. BAR SIZE: #5 #8 BAR SPACING (ST): 1.00 in 18.00 in REINFORCEMENT RATIO: 0.14 % 5.00 % CLEAR COVER (CT): 1.50 in CLEAR DISTANCE BETWEEN BAR LAYERS (SL): 1.00 in EXAMPLES \| 453 \| 8. Click on Beam Stirrups from Reinforcement Criteria to display the Beam Stirrups dialog box. • Enter the following: Min. Max. BAR SIZE: #3 #5 BAR SPACING (S): 6.00 in 18.00 in NUMBER OF LEGS: 2 6 SIDE COVER – CLEAR (CS): 1.50 in FIRST STIRRUP FROM FOS (S1): 3.00 in EXAMPLES \| 454 \| 9. Click on Load Cases from Load Case/Combo. to display the Load Cases dialog box. • Enter the following: CASE A: Dead CASE B: Live EXAMPLES \| 455 \| 10. Click on Load Combinations from Load Case/Combo. to display the Load Combinations dialog box. • Enter the following load combination shown in the figure below: EXAMPLES \| 456 \| 8.5.3. Assigning Spans 11. From the Ribbon, select Spans command. • In the left panel, select Slab and enter the following: THICKNESS (T): 6.00 in WIDTH – LEFT (L): 11.00 ft WIDTH – RIGHT (R): 11.00 ft • Apply to all spans as shown in the figure below. EXAMPLES \| 457 \| • In the left panel, select Beam and enter the following: WIDTH (W): 14.00 in DEPTH (D): 20.00 in OFFSET (S): 0.00 in • Apply to all spans as shown in the figure below. EXAMPLES \| 458 \| 8.5.4. Assigning Supports 12. From the Ribbon, select Supports command. • In the left panel, select Column and enter the following for COLUMN – ABOVE and COLUMN – BELOW: TYPE: Rectangular HEIGHT: 12.00 ft c1: 18.00 in c2: 18.00 in FAR END CONDITION: Fixed CHECK PUNCHING SHEAR: Yes INCREASE GAMMAF: No • Apply to all supports as shown in the figure below. • Notice how the cantilevers adjusted to the column faces when the exterior columns are assigned. EXAMPLES \| 459 \| EXAMPLES \| 460 \| For exterior transverse beams: • In the left panel, select Beam and enter the following: WIDTH (W): 14.00 in DEPTH (D): 27.00 in OFFSET (S): 0.00 in • Apply to all exterior supports as shown in the figure below. EXAMPLES \| 461 \| For interior transverse beams: • In the left panel, select Beam and enter the following: WIDTH (W): 14.00 in DEPTH (D): 20.00 in OFFSET (S): 0.00 in • Apply to all interior supports as shown in the figure below. EXAMPLES \| 462 \| 8.5.5. Assigning Loads 13. From the Ribbon, select Loads command. • In the left panel, select Area Load then select A-DEAD from LOAD CASE and enter the following: W: 84.30 psf • Apply to all spans as shown in the figure below. EXAMPLES \| 463 \| • Select B-LIVE from LOAD CASE and enter the following: W: 100.00 psf • Apply to all spans as shown in the figure below. EXAMPLES \| 464 \| • Also, you can click on the VIEW 3D icon from View Controls (top right of the active viewport) to get a better view of the applied loads. EXAMPLES \| 465 \| 8.5.6. Solving 14. From the Ribbon, select Solve command. For Design Options: • Uncheck DISTRIBUTE SHEAR TO SLAB STRIPS for SHEAR DESIGN section. • Uncheck BEAM T-SECTION DESIGN for BEAM DESIGN section. • Leave all the other Design Options to their default settings. EXAMPLES \| 466 \| For Deflection Options: • Leave all Design Options to their default settings. EXAMPLES \| 467 \| • Click on the Run button. • The spSlab Solver window is displayed and the solver messages are listed. After the solution is done, the design will be performed and then the focus will immediately be passed to the Results scope. EXAMPLES \| 468 \| 8.5.7. Viewing and Printing Results 15. After a successful run, results can be viewed by selecting Internal Forces, Moment Capacity, Shear Capacity, Deflection, or Reinforcement from the left panel. EXAMPLES \| 469 \| EXAMPLES \| 470 \| EXAMPLES \| 471 \| EXAMPLES \| 472 \| EXAMPLES \| 473 \| 16. Results can be also viewed in table format by selecting the Tables command from the Ribbon. EXAMPLES \| 474 \| 17. Results can be printed or exported in different formats by selecting the Reporter command from the Ribbon. \| 475 \| APPENDIX A.1. Default Load Case and Combination Factors .....................................................................476 A.1.1. For ACI 318-14/11 .................................................................................................... 477 A.1.2. For ACI 318-08/05/02 ............................................................................................... 478 A.1.3. For ACI 318-99 ......................................................................................................... 479 A.1.4. For CSA A23.3-14 ..................................................................................................... 480 A.1.5. For CSA A23.3-04 ..................................................................................................... 481 A.1.5. For CSA A23.3-94 ..................................................................................................... 482 A.2. Conversion Factors - English to SI .....................................................................................483 A.3. Conversion Factors - SI to English .....................................................................................484 A.4. Technical Resources ...........................................................................................................485 A.5. Contact Information ............................................................................................................486 CHAPTER A APPENDIX \| 476 \| A.1. Default Load Case and Combination Factors The program allows defining up to 50 load combinations. The user has full control over the combinations. The program contains predefined (built into the program) default primary load combinations for the supported codes. These default combinations are created when starting a new project. APPENDIX \| 477 \| A.1.1. For ACI 318-14/11 For the ACI 318-14 and ACI 318-11 codes, the default combinations of the Self-weight (SW), Dead (D), Live (L), Snow (S), Wind (W) and Earthquake (E) loads considered by the program are1: Ultimate Load Combinations – ACI 318 – 14 / 11 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.40 1.40 SW, D - U2 1.20 1.20 1.60 0.50 SW, D, L S U3 1.20 1.20 1.00 1.60 SW, D, S L U4 1.20 1.20 1.60 0.50 SW, D, S W U5 1.20 1.20 1.60 -0.50 SW, D, S W U6 1.20 1.20 1.00 0.50 1.00 SW, D, W L, S U7 1.20 1.20 1.00 0.50 -1.00 SW, D, W L, S U8 1.20 1.20 1.00 0.20 1.00 SW, D, E L, S U9 1.20 1.20 1.00 0.20 -1.00 SW, D, E L, S U10 0.90 0.90 1.00 SW, D, W - U11 0.90 0.90 -1.00 SW, D, W - U12 0.90 0.90 1.00 SW, D, E - U13 0.90 0.90 -1.00 SW, D, E - 1 ACI 318-14, 5.3.1; ACI 318-11, 9.2; (assuming W and E based on ultimate-level forces) APPENDIX \| 478 \| A.1.2. For ACI 318-08/05/02 For the ACI 318-08, ACI 318-05, and ACI 318-02 codes, the default combinations of the Self-weight (SW), Dead (D), Live (L), Snow (S), Wind (W) and Earthquake (E) loads considered by the program are2: Ultimate Load Combinations – ACI 318 – 08 / 05 / 02 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.40 1.40 SW, D - U2 1.20 1.20 1.60 0.50 SW, D, L S U3 1.20 1.20 1.00 1.60 SW, D, S L U4 1.20 1.20 1.60 0.80 SW, D, S W U5 1.20 1.20 1.60 -0.80 SW, D, S W U6 1.20 1.20 1.00 0.50 1.60 SW, D, W L, S U7 1.20 1.20 1.00 0.50 -1.60 SW, D, W L, S U8 1.20 1.20 1.00 0.20 1.00 SW, D, E L, S U9 1.20 1.20 1.00 0.20 -1.00 SW, D, E L, S U10 0.90 0.90 1.60 SW, D, W - U11 0.90 0.90 -1.60 SW, D, W - U12 0.90 0.90 1.00 SW, D, E - U13 0.90 0.90 -1.00 SW, D, E - 2 ACI 318-08, 9.2; ACI 318-05, 9.2; ACI 318-02, 9.2 (assuming W based on service-level wind load and E based on ultimate-level forces) APPENDIX \| 479 \| A.1.3. For ACI 318-99 For the ACI 318-99 code, the default combinations of the Self-weight (SW), Dead (D), Live (L), Wind (W) and Earthquake (E) loads considered by the program are3: Ultimate Load Combinations – ACI 318 – 99 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.400 1.400 1.700 SW, D, L - U2 1.050 1.050 1.275 1.275 SW, D, L, W - U3 1.050 1.050 1.275 -1.275 SW, D, L, W - U4 1.050 1.050 1.275 SW, D, W - U5 1.050 1.050 -1.275 SW, D, W - U6 1.050 1.050 1.275 1.430 SW, D, L, E - U7 1.050 1.050 1.275 -1.430 SW, D, L, E - U8 1.050 1.050 1.430 SW, D, E - U9 1.050 1.050 -1.430 SW, D, E - U10 0.900 0.900 1.300 SW, D, W - U11 0.900 0.900 -1.300 SW, D, W - U12 0.900 0.900 1.430 SW, D, E - U13 0.900 0.900 -1.430 SW, D, E - 3 ACI 318-99, 9.2 APPENDIX \| 480 \| A.1.4. For CSA A23.3-14 For the CSA A23.3-14 code load combinations are compliant with 2015 NBCC. The default combinations of the Self-weight (SW), Dead (D), Live (L), Snow (S), Wind (W) and Earthquake (E) loads considered by the program are4: Ultimate Load Combinations – CSA A23.3 – 14 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.40 1.40 SW, D - U2 1.25 1.25 1.50 1.00 SW, D, L S U3 0.90 0.90 1.50 1.00 SW, D, L S U4 1.25 1.25 1.50 0.40 SW, D, L W U5 1.25 1.25 1.50 -0.40 SW, D, L W U6 0.90 0.90 1.50 0.40 SW, D, L W U7 0.90 0.90 1.50 -0.40 SW, D, L W U8 1.25 1.25 1.00 1.50 SW, D, S L U9 0.90 0.90 1.00 1.50 SW, D, S L U10 1.25 1.25 1.50 0.40 SW, D, S W U11 1.25 1.25 1.50 -0.40 SW, D, S W U12 0.90 0.90 1.50 0.40 SW, D, S W U13 0.90 0.90 1.50 -0.40 SW, D, S W U14 1.25 1.25 0.50 1.40 SW, D, W L U15 1.25 1.25 0.50 -1.40 SW, D, W L U16 1.25 1.25 0.50 1.40 SW, D, W S U17 1.25 1.25 0.50 -1.40 SW, D, W S U18 0.90 0.90 0.50 1.40 SW, D, W L U19 0.90 0.90 0.50 -1.40 SW, D, W L U20 0.90 0.90 0.50 1.40 SW, D, W S U21 0.90 0.90 0.50 -1.40 SW, D, W S U22 1.00 1.00 0.50 0.25 1.00 SW, D, E L, S U23 1.00 1.00 0.50 0.25 -1.00 SW, D, E L, S 4 CSA A23.3-14, 8.3.2, Annex C, Table C1 a); NBCC 2015, Table 4.1.3.2.-A APPENDIX \| 481 \| A.1.5. For CSA A23.3-04 For the CSA A23.3-04 code load combinations are compliant with 2005 NBCC. The default combinations of the Self-weight (SW), Dead (D), Live (L), Snow (S), Wind (W) and Earthquake (E) loads considered by the program are5: Ultimate Load Combinations – CSA A23.3 – 04 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.40 1.40 SW, D - U2 1.25 1.25 1.50 0.50 SW, D, L S U3 0.90 0.90 1.50 0.50 SW, D, L S U4 1.25 1.25 1.50 0.40 SW, D, L W U5 1.25 1.25 1.50 -0.40 SW, D, L W U6 0.90 0.90 1.50 0.40 SW, D, L W U7 0.90 0.90 1.50 -0.40 SW, D, L W U8 1.25 1.25 0.50 1.50 SW, D, S L U9 0.90 0.90 0.50 1.50 SW, D, S L U10 1.25 1.25 1.50 0.40 SW, D, S W U11 1.25 1.25 1.50 -0.40 SW, D, S W U12 0.90 0.90 1.50 0.40 SW, D, S W U13 0.90 0.90 1.50 -0.40 SW, D, S W U14 1.25 1.25 0.50 1.40 SW, D, W L U15 1.25 1.25 0.50 -1.40 SW, D, W L U16 1.25 1.25 0.50 1.40 SW, D, W S U17 1.25 1.25 0.50 -1.40 SW, D, W S U18 0.90 0.90 0.50 1.40 SW, D, W L U19 0.90 0.90 0.50 -1.40 SW, D, W L U20 0.90 0.90 0.50 1.40 SW, D, W S U21 0.90 0.90 0.50 -1.40 SW, D, W S U22 1.00 1.00 0.50 0.25 1.00 SW, D, E L, S U23 1.00 1.00 0.50 0.25 -1.00 SW, D, E L, S 5 CSA A23.3-04, 8.3.2, Annex C, Table C1 a); NBCC 2005, Table 4.1.3.2.-A APPENDIX \| 482 \| A.1.6. For CSA A23.3-94 For the CSA A23.3-94 code, the default combinations of the Self-weight (SW), Dead (D), Live (L), Wind (W), Earthquake (E), and Snow (S) loads considered by the program are6: Ultimate Load Combinations – CSA A23.3 – 94 Load Combo Self-weight SW Dead D Live L Snow S Wind W EQ E Principal Loads Companion Loads U1 1.25 1.25 SW, D - U2 1.25 1.25 1.50 1.50 SW, D, L S 1.25 1.25 1.50 1.50 SW, D, S L U3 0.85 0.85 1.50 1.50 SW, D, L S 0.85 0.85 1.50 1.50 SW, D, S L U4 1.25 1.25 1.05 1.05 1.05 SW, D, L, W S 1.25 1.25 1.05 1.05 1.05 SW, D, S, W L U5 1.25 1.25 1.05 1.05 -1.05 SW, D, L, W S 1.25 1.25 1.05 1.05 -1.05 SW, D, S, W L U6 0.85 0.85 1.05 1.05 1.05 SW, D, L, W S 0.85 0.85 1.05 1.05 1.05 SW, D, S, W L U7 0.85 0.85 1.05 1.05 -1.05 SW, D, L, W S 0.85 0.85 1.05 1.05 -1.05 SW, D, S, W L U8 1.25 1.25 1.50 SW, D, W - U9 1.25 1.25 -1.50 SW, D, W - U10 0.85 0.85 1.50 SW, D, W - U11 0.85 0.85 -1.50 SW, D, W - U12 1.00 1.00 1.00 SW, D, E - U13 1.00 1.00 -1.00 SW, D, E - U14 1.00 1.00 0.50 0.50 1.00 SW, D, L, E S 1.00 1.00 0.50 0.50 1.00 SW, D, S, E L U15 1.00 1.00 0.50 0.50 -1.00 SW, D, L, E S 1.00 1.00 0.50 0.50 -1.00 SW, D, S, E L 6 CSA A23.3-94, 8.3.2 (assuming occupancies other than storage and assembly); NBCC 1995, 4.1.3.2 APPENDIX \| 483 \| A.2. Conversion Factors – English to SI To convert from To Multiply by in. m (1,000 mm) 0.025400 ft m 0.304800 lb N (0.001 kN) 4.448222 kip (1,000 lbs) kN 4.448222 plf (lb/ft) N/m 14.593904 psi (lb/in.2) kPa 6.894757 ksi (kips/in.2) MPa 6.894757 psf (lb/ft2) N/m2 (Pa) 47.88026 pcf (lb/ft3) kg/m3 16.018460 ft-kips kN × m 1.355818 APPENDIX \| 484 \| A.3. Conversion Factors – SI to English To convert from To Multiply by m (1,000 mm) in. 39.37008 m ft 3.28084 N (0.001 kN) lb 0.224809 kN kip (1,000 lbs) 0.224809 kN/m plf (lb/ft) 68.52601 MPa psi (lb/in.2) 145.0377 MPa ksi (kips/in.2) 0.145038 kN/m2 (kPa) psf (lb/ft2) 20.88555 kg/m3 pcf (lb/ft3) 0.062428 kN × m ft-kips 0.737562 APPENDIX \| 485 \| A.4. Technical Resources APPENDIX \| 486 \| A.5. Contact Information Web: www.StructurePoint.org E-mail: info@StructurePoint.org support@StructurePoint.org licensing@StructurePoint.org Mailing Address: 1520 Artaius Pkwy #44 Libertyville, IL 60048 USA Phone: +1- 847- 966- 4357 Fax: +1- 847- 966- 1542
6268	https://ntrs.nasa.gov/api/citations/19980210404/downloads/19980210404.pdf	N AC_,-coJ lh'-..3 ., DYNAMICS OF VARIABLE MASS SYSTEMS _ _ " . CU+ _ -NASA/CR--I , ,__ 208246 NASA RESEARCH GRANT NO. NAG-2-4003 FINAL TECHNICAL REPORT Submitted by DEPARTMENT OF MECHANICAL AND AERONAUTICAL ENGINEERING UNIVERSITY OF CALIFORNIA ONE SHIELDS AVENUE DAVIS, CA 95616 Principal Investigator: Fidelis O. Eke Period: DECEMBER 1, 1993 TO MARCH 15, 1998 DYNAMICS OF VARIABLE MASS SYSTEMS ABSTRACT This report presents the results of an investigation of the effects of mass loss on the attitude behavior of spinning bodies in flight. The principal goal is to determine whether there arc circumstances under which the motion of variable mass systems can become unstable in the sense that their transverse angular velocities become unbounded. Obviously, results from a study if this kind would find immediate application in the aerospace field. " The first part of this study feann_s a complete and mathematically rigorous derivation of a set of equations that govern both the translational and rotational motions of general variable mass systems. The remainder of the study is then devoted to the application of the equations obtained to a systematic investigation of the effect of various mass loss scenarios on the dynamics of increasingly complex models of variable mass systems. It is found that mass loss can have a major impact on the dynamics of mechanical systems,includinga possible change inthe systems stability picture. Factorssuch as nozzle geometry, combustion chamber geometry, propellant's initial shape, size and relative mass, and propellant location can all have important influences on the system's dynamic behavior. The relative importanceof theseparameterson-system motion are quantified in a way that isusefulfordesignpurposes. ii ACKNOWLEDGMENT Although there is only one official author listed on the cover of this document, the material contained in this report is based on research work performed by several graduate students who were supported under this NASA grant. Hence, I wish to acknowledge the contributions of the following individuals: Dr. Song-Min Wang; Dr. Tai-Chien Mao, whose Ph.D dissertation work was completely supported by this grant; Mr. Enrique Cervantes, whose M.S thesis work was based on this topic and was supported by the grant; Mrs Asayo Harvey, whose M.S thesis was partially supported by this grant; and Mr. Jason Mudge, whose Ph.D research work was supported by this grant until the expiration of the grant. Each of these individuals made important contributions to the final outcome of this research project. I take this opportunity to express my immense gratitude to the NASA Technical Officer in charge of this project, Dr. Kajal Gupta of NASA Dryden, for his frequent and invaluable advice, support and inspiration throughout the duration of the project. His kindness has had a profound positive influence not only on the outcome of this work but on my whole career. I also wish to thank Mr. Stevcn Ycc, the Grants Officer, for his patience, understanding, and assistance with paperwork on several occasions. Finally, I am most grateful to NASA for the generous support of my research effort through this grant No. NAG-2-4003. iii TABLE OF CONTENTS Chapter 1 1.1 1.2 1.3 1.4 Abstract Acknowledgments Introduction Background Early Studies of Variable Mass systems The Star 48 Problem This Work ii iii 1 1 2 4 5 Chapter 2 2.1 2.2 2.3 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5 Chapter 3 3.1 3.2 Chapter 4 4.1 4.2 4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.4 4.4.1 4.4.2 4.4.3 Equations of Motion of Variable Mass Systems Model Description Equation Formulation Strategy Dynamical Equations Useful KinematicalQuantities Generalized Inertia Forces Generalized Active Forces Equations of Motion Another Form of the Equation of Attitude Motion Effects of Mass Variation Translational Motion Rotational Motion Attitude Motions of a Variable Mass Cylinder Introduction Attitude Equations for the Cylinder Attitude Stability The Uniform Burning Cylinder The End Burning Cylinder Radial Burn Centripetal Burn Tune dependence of Motion and inertia Properties End Burning Case Centrifugal Burn Centripetal Bum 7 7 9 13 14 18 21 23 28 31 31 34 39 39 40 43 47 48 52 60 64 65 68 71 iv Chapter 5 5.1 5.2 5.3 5.3.1 5.3.2 5.4 5.4.1 5.4.2 Chapter 6 Motion of Axisymmetric Bodies with Mass Loss Governing Equations Non-Dimensional Equations of Motion Centrifugal Bum Spin Rate Analysis Transverse Rate Analysis Uniform Burn Spin Rate Analysis Transverse Ram Analysis Forced Motion of Axisymmetric Systems 6.1 Equations of Attitude Motion 6.2 Change of the Independent Variable 6.3 Special Case - Rigid Body Motion Chapter 7 Conclusion References and Bibliography 74 75 80 87 92 94 101 102 104 105 110 113 116 118 CHAPTER 1 INTRODUCTION 1.1 Background Studies of free and forced motions of spinning rigid bodies of various geometries have been, and continue to be documented in great detail in the literature. Such studies have led to the development of important scientific instruments (gyroscopes, etc.) and to the concept of spin stabilization of modern spacecraft. By contrast, variable mass systems have receivedrelatively little attention intheliterature, though they play an equally importantroleinmodem technology, especially inspace flight. The expression"variable mass system,"asused inthecontextof this document, refers tomechanical systems that loseand/orgain mass while inmotion. Examples of such devicesabound inthe engineeringliterature. They includecomplex systems such as aircraft, rockets, and moving robotspickingup or letting go of objects, as well as simplersystems such aswater sprinkler systems or an inflated balloonwith airlossthroughone or more holes. Variablemass systems can be dividedintotwo classes: thosewith continuousmass variation and those with discrete mass variation. Rockets, for example, fall in the continuous variable mass class; and, robots picking up or releasing objects, or a moving vehicle dropping off some of its payload in discrete chunks, fall in the discrete variable mass system class. Because systems with discrete mass variation can be analyzed using well known principles of multi-rigid-body dynamics, thefocus of theanalysis presentedin thisstudyison systems with continuouslyvaryingmass. Q 2 It is clear fi'om intuition that when the net change in mass 0 a system, as well as its mass variation rate are small, it is unnecessary to account for the change in mass in the study of the system's motion. For example, automobiles arc in fact variable mass systems, yet, no one takes mass variation into account in handling and performance studies of ordinary automobiles. The reason is that the rate of mass variation is viewed as negligible and rightly so. On the other hand, a system that undergoes a substantial change in mass, especially if this occurs in a short period of time, will definitely require that mass variation be accounted for in the study of its motion; otherwise, any predicted response of the system will be far removed from its true behavior. The focus of this study is on the determination of the impact of mass variation on the dynamic behavior of systems with substantial change in mass .. 1.2 Early Studies of Variable Mass Systems Scientific study of variable mass systems has been in progress for more than two hundred years; so developments in this field have a long though sporadic history. The earliest recorded work on the dynamics of bodies with varying mass was performed in the 18th century by BemouUi (1738). He was then studying the forces acting on a liquid jet propelled hydroreactive ship-an ancient application of the principle of jet propulsion. He actually derived what may be referred to as the equation of motion in this special case. The Czech scientist and inventor, George von Buquoy (1781-1851), was the first to pose the general problem of the dynamics of systems with varying mass. _ 1812, he obtained his "motion formula" for such systems, and went on to solve a large number of examples based on his formula, von Buquoy's work can be said to mark the birth of the theory of the dynamics of systems with varying mass. In the mean time, William Moore worked out his mathematical theory of rocket motion in England in 1813, and, in 1819, Poisson rook a rathermodern approach and derived the equations of motion of variable mass systems based on Lagrange's general formula. In their book published in 1856, Talt and Steele included a section on mass variation. They postulated that mass variation produced small continuous impacts or impulsive forces on systems, and thus resulted in continuous change of velocity. This work was followed, several years later, by that of Meshchersldi, whose work spanned the period 1897 to 1904. He essentially laid the foundation for the development of variable mass dynamics as a special discipline of mechanics. He devoted his 160 page master's thesis to exploring a large array of issues relevant to variable mass dynamics m from the derivation of equations of motion to the solution of a series of problems in the field. All of these early investigations of variable mass systems were limited in one way. They were only concerned with the study of the translational motion of such systems. The issue of rotational motion of such systems was not addressed until the mid 1940's. The second world war brought with it a resurgence of interest and activity in the dynamics of variable mass systems, mostly in connection with rocketry. At this time, wanslational motion of such systems was relatively well understood and the main focus of research in variable mass dynamics began to shift to the attitude motion of such systems. Some of the scientific giants of this new era include Rosser et al. (1947), Gantmacher and Levin (1947), Rankin (1949), Ellis and McArthur (1959). The equations of rotational motion derived by these investigators are quite similar, and have forms similar to Euler's equations for rigid bodies, with extra terms that account for mass variability. Thomson stands out as a major contributor to this field through his book (1961) and the companion papers (1965, 1966). He derived several versions of the equations of motion of variable mass systems and his work gives a great deal of physical insight into the behavior of rockets. In his study of the transverse attitude motion of a non-spinning axisymmetric rocket, he showed that transverse rotation rate depends on the ratio43f the distance of the system mass center from nozzle exit, to the transverse radius of gyration of the rocket. If thisratiois greater than one (the most common case), the transverse angular velocity decreases with time; and when the ratio is less than one, the transverse angular velocity increases with time. Warner and Snyder (1968) brought some refinements to Thomson's work and pointed out how various simplifying assumptions can lead to drastically different motion predictions. Meirovitch (1970) moved work on variable mass systems one step further by considering the impact of mass variation on variable mass rockets. 4 1.3 The Star 48 Problem Flaws in current understanding of the dynamics of variable mass systems was brought to light in the early 80's, when several space missions with upper stages powered by the Star 48 solid rocket motor were observed to exhibit anomalous behavior. Unexpected and unexplainable rapid growth in cone angle occurred near the end of the motor bum. The output of a typical rate gyro mounted on one of such flights is shown in Fig. 1. We note from this figure that the flight is uneventful until about two thirds into the motor bum, when there begins an exponential growth in transverse rate, and thus in nutation angle. The Star 48 is the first solid rocket motor known to produce such anomaly, and it differs from its predecessors mainly in its much larger size and the existence of a submerged nozzle construction. The Star 48 problem sparked another flurry of investigations [ Eke (1983), Meyer (1983), Mingori and Yam (1986), Flandro et al. (1987), Cochran and Kang (1991)] into the behavior of variable mass systems, and is one of the main factors that motivated this work. In his Ph. D dissertation work, part of which was supported by this project, Wang (1993) [ see also Eke and Wang (1995)] modeled rocket type variable mass systems as a simple cylinder of varying mass, and produced elaborate closed-form expressions that describe the attitude motion of such systems for various burn geometries. The study showed that certain propellant bum scenarios can actually cause the transverse rates of short and large rocket systems to diverge in a manner similar to that observed on the Star 48 flights. 0 10 20 40 NE R 161¢11_ m¢ , 1 L.AI 'li llll 'II! ii ii i Fig. 1.1 Transverse Rate vs. Time for a Typical Star 48 Flight 1.4 This Work This study is an extension of Wang's work. Like Wang's work (1993), this study utilizes modern mathematical tools in the study of the dynamics of variable mass systems, with the general intent of making valuable contributions to the field and enhancing current understanding of the dynamic behavior of such systems. Mostinvestigators thathaveattempted thederivation of dynamical equations for variable mass systems have relied heavily on heuristics. In the next chapter, we present complete and mathematically rigorous derivations of both the translational and rotational equations of motion of variable mass systems. The remainder of the study applies the equations of chapter 2 to increasingly complex models of various classes of variable mass systems, extracting and presenting a wealth of new information on the attitude dynamics of such systems. Throughout this document, equations and figures are numbered in the form (a.b), where the first number a represents the chapter in which the item appears, and the second number b is the actual item number within the chapter. To refer to an item, the number b is used if the referencing is done in the same chapter in which the item appears; otherwise, the full label (a.b) is utilized. 6 7 CHAPTER 2 EQUATIONS OF MOTION OF VARIABLE MASS SYSTEMS This chapter begins with a description of the model used to characterize variable mass systems in this study. This model is general enough to represent a wide variety of physical systems that gain or lose mass while subjected to general three-dimensional motion. The complete equations of motion for both rotational and translational motion of such systems are derived using one of the methods of analytical dynamics - Kane's formalism (Kane and Levinson 1985). The merit of this approach is its efficiency. It produces the equations of transitional and rotational motion in one mathematically rigorous step, and makes it possible to clarify a lot of conceptual issues in the derivation, that have been very difficult to do in previous work. The equations of motion that are derived are then compared with those obtained by Wang (1993) and others, who used the Newton-Euler approach. 2.1 Model Description The system of interest is shown, in its most general form, in Fig. 1. It is determined by the closed surface, B, and its contents. The contents of B at any given instant can be solid (R), fluid (G), or a mixture of both. B and its contents undergo general three-dimensional motion in space, and matter can flow continuously in and out of B 8 during this motion. For example, parts of R can "dissolve" into G by combustion or other processes; and, some of such products of combustion can then flow across the boundary B. At any given instant of time, only the surface B and whatever happens to be inside it at the instant constitute "the system" for that instant. Thus, the system under consideration here evolves continuously, both as regards its location in space, and its material constitution - and hence its mass. We will use the symbol S to designate this system. N Fig. 2.1 VariableMass System 9 The derivation of equations of motion for a system such as the one described above is not as straightforward as it would normally be for a system of particles and/or rigid bodies of constant mass. The reason is that the basic principles of dynamics, such as Ncwton-Euler equations and Lagrange's equations are only valid when applied to a definite set of particles or rigid bodies. Two choices are then possible at this point. One, is to seek or develop new formalisms that would be valid for variable mass systems; another solution is to model variable mass systems in a way that allows them to be viewed as constant mass systems, and thus make them amenable to treatment by existing principles of dynamics. This latter approach is the choice adopted in this study. 2.2 Equation Formulation Strategy Before formulating the dynamical equations for the system of Fig. 1, we start by temporarily restricting the system in some important ways. First, we assume that the closed boundary, B, of the system maintains a constant shape, and thus encloses a region of constant volume at aU times. In other words, B is taken to be a rigid massless shell. As further help in the equation derivation process, we introduce the concept of constant mass systems associated with the variable mass system under study. We consider once more the system as shown in Fig. 1, keeping in mind that the outer shell is now of constant shape. At some instant of time, t1, there is a definite set of material particles inside B. Let us assume that this set of particles is contained in a closed elastic container, B1, that is identical to B at time tj. In fact, we take the viewpoint that BI has always enclosed the exact particles that ended up in B at time tl, and that B_ will continue to delimit these particles. Obviously, subsequent to time tj, the shape of B_ will deviate from that of B if it is to continue to delimit the particles that were in B at time t_, since some (or even all) of these particles may have exited B. Similarly, prior to time t_, the shape of B 1 was quite different from that of B, since only some, or maybe none, of the particles inside B1 then were also inside B. The shape of B_ is thus seen to vary with time, becoming identical to, and containing the same amount of matter as B at time h- We note however, that B_ and its contents maintain the same mass at all times. We shall represent 131 and its contents with the symbol S_, which will be referred to henceforth as the constant mass system associated with the variable mass system S at time h. Similarly, we can def'me $2, $3, etc., as constant mass systems associated with S at times h, tj, etc. Furthermore, we assume that there exists a special subset of the particles of S that remains within B throughout the interval of time of interest in this study. In fact, this set is further assumed to constitute a rigid body, R, that is rigidly connected to B. The equations of motion of any one of the constant mass systems described above can be formulated using any of the classical approaches (Newton's Second Law, Lagrange's Equations, Kane's Equations, etc.) since each system is of constant mass. Suppose, for example, that Kane's formalism is applied to the constant mass system _ and yields + = ..... nk) (2.1) k r k =0 (r 1 2, 10 where/7, is the generalized active force on the system, F, is the generalized inertia force, nk is the number of degrees of freedom of , and the subscript k simply indicates that we are dealing with the system . Assuming that the motion of the fluid particles of , relative to the rigid part of this constant mass system is known, _ has six degrees of freedom nk =6. Eqs. (1) have the explicit form 11 u,=fl, ql, /6, (2.2) uI ..... thi , r=1,2 .... ,6 where t is time, q)and u) are, respectively, the r - th generalized coordinate and generalized speed of the constant mass system ,,. In other words, the time derivative of a given generalized speed of ,, is a function of generalized speeds and generalized coordinates of S., and possibly time. Eqs. (2) can be supplemented with kinematical equations of the form .° q, = g, [ul ..... ut, ql, ,q_, r =1,2 ..... 6 (2.3) Eqs. (2) and (3) constitute the equations of motion of , and can be written as =/lr _Yl,'",Yl2, ,j r =1,2, .... 12 (2.4) where y,' is a generalized motion variable (generalized speed or generalized coordinate) of Next, we imagine that the set of differential equations (4) is solved for y: as functions of time. We note then that y_kl,. are the generalized motii_n variables for the variable mass system S at time t = t,. We now consider some quantity v e that characterizes the motion of k in some way, and is therefore obtainable from the generalized motion variables y_t, or is simply one of them. An example of a good candidate for _ is the magnitude of the velocity of the mass center of . Another example is a component of the angular velocity of the rigid body B that is a part of . We will call _ a characteristic motion variable for _k" Once y is known for all times, v k is known, and can be plotted as a " 12 function of time. Now, suppose we solve Eqs. (4) for various values of k, and thus determine Xt for several constant mass systems; and all of these V. t (k = 1, 2 ..... m) are plotted as functions of time on the same scale, but staggered for clarity as shown in Fig. 2. Now, we consider the motion characteristic, v, of the variable mass system S, that corresponds to v k. If, for example, v_ represents the speed of the mass center of , then v would represent the speed of the mass center of S, keeping in mind that different material particles make up S at different times. Because the dynamic behavior of S at some instant of time tj is, in fact, the dynamic behavior at time t] of ., it is clear that v is given as a function of time by the curve labeled F in Fig. 2; or, more precisely, the projection of F onto the plane gz-t (see v in Fig. 2). The task before us can be viewed as the determination of an efficient method for generating the differential equations whose solutions lead directly to the curve v of Fig. 2. The route to accomplishing this task will now be delineated. We consider once more Eqs. (4), which are the equations of motion for the constant mass system L,. First, we note that this equation has the same form for various values of k. The only items that change with k are system parameters such as mass and moments of inertia. Setting t = tk in Eqs. (4) yields a set of algebraic equations that produce y t/. (r=1,2,...,12); these quantities are equal respectively to y/tj), Y2(t,) .... y12(tj), the time derivatives of the corresponding generalized motion variables for S evaluated at time t = tk. One of these (or some function of these) represents the slope of cul"ve v of Fig. 2 at t = t and is plotted as point Pk in Fig. 3. The above process can be repeated for The system S t at time tj, the system _ at time t2, etc., and the relevant results are used to complete the plot of Fig. 3. The equations of curves such as P in Fig. 3 are the equations of motion of the variable mass system. Since the points of such curves axe generated from the equations of motion of the various constant mass systems, the equations of motion of the variable mass system S, have exactly the same form as the equations of motion of a typical constam mass system. However, to apply such equations correctly to variable mass systems, care must be taken to interpret mass and inertia parameters correctly. At any given instant of time, these parameters take on their values for the corresponding constant mass system for the instant under consideration. 13 Fig. 2.2 Characteristic Variables 2.3 Dynamical Equations To obtain the dynamical equations of the full variable mass system, S, all we need to do is to derive the dynamical equations for a typical constant mass system. We will now do this using Kane's equations given in Eq. (1). 14 p, Fig. 2.3 Graph ofy,(O versus time 2.3.1 Useful Kinematical Quantities In order to keep the mathematical developments that are going to follow relatively compact, and to avoid symbol definitions that are interspersed throughout the document, we give and define, in Table 1, a set of symbols that will be used repeatedly in the remainder of this document. A typical constant mass system, _, is shown in Fig. 4. A possible choice of generalized speeds, Ur, for the system is ¢a.br(r =1,2,3) u, /v .br_3(r 4, 5, 6) (2.5) Table 2.1 -List of Symbols A symbol in boldface type signifies a vector, and a boldface symbol with a tilde above it represents a dyadic. Several of the symbols defined here are shown in Fig. 4. 15 • N-an inertial reference frame • B - the boundary of the variable mass system • R - the solid portion of the variable mass system • S - the variable mass system enclosed by its boundary B • S - mass center of the system • O - an arbitrary point of R • P-a generic point of the system • v-velocity of P in an inertial reference frame • co - angular velocity of R in an inertial reference frame • a - acceleration of P in an inertial reference frame • Vr / _ - velocity / acceleration of P relative to R • v ° / a ° - velocity / acceleration of O in an inertial frame • vS / as - velocity/acceleration of S in an inertial frame • v_ / a_-velocity / acceleration of S relative to R • bi 0=1,2,3) - a dextral set of mutually perpendicular unit vectors fixed in R • r- position vector from O to P (see Fig. 3) • r - position vectcx from O to S (see Fig. 3) • a subscript, Ur, attached to any v or c0 indicates the corresponding partial velocity • p - position vector from S to P • I,-central inertia dyadic of the system 16 N b Fig. 2.4 Constant Mass System, By virtue of (5) above, ¢.0 = Ulbl + u2b2 + u3b3 (2.6) and v S = U4bl + usb2 + u663 (2.7) 17 The velocity of a generic point, P, of the system is v =v°+co×r+Vr (2.8) Because the motion of the fluid particles relative to R is assumed to be described by known functions of time, the partial velocities [see Kane and Levinson (1985)] of P are given by .o --V_r + VUr O)Ur × r (2.9) Now, thevelocity of the system mass center can be expressed as vS" = vO + cox r + vS" (2.10) and so, the corresponding partial velocities are S _ = v° r_ + tour x r VUr (2.11) From (11), v_, = Vu, - cou, x r" (2.12) 18 so that (9) and (12) give S ° VUr "- VUr + t.OUr× p (2.13) where p is the position vector from S to P. The acceleration of the generic point, P, of S is a=a°+ otx r+ cox(cox r)+ 2 oax Vr + ar (2.14) S_fl_y, aS" = aO + ¢x × r" + co x (c0 × r') + 2 o× vrS"+ a_" (2.15) (14) and (15) can be combined to give a=aS+otxp+t.ox(oxp)+2o3x(vr-vrS)+(ar-_ ) (2.16) 2.3.2 Generalized Inertia Forces The contribution of matter contained in an elementary parallelepiped located at P (see Fig. 4) to the generalized inertia forces of the system _ is 19 (Fr)e = -dm a-Vu,. (2.17) where dm is the mass of matter inside the paraUelepiped. (13), (16) and (17) yield = s (F;)p -dm[aS+o_xp+co×(mxp)+2cox(v,-v,.S)+(a,.-a; )].v_; S -dm[aS+(x×p+cox(coxp)+2mx(vr-vS)+(a,-ar. )]" (COu,.× p ) (2.18) Hence, the generalized inertia force on _ corresponding to the generalized speed, Ur, is F;=Vur. p aS + s s +axp+tex(texp) 2rex Vr-V r + ar-a r dV S X (0 X X + a r - a r -{our-p px +otxp+m p)+2m Vr-V r dV (2.19) We note here that the integrals in (19) are volume integrals that are taken over the region enclosed by 1_. We recall that I_ always contains all the material particles found inside B at time tk. I_ coincides with B at the instant t_, but is different from B at any other time. We now return tO the determination of the generalized inertia force, F;. From ., Fig. 3 and the definition of mass center, 20 fB pdm = fB (r-r)dm = 0 k k (2.20) Similarly, f_ _ s. k k (2.21) Using (20) and (21) together with the facts that f_[××] . k (2.22) and (2.23) (19) is reduced to × (2.24) Hence, for r = 1, 2, and 3, . 21 (2.25) and,for r = 4, 5, and 6, F;=-m vSu_.a S (2.26) 2.3.3 Generalized Active Forces The force per unit volume, F P, acting on the generic particle P of the system can be written as F/'= Fee + F/P (2.27) where F_ comes from forces external to the system, and F/t' is from the internal forces acting on P. The generalized active force, Fr, for the system has the form -22 + (2.28) Now, if the internal force exerted on any particle P of the system . by another particle Q of _ is assumed to act along the line connecting P and Q, then fB F_i dV=O (2.29) k and (2.30) In that case, the generalized active force can be written as s. F + C0Ur" M (2.31) Fr = Vu r where F is the resultant external force on the system, and M is the sum of the moments of all the external forces on the system about the system mass center. (5) and (31) then give 23 and F r = O)Ur. $ . Fr = Vur F M for r= 1,2,3 (2.32) for r = 4, 5, 6 (2.33) 2.3.4 Equations of Motion Recalling that Kane's equations have the form given in (1), we now combine (25) and (32) to obtain the equation of attitude motion as (2.34) Similarly, the equation of translational motion comes from (26) and (33), and is m a s = F (2.35) At this point, there are two main obstacles that prevent (34) and (35) from being useful as given above. First, the motion of fluid particles within I_ was assumed known. This means that both the velocity and acceleration fields within I_ are known as functions of time. In reality, none of these functions is known, but reasonable guesses can be made for the velocity field within B; that is, the velocity distribution inside ] for the instant t = t, when the constant mass system coincides with the variable mass system S. On the other hand, we have no handle over the velocity distribution of any particles outside B; and some such particles would normally he within ]k at instants of time different from tt. The other problem is that we have no way of estimating the acceleration field - not even within B. To circumvent these problems, two imlx_rtant measures are taken. One is that attempts are made to convert all accelerations that appear explicitly in (34) and (35) to time derivatives of velocity. If done properly, this would take care of the second problem above. The other measure is that ways are found to convert any volume integral that contains a velocity term, and is taken over the region 1_ to a volume integral over B, where velocities can be estimated. This measure would resolve the first problem. We now show how these measures can be applied to Eq. (35). Substituting for as from (15), (35)becomes In Io . ( ]-F a +axr +cox mxr" xv_+a_ (2.36) 24 It is convenient to express (36) as m ,. (2.37) We then re-write the last term on the left hand side of (37) as 25 Rdvr Rd LkP ardV = fBk P --" dV =-_fBk P VrdV (2.38) where the left superscript, R, indicates the reference frame in which the time derivative is taken. Next, we substitute (38) into (37) and evaluate the new expression at time tk, when -k is identical with S, and obtain .. m • a°+otxr " p v r dV = F(tk) +2. p(mxv )dV+ , t=tk t=tk (2.39) Because the last term on the left hand side of (39) contains a time derivative outside the integral sign, the limit of the integral cannot he changed to B as was done for the previous tenn. We are thus faced with the problem of taking the integral of an expression containing a velocity term over a region in which the velocity field is not known - the trn'st problem .. discussed above. Fortunately, this dilemma can be resolved by means of the Reynolds Transport Theorem, which gives in this case, i_I_d-tfB ] [RdfB fs ] P "rdV = -k p Yr dV + p Yr (V, " !11)d k t = t k t = t k (2.40) where S is the surface area of B, and n is a unit vector normal to B and pointing outwards. Eqs. (39) and (40) thus give 26 0('x')" l t=tk t=tk + p + (,,..) =E(, Jt = t k t= t k (2.41) where m is the mass of_ at ILrne t_; that is, the mass of S at time t_. Because (41) has the same form for all values of tk, the equation of translational motion for the full variable mass system is (2.42) The equation of rotational motion for the variable mass system is obtained in a similar way from (34). Ftrst, we have that (2.43) " 27 Then we invoke Reynolds Transport Theorem once again to convert (43) to t=t k ,o I I t= tlc (2.44) Thus, the equation of rotational motion becomes [from (34) & (44)] k (2.45) where the inertia dyadic that appears in (45) is that of matter that is within B at the instant under consideration. The vector equations of motion, (42) and (45), obtained through the application of Kane's equations, are identical to those obtained by other authors [ Meirovitch (1970), Wang (1993)] using Newton-Euler formulation. 2.3.5 Another form of the Equation of Attitude Motion 28 Another very useful form of the vector equation of rotational motion can be obtained as follows. First, we go back to the equation of attitude motion of a representative constant mass system as given in Eq. (34), and consider the expression Rd Rd A---ife k [p × (W x p)] din= ---_fBk p [p ×(× p)] dV (2.46) This expression can be expanded in the following way : A=- k[px(toxp)]dm= - [px(wxp)]dm k = fBk[vrX(t_x p)] dm + fB k [p x (txx p)] din+ fBk[PX(OX vr) ] dm (2.47) The last team on the fight hand side of (47) can be added and subtracted from (47), then by using the following equality Iv,x(co xp)]-[p x(co xvr)]=[co x(v,xp?] (2.48) we have 29 k k (2.49) On the other hand, A, as given in (47), can be evaluated at t = t_ and expanded using ,, Reynolds Transport Theorem to yield ( "' / A_, = _f,, [p × =× p)],,,,, + J',,, p × _= × p) _ t= tk '"(T.,,,) + J's" [p× °× p)] (v,.n),} It = tk [(Rail ¢o+i.ot+fspp×(mxp)aS } =[I-dT) "='k (2.50) We then have, from (49) and (50), that dV -((Rd]).ei+fsppx(o_xp) 1 (3.2) where and Fc=-2 fB P(mXvv)dV ed rL=- ?fB pv, aV FT=-fsPVr(V,.'n)dS (3.3) (3.4) (3.5) If v, is identically zero everywhere, then F c = FL = Fr = 0, and we recover the equations of translational motion for a rigid body. Thus, mass variation appears to augment the "external forces" on an "equivalent rigid body" by three terms. Fc is often referred to in the literature as the Coriolis force, since it derives from the Coriolis component of the acceleration. FL is the rate at which the system's linear momentum relative to B decreases with time because of particle motion inside B. FT represents the rate at which relative linear momentum is lost across the boundary B, and is often referred to as the thrust vector in rocket applications. If those particles of the system that can move relative to B are allowed to move within B but do not cross the boundary B, then FT becomes zero. F T would be non-zero but negligible if either a very small percentage of the system's particles is allowed to cross B, or those particles that cross B do so at very slow rate. In other words, the thrust vector can be neglected whenever the amount of matter that is lost or gained per unit time is small. What matters is the rate, not the total amount of matter lost or gained. Note that we can use Reynold's transport theorem to expand the Coriolis force into 33 where and prdV pv, dV=-2 mx =-2m _ prdV+ pr(vr.n)dS =Fcl+Fc2 Rd_ Fcl =-2 cox _ prdV Fc2 =- 2 ¢o x _ p r (Vr.n) dS (3.6) (3.7) (3.8) There arc situations where Yr Can be considered negligible inside the boundary B but not at an exit from B. An example is an inflated balloon with a hole. Gas motion inside the balloon is hardly noticeable while gas exit velocity at the hole is substantial. In cases like this, Ft. and the first term of F c (Fc_)are negligible, but Fc_ as well as the thrust vector survive. As a matter of fact, this is not an unreasonable assumption for praclical systems such as rockets, and can provide a way of rendering the equations of such systems tractable, since the details of internal gas flow can be neglected. Even if v, is not negligible "" 34 within B, but the particles within B can attain some type of steady state in their motion relative to B, then F L and Fcj are once more negligible. After ignition, rocket systems quickly attain an approximate steady state, and so, F L and Fcl can be ignored for such systems. In summary, the vector equation of translational motion for variable mass systems can, in many cases be reduced to m a°+ ot x r + m x (a) x r')] = F + FT +Fc2 (3.9) where F is the external force, F r is the thrust vector, and Fc2 is part of the Coriolis force and is given by (8). F c and F r can be dropped completely only when the rate of mass loss is negligible. 3.2 Rotational Motion Two forms of the vector equation of attitude motion were developed in chapter 2 [(2.45) and (2.52)1, and are reproduced below. -fBt (P x vr) dV + fs P (P x Vr)(Vr'n) dS = M + P (3.10) 35 (3.11) Both of these equations reduce to rigid body rotational equations if Vr = 0 within B as well as on B. Thus (10) can be written as I-a+texl-te=M+M c+MH+M T (3.12) where (3.13) is the Coriolis moment, Rd MH = --_fB p(p x vr) dV (3.14) represents the rate of decrease of the system's relative angular momentum inside B, and 36 M,--fsp(pxq(v,.) s (3.15) is the rate of loss of relative angular momentum across the bounda, and is also equal to the moment of the thrust vector about the system mass center. In those situations where it is reasonable to assume that the motion of the fluid phase relative to the solid phase has axial symmetry, and when there is no whirling motion, we have that fBk p(pxv,.)dV = Is p(pxv,,)dg-0 (3.16) This is so because, for every particle P of the system with position vector p and relative velocity Vr, there exists another particle P' of position vector p' and relative velocity V'r such that the vectors pXVr and p'Xv'rhave the same magnitude but oPtx_site directions. And, since axisymmetry also implies that the immediate neighborhoods of P and P' have the same mass density, Eq. (3.16) follows. This equation immediately leads to M. = M r = 0 (3.17) We arrive at the same conclusion for systems where Vr can be considered negligible inside the boundary B but not at an exit from B, such as rockets. M x is also negligible for steady state relative motion of the fluid particles. From (10) and (11) the Coriolis moment can be expanded to " 37 kp mx x (3.18) The last term on the right hand side of (18) vanishes for axisymmetric motion as well as for negligible internal flow. The second term on the right hand side of (18) is often referred to as the jet damping moment, because it has been shown [see Dryer (1963)] to have attenuating effects on the angular rates in some types of rocket systems. To evaluate this surface integral, it is necessary to know the geometric shape of those parts of the boundary where particles are allowed to exit or enter the system, as well as the velocity profile at these locations. This moment is thus very much dependent on the system's geometry. In the case of rockets, for example, the longitudinal dimension of the combustion chamber (for solid rockets) as well as the exit nozzle radius have much to do with the impact of this term on the system's attitude motion. The first term on the right hand side of (18) captures the contribution of inertia variation. We thus conclude that jet damping moment and the moment due to the changes in inertia properties have dominant effects on the attitude dynamics of the system. The relative importance and the interaction between these two moments, essentially determine the character of the attitude dynamics of variable mass systems. There is advantage in using the reduced form of (11) for the study of rotational motion. If the assumptions of symmetric and/or negligible internal motion are made, the most "troublesome" terms will have dropped out. The only term in this equation that would contain Vr is a surface integral over B. Vr'n is zero everywhere on the surface of B except at those places where fluid particles can enter or leave the region delimited by B (the nozzle exit plane in the case of 38 rockets). In general, Vr.n can be approximated relatively well at ..these locations; hence the surface integral can in fact be evaluated in closed form. In the study of the dynamic behavior of variable mass mechanical systems, the most important forces appear to be the thrust vector and the Coriolis force. The jet damping moment and the moment due to inertia variation are the dominant moments. These quantities should be included in any dynamic studies of variable mass systems to ensure that meaningful predictions can be made from the analysis. 39 CHAPTER 4 ATTITUDE MOTIONS OF A VARIABLE MASS CYLINDER 4.1 Introduction In this chapter, we begin to narrow the scope of our study to an important and useful class of variable mass systems - space rockets. We start this process with an in-depth study of the attitude motions of a variable mass system that is initially a solid right circular cylinder, which loses its mass continuously through combustion as it moves around in space. The cylinder problem is a very important and useful problem for several reasons. First, it models rocket-type systems in a simple enough way that the equations of motion become relatively tractable. Generally, space vehicles are designed to be more or less axisymmetric is shape, and, a cylinder is actually a good - albeit rough -approximation for such systems. Hence, a thorough study of the variable mass cylinder problem is a useful exercise in that it provides a means of performing tractable analytical studies of rocket-type systems, and can lead to great insight into the dynamic behavior of this class of variable mass systems. The idea of using a cylinder to study the a_tude behavior of variable mass systems was originated by Eke and Wang (1995). In their work, they solved completely in closed form, theequationsof motion of a variable mass cylinder forseveralburn scenarios.They then extracted useful qualitative as wcU as quantitative information about the attitude behaviorof thecylinderfi'omtheanalytical expressions of thesolutions of theequationsof motion. The approach here is different. The strategy here is to develop an analytical method that yields a wealth of fundamental information about the behavior of the system without actually solving the equations of motion. The main motiv..ation for this approach is that such a strategy can be applied to more complicated models of variable mass systems for which closed form solutions cannot be found. 40 4.2 Attitude Equations for the Cylinder Consider a variable mass system of the rocket type that is initially a right circular cylinder as shown in dotted lines in Fig. 1. This system is given an initial angular velocity 0%, and is then allowed to move freely in a torque-free environment as it loses its mass continuously through combustion or similar Imx:esses. The intent is to perform qualitative and quantitative studies of the attitude motions of such a system. To begin, we define as .. control region, the space enclosed in dotted lines in Fig. 1. Matter within this region at any given instant is considered part of the system at that instant. The attitude motions of such a system are governed by Eq. (2.52), which can be simplified, as discussed in chapter 3, by making the following assumptions: (a) the burning process proceeds in such a way that the solid unburned part of the cylinder remains symmetrical about the original cylinder axis at all times; (b) the fluid products of combustion move in an axisymmetric manner relative to the solid unburned part of the 41 cylinder;, (c) whirling motion of the products of combustion relative m the solid portion of the system is negligible. With these assumptions, (2.52) reduces to (/ [. C°x4/-n)_s M=I'a+_xI'+ t dt ]'°+ fBP x (4.1) . :,,:ii;i:i:i-::!S!7:!:!:!-!.". .................................... ;::".... "'-. ----i :::::::::::::::::::::::::::: _ ......................-..................... • / Fig. 4.1 Variable Mass Cylinder The axisymmetry assumption allows us to express the instantaneous central inertia dyadic of the system as "I = l(blbi +b2b2)+ Izb3b 3 (4.2) .° where bi, b2, I)3 are unit vectors fixed in B and directed as shown in Fig. 1; and I and I, are respectively the transverse and axial central inertia scalars of the system. If the inertial angular velocity of B at some general instant of time is taken to be ¢0= (.0xb1 + o_b2+ (.0zb3 (4.3) 42 then, the angular acceleration is IX= (oxb 1 + _oyb2 + co.zb3 (4.4) and the various terms of Eq. (1) become x i . m= (tz- t)a (a bl -(_t i ) " ca = i (o l + o_b 2) + Jz b 3 (4.5) (4.6) (4.7) In order to evaluate the last term of (1), an assumption must be made concerning the velocity field of the fluid particles of the system at the surface of the control region. Here, we assume that the vector Vr is zero everywhere on B except at one of the ends of the cylinder--the right end, say. And there, " vr "n = u(r) (4.8) because of our axisymmetry assumption. If it is further assumed that 43 then u(r) = u = constant (4.9) fB "o{P×(O_P)(Vr'n)] dS =-m (e2+--R2)(a_ z + b 2)+ 21--R2_b3 (4.10) where m is the instantaneous mass of the system, and l(t)is the distance of the mass center from the right face of the shell, B. By substituting (5), (6), (7) and (10) into (1) and setting M to zero, we obtain the scalar equations that describe the torque-free attitude motion of the variable mass cylinder;, and they are ,,,,(,, (4.11) (4.12) (4.13) 4.3 Attitude Stability Multiplying (11) by cox, (12) by (0y and adding the two resulting equations leads to I I &-(to2x + 0_y)+ i-6a (£ 2 + R 2/4) 2 + 0_y)=0 (4.14) We nowlet 0 = iz-ria R 2/2 (4.15) 44 = i- rh(l 2 + R 2 / 4) (4.16) and o)t = (o2x+ o_y (4.17) (13) and (14) can then be written more compactly as do)z I_---+ 0 _ = 0 and I d¢ + 20 ¢ot=O (4.18) (4.19) If the functions 0(t) and ¢(t) are known, (18) and (19) can serve as basis for the determination of the essential features of the rotational motion of the variable mass cylinder. We note that these equations are uncoupled, so that the spin rate has no effect on the wansvcrse angular rates. We also note that ff O(t) = O, then the spin rate remains constant at its initial value throughout. Similarly, if ¢(t) = 0, the magnitude of the transverse angular velocity does not change during the burn. Equations (18) and (19) lead to .--.o.xpE-I (4.20) 45 [ f 'dtl o_t= toexp -2 l(t) ] (4.21) or c0xy = o)xy 0 exp (4.22) where foxy = _ is the magnitude of the transverse angular velocity, and the subscript 0 indicates initial value; that is, value at ignition. We shall use the subscript .o to indicate values at burnout. Looking back at (15) and (16), we observe that each of the functions 0 and q is made up of two parts. For example, - 1 + _2 (4.23) where _1 = l (4.24) and = - th (t 2 + R 2 / 4) (4.25) Because we are concerned here with a situation where mass is being lost, both 1[? and rh are negative. Hence, _1 remains negative throughout the burn while _ is positive throughout. Thus, _ can be positive, negative or zero. _1 is contributed by the change in system transverse inertia, and because it is always negative, it will tend to cause ¢Oxy to diverge [see (22)]. On the other hand, 02 is, in this case, the so-called jet damping term, and it does in fact attenuate ¢0xy by virtue of the fact that it is always positive. Although 0t is determined by the bum geometry of the cylinder, 02 depends on the size of the cylinder as well as the velocity distribution across its exit plane. All the statements made above about O(t) apply equally to O(t) since (15) and (16) have the same form, and (20) is similar to (22). .. It is possible to extract important qualitative information about the attitude motions of the system from a detailed examination of the functions 0(t) and 0(t). From (20) and (22), it is clear that for 0(t) and (_t) > 0, both oh and oxy approach zero from any initial conditions. On the other hand, for 0(t) and (t) < 0, both oh and taxy diverge. Profound changes in angular velocity can occur when variations are made in 0(t) and (t). To make further progress with our study of the attitude hehavior of the variable mass cylinder, we will now proceed with a close study of the functions 0(t) and (t). These functions require that the system mass, inertia scalars and some geometric properties be known as functions of time; and these parameters can be determined if the burn scenario is known. We will therefore consider, as did Eke and Wang(1995), four burn scenarios : the uniform burn, the end burn, the radial or centrifugal burn, and the centripetal or anti-radial burn. In uniform burn, the assumption is that the cylinder is ignited simultaneously everywhere inside it at time to. It then burns in the same uniform manner everywhere, so that the external dimensions remain the same throughout, but the density of matter within the cylinder decreases uniformly in the same manner everywhere within the cylinder. The density is thus the same function of time everywhere inside the cylinder, and the products of combustion are expelled at one end of the cylinder. An end burning cylinder burns from one of its ends to the other, in such a way that the intermediate shape of the system is always a cylinder of the same radius but decreasing length. Once more, the products of combustion are expelled from one of the ends - the burning end. In radial burn, the axis of 46 ° 47 the cylinder is ignited at time to, and the system then burns radially outwards in such a way thattheintermediate shape isa cylindrical pipe. Counter-radial or centripetal burn isthe reverseof theradial burn. The cylinder isignited atits periphery(butnot theends) and burnsradially inwards,with theintermediate shape lacing a cylinder of constantlengthbut of decreasing radius. 4.3.1 The Uniform Burning Cylinder -I--i U Fig. 4.2 Cylinder in Uniform Burn We now considerthe case ofa cylinderinuniform burn with uniform velocity profile attheexitplane,as shown inFig.2 above. With thedimensions shown in the figure, the inertia scalars are, in this case, I= m (R2/4 + h2/3 ) (4.26) and • 48 nencc Iz = m R2/2 (4.27) (4.28) and From (15) and (29), we have that \|, = lil R2/2 0(t) = 0 (4.29) (4.30) at all times, and so, the spin rate, ok, will remain constant at its initial value. From (16) and (28), and recalling that t in (16) is the same as h in Fig. 2 for this case, O(t) = - 32-fia h 2 > 0 (4.31) Because ¢(0 > 0 at all times during the burn, the magnitude of the transverse angular velocity, arxy, decreases exponentially to zero. 4.3.2 The End Burning Cylinder Fig. 3 shows the dimensions and intermediate shape of a cylinder in end bum. The auxiliary spatial coordinate z is employed to characterize unburned propellant inertia properties, including mass, moments of inertia and the location of center of mass. The mass and inertia properties and their time derivatives are '" 49 m=2p_R2z (4.32) fia=2p_R22 (4.33) I = m (R2/4 ÷ z2/3) (4.34) (4.35) Iz = m R2/2 (4.36) iz = rh R2/2 (4.37) L h •----2 h-z---- .. U Fig. 4.3 Cylinder in End Burn Furthermore, 50 l = (2h-2z) +z =2h-z (4.38) so, from (16) and (38), (t) = rh ( R2/4 + z2 )-lh[ (2 h - z )2 + R2/4 ] = thh2[ 4(z/h ) -4] (4.39) At ignition t = to and z = h, so that = 0 (4.40) At burnout t = tand z = 0; hence ¢ =-4 rh h2>0 (4.41) From the mass continuity equation, we also have rh=-pgR2u<0 (4.42) Based on the assumption of uniform velocity profile at the exit pla,l.e, we conclude that th is constant dtLring the time interval [to, t]. We also know that r is negative in this interval. From (39), we have that dO/dt = 4 h th _ > 0 (4.43) 51 between ignition and burnout. Thus, ¢(t) increases monotonically from 0 to the positive value given in (41). We have a situation where q)(t) has the form shown in Fig. 4. ¢(t) _t Fig. 4.4 The shape of (t) for End Bum We conclude from (40), (41) and (43) that the transverse angular rate, c0xy, stays approximately constant during the initial phase of the burn, then decreases exponentially to zero. As for the spin rate, (15) and (37) indicate that 0(t) = 0 forall t, so that the spin rate is again constant. Having established that the transverse angular velocity of an end burning cylinder is always damped, our next task is to examine the effect of the cylinder's initial geometric configuration on the lateral response. To accomplish this task, let us define W = t) / I(t) (4.44) It is clear from (22) that this factor actually governs the speed with which the transverse angular speed approaches zero from any initial value. The larger the magnitude of _P, the higher the rate of decay of lateral motion. For end bum, " " 52 _p==fia(4hz-4h 2) I m (R2[4 + z2/3 ) (4.45) By virtue of (32) and (42), (45) can be rewritten as F= 2u(1 -Z/h) (4.46) z[1/4(R/h + 1/3(Z/h_] To compare the behavior of cylinders of the same length but different initial external radii, we allow R to vary while keeping z, h and u invariant. A close inspection of (46) indicates that increasing R/h results in smaller values for F. Hence, the transverse angular velocity magnitude, 0Jxy, converges more slowly as the ratio R/h is increased. In other words, the transverse motion of prolate cylinders in end bum is highly dahaped while that of oblate cylinder is only very slightly damped. In the limiting case, as R/h -- , F = 0, the motion is totally undamped as in the torque-free motion of a rigid cylinder. We conclude then that the amplitude of lateral oscillation of a variable mass cylinder in end bum can be influenced by the choice of its initial shape. Fig. 5 is obtained by numerical integration of (22) and demonstrates all the above properties. .. 4.3.3 Radial Burn We recall that in radial or centrifugal burn, the cylinder starts burning on its axis, and the burn propagates radially outwards with time. The intermediate configuration of unburned propellant is a hollow cylinder, as shown in Fig. 6, and the mass and inertia properties in terms of auxiliary spatial coordinate r have the form 53 1 0.9 0.8 0.2 0.1 0 0 R/h Fig. 4.5 Transverse Angular Rate for End Burn P CV [. h , h I-"I u .......... m .............. B ........... m ............. '!........ .° 0 Fig. 4.6 Cylinder in Radial Burn 54 m =20 hl_(R2-r 2 } (4.47) I = m [1/4(R2 + r2) + h2/3] (4.48) Iz= m (R2 + r2)/2 (4.49) Hence rh=-4phnrf (4.50) It is relatively easy to show that (4.51) and i7 =m r2 (4.52) To look at the behavior of the spin rate, we examine 0(t) = Iz-(rh / 2) R 2 = rh a2[ ( r/R) 2 -1/..2] (4.53) Assuming that at time t = 0, r - 0, we have 0(t) It = 0 = - rh R2/2 > 0 (4.54) 55 and at the end of the burn ( t --_ too, r = R), 0(t) ] t _ t = ril R2]2 <0 (4.55) From (53), we have that dO/dt = 2 fia r f < 0 (4.56) between ignition and burnout. Because r varies from 0 to R in the time interval [ 0, t] and from the physics of the problem, it is clear that f" > 0 in the given interval. We thus have a situation where 0(t) has the form shown in Fig. 7 o(t) .° Fig. 4.7 Shape of 0(t) for Radial Burn The switchover point, t, at which the sign of 0(t) changes from positive to negative is obtained by setting 0(t) = 0 in (53) : r/R = __ = 0.707 (4.57) 56 This implies that the spin rate decreases in the first stage of the bum, when r / R < 0.707, but increases exponentially in the later phase of the burn when r / R > 0.707; a fact that is clearly support_ by the plot shown in Fig. 8. This plot is obtained from numerical integration of (20) for centrifugal bum. 0.8 0.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1 r/R .? Fig. 4.8 Spin Rate Curve for Centrifugal Bum We now discuss the evolution of the magnitude of the transverse angular velocity, foxy. (t)=Z-rh(12+g2/4)=fnR2[(1/2)(r/R)2-(2/3)(h/R)2-1/4] (4.58) Hence (t)lt= 0 =4,0> 0 (4.59) 57 and ,..=th a211t4-(2 t3)(h/a ] (4.60) From (58), d/dt =rhrf <0 (4.61) The sign of 0- depends on what can be described as the cylinder's "shape factor"-the ratio of the initial value of its diameter to its length. Hence, whether xy is bounded or unbounded depends on this shape factor. To determine the value of R / h that separates the "stable" region of _xy from the "unstable" region, we set 0(t) to zero in (60), and obtain R / h - ¢KT-'J -- 1.63 (4.62) Relevant "pictures" for _t) are shown in Fig. 9. O(t) 1.63 I I _ t Fig. 4.9 Shape of (t) for Radial Bum 58 Hence, the magnitude of the u'ansversc angular velocity of a cylinder in radial burn always decreases with time in the early phases of the burn. If the ratio of the unburned cylinder's diameter to its length (shape factor) is less than that of 1.63, the decrease of xy continues all the way to burnout. On the other hand, if the cylinder's shape factor is greater than 1.63, then there comes a time during the burn when o)xy levels off, and then starts increasing exponentially through burnout. The initial shape of'the cylinder thus has a critical effect on the cylinder's lateral attitude motion for radial burn. Pencil-shaped cylinders will tend to be stable in radial burn while hamburger-shaped cylinders will tend to be unstable.We note here that thesame shape factor thatdeterminesstability or instability in radialbum was shown to influence the rateof convergence of t0xy for end-burning cylinders. Itwould be usefultodetermine theonsetofinstability of O_xyfor cases where R /h > 1.63. This can be done by determining the value of r/R at which the sign of 0(0 changes from positive to negativein(58).We thus have,from (58), that +I (4.63) YR[,=0 V3 R, 2 So, the onset of instability for O)xy depends again on the shape factor. Once a value is specified for the shape factor (R / h), the value of r / R at the beginning of instability can be determined. For example, for 2 I R 5 r 0.913 = 0.744 = o 0.707 (4.64) 59 Weconclude, from (64), that (Oxy begins to diverge much earlier for cylinders with higher values of the shape factor (fat and short cylinders) than for cylinders with lower values of R / h. Numerically obtained plots of the transverse angular velocity are shown in Fig. 10, and are totally consistent with the above inferences. Furthermore, the above analyses captttre a great deal more of the vital characteristics of radial burn than do these plots and all previous work on the cylinder. 0.4 0.2 0 0 unstable region Gable region 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 r/R Fig. 4.10 Transverse Angular Rate for Radial Burn 60 4.3.4 Centripetal Burn In centripetal bum, the cylinder bums radially inwards without changing its length. The intermediate shape is a solid cylinder of diminishing radius, as shown in Fig. 11. In this case, we have that rn = 2 p h _ r2 (4.65) and I_, = m r2 / 2 (4.66) I=m(r2/4+h2/3) (4.67) !!iiiiii!iiii !i_iiii!i!iiiiiiiiii!iiiiii!iii:iiil i!i i:i! il iiiii__ii!i!ii!iiiiiiii!ii ii!iiiii ___! !! !! !i !! !!!! !! !! ! Fig. 4. I 1 Cylinder in Centripetal Bum 61 The time derivatives are h=4phnrf (4.68) l, = rh rz (4.69) and i = rh ( r2/2 + h2/3 ) (4.70) For a study of the spin rate, we have e(t) : \|z-t1 / 2){th R 2) = rh (r2-R 2 / 2) (4.71) 00 = rh R2 / 2 < 0 (4.72) e. =-_h(R2/2)>0 (4.73) and from (7I), d0/dt = 2 rh r f > 0 (4.74) since f-< 0 in this case. We thus have a situation where 0(t)has the form shown in Fig. 12 62 0(t) .° Fig. 4.12 Shape of 0(t) for Centripetal Burn Hence the spin ram increases h-tidally, attains a peak value then de,creases rapidly with the burn. The time of attaining the peak value of the spin rate corresponds to 0(t) = 0. That is r/R = 1/-= 0.707 (4.75) This trend is validated by the plot of Fig. 13 below. 1.6 1.4 1.2 q 3 0.8 3 0.6 0.4 0.2 0 0 ! , ! ! ! ! , ii ii iiiii!!iiill iiiiiiiiiiiiiii iiiiiiiiil ..... ................................. .............................. ......... ............... ....... !i......... iii 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -r/R Fig. 4.13 Spin Rate of C_ntripetal Burn [Numerical Solution] To assess the evolution of the transverse angular velocity magnitude, we find _t) = I-rh(t 2+R2/4)=rn R2 [(1/ 2) (r / R) 2-(2/3)(h/R) 2-1/4] (4.76) 63 which is exactly the same as (58) for radial bum. ¢_0= rh a2[1/4-(2/3}(h/R) 2] (4.77) ¢p.=-rh R2[1/4 +(2 / 3)(h/R)2] > 0 (4.78) and from (76) d_/dt = ria r f > 0 (4.79) From (77), we determine that t_o = 0 when R / h = 1.63, _P0> 0 for R / h < 1.63, and _0 < 0 for R / h > 1.63. We conclude from all of this that (a) as long as the ratio of the cylinder's initial diameter to its length is less than 1.63, the transverse angular velocity magnitude decreases as the bum progress; (b) if the ratio R / h is greater than 1.63, the magnitude of the transverse angular velocity increases initially until it reaches a peak value, then it decreases with time for the reminder of the bum. Thus foxy never really becomes unbounded for centripetal bum. These facts are demonstrated below in Fig. 14. Our findings so far indicate that the stability of rotational motion of a variable mass cylinder depends on two factors : the bum pattern and the cylinder's shape factor. Both the spin rate and transverse angular velocity are bounded for all shape factors when the cylinder is subjected to uniform, end, or centripetal bum. For radial or centrifugal bum, the spin rate is always unbounded, while the transverse angular speed is bounded for the shape factor(R / h) that are less then \|.63, and unbounded for shape factors that are greater than 1.63. All the results obtained in this section arc totally consistent with, and augment considerably, earlier results by Eke and Wang(1995). 64 1.4, , i t i ! i !.... i ; ' l "'" , :............. ', •........... ,.............. i ........... ................................ ........................... ................. .... ............................. ...................... ° ol .............. i ................ ................ ...................... 0 0. I 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -r/P, Fig. 4.14 Transverse Angular Rate of Centripetal Bum [Numerical Results] 4.4 Time Dependence of Motion and Inertia Properties Up to this point, all variables and parameters of interest in this study have been def'med or determined as functions of the auxiliary variable z or r. This approach turned out to be a very astute and efficient analytical strategy. Through this choice, we have been able to avoid some of the myriad problems encountered by previous investigators of the dynamics of variable mass systems. For example, practicaUy all previous investigators worked directly with time as independent variable. Hence, to make progress with the equations of motion [ for example (11)-(13)], it was necessary to specify the time dependence of the mass (m), the location of center of mass (t) and moments of inertia ( I, Iz ). These investigators simply made assumptions such as linear mass variation -m = m0-a t, and made "reasonable" guesses for the coefficients. They made similar guesses for l, I and Iz. The problem is that these inertia quantities are interrelated; a fact that is generally not correctly reflected in the guesses for the independent time functions assigned to these quantities. All such difficulty is avoided by the clean analytical scheme of judiciously introducing the auxiliary variables z or r, as done in this study. Although avoiding the explicit use of time in our analysis so far has been quite beneficial from analytical point of view, it is in fact still desirable to know how the angular rates and even the inertia properties vary with time. For example, this enables us to compare analytically predicted trends with experimental results, obtained through on-board instrumentation. In this section, we show how some of the results obtained so far can be converted to time functions. 65 4.4.1 End Burning Case We now reexamine some of the results obtained for the end burning cylinder. From (33) and (42), we have that .. :=-u/2 (4.80) or 66 (4.81) which leads to z = - (u / 2) t + h (4.82) The inertia properties can now be converted to time functions simply by substituting (82) into their expressions. Thus, from (32) and (82) m = 2 p _ R2 z = 2 p x R2(h-lu t)= mo+fiat (4.83) where mo = 2 p _ R 2 h (4.84) and th is as given in (42). Similarly, I=m( R2/4 + z2/3)=(mo+fiat)[R2/4+( h-ut/2)2/3] =Io+Pt-Qt2+S t3 (4.85) where (4.86) P= rh(+ h 2) (4.87) Q = 2L- fia h u (4.88) 67 and S =1-_-2 rhu2 (4.89) The value of t,i. e. the burnout time, is obtained by setting z = 0 in (82) : L.=2h/u (4.90) which is independent of the radius of the cylinder. The instantaneous transverse central radius of gyration, k, is given by I=mk 2 (4.91) So, fi'om (85) R_ 2 2 k 2 + hZ_hu (4.92) Since (4.93) k2 = k_-h_t t +-t 2 (4.94) For the time interval 0 < t < 2 h / u, k 2 is a decreasing quadratic function of time. At burnout 68 k = R2/4 (4.95) The axial moment of inertia is Iz=m R2 = 2LR2(m o +fiat) (4.96) and the axial radius of gyration is a constant. Since fia and u am constants, we see that the transverse moment of inertia is a cubic function of time while both the axial inertia and the mass are linear functions of time for the end bum assumption. 4.4.2 Centrifugal Burn For centrifugal bum, we have, from (47), that rn =-2 p h _ d(r2) -dt p_R2u (4.97) Hence d(:) =R2u dt 2h (4.98) Integrating both sides over appropriate limits, we have r2=R2ut (4.99) 9h 69 Hence, m = 2 p h n( R 2-r2 )= 2 p h _ [R2-(R2 u t}/(2 h)] = mo + fiat (4.100) i=mlRZ+r2 +h21 4 3/ =Io+Bt+At 2 =(mo+fiat}(4R2+8R--_h t+h2 } 3 (4.101) where Io = mo (R2/4 + h2/3) (4.102) A = (fia R2 uY(8 h) (4.103) B= fia h2/3 (4.104) Since both A and B are negative, it is clear from (101) that I is a decreasing quadratic function of time. Note also that i=B+2At<0 (4.105) From (49) 70 +R2ut I - ml ÷r l- mo÷m0tR2 ) =Izo+2At 2 (4.106) and where IT=4At(£ 2 + R21/4) (COx bx + COyby)+ (R2/2)COzbz B where _e is the axial distance between F and the nozzle exit plane. By combining (1) and (13)-(16), we have that the scalar equations of attitude motion for the system are I rbx+( J-I )oyCOz+[ l-thf(l 2 + R2/4)]C0x=0 (5.17) _,-( J-_ )__ +[ i-mt( _: +_/4 )] .=0 (5.18) (5.19) where I and J are given by (9) and (6) respectively, and [see (9)] 80 i=if+_tc2+2gc_: (5.20) j = .if (5.21) 5.2 Non-Dimensional Equations of Motion In this section, we render the equations of attitude motion very compact through the introduction of a set of dimensionless parameters. This allows us to perform further studies of these equations in a way that provides great insight into the most important factors that govern the behavior of the system. We start by introducing the non-dimensional time. For the system of Fig. 1, the instantaneous mass.of the cylindrical propellant is mf = mr0 - m t (5.22) where = - rh (5.23) 81 andis takento beapositiveconstant thatrepresents therateof fuelconsumption.We assume thatt = 0 at ignition, when mf = mfo, and that at burnout, t = tb and mf = 0. Hence, (22) gives, at burnout, tb = mr0/_ . (5.24) We define a dimensionless time, x, as x=t= m t=ott (5.25) tb mf0 Thus, z= 0 at ignition, and x = 1 at burnout. Also, the quantity otl has the dimension of time and is an appropriate scaling factor for time. We have from (25) that = ot d (5.26) dx and the equations of motion, (17)-(19), can be re-written as -(J-I) 0_ oz-0t[I'-m' (_2 + R21/4)] Ox Iot (5.27) -82 r /A ,.; : (J- I)COx _-a Lf- m' (gz + R_/4j] oh ' (5.28) _= - [J'-m' R21/2] 0_ (5.29) J where the prime indicates derivative with respect to "1:, and the inertia properties I, J and mf are assumed to be expressed in terms of "t:. In the next phase of the non-dimensionalization process, we use mr0 and R as scale factors for mass and length respectively, so that the scaring factor for inertia scalars becomes mm R 2. It is also convenient to introduce, at this point, the following non-dimensional parameters : ]tl = ks / R, _'2 = ksz / R, _t = Ms / mo, 8 = L / R t 1_=R1/a, _1 =gl/R, _ = 2/R / (5.30) where ks and k,z are, respectively the transverse and axial radius of gyration of S. We now divide the numerator and denominator of the right hand side of (27) by mm R 2 to obtain _ ____/_ I + O_= mr°R2 m_R2taY°z-mr°R2 mr°R2 I 0_ mfo R 2 (5.31) 83 If both sides of (31) are divided by or, we get _=-(_-l)o._,COz -{i'-mf°L_R! +-4"]} _t (5.32) Similarly, (5.33) and ( _' m' -' -meo T _z %= (5.34) 84 where COx,Oy, O)zare non-dimensionalized angular velocity components, and I, J are ,, normalized inertia scalars. Following a strategy similar to that utilized in chapter 3, we let (5.35) and o(_)= F_-m-_2 mfo 2 (5.36) m E Next, we multiply (32) by COx and (33) by % and add : _,r (5.37) (34) becomes d_z=-_. _ (5.38) 85 (37) and (38) lead to n i cot = O_toexp (5.39) and (fo I J (5.40) where As in the case of the cylinder, the sign of the functions _ and 0 determine whether the angular rates grow or decay with time. A positive sign implies decay and a negative sign signals divergence. From Fig. 1 mr0 = Pro _ R2 L (5.42) 86 where Pro is the mass density ofF. Continuity equation gives _0 R1 rh = fiat =-_ = 2 x O u(r) r dr (5.43) We recall here that p is the density of the fluid products of combustion at the exit plane. For a uniform velocity profile at the exit plane, u(t) is constant, so that "I ria = rhf=-_ =-p 7t R2t u (5.44) We then have an alternate expression for ot as (5.45) where rl is the ratio p / Of0. Note also that R 11 (5.46) " 87 To make further progress with our analysis, we will now focus attention on two specific burn scenarios for F that are closest to what occurs in real rockets. 5.3 Centrifugal Burn The fuel burns radially outwards in a symmetric manner (see Fig. 2) so that F remains fixed in S; however, B still moves toward S. From Fig. 2, mf=pfxL(R2-r 2) (5.47) '_2 C Fig. 5.2 Centrifugal Burn 88 From (47) and (44) dmf dt __=_pfxLd(r2) dt =-pxR_u (5.48) Hence fo r2 _o t d(r2)=_ dt (5.49) and r2=_R12ut=_R12u x L Lot (5.50) From(50)and(45) 2 x2- = rl =rex =x R 2 --_-X 0_ (5.51) _____L =Pf _L(R2_ :) mr0 = 1--rL I-_ pf _r L R 2 R 2 = (5.52) 89 and so, moR 2 rno[ 4R 2 12R2J • + 4 12 12 4 (5.53) fromwhich we have 12 2 (5.54) Similarly, )-f=Jf =_(R2+,.2) mnR2 mO--2R2 -=('l-'r)-_ =_ (5.55) and (5.56) 90 Because (see Fig. 2) c=12+LI2 (5.57) we have C'=0 (5.58) and 2gcc'=0 (5.59) From (10) and (57) I.tc 2 _(mf/mo)(Ms/mo)(t2+L/2 _ (1-x)_ /_ , 2 moR 2 mftmo+Ms/mo" R -(fL-xx+_)_°2"2 _) (5.6O) so that I,I,' c 2 _r2 ( _)2 mo R2-(I-x+V) 2 _+ (5.61) 91 (20) yields .... g'c 2 2 _t'c c' I =If+ _---moR 2 mo R2 (5.62) We thus have, from (62), (54), (61) and (59) that ( i' =_8"_t_ __ 82+ 12 2 (V+I ,02 (5.63) Similarly, (21) and (56) give JB _l J =-Jf =-x (5.64) Now (see Fig. 2), R R (5.65) and (52) leads to " 92 m (5.66) Hence, (5.67) and (35), (63) and (67) yield (5.68) In the same way, (36), (64) and (66) give (5.69) 5.3.1 Spin Rate Analysis From (69), the function 0(x) is a linear function of "1:; it has a slope of-1 and is _/ _/ positive and equal to 2 at q: = 0, and is equal to 2 - 1 at "t: = 1. Fig. 3 shows O for various values of 13. 93 0 I 1 Fig. 5.3 The function 0 for centrifugal burn We conclude from the figure that the spin rate always decreases at the beginning of the burn. If 13>- ":2", the spin rate continues to decrease all the way to burnout. For 13< _2, the spin rate bottoms out during the burn, and then increases during the later phases of the burn. The turning point for this latter case occurs at o +132/ =-2=0 (5.69) That is, at x = 132/2 (5.70) 94 This result is consistent with, and augments that obtained for the cylinder, for which 13 = 1 [ see (4.57) and Fig. 4.8]. The same result is validated by Fig. 4, obtained by numerical integration of (40). -0.0 0.2 0.4 0.6 0.8 T (norm\|ized time) .0 ¥=2 ct = 0.01 8= 10 6t=2 "t = 1.2 y2=l Fig. 5.4 Centrifugal burn" the effect of expansion ratio I on spin rate 5.3.2 Transverse Rate Analysis From (68), the slope of the function ¢('t) is 0' = 1 2V2 (82+ --2 OOrg 1--g)3 "-2) 2 (5.71) and remains negative throughout the burn. 95 At ignition ( "1: = 0 ), T) V2 ¢o= 82+ 6f+8,8+ (82 + 2)2 (/+ 1)2 (5.72) and atburnout ( "I: = 1 ), 1 82 +82+6 51 5 + --21" --(52 + 2) (5.73) It is clear from (72) and (73) that % > (h, and that it is possible for any of these quantities t to be negative or positive. These facts, coupled with the fact that 0 is negative, mean that the three scenarios depicted in Fig. 5 ale possible, depending on the values assigned to the parameters. ° II1 Fig. 5.5 The function _ for centrifugal burn 96 We conclude from Fig. 5thatthemagnitude of the transverse angular velocity for centrifugal bum can (i) decrease from ignition to burnout, (ii) decrease initially, then diverge later, or (iii) increase from ignition to burnout, depending on system parameters. Factors such as the distance of the exit plane from the combustion chamber (51), shape .. factor of the propellant grain (5), location of the propellant grain (82), nozzle expansion ratio ([5), and the relative amount of propellant in the system (), all affect the transverse angular velocity. The main difference between the general axisymmetric system being considered here and the variable mass cylinder considered in chapter 4 is that case (iii) in Fig. 5 does not exist for cylinders. In other words, the transverse angular velocity of a variable mass cylinder always decreases initially, while a general axisymmewic system can exhibit divergent transverse rate throughout the propellant bum. Fig. 6 shows the transverse angular velocity magnitude as obtained by numerical integration of (39). D m-= I0 o_ = 0.01 5t=3 TI = 1.2 y2=l Fig. 5.6 Centrifugal bum : the effect of various 5 on cross-spin 97 5.4 Uniform Burn We recall that in uniform bum, the dimensions of F remain unchanged, while its density decreases continuously with time - see Fig. 7. In this case., mf has the same expression as mr0 in (42). This expression and (44) lead to ! 12 -iiiiii!iii!i!iiiiiiiii iiiiiiiiiiiiiii!iiiiiii iiiiiiiiiiiiiiiiiiiiiiii_iiiii_i!iii!iiiiiiiiiiiiiiiiiii_i_i_ii_i_i!!!ii ! "1 U Fig. 5.7 Uniform Bum dm(=l_fnLR2=-& Pn R12u (5.74) Hence :)f 2 t f0 d(pf)=-p R2LJo dt (5.75) 98 so R12ut Pf= Pro-P R2L (5.76) From (74) and (76) mf= Pfo R 2 L-p x R 2 u t (5.77) rne = peon R2L-p _ R_ut mfo Pro _ L R2 = 1--(zt= 1-x (5.78) and so, m,, R 2 mo [.4 R 2 12 R2J 12 (5.79) from which we have (5.80) f Jf mr R2 -1-x mo R 2 mo 2 R 2 2 (5.81) 99 and j =1. 2 (5.82) As before (see Fig. 7), c =g2+L/2 (5.83) c'=O (5.84) and 2p, cc'=0 (5.85) From (10) and (83) _tc2 (mf/mo)(IVls/mo). (g2+L/2) 2 (l-f) {82+ mo R2 = mf/mo + Ms/mo R -- (1"--+q/) )2 (5.86) and t'c2 V2 ( 2_)2 m,,R-----2 = (I_,+V) 2 _2+ (5.87) We thus have, from (62), (80), (85) and (87) that .? 100 (V"+ 1 - x)2 (5.88) Similarly, (21) and (82) give --i --t J =-Jf =-1/2 (5.89) Now (see Fig. 7), and (78) leads to Hence, ge = 'J" + 9.-R = _ R R +" 7 m' mid = m' -=-1 (5.9O) (5.91) (5.92) and (35), (88) and (92) yield (')=q52+_2+_518+6 -4 I" (+i-,t)2(82+-8-2-2} 2 (5.93) In the same way, (36), (89)and(91)give I01 (5.94) 5.4.1 Spin Rate Analysis From (94), the function 0(x) is a constant. Fig. 8 shows O for various values of O 13>1 13=1 13<1 Fig. 5.8 The function 0 for uniform burn If 13> 1, the spin continues to decrease all the way to burnout. For 13= 1, the spin rate always equals its initial value. For 13< 1, the spin rate increases from the beginning to the end of the bum. The same result is validated by Fig. 9, obtained by numerical integration of (40) tD u'1 oq (:3 -0.0 / 0.7 0.4 0.6 0,8 1.0 T (norm_t_ized lime) ¥=2 Ot= 0.01 8= 10 8t=2 82=3 Yt = 1.2 y2=l Fig. 5.9 Uniform Bum" the effect of expansion ratio 13 on spin rate 102 5.4.2 Transverse Rate Analysis From (93), the slope of the function ¢(x) is , 2_ 2 (2 + _2)2 ¢ =-(v + 3 (5.95) and stays negative throughout the bum. At ignition ( '1; = 0 ), (5.96) I03 and at burnout ( "t: = 1 ), (5.97) Hence _b0> _l, always, and either of these quantities can be negative or positive. Since we t also have that _ is always negative, we have the same scenario obtained for centrifugal bum. Numerical simulation results provided in Fig. lO below confmn these inferences. o LO Lo .-2-0 _xyO o o" -0.0 / 20 A / / / I[ 0.2 0.4 0.6 0.8 L.O T ¥--15 B"I = 0.01 Si=3 Yt = 1.2 72=1 Fig. 5.10 Uniform Bum" the effect of 8 on cross-spin 104 CHAPTER 6 FORCED MOTION OF AXISYMMETRIC SYSTEMS All the development of the last two chapters have been restricted to systems subjected to zero external torque. The only forcing function present has been a perfectly axial thrust, whose line of action passes through the system mass center at all times, and thus produces no couple on the system. Great pains are taken in the design of real rockets to minimize thrust misalignments. Yet, it is unrealistic to expect that the sys_m mass center, which generally moves during rocket burn, will remain positioned on the line of action of the resultant thrust vector throughout the propellant bum. Thus, torques due to .? thrust misalignment are often inevitable. There are also other potential sources of extranexms torques on a rocket system in flight, including aerodynamic and gravitational moments. For this reason, we will, in this chapter, develop relationships that describe the effects of externally applied moments on the attitude behavior of the same axisymmetric variable mass system that was studied in the previous chapter. 6.1 Equations of Attitude Motion 105 The equations of forced attitude motion of the system of Fig. 5.1 are really the same as (5.17)-(5.19), except that the right hand side of each of these equation is non-zero. Thus, we have I ¢bx+( J-I ) toy ¢X_z +[ I-tilf( 2 + R12/4)] 0)x__ Mx (6.1) I ¢by-( J-I ) COx ¢Oz+[ t-rhf( l 2 + R2/4 )] o -- My .? J tbz + ( Jr- lilf R2/2 ) t_ = Mz (6.2) (6.3) The forcing functions Mx, My and Mz are, in general, functions of time. This is so because at least the contributions from thrust misalignments vary with time since the system mass center moves relative to the system base structure. We start the solution process by letting _,2 = {Jr- rhfR12/2) / J (6.4) and Mz = Mz / J (6.5) " 106 (3) then becomes dct_z + X2(t) c0z = Mz (6.6) dt The general solution of (6) is (6.7) This solution can be thought of as the sum of two components : a first term that represents the homogenous part due to initial condition, and a second term - the particular solution -that is contributed by the forcing function. If J, jr, rhf and Mz are known functions of time, and the initial spin rate is known, (7) determines uniquely the spin rate ¢x_z of the body S. Note that ¢Ozdepends only on 2L2and Mz, and is independent of Mx and My. This is true because of our assumption of axisymmetry. We are now ready to investigate the transverse angular velocity components of the system. To this end, we introduce the following complex quantities" ¢.Oxy = COx + j oh, (6.8) Mxy = Mx + j My (6.9) 107 where We also define j = (6.10) )'.l : [i-tiaf(t2 + R2/4)]/I (6.11) o=(J-I)/I (6.12) and Mxy = Mxy / I (6.13) Adding j ILrnes (2) to (1), we obtain I ¢J_xy- j (J - I) 0.z(j ¢0y + ¢.0x}+ ),.1 ¢t_xy= Mxy (6.14) Or .. 108 d_xy dt + P2(t) tOxy = Mxy (6.15) where P2(t) = Xl-o (J)z (6.16) The solution to (16) is (6.17) or equivalently, cox + j toy = {O_o + j toyo + _ot [Mx(s)/I +j My (s)/I] {ex_F (s)]} [cosO(s)-j sinO(s)] ds} {exp [- F (s)]} [cosO(t)+ j sine(t}] (6.18) where (6.19) and . 109 (6.20) The functionsI_t) and O(t), which have fundarncntally different pro_rfics,govern the character ofthe transverse motion of thevariable mass system under consideration. _t) determinesthe amplitudeof theangularmotion and O(t)affects motion frequency. Explicit formula for COxand COycan be obtainedfrom 08) by separating therealand imaginary parts of theexpression; they arc: cox(t) = e- r (t)[ox o cosO(t)-coyo sinO(t)] + (6.21) f e[r(')-r (t)]{M, (s) cos_O(s)-O(t)] + My (s)silO(s)-O(t)]} ds I and t)= e-r (t)[_osinO(O +_o co,O(t)] + e[r(s)-r (t)]{My (s)codO(s)-O(t)]-Mx Is)silO(s)-O(t} (is I (6.22) We observe that in general the spin torque, Mz has no effect on the transverse angular velocity, nor is the spin rate affected by any of the transverse torques. 110 6.2 Change of the Independent Variable We now introduce a new independent variable in order to simplify somewhat the equations of attitude motion and their solution. We exploit here, a transformation defined earlier and first proposed by Jarmolow (1957) : e(t)= _ d_= COz d¢ (6.23) 0, has the dimension of angle (radians) and is related to the spin angle. It will be used as the new measure of time. Note that -? dO = Ct_z { dt (6.24) SO d =¢.Oz (J" - I) d (6.25) dt I dO The starred quantities are assumed to be expressed in terms of the new variable O. This is indeed possible because (23) gives 0 as a function of t, and can be inverted to yield the " 111 expression oft as a function of 0. Equations (1) and (2) can be rc-writlvn in terms of O as follows : -.t_ 2 + O)z COx = l_x (6.26) dO ¢Oz(J-I)dcth'-(dO J-I )COx¢Oz+[dIkdO dm_d_O (t:2+ R21/4)] (J- I) o_z _ = l_y (6.27) Dividing through by (J - I) (Oz, one obtains [. . /4)] d__+coy + dJS__dmf (te2 + R 2 (tk _ 1_ X de Lde de I" 0)z (J - I') (6.28) and ¢t_t+ R21/4 COy l_y dO LdO dO I (t_z(J - I) (6.29) Then, multiplying (29) by j and adding to (28), we have dO (6.30) 112 wh¢l'c ,(o)=,(o) +j_o) (6.31) "" I/i _ =[di-__(: +R_/ kdO dO (6.32) and (6.33) if w d_o:(o)as (6.34) then the solution of (30)for COxand COyin_rms of thenew variable 9 can be written explicitly as COx (0)= (o cosO-: sinO)e -r (0) .+ e [r" (s)- r" (o)] _z'-?i [()co__o)+g(s) si_s-o)] e (6.35) 113 , (o)= [,_o ,_o + co,o] e r" (o) + fo° c[l"{s!-I'_{O)][I_fy(s)c°s(s-O)-l_x(s)sin(s-O)]dsc0z(J -I) (6.36) In order to obtain explicit expressions for the known functions such as I, J, Mx and My, the function t(O) must be determined. This may or may not be a trivial task. The integrand of (23) is a rational function of time whose integral may not have a closed form analytical expression. However, a numerical solution is always possible, so we can be sure that the complete solution sketched above can always be obtained by numerical means - at least. 6.3 Special Case -Rigid Body Motion The above solution can be used to investigate the altitude motion of an axisymmetric rigid body simply by setting (32)and (34)then yield dI = d.r = dmf = 0 (6.37) dO dO dO F(O)=0 ' (6.38) and (35) and (36) simplifyas follows 114 cox (0) =(¢oxo ¢osO-o._o sinO) + .. 1 [l_x (s)cos(s-O)+ l_y (s)sin{s-0)] ds ¢Oz(J" - I') (6.39) coy (0) = [¢OxosinO + ¢OyocosO] + I; O)z(J - I) (6.40) In this form, the physical significance of the various parameters of the problem is not so .? obvious. Simpler and more revealing expressions for COxand o. b, are obtained for the case of constant body-fixed torques. For instance, ff we take Mx and My to be constants, and Mz = 0, (39) and (40) further simplify to ¢os(o)-,,,yo sin(O)] + 1 {l_x sin{O)-l_y[1-cos(O)]} (J" - I)¢OzO (6.41) y(O) ;[¢Oyo cos(O) +¢Oxo sin(O)] + 1 {l_y sin(O) + l_x[1-cos{O)]. (J - I) ¢OzO (6.42) where 115 0 (t)= - )zO t (6.43) We noteherethatthespinrate remains constant at its initial value, as can be seen from (4), (5) and (7). (41) and (42) indicate that transverse velocities increase with u'ansverse torques and decrease with the inertia difference, J - I, and initial spin rate. Spinning a rigid body about one of its principal axes provides the so-called spin rigidity or gyroscopic stiffness; that is, a resistance to external disturbances during motion. This is the well known advantage of spin stabilization. In a spin stabilized system, the transverse angular rates COxand cA/ axe usually the result of unforeseen disturbance, and are generally undesirable. If Nix and My are viewed as constants, body-fixed disturbance torques in (41) and (42) above, then, these equations show clearly that high spin rates will reduce the effects of the disturbance torques on vehicle motion. This indicates that the nominal spin rate of a spin stabilized vehicle should be given the highest value compatible with the vehicle's structural integrity. A fact that agrees with physical intuition. It can thus be said that as the spin rate increases, the stability of a given system increases as measured by the reduction in angular rates introduced by a given disturbance. For a single spinning body, (41) and (42) also show that gyroscopic stiffness depends in a crucial way on the system inertia property, J - I, or the difference between the spin and the transverse moment of inertia of the body. Spin rigidity vanishes as the inertia difference tendstozero.Inotherwords, bodies with spherical symmetry can have no gyroscopicstiffness, irrespective of thevaluethe spinrate.On the otherhand, heavily prolate or oblatebodies arcseen tobe quitestable inspin. 116 CHAPTER 7 CONCLUSION The study presented in this report deals with the dynamic behavior of spinning bodies that lose mass while in motion. The study began with the determination of the complete equations of such systems using one of the most powerful methods of analytical dynamics m Kane's formalism. The resulting equations are then applied to the study of the attitude dynamics of space-based variable mass systems using mathematical models that vary from a simple cylinder to general axisymmetric systems. The emphasis of these efforts was on information extraction. Special mathematical techniques were used to extract as much information as possible about the motion of the system, without actually attempting a full solution of the equations of motion. .: In the first example presented, a variable mass rocket system is represented by a simple cylinder with four possible mass loss scenarios. Results obtained indicate that such a system can fly stably in a torque free environment if the ratio of the radius m the length is small. Large and short cylindrical vehicles can have stability problems, especially if they are subjected to radial bum. This study goes far beyond previous work in this area by clearly identifying the exact inertia conditions and the precise stages of vehicle motion at which the character of motion begins to change. For cases where there is stability issue, limiting values of motion parameters are determined. Attention was next mined to a slightly more complicated model --a general axisymmetric variable mass system. Although important differences were found in the details of the motion of the general axisymmetric system as compared to that of the simple cylinder, the main stability results did not change. It was found in this second example, that in addition to the shape factor of the system propellant, the location of the propellant grain within the system as well as the nozzle expansion ratio all have significant effects on the attitude motion of a rocket. In the last section of this study, the combined effect of mass loss and external transverse torques on system motion is examined, and solutions to the equations of motion in this case are presented. Overall, this study augments considerably, previous knowledge of the dynamics of variable mass systems and gives great insight into the effects of mass variation on the attitude behavior of spinning bodies. 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6270	https://physics.stackexchange.com/questions/451014/photon-energy-comes-in-packets	quantum mechanics - Photon energy comes in packets - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Photon energy comes in packets Ask Question Asked 6 years, 9 months ago Modified6 years, 9 months ago Viewed 970 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. From the HyperPhysics page on the Photoelectric Effect: According to the Planck hypothesis, all electromagnetic radiation is quantized and occurs in finite "bundles" of energy which we call photons. The quantum of energy for a photon is not Planck's constant h h itself, but the product of h h and the frequency. The quantization implies that a photon of blue light of given frequency or wavelength will always have the same size quantum of energy. I still don't understand how does a photon look like or how does light/EM waves are quantized. If in some unbounded system (not in a box, but vacuum with no boundaries), I have "light" and nothing else, and the energy of the system is E 0 E 0, How do I know how many photons exist in that system? One way to look at it is: E 0=ℏ ω 0 E 0=ℏ ω 0 it would mean there is one photon with energy of ℏ ω 0 ℏ ω 0. However, one could also see it as E 0=2 ℏ ω 1 E 0=2 ℏ ω 1 where ω 1=ω 0/2 ω 1=ω 0/2. And it would mean that there are two photons with frequency ω 1 ω 1. In other words, how can energy be quantized/discrete when frequency itself is continuous. I know there are similar post out there but I couldn't really understand it. In some of the post, they mentioned the photoelectric effects, but from what I understood, it just meant that light contains energy, and energy is transferable. There are also answers on Planck hypothesis, for me that is just the mathematical formulation to fit the experimental data. But the intuition behind "light is quantized" is still a mystery to me. quantum-mechanics energy photons quantization Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Dec 29, 2018 at 17:44 Kyle Kanos 29.2k 41 41 gold badges 70 70 silver badges 138 138 bronze badges asked Dec 29, 2018 at 15:52 KYSKYS 389 1 1 silver badge 7 7 bronze badges 7 you are confusing classical electromagnetic waves, with the photon. Classical waves emerge from a superposition of an ensemble of photons with frequency nu. See my answers here physics.stackexchange.com/questions/444917/… and here physics.stackexchange.com/questions/449021/…anna v –anna v 2018-12-29 17:35:26 +00:00 Commented Dec 29, 2018 at 17:35 You’re not taking photons as individual particles serious enough. The energy comes from how fast the photon oscillates as it travels along at the speed of light. It’s linear momentum is separate and much smaller than its oscillating momentum. A blue photon oscillates much faster than a red photon.Bill Alsept –Bill Alsept 2018-12-29 19:03:42 +00:00 Commented Dec 29, 2018 at 19:03 @annav From what I understood, are you saying that if I have some wavefunction that represent the electromagnetic wave, the Fourier decomposition of that wavefunction encodes all the possible photon that exist in that particular electromagnetic wave? And since we will decompose a wavefunction in terms of infinitely many sinusoidal waves with infinitely different frequency, then we would have infinitely many photons for a specific electromagnetic wave? So, experimentally, its only possible to make one photon when we have an unimaginably accurate frequency. Is my understanding correct?KYS –KYS 2018-12-30 03:27:57 +00:00 Commented Dec 30, 2018 at 3:27 No, it is not correct. If you are a physicist or an aspiring physicist you have to understand that the theories of physics are models, i.e have extra laws like axioms which pick up the solutions that fit data. These models have specific variables and frameworks: classical Maxwell equations describe the macroscopic light behavior, when the light intensity is enough for our eyes to see(rough definition). When the intensity falls, the framework is not longer classical but quantum mechanical. and light is seen experimentally(in the double slit experiment for example) to be composed of anna v –anna v 2018-12-30 05:10:23 +00:00 Commented Dec 30, 2018 at 5:10 single particle impacts which add up to give the classical wave interference pattern. These are the photons, zero mass , point particles , of energy=hnu, spin +/-1 to its direction of motion, and part of the elementary particle table.en.wikipedia.org/wiki/Standard_Model . As a quantum entity it obeys a quantum mechanical wave equations, where what is waving is the probability of finding the photon at an (x,y,z) point in space. This as I show in the answers, are complex wave functions which are a solution of quantized maxwell's equations .anna v –anna v 2018-12-30 05:16:13 +00:00 Commented Dec 30, 2018 at 5:16 \|Show 2 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Energy as such is not quantized, we can see that from the spectrum of a black body: It is a continuous spectrum, so all values of frequency and thus energy are possible. However, the energy of photons which are emitted by atoms when an electron in an atom's hull switches from a higher to a lower level, is only permitted to attain some particular values. The origin of this is the fact that an atom is a bounded system so the electron in its hull is not free. A free electron's energy could again attain any value. Now, to describe the discrete energy levels in an atom it was necessary to have a model where electromagnetic waves can carry some characteristic energy, the photon model. With a classical wave model, such discrete energy levels cannot be described. That doesn't mean, however, that any system shows discrete energy levels, only bound states like atoms do so. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Dec 29, 2018 at 16:04 PhotonPhoton 5,067 2 2 gold badges 18 18 silver badges 30 30 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Your question relates to Planck's original work where he defined the Planck constant. First, you must build on thermodynamics which tells us how systems in equilibrium distribute energy among themselves. So, a system with fixed energy could have one high energy photon, or it could have two photons with half as much energy - but neither system is in thermal equilibrium. Given abundant mechanisms for redistributing the energy into other frequencies the distribution of the photons will be the so-called black-body radiation spectrum. If light was not quantized then thermal effects would favor all energy making its way into the higher frequencies. If, however, the amount of energy in a given frequency is quantized then the energy density will peak at a finite frequency. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Dec 29, 2018 at 18:07 Paul YoungPaul Young 3,588 16 16 silver badges 36 36 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Radiation appears quantized only in confinement. The same is for electrons- they appear to move like waves in confinement. In confinement there are many participants and very small space to move. It could also be a small number, but moving very fast in a small space, giving the same effect. As a result of confinement, the Coulomb inverse square changes to Hook's law- the space spring law. Hook's law is seen in every vibratory system.(The proof is simple, take three equally spaced points on a line. Give the middle a nudge keeping the ends fixed and you see the force;f=k/r^2 changing to f=Kr in the limit of small displacement). The result of vibration is that the whole force field becomes harmonic causing an electron to behave as a wave. And the energy levels to become discrete(characteristic of vibrating systems), ending in light behaving as quanta or particles as Einstein found in his photon work. But when in the open, electrons behave as particles with a well defined path- as one can see in numerous experiments, ranging from the vacuum tube to the cloud chamber. And photons/radiation revert back to the wave behavior, as Huygens found. The easiest way to imagine a photon is think of it as the probability of finding that photon energy at a point. This way one can avoid questions about size and the rest. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Dec 30, 2018 at 8:54 answered Dec 30, 2018 at 8:48 RiadRiad 609 5 5 silver badges 9 9 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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6271	https://www.waterboards.ca.gov/waterrights/water_issues/programs/bay_delta/california_waterfix/exhibits/docs/dd_jardins/DDJ-148%20ASCE%207-10.pdf	Minimum Design Loads for Buildings and Other Structures This document uses both the International System of Units (SI) and customary units A S C E S TA N D A R D ASCE/SEI 7–10 A S C E S TA N D A R D ASCE/SEI 7-10 American Society of Civil Engineers Minimum Design Loads for Buildings and Other Structures This document uses both the International System of Units (SI) and customary units. Library of Congress Cataloging-in-Publication Data Minimum design loads for buildings and other structures. p. cm. “ASCE Standard ASCE/SEI 7-10.” Includes bibliographical references and index. ISBN 978-0-7844-1085-1 (alk. paper) 1. Structural engineering–Standards–United States. 2. Buildings–Standards–United States. 3. Strains and stresses. 4. Standards, Engineering–United States. I. American Society of Civil Engineers. TH851.M56 2010 624.1′75021873—dc22 2010011011 Published by American Society of Civil Engineers 1801 Alexander Bell Drive Reston, Virginia 20191 www.pubs.asce.org This standard was developed by a consensus standards development process which has been accredited by the American National Standards Institute (ANSI). Accreditation by ANSI, a voluntary accreditation body representing public and private sector standards development organizations in the U.S. and abroad, signiﬁ es that the standards devel-opment process used by ASCE has met the ANSI requirements for openness, balance, consensus, and due process. 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ISBN 978-0-7844-1085-1 Manufactured in the United States of America. 18 17 16 15 14 13 12 11 10 1 2 3 4 5 iii STANDARDS In 2003, the Board of Direction approved the revision to the ASCE Rules for Standards Committees to govern the writing and maintenance of standards developed by the Society. All such standards are developed by a consensus standards process managed by the Society’s Codes and Standards Committee (CSC). The consensus process includes balloting by a balanced standards committee made up of Society members and nonmembers, balloting by the member-ship of the Society as a whole, and balloting by the public. All standards are updated or reafﬁ rmed by the same process at intervals not exceeding ﬁ ve years. The following standards have been issued: ANSI/ASCE 1-82 N-725 Guideline for Design and Analysis of Nuclear Safety Related Earth Structures ASCE/EWRI 2-06 Measurement of Oxygen Transfer in Clean Water ANSI/ASCE 3-91 Standard for the Structural Design of Composite Slabs and ANSI/ASCE 9-91 Standard Practice for the Construction and Inspection of Composite Slabs ASCE 4-98 Seismic Analysis of Safety-Related Nuclear Structures Building Code Requirements for Masonry Structures (ACI 530-02/ASCE 5-02/TMS 402-02) and Speciﬁ cations for Masonry Structures (ACI 530.1-02/ASCE 6-02/TMS 602-02) ASCE/SEI 7-10 Minimum Design Loads for Buildings and Other Structures SEI/ASCE 8-02 Standard Speciﬁ cation for the Design of Cold-Formed Stainless Steel Structural Members ANSI/ASCE 9-91 listed with ASCE 3-91 ASCE 10-97 Design of Latticed Steel Transmission Structures SEI/ASCE 11-99 Guideline for Structural Condition Assessment of Existing Buildings ASCE/EWRI 12-05 Guideline for the Design of Urban Subsurface Drainage ASCE/EWRI 13-05 Standard Guidelines for Installation of Urban Subsurface Drainage ASCE/EWRI 14-05 Standard Guidelines for Operation and Maintenance of Urban Subsurface Drainage ASCE 15-98 Standard Practice for Direct Design of Buried Precast Concrete Pipe Using Standard Installations (SIDD) ASCE 16-95 Standard for Load Resistance Factor Design (LRFD) of Engineered Wood Construction ASCE 17-96 Air-Supported Structures ASCE 18-96 Standard Guidelines for In-Process Oxygen Transfer Testing ASCE 19-96 Structural Applications of Steel Cables for Buildings ASCE 20-96 Standard Guidelines for the Design and Installation of Pile Foundations ANSI/ASCE/T&DI 21-05 Automated People Mover Standards—Part 1 ANSI/ASCE/T&DI 21.2-08 Automated People Mover Standards—Part 2 ANSI/ASCE/T&DI 21.3-08 Automated People Mover Standards—Part 3 ANSI/ASCE/T&DI 21.4-08 Automated People Mover Standards—Part 4 SEI/ASCE 23-97 Speciﬁ cation for Structural Steel Beams with Web Openings ASCE/SEI 24-05 Flood Resistant Design and Construction ASCE/SEI 25-06 Earthquake-Actuated Automatic Gas Shutoff Devices ASCE 26-97 Standard Practice for Design of Buried Precast Concrete Box Sections ASCE 27-00 Standard Practice for Direct Design of Precast Concrete Pipe for Jacking in Trenchless Construction ASCE 28-00 Standard Practice for Direct Design of Precast Concrete Box Sections for Jacking in Trenchless Construction ASCE/SEI/SFPE 29-05 Standard Calculation Methods for Structural Fire Protection SEI/ASCE 30-00 Guideline for Condition Assessment of the Building Envelope SEI/ASCE 31-03 Seismic Evaluation of Existing Buildings SEI/ASCE 32-01 Design and Construction of Frost-Protected Shallow Foundations EWRI/ASCE 33-01 Comprehensive Transboundary International Water Quality Management Agreement iv STANDARDS EWRI/ASCE 34-01 Standard Guidelines for Artiﬁ cial Recharge of Ground Water EWRI/ASCE 35-01 Guidelines for Quality Assurance of Installed Fine-Pore Aeration Equipment CI/ASCE 36-01 Standard Construction Guidelines for Microtunneling SEI/ASCE 37-02 Design Loads on Structures during Construction CI/ASCE 38-02 Standard Guideline for the Collection and Depiction of Existing Subsurface Utility Data EWRI/ASCE 39-03 Standard Practice for the Design and Operation of Hail Suppression Projects ASCE/EWRI 40-03 Regulated Riparian Model Water Code ASCE/SEI 41-06 Seismic Rehabilitation of Existing Buildings ASCE/EWRI 42-04 Standard Practice for the Design and Operation of Precipitation Enhancement Projects ASCE/SEI 43-05 Seismic Design Criteria for Structures, Systems, and Components in Nuclear Facilities ASCE/EWRI 44-05 Standard Practice for the Design and Operation of Supercooled Fog Dispersal Projects ASCE/EWRI 45-05 Standard Guidelines for the Design of Urban Stormwater Systems ASCE/EWRI 46-05 Standard Guidelines for the Installation of Urban Stormwater Systems ASCE/EWRI 47-05 Standard Guidelines for the Operation and Maintenance of Urban Stormwater Systems ASCE/SEI 48-05 Design of Steel Transmission Pole Structures ASCE/EWRI 50-08 Standard Guideline for Fitting Saturated Hydraulic Conductivity Using Probability Density Functions ASCE/EWRI 51-08 Standard Guideline for Calculating the Effective Saturated Hydraulic Conductivity ASCE/SEI 52-10 Design of Fiberglass-Reinforced Plastic (FRP) Stacks ASCE/G-I 53-10 Compaction Grouting Consensus Guide v FOREWORD The material presented in this standard has been prepared in accordance with recognized engineering principles. This standard should not be used without ﬁ rst securing competent advice with respect to its suitability for any given application. The publication of the material contained herein is not intended as a representation or warranty on the part of the American Society of Civil Engineers, or of any other person named herein, that this information is suitable for any general or particular use or promises freedom from infringement of any patent or patents. Anyone making use of this information assumes all liability from such use. In the margin of Chapters 1 through 23, a bar has been placed to indicate a substantial technical revision in the standard from the 2005 edition. Because of the reorganization of the wind provisions, these bars are not used in Chapters 26 through 31. Likewise, bars are not used to indicate changes in any parts of the Commentary. vii ACKNOWLEDGMENTS The American Society of Civil Engineers (ASCE) acknowledges the work of the Minimum Design Loads on Buildings and Other Structures Standards Committee of the Codes and Standards Activities Division of the Structural Engineering Institute. This group comprises individuals from many backgrounds, including consulting engineering, research, construc-tion industry, education, government, design, and private practice. This revision of the standard began in 2006 and incorporates information as described in the commentary. This standard was prepared through the consensus standards process by balloting in compliance with procedures of ASCE’s Codes and Standards Activities Committee. Those individuals who serve on the Standards Committee are: Voting Members Donald Dusenberry, P.E., F.ASCE, Chair Robert E. Bachman, P.E., M.ASCE, Vice-Chair James R. Harris, Ph.D., P.E., M.ASCE, Past-Chair James G. Soules, P.E., S.E., F.ASCE, Secretary James R. Cagley, P.E., M.ASCE Dominic Campi, M.ASCE Jay H. Crandell, P.E., M.ASCE James M. Fisher, Ph.D., P.E., M.ASCE Nathan C. Gould, P.E., M.ASCE Lawrence G. Grifﬁ s, P.E., M.ASCE Ronald O. Hamburger, P.E. John D. Hooper, M.ASCE Daniel G. Howell, P.E., M.ASCE Richart Kahler, P.E., M.ASCE John R. Kissell, P.E., M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Robert B. Paullus Jr., P.E., M.ASCE Timothy A. Reinhold, P.E., M.ASCE John G. Tawresey, P.E., M.ASCE Harry B. Thomas, P.E., M.ASCE Thomas R. Tyson, P.E., M.ASCE Peter J G. Willse, P.E., M.ASCE Alan Carr Majed A. Dabdoub Mo A. Madani Jonathan C. Siu, P.E., M.ASCE Christos V. Tokas Finley A. Charney, F.ASCE Ronald A. Cook, Ph.D., P.E., M.ASCE Bruce R. Ellingwood, Ph.D., P.E., F.ASCE Theodore V. Galambos, Ph.D., P.E., NAE, Dist.M.ASCE Robert D. Hanson, Ph.D., P.E., F.ASCE Neil M. Hawkins, Ph.D., M.ASCE Marc L. Levitan, A.M.ASCE Timothy W. Mays, A.M.ASCE Therese P. Mc Allister, P.E. Michael O’Rourke, Ph.D., P.E., M.ASCE Andrew S. Whittaker, Ph.D., S.E., M.ASCE David G. Brinker, P.E., M.ASCE Bradford K. Douglas, P.E., M.ASCE Gary J. Ehrlich, P.E., M.ASCE Satyendra K. Ghosh, M.ASCE Dennis W. Graber, P.E., L.S., M.ASCE Kurt D. Gustafson, P.E., F.ASCE Jason J. Krohn, P.E., M.ASCE Bonnie E. Manley, P.E., M.ASCE Joseph J. Messersmith Jr., P.E., M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE Thomas D. Skaggs, P.E., M.ASCE Brian E. Trimble, P.E., M.ASCE Eric H. Wey, P.E., M.ASCE Distinguished Members Jack E. Cermak, Ph.D., P.E., NAE, Hon.M.ASCE Gilliam S. Harris, P.E., F.ASCE Nicholas Isyumov, P.E., F.ASCE Kathleen F. Jones Kishor C. Mehta, Ph.D., P.E., NAE, Dist.M.ASCE Lawrence D. Reaveley, P.E., M. ASCE Emil Simiu, Ph.D., P.E., F.ASCE Yi Kwei Wen, Ph.D., M.ASCE Associate Members Farid Alfawakhiri, P.E., M.ASCE Leonel I. Almanzar, P.E., M.ASCE Iyad M. Alsamsam, Ph.D., P.E., S.E., M.ASCE Bibo Bahaa Charles C. Baldwin, P.E., M.ASCE Philip R. Brazil, S.E., M.ASCE Ray A. Bucklin, Ph.D., P.E., M.ASCE Alexander Bykovtsev, P.E., M.ASCE James Carlson Anthony C. Cerino, P.E. Robert N. Chittenden, P.E., F.ASCE Adam Cone, S.M.ASCE William L. Coulbourne, P.E., M.ASCE Charles B. Crouse, Ph.D., P.E., M.ASCE Mukti L. Das, Ph.D., P.E., F.ASCE Richard J. Davis, P.E., M.ASCE Yong Deng, Ph.D., M.ASCE David H. Devalve, P.E., M.ASCE Ryan J. Dexter, P.E. Richard M. Drake, S.E., M.ASCE viii ACKNOWLEDGMENTS John F. Duntemann, P.E., M.ASCE Sam S. Eskildsen, A.M.ASCE Mohammed M. Ettouney, M.ASCE David A. Fanella, Ph.D., P.E., F.ASCE Lawrence Fischer, P.E., M.ASCE Donna L.R. Friis, P.E., M.ASCE Amir S.J. Gilani, P.E., S.E., M.ASCE David E. Gloss, P.E., M.ASCE Charles B. Goldsmith David S. Gromala, P.E., M.ASCE Reza Hassanli, S.M.ASCE Todd R. Hawkinson, P.E., M.ASCE Mark J. Henry, P.E., M.ASCE Mark A. Hershberg, P.E., S.E., M.ASCE Joseph R. Hetzel, P.E., M.ASCE Thomas B. Higgins, P.E., S.E., M.ASCE Xiapin Hua, P.E., S.E., M.ASCE Mohammad Iqbal, Ph.D., P.E., S.E., F.ASCE Christopher P. Jones, P.E., M.ASCE Mohammad R. Karim Volkan Kebeli, A.M.ASCE Jon P. Kiland, P.E., S.E., M.ASCE Lionel A. Lemay, P.E., M.ASCE Philip Line, M.ASCE Scott A. Lockyear, A.M.ASCE John V. Loscheider, P.E., M.ASCE David K. Low, P.E., M.ASCE Mustafa A. Mahamid, Ph.D., P.E., M.ASCE Lance Manuel, Ph.D., P.E., M.ASCE Shalva M. Marjanishvili, P.E., S.E., M.ASCE Andrew F. Martin, P.E., M.ASCE Scott E. Maxwell, P.E., S.E., M.ASCE Dennis McCreary, P.E., M.ASCE Kevin Mcosker J. S. Mitchell Kit Miyamoto, P.E., S.E., F.ASCE Rudy Mulia, P.E., M.ASCE Javeed Munshi, P.E., M.ASCE Frank A. Nadeau, M.ASCE Joe N. Nunnery, P.E., M.ASCE Robert F. Oleck Jr., P.E., M.ASCE George N. Olive, M.ASCE Frank K.H. Park, P.E., A.M.ASCE Alan B. Peabody, P.E., M.ASCE David Pierson, P.E., M.ASCE David O. Prevatt, P.E., M.ASCE James A. Rossberg, P.E., M.ASCE Scott A. Russell, P.E., M.ASCE Fahim Sadek, Ph.D., M.ASCE Jerry R. Salmon, M.ASCE Jeremy T. Salmon, A.M.ASCE Phillip J. Samblanet, P.E., M.ASCE William Scott, P.E., M.ASCE Gary Searer Thomas L. Smith Jean Smith Alexis Spyrou, P.E., M.ASCE Theodore Stathopoulos, Ph.D., P.E., F.ASCE David A. Steele, P.E., M.ASCE Sayed Stoman, P.E., S.E., M.ASCE Yelena K. Straight, A.M.ASCE Lee Tedesco, Aff.M.ASCE Jason J. Thompson Mai Tong David P. Tyree, P.E., M.ASCE Victoria B. Valentine, P.E., M.ASCE Miles E. Waltz, P.E., M.ASCE Terence A. Weigel, Ph.D., P.E., M.ASCE Peter Wrenn, P.E., M.ASCE Tom C. Xia, P.E., M.ASCE Bradley Young, M.ASCE Subcommittee on Atmospheric Ice Loads Alan B. Peabody, P.E., M.ASCE, Chair Jamey M. Bertram, P.E., M.ASCE David G. Brinker, P.E., M.ASCE Joseph A. Catalano, A.M.ASCE Maggie Emery Karen Finstad Asim K. Haldar Kathleen F. Jones Jack N. Lott Lawrence M. Slavin, A.M.ASCE Ronald M. Thorkildson, A.M.ASCE Subcommittee on Dead and Live Loads Thomas R. Tyson, P.E., M.ASCE, Chair Adam W. Dayhoff, A.M.ASCE John V. Loscheider, P.E., M.ASCE Mustafa A. Mahamid, Ph.D., P.E., M.ASCE Frank A. Nadeau, M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE John G. Tawresey, P.E., M.ASCE Harry B. Thomas, P.E., M.ASCE Subcommittee on Flood Loads Christopher P. Jones, P.E., M. ASCE, Chair Subcommittee for General Structural Requirements Ronald O. Hamburger, P.E., Chair Farid Alfawakhiri, P.E., M.ASCE Iyad M. Alsamsam, Ph.D., P.E., S.E., M.ASCE Philip R. Brazil, S.E., M.ASCE Dominic Campi, M.ASCE Theodore V. Galambos, Ph.D., P.E., NAE, Dist.M.ASCE Satyendra K. Ghosh, M.ASCE Nathan C. Gould, P.E., M.ASCE James R. Harris, Ph.D., P.E., M.ASCE Todd R. Hawkinson, P.E., M.ASCE Thomas F. Heausler, P.E., M.ASCE Jason J. Krohn, P.E., M.ASCE Philip Line, M.ASCE Timothy W. Mays, A.M.ASCE Therese P. Mc Allister, P.E. Brian J. Meacham Timothy A. Reinhold, P.E., M.ASCE Jonathan C. Siu, P.E., M.ASCE James G. Soules, P.E., S.E., F.ASCE Peter J. Vickery, M.ASCE Subcommittee on Seismic Loads John D. Hooper, M.ASCE, Chair Dennis A. Alvarez, P.E., M.ASCE Victor D. Azzi, P.E., M.ASCE Robert E. Bachman, P.E., M.ASCE David R. Bonneville, M.ASCE Philip R. Brazil, S.E., M.ASCE Philip Caldwell ix ACKNOWLEDGMENTS Dominic Campi, M.ASCE James A. Carlson Finley A. Charney, F.ASCE Robert N. Chittenden, P.E., F.ASCE Charles B. Crouse, Ph.D., P.E., M.ASCE Bradford K. Douglas, P.E., M.ASCE Satyendra K. Ghosh, M.ASCE John D. Gillengerten Nathan C. Gould, P.E., M.ASCE Ronald O. Hamburger, P.E. Robert D. Hanson, Ph.D., P.E., F.ASCE James R. Harris, Ph.D., P.E., M.ASCE John L. Harris III, P.E., S.E., M.ASCE Ronald W. Haupt, P.E., M.ASCE Neil M. Hawkins, Ph.D., M.ASCE Thomas F. Heausler, P.E., M.ASCE Douglas G. Honegger, M.ASCE Y. Henry Huang, P.E., M.ASCE William V. Joerger, M.ASCE Martin W. Johnson, P.E., M.ASCE Richart Kahler, P.E., M.ASCE Dominic J. Kelly, P.E., M.ASCE Jon P. Kiland, P.E., S.E., M.ASCE Charles A. Kircher, Ph.D., P.E., M.ASCE Vladimir G. Kochkin, A.M.ASCE James S. Lai, P.E., F.ASCE Edgar V. Leyendecker Philip Line, M.ASCE John V. Loscheider, P.E., M.ASCE Nicolas Luco, A.M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Bonnie E. Manley, P.E., M.ASCE Igor Marinovic, P.E., M.ASCE Scott E. Maxwell, P.E., S.E., M.ASCE Kit Miyamoto, P.E., S.E., F.ASCE Rudy Mulia, P.E., S.E., M.ASCE Bernard F. Murphy, P.E., M.ASCE Frank A. Nadeau, M.ASCE Corey D. Norris, P.E., M.ASCE Robert B. Paullus Jr., P.E., M.ASCE Robert G. Pekelnicky, P.E., S.E., M.ASCE Maurice S. Power, M.ASCE James A. Rossberg, P.E., M.ASCE Rafael G. Sabelli, P.E., S.E., M.ASCE Phillip J. Samblanet, P.E., M.ASCE William Scott, P.E., M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE John F. Silva, S.E., M.ASCE Jonathan C. Siu, P.E., M.ASCE Jean Smith James G. Soules, P.E., S.E., F.ASCE Harold O. Sprague Jr., P.E., F.ASCE Bill Staehlin Sayed Stoman, P.E., S.E., M.ASCE Jason J. Thompson Christos V. Tokas Mai Tong Victoria B. Valentine, P.E., M.ASCE Miroslav Vejvoda, P.E., F.ASCE Miles E. Waltz, P.E., M.ASCE Eric H. Wey, P.E., M.ASCE Andrew S. Whittaker, Ph.D., S.E., M.ASCE Ben Youseﬁ , P.E., S.E., M.ASCE Seismic Task Committee on Ground Motions Charles B. Crouse, Ph.D., P.E., M.ASCE, Chair Robert E. Bachman, P.E., M.ASCE Finley A. Charney, F.ASCE Neil M. Hawkins, Ph.D., M.ASCE John D. Hooper, M.ASCE Edgar V. Leyendecker Nicolas Luco, A.M.ASCE Maurice S. Power, M.ASCE William Scott, P.E., M.ASCE Andrew S. Whittaker, Ph.D., S.E., M.ASCE Seismic Task Committee on General Provisions Jon P. Kiland, P.E., S.E., M.ASCE, Chair Robert E. Bachman, P.E., M.ASCE David R. Bonneville, M.ASCE Philip R. Brazil, S.E., M.ASCE Dominic Campi, M.ASCE Finley A. Charney, F.ASCE Satyendra K. Ghosh, M.ASCE John D. Gillengerten Nathan C. Gould, P.E., M.ASCE Ronald O. Hamburger, P.E. James R. Harris, Ph.D., P.E., M.ASCE John L. Harris III, P.E., S.E., M.ASCE John R. Hayes Jr., Ph.D., P.E., M.ASCE Thomas F. Heausler, P.E., M.ASCE John D. Hooper, M.ASCE Martin W. Johnson, P.E., M.ASCE Dominic J. Kelly, P.E., M.ASCE Ryan A. Kersting, A.M.ASCE Philip Line, M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Bonnie E. Manley, P.E., M.ASCE Kit Miyamoto, P.E., S.E., F.ASCE Rudy Mulia, P.E., S.E., M.ASCE Robert G. Pekelnicky, P.E., S.E., M.ASCE Rafael G. Sabelli, P.E., S.E., M.ASCE William Scott, P.E., M.ASCE Eric H. Wey, P.E., M.ASCE Andrew S. Whittaker, Ph.D., S.E., M.ASCE Ben Youseﬁ , P.E., S.E., M.ASCE Seismic Task Committee on Foundations / Site Conditions Martin W. Johnson, P.E., M.ASCE, Chair Robert N. Chittenden, P.E., F. ASCE Charles B. Crouse, Ph.D., P.E., M.ASCE Neil M. Hawkins, Ph.D., M.ASCE Dominic J. Kelly, P.E., M.ASCE Maurice S. Power, M.ASCE Eric H. Wey, P.E., M.ASCE Seismic Task Committee on Concrete Neil M. Hawkins, Ph.D., M.ASCE, Chair x ACKNOWLEDGMENTS Satyendra K. Ghosh, M.ASCE John R. Hayes Jr., Ph.D., P.E., M.ASCE Jon P. Kiland, P.E., S.E., M.ASCE John F. Silva, S.E., M.ASCE Miroslav Vejvoda, P.E., F.ASCE Ben Youseﬁ , P.E., S.E., M.ASCE Seismic Task Committee on Masonry Jason J. Thompson, Chair Robert N. Chittenden, P.E., F.ASCE Jon P. Kiland, P.E., S.E., M.ASCE Seismic Task Committee on Steel & Composite Structures Rafael G. Sabelli, P.E., S.E., M.ASCE, Chair Thomas F. Heausler, P.E., M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Bonnie E. Manley, P.E., M.ASCE William Scott, P.E., M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE Seismic Task Committee on Wood Philip Line, M.ASCE, Chair Philip R. Brazil, S.E., M.ASCE Robert N. Chittenden, P.E., F.ASCE Bradford K. Douglas, P.E., M.ASCE Vladimir G. Kochkin, A.M.ASCE Bonnie E. Manley, P.E., M.ASCE Jonathan C. Siu, P.E., M.ASCE Miles E. Waltz, P.E., M.ASCE Ben Youseﬁ , P.E., S.E., M.ASCE Seismic Task Committee on Non-Structural Components John F. Silva, S.E., M.ASCE, Chair Dennis A. Alvarez, P.E., M.ASCE Robert E. Bachman, P.E., M.ASCE David R. Bonneville, M.ASCE Philip J. Caldwell, A.M.ASCE James Carlson John D. Gillengerten Nathan C. Gould, P.E., M.ASCE Ronald W. Haupt, P.E., M.ASCE Thomas F. Heausler, P.E., M.ASCE Douglas G. Honegger, M.ASCE Francis E. Jehrio William V. Joerger, M.ASCE Richard Lloyd, A.M.ASCE Michael Mahoney Kit Miyamoto, P.E., S.E., F.ASCE Rudy Mulia, P.E., S.E., M.ASCE William Scott, P.E., M.ASCE Jean Smith James G. Soules, P.E., S.E., F.ASCE Harold O. Sprague Jr., P.E., F.ASCE Bill Staehlin Chris Tokas Victoria B. Valentine, P.E., M.ASCE Eric H. Wey, P.E., M.ASCE Paul R. Wilson, P.E., M.ASCE Seismic Task Committee on Administrative and QA Provisions Jonathan C. Siu, P.E., M.ASCE, Chair Robert E. Bachman, P.E., M.ASCE Philip R. Brazil, S.E., M.ASCE John D. Hooper, M.ASCE Jon P. Kiland, P.E., S.E., M.ASCE Robert G. Pekelnicky, P.E., S.E., M.ASCE John F. Silva, S.E., M.ASCE Seismic Task Committee on Seismic Isolation and Damping Andrew S. Whittaker, Ph.D., S.E., M.ASCE, Chair Robert E. Bachman, P.E., M.ASCE Finley A. Charney, F.ASCE Robert D. Hanson, Ph.D., P.E., F.ASCE Martin W. Johnson, P.E., M.ASCE Charles A. Kircher, Ph.D., P.E., M.ASCE Kit Miyamoto, P.E., S.E., F.ASCE Seismic Task Committee on Non-Building Structures James G. Soules, P.E., S.E., F.ASCE, Chair Victor D. Azzi, P.E., M.ASCE Robert E. Bachman, P.E., M.ASCE Philip J. Caldwell, A.M.ASCE Charles B. Crouse, Ph.D., P.E., M.ASCE Ronald W. Haupt, P.E., M.ASCE Thomas F. Heausler, P.E., M.ASCE Douglas G. Honegger, M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Rudy Mulia, P.E., S.E., M.ASCE William Scott, P.E., M.ASCE John F. Silva, S.E., M.ASCE Harold O. Sprague Jr., P.E., F.ASCE Sayed Stoman, P.E., S.E., M.ASCE Eric H. Wey, P.E., M.ASCE Subcommittee on Snow and Rain Loads Michael O’Rourke, Ph.D., P.E., M.ASCE, Chair Timothy J. Allison, A.M.ASCE John Cocca, A.M.ASCE Bradford K. Douglas, P.E., M.ASCE John F. Duntemann, P.E., M.ASCE Gary J. Ehrlich, P.E., M.ASCE James M. Fisher, Ph.D., P.E., M.ASCE James R. Harris, Ph.D., P.E., M.ASCE Thomas B. Higgins, P.E., S.E., M.ASCE Daniel G. Howell, P.E., M.ASCE Nicholas Isyumov, P.E., F.ASCE Scott A. Lockyear, A.M.ASCE Ian Mackinlay, Aff.M.ASCE Joe N. Nunnery, P.E., M.ASCE George N. Olive, M.ASCE Michael F. Pacey, P.E., M.ASCE David B. Peraza, P.E., M.ASCE Mark K. Radmaker, P.E. Scott A. Russell, P.E., M.ASCE Ronald L. Sack, Ph.D., P.E., F.ASCE Joseph D. Scholze, P.E., M.ASCE Gary L. Schumacher, P.E., M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE xi ACKNOWLEDGMENTS Daniel J. Walker, P.E., M.ASCE Peter Wrenn, P.E., M.ASCE Subcommittee on Strength Criteria Bruce R. Ellingwood, Ph.D., P.E., M.ASCE, Chair Therese P. McAllister, P.E. Iyad M. Alsamsam, Ph.D., P.E., S.E., M.ASCE Charles C. Baldwin, P.E., M.ASCE Theodore V. Galambos, Ph.D., P.E., NAE, Dist.M.ASCE David S. Gromala, P.E., M.ASCE Ronald O. Hamburger, P.E. James R. Harris, Ph.D., P.E., M.ASCE Nestor R. Iwankiw, P.E., M.ASCE John V. Loscheider, P.E., M.ASCE Sanjeev R. Malushte, P.E., S.E., F.ASCE Clarkson W. Pinkham, P.E., F.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE James G. Soules, P.E., S.E., F.ASCE Jason J. Thompson Yi Kwei Wen, Ph.D., M.ASCE Subcommittee on Wind Loads Voting Members Ronald A. Cook, Ph.D., P.E., M.ASCE, Chair Gary Y.K. Chock, M.ASCE Jay H. Crandell, P.E., M.ASCE Bradford K. Douglas, P.E., M.ASCE Charles Everly, P.E., CBO Charles B. Goldsmith Dennis W. Graber, P.E., L.S., M.ASCE Lawrence G. Grifﬁ s, P.E., M.ASCE Gilliam S. Harris, P.E., F.ASCE Peter A. Irwin, Ph.D., P.Eng, F.ASCE Ahsan Kareem, Ph.D., M.ASCE Marc L. Levitan, A.M.ASCE Mo A.F. Madani Joseph J. Messersmith Jr., P.E., M.ASCE Jon A. Peterka, P.E., M.ASCE Timothy A. Reinhold, P.E., M.ASCE Donald R. Scott, P.E., M.ASCE Emil Simiu, Ph.D., P.E., F.ASCE Douglas A. Smith, P.E., M.ASCE Thomas L. Smith Thomas E. Stafford Theodore Stathopoulos, Ph.D., P.E., F.ASCE Peter J. Vickery, M.ASCE Robert J. Wills, P.E., M.ASCE Associate Members Timothy J. Allison, A.M.ASCE Roberto H. Behncke, Aff.M.ASCE Daryl W. Boggs, P.E., M.ASCE William L. Coulbourne, P.E., M.ASCE Richard J. Davis, P.E., M.ASCE Joffrey Easley, P.E., M.ASCE Gary J. Ehrlich, P.E., M.ASCE Donna L.R. Friis, P.E., M.ASCE Jon K. Galsworthy, P.E., M.ASCE Gerald L. Hatch, P.E., L.S., M.ASCE Mark J. Henry, P.E., M.ASCE Joseph R. Hetzel, P.E., M.ASCE Thomas B. Higgins, P.E., S.E., M.ASCE Nicholas Isyumov, P.E., F.ASCE Anurag Jain, Ph.D., P.E., M.ASCE Edward L. Keith, P.E., M.ASCE Robert Konz, P.E., M.ASCE Edward M. Laatsch, P.E., M.ASCE Philip Line, M.ASCE Scott A. Lockyear, A.M.ASCE John V. Loscheider, P.E., M.ASCE Andrew F. Martin, P.E., M.ASCE Patrick W. McCarthy, P.E., M.ASCE Kishor C. Mehta, Ph.D., P.E., NAE, Dist.M.ASCE George N. Olive, M.ASCE Robert B. Paullus Jr., P.E., M.ASCE Rick Perry William C. Rosencutter, P.E., M.ASCE William L. Shoemaker, Ph.D., P.E., M.ASCE Peter J G. Willse, P.E., M.ASCE Tom C. Xia, P.E., M.ASCE xiii DEDICATION Thomas R. Tyson, P.E., S.E. The members of the Minimum Design Loads for Buildings and Other Structures Standards Committee of the Structural Engineering Institute respectfully dedicate this Standard in the memory of Thomas R. Tyson, P.E., S.E., who passed away on December 19, 2009. His structural engineering expertise complemented his dedication to our profession, and these qualities guided the members of the Live Load Subcommittee, which he chaired during the prepara-tion of this Standard. His practical advice, quick smile, and good nature will be greatly missed. xv CONTENTS Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii 1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Deﬁ nitions and Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.1 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.2 Symbols and Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Basic Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3.1 Strength and Stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3.1.1 Strength Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3.1.2 Allowable Stress Procedures . . . . . . . . . . . . . . . . . . . . . . 3 1.3.1.3 Performance-Based Procedures . . . . . . . . . . . . . . . . . . . . 3 1.3.2 Serviceability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3.3 Self-Straining Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3.4 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3.5 Counteracting Structural Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 General Structural Integrity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.1 Load Combinations of Integrity Loads . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.1.1 Strength Design Notional Load Combinations . . . . . . . . 4 1.4.1.2 Allowable Stress Design Notional Load Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.2 Load Path Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.3 Lateral Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.4 Connection to Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.5 Anchorage of Structural Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4.6 Extraordinary Loads and Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5 Classiﬁ cation of Buildings and Other Structures . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.1 Risk Categorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.2 Multiple Risk Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.3 Toxic, Highly Toxic, and Explosive Substances . . . . . . . . . . . . . . . . . 5 1.6 Additions and Alterations to Existing Structures . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.7 Load Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.8 Consensus Standards and Other Referenced Documents . . . . . . . . . . . . . . . . . . 6 2 Combinations of Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Combining Factored Loads Using Strength Design . . . . . . . . . . . . . . . . . . . . . . 7 2.3.1 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3.2 Basic Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3.3 Load Combinations Including Flood Load . . . . . . . . . . . . . . . . . . . . . 7 2.3.4 Load Combinations Including Atmospheric Ice Loads . . . . . . . . . . . . 8 2.3.5 Load Combinations Including Self-Straining Loads . . . . . . . . . . . . . . 8 2.3.6 Load Combinations for Nonspeciﬁ ed Loads . . . . . . . . . . . . . . . . . . . . 8 CONTENTS xvi 2.4 Combining Nominal Loads Using Allowable Stress Design . . . . . . . . . . . . . . . 8 2.4.1 Basic Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.4.2 Load Combinations Including Flood Load . . . . . . . . . . . . . . . . . . . . . 9 2.4.3 Load Combinations Including Atmospheric Ice Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.4.4 Load Combinations Including Self-Straining Loads . . . . . . . . . . . . . . 9 2.5 Load Combinations for Extraordinary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.1 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.2 Load Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.2.1 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.2.2 Residual Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.3 Stability Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 Dead Loads, Soil Loads, and Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.1 Dead Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.1.1 Deﬁ nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.1.2 Weights of Materials and Constructions . . . . . . . . . . . . . . . . . . . . . . . 11 3.1.3 Weight of Fixed Service Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2 Soil Loads and Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2.1 Lateral Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2.2 Uplift on Floors and Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 4 Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.1 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.2 Loads Not Speciﬁ ed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.3 Uniformly Distributed Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.3.1 Required Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.3.2 Provision for Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.3.3 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.4 Concentrated Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.5 Loads on Handrail, Guardrail, Grab Bar, Vehicle Barrier Systems, and Fixed Ladders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.5.1 Loads on Handrail and Guardrail Systems . . . . . . . . . . . . . . . . . . . . . 14 4.5.2 Loads on Grab Bar Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.5.3 Loads on Vehicle Barrier Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.5.4 Loads on Fixed Ladders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.6 Impact Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.6.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.6.2 Elevators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.6.3 Machinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.7 Reduction in Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.7.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.7.2 Reduction in Uniform Live Loads. . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.7.3 Heavy Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.7.4 Passenger Vehicle Garages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.7.5 Assembly Uses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.7.6 Limitations on One-Way Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.8 Reduction in Roof Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.8.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.8.2 Flat, Pitched, and Curved Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.8.3 Special Purpose Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9 Crane Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9.2 Maximum Wheel Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 CONTENTS xvii 4.9.3 Vertical Impact Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9.4 Lateral Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9.5 Longitudinal Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 5 Flood Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3 Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3.1 Design Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3.2 Erosion and Scour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3.3 Loads on Breakaway Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.4 Loads During Flooding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4.1 Load Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4.2 Hydrostatic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4.3 Hydrodynamic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4.4 Wave Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4.4.1 Breaking Wave Loads on Vertical Pilings and Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.4.4.2 Breaking Wave Loads on Vertical Walls . . . . . . . . . . . . . 23 5.4.4.3 Breaking Wave Loads on Nonvertical Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.4.4.4 Breaking Wave Loads from Obliquely Incident Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 5.4.5 Impact Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 5.5 Consensus Standards and Other Referenced Documents . . . . . . . . . . . . . . . . . . 25 6 Reserved for Future Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 7 Snow Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.1 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.2 Ground Snow Loads, pg . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.3 Flat Roof Snow Loads, pf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.3.1 Exposure Factor, Ce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.3.2 Thermal Factor, Ct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.3.3 Importance Factor, Is . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.3.4 Minimum Snow Load for Low-Slope Roofs, pm . . . . . . . . . . . . . . . . 29 7.4 Sloped Roof Snow Loads, ps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4.1 Warm Roof Slope Factor, Cs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4.2 Cold Roof Slope Factor, Cs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4.3 Roof Slope Factor for Curved Roofs . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4.4 Roof Slope Factor for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.4.5 Ice Dams and Icicles Along Eaves . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.5 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.5.1 Continuous Beam Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.5.2 Other Structural Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.6 Unbalanced Roof Snow Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.6.1 Unbalanced Snow Loads for Hip and Gable Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.6.2 Unbalanced Snow Loads for Curved Roofs . . . . . . . . . . . . . . . . . . . . 32 7.6.3 Unbalanced Snow Loads for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.6.4 Unbalanced Snow Loads for Dome Roofs . . . . . . . . . . . . . . . . . . . . . 32 CONTENTS xviii 7.7 Drifts on Lower Roofs (Aerodynamic Shade) . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.7.1 Lower Roof of a Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.7.2 Adjacent Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.8 Roof Projections and Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.9 Sliding Snow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.10 Rain-On-Snow Surcharge Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.11 Ponding Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.12 Existing Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 8 Rain Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8.1 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8.2 Roof Drainage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8.3 Design Rain Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8.4 Ponding Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8.5 Controlled Drainage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 9 Reserved for Future Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 10 Ice Loads—Atmospheric Icing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.1.1 Site-Speciﬁ c Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.1.2 Dynamic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.1.3 Exclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 10.3 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4 Ice Loads Due to Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.1 Ice Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.2 Nominal Ice Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.3 Height Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.4 Importance Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.5 Topographic Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 10.4.6 Design Ice Thickness for Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . 48 10.5 Wind on Ice-Covered Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 10.5.1 Wind on Ice-Covered Chimneys, Tanks, and Similar Structures . . . . 49 10.5.2 Wind on Ice-Covered Solid Freestanding Walls and Solid Signs . . . 49 10.5.3 Wind on Ice-Covered Open Signs and Lattice Frameworks . . . . . . . 49 10.5.4 Wind on Ice-Covered Trussed Towers . . . . . . . . . . . . . . . . . . . . . . . . 49 10.5.5 Wind on Ice-Covered Guys and Cables . . . . . . . . . . . . . . . . . . . . . . . 49 10.6 Design Temperatures for Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 10.7 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 10.8 Design Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 10.9 Consensus Standards and Other Referenced Documents . . . . . . . . . . . . . . . . . . 50 11 Seismic Design Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.1.2 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.1.3 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.1.4 Alternate Materials and Methods of Construction . . . . . . . . . . . . . . . 57 11.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.3 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 11.4 Seismic Ground Motion Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 11.4.1 Mapped Acceleration Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 11.4.2 Site Class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 11.4.3 Site Coefﬁ cients and Risk-Targeted Maximum Considered Earthquake (MCER) Spectral Response Acceleration Parameters . . . 65 CONTENTS xix 11.4.4 Design Spectral Acceleration Parameters . . . . . . . . . . . . . . . . . . . . . . 65 11.4.5 Design Response Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 11.4.6 Risk-Targeted Maximum Considered (MCER) Response Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 11.4.7 Site-Speciﬁ c Ground Motion Procedures . . . . . . . . . . . . . . . . . . . . . . 67 11.5 Importance Factor and Risk Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 11.5.1 Importance Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 11.5.2 Protected Access for Risk Category IV . . . . . . . . . . . . . . . . . . . . . . . . 67 11.6 Seismic Design Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 11.7 Design Requirements for Seismic Design Category A . . . . . . . . . . . . . . . . . . . . 68 11.8 Geologic Hazards and Geotechnical Investigation . . . . . . . . . . . . . . . . . . . . . . . 68 11.8.2 Geotechnical Investigation Report Requirements for Seismic Design Categories C through F . . . . . . . . 68 11.8.3 Additional Geotechnical Investigation Report Requirements for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . 68 12 Seismic Design Requirements for Building Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 12.1 Structural Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 12.1.1 Basic Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 12.1.2 Member Design, Connection Design, and Deformation Limit . . . . . . 71 12.1.3 Continuous Load Path and Interconnection . . . . . . . . . . . . . . . . . . . . 71 12.1.4 Connection to Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 12.1.5 Foundation Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 12.1.6 Material Design and Detailing Requirements . . . . . . . . . . . . . . . . . . . 72 12.2 Structural System Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 12.2.1 Selection and Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 12.2.2 Combinations of Framing Systems in Different Directions . . . . . . . . 72 12.2.3 Combinations of Framing Systems in the Same Direction . . . . . . . . . 72 12.2.3.1 R, Cd, and Ω0 Values for Vertical Combinations . . . . . . . 72 12.2.3.2 Two Stage Analysis Procedure . . . . . . . . . . . . . . . . . . . . . 72 12.2.3.3 R, Cd, and Ω0 Values for Horizontal Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 12.2.4 Combination Framing Detailing Requirements . . . . . . . . . . . . . . . . . . 78 12.2.5 System Speciﬁ c Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 12.2.5.1 Dual System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 12.2.5.2 Cantilever Column Systems . . . . . . . . . . . . . . . . . . . . . . . 78 12.2.5.3 Inverted Pendulum-Type Structures . . . . . . . . . . . . . . . . . 78 12.2.5.4 Increased Structural Height Limit for Steel Eccentrically Braced Frames, Steel Special Concentrically Braced Frames, Steel Buckling-restrained Braced Frames, Steel Special Plate Shear Walls and Special Reinforced Concrete Shear Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 12.2.5.5 Special Moment Frames in Structures Assigned to Seismic Design Categories D through F . . . . . . . . . . . . . 79 12.2.5.6 Steel Ordinary Moment Frames . . . . . . . . . . . . . . . . . . . . 79 12.2.5.7 Steel Intermediate Moment Frames . . . . . . . . . . . . . . . . . 79 12.2.5.8 Shear Wall-Frame Interactive Systems . . . . . . . . . . . . . . 80 12.3 Diaphragm Flexibility, Conﬁ guration Irregularities, and Redundancy . . . . . . . . 80 12.3.1 Diaphragm Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 12.3.1.1 Flexible Diaphragm Condition . . . . . . . . . . . . . . . . . . . . . 80 12.3.1.2 Rigid Diaphragm Condition . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.1.3 Calculated Flexible Diaphragm Condition . . . . . . . . . . . 81 CONTENTS xx 12.3.2 Irregular and Regular Classiﬁ cation . . . . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.2.1 Horizontal Irregularity . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.2.2 Vertical Irregularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.3 Limitations and Additional Requirements for Systems with Structural Irregularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.3.1 Prohibited Horizontal and Vertical Irregularities for Seismic Design Categories D through F . . . . . . . . . . . . . 81 12.3.3.2 Extreme Weak Stories . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 12.3.3.3 Elements Supporting Discontinuous Walls or Frames . . . 82 12.3.3.4 Increase in Forces Due to Irregularities for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . 82 12.3.4 Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 12.3.4.1 Conditions Where Value of ρ is 1.0 . . . . . . . . . . . . . . . . . 83 12.3.4.2 Redundancy Factor, ρ, for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . 84 12.4 Seismic Load Effects and Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 12.4.1 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 12.4.2 Seismic Load Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 12.4.2.1 Horizontal Seismic Load Effect . . . . . . . . . . . . . . . . . . . . 84 12.4.2.2 Vertical Seismic Load Effect . . . . . . . . . . . . . . . . . . . . . . 86 12.4.2.3 Seismic Load Combinations . . . . . . . . . . . . . . . . . . . . . . 86 12.4.3 Seismic Load Effect Including Overstrength Factor . . . . . . . . . . . . . . 86 12.4.3.1 Horizontal Seismic Load Effect with Overstrength Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 12.4.3.2 Load Combinations with Overstrength Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.4.3.3 Allowable Stress Increase for Load Combinations with Overstrength . . . . . . . . . . . . . . . . . . . 87 12.4.4 Minimum Upward Force for Horizontal Cantilevers for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.5 Direction of Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.5.1 Direction of Loading Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.5.2 Seismic Design Category B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.5.3 Seismic Design Category C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 12.5.4 Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . 88 12.6 Analysis Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 12.7 Modeling Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 12.7.1 Foundation Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 12.7.2 Effective Seismic Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 12.7.3 Structural Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 12.7.4 Interaction Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 12.8 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 12.8.1 Seismic Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 12.8.1.1 Calculation of Seismic Response Coefﬁ cient . . . . . . . . . 89 12.8.1.2 Soil Structure Interaction Reduction . . . . . . . . . . . . . . . . 90 12.8.1.3 Maximum Ss Value in Determination of Cs . . . . . . . . . . 90 12.8.2 Period Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 12.8.2.1 Approximate Fundamental Period . . . . . . . . . . . . . . . . . . 90 12.8.3 Vertical Distribution of Seismic Forces. . . . . . . . . . . . . . . . . . . . . . . . 91 12.8.4 Horizontal Distribution of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 12.8.4.1 Inherent Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 12.8.4.2 Accidental Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 12.8.4.3 Ampliﬁ cation of Accidental Torsional Moment . . . . . . . 91 CONTENTS xxi 12.8.5 Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 12.8.6 Story Drift Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 12.8.6.1 Minimum Base Shear for Computing Drift . . . . . . . . . . . 92 12.8.6.2 Period for Computing Drift . . . . . . . . . . . . . . . . . . . . . . . 93 12.8.7 P-Delta Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 12.9 Modal Response Spectrum Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.1 Number of Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.2 Modal Response Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.3 Combined Response Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.4 Scaling Design Values of Combined Response . . . . . . . . . . . . . . . . . 94 12.9.4.1 Scaling of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.4.2 Scaling of Drifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.5 Horizontal Shear Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.6 P-Delta Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.9.7 Soil Structure Interaction Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.10 Diaphragms, Chords, and Collectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.10.1 Diaphragm Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.10.1.1 Diaphragm Design Forces . . . . . . . . . . . . . . . . . . . . . . . . 94 12.10.2 Collector Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 12.10.2.1 Collector Elements Requiring Load Combinations with Overstrength Factor for Seismic Design Categories C through F . . . . . . . . . . . . . . . . . . . . . . . . . . 95 12.11 Structural Walls and Their Anchorage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.11.1 Design for Out-of-Plane Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.11.2 Anchorage of Structural Walls and Transfer of Design Forces into Diaphragms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.11.2.1 Wall Anchorage Forces . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.11.2.2 Additional Requirements for Diaphragms in Structures Assigned to Seismic Design Categories C through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.12 Drift And Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 12.12.1 Story Drift Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 12.12.1.1 Moment Frames in Structures Assigned to Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . 97 12.12.2 Diaphragm Deﬂ ection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 12.12.3 Structural Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 12.12.4 Members Spanning between Structures . . . . . . . . . . . . . . . . . . . . . . . 98 12.12.5 Deformation Compatibility for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13 Foundation Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.1 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.2 Materials of Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.3 Foundation Load-Deformation Characteristics . . . . . . . . . . . . . . . . . . 98 12.13.4 Reduction of Foundation Overturning . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.5 Requirements for Structures Assigned to Seismic Design Category C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.5.1 Pole-Type Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 12.13.5.2 Foundation Ties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 12.13.5.3 Pile Anchorage Requirements . . . . . . . . . . . . . . . . . . . . . 99 12.13.6 Requirements for Structures Assigned to Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 12.13.6.1 Pole-Type Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 12.13.6.2 Foundation Ties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 CONTENTS xxii 12.13.6.3 General Pile Design Requirement . . . . . . . . . . . . . . . . . . 99 12.13.6.4 Batter Piles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 12.13.6.5 Pile Anchorage Requirements . . . . . . . . . . . . . . . . . . . . . 99 12.13.6.6 Splices of Pile Segments . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.13.6.7 Pile Soil Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.13.6.8 Pile Group Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.14 Simpliﬁ ed Alternative Structural Design Criteria for Simple Bearing Wall or Building Frame Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.14.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.14.1.1 Simpliﬁ ed Design Procedure . . . . . . . . . . . . . . . . . . . . . . 100 12.14.1.2 Reference Documents. . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12.14.1.3 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12.14.1.4 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12.14.2 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 12.14.3 Seismic Load Effects and Combinations . . . . . . . . . . . . . . . . . . . . . . . 104 12.14.3.1 Seismic Load Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 12.14.3.2 Seismic Load Effect Including a 2.5 Overstrength Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 12.14.4 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.4.1 Selection and Limitations . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.4.2 Combinations of Framing Systems . . . . . . . . . . . . . . . . . 106 12.14.5 Diaphragm Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.6 Application of Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.7 Design and Detailing Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.7.1 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 12.14.7.2 Openings or Reentrant Building Corners . . . . . . . . . . . . 107 12.14.7.3 Collector Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 12.14.7.4 Diaphragms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 12.14.7.5 Anchorage of Structural Walls . . . . . . . . . . . . . . . . . . . . . 107 12.14.7.6 Bearing Walls and Shear Walls . . . . . . . . . . . . . . . . . . . . 108 12.14.7.7 Anchorage of Nonstructural Systems . . . . . . . . . . . . . . . 108 12.14.8 Simpliﬁ ed Lateral Force Analysis Procedure . . . . . . . . . . . . . . . . . . . 108 12.14.8.1 Seismic Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 12.14.8.2 Vertical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 12.14.8.3 Horizontal Shear Distribution . . . . . . . . . . . . . . . . . . . . . 108 12.14.8.4 Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12.14.8.5 Drift Limits and Building Separation . . . . . . . . . . . . . . . 109 13 Seismic Design Requirements for Nonstructural Components . . . . . . . . . . . . . . . . . . . . . . 111 13.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.2 Seismic Design Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.3 Component Importance Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.4 Exemptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.5 Application of Nonstructural Component Requirements to Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.1.6 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 13.1.7 Reference Documents Using Allowable Stress Design . . . . . . . . . . . 112 13.2 General Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 13.2.1 Applicable Requirements for Architectural, Mechanical, and Electrical Components, Supports, and Attachments . . . . . . . . . . . . . . 112 13.2.2 Special Certiﬁ cation Requirements for Designated Seismic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 CONTENTS xxiii 13.2.3 Consequential Damage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.2.4 Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.2.5 Testing Alternative for Seismic Capacity Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.2.6 Experience Data Alternative for Seismic Capacity Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.2.7 Construction Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.3 Seismic Demands on Nonstructural Components . . . . . . . . . . . . . . . . . . . . . . . . 113 13.3.1 Seismic Design Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 13.3.2 Seismic Relative Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 13.3.2.1 Displacements within Structures . . . . . . . . . . . . . . . . . . . 114 13.3.2.2 Displacements between Structures . . . . . . . . . . . . . . . . . . 114 13.4 Nonstructural Component Anchorage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.1 Design Force in the Attachment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.2 Anchors in Concrete or Masonry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.2.1 Anchors in Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.2.2 Anchors in Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.2.3 Post-Installed Anchors in Concrete and Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.3 Installation Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.4 Multiple Attachments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.5 Power Actuated Fasteners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.4.6 Friction Clips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.5 Architectural Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.2 Forces and Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.3 Exterior Nonstructural Wall Elements and Connections . . . . . . . . . . . 116 13.5.4 Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.5 Out-of-Plane Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.6 Suspended Ceilings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.6.1 Seismic Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.5.6.2 Industry Standard Construction for Acoustical Tile or Lay-in Panel Ceilings . . . . . . . . . . . . . . . . . . . . . . . . . 117 13.5.6.3 Integral Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.7 Access Floors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.7.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.7.2 Special Access Floors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.8 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.8.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.5.8.2 Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13.5.9 Glass in Glazed Curtain Walls, Glazed Storefronts, and Glazed Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13.5.9.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13.5.9.2 Seismic Drift Limits for Glass Components . . . . . . . . . . 119 13.6 Mechanical and Electrical Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13.6.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13.6.2 Component Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.6.3 Mechanical Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.6.4 Electrical Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.6.5 Component Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.6.5.1 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.6.5.2 Design for Relative Displacement . . . . . . . . . . . . . . . . . . 122 13.6.5.3 Support Attachment to Component . . . . . . . . . . . . . . . . . 122 CONTENTS xxiv 13.6.5.4 Material Detailing Requirements . . . . . . . . . . . . . . . . . . . 122 13.6.5.5 Additional Requirements . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.6.5.6 Conduit, Cable Tray, and Other Electrical Distribution Systems (Raceways) . . . . . . . . . . . . . . . . . . 123 13.6.6 Utility and Service Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 13.6.7 Ductwork . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 13.6.8 Piping Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 13.6.8.1 ASME Pressure Piping Systems . . . . . . . . . . . . . . . . . . . 124 13.6.8.2 Fire Protection Sprinkler Piping Systems . . . . . . . . . . . . 124 13.6.8.3 Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 13.6.9 Boilers and Pressure Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 13.6.10 Elevator and Escalator Design Requirements . . . . . . . . . . . . . . . . . . . 125 13.6.10.1 Escalators, Elevators, and Hoistway Structural System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 13.6.10.2 Elevator Equipment and Controller Supports and Attachments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 13.6.10.3 Seismic Controls for Elevators . . . . . . . . . . . . . . . . . . . . 125 13.6.10.4 Retainer Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 13.6.11 Other Mechanical and Electrical Components . . . . . . . . . . . . . . . . . . 125 14 Material Speciﬁ c Seismic Design and Detailing Requirements . . . . . . . . . . . . . . . . . . . . . 127 14.0 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1 Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.1 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.2 Structural Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.2.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.2.2 Seismic Requirements for Structural Steel Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.3 Cold-Formed Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.3.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.3.2 Seismic Requirements for Cold-Formed Steel Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 14.1.3.3 Modiﬁ cations to AISI S110 . . . . . . . . . . . . . . . . . . . . . . . 128 14.1.4 Cold-Formed Steel Light-Frame Construction . . . . . . . . . . . . . . . . . . 128 14.1.4.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 14.1.4.2 Seismic Requirements for Cold-Formed Steel Light-Frame Construction . . . . . . . . . . . . . . . . . . . . . . . . 128 14.1.4.3 Prescriptive Cold-Formed Steel Light-Frame Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.1.5 Steel Deck Diaphragms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.1.6 Steel Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.1.7 Additional Detailing Requirements for Steel Piles in Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2.1 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2.2 Modiﬁ cations to ACI 318 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2.2.1 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2.2.2 ACI 318, Section 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.2.2.3 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 14.2.2.4 Intermediate Precast Structural Walls . . . . . . . . . . . . . . . 130 14.2.2.5 Wall Piers and Wall Segments . . . . . . . . . . . . . . . . . . . . . 130 14.2.2.6 Special Precast Structural Walls . . . . . . . . . . . . . . . . . . . 130 14.2.2.7 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 CONTENTS xxv 14.2.2.8 Detailed Plain Concrete Shear Walls . . . . . . . . . . . . . . . . 130 14.2.2.9 Strength Requirements for Anchors . . . . . . . . . . . . . . . . . 131 14.2.3 Additional Detailing Requirements for Concrete Piles . . . . . . . . . . . . 131 14.2.3.1 Concrete Pile Requirements for Seismic Design Category C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 14.2.3.2 Concrete Pile Requirements for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . . . . 132 14.3 Composite Steel And Concrete Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.3.1 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.3.2 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.3.3 Seismic Requirements for Composite Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.3.4 Metal-Cased Concrete Piles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4 Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4.1 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4.2 R factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4.3 Modiﬁ cations to Chapter 1 of TMS 402/ACI 530/ASCE 5 . . . . . . . . 134 14.4.3.1 Separation Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4.4 Modiﬁ cations to Chapter 2 of TMS 402/ACI 530/ASCE 5 . . . . . . . . 134 14.4.4.1 Stress Increase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4.4.2 Reinforcement Requirements and Details . . . . . . . . . . . . 134 14.4.5 Modiﬁ cations to Chapter 3 of TMS 402/ACI 530/ASCE 5 . . . . . . . . 135 14.4.5.1 Anchoring to Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 14.4.5.2 Splices in Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . 135 14.4.5.3 Coupling Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 14.4.5.4 Deep Flexural Members . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.4.5.5 Walls with Factored Axial Stress Greater Than 0.05 fm′. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.4.5.6 Shear Keys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.4.6 Modiﬁ cations to Chapter 6 of TMS 402/ACI 530/ASCE 5 . . . . . . . . 136 14.4.6.1 Corrugated Sheet Metal Anchors . . . . . . . . . . . . . . . . . . . 136 14.4.7 Modiﬁ cations to TMS 602/ACI 530.1/ASCE 6 . . . . . . . . . . . . . . . . . 136 14.4.7.1 Construction Procedures. . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.5 Wood . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.5.1 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.5.2 Framing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 15 Seismic Design Requirements for Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . 139 15.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 15.1.1 Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 15.1.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 15.1.3 Structural Analysis Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . 139 15.2 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 15.3 Nonbuilding Structures Supported by Other Structures . . . . . . . . . . . . . . . . . . . 139 15.3.1 Less Than 25 percent Combined Weight Condition . . . . . . . . . . . . . . 139 15.3.2 Greater Than or Equal to 25 Percent Combined Weight Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 15.3.3 Architectural, Mechanical, and Electrical Components . . . . . . . . . . . 140 15.4 Structural Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 15.4.1 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 15.4.1.1 Importance Factor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 15.4.2 Rigid Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 15.4.3 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 CONTENTS xxvi 15.4.4 Fundamental Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 15.4.5 Drift Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.6 Materials Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.7 Deﬂ ection Limits and Structure Separation . . . . . . . . . . . . . . . . . . . . 145 15.4.8 Site-Speciﬁ c Response Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.9 Anchors in Concrete or Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.9.1 Anchors in Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.9.2 Anchors in Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.4.9.3 Post-Installed Anchors in Concrete and Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.5 Nonbuilding Structures Similar to Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.5.2 Pipe Racks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.5.2.1 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 15.5.3 Steel Storage Racks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.5.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.5.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.5.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.5.3.4 Alternative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.5.3.5 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.3.6 Operating Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.3.7 Vertical Distribution of Seismic Forces . . . . . . . . . . . . . . 147 15.5.3.8 Seismic Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.4 Electrical Power Generating Facilities . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.4.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.4.2 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.5 Structural Towers for Tanks and Vessels . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.6 Piers and Wharves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.6.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.5.6.2 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.6 General Requirements for Nonbuilding Structures Not Similar to Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 15.6.1 Earth-Retaining Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 15.6.2 Stacks and Chimneys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 15.6.3 Amusement Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 15.6.4 Special Hydraulic Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.6.4.1 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.6.5 Secondary Containment Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.6.5.1 Freeboard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.6.6 Telecommunication Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.7 Tanks and Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.7.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.7.2 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.7.3 Strength and Ductility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 15.7.4 Flexibility of Piping Attachments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 15.7.5 Anchorage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 15.7.6 Ground-Supported Storage Tanks for Liquids . . . . . . . . . . . . . . . . . . 152 15.7.6.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 15.7.7 Water Storage and Water Treatment Tanks and Vessels . . . . . . . . . . . 155 15.7.7.1 Welded Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 15.7.7.2 Bolted Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 15.7.7.3 Reinforced and Prestressed Concrete . . . . . . . . . . . . . . . . 155 CONTENTS xxvii 15.7.8 Petrochemical and Industrial Tanks and Vessels Storing Liquids . . . 155 15.7.8.1 Welded Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 15.7.8.2 Bolted Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 15.7.8.3 Reinforced and Prestressed Concrete . . . . . . . . . . . . . . . . 155 15.7.9 Ground-Supported Storage Tanks for Granular Materials . . . . . . . . . 156 15.7.9.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.9.2 Lateral Force Determination . . . . . . . . . . . . . . . . . . . . . . 156 15.7.9.3 Force Distribution to Shell and Foundation . . . . . . . . . . 156 15.7.9.4 Welded Steel Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.9.5 Bolted Steel Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.9.6 Reinforced Concrete Structures Reinforced concrete structures for the storage of granular materials shall be designed in accordance with the seismic force requirements of this standard and the requirements of ACI 313. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.9.7 Prestressed Concrete Structures . . . . . . . . . . . . . . . . . . . . 156 15.7.10 Elevated Tanks and Vessels for Liquids and Granular Materials . . . . 156 15.7.10.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.10.2 Effective Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 15.7.10.3 P-Delta Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 15.7.10.4 Transfer of Lateral Forces into Support Tower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 15.7.10.5 Evaluation of Structures Sensitive to Buckling Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 15.7.10.6 Welded Steel Water Storage Structures . . . . . . . . . . . . . . 157 15.7.10.7 Concrete Pedestal (Composite) Tanks . . . . . . . . . . . . . . . 157 15.7.11 Boilers and Pressure Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 15.7.11.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 15.7.11.2 ASME Boilers and Pressure Vessels . . . . . . . . . . . . . . . . 158 15.7.11.3 Attachments of Internal Equipment and Refractory . . . . 158 15.7.11.4 Coupling of Vessel and Support Structure . . . . . . . . . . . . 158 15.7.11.5 Effective Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 15.7.11.6 Other Boilers and Pressure Vessels . . . . . . . . . . . . . . . . . 158 15.7.11.7 Supports and Attachments for Boilers and Pressure Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 15.7.12 Liquid and Gas Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 15.7.12.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 15.7.12.2 ASME Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 15.7.12.3 Attachments of Internal Equipment and Refractory . . . . 159 15.7.12.4 Effective Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 15.7.12.5 Post and Rod Supported . . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.7.12.6 Skirt Supported . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.7.13 Refrigerated Gas Liquid Storage Tanks and Vessels. . . . . . . . . . . . . . 160 15.7.13.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.7.14 Horizontal, Saddle Supported Vessels for Liquid or Vapor Storage . . . 160 15.7.14.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.7.14.2 Effective Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.7.14.3 Vessel Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 16 Seismic Response History Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 16.1 Linear Response History Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 16.1.1 Analysis Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 16.1.2 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 CONTENTS xxviii 16.1.3 Ground Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 16.1.3.1 Two-Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . 161 16.1.3.2 Three-Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . 161 16.1.4 Response Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 16.1.5 Horizontal Shear Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 16.2 Nonlinear Response History Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 16.2.1 Analysis Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 16.2.2 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 16.2.3 Ground Motion and Other Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 16.2.4 Response Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 16.2.4.1 Member Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 16.2.4.2 Member Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 16.2.4.3 Story Drift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 16.2.5 Design Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 17 Seismic Design Requirements for Seismically Isolated Structures . . . . . . . . . . . . . . . . . . 165 17.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 17.1.1 Variations in Material Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 17.1.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 17.1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 17.2 General Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.1 Importance Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.2 MCER Spectral Response Acceleration Parameters, SMS and SM1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.3 Conﬁ guration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4 Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.1 Environmental Conditions . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.2 Wind Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.3 Fire Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.4 Lateral Restoring Force . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.5 Displacement Restraint. . . . . . . . . . . . . . . . . . . . . . . . . . . 167 17.2.4.6 Vertical-Load Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.4.7 Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.4.8 Inspection and Replacement . . . . . . . . . . . . . . . . . . . . . . 168 17.2.4.9 Quality Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.5 Structural System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.5.1 Horizontal Distribution of Force . . . . . . . . . . . . . . . . . . . 168 17.2.5.2 Building Separations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.5.3 Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.2.6 Elements of Structures and Nonstructural Components . . . . . . . . . . . 168 17.2.6.1 Components at or above the Isolation Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.2.6.2 Components Crossing the Isolation Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.2.6.3 Components below the Isolation Interface . . . . . . . . . . . 169 17.3 Ground Motion for Isolated Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.3.1 Design Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.3.2 Ground Motion Histories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.4 Analysis Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.4.1 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.4.2 Dynamic Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.4.2.1 Response-Spectrum Procedure . . . . . . . . . . . . . . . . . . . . . 169 17.4.2.2 Response-History Procedure . . . . . . . . . . . . . . . . . . . . . . 170 CONTENTS xxix 17.5 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 17.5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 17.5.2 Deformation Characteristics of the Isolation System . . . . . . . . . . . . . 170 17.5.3 Minimum Lateral Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 17.5.3.1 Design Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 17.5.3.2 Effective Period at Design Displacement . . . . . . . . . . . . 170 17.5.3.3 Maximum Displacement . . . . . . . . . . . . . . . . . . . . . . . . . 170 17.5.3.4 Effective Period at Maximum Displacement . . . . . . . . . . 171 17.5.3.5 Total Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 17.5.4 Minimum Lateral Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 17.5.4.1 Isolation System and Structural Elements below the Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 17.5.4.2 Structural Elements above the Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.5.4.3 Limits on Vs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.5.5 Vertical Distribution of Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.5.6 Drift Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.6 Dynamic Analysis Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.6.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.6.2 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.6.2.1 Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 17.6.2.2 Isolated Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.6.3 Description of Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.6.3.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.6.3.2 Input Earthquake . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.6.3.3 Response-Spectrum Procedure . . . . . . . . . . . . . . . . . . . . . 173 17.6.3.4 Response-History Procedure . . . . . . . . . . . . . . . . . . . . . . 173 17.6.4 Minimum Lateral Displacements and Forces . . . . . . . . . . . . . . . . . . . 174 17.6.4.1 Isolation System and Structural Elements below the Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 17.6.4.2 Structural Elements above the Isolation System . . . . . . . 174 17.6.4.3 Scaling of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 17.6.4.4 Drift Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 17.7 Design Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8 Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8.2 Prototype Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8.2.1 Record . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8.2.2 Sequence and Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 17.8.2.3 Units Dependent on Loading Rates . . . . . . . . . . . . . . . . . 175 17.8.2.4 Units Dependent on Bilateral Load . . . . . . . . . . . . . . . . . 176 17.8.2.5 Maximum and Minimum Vertical Load . . . . . . . . . . . . . 176 17.8.2.6 Sacriﬁ cial Wind-Restraint Systems . . . . . . . . . . . . . . . . . 176 17.8.2.7 Testing Similar Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 17.8.3 Determination of Force-Deﬂ ection Characteristics . . . . . . . . . . . . . . . 176 17.8.4 Test Specimen Adequacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 17.8.5 Design Properties of the Isolation System . . . . . . . . . . . . . . . . . . . . . 177 17.8.5.1 Maximum and Minimum Effective Stiffness . . . . . . . . . 177 17.8.5.2 Effective Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 18 Seismic Design Requirements for Structures with Damping Systems . . . . . . . . . . . . . . . . 179 18.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 18.1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 CONTENTS xxx 18.1.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 18.1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 18.2 General Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.1 Seismic Design Category A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.2 System Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.2.1 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . 182 18.2.2.2 Damping System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.3 Ground Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.3.1 Design Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.3.2 Ground Motion Histories . . . . . . . . . . . . . . . . . . . . . . . . . 182 18.2.4 Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.2.4.1 Nonlinear Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.2.4.2 Response-Spectrum Procedure . . . . . . . . . . . . . . . . . . . . . 183 18.2.4.3 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . 183 18.2.5 Damping System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.2.5.1 Device Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.2.5.2 Multiaxis Movement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.2.5.3 Inspection and Periodic Testing . . . . . . . . . . . . . . . . . . . . 183 18.2.5.4 Quality Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 18.3 Nonlinear Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 18.3.1 Nonlinear Response-History Procedure . . . . . . . . . . . . . . . . . . . . . . . 184 18.3.1.1 Damping Device Modeling . . . . . . . . . . . . . . . . . . . . . . . 184 18.3.1.2 Response Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 18.3.2 Nonlinear Static Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 18.4 Response-Spectrum Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 18.4.1 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 18.4.2 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 18.4.2.1 Seismic Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 18.4.2.2 Modal Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 18.4.2.3 Modal Participation Factor . . . . . . . . . . . . . . . . . . . . . . . . 185 18.4.2.4 Fundamental Mode Seismic Response Coefﬁ cient . . . . . 185 18.4.2.5 Effective Fundamental Mode Period Determination . . . . 185 18.4.2.6 Higher Mode Seismic Response Coefﬁ cient . . . . . . . . . . 186 18.4.2.7 Design Lateral Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 18.4.3 Damping System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 18.4.3.1 Design Earthquake Floor Deﬂ ection . . . . . . . . . . . . . . . . 186 18.4.3.2 Design Earthquake Roof Displacement . . . . . . . . . . . . . . 186 18.4.3.3 Design Earthquake Story Drift . . . . . . . . . . . . . . . . . . . . . 186 18.4.3.4 Design Earthquake Story Velocity . . . . . . . . . . . . . . . . . . 186 18.4.3.5 Maximum Considered Earthquake Response . . . . . . . . . 187 18.5 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 18.5.1 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 18.5.2 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 18.5.2.1 Seismic Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 18.5.2.2 Fundamental Mode Base Shear . . . . . . . . . . . . . . . . . . . . 187 18.5.2.3 Fundamental Mode Properties . . . . . . . . . . . . . . . . . . . . . 187 18.5.2.4 Fundamental Mode Seismic Response Coefﬁ cient . . . . . 188 18.5.2.5 Effective Fundamental Mode Period Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 18.5.2.6 Residual Mode Base Shear . . . . . . . . . . . . . . . . . . . . . . . 188 18.5.2.7 Residual Mode Properties . . . . . . . . . . . . . . . . . . . . . . . . 188 18.5.2.8 Residual Mode Seismic Response Coefﬁ cient . . . . . . . . 188 18.5.2.9 Design Lateral Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 CONTENTS xxxi 18.5.3 Damping System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 18.5.3.1 Design Earthquake Floor Deﬂ ection . . . . . . . . . . . . . . . . 189 18.5.3.2 Design Earthquake Roof Displacement . . . . . . . . . . . . . . 189 18.5.3.3 Design Earthquake Story Drift . . . . . . . . . . . . . . . . . . . . . 189 18.5.3.4 Design Earthquake Story Velocity . . . . . . . . . . . . . . . . . . 189 18.5.3.5 Maximum Considered Earthquake Response . . . . . . . . . 190 18.6 Damped Response Modiﬁ cation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 18.6.1 Damping Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 18.6.2 Effective Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 18.6.2.1 Inherent Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 18.6.2.2 Hysteretic Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 18.6.2.3 Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 18.6.3 Effective Ductility Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 18.6.4 Maximum Effective Ductility Demand . . . . . . . . . . . . . . . . . . . . . . . . 192 18.7 Seismic Load Conditions and Acceptance Criteria . . . . . . . . . . . . . . . . . . . . . . . 192 18.7.1 Nonlinear Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 18.7.1.1 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . 192 18.7.1.2 Damping Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 18.7.1.3 Combination of Load Effects . . . . . . . . . . . . . . . . . . . . . . 193 18.7.1.4 Acceptance Criteria for the Response Parameters of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 18.7.2 Response-Spectrum and Equivalent Lateral Force Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 18.7.2.1 Seismic Force-Resisting System . . . . . . . . . . . . . . . . . . . 193 18.7.2.2 Damping System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 18.7.2.3 Combination of Load Effects . . . . . . . . . . . . . . . . . . . . . . 193 18.7.2.4 Modal Damping System Design Forces . . . . . . . . . . . . . 193 18.7.2.5 Seismic Load Conditions and Combination of Modal Responses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 18.7.2.6 Inelastic Response Limits . . . . . . . . . . . . . . . . . . . . . . . . 195 18.8 Design Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 18.9 Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 18.9.1 Prototype Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 18.9.1.1 Data Recording . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 18.9.1.2 Sequence and Cycles of Testing . . . . . . . . . . . . . . . . . . . 196 18.9.1.3 Testing Similar Devices . . . . . . . . . . . . . . . . . . . . . . . . . . 196 18.9.1.4 Determination of Force-Velocity-Displacement Characteristics . . . . . . . . . 196 18.9.1.5 Device Adequacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 18.9.2 Production Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 19 Soil–Structure Interaction for Seismic Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 19.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 19.2 Equivalent Lateral Force Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 19.2.1 Base Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 19.2.1.1 Effective Building Period . . . . . . . . . . . . . . . . . . . . . . . . . 199 19.2.1.2 Effective Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 19.2.2 Vertical Distribution of Seismic Forces. . . . . . . . . . . . . . . . . . . . . . . . 201 19.2.3 Other Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 19.3 Modal Analysis Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 19.3.1 Modal Base Shears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 19.3.2 Other Modal Effects. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 19.3.3 Design Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 CONTENTS xxxii 20 Site Classiﬁ cation Procedure for Seismic Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.1 Site Classiﬁ cation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.2 Site Response Analysis for Site Class F Soil . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3 Site Class Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3.1 Site Class F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3.2 Soft Clay Site Class E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3.3 Site Classes C, D, and E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3.4 Shear Wave Velocity for Site Class B . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.3.5 Shear Wave Velocity for Site Class A . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.4 Deﬁ nitions of Site Class Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 20.4.1 v _ s, Average Shear Wave Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 20.4.2 N _ , Average Field Standard Penetration Resistance and N _ ch, Average Standard Penetration Resistance for Cohesionless Soil Layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 20.4.3 s _ u, Average Undrained Shear Strength . . . . . . . . . . . . . . . . . . . . . . . . 204 21 Site-Speciﬁ c Ground Motion Procedures for Seismic Design . . . . . . . . . . . . . . . . . . . . . . 207 21.1 Site Response Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 21.1.1 Base Ground Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 21.1.2 Site Condition Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 21.1.3 Site Response Analysis and Computed Results . . . . . . . . . . . . . . . . . 207 21.2 Risk-Targeted Maximum Considered Earthquake (Mcer) Ground Motion Hazard Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 21.2.1 Probabilistic (MCER) Ground Motions . . . . . . . . . . . . . . . . . . . . . . . 208 21.2.1.1 Method 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 21.2.1.2 Method 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 21.2.2 Deterministic (MCER) Ground Motions . . . . . . . . . . . . . . . . . . . . . . . 208 21.2.3 Site-Speciﬁ c MCER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 21.3 Design Response Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 21.4 Design Acceleration Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 21.5 Maximum Considered Earthquake Geometric Mean (Mceg) Peak Ground Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 21.5.1 Probabilistic MCEG Peak Ground Acceleration . . . . . . . . . . . . . . . . . 209 21.5.2 Deterministic MCEG Peak Ground Acceleration . . . . . . . . . . . . . . . . 209 21.5.3 Site-Speciﬁ c MCEG Peak Ground Acceleration . . . . . . . . . . . . . . . . . 209 22 Seismic Ground Motion Long-Period Transition and Risk Coefﬁ cient Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 23 Seismic Design Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 23.1 Consensus Standards and Other Reference Documents . . . . . . . . . . . . . . . . . . . 233 26 Wind Loads: General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 26.1 Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 26.1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 26.1.2 Permitted Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 26.1.2.1 Main Wind-Force Resisting System (MWFRS) . . . . . . . 241 26.1.2.2 Components and Cladding . . . . . . . . . . . . . . . . . . . . . . . . 241 26.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 26.3 Symbols and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 26.4 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 26.4.1 Sign Convention. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 26.4.2 Critical Load Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 26.4.3 Wind Pressures Acting on Opposite Faces of Each Building Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 CONTENTS xxxiii 26.5 Wind Hazard Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.5.1 Basic Wind Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.5.2 Special Wind Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.5.3 Estimation of Basic Wind Speeds from Regional Climatic Data . . . . 246 26.5.4 Limitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.6 Wind Directionality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.7 Exposure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.7.1 Wind Directions and Sectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.7.2 Surface Roughness Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 26.7.3 Exposure Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 26.7.4 Exposure Requirements. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 26.7.4.1 Directional Procedure (Chapter 27) . . . . . . . . . . . . . . . . . 251 26.7.4.2 Envelope Procedure (Chapter 28) . . . . . . . . . . . . . . . . . . 251 26.7.4.3 Directional Procedure for Building Appurtenances and Other Structures (Chapter 29) . . . . . . . . . . . . . . . . . . 251 26.7.4.4 Components and Cladding (Chapter 30) . . . . . . . . . . . . . 251 26.8 Topographic Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 26.8.1 Wind Speed-Up over Hills, Ridges, and Escarpments . . . . . . . . . . . . 251 26.8.2 Topographic Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.9 Gust-Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.9.2 Frequency Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.9.2.1 Limitations for Approximate Natural Frequency . . . . . . 254 26.9.3 Approximate Natural Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.9.4 Rigid Buildings or Other Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 254 26.9.5 Flexible or Dynamically Sensitive Buildings or Other Structures . . . 255 26.9.6 Rational Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.9.7 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.10 Enclosure Classiﬁ cation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.10.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.10.2 Openings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.10.3 Protection of Glazed Openings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 26.10.3.1 Wind-borne Debris Regions . . . . . . . . . . . . . . . . . . . . . . . 255 26.10.3.2 Protection Requirements for Glazed Openings . . . . . . . . 257 26.10.4 Multiple Classiﬁ cations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 26.11 Internal Pressure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 26.11.1 Internal Pressure Coefﬁ cients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 26.11.1.1 Reduction Factor for Large Volume Buildings, Ri . . . . . 257 27 Wind Loads on Buildings—MWFRS (Directional Procedure) . . . . . . . . . . . . . . . . . . . . . . 259 27.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.1.1 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.1.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.1.3 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.1.4 Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Part 1: Enclosed, Partially Enclosed, and Open Buildings of All Heights . . . . . . . . . . . . . 259 27.2 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.2.1 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 259 27.3 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 259 27.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 27.4 Wind Loads—Main Wind Force-Resisting System . . . . . . . . . . . . . . . . . . . . . . . 260 27.4.1 Enclosed and Partially Enclosed Rigid Buildings . . . . . . . . . . . . . . . . 260 27.4.2 Enclosed and Partially Enclosed Flexible Buildings . . . . . . . . . . . . . 262 CONTENTS xxxiv 27.4.3 Open Buildings with Monoslope, Pitched, or Troughed Free Roofs . 262 27.4.4 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 27.4.5 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 27.4.6 Design Wind Load Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 27.4.7 Minimum Design Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 Part 2: Enclosed Simple Diaphragm Buildings with h ≤ 160 ft (48.8 m) . . . . . . . . . . . . . 272 27.5 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.5.1 Design Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.5.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.5.3 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 272 27.5.4 Diaphragm Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.6 Wind Loads—Main Wind Force-Resisting System . . . . . . . . . . . . . . . . . . . . . . . 273 27.6.1 Wall and Roof Surfaces—Class 1 and 2 Buildings . . . . . . . . . . . . . . 273 27.6.2 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 27.6.3 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 28 Wind Loads on Buildings—MWFRS (Envelope Procedure) . . . . . . . . . . . . . . . . . . . . . . . 297 28.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.1.1 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.1.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.1.3 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.1.4 Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 Part 1: Enclosed and Partially Enclosed Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . 297 28.2 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.2.1 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 297 28.3 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 28.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 297 28.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 28.4 Wind Loads—Main Wind-Force Resisting System . . . . . . . . . . . . . . . . . . . . . . . 298 28.4.1 Design Wind Pressure for Low-Rise Buildings . . . . . . . . . . . . . . . . . 298 28.4.1.1 External Pressure Coefﬁ cients (GCpf) . . . . . . . . . . . . . . . 298 28.4.2 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 28.4.3 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 28.4.4 Minimum Design Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Part 2: Enclosed Simple Diaphragm Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.5 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.5.1 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 302 28.6 Wind Loads—Main Wind-Force Resisting System . . . . . . . . . . . . . . . . . . . . . . . 302 28.6.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.6.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.6.3 Design Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 28.6.4 Minimum Design Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 29 Wind Loads on Other Structures and Building Appurtenances—MWFRS . . . . . . . . . . . . 307 29.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.1.1 Structure Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.1.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.1.3 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.1.4 Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.2 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.2.1 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 307 29.3 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 307 29.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 CONTENTS xxxv 29.4 Design Wind Loads—Solid Freestanding Walls and Solid Signs . . . . . . . . . . . . 308 29.4.1 Solid Freestanding Walls and Solid Freestanding Signs . . . . . . . . . . . 308 29.4.2 Solid Attached Signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 29.5 Design Wind Loads—Other Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 29.5.1 Rooftop Structures and Equipment for Buildings with h ≤ 60 ft (18.3 m). . . . . 308 29.6 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 29.7 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 29.8 Minimum Design wind Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 30 Wind Loads—Components and Cladding (C&C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1.1 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1.2 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1.3 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1.4 Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.1.5 Air-Permeable Cladding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.2 General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 30.2.1 Wind Load Parameters Speciﬁ ed in Chapter 26 . . . . . . . . . . . . . . . . . 315 30.2.2 Minimum Design Wind Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 30.2.3 Tributary Areas Greater than 700 ft2 (65 m2) . . . . . . . . . . . . . . . . . . 316 30.2.4 External Pressure Coefﬁ cients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 30.3 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 30.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 316 30.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Part 1: Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 30.4 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 30.4.1 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 30.4.2 Design Wind Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Part 2: Low-Rise Buildings (Simpliﬁ ed) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 30.5 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 30.5.1 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 30.5.2 Design Wind Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 Part 3: Buildings with h > 60 ft (18.3 m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 30.6 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 30.6.1 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 30.6.2 Design Wind Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 Part 4: Buildings with h ≤ 160 ft (48.8 M) (Simpliﬁ ed). . . . . . . . . . . . . . . . . . . . . . . . . . . 321 30.7 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 30.7.1 Wind Loads—Components And Cladding . . . . . . . . . . . . . . . . . . . . . 321 30.7.1.1 Wall and Roof Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 321 30.7.1.2 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 30.7.1.3 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 Part 5: Open Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 30.8 Building Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 30.8.1 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 30.8.2 Design Wind Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 Part 6: Building Appurtenances and Rooftop Structures and Equipment . . . . . . . . . . . . . . 332 30.9 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 30.10 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 30.11 Rooftop Structures and Equipment for Buildings with h ≤ 60 ft (18.3 m). . . . . 334 31 Wind Tunnel Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.2 Test Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 CONTENTS xxxvi 31.3 Dynamic Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.4 Load Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.4.1 Mean Recurrence Intervals of Load Effects . . . . . . . . . . . . . . . . . . . . 357 31.4.2 Limitations on Wind Speeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.4.3 Limitations on Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.5 Wind-Borne Debris . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 Appendix 11A Quality Assurance Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 11A.1 Quality Assurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 11A.1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 11A.1.2 Quality Assurance Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 11A.1.2.1 Details of Quality Assurance Plan . . . . . . . . . . . . . . . . . . 359 11A.1.2.2 Contractor Responsibility . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3 Special Inspection and Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.1 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.2 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.3 Structural Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.4 Prestressed Concrete. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.5 Structural Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.6 Structural Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.7 Structural Wood . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 11A.1.3.8 Cold-Formed Steel Framing . . . . . . . . . . . . . . . . . . . . . . . 361 11A.1.3.9 Architectural Components . . . . . . . . . . . . . . . . . . . . . . . . 361 11A.1.3.10 Mechanical and Electrical Components . . . . . . . . . . . . . . 361 11A.1.3.11 Seismic Isolation System . . . . . . . . . . . . . . . . . . . . . . . . . 361 11A.2 Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 11A.2.1 Reinforcing and Prestressing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 11A.2.1.1 Certiﬁ ed Mill Test Reports . . . . . . . . . . . . . . . . . . . . . . . . 361 11A.2.1.2 ASTM A615 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . 362 11A.2.1.3 Welding of ASTM A615 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.2.2 Structural Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.2.3 Structural Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.2.4 Structural Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.2.5 Seismic-Isolated Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.2.6 Mechanical and Electrical Equipment . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.3 Structural Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 11A.4 Reporting and Compliance Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Appendix 11B Existing Building Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 11B.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 11B.2 Structurally Independent Additions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 11B.3 Structurally Dependent Additions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 11B.4 Alterations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 11B.5 Change of Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 Appendix C Serviceability Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C. Serviceability Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.1 Deﬂ ection, Vibration, and Drift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.1.1 Vertical Deﬂ ections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.1.2 Drift of Walls and Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.1.3 Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.2 Design for Long-Term Deﬂ ection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.3 Camber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 CONTENTS xxxvii C.4 Expansion and Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 C.5 Durability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 Appendix D Buildings Exempted from Torsional Wind Load Cases . . . . . . . . . . . . . . . . . . . . . 367 D1.0 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 D1.1 One and Two Story Buildings Meeting the Following Requirements . . . . . . . . 367 D1.2 Buildings Controlled by Seismic Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 D1.2.1 Buildings with Diaphragms at Each Level that Are Not Flexible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 D1.2.2 Buildings with Diaphragms at Each Level that Are Flexible . . . . . . . 367 D1.3 Buildings Classiﬁ ed as Torsionally Regular under Wind Load. . . . . . . . . . . . . . 367 D1.4 Buildings with Diaphragms that are Flexible and Designed for Increased Wind Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 D1.5 Class 1 and Class 2 Simple Diaphragm Buildings (h ≤ 160 ft.) Meeting the Following Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 D1.5.1 Case A – Class 1 and Class 2 Buildings . . . . . . . . . . . . . . . . . . . . . . . 367 D1.5.2 Case B – Class 1 and Class 2 Buildings . . . . . . . . . . . . . . . . . . . . . . . 368 D1.5.3 Case C – Class 1 and Class 2 Buildings . . . . . . . . . . . . . . . . . . . . . . . 368 D1.5.4 Case D – Class 1 and Class 2 Buildings . . . . . . . . . . . . . . . . . . . . . . . 368 D1.5.5 Case E – Class 1 and Class 2 Buildings . . . . . . . . . . . . . . . . . . . . . . . 368 D1.5.6 Case F – Class 1 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 Commentary to American Society of Civil Engineers/Structural Engineering Institute Standard 7-10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 C1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 C1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 C1.3 Basic Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 C1.3.1 Strength and Stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 C1.3.1.3 Performance-Based Procedures . . . . . . . . . . . . . . . . . . . . 375 C1.3.2 Serviceability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 C1.3.3 Self-Straining Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 C1.4 General Structural Integrity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 C1.5 Classiﬁ cation of Buildings and Other Structures . . . . . . . . . . . . . . . . . . . . . . . . . 380 C1.5.1 Risk Categorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 C1.5.3 Toxic, Highly Toxic, and Explosive Substances . . . . . . . . . . . . . . . . . 382 C1.7 Load Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 C2 Combinations of Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 C2.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 C2.2 Symbols and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 C2.3 Combining Factored Loads Using Strength Design . . . . . . . . . . . . . . . . . . . . . . 387 C2.3.1 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 C2.3.2 Basic Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 C2.3.3 Load Combinations Including Flood Load . . . . . . . . . . . . . . . . . . . . . 389 C2.3.4 Load Combinations Including Atmospheric Ice Loads . . . . . . . . . . . . 389 C2.3.5 Load Combinations Including Self-Straining Loads . . . . . . . . . . . . . . 389 C2.3.6 Load Combinations for Nonspeciﬁ ed Loads . . . . . . . . . . . . . . . . . . . . 390 C2.4 Combining Nominal Loads Using Allowable Stress Design . . . . . . . . . . . . . . . 391 C2.4.1 Basic Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 C2.4.2 Load Combinations Including Flood Load . . . . . . . . . . . . . . . . . . . . . 393 C2.4.3 Load Combinations Including Atmospheric Ice Loads . . . . . . . . . . . . 393 C2.4.4 Load Combinations Including Self-Straining Loads . . . . . . . . . . . . . . 393 CONTENTS xxxviii C2.5 Load Combinations for Extraordinary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 C3 Dead Loads, Soil Loads, and Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 C3.1.2 Weights of Materials and Constructions . . . . . . . . . . . . . . . . . . . . . . . 397 C3.2 Soil Loads and Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 C3.2.1 Lateral Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 C3.2.2 Uplift on Floors and Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 C4 Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 C4.3 Uniformly Distributed Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 C4.3.1 Required Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 C4.3.2 Provision for Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 C4.3.3 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 C4.4 Concentrated Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.5 Loads on Handrail, Guardrail, Grab Bar, and Vehicle Barrier Systems, and Fixed Ladders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.5.1 Loads on Handrail and Guardrail Systems . . . . . . . . . . . . . . . . . . . . . 409 C4.5.2 Loads on Grab Bar Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.5.3 Loads on Vehicle Barrier Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.5.4 Loads on Fixed Ladders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.6 Impact Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 C4.7 Reduction In Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 C4.7.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 C4.7.3 Heavy Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 C4.7.4 Passenger Vehicle Garages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 C4.7.6 Limitations on One-Way Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C4.8 Reduction In Roof Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C4.8.2 Flat, Pitched, and Curved Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C4.8.3 Special Purpose Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C4.9 Crane Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C5 Flood Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 C5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 C5.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 C5.3 Design Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 C5.3.1 Design Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 C5.3.2 Erosion and Scour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 C5.3.3 Loads on Breakaway Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 C5.4.1 Load Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 C5.4.2 Hydrostatic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 C5.4.3 Hydrodynamic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 C5.4.4 Wave Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 C5.4.4.2 Breaking Wave Loads on Vertical Walls . . . . . . . . . . . . . 418 C5.4.5 Impact Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 C7 Snow Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 C7.0 Snow Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 C7.2 Ground Snow Loads, pg . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 C7.3 Flat-Roof Snow Loads, pf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 C7.3.1 Exposure Factor, Ce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 C7.3.2 Thermal Factor, Ct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 CONTENTS xxxix C7.3.3 Importance Factor, Is . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 C7.3.4 Minimum Snow Load for Low-Slope Roofs, pm . . . . . . . . . . . . . . . . 429 C7.4 Sloped Roof Snow Loads, ps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 C7.4.3 Roof Slope Factor for Curved Roofs . . . . . . . . . . . . . . . . . . . . . . . . . 430 C7.4.4 Roof Slope Factor for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 C7.4.5 Ice Dams and Icicles Along Eaves . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 C7.5 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 C7.6 Unbalanced Roof Snow Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 C7.6.1 Unbalanced Snow Loads for Hip and Gable Roofs . . . . . . . . . . . . . . 431 C7.6.2 Unbalanced Snow Loads for Curved Roofs . . . . . . . . . . . . . . . . . . . . 431 C7.6.3 Unbalanced Snow Loads for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 C7.6.4 Unbalanced Snow Loads for Dome Roofs . . . . . . . . . . . . . . . . . . . . . 432 C7.7 Drifts on Lower Roofs (Aerodynamic Shade) . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 C7.7.2 Adjacent Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 C7.8 Roof Projections and Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 C7.9 Sliding Snow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 C7.10 Rain-on-Snow Surcharge Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 C7.11 Ponding Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 C7.12 Existing Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 C7.13 Other Roofs and Sites. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 C8 Rain Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 C8.1 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 C8.2 Roof Drainage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 C8.3 Design Rain Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 C8.4 Ponding Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 C8.5 Controlled Drainage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 C10 Ice Loads—Atmospheric Icing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 C10.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 C10.1.1 Site-Speciﬁ c Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 C10.1.2 Dynamic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 C10.1.3 Exclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 C10.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 C10.4 Ice Loads Due to Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 C10.4.1 Ice Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 C10.4.2 Nominal Ice Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 C10.4.4 Importance Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460 C10.4.6 Design Ice Thickness for Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . 461 C10.5 Wind on Ice-Covered Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 C10.5.5 Wind on Ice-Covered Guys and Cables . . . . . . . . . . . . . . . . . . . . . . . 461 C10.6 Design Temperatures for Freezing Rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 C10.7 Partial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 C11 Seismic Design Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 C11.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 C11.1.1 Purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 C11.1.3 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 C11.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 CONTENTS xl C11.4 Seismic Ground Motion Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476 C11.7 Design Requirements for Seismic Design Category A . . . . . . . . . . . . . . . . . . . . 477 C11.8.2 Geotechnical Investigation Report Requirements for Seismic Design Categories C through F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 C11.8.3 Additional Geotechnical Investigation Report Requirements for Seismic Design Categories D through F . . . . . . . . . . . . . . . . . . . . . . . 477 C12 Seismic Design Requirements for Building Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 C12.3.3.3 Elements Supporting Discontinuous Walls or Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 C12.3.4 Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 C12.4.3 Seismic Load Effect Including Overstrength Factor . . . . . . . . . . . . . . 479 C12.6 Analysis Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 C12.7.1 Foundation Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 C12.8.4.1 Inherent Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 C12.11.2 Anchorage of Structural Walls and Transfer of Design Forces into Diaphragms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 C13 Seismic Design Requirements for Nonstructural Components . . . . . . . . . . . . . . . . . . . . . . 483 C13.0 Seismic Design Requirements for Nonstructural Components . . . . . . . . . . . . . . 483 C13.1.4 Exemptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483 C13.2.2 Special Certiﬁ cation Requirements for Designated Seismic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483 C13.3.2 Seismic Relative Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 C13.4.2.3 Post-Installed Anchors in Concrete and Masonry . . . . . . 484 C13.4.6 Friction Clips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 C13.5.9 Glass in Glazed Curtain Walls, Glazed Storefronts, and Glazed Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 C13.6 Mechanical and Electrical Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 C13.6.5.5 Additional Requirements . . . . . . . . . . . . . . . . . . . . . . . . . 485 C13.6.5.6 Conduit, Cable Tray, and Other Electrical Distribution Systems (Raceways) . . . . . . . . . . . . . . . . . . 486 C13.6.8 Piping Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 C13.6.8.1 ASME Pressure Piping Systems . . . . . . . . . . . . . . . . . . . 488 C13.6.8.2 Fire Protection Sprinkler Piping Systems . . . . . . . . . . . . 488 C13.6.8.3 Exceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 C14 Material-Speciﬁ c Seismic Design and Detailing Requirements . . . . . . . . . . . . . . . . . . . . . 489 C14.2 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.1 ACI 318, Section 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.3 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.4 Wall Piers and Wall Segments . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.6 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 C14.2.2.7 Intermediate Precast Structural Walls . . . . . . . . . . . . . . . 489 C14.2.2.8 Detailed Plain Concrete Shear Walls . . . . . . . . . . . . . . . . 490 C14.2.2.9 Strength Requirements for Anchors . . . . . . . . . . . . . . . . . 490 C14.2.3 Additional Detailing Requirements for Concrete Piles . . . . . . . . . . . . 490 C14.4 Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 C15 Seismic Design Requirements for Nonbuilding Structures . . . . . . . . . . . . . . . . . . . . . . . . . 493 C15.0 Seismic Design Requirements for Nonbuilding Structures . . . . . . . . . . . . . . . . . 493 C15.1.3 Structural Analysis Procedure Selection . . . . . . . . . . . . . . . . . . . . . . . 493 CONTENTS xli C15.2 Reference Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 C15.4.4 Fundamental Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 C15.4.9.3 Post-Installed Anchors in Concrete and Masonry . . . . . . 496 C15.6.5 Secondary Containment Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 C15.6.6 Telecommunication Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 C15.7 Tanks and Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 C15.7.2 Design Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 C15.7.6 Ground-Supported Storage Tanks for Liquids . . . . . . . . . . . . . . . . . . 497 C15.7.8.2 Bolted Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 C15.7.13 Refrigerated Gas Liquid Storage Tanks and Vessels . . . . . . . . . . . . . . 498 C19 Soil–Structure Interaction for Seismic Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 C19 Soil–Structure Interaction for Seismic Design . . . . . . . . . . . . . . . . . . . . . . . . . . 501 C22 Seismic Ground Motion, Long-Period Transition and Risk Coefﬁ cient Maps . . . . . . . . . 503 C26 Wind Loads—General Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 C26.1.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 C26.1.2 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 C26.2 Deﬁ nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 C26.3 Symbols and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 C26.4.3 Wind Pressures Acting on Opposite Faces of Each Building Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 C26.5.1 Basic Wind Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 C26.5.2 Special Wind Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 C26.5.3 Estimation of Basic Wind Speeds from Regional Climatic Data . . . . 512 C26.5.4 Limitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 C26.6 Wind Directionality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 C26.7 Exposure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514 C26.7.4 Exposure Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 C26.8 Topographic Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518 C26.9 Gust Effect Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519 C26.10 Enclosure Classiﬁ cation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 C26.11 Internal Pressure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526 Additional References of Interest. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 C27 Wind Loads on Buildings—MWFRS Directional Procedure . . . . . . . . . . . . . . . . . . . . . . . 547 Part 1: Enclosed, Partially Enclosed, and Open Buildings of All Heights . . . . . . . . . . . . . 547 C27.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 547 27.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 C27.4.1 Enclosed and Partially Enclosed Rigid Buildings . . . . . . . . . . . . . . . . 550 C27.4.3 Open Buildings with Monoslope, Pitched, or Troughed Free Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552 C27.4.6 Design Wind Load Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552 C27.4.7 Minimum Design Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553 Part 2: Enclosed Simple Diaphragm Buildings with h ≤ 160 ft . . . . . . . . . . . . . . . . . . . . . 553 C27.6.1 Wall and Roof Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 C27.6.2 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555 C27.6.3 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555 C28 Wind Loads on Buildings—MWFRS (Envelope Procedure) . . . . . . . . . . . . . . . . . . . . . . . 557 Part 1: Enclosed and Partially Enclosed Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . 557 C28.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 557 CONTENTS xlii C28.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557 C28.4.4 Minimum Design Wind Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 Part 2: Enclosed Simple Diaphragm Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . . . . 560 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 C29 Wind Loads (MWFRS)—Other Structures and Building Appurtenances . . . . . . . . . . . . . 563 C29.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 563 C29.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563 C29.4.2 Solid Attached Signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563 C29.6 Rooftop Structures and Equipment for Buildings with h ≤ 60 ft . . . . . . . . . . . . 564 C29.7 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 C29.9 Minimum Design Wind Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 C30 Wind Loads—Components and Cladding (C&C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569 C30.1.5 Air-Permeable Cladding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569 C30.3.1 Velocity Pressure Exposure Coefﬁ cient . . . . . . . . . . . . . . . . . . . . . . . 570 C30.3.2 Velocity Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 570 Part 1: Low-Rise Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 Part 3: Buildings With h > 60 ft (18.3 m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 Part 4: Buildings with h ≤ 160 ft (Simpliﬁ ed) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 C30.7.1.2 Parapets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 C30.7.1.3 Roof Overhangs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 Part 5: Open Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573 C31 Wind Tunnel Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575 C31.4.1 Mean Recurrence Intervals of Load Effects . . . . . . . . . . . . . . . . . . . . 576 C31.4.2 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 Commentary Appendix C Serviceability Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 CC. Serviceability Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 CC.1.1 Vertical Deﬂ ections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 CC.1.2 Drift of Walls and Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 CC.1.3 Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 CC.2 Design for Long-Term Deﬂ ection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 CC.3 Camber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582 CC.4 Expansion and Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582 CC.5 Durability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582 Commentary Chapter: Appendix D Buildings Exempted from Torsional Wind Load Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595 1 Chapter 1 GENERAL judged either to be no longer useful for its intended function (serviceability limit state) or to be unsafe (strength limit state). LOAD EFFECTS: Forces and deformations produced in structural members by the applied loads. LOAD FACTOR: A factor that accounts for deviations of the actual load from the nominal load, for uncertainties in the analysis that transforms the load into a load effect, and for the probability that more than one extreme load will occur simultaneously. LOADS: Forces or other actions that result from the weight of all building materials, occupants and their possessions, environmental effects, differential movement, and restrained dimensional changes. Permanent loads are those loads in which variations over time are rare or of small magnitude. All other loads are variable loads (see also “nominal loads”). NOMINAL LOADS: The magnitudes of the loads speciﬁ ed in this standard for dead, live, soil, wind, snow, rain, ﬂ ood, and earthquake. NOMINAL STRENGTH: The capacity of a structure or member to resist the effects of loads, as determined by computations using speciﬁ ed material strengths and dimensions and formulas derived from accepted principles of structural mechanics or by ﬁ eld tests or laboratory tests of scaled models, allowing for modeling effects and differences between laboratory and ﬁ eld conditions. OCCUPANCY: The purpose for which a building or other structure, or part thereof, is used or intended to be used. OTHER STRUCTURES: Structures, other than buildings, for which loads are speciﬁ ed in this standard. P-DELTA EFFECT: The second order effect on shears and moments of frame members induced by axial loads on a laterally displaced building frame. RESISTANCE FACTOR: A factor that accounts for deviations of the actual strength from the nominal strength and the manner and consequences of failure (also called “strength reduction factor”). RISK CATEGORY: A categorization of buildings and other structures for determination of ﬂ ood, wind, snow, ice, and earthquake loads based on the risk associated with unacceptable performance. See Table 1.5-1. STRENGTH DESIGN: A method of proportion-ing structural members such that the computed forces produced in the members by the factored loads do not 1.1 SCOPE This standard provides minimum load requirements for the design of buildings and other structures that are subject to building code requirements. Loads and appropriate load combinations, which have been developed to be used together, are set forth for strength design and allowable stress design. For design strengths and allowable stress limits, design speciﬁ cations for conventional structural materials used in buildings and modiﬁ cations contained in this standard shall be followed. 1.2 DEFINITIONS AND NOTATIONS 1.2.1 Deﬁ nitions The following deﬁ nitions apply to the provisions of the entire standard. ALLOWABLE STRESS DESIGN: A method of proportioning structural members such that elastically computed stresses produced in the members by nominal loads do not exceed speciﬁ ed allowable stresses (also called “working stress design”). AUTHORITY HAVING JURISDICTION: The organization, political subdivision, ofﬁ ce, or individual charged with the responsibility of adminis-tering and enforcing the provisions of this standard. BUILDINGS: Structures, usually enclosed by walls and a roof, constructed to provide support or shelter for an intended occupancy. DESIGN STRENGTH: The product of the nominal strength and a resistance factor. ESSENTIAL FACILITIES: Buildings and other structures that are intended to remain operational in the event of extreme environmental loading from ﬂ ood, wind, snow, or earthquakes. FACTORED LOAD: The product of the nominal load and a load factor. HIGHLY TOXIC SUBSTANCE: As deﬁ ned in 29 CFR 1910.1200 Appendix A with Amendments as of February 1, 2000. IMPORTANCE FACTOR: A factor that accounts for the degree of risk to human life, health, and welfare associated with damage to property or loss of use or functionality. LIMIT STATE: A condition beyond which a structure or member becomes unﬁ t for service and is CHAPTER 1 GENERAL 2 exceed the member design strength (also called “load and resistance factor design”). TEMPORARY FACILITIES: Buildings or other structures that are to be in service for a limited time and have a limited exposure period for environ-mental loadings. TOXIC SUBSTANCE: As deﬁ ned in 29 CFR 1910.1200 Appendix A with Amendments as of February 1, 2000. 1.1.2 Symbols and Notations Fx A minimum design lateral force applied to level x of the structure and used for purposes of evaluating structural integrity in accordance with Section 1.4.2. Wx The portion of the total dead load of the struc-ture, D, located or assigned to Level x. D Dead load. L Live load. Lr Roof live load. N Notional load used to evaluate conformance with minimum structural integrity criteria. R Rain load. S Snow load. 1.3 BASIC REQUIREMENTS 1.3.1 Strength and Stiffness Buildings and other structures, and all parts thereof, shall be designed and constructed with adequate strength and stiffness to provide structural stability, protect nonstructural components and systems from unacceptable damage, and meet the serviceability requirements of Section 1.3.2. Acceptable strength shall be demonstrated using one or more of the following procedures: a. the Strength Procedures of Section 1.3.1.1, b. the Allowable Stress Procedures of Section 1.3.1.2, or c. subject to the approval of the authority having jurisdiction for individual projects, the Performance-Based Procedures of Section 1.3.1.3. Table 1.5-1 Risk Category of Buildings and Other Structures for Flood, Wind, Snow, Earthquake, and Ice Loads Use or Occupancy of Buildings and Structures Risk Category Buildings and other structures that represent a low risk to human life in the event of failure I All buildings and other structures except those listed in Risk Categories I, III, and IV II Buildings and other structures, the failure of which could pose a substantial risk to human life. Buildings and other structures, not included in Risk Category IV, with potential to cause a substantial economic impact and/or mass disruption of day-to-day civilian life in the event of failure. Buildings and other structures not included in Risk Category IV (including, but not limited to, facilities that manufacture, process, handle, store, use, or dispose of such substances as hazardous fuels, hazardous chemicals, hazardous waste, or explosives) containing toxic or explosive substances where their quantity exceeds a threshold quantity established by the authority having jurisdiction and is sufﬁ cient to pose a threat to the public if released. III Buildings and other structures designated as essential facilities. Buildings and other structures, the failure of which could pose a substantial hazard to the community. Buildings and other structures (including, but not limited to, facilities that manufacture, process, handle, store, use, or dispose of such substances as hazardous fuels, hazardous chemicals, or hazardous waste) containing sufﬁ cient quantities of highly toxic substances where the quantity exceeds a threshold quantity established by the authority having jurisdiction to be dangerous to the public if released and is sufﬁ cient to pose a threat to the public if released.a Buildings and other structures required to maintain the functionality of other Risk Category IV structures. IV a Buildings and other structures containing toxic, highly toxic, or explosive substances shall be eligible for classiﬁ cation to a lower Risk Category if it can be demonstrated to the satisfaction of the authority having jurisdiction by a hazard assessment as described in Section 1.5.2 that a release of the substances is commensurate with the risk associated with that Risk Category. MINIMUM DESIGN LOADS 3 It shall be permitted to use alternative procedures for different parts of a structure and for different load combinations, subject to the limitations of Chapter 2. Where resistance to extraordinary events is considered, the procedures of Section 2.5 shall be used. 1.3.1.1 Strength Procedures Structural and nonstructural components and their connections shall have adequate strength to resist the applicable load combinations of Section 2.3 of this Standard without exceeding the applicable strength limit states for the materials of construction. 1.3.1.2 Allowable Stress Procedures Structural and nonstructural components and their connections shall have adequate strength to resist the applicable load combinations of Section 2.4 of this Standard without exceeding the applicable allowable stresses for the materials of construction. 1.3.1.3 Performance-Based Procedures Structural and nonstructural components and their connections shall be demonstrated by analysis or by a combination of analysis and testing to provide a reliability not less than that expected for similar components designed in accordance with the Strength Procedures of Section 1.3.1.1 when subject to the inﬂ uence of dead, live, environmental, and other loads. Consideration shall be given to uncertainties in loading and resistance. 1.3.1.3.1 Analysis Analysis shall employ rational methods based on accepted principles of engineering mechanics and shall consider all signiﬁ cant sources of deformation and resistance. Assumptions of stiffness, strength, damping, and other properties of components and connections incorporated in the analysis shall be based on approved test data or referenced Standards. 1.3.1.3.2 Testing Testing used to substantiate the performance capability of structural and nonstructural components and their connections under load shall accurately represent the materials, conﬁ guration, construction, loading intensity, and boundary condi-tions anticipated in the structure. Where an approved industry standard or practice that governs the testing of similar components exists, the test program and determination of design values from the test program shall be in accordance with those industry standards and practices. Where such standards or practices do not exist, specimens shall be constructed to a scale similar to that of the intended application unless it can be demonstrated that scale effects are not signiﬁ cant to the indicated performance. Evaluation of test results shall be made on the basis of the values obtained from not less than 3 tests, provided that the deviation of any value obtained from any single test does not vary from the average value for all tests by more than 15%. If such deviaton from the average value for any test exceeds 15%, then additional tests shall be performed until the deviation of any test from the average value does not exceed 15% or a minimum of 6 tests have been performed. No test shall be eliminated unless a rationale for its exclusion is given. Test reports shall document the location, the time and date of the test, the characteristics of the tested specimen, the laboratory facilities, the test conﬁ gura-tion, the applied loading and deformation under load, and the occurrence of any damage sustained by the specimen, together with the loading and deformation at which such damage occurred. 1.3.1.3.3 Documentation The procedures used to demonstrate compliance with this section and the results of analysis and testing shall be documented in one or more reports submitted to the authority having jurisdiction and to an independent peer review. 1.3.1.3.4 Peer Review The procedures and results of analysis, testing, and calculation used to demonstrate compliance with the requirements of this section shall be subject to an independent peer review approved by the authority having jurisdiction. The peer review shall comprise one or more persons having the necessary expertise and knowledge to evaluate compliance, including knowledge of the expected performance, the structural and component behavior, the particular loads considered, structural analysis of the type performed, the materials of construction, and laboratory testing of elements and components to determine structural resistance and performance characteristics. The review shall include the assump-tions, criteria, procedures, calculations, analytical models, test setup, test data, ﬁ nal drawings, and reports. Upon satisfactory completion, the peer review shall submit a letter to the authority having jurisdic-tion indicating the scope of their review and their ﬁ ndings. 1.3.2 Serviceability Structural systems, and members thereof, shall be designed to have adequate stiffness to limit deﬂ ec-tions, lateral drift, vibration, or any other deforma-tions that adversely affect the intended use and performance of buildings and other structures. CHAPTER 1 GENERAL 4 1.3.3 Self-Straining Forces Provision shall be made for anticipated self-straining forces arising from differential settlements of foundations and from restrained dimensional changes due to temperature, moisture, shrinkage, creep, and similar effects. 1.3.4 Analysis Load effects on individual structural members shall be determined by methods of structural analysis that take into account equilibrium, general stability, geometric compatibility, and both short- and long-term material properties. Members that tend to accumulate residual deformations under repeated service loads shall have included in their analysis the added eccen-tricities expected to occur during their service life. 1.3.5 Counteracting Structural Actions All structural members and systems, and all components and cladding in a building or other structure, shall be designed to resist forces due to earthquake and wind, with consideration of overturn-ing, sliding, and uplift, and continuous load paths shall be provided for transmitting these forces to the foundation. Where sliding is used to isolate the elements, the effects of friction between sliding elements shall be included as a force. Where all or a portion of the resistance to these forces is provided by dead load, the dead load shall be taken as the minimum dead load likely to be in place during the event causing the considered forces. Consideration shall be given to the effects of vertical and horizontal deﬂ ections resulting from such forces. 1.4 GENERAL STRUCTURAL INTEGRITY All structures shall be provided with a continuous load path in accordance with the requirements of Section 1.4.1 and shall have a complete lateral force-resisting system with adequate strength to resist the forces indicated in Section 1.4.2. All members of the structural system shall be connected to their support-ing members in accordance with Section 1.4.3. Structural walls shall be anchored to diaphragms and supports in accordance with Section 1.4.4. The effects on the structure and its components due to the forces stipulated in this section shall be taken as the notional load, N, and combined with the effects of other loads in accordance with the load combinations of Section of Section 1.4.1. Where material resistance is depen-dent on load duration, notional loads are permitted to be taken as having a duration of 10 minutes. Structures designed in conformance with the requirements of this Standard for Seismic Design Categories B, C, D, E, or F shall be deemed to comply with the requirements of Sections 1.4.1, 1.4.2, 1.4.3, 1.4.4 and 1.4.5. 1.4.1 Load Combinations of Integrity Loads The notional loads, N, speciﬁ ed in Sections 1.4.2 through 1.4.5 shall be combined with dead and live loads in accordance with Section 1.4.1.1 for strength design and 1.4.1.2 for allowable stress design. 1.4.1.1 Strength Design Notional Load Combinations a. 1.2D + 1.0N + L + 0.2S b. 0.9D + 1.0N 1.4.1.2 Allowable Stress Design Notional Load Combinations a. D 0.7N b. D + 0.75(0.7N) + 0.75L+ 0.75(Lr or S or R) c. 0.6D + 0.7N 1.4.2 Load Path Connections All parts of the structure between separation joints shall be interconnected to form a continuous path to the lateral force-resisting system, and the connections shall be capable of transmitting the lateral forces induced by the parts being connected. Any smaller portion of the structure shall be tied to the remainder of the structure with elements having strength to resist a force of not less than 5% of the portion’s weight. 1.4.3 Lateral Forces Each structure shall be analyzed for the effects of static lateral forces applied independently in each of two orthogonal directions. In each direction, the static lateral forces at all levels shall be applied simultane-ously. For purposes of analysis, the force at each level shall be determined using Eq. 1.4-1 as follows: Fx = 0.01 Wx (1.4-1) where Fx = the design lateral force applied at story x and Wx = the portion of the total dead load of the struc-ture, D, located or assigned to level x. Structures explicitly designed for stability, including second-order effects, shall be deemed to comply with the requirements of this section. 1.4.4 Connection to Supports A positive connection for resisting a horizontal force acting parallel to the member shall be provided MINIMUM DESIGN LOADS 5 for each beam, girder, or truss either directly to its supporting elements or to slabs designed to act as diaphragms. Where the connection is through a diaphragm, the member’s supporting element shall also be connected to the diaphragm. The connection shall have the strength to resist a force of 5 percent of the unfactored dead load plus live load reaction imposed by the supported member on the supporting member. 1.4.5 Anchorage of Structural Walls Walls that provide vertical load bearing or lateral shear resistance for a portion of the structure shall be anchored to the roof and all ﬂ oors and members that provide lateral support for the wall or that are supported by the wall. The anchorage shall provide a direct connection between the walls and the roof or ﬂ oor construction. The connections shall be capable of resisting a strength level horizontal force perpen-dicular to the plane of the wall equal to 0.2 times the weight of the wall tributary to the connection, but not less than 5 psf (0.24 kN/m2). 1.4.6 Extraordinary Loads and Events When considered, design for resistance to extraordinary loads and events shall be in accordance with the procedures of Section 2.5. 1.5 CLASSIFICATION OF BUILDINGS AND OTHER STRUCTURES 1.5.1 Risk Categorization Buildings and other structures shall be classiﬁ ed, based on the risk to human life, health, and welfare associated with their damage or failure by nature of their occupancy or use, according to Table 1.5-1 for the purposes of applying ﬂ ood, wind, snow, earth-quake, and ice provisions. Each building or other structure shall be assigned to the highest applicable risk category or categories. Minimum design loads for structures shall incorporate the applicable importance factors given in Table 1.5-2, as required by other sections of this Standard. Assignment of a building or other structure to multiple risk categories based on the type of load condition being evaluated (e.g., snow or seismic) shall be permitted. When the building code or other referenced standard speciﬁ es an Occupancy Category, the Risk Category shall not be taken as lower than the Occu-pancy Category speciﬁ ed therein. 1.5.2 Multiple Risk Categories Where buildings or other structures are divided into portions with independent structural systems, the classiﬁ cation for each portion shall be permitted to be determined independently. Where building systems, such as required egress, HVAC, or electrical power, for a portion with a higher risk category pass through or depend on other portions of the building or other structure having a lower risk category, those portions shall be assigned to the higher risk category. 1.5.3 Toxic, Highly Toxic, and Explosive Substances Buildings and other structures containing toxic, highly toxic, or explosive substances are permitted to be classiﬁ ed as Risk Category II structures if it can be demonstrated to the satisfaction of the authority having jurisdiction by a hazard assessment as part of an overall risk management plan (RMP) that a release of the toxic, highly toxic, or explosive substances is not sufﬁ cient to pose a threat to the public. To qualify for this reduced classiﬁ cation, the owner or operator of the buildings or other structures Table 1.5-2 Importance Factors by Risk Category of Buildings and Other Structures for Snow, Ice, and Earthquake Loadsa Risk Category from Table 1.5-1 Snow Importance Factor, Is Ice Importance Factor—Thickness, Ii Ice Importance Factor—Wind, Iw Seismic Importance Factor, Ie I 0.80 0.80 1.00 1.00 II 1.00 1.00 1.00 1.00 III 1.10 1.25 1.00 1.25 IV 1.20 1.25 1.00 1.50 aThe component importance factor, Ip, applicable to earthquake loads, is not included in this table because it is dependent on the importance of the individual component rather than that of the building as a whole, or its occupancy. Refer to Section 13.1.3. CHAPTER 1 GENERAL 6 containing the toxic, highly toxic, or explosive substances shall have an RMP that incorporates three elements as a minimum: a hazard assessment, a prevention program, and an emergency response plan. As a minimum, the hazard assessment shall include the preparation and reporting of worst-case release scenarios for each structure under consider-ation, showing the potential effect on the public for each. As a minimum, the worst-case event shall include the complete failure (instantaneous release of entire contents) of a vessel, piping system, or other storage structure. A worst-case event includes (but is not limited to) a release during the design wind or design seismic event. In this assessment, the evalua-tion of the effectiveness of subsequent measures for accident mitigation shall be based on the assumption that the complete failure of the primary storage structure has occurred. The offsite impact shall be deﬁ ned in terms of population within the potentially affected area. To qualify for the reduced classiﬁ cation, the hazard assessment shall demonstrate that a release of the toxic, highly toxic, or explosive substances from a worst-case event does not pose a threat to the public outside the property boundary of the facility. As a minimum, the prevention program shall consist of the comprehensive elements of process safety management, which is based upon accident prevention through the application of management controls in the key areas of design, construction, operation, and maintenance. Secondary containment of the toxic, highly toxic, or explosive substances (including, but not limited to, double wall tank, dike of sufﬁ cient size to contain a spill, or other means to contain a release of the toxic, highly toxic, or explo-sive substances within the property boundary of the facility and prevent release of harmful quantities of contaminants to the air, soil, ground water, or surface water) are permitted to be used to mitigate the risk of release. Where secondary containment is provided, it shall be designed for all environmental loads and is not eligible for this reduced classiﬁ cation. In hurricane-prone regions, mandatory practices and procedures that effectively diminish the effects of wind on critical structural elements or that alterna-tively protect against harmful releases during and after hurricanes are permitted to be used to mitigate the risk of release. As a minimum, the emergency response plan shall address public notiﬁ cation, emergency medical treatment for accidental exposure to humans, and procedures for emergency response to releases that have consequences beyond the property boundary of the facility. The emergency response plan shall address the potential that resources for response could be compromised by the event that has caused the emergency. 1.6 ADDITIONS AND ALTERATIONS TO EXISTING STRUCTURES When an existing building or other structure is enlarged or otherwise altered, structural members affected shall be strengthened if necessary so that the factored loads deﬁ ned in this document will be supported without exceeding the speciﬁ ed design strength for the materials of construction. When using allowable stress design, strengthening is required when the stresses due to nominal loads exceed the speciﬁ ed allowable stresses for the materials of construction. 1.7 LOAD TESTS A load test of any construction shall be conducted when required by the authority having jurisdiction whenever there is reason to question its safety for the intended use. 1.8 CONSENSUS STANDARDS AND OTHER REFERENCED DOCUMENTS This section lists the consensus standards and other documents that are adopted by reference within this chapter: OSHA Occupational Safety and Health Administration 200 Constitution Avenue, NW Washington, DC 20210 29 CFR 1910.1200 Appendix A with Amendments as of February 1, 2000. Section 1.2 OSHA Standards for General Industry, 29 CFR (Code of Federal Regulations) Part 1910.1200 Appendix A, United States Department of Labor, Occupational Safety and Health Administration, Washington, DC, 2005 7 Chapter 2 COMBINATIONS OF LOADS 5. 1.2D + 1.0E + L + 0.2S 6. 0.9D + 1.0W 7. 0.9D + 1.0E EXCEPTIONS: 1. The load factor on L in combinations 3, 4, and 5 is permitted to equal 0.5 for all occupancies in which Lo in Table 4-1 is less than or equal to 100 psf, with the exception of garages or areas occupied as places of public assembly. 2. In combinations 2, 4, and 5, the companion load S shall be taken as either the ﬂ at roof snow load (pf) or the sloped roof snow load (ps). Where ﬂ uid loads F are present, they shall be included with the same load factor as dead load D in combinations 1 through 5 and 7. Where load H are present, they shall be included as follows: 1. where the effect of H adds to the primary variable load effect, include H with a load factor of 1.6; 2. where the effect of H resists the primary variable load effect, include H with a load factor of 0.9 where the load is permanent or a load factor of 0 for all other conditions. Effects of one or more loads not acting shall be investigated. The most unfavorable effects from both wind and earthquake loads shall be investigated, where appropriate, but they need not be considered to act simultaneously. Refer to Section 12.4 for speciﬁ c deﬁ nition of the earthquake load effect E.1 Each relevant strength limit state shall be investigated. 2.3.3 Load Combinations Including Flood Load When a structure is located in a ﬂ ood zone (Section 5.3.1), the following load combinations shall be considered in addition to the basic combinations in Section 2.3.2: 1. In V-Zones or Coastal A-Zones, 1.0W in combina-tions 4 and 6 shall be replaced by 1.0W + 2.0Fa. 2. In noncoastal A-Zones, 1.0W in combinations 4 and 6 shall be replaced by 0.5W + 1.0Fa. 2.1 GENERAL Buildings and other structures shall be designed using the provisions of either Section 2.3 or 2.4. Where elements of a structure are designed by a particular material standard or speciﬁ cation, they shall be designed exclusively by either Section 2.3 or 2.4. 2.2 SYMBOLS Ak = load or load effect arising from extra ordinary event A D = dead load Di = weight of ice E = earthquake load F = load due to ﬂ uids with well-deﬁ ned pressures and maximum heights Fa = ﬂ ood load H = load due to lateral earth pressure, ground water pressure, or pressure of bulk materials L = live load Lr = roof live load R = rain load S = snow load T = self-straining load W = wind load Wi = wind-on-ice determined in accordance with Chapter 10 2.3 COMBINING FACTORED LOADS USING STRENGTH DESIGN 2.3.1 Applicability The load combinations and load factors given in Section 2.3.2 shall be used only in those cases in which they are speciﬁ cally authorized by the appli-cable material design standard. 2.3.2 Basic Combinations Structures, components, and foundations shall be designed so that their design strength equals or exceeds the effects of the factored loads in the following combinations: 1. 1.4D 2. 1.2D + 1.6L + 0.5(Lr or S or R) 3. 1.2D + 1.6(Lr or S or R) + (L or 0.5W) 4. 1.2D + 1.0W + L + 0.5(Lr or S or R) 1 The same E from Sections 1.4 and 12.4 is used for both Sections 2.3.2 and 2.4.1. Refer to the Chapter 11 Commentary for the Seismic Provisions. CHAPTER 2 COMBINATIONS OF LOADS 8 2.3.4. Load Combinations Including Atmospheric Ice Loads When a structure is subjected to atmospheric ice and wind-on-ice loads, the following load combina-tions shall be considered: 1. 0.5(Lr or S or R) in combination 2 shall be replaced by 0.2Di + 0.5S. 2. 1.0W + 0.5(Lr or S or R) in combination 4 shall be replaced by Di + Wi + 0.5S. 3. 1.0W in combination 6 shall be replaced by Di + Wi. 2.3.5 Load Combinations Including Self-Straining Loads Where applicable, the structural effects of load T shall be considered in combination with other loads. The load factor on load T shall be established consid-ering the uncertainty associated with the likely magnitude of the load, the probability that the maximum effect of T will occur simultaneously with other applied loadings, and the potential adverse consequences if the effect of T is greater than assumed. The load factor on T shall not have a value less than 1.0. 2.3.6 Load Combinations for Nonspeciﬁ ed Loads Where approved by the Authority Having Jurisdiction, the Responsible Design Professional is permitted to determine the combined load effect for strength design using a method that is consistent with the method on which the load combination require-ments in Section 2.3.2 are based. Such a method must be probability-based and must be accompanied by documentation regarding the analysis and collection of supporting data that is acceptable to the Authority Having Jurisdiction. 2.4 COMBINING NOMINAL LOADS USING ALLOWABLE STRESS DESIGN 2.4.1 Basic Combinations Loads listed herein shall be considered to act in the following combinations; whichever produces the most unfavorable effect in the building, foundation, or structural member being considered. Effects of one or more loads not acting shall be considered. 1. D 2. D + L 3. D + (Lr or S or R) 4. D + 0.75L + 0.75(Lr or S or R) 5. D + (0.6W or 0.7E) 6a. D + 0.75L + 0.75(0.6W) + 0.75(Lr or S or R) 6b. D + 0.75L + 0.75(0.7E) + 0.75S 7. 0.6D + 0.6W 8. 0.6D + 0.7E EXCEPTIONS: 1. In combinations 4 and 6, the companion load S shall be taken as either the ﬂ at roof snow load (pf) or the sloped roof snow load (ps). 2. For nonbuilding structures, in which the wind load is determined from force coefﬁ cients, Cf, identiﬁ ed in Figures 29.5-1, 29.5-2 and 29.5-3 and the projected area contributing wind force to a founda-tion element exceeds 1,000 square feet on either a vertical or a horizontal plane, it shall be permitted to replace W with 0.9W in combination 7 for design of the foundation, excluding anchorage of the structure to the foundation. 3. It shall be permitted to replace 0.6D with 0.9D in combination 8 for the design of Special Reinforced Masonry Shear Walls, where the walls satisfy the requirement of Section 14.4.2. Where ﬂ uid loads F are present, they shall be included in combinations 1 through 6 and 8 with the same factor as that used for dead load D. Where load H is present, it shall be included as follows: 1. where the effect of H adds to the primary variable load effect, include H with a load factor of 1.0; 2. where the effect of H resists the primary variable load effect, include H with a load factor of 0.6 where the load is permanent or a load factor of 0 for all other conditions. The most unfavorable effects from both wind and earthquake loads shall be considered, where appropriate, but they need not be assumed to act simultaneously. Refer to Section 1.4 and 12.4 for the speciﬁ c deﬁ nition of the earthquake load effect E.2 Increases in allowable stress shall not be used with the loads or load combinations given in this standard unless it can be demonstrated that such an increase is justiﬁ ed by structural behavior caused by rate or duration of load. 2 The same E from Sections 1.4 and 12.4 is used for both Sections 2.3.2 and 2.4.1. Refer to the Chapter 11 Commentary for the Seismic Provisions. MINIMUM DESIGN LOADS 9 2.4.2 Load Combinations Including Flood Load When a structure is located in a ﬂ ood zone, the following load combinations shall be considered in addition to the basic combinations in Section 2.4.1: 1. In V-Zones or Coastal A-Zones (Section 5.3.1), 1.5Fa shall be added to other loads in combinations 5, 6, and 7, and E shall be set equal to zero in 5 and 6. 2. In non-coastal A-Zones, 0.75Fa shall be added to combinations 5, 6, and 7, and E shall be set equal to zero in 5 and 6. 2.4.3 Load Combinations Including Atmospheric Ice Loads When a structure is subjected to atmospheric ice and wind-on-ice loads, the following load combina-tions shall be considered: 1. 0.7Di shall be added to combination 2. 2. (Lr or S or R) in combination 3 shall be replaced by 0.7Di + 0.7Wi + S. 3. 0.6W in combination 7 shall be replaced by 0.7Di + 0.7Wi. 2.4.4 Load Combinations Including Self-Straining Loads Where applicable, the structural effects of load T shall be considered in combination with other loads. Where the maximum effect of load T is unlikely to occur simultaneously with the maximum effects of other variable loads, it shall be permitted to reduce the magnitude of T considered in combination with these other loads. The fraction of T considered in combination with other loads shall not be less than 0.75. 2.5 LOAD COMBINATIONS FOR EXTRAORDINARY EVENTS 2.5.1 Applicability Where required by the owner or applicable code, strength and stability shall be checked to ensure that structures are capable of withstanding the effects of extraordinary (i.e., low-probability) events, such as ﬁ res, explosions, and vehicular impact without disproportionate collapse. 2.5.2 Load Combinations 2.5.2.1 Capacity For checking the capacity of a structure or structural element to withstand the effect of an extraordinary event, the following gravity load combination shall be considered: (0.9 or 1.2)D + Ak + 0.5L + 0.2S (2.5-1) in which Ak = the load or load effect resulting from extraordinary event A. 2.5.2.2 Residual Capacity For checking the residual load-carrying capacity of a structure or structural element following the occurrence of a damaging event, selected load-bearing elements identiﬁ ed by the Responsible Design Professional shall be notionally removed, and the capacity of the damaged structure shall be evaluated using the following gravity load combination: (0.9 or 1.2)D + 0.5L + 0.2(Lr or S or R) (2.5-2) 2.5.3 Stability Requirements Stability shall be provided for the structure as a whole and for each of its elements. Any method that considers the inﬂ uence of second-order effects is permitted. 11 Chapter 3 DEAD LOADS, SOIL LOADS, AND HYDROSTATIC PRESSURE 3.1.3 Weight of Fixed Service Equipment In determining dead loads for purposes of design, the weight of ﬁ xed service equipment, such as plumbing stacks and risers, electrical feeders, and heating, ventilating, and air conditioning systems shall be included. 3.2 SOIL LOADS AND HYDROSTATIC PRESSURE 3.2.1 Lateral Pressures In the design of structures below grade, provision shall be made for the lateral pressure of adjacent soil. If soil loads are not given in a soil investigation report approved by the authority having jurisdiction, then the soil loads speciﬁ ed in Table 3.2-1 shall be used as the 3.1 DEAD LOADS 3.1.1 Deﬁ nition Dead loads consist of the weight of all materials of construction incorporated into the building includ-ing, but not limited to, walls, ﬂ oors, roofs, ceilings, stairways, built-in partitions, ﬁ nishes, cladding, and other similarly incorporated architectural and struc-tural items, and ﬁ xed service equipment including the weight of cranes. 3.1.2 Weights of Materials and Constructions In determining dead loads for purposes of design, the actual weights of materials and constructions shall be used provided that in the absence of deﬁ nite information, values approved by the authority having jurisdiction shall be used. Table 3.2-1 Design Lateral Soil Load Description of Backﬁ ll Material Uniﬁ ed Soil Classiﬁ cation Design Lateral Soil Loada psf per foot of depth (kN/m2 per meter of depth) Well-graded, clean gravels; gravel–sand mixes GW 35 (5.50)b Poorly graded clean gravels; gravel–sand mixes GP 35 (5.50)b Silty gravels, poorly graded gravel–sand mixes GM 35 (5.50)b Clayey gravels, poorly graded gravel-and-clay mixes GC 45 (7.07)b Well-graded, clean sands; gravelly–sand mixes SW 35 (5.50)b Poorly graded clean sands; sand–gravel mixes SP 35 (5.50)b Silty sands, poorly graded sand–silt mixes SM 45 (7.07)b Sand–silt clay mix with plastic ﬁ nes SM–SC 85 (13.35)c Clayey sands, poorly graded sand–clay mixes SC 85 (13.35)c Inorganic silts and clayey silts ML 85 (13.35)c Mixture of inorganic silt and clay ML–CL 85 (13.35)c Inorganic clays of low to medium plasticity CL 100 (15.71) Organic silts and silt–clays, low plasticity OL d Inorganic clayey silts, elastic silts MH d Inorganic clays of high plasticity CH d Organic clays and silty clays OH d aDesign lateral soil loads are given for moist conditions for the speciﬁ ed soils at their optimum densities. Actual ﬁ eld conditions shall govern. Submerged or saturated soil pressures shall include the weight of the buoyant soil plus the hydrostatic loads. cFor relatively rigid walls, as when braced by ﬂ oors, the design lateral soil load shall be increased for sand and gravel type soils to 60 psf (9.43 kN/m2) per foot (meter) of depth. Basement walls extending not more than 8 ft (2.44 m) below grade and supporting light ﬂ oor systems are not considered as being relatively rigid walls. dFor relatively rigid walls, as when braced by ﬂ oors, the design lateral load shall be increased for silt and clay type soils to 100 psf (15.71 kN/m2) per foot (meter) of depth. Basement walls extending not more than 8 ft (2.44 m) below grade and supporting light ﬂ oor systems are not considered as being relatively rigid walls. bUnsuitable as backﬁ ll material. CHAPTER 3 DEAD LOADS, SOIL LOADS, AND HYDROSTATIC PRESSURE 12 minimum design lateral loads. Due allowance shall be made for possible surcharge from ﬁ xed or moving loads. When a portion or the whole of the adjacent soil is below a free-water surface, computations shall be based upon the weight of the soil diminished by buoyancy, plus full hydrostatic pressure. The lateral pressure shall be increased if soils with expansion potential are present at the site as determined by a geotechnical investigation. 3.2.2 Uplift on Floors and Foundations In the design of basement ﬂ oors and similar approximately horizontal elements below grade, the upward pressure of water, where applicable, shall be taken as the full hydrostatic pressure applied over the entire area. The hydrostatic load shall be measured from the underside of the construction. Any other upward loads shall be included in the design. Where expansive soils are present under founda-tions or slabs-on-ground, the foundations, slabs, and other components shall be designed to tolerate the movement or resist the upward loads caused by the expansive soils, or the expansive soil shall be removed or stabilized around and beneath the structure. 13 Chapter 4 LIVE LOADS with a method approved by the authority having jurisdiction. 4.3 UNIFORMLY DISTRIBUTED LIVE LOADS 4.3.1 Required Live Loads The live loads used in the design of buildings and other structures shall be the maximum loads expected by the intended use or occupancy, but shall in no case be less than the minimum uniformly distributed unit loads required by Table 4-1, including any permis-sible reduction. 4.3.2 Provision for Partitions In ofﬁ ce buildings or other buildings where partitions will be erected or rearranged, provision for partition weight shall be made, whether or not partitions are shown on the plans. Partition load shall not be less than 15 psf (0.72 kN/m2). EXCEPTION: A partition live load is not required where the minimum speciﬁ ed live load exceeds 80 psf (3.83 kN/m2). 4.3.3 Partial Loading The full intensity of the appropriately reduced live load applied only to a portion of a structure or member shall be accounted for if it produces a more unfavorable load effect than the same intensity applied over the full structure or member. Roof live loads shall be distributed as speciﬁ ed in Table 4-1. 4.4 CONCENTRATED LIVE LOADS Floors, roofs, and other similar surfaces shall be designed to support safely the uniformly distributed live loads prescribed in Section 4.3 or the concentrated load, in pounds or kilonewtons (kN), given in Table 4-1, whichever produces the greater load effects. Unless otherwise speciﬁ ed, the indicated concentration shall be assumed to be uniformly distributed over an area 2.5 ft (762 mm) by 2.5 ft (762 mm) and shall be located so as to produce the maximum load effects in the members. 4.1 DEFINITIONS FIXED LADDER: A ladder that is permanently attached to a structure, building, or equipment. GRAB BAR SYSTEM: A bar and associated anchorages and attachments to the structural system, for the support of body weight in locations such as toilets, showers, and tub enclosures. GUARDRAIL SYSTEM: A system of compo-nents, including anchorages and attachments to the structural system, near open sides of an elevated surface for the purpose of minimizing the possibility of a fall from the elevated surface by people, equip-ment, or material. HANDRAIL SYSTEM: A rail grasped by hand for guidance and support, and associated anchorages and attachments to the structural system. HELIPAD: A structural surface that is used for landing, taking off, taxiing, and parking of helicopters. LIVE LOAD: A load produced by the use and occupancy of the building or other structure that does not include construction or environmental loads, such as wind load, snow load, rain load, earthquake load, ﬂ ood load, or dead load. ROOF LIVE LOAD: A load on a roof produced (1) during maintenance by workers, equipment, and materials and (2) during the life of the structure by movable objects, such as planters or other similar small decorative appurtenances that are not occupancy related. SCREEN ENCLOSURE: A building or part thereof, in whole or in part self-supporting, having walls and a roof of insect or sun screening using ﬁ berglass, aluminum, plastic, or similar lightweight netting material, which enclose an occupancy or use such as outdoor swimming pools, patios or decks, and horticultural and agricultural production facilities. VEHICLE BARRIER SYSTEM: A system of components, including anchorages and attachments to the structural system near open sides or walls of garage ﬂ oors or ramps, that acts as a restraint for vehicles. 4.2 LOADS NOT SPECIFIED For occupancies or uses not designated in this chapter, the live load shall be determined in accordance CHAPTER 4 LIVE LOADS 14 4.5 LOADS ON HANDRAIL, GUARDRAIL, GRAB BAR, VEHICLE BARRIER SYSTEMS, AND FIXED LADDERS 4.5.1 Loads on Handrail and Guardrail Systems All handrail and guardrail systems shall be designed to resist a single concentrated load of 200 lb (0.89 kN) applied in any direction at any point on the handrail or top rail and to transfer this load through the supports to the structure to produce the maximum load effect on the element being considered. Further, all handrail and guardrail systems shall be designed to resist a load of 50 lb/ft (pound-force per linear foot) (0.73 kN/m) applied in any direction along the handrail or top rail. This load need not be assumed to act concurrently with the load speciﬁ ed in the preceding paragraph, and this load need not be considered for the following occupancies: 1. One- and two-family dwellings. 2. Factory, industrial, and storage occupancies, in areas that are not accessible to the public and that serve an occupant load not greater than 50. Intermediate rails (all those except the handrail), and panel ﬁ llers shall be designed to withstand a horizontally applied normal load of 50 lb (0.22 kN) on an area not to exceed 12 in. by 12 in. (305 mm by 305 mm) including openings and space between rails and located so as to produce the maximum load effects. Reactions due to this loading are not required to be superimposed with the loads speciﬁ ed in either preceding paragraph. 4.5.2 Loads on Grab Bar Systems Grab bar systems shall be designed to resist a single concentrated load of 250 lb (1.11 kN) applied in any direction at any point on the grab bar to produce the maximum load effect. 4.5.3 Loads on Vehicle Barrier Systems Vehicle barrier systems for passenger vehicles shall be designed to resist a single load of 6,000 lb (26.70 kN) applied horizontally in any direction to the barrier system, and shall have anchorages or attach-ments capable of transferring this load to the struc-ture. For design of the system, the load shall be assumed to act at heights between 1 ft 6 in. (460 mm) and 2 ft 3 in. (686 mm) above the ﬂ oor or ramp surface, selected to produce the maximum load effect. The load shall be applied on an area not to exceed 12 in. by 12 in. (305 mm by 305 mm) and located so as to produce the maximum load effects. This load is not required to act concurrently with any handrail or guardrail system loadings speciﬁ ed in Section 4.5.1. Vehicle barrier systems in garages accommodating trucks and buses shall be designed in accordance with AASHTO LRFD Bridge Design Speciﬁ cations. 4.5.4 Loads on Fixed Ladders The minimum design live load on ﬁ xed ladders with rungs shall be a single concentrated load of 300 lb (1.33 kN), and shall be applied at any point to produce the maximum load effect on the element being considered. The number and position of additional concentrated live load units shall be a minimum of 1 unit of 300 lb (1.33 kN) for every 10 ft (3.05 m) of ladder height. Where rails of ﬁ xed ladders extend above a ﬂ oor or platform at the top of the ladder, each side rail extension shall be designed to resist a single concentrated live load of 100 lb (0.445 kN) in any direction at any height up to the top of the side rail extension. Ship ladders with treads instead of rungs shall have minimum design loads as stairs, deﬁ ned in Table 4-1. 4.6 IMPACT LOADS 4.6.1 General The live loads speciﬁ ed in Sections 4.3 through 4.5 shall be assumed to include adequate allowance for ordinary impact conditions. Provision shall be made in the structural design for uses and loads that involve unusual vibration and impact forces. 4.6.2 Elevators All elements subject to dynamic loads from elevators shall be designed for impact loads and deﬂ ection limits prescribed by ASME A17.1. 4.6.3 Machinery For the purpose of design, the weight of machin-ery and moving loads shall be increased as follows to allow for impact: (1) light machinery, shaft- or motor-driven, 20 percent; and (2) reciprocating machinery or power-driven units, 50 percent. All percentages shall be increased where speciﬁ ed by the manufacturer. 4.7 REDUCTION IN LIVE LOADS 4.7.1 General Except for roof uniform live loads, all other minimum uniformly distributed live loads, Lo in MINIMUM DESIGN LOADS 15 Table 4-1, shall be permitted to be reduced in accordance with the requirements of Sections 4.7.2 through 4.7.6. 4.7.2 Reduction in Uniform Live Loads Subject to the limitations of Sections 4.7.3 through 4.7.6, members for which a value of KLLAT is 400 ft2 (37.16 m2) or more are permitted to be designed for a reduced live load in accordance with the following formula: L L K A o LL T = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 25 15 . (4.7-1) In SI: L L K A o LL T = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 25 4 57 . . where L = reduced design live load per ft2 (m2) of area supported by the member Lo = unreduced design live load per ft2 (m2) of area supported by the member (see Table 4-1) KLL = live load element factor (see Table 4-2) AT = tributary area in ft2 (m2) L shall not be less than 0.50Lo for members supporting one ﬂ oor and L shall not be less than 0.40Lo for members supporting two or more ﬂ oors. EXCEPTION: For structural members in one- and two-family dwellings supporting more than one ﬂ oor load, the following ﬂ oor live load reduction shall be permitted as an alternative to Eq. 4.7-1: L = 0.7 × (Lo1 + Lo2 + …) Lo1, Lo2, … are the unreduced ﬂ oor live loads appli-cable to each of multiple supported story levels regardless of tributary area. The reduced ﬂ oor live load effect, L, shall not be less than that produced by the effect of the largest unreduced ﬂ oor live load on a given story level acting alone. 4.7.3 Heavy Live Loads Live loads that exceed 100 lb/ft2 (4.79 kN/m2) shall not be reduced. EXCEPTION: Live loads for members supporting two or more ﬂ oors shall be permitted to be reduced by 20 percent. 4.7.4 Passenger Vehicle Garages The live loads shall not be reduced in passenger vehicle garages. EXCEPTION: Live loads for members supporting two or more ﬂ oors shall be permitted to be reduced by 20 percent. 4.7.5 Assembly Uses Live loads shall not be reduced in assembly uses. 4.7.6 Limitations on One-Way Slabs The tributary area, AT, for one-way slabs shall not exceed an area deﬁ ned by the slab span times a width normal to the span of 1.5 times the slab span. 4.8 REDUCTION IN ROOF LIVE LOADS 4.8.1 General The minimum uniformly distributed roof live loads, Lo in Table 4-1, are permitted to be reduced in accor-dance with the requirements of Sections 4.8.2 and 4.8.3. 4.8.2 Flat, Pitched, and Curved Roofs Ordinary ﬂ at, pitched, and curved roofs, and awning and canopies other than those of fabric construction supported by a skeleton structure, are permitted to be designed for a reduced roof live load, as speciﬁ ed in Eq. 4.8-1 or other controlling combinations of loads, as speciﬁ ed in Chapter 2, whichever produces the greater load effect. In structures such as greenhouses, where special scaffolding is used as a work surface for workers and materials during maintenance and repair operations, a lower roof load than speciﬁ ed in Eq. 4.8-1 shall not be used unless approved by the authority having jurisdiction. On such structures, the minimum roof live load shall be 12 psf (0.58 kN/m2). Lr = LoR1R2 where 12 ≤ Lr ≤ 20 (4.8-1) In SI: Lr = LoR1R2 where 0.58 ≤ Lr ≤ 0.96 where Lr = reduced roof live load per ft2 (m2) of horizontal projection supported by the member Lo = unreduced design roof live load per ft2 (m2) of horizontal projection supported by the member (see Table 4-1) The reduction factors R1 and R2 shall be deter-mined as follows: 1 for A T ≤ 200 ft2 R1 = 1.2 − 0.001At for 200 ft2 < A T < 600 ft2 0.6 for A T ≥ 600 ft2 CHAPTER 4 LIVE LOADS 16 in SI: 1 for A T ≤ 18.58 m2 R1 = 1.2 − 0.011At for 18.58 m2 < A T < 55.74 m2 0.6 for A T ≥ 55.74 m2 where AT = tributary area in ft2 (m2) supported by the member and 1 for F ≤ 4 R2 = 1.2 − 0.05F for 4 < F < 12 0.6 for F ≥ 12 where, for a pitched roof, F = number of inches of rise per foot (in SI: F = 0.12 × slope, with slope expressed in percentage points) and, for an arch or dome, F = rise-to-span ratio multiplied by 32. 4.8.3 Special Purpose Roofs Roofs that have an occupancy function, such as roof gardens, assembly purposes, or other special purposes are permitted to have their uniformly distributed live load reduced in accordance with the requirements of Section 4.7. 4.9 CRANE LOADS 4.9.1 General The crane live load shall be the rated capacity of the crane. Design loads for the runway beams, including connections and support brackets, of moving bridge cranes and monorail cranes shall include the maximum wheel loads of the crane and the vertical impact, lateral, and longitudinal forces induced by the moving crane. 4.9.2 Maximum Wheel Load The maximum wheel loads shall be the wheel loads produced by the weight of the bridge, as applicable, plus the sum of the rated capacity and the weight of the trolley with the trolley positioned on its runway at the location where the resulting load effect is maximum. 4.9.3 Vertical Impact Force The maximum wheel loads of the crane shall be increased by the percentages shown in the following text to determine the induced vertical impact or vibration force: Monorail cranes (powered) 25 Cab-operated or remotely operated bridge cranes (powered) 25 Pendant-operated bridge cranes (powered) 10 Bridge cranes or monorail cranes with hand-geared bridge, trolley, and hoist 0 4.9.4 Lateral Force The lateral force on crane runway beams with electrically powered trolleys shall be calculated as 20 percent of the sum of the rated capacity of the crane and the weight of the hoist and trolley. The lateral force shall be assumed to act horizontally at the traction surface of a runway beam, in either direction perpendicular to the beam, and shall be distributed with due regard to the lateral stiffness of the runway beam and supporting structure. 4.9.5 Longitudinal Force The longitudinal force on crane runway beams, except for bridge cranes with hand-geared bridges, shall be calculated as 10 percent of the maximum wheel loads of the crane. The longitudinal force shall be assumed to act horizontally at the traction surface of a runway beam in either direction parallel to the beam. 4.10 CONSENSUS STANDARDS AND OTHER REFERENCED DOCUMENTS This section lists the consensus standards and other documents that are adopted by reference within this chapter: AASHTO American Association of State Highway and Transportation Ofﬁ cials 444 North Capitol Street, NW, Suite 249 Washington, DC 20001 Sections 4.4.3, Table 4-1 AASHTO LRFD Bridge Design Speciﬁ cations, 4th edition, 2007, with 2008 Interim Revisions Sections 4.5.3, Table 4-1 ASME American Society of Mechanical Engineers Three Park Avenue New York, NY 10016-5900 ASME A17.1 Section 4.6.2 American National Standard Safety Code for Elevators and Escalators, 2007. MINIMUM DESIGN LOADS 17 Table 4-1 Minimum Uniformly Distributed Live Loads, Lo, and Minimum Concentrated Live Loads Occupancy or Use Uniform psf (kN/m2) Conc. lb (kN) Apartments (see Residential) Access ﬂ oor systems Ofﬁ ce use 50 (2.4) 2,000 (8.9) Computer use 100 (4.79) 2,000 (8.9) Armories and drill rooms 150 (7.18)a Assembly areas and theaters Fixed seats (fastened to ﬂ oor) 60 (2.87)a Lobbies 100 (4.79)a Movable seats 100 (4.79)a Platforms (assembly) 100 (4.79)a Stage ﬂ oors 150 (7.18)a Balconies and decks 1.5 times the live load for the occupancy served. Not required to exceed 100 psf (4.79 kN/m2) Catwalks for maintenance access 40 (1.92) 300 (1.33) Corridors First ﬂ oor 100 (4.79) Other ﬂ oors, same as occupancy served except as indicated Dining rooms and restaurants 100 (4.79)a Dwellings (see Residential) Elevator machine room grating (on area of 2 in. by 2 in. (50 mm by 50 mm)) 300 (1.33) Finish light ﬂ oor plate construction (on area of 1 in. by 1 in. (25 mm by 25 mm)) 200 (0.89) Fire escapes 100 (4.79) On single-family dwellings only 40 (1.92) Fixed ladders See Section 4.5 Garages Passenger vehicles only 40 (1.92)a,b,c Trucks and buses c Handrails, guardrails, and grab bars See Section 4.5 Helipads 60 (2.87)d,e Nonreducible e,f,g Hospitals Operating rooms, laboratories 60 (2.87) 1,000 (4.45) Patient rooms 40 (1.92) 1,000 (4.45) Corridors above ﬁ rst ﬂ oor 80 (3.83) 1,000 (4.45) Hotels (see Residential) Libraries Reading rooms 60 (2.87) 1,000 (4.45) Stack rooms 150 (7.18)a,h 1,000 (4.45) Corridors above ﬁ rst ﬂ oor 80 (3.83) 1,000 (4.45) Manufacturing Light 125 (6.00)a 2,000 (8.90) Heavy 250 (11.97)a 3,000 (13.40) Continued CHAPTER 4 LIVE LOADS 18 Occupancy or Use Uniform psf (kN/m2) Conc. lb (kN) Ofﬁ ce buildings File and computer rooms shall be designed for heavier loads based on anticipated occupancy Lobbies and ﬁ rst-ﬂ oor corridors 100 (4.79) 2,000 (8.90) Ofﬁ ces 50 (2.40) 2,000 (8.90) Corridors above ﬁ rst ﬂ oor 80 (3.83) 2,000 (8.90) Penal institutions Cell blocks 40 (1.92) Corridors 100 (4.79) Recreational uses Bowling alleys, poolrooms, and similar uses Dance halls and ballrooms Gymnasiums Reviewing stands, grandstands, and bleachers Stadiums and arenas with ﬁ xed seats (fastened to the ﬂ oor) 75 (3.59)a 100 (4.79)a 100 (4.79)a 100 (4.79)a,k 60 (2.87)a,k Residential One- and two-family dwellings Uninhabitable attics without storage 10 (0.48)l Uninhabitable attics with storage 20 (0.96)m Habitable attics and sleeping areas 30 (1.44) All other areas except stairs 40 (1.92) All other residential occupancies Private rooms and corridors serving them 40 (1.92) Public roomsa and corridors serving them 100 (4.79) Roofs Ordinary ﬂ at, pitched, and curved roofs 20 (0.96)n Roofs used for roof gardens 100 (4.79) Roofs used for assembly purposes Same as occupancy served Roofs used for other occupancies o o Awnings and canopies Fabric construction supported by a skeleton structure 5 (0.24) nonreducible 300 (1.33) applied to skeleton structure Screen enclosure support frame 5 (0.24) nonreducible and applied to the roof frame members only, not the screen 200 (0.89) applied to supporting roof frame members only All other construction 20 (0.96) Primary roof members, exposed to a work ﬂ oor Single panel point of lower chord of roof trusses or any point along primary structural members supporting roofs over manufacturing, storage warehouses, and repair garages 2,000 (8.9) All other primary roof members 300 (1.33) All roof surfaces subject to maintenance workers 300 (1.33) Schools Classrooms 40 (1.92) 1,000 (4.45) Corridors above ﬁ rst ﬂ oor 80 (3.83) 1,000 (4.45) First-ﬂ oor corridors 100 (4.79) 1,000 (4.45) Scuttles, skylight ribs, and accessible ceilings 200 (0.89) Sidewalks, vehicular driveways, and yards subject to trucking 250 (11.97)a,p 8,000 (35.60)q Stairs and exit ways 100 (4.79) 300r One- and two-family dwellings only 40 (1.92) 300r Table 4-1 (Continued) MINIMUM DESIGN LOADS 19 Occupancy or Use Uniform psf (kN/m2) Conc. lb (kN) Storage areas above ceilings 20 (0.96) Storage warehouses (shall be designed for heavier loads if required for anticipated storage) Light 125 (6.00)a Heavy 250 (11.97)a Stores Retail First ﬂ oor 100 (4.79) 1,000 (4.45) Upper ﬂ oors 75 (3.59) 1,000 (4.45) Wholesale, all ﬂ oors 125 (6.00)a 1,000 (4.45) Vehicle barriers See Section 4.5 Walkways and elevated platforms (other than exit ways) 60 (2.87) Yards and terraces, pedestrian 100 (4.79)a a Live load reduction for this use is not permitted by Section 4.7 unless speciﬁ c exceptions apply. b Floors in garages or portions of a building used for the storage of motor vehicles shall be designed for the uniformly distributed live loads of Table 4-1 or the following concentrated load: (1) for garages restricted to passenger vehicles accommodating not more than nine passengers, 3,000 lb (13.35 kN) acting on an area of 4.5 in. by 4.5 in. (114 mm by 114 mm); and (2) for mechanical parking structures without slab or deck that are used for storing passenger vehicles only, 2,250 lb (10 kN) per wheel. c Design for trucks and buses shall be per AASHTO LRFD Bridge Design Speciﬁ cations; however, provisions for fatigue and dynamic load allowance are not required to be applied. d Uniform load shall be 40 psf (1.92 kN/m2) where the design basis helicopter has a maximum take-off weight of 3,000 lbs (13.35 kN) or less. This load shall not be reduced. e Labeling of helicopter capacity shall be as required by the authority having jurisdiction. f Two single concentrated loads, 8 ft (2.44 m) apart shall be applied on the landing area (representing the helicopter’s two main landing gear, whether skid type or wheeled type), each having a magnitude of 0.75 times the maximum take-off weight of the helicopter and located to produce the maximum load effect on the structural elements under consideration. The concentrated loads shall be applied over an area of 8 in. by 8 in. (200 mm by 200 mm) and shall not be concurrent with other uniform or concentrated live loads. gA single concentrated load of 3,000 lbs (13.35 kN) shall be applied over an area 4.5 in. by 4.5 in. (114 mm by 114 mm), located so as to produce the maximum load effects on the structural elements under consideration. The concentrated load need not be assumed to act concurrently with other uniform or concentrated live loads. h The loading applies to stack room ﬂ oors that support nonmobile, double-faced library book stacks subject to the following limitations: (1) The nominal book stack unit height shall not exceed 90 in. (2,290 mm); (2) the nominal shelf depth shall not exceed 12 in. (305 mm) for each face; and (3) parallel rows of double-faced book stacks shall be separated by aisles not less than 36 in. (914 mm) wide. k In addition to the vertical live loads, the design shall include horizontal swaying forces applied to each row of the seats as follows: 24 lb per linear ft of seat applied in a direction parallel to each row of seats and 10 lb per linear ft of seat applied in a direction perpendicular to each row of seats. The parallel and perpendicular horizontal swaying forces need not be applied simultaneously. l Uninhabitable attic areas without storage are those where the maximum clear height between the joist and rafter is less than 42 in. (1,067 mm), or where there are not two or more adjacent trusses with web conﬁ gurations capable of accommodating an assumed rectangle 42 in. (1,067 mm) in height by 24 in. (610 mm) in width, or greater, within the plane of the trusses. This live load need not be assumed to act concurrently with any other live load requirement. m Uninhabitable attic areas with storage are those where the maximum clear height between the joist and rafter is 42 in. (1,067 mm) or greater, or where there are two or more adjacent trusses with web conﬁ gurations capable of accommodating an assumed rectangle 42 in. (1,067 mm) in height by 24 in. (610 mm) in width, or greater, within the plane of the trusses. At the trusses, the live load need only be applied to those portions of the bottom chords where both of the following conditions are met: i. The attic area is accessible from an opening not less than 20 in. (508 mm) in width by 30 in. (762 mm) in length that is located where the clear height in the attic is a minimum of 30 in. (762 mm); and ii. The slope of the truss bottom chord is no greater than 2 units vertical to 12 units horizontal (9.5% slope). The remaining portions of the bottom chords shall be designed for a uniformly distributed nonconcurrent live load of not less than 10 lb/ft2 (0.48 kN/m2). n Where uniform roof live loads are reduced to less than 20 lb/ft2 (0.96 kN/m2) in accordance with Section 4.8.1 and are applied to the design of structural members arranged so as to create continuity, the reduced roof live load shall be applied to adjacent spans or to alternate spans, whichever produces the greatest unfavorable load effect. o Roofs used for other occupancies shall be designed for appropriate loads as approved by the authority having jurisdiction. p Other uniform loads in accordance with an approved method, which contains provisions for truck loadings, shall also be considered where appropriate. q The concentrated wheel load shall be applied on an area of 4.5 in. by 4.5 in. (114 mm by 114 mm). r Minimum concentrated load on stair treads (on area of 2 in. by 2 in. [50 mm by 50 mm]) is to be applied nonconcurrent with the uniform load. Table 4-1 (Continued) CHAPTER 4 LIVE LOADS 20 Table 4-2 Live Load Element Factor, KLL Element KLL a Interior columns 4 Exterior columns without cantilever slabs 4 Edge columns with cantilever slabs 3 Corner columns with cantilever slabs 2 Edge beams without cantilever slabs 2 Interior beams 2 All other members not identiﬁ ed, including: 1 Edge beams with cantilever slabs Cantilever beams One-way slabs Two-way slabs Members without provisions for continuous shear transfer normal to their span a In lieu of the preceding values, KLL is permitted to be calculated. 21 Chapter 5 FLOOD LOADS community’s FIRM; or (2) the ﬂ ood corresponding to the area designated as a Flood Hazard Area on a community’s Flood Hazard Map or otherwise legally designated. DESIGN FLOOD ELEVATION (DFE): The elevation of the design ﬂ ood, including wave height, relative to the datum speciﬁ ed on a community’s ﬂ ood hazard map. FLOOD HAZARD AREA: The area subject to ﬂ ooding during the design ﬂ ood. FLOOD HAZARD MAP: The map delineating Flood Hazard Areas adopted by the authority having jurisdiction. FLOOD INSURANCE RATE MAP (FIRM): An ofﬁ cial map of a community on which the Federal Insurance and Mitigation Administration has delin-eated both special ﬂ ood hazard areas and the risk premium zones applicable to the community. SPECIAL FLOOD HAZARD AREA (AREA OF SPECIAL FLOOD HAZARD): The land in the ﬂ oodplain subject to a 1 percent or greater chance of ﬂ ooding in any given year. These areas are delineated on a community’s FIRM as A-Zones (A, AE, A1-30, A99, AR, AO, or AH) or V-Zones (V, VE, VO, or V1-30). 5.3 DESIGN REQUIREMENTS 5.3.1 Design Loads Structural systems of buildings or other structures shall be designed, constructed, connected, and anchored to resist ﬂ otation, collapse, and permanent lateral displacement due to action of ﬂ ood loads associated with the design ﬂ ood (see Section 5.3.3) and other loads in accordance with the load combina-tions of Chapter 2. 5.3.2 Erosion and Scour The effects of erosion and scour shall be included in the calculation of loads on buildings and other structures in ﬂ ood hazard areas. 5.3.3 Loads on Breakaway Walls Walls and partitions required by ASCE/SEI 24 to break away, including their connections to the structure, shall be designed for the largest of the 5.1 GENERAL The provisions of this section apply to buildings and other structures located in areas prone to ﬂ ooding as deﬁ ned on a ﬂ ood hazard map. 5.2 DEFINITIONS The following deﬁ nitions apply to the provisions of this chapter: APPROVED: Acceptable to the authority having jurisdiction. BASE FLOOD: The ﬂ ood having a 1 percent chance of being equaled or exceeded in any given year. BASE FLOOD ELEVATION (BFE): The elevation of ﬂ ooding, including wave height, having a 1 percent chance of being equaled or exceeded in any given year. BREAKAWAY WALL: Any type of wall subject to ﬂ ooding that is not required to provide structural support to a building or other structure and that is designed and constructed such that, under base ﬂ ood or lesser ﬂ ood conditions, it will collapse in such a way that: (1) it allows the free passage of ﬂ oodwaters, and (2) it does not damage the structure or supporting foundation system. COASTAL A-ZONE: An area within a special ﬂ ood hazard area, landward of a V-Zone or landward of an open coast without mapped V-Zones. To be classiﬁ ed as a Coastal A-Zone, the principal source of ﬂ ooding must be astronomical tides, storm surges, seiches, or tsunamis, not riverine ﬂ ooding, and the potential for breaking wave heights greater than or equal to 1.5 ft (0.46 m) must exist during the base ﬂ ood. COASTAL HIGH HAZARD AREA (V-ZONE): An area within a Special Flood Hazard Area, extending from offshore to the inland limit of a primary frontal dune along an open coast, and any other area that is subject to high-velocity wave action from storms or seismic sources. This area is desig-nated on Flood Insurance Rate Maps (FIRMs) as V, VE, VO, or V1-30. DESIGN FLOOD: The greater of the following two ﬂ ood events: (1) the Base Flood, affecting those areas identiﬁ ed as Special Flood Hazard Areas on the CHAPTER 5 FLOOD LOADS 22 following loads acting perpendicular to the plane of the wall: 1. The wind load speciﬁ ed in Chapter 26. 2. The earthquake load speciﬁ ed in Chapter 12. 3. 10 psf (0.48 kN/m2). The loading at which breakaway walls are intended to collapse shall not exceed 20 psf (0.96 kN/m2) unless the design meets the following conditions: 1. Breakaway wall collapse is designed to result from a ﬂ ood load less than that which occurs during the base ﬂ ood. 2. The supporting foundation and the elevated portion of the building shall be designed against collapse, permanent lateral displacement, and other struc-tural damage due to the effects of ﬂ ood loads in combination with other loads as speciﬁ ed in Chapter 2. 5.4 LOADS DURING FLOODING 5.4.1 Load Basis In ﬂ ood hazard areas, the structural design shall be based on the design ﬂ ood. 5.4.2 Hydrostatic Loads Hydrostatic loads caused by a depth of water to the level of the DFE shall be applied over all surfaces involved, both above and below ground level, except that for surfaces exposed to free water, the design depth shall be increased by 1 ft (0.30 m). Reduced uplift and lateral loads on surfaces of enclosed spaces below the DFE shall apply only if provision is made for entry and exit of ﬂ oodwater. 5.4.3 Hydrodynamic Loads Dynamic effects of moving water shall be determined by a detailed analysis utilizing basic concepts of ﬂ uid mechanics. EXCEPTION: Where water velocities do not exceed 10 ft/s (3.05 m/s), dynamic effects of moving water shall be permitted to be converted into equivalent hydrostatic loads by increasing the DFE for design purposes by an equivalent surcharge depth, dh, on the headwater side and above the ground level only, equal to d aV g h = 2 2 (5.4-1) where V = average velocity of water in ft/s (m/s) g = acceleration due to gravity, 32.2 ft/s2 (9.81 m/s2) a = coefﬁ cient of drag or shape factor (not less than 1.25) The equivalent surcharge depth shall be added to the DFE design depth and the resultant hydrostatic pressures applied to, and uniformly distributed across, the vertical projected area of the building or structure that is perpendicular to the ﬂ ow. Surfaces parallel to the ﬂ ow or surfaces wetted by the tail water shall be subject to the hydrostatic pressures for depths to the DFE only. 5.4.4 Wave Loads Wave loads shall be determined by one of the following three methods: (1) by using the analytical procedures outlined in this section, (2) by more advanced numerical modeling procedures, or (3) by laboratory test procedures (physical modeling). Wave loads are those loads that result from water waves propagating over the water surface and striking a building or other structure. Design and construction of buildings and other structures subject to wave loads shall account for the following loads: waves breaking on any portion of the building or structure; uplift forces caused by shoaling waves beneath a building or structure, or portion thereof; wave runup striking any portion of the building or structure; wave-induced drag and inertia forces; and wave-induced scour at the base of a building or structure, or its foundation. Wave loads shall be included for both V-Zones and A-Zones. In V-Zones, waves are 3 ft (0.91 m) high, or higher; in coastal ﬂ oodplains landward of the V-Zone, waves are less than 3 ft high (0.91 m). Nonbreaking and broken wave loads shall be calculated using the procedures described in Sections 5.4.2 and 5.4.3 that show how to calculate hydrostatic and hydrodynamic loads. Breaking wave loads shall be calculated using the procedures described in Sections 5.4.4.1 through 5.4.4.4. Breaking wave heights used in the procedures described in Sections 5.4.4.1 through 5.4.4.4 shall be calculated for V-Zones and Coastal A-Zones using Eqs. 5.4-2 and 5.4-3. Hb = 0.78ds (5.4-2) where Hb = breaking wave height in ft (m) ds = local still water depth in ft (m) MINIMUM DESIGN LOADS 23 The local still water depth shall be calculated using Eq. 5.4-3, unless more advanced procedures or laboratory tests permitted by this section are used. ds = 0.65(BFE – G) (5.4-3) where BFE = BFE in ft (m) G = ground elevation in ft (m) 5.4.4.1 Breaking Wave Loads on Vertical Pilings and Columns The net force resulting from a breaking wave acting on a rigid vertical pile or column shall be assumed to act at the still water elevation and shall be calculated by the following: FD = 0.5γwCDDHb 2 (5.4-4) where FD = net wave force, in lb (kN) γw = unit weight of water, in lb per cubic ft (kN/m3), = 62.4 pcf (9.80 kN/m3) for fresh water and 64.0 pcf (10.05 kN/m3) for salt water CD = coefﬁ cient of drag for breaking waves, = 1.75 for round piles or columns and = 2.25 for square piles or columns D = pile or column diameter, in ft (m) for circular sections, or for a square pile or column, 1.4 times the width of the pile or column in ft (m) Hb = breaking wave height, in ft (m) 5.4.4.2 Breaking Wave Loads on Vertical Walls Maximum pressures and net forces resulting from a normally incident breaking wave (depth-limited in size, with Hb = 0.78ds) acting on a rigid vertical wall shall be calculated by the following: Pmax = Cpγwds + 1.2γwds (5.4-5) and Ft = 1.1Cpγwds 2 + 2.4γwds 2 (5.4-6) where Pmax = maximum combined dynamic (Cpγwds) and static (1.2γwds) wave pressures, also referred to as shock pressures in lb/ft2 (kN/m2) Ft = net breaking wave force per unit length of structure, also referred to as shock, impulse, or wave impact force in lb/ft (kN/m), acting near the still water elevation Cp = dynamic pressure coefﬁ cient (1.6 < Cp < 3.5) (see Table 5.4-1) Table 5.4-1 Value of Dynamic Pressure Coefﬁ cient, Cp Risk Categorya Cp I 1.6 II 2.8 III 3.2 IV 3.5 a For Risk Category, see Table 1.5-1. γw = unit weight of water, in lb per cubic ft (kN/m3), = 62.4 pcf (9.80 kN/m3) for fresh water and 64.0 pcf (10.05 kN/m3) for salt water ds = still water depth in ft (m) at base of building or other structure where the wave breaks This procedure assumes the vertical wall causes a reﬂ ected or standing wave against the waterward side of the wall with the crest of the wave at a height of 1.2ds above the still water level. Thus, the dynamic static and total pressure distributions against the wall are as shown in Fig. 5.4-1. This procedure also assumes the space behind the vertical wall is dry, with no ﬂ uid balancing the static component of the wave force on the outside of the wall. If free water exists behind the wall, a portion of the hydrostatic component of the wave pressure and force disappears (see Fig. 5.4-2) and the net force shall be computed by Eq. 5.4-7 (the maximum combined wave pressure is still computed with Eq. 5.4-5). Ft = 1.1Cpγwds 2 + 1.9γwds 2 (5.4-7) where Ft = net breaking wave force per unit length of structure, also referred to as shock, impulse, or wave impact force in lb/ft (kN/m), acting near the still water elevation Cp = dynamic pressure coefﬁ cient (1.6 < Cp < 3.5) (see Table 5.4-1) γw = unit weight of water, in lb per cubic ft (kN/m3), = 62.4 pcf (9.80 kN/m3) for fresh water and 64.0 pcf (10.05 kN/m3) for salt water ds = still water depth in ft (m) at base of building or other structure where the wave breaks 5.4.4.3 Breaking Wave Loads on Nonvertical Walls Breaking wave forces given by Eqs. 5.4-6 and 5.4-7 shall be modiﬁ ed in instances where the walls or surfaces upon which the breaking waves act are CHAPTER 5 FLOOD LOADS 24 nonvertical. The horizontal component of breaking wave force shall be given by Fnv = Ft sin2α (5.4-8) where Fnv = horizontal component of breaking wave force in lb/ft (kN/m) Ft = net breaking wave force acting on a vertical surface in lb/ft (kN/m) α = vertical angle between nonvertical surface and the horizontal 5.4.4.4 Breaking Wave Loads from Obliquely Incident Waves Breaking wave forces given by Eqs. 5.4-6 and 5.4-7 shall be modiﬁ ed in instances where waves are obliquely incident. Breaking wave forces from non-normally incident waves shall be given by Foi = Ft sin2α (5.4-9) where Foi = horizontal component of obliquely incident breaking wave force in lb/ft (kN/m) Ft = net breaking wave force (normally incident waves) acting on a vertical surface in lb/ft (kN/m) α = horizontal angle between the direction of wave approach and the vertical surface 5.4.5 Impact Loads Impact loads are those that result from debris, ice, and any object transported by ﬂ oodwaters striking against buildings and structures, or parts thereof. Impact loads shall be determined using a rational approach as concentrated loads acting horizontally at the most critical location at or below the DFE. Vertical Wall Crest of reflected wave Dynamic pressure 1.2 ds Crest of incident wave 0.55 ds Stillwater level ds Hydrostatic pressure Ground elevation FIGURE 5.4-1 Normally Incident Breaking Wave Pressures against a Vertical Wall (Space behind Vertical Wall is Dry). MINIMUM DESIGN LOADS 25 Vertical Wall Crest of reflected wave Dynamic pressure 1.2 ds Crest of incident wave 0.55 ds Stillwater level ds Net hydrostatic pressure Ground elevation FIGURE 5.4-2 Normally Incident Breaking Wave Pressures against a Vertical Wall (Still Water Level Equal on Both Sides of Wall). 5.5 CONSENSUS STANDARDS AND OTHER REFERENCED DOCUMENTS This section lists the consensus standards and other documents that are adopted by reference within this chapter: ASCE/SEI American Society of Civil Engineers Structural Engineering Institute 1801 Alexander Bell Drive Reston, VA 20191-4400 ASCE/SEI 24 Section 5.3.3 Flood Resistant Design and Construction, 1998 27 Chapter 6 RESERVED FOR FUTURE PROVISIONS WLSC, the wind load provisions of ASCE 7 are presented in Chapters 26 through 31 as opposed to prior editions wherein the wind load provisions were contained in a single section (previously Chapter 6). In preparing the wind load provisions contained within this standard, the Wind Load Subcommittee (WLSC) of ASCE 7 established as one of its primary goals the improvement of the clarity and use of the standard. As a result of the efforts of the 29 Chapter 7 SNOW LOADS designated CS in Fig. 7-1. Ground snow loads for sites at elevations above the limits indicated in Fig. 7-1 and for all sites within the CS areas shall be approved by the authority having jurisdiction. Ground snow load determination for such sites shall be based on an extreme value statistical analysis of data available in the vicinity of the site using a value with a 2 percent annual probability of being exceeded (50-year mean recurrence interval). Snow loads are zero for Hawaii, except in mountainous regions as determined by the authority having jurisdiction. 7.3 FLAT ROOF SNOW LOADS, pf The ﬂ at roof snow load, pf, shall be calculated in lb/ft2 (kN/m2) using the following formula: pf = 0.7CeCt Ispg (7.3-1) 7.3.1 Exposure Factor, Ce The value for Ce shall be determined from Table 7-2. 7.3.2 Thermal Factor, Ct The value for Ct shall be determined from Table 7-3. 7.3.3 Importance Factor, Is The value for Is shall be determined from Table 1.5-2 based on the Risk Category from Table 1.5-1. 7.3.4 Minimum Snow Load for Low-Slope Roofs, pm A minimum roof snow load, pm, shall only apply to monoslope, hip and gable roofs with slopes less than 15°, and to curved roofs where the vertical angle from the eaves to the crown is less than 10°. The minimum roof snow load for low-slope roofs shall be obtained using the following formula: Where pg is 20 lb/ft2 (0.96 kN/m2) or less: pm = Ispg (Importance Factor times pg) Where pg exceeds 20 lb/ft2 (0.96 kN/m2): pm = 20 (Is ) (20 lb/ft2 times Importance Factor) This minimum roof snow load is a separate uniform load case. It need not be used in determining 7.1 SYMBOLS Ce = exposure factor as determined from Table 7-2 Cs = slope factor as determined from Fig. 7-2 Ct = thermal factor as determined from Table 7-3 h = vertical separation distance in feet (m) between the edge of a higher roof including any parapet and the edge of a lower adjacent roof excluding any parapet hb = height of balanced snow load determined by dividing ps by γ, in ft (m) hc = clear height from top of balanced snow load to (1) closest point on adjacent upper roof, (2) top of parapet, or (3) top of a projection on the roof, in ft (m) hd = height of snow drift, in ft (m) ho = height of obstruction above the surface of the roof, in ft (m) Is = importance factor as prescribed in Section 7.3.3 lu = length of the roof upwind of the drift, in ft (m) pd = maximum intensity of drift surcharge load, in lb/ft2 (kN/m2) pf = snow load on ﬂ at roofs (“ﬂ at” = roof slope ≤ 5°), in lb/ft2 (kN/m2) pg = ground snow load as determined from Fig. 7-1 and Table 7-1; or a site-speciﬁ c analysis, in lb/ft2 (kN/m2) pm = minimum snow load for low-slope roofs, in lb/ft2 (kN/m2) ps = sloped roof (balanced) snow load, in lb/ft2 (kN/m2) s = horizontal separation distance in feet (m) between the edges of two adjacent buildings S = roof slope run for a rise of one θ = roof slope on the leeward side, in degrees w = width of snow drift, in ft (m) W = horizontal distance from eave to ridge, in ft (m) γ = snow density, in lb/ft3 (kN/m3) as determined from Eq. 7.7-1 7.2 GROUND SNOW LOADS, pg Ground snow loads, pg, to be used in the determina-tion of design snow loads for roofs shall be as set forth in Fig. 7-1 for the contiguous United States and Table 7-1 for Alaska. Site-speciﬁ c case studies shall be made to determine ground snow loads in areas CHAPTER 7 SNOW LOADS 30 Table 7-1 Ground Snow Loads, pg, for Alaskan Locations pg pg pg Location lb/ft2 kN/m2 Location lb/ft2 kN/m2 Location lb/ft2 kN/m2 Adak 30 1.4 Galena 60 2.9 Petersburg 150 7.2 Anchorage 50 2.4 Gulkana 70 3.4 St. Paul 40 1.9 Angoon 70 3.4 Homer 40 1.9 Seward 50 2.4 Barrow 25 1.2 Juneau 60 2.9 Shemya 25 1.2 Barter 35 1.7 Kenai 70 3.4 Sitka 50 2.4 Bethel 40 1.9 Kodiak 30 1.4 Talkeetna 120 5.8 Big Delta 50 2.4 Kotzebue 60 2.9 Unalakleet 50 2.4 Cold Bay 25 1.2 McGrath 70 3.4 Valdez 160 7.7 Cordova 100 4.8 Nenana 80 3.8 Whittier 300 14.4 Fairbanks 60 2.9 Nome 70 3.4 Wrangell 60 2.9 Fort Yukon 60 2.9 Palmer 50 2.4 Yakutat 150 7.2 Table 7-2 Exposure Factor, Ce Terrain Category Exposure of Roofa Fully Exposed Partially Exposed Sheltered B (see Section 26.7) 0.9 1.0 1.2 C (see Section 26.7) 0.9 1.0 1.1 D (see Section 26.7) 0.8 0.9 1.0 Above the treeline in windswept mountainous areas. 0.7 0.8 N/A In Alaska, in areas where trees do not exist within a 2-mile (3-km) radius of the site. 0.7 0.8 N/A The terrain category and roof exposure condition chosen shall be representative of the anticipated conditions during the life of the structure. An exposure factor shall be determined for each roof of a structure. aDeﬁ nitions: Partially Exposed: All roofs except as indicated in the following text. Fully Exposed: Roofs exposed on all sides with no shelterb afforded by terrain, higher structures, or trees. Roofs that contain several large pieces of mechanical equipment, parapets that extend above the height of the balanced snow load (hb), or other obstructions are not in this category. Sheltered: Roofs located tight in among conifers that qualify as obstructions. bObstructions within a distance of 10ho provide “shelter,” where ho is the height of the obstruction above the roof level. If the only obstructions are a few deciduous trees that are leaﬂ ess in winter, the “fully exposed” category shall be used. Note that these are heights above the roof. Heights used to establish the Exposure Category in Section 26.7 are heights above the ground. Table 7-3 Thermal Factor, Ct Thermal Conditiona Ct All structures except as indicated below 1.0 Structures kept just above freezing and others with cold, ventilated roofs in which the thermal resistance (R-value) between the ventilated space and the heated space exceeds 25 °F × h × ft2/Btu (4.4 K × m2/W). 1.1 Unheated and open air structures 1.2 Structures intentionally kept below freezing 1.3 Continuously heated greenhousesb with a roof having a thermal resistance (R-value) less than 2.0 °F × h × ft2/Btu (0.4 K × m2/W) 0.85 aThese conditions shall be representative of the anticipated conditions during winters for the life of the structure. bGreenhouses with a constantly maintained interior temperature of 50 °F (10 °C) or more at any point 3 ft above the ﬂ oor level during winters and having either a maintenance attendant on duty at all times or a temperature alarm system to provide warning in the event of a heating failure. MINIMUM DESIGN LOADS 31 or in combination with drift, sliding, unbalanced, or partial loads. 7.4 SLOPED ROOF SNOW LOADS, ps Snow loads acting on a sloping surface shall be assumed to act on the horizontal projection of that surface. The sloped roof (balanced) snow load, ps, shall be obtained by multiplying the ﬂ at roof snow load, pf, by the roof slope factor, Cs: ps = Cspf (7.4-1) Values of Cs for warm roofs, cold roofs, curved roofs, and multiple roofs are determined from Sections 7.4.1 through 7.4.4. The thermal factor, Ct, from Table 7-3 determines if a roof is “cold” or “warm.” “Slippery surface” values shall be used only where the roof’s surface is unobstructed and sufﬁ cient space is avail-able below the eaves to accept all the sliding snow. A roof shall be considered unobstructed if no objects exist on it that prevent snow on it from sliding. Slippery surfaces shall include metal, slate, glass, and bituminous, rubber, and plastic membranes with a smooth surface. Membranes with an imbedded aggregate or mineral granule surface shall not be considered smooth. Asphalt shingles, wood shingles, and shakes shall not be considered slippery. 7.4.1 Warm Roof Slope Factor, Cs For warm roofs (Ct ≤ 1.0 as determined from Table 7-3) with an unobstructed slippery surface that will allow snow to slide off the eaves, the roof slope factor Cs shall be determined using the dashed line in Fig. 7-2a, provided that for nonventilated warm roofs, their thermal resistance (R-value) equals or exceeds 30 ft2 hr °F/Btu (5.3 °C m2/W) and for warm venti-lated roofs, their R-value equals or exceeds 20 ft2 hr °F/Btu (3.5 °C m2/W). Exterior air shall be able to circulate freely under a ventilated roof from its eaves to its ridge. For warm roofs that do not meet the aforementioned conditions, the solid line in Fig. 7-2a shall be used to determine the roof slope factor Cs. 7.4.2 Cold Roof Slope Factor, Cs Cold roofs are those with a Ct > 1.0 as deter-mined from Table 7-3. For cold roofs with Ct = 1.1 and an unobstructed slippery surface that will allow snow to slide off the eaves, the roof slope factor Cs shall be determined using the dashed line in Fig. 7-2b. For all other cold roofs with Ct = 1.1, the solid line in Fig. 7-2b shall be used to determine the roof slope factor Cs. For cold roofs with Ct = 1.2 and an unob-structed slippery surface that will allow snow to slide off the eaves, the roof slope factor Cs shall be determined using the dashed line on Fig. 7-2c. For all other cold roofs with Ct = 1.2, the solid line in Fig. 7-2c shall be used to determine the roof slope factor Cs. 7.4.3 Roof Slope Factor for Curved Roofs Portions of curved roofs having a slope exceeding 70° shall be considered free of snow load (i.e., Cs = 0). Balanced loads shall be determined from the balanced load diagrams in Fig. 7-3 with Cs determined from the appropriate curve in Fig. 7-2. 7.4.4 Roof Slope Factor for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs Multiple folded plate, sawtooth, or barrel vault roofs shall have a Cs = 1.0, with no reduction in snow load because of slope (i.e., ps = pf). 7.4.5 Ice Dams and Icicles Along Eaves Two types of warm roofs that drain water over their eaves shall be capable of sustaining a uniformly distributed load of 2pf on all overhanging portions: those that are unventilated and have an R-value less than 30 ft2 hr °F/Btu (5.3 °C m2/W) and those that are ventilated and have an R-value less than 20 ft2 hr °F/ Btu (3.5 °C m2/W). The load on the overhang shall be based upon the ﬂ at roof snow load for the heated portion of the roof up-slope of the exterior wall. No other loads except dead loads shall be present on the roof when this uniformly distributed load is applied. 7.5 PARTIAL LOADING The effect of having selected spans loaded with the balanced snow load and remaining spans loaded with half the balanced snow load shall be investigated as follows: 7.5.1 Continuous Beam Systems Continuous beam systems shall be investigated for the effects of the three loadings shown in Fig. 7-4: Case 1: Full balanced snow load on either exterior span and half the balanced snow load on all other spans. Case 2: Half the balanced snow load on either exterior span and full balanced snow load on all other spans. Case 3: All possible combinations of full balanced snow load on any two adjacent spans and half the balanced snow load on all other spans. For this case there will be (n –1) possible combinations where n equals the number of spans in the continu-ous beam system. CHAPTER 7 SNOW LOADS 32 If a cantilever is present in any of the above cases, it shall be considered to be a span. Partial load provisions need not be applied to structural members that span perpendicular to the ridgeline in gable roofs with slopes of 2.38˚ (½ on 12) and greater. 7.5.2 Other Structural Systems Areas sustaining only half the balanced snow load shall be chosen so as to produce the greatest effects on members being analyzed. 7.6 UNBALANCED ROOF SNOW LOADS Balanced and unbalanced loads shall be analyzed separately. Winds from all directions shall be accounted for when establishing unbalanced loads. 7.6.1 Unbalanced Snow Loads for Hip and Gable Roofs For hip and gable roofs with a slope exceeding 7 on 12 (30.2°) or with a slope less than 2.38° (½ on 12) unbalanced snow loads are not required to be applied. Roofs with an eave to ridge distance, W, of 20 ft (6.1 m) or less, having simply supported prismatic members spanning from ridge to eave shall be designed to resist an unbalanced uniform snow load on the leeward side equal to Ipg. For these roofs the windward side shall be unloaded. For all other gable roofs, the unbalanced load shall consist of 0.3ps on the windward side, ps on the leeward side plus a rectangular surcharge with magnitude hdγ/ S and horizontal extent from the ridge 8 3 Shd / where hd is the drift height from Fig. 7-9 with lu equal to the eave to ridge distance for the windward portion of the roof, W. For W less than 20 ft (6.1 m), use W = lu = 20 ft in Fig 7-9. Balanced and unbalanced loading diagrams are presented in Fig. 7-5. 7.6.2 Unbalanced Snow Loads for Curved Roofs Portions of curved roofs having a slope exceeding 70° shall be considered free of snow load. If the slope of a straight line from the eaves (or the 70° point, if present) to the crown is less than 10° or greater than 60°, unbalanced snow loads shall not be taken into account. Unbalanced loads shall be determined according to the loading diagrams in Fig. 7-3. In all cases the windward side shall be considered free of snow. If the ground or another roof abuts a Case II or Case III (see Fig. 7-3) curved roof at or within 3 ft (0.91 m) of its eaves, the snow load shall not be decreased between the 30° point and the eaves, but shall remain constant at the 30° point value. This distribution is shown as a dashed line in Fig. 7-3. 7.6.3 Unbalanced Snow Loads for Multiple Folded Plate, Sawtooth, and Barrel Vault Roofs Unbalanced loads shall be applied to folded plate, sawtooth, and barrel-vaulted multiple roofs with a slope exceeding 3/8 in./ft (1.79°). According to Section 7.4.4, Cs = 1.0 for such roofs, and the balanced snow load equals pf. The unbalanced snow load shall increase from one-half the balanced load at the ridge or crown (i.e., 0.5pf) to two times the balanced load given in Section 7.4.4 divided by Ce at the valley (i.e., 2pf/Ce). Balanced and unbalanced loading diagrams for a sawtooth roof are presented in Fig. 7-6. However, the snow surface above the valley shall not be at an elevation higher than the snow above the ridge. Snow depths shall be determined by dividing the snow load by the density of that snow from Eq. 7.7-1, which is in Section 7.7.1. 7.6.4 Unbalanced Snow Loads for Dome Roofs Unbalanced snow loads shall be applied to domes and similar rounded structures. Snow loads, deter-mined in the same manner as for curved roofs in Section 7.6.2, shall be applied to the downwind 90° sector in plan view. At both edges of this sector, the load shall decrease linearly to zero over sectors of 22.5° each. There shall be no snow load on the remaining 225° upwind sector. 7.7 DRIFTS ON LOWER ROOFS (AERODYNAMIC SHADE) Roofs shall be designed to sustain localized loads from snowdrifts that form in the wind shadow of (1) higher portions of the same structure and (2) adjacent structures and terrain features. 7.7.1 Lower Roof of a Structure Snow that forms drifts comes from a higher roof or, with the wind from the opposite direction, from the roof on which the drift is located. These two kinds of drifts (“leeward” and “windward” respectively) are shown in Fig. 7-7. The geometry of the surcharge load due to snow drifting shall be approximated by a triangle as shown in Fig. 7-8. Drift loads shall be superimposed on the balanced snow load. If hc/hb is less than 0.2, drift loads are not required to be applied. For leeward drifts, the drift height hd shall be determined directly from Fig. 7-9 using the length of the upper roof. For windward drifts, the drift height shall be determined by substituting the length of the MINIMUM DESIGN LOADS 33 lower roof for lu in Fig. 7-9 and using three-quarters of hd as determined from Fig. 7-9 as the drift height. The larger of these two heights shall be used in design. If this height is equal to or less than hc, the drift width, w, shall equal 4hd and the drift height shall equal hd. If this height exceeds hc, the drift width, w, shall equal 4hd 2/hc and the drift height shall equal hc. However, the drift width, w, shall not be greater than 8hc. If the drift width, w, exceeds the width of the lower roof, the drift shall be truncated at the far edge of the roof, not reduced to zero there. The maximum intensity of the drift surcharge load, pd, equals hdγ where snow density, γ, is deﬁ ned in Eq. 7.7-1: γ = 0.13pg + 14 but not more than 30 pcf (7.7-1) (in SI: γ = 0.426pg + 2.2, but not more than 4.7 kN/m3) This density shall also be used to determine hb by dividing ps by γ (in SI: also multiply by 102 to get the depth in m). 7.7.2 Adjacent Structures If the horizontal separation distance between adjacent structures, s, is less than 20 ft (6.1 m) and less than six times the vertical separation distance (s < 6h), then the requirements for the leeward drift of Section 7.7.1 shall be used to determine the drift load on the lower structure. The height of the snow drift shall be the smaller of hd, based upon the length of the adjacent higher structure, and (6h – s)/6. The horizontal extent of the drift shall be the smaller of 6hd or (6h – s). For windward drifts, the requirements of Section 7.7.1 shall be used. The resulting drift is permitted to be truncated. 7.8 ROOF PROJECTIONS AND PARAPETS The method in Section 7.7.1 shall be used to calculate drift loads on all sides of roof projections and at parapet walls. The height of such drifts shall be taken as three-quarters the drift height from Fig. 7-9 (i.e., 0.75hd). For parapet walls, lu shall be taken equal to the length of the roof upwind of the wall. For roof projec-tions, lu shall be taken equal to the greater of the length of the roof upwind or downwind of the projection. If the side of a roof projection is less than 15 ft (4.6 m) long, a drift load is not required to be applied to that side. 7.9 SLIDING SNOW The load caused by snow sliding off a sloped roof onto a lower roof shall be determined for slippery upper roofs with slopes greater than ¼ on 12, and for other (i.e., nonslippery) upper roofs with slopes greater than 2 on 12. The total sliding load per unit length of eave shall be 0.4pfW, where W is the horizontal distance from the eave to ridge for the sloped upper roof. The sliding load shall be distrib-uted uniformly on the lower roof over a distance of 15 ft (4.6 m) from the upper roof eave. If the width of the lower roof is less than 15 ft (4.6 m), the sliding load shall be reduced proportionally. The sliding snow load shall not be further reduced unless a portion of the snow on the upper roof is blocked from sliding onto the lower roof by snow already on the lower roof. For separated structures, sliding loads shall be considered when h/s > 1 and s < 15 ft (4.6 m). The horizontal extent of the sliding load on the lower roof shall be 15 – s with s in feet (4.6 – s with s in meters), and the load per unit length shall be 0.4 pf W (15 – s)/15 with s in feet (0.4pfW (4.6 – s)/4.6 with s in meters). Sliding loads shall be superimposed on the balanced snow load and need not be used in combina-tion with drift, unbalanced, partial, or rain-on-snow loads. 7.10 RAIN-ON-SNOW SURCHARGE LOAD For locations where pg is 20 lb/ft2 (0.96 kN/m2) or less, but not zero, all roofs with slopes (in degrees) less than W/50 with W in ft (in SI: W/15.2 with W in m) shall include a 5 lb/ft2 (0.24 kN/m2) rain-on-snow surcharge load. This additional load applies only to the sloped roof (balanced) load case and need not be used in combination with drift, sliding, unbalanced, minimum, or partial loads. 7.11 PONDING INSTABILITY Roofs shall be designed to preclude ponding instabil-ity. For roofs with a slope less than ¼ in./ft (1.19˚) and roofs where water can be impounded, roof deﬂ ections caused by full snow loads shall be evalu-ated when determining the likelihood of ponding instability (see Section 8.4). 7.12 EXISTING ROOFS Existing roofs shall be evaluated for increased snow loads caused by additions or alterations. Owners or agents for owners of an existing lower roof shall be advised of the potential for increased snow loads where a higher roof is constructed within 20 ft (6.1 m). See footnote to Table 7-2 and Section 7.7.2. CHAPTER 7 SNOW LOADS 34 FIGURE 7-1 Ground Snow Loads, Pg, for the United States (Lb/Ft2). MINIMUM DESIGN LOADS 35 FIGURE 7-1. (Continued) CHAPTER 7 SNOW LOADS 36 FIGURE 7-2 Graphs for Determining Roof Slope Factor Cs, for Warm and Cold Roofs (See Table 7-3 for Ct Deﬁ nitions). MINIMUM DESIGN LOADS 37 FIGURE 7-3 Balanced and Unbalanced Loads for Curved Roofs. CHAPTER 7 SNOW LOADS 38 FIGURE 7-4 Partial Loading Diagrams for Continuous Beams. MINIMUM DESIGN LOADS 39 W 1 S p d e c n a l a B s Unbalanced W < 20 ft with roof rafter system I pg Unbalanced Other S h 3 8 d S γ h d ps 0.3 ps Note: Unbalanced loads need not be considered for θ > 30.2° (7 on 12) or for θ ≤ 2.38° (1/2 on 12). FIGURE 7-5 Balanced and Unbalanced Snow Loads for Hip and Gable Roofs. CHAPTER 7 SNOW LOADS 40 2 pf/Ce ∗ pf 0.5 pf Balanced Load 0 Unbalanced Load May be somewhat less; see Section 7.6.3 0 FIGURE 7-6 Balanced and Unbalanced Snow Loads for a Sawtooth Roof. FIGURE 7-7 Drifts Formed at Windward and Leeward Steps. MINIMUM DESIGN LOADS 41 FIGURE 7-8 Conﬁ guration of Snow Drifts on Lower Roofs. FIGURE 7-9 Graph and Equation for Determining Drift Height, hd. 43 Chapter 8 RAIN LOADS If the secondary drainage systems contain drain lines, such lines and their point of discharge shall be separate from the primary drain lines. 8.4 PONDING INSTABILITY “Ponding” refers to the retention of water due solely to the deﬂ ection of relatively ﬂ at roofs. Susceptible bays shall be investigated by structural analysis to assure that they possess adequate stiffness to preclude progressive deﬂ ection (i.e., instability) as rain falls on them or meltwater is created from snow on them. Bays with a roof slope less than 1/4 in./ft., or on which water is impounded upon them (in whole or in part) when the primary drain system is blocked, but the secondary drain system is functional, shall be designated as susceptible bays. Roof surfaces with a slope of at least 1/4 in. per ft (1.19º) towards points of free drainage need not be considered a susceptible bay. The larger of the snow load or the rain load equal to the design condition for a blocked primary drain system shall be used in this analysis. 8.5 CONTROLLED DRAINAGE Roofs equipped with hardware to control the rate of drainage shall be equipped with a secondary drainage system at a higher elevation that limits accumulation of water on the roof above that elevation. Such roofs shall be designed to sustain the load of all rainwater that will accumulate on them to the elevation of the secondary drainage system plus the uniform load caused by water that rises above the inlet of the secondary drainage system at its design ﬂ ow (deter-mined from Section 8.3). Such roofs shall also be checked for ponding instability (determined from Section 8.4). 8.1 SYMBOLS R = rain load on the undeﬂ ected roof, in lb/ft2 (kN/m2). When the phrase “undeﬂ ected roof” is used, deﬂ ections from loads (including dead loads) shall not be considered when determining the amount of rain on the roof. ds = depth of water on the undeﬂ ected roof up to the inlet of the secondary drainage system when the primary drainage system is blocked (i.e., the static head), in in. (mm). dh = additional depth of water on the undeﬂ ected roof above the inlet of the secondary drainage system at its design ﬂ ow (i.e., the hydraulic head), in in. (mm). 8.2 ROOF DRAINAGE Roof drainage systems shall be designed in accor-dance with the provisions of the code having jurisdic-tion. The ﬂ ow capacity of secondary (overﬂ ow) drains or scuppers shall not be less than that of the primary drains or scuppers. 8.3 DESIGN RAIN LOADS Each portion of a roof shall be designed to sustain the load of all rainwater that will accumulate on it if the primary drainage system for that portion is blocked plus the uniform load caused by water that rises above the inlet of the secondary drainage system at its design ﬂ ow. R = 5.2(ds + dh) (8.3-1) In SI: R = 0.0098(ds + dh) 45 Chapter 9 RESERVED FOR FUTURE PROVISIONS separate sections and to relocate provisions into their most logical new sections. The provisions for buildings and nonbuilding structures are now distinctly separate as are the provisions for nonstructural components. Less commonly used provisions, such as those for seismi-cally isolated structures, have also been located in their own distinct chapter. We hope that the users of ASCE 7 will ﬁ nd the reformatted seismic provisions to be a signiﬁ cant improvement in organization and presentation over prior editions and will be able to more quickly locate applicable provisions. Table C11-1, located in Commentary Chapter C11 of the 2005 edition of ASCE 7 was provided to assist users in locating provisions between the 2002 and 2005 editions of the standard. Table C11-1 is not included in this edition of the standard. In preparing the seismic provisions contained within this standard, the Seismic Task Committee of ASCE 7 established a Scope and Format Subcommittee to review the layout and presentation of the seismic provisions and to make recommendations to improve the clarity and use of the standard. As a result of the efforts of this subcommittee, the seismic provisions of ASCE 7 are presented in Chapters 11 through 23 and Appendices 11A and 11B, as opposed to prior editions wherein the seismic provisions were pre-sented in a single section (previously Section 9). Of foremost concern in the reformat effort was the organization of the seismic provisions in a logical sequence for the general structural design community and the clariﬁ cation of the various headings to more accurately reﬂ ect their content. Accomplishing these two primary goals led to the decision to create 13 47 Chapter 10 ICE LOADS—ATMOSPHERIC ICING caused or enhanced by an ice accretion on a ﬂ exible structural member, component, or appurtenance are not covered in this section. 10.1.3 Exclusions Electric transmission systems, communications towers and masts, and other structures for which national standards exist are excluded from the requirements of this section. Applicable standards and guidelines include the NESC, ASCE Manual 74, and ANSI/EIA/TIA-222. 10.2 DEFINITIONS The following deﬁ nitions apply only to the provisions of this chapter. COMPONENTS AND APPURTENANCES: Nonstructural elements that may be exposed to atmospheric icing. Examples are ladders, handrails, antennas, waveguides, Radio Frequency (RF) trans-mission lines, pipes, electrical conduits, and cable trays. FREEZING RAIN: Rain or drizzle that falls into a layer of subfreezing air at the earth’s surface and freezes on contact with the ground or an object to form glaze ice. GLAZE: Clear high-density ice. HOARFROST: An accumulation of ice crystals formed by direct deposition of water vapor from the air onto an object. ICE-SENSITIVE STRUCTURES: Structures for which the effect of an atmospheric icing load governs the design of part or all of the structure. This includes, but is not limited to, lattice structures, guyed masts, overhead lines, light suspension and cable-stayed bridges, aerial cable systems (e.g., for ski lifts and logging operations), amusement rides, open catwalks and platforms, ﬂ agpoles, and signs. IN-CLOUD ICING: Occurs when supercooled cloud or fog droplets carried by the wind freeze on impact with objects. In-cloud icing usually forms rime, but may also form glaze. RIME: White or opaque ice with entrapped air. SNOW: Snow that adheres to objects by some combination of capillary forces, freezing, and sintering. 10.1 GENERAL Atmospheric ice loads due to freezing rain, snow, and in-cloud icing shall be considered in the design of ice-sensitive structures. In areas where records or experience indicate that snow or in-cloud icing produces larger loads than freezing rain, site-speciﬁ c studies shall be used. Structural loads due to hoarfrost are not a design consideration. Roof snow loads are covered in Chapter 7. 10.1.1 Site-Speciﬁ c Studies Mountainous terrain and gorges shall be exam-ined for unusual icing conditions. Site-speciﬁ c studies shall be used to determine the 50-year mean recur-rence interval ice thickness, concurrent wind speed, and concurrent temperature in 1. Alaska. 2. Areas where records or experience indicate that snow or in-cloud icing produces larger loads than freezing rain. 3. Special icing regions shown in Figs. 10-2, 10-4, and 10-5. 4. Mountainous terrain and gorges where examination indicates unusual icing conditions exist. Site-speciﬁ c studies shall be subject to review and approval by the authority having jurisdiction. In lieu of using the mapped values, it shall be permitted to determine the ice thickness, the concur-rent wind speed, and the concurrent temperature for a structure from local meteorological data based on a 50-year mean recurrence interval provided that 1. The quality of the data for wind and type and amount of precipitation has been taken into account. 2. A robust ice accretion algorithm has been used to estimate uniform ice thicknesses and concurrent wind speeds from these data. 3. Extreme-value statistical analysis procedures acceptable to the authority having jurisdiction have been employed in analyzing the ice thickness and concurrent wind speed data. 4. The length of record and sampling error have been taken into account. 10.1.2 Dynamic Loads Dynamic loads, such as those resulting from gallop-ing, ice shedding, and aeolian vibrations, that are CHAPTER 10 ICE LOADS—ATMOSPHERIC ICING 48 10.3 SYMBOLS As = surface area of one side of a ﬂ at plate or the projected area of complex shapes Ai = cross-sectional area of ice D = diameter of a circular structure or member as deﬁ ned in Chapter 29, in ft (m) Dc = diameter of the cylinder circumscribing an object fz = factor to account for the increase in ice thick-ness with height Ii = importance factor for ice thickness from Table 1.5-2 based on the Risk Category from Table 1.5-1 Iw = importance factor for concurrent wind pressure from Table 1.5-2 based on the Risk Category from Table 1.5-1 Kzt = topographic factor as deﬁ ned in Chapter 26 qz = velocity pressure evaluated at height z above ground, in lb/ft2 (N/m2) as deﬁ ned in Chapter 29 r = radius of the maximum cross-section of a dome or radius of a sphere t = nominal ice thickness due to freezing rain at a height of 33 ft (10 m) from Figs. 10-2 through 10-6 in inches (mm) td = design ice thickness in in. (mm) from Eq. 10.4-5 Vc = concurrent wind speed mph (m/s) from Figs. 10-2 through 10-6 Vi = volume of ice z = height above ground in ft (m) ∈ = solidity ratio as deﬁ ned in Chapter 29 10.4 ICE LOADS DUE TO FREEZING RAIN 10.4.1 Ice Weight The ice load shall be determined using the weight of glaze ice formed on all exposed surfaces of structural members, guys, components, appurtenances, and cable systems. On structural shapes, prismatic members, and other similar shapes, the cross-sectional area of ice shall be determined by Ai = πtd(Dc + td) (10.4-1) Dc is shown for a variety of cross-sectional shapes in Fig. 10-1. On ﬂ at plates and large three-dimensional objects such as domes and spheres, the volume of ice shall be determined by Vi = πtdAs (10.4-2) For a ﬂ at plate As shall be the area of one side of the plate, for domes and spheres As shall be deter-mined by As = πr2 (10.4-3) It is acceptable to multiply Vi by 0.8 for vertical plates and 0.6 for horizontal plates. The ice density shall be not less than 56 pcf (900 kg/m3). 10.4.2 Nominal Ice Thickness Figs. 10-2 through 10-6 show the equivalent uniform radial thicknesses t of ice due to freezing rain at a height of 33 ft (10 m) over the contiguous 48 states and Alaska for a 50-year mean recurrence interval. Also shown are concurrent 3-s gust wind speeds. Thicknesses for Hawaii, and for ice accretions due to other sources in all regions, shall be obtained from local meteorological studies. 10.4.3 Height Factor The height factor fz used to increase the radial thickness of ice for height above ground z shall be determined by fz = z 33 0 10 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . for 0 ft < z ≤ 900 ft (10.4-4) fz = 1.4 for z > 900 ft In SI: fz = z 10 0 10 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . for 0 m < z ≤ 275 m fz = 1.4 for z > 275 m 10.4.4 Importance Factors Importance factors to be applied to the radial ice thickness and wind pressure shall be determined from Table 1.5-2 based on the Risk Category from Table 1.5-1. The importance factor Ii shall be applied to the ice thickness, not the ice weight, because the ice weight is not a linear function of thickness. 10.4.5 Topographic Factor Both the ice thickness and concurrent wind speed for structures on hills, ridges, and escarpments are higher than those on level terrain because of wind speed-up effects. The topographic factor for the concurrent wind pressure is Kzt and the topographic factor for ice thickness is (Kzt)0.35, where Kzt is obtained from Eq. 26.8-1. 10.4.6 Design Ice Thickness for Freezing Rain The design ice thickness td shall be calculated from Eq. 10.4-5. td = 2.0tIifz(Kzt)0.35 (10.4-5) MINIMUM DESIGN LOADS 49 10.5 WIND ON ICE-COVERED STRUCTURES Ice accreted on structural members, components, and appurtenances increases the projected area of the structure exposed to wind. The projected area shall be increased by adding td to all free edges of the projected area. Wind loads on this increased projected area shall be used in the design of ice-sensitive structures. Figs. 10-2 to 10-6 include 3-s gust wind speeds at 33 ft (10 m) above grade that are concurrent with the ice loads due to freezing rain. Wind loads shall be calculated in accordance with Chapters 26 through 31 as modiﬁ ed by Sections 10.5.1 through 10.5.5. 10.5.1 Wind on Ice-Covered Chimneys, Tanks, and Similar Structures Force coefﬁ cients Cf for structures with square, hexagonal, and octagonal cross-sections shall be as given in Fig. 29.5-1. Force coefﬁ cients Cf for struc-tures with round cross-sections shall be as given in Fig. 29.5-1 for round cross-sections with D√qz ≤ 2.5 for all ice thicknesses, wind speeds, and structure diameters. 10.5.2 Wind on Ice-Covered Solid Freestanding Walls and Solid Signs Force coefﬁ cients Cf shall be as given in Fig. 29.4 based on the dimensions of the wall or sign including ice. 10.5.3 Wind on Ice-Covered Open Signs and Lattice Frameworks The solidity ratio ∈ shall be based on the projected area including ice. The force coefﬁ cient Cf for the projected area of ﬂ at members shall be as given in Fig. 29.5-2. The force coefﬁ cient Cf for rounded members and for the additional projected area due to ice on both ﬂ at and rounded members shall be as given in Fig. 29.5-2 for rounded members with D√qz ≤ 2.5 for all ice thicknesses, wind speeds, and member diameters. 10.5.4 Wind on Ice-Covered Trussed Towers The solidity ratio ∈ shall be based on the projected area including ice. The force coefﬁ cients Cf shall be as given in Fig. 29.5-3. It is acceptable to reduce the force coefﬁ cients Cf for the additional projected area due to ice on both round and ﬂ at members by the factor for rounded members in Note 3 of Fig. 29.5-3. 10.5.5 Wind on Ice-Covered Guys and Cables The force coefﬁ cient Cf (as deﬁ ned in Chapter 29) for ice-covered guys and cables shall be 1.2. 10.6 Design Temperatures for Freezing Rain The design temperatures for ice and wind-on-ice due to freezing rain shall be either the temperature for the site shown in Figs. 10-7 and 10-8 or 32°F (0°C), whichever gives the maximum load effect. The temperature for Hawaii shall be 32°F (0°C). For temperature sensitive structures, the load shall include the effect of temperature change from everyday conditions to the design temperature for ice and wind-on-ice. These temperatures are to be used with ice thicknesses for all mean recurrence intervals. The design temperatures are considered to be concurrent with the design ice load and the concurrent wind load. 10.7 PARTIAL LOADING The effects of a partial ice load shall be considered when this condition is critical for the type of structure under consideration. It is permitted to consider this to be a static load. 10.8 DESIGN PROCEDURE 1. The nominal ice thickness, t, the concurrent wind speed, Vc, and the concurrent temperature for the site shall be determined from Figs. 10-2 to 10-8 or a site-speciﬁ c study. 2. The topographic factor for the site, Kzt, shall be determined in accordance with Section 10.4.5. 3. The importance factor for ice thickness, Ii, shall be determined in accordance with Section 10.4.4. 4. The height factor, fz, shall be determined in accordance with Section 10.4.3 for each design segment of the structure. 5. The design ice thickness, td, shall be determined in accordance with Section 10.4.6, Eq. 10.4-5. 6. The weight of ice shall be calculated for the design ice thickness, td, in accordance with Section 10.4.1. 7. The velocity pressure qz for wind speed Vc shall be determined in accordance with Section 29.3 using the importance factor for concurrent wind pressure Iw determined in accordance with Section 10.4.4. 8. The wind force coefﬁ cients Cf shall be deter-mined in accordance with Section 10.5. 9. The gust effect factor shall be determined in accordance with Section 26.9. CHAPTER 10 ICE LOADS—ATMOSPHERIC ICING 50 10. The design wind force shall be determined in accordance with Chapter 29. 11. The iced structure shall be analyzed for the load combinations in either Section 2.3 or 2.4. 10.9 CONSENSUS STANDARDS AND OTHER REFERENCED DOCUMENTS This section lists the consensus standards and other documents that are adopted by reference within this chapter: ASCE American Society of Civil Engineers 1801 Alexander Bell Drive Reston, VA 20191 ASCE Manual 74 Section 10.1.3 Guidelines for Electrical Transmission Line Structural Loading, 1991 ANSI American National Standards Institute 25 West 43rd Street, 4th Floor New York, NY 10036 ANSI/EIA/TIA-222 Section 10.1.3 Structural Standards for Steel Antenna Towers and Antenna Supporting Structures, 1996 IEEE 445 Hoes Lane Piscataway, NJ 08854-1331 NESC Section 10.1.3 National Electrical Safety Code, 2001 MINIMUM DESIGN LOADS 51 FIGURE 10-1 Characteristic Dimension Dc for Calculating the Ice Area for a Variety of Cross-Sectional Shapes. CHAPTER 10 ICE LOADS—ATMOSPHERIC ICING 52 FIGURE 10-2 Equivalent Radial Ice Thicknesses Due to Freezing Rain with Concurrent 3-Second Gust Speeds, for a 50-Year Mean Recurrence Interval. MINIMUM DESIGN LOADS 53 FIGURE 10-2 (Continued) CHAPTER 10 ICE LOADS—ATMOSPHERIC ICING 54 FIGURE 10-3 Lake Superior Detail. FIGURE 10-4 Fraser Valley Detail. FIGURE 10-5 Columbia River Gorge Detail. FIGURE 10-6 50-Yr Mean Recurrence Interval Uniform Ice Thicknesses Due to Freezing Rain with Concurrent 3-Second Gust Speeds: Alaska. FIGURE 10-7 Temperatures Concurrent with Ice Thicknesses Due to Freezing Rain: Contiguous 48 States. FIGURE 10-8 Temperatures Concurrent with Ice Thicknesses Due to Freezing Rain: Alaska. 57 Chapter 11 SEISMIC DESIGN CRITERIA vehicular bridges, electrical transmission towers, hydraulic structures, buried utility lines and their appurtenances, and nuclear reactors. 5. Piers and wharves that are not accessible to the general public. 11.1.3 Applicability Structures and their nonstructural components shall be designed and constructed in accordance with the requirement of the following sections based on the type of structure or component: a. Buildings: Chapter 12 b. Nonbuilding Structures: Chapter 15 c. Nonstructural Components: Chapter 13 d. Seismically Isolated Structures: Chapter 17 e. Structures with Damping Systems: Chapter 18 Buildings whose purpose is to enclose equipment or machinery and whose occupants are engaged in maintenance or monitoring of that equipment, machinery or their associated processes shall be permitted to be classiﬁ ed as nonbuilding structures designed and detailed in accordance with Section 15.5 of this standard. 11.1.4 Alternate Materials and Methods of Construction Alternate materials and methods of construction to those prescribed in the seismic requirements of this standard shall not be used unless approved by the authority having jurisdiction. Substantiating evidence shall be submitted demonstrating that the proposed alternate, for the purpose intended, will be at least equal in strength, durability, and seismic resistance. 11.2 DEFINITIONS The following deﬁ nitions apply only to the seismic requirements of this standard. ACTIVE FAULT: A fault determined to be active by the authority having jurisdiction from properly substantiated data (e.g., most recent mapping of active faults by the United States Geological Survey). ADDITION: An increase in building area, aggregate ﬂ oor area, height, or number of stories of a structure. 11.1 GENERAL 11.1.1 Purpose Chapter 11 presents criteria for the design and construction of buildings and other structures subject to earthquake ground motions. The speciﬁ ed earth-quake loads are based upon post-elastic energy dissipation in the structure, and because of this fact, the requirements for design, detailing, and construc-tion shall be satisﬁ ed even for structures and members for which load combinations that do not contain earthquake loads indicate larger demands than combinations that include earthquake loads. Minimum requirements for quality assurance for seismic force-resisting systems are set forth in Appendix 11A. 11.1.2 Scope Every structure, and portion thereof, including nonstructural components, shall be designed and constructed to resist the effects of earthquake motions as prescribed by the seismic requirements of this standard. Certain nonbuilding structures, as described in Chapter 15, are also within the scope and shall be designed and constructed in accordance with the requirements of Chapter 15. Requirements concerning alterations, additions, and change of use are set forth in Appendix 11B. Existing structures and alterations to existing structures need only comply with the seismic requirements of this standard where required by Appendix 11B. The following structures are exempt from the seismic requirements of this standard: 1. Detached one- and two-family dwellings that are located where the mapped, short period, spectral response acceleration parameter, SS, is less than 0.4 or where the Seismic Design Category determined in accordance with Section 11.6 is A, B, or C. 2. Detached one- and two-family wood-frame dwellings not included in Exception 1 with not more than two stories above grade plane, satisfying the limitations of and constructed in accordance with the IRC. 3. Agricultural storage structures that are intended only for incidental human occupancy. 4. Structures that require special consideration of their response characteristics and environment that are not addressed in Chapter 15 and for which other regulations provide seismic criteria, such as CHAPTER 11 SEISMIC DESIGN CRITERIA 58 ALTERATION: Any construction or renovation to an existing structure other than an addition. APPENDAGE: An architectural component such as a canopy, marquee, ornamental balcony, or statuary. APPROVAL: The written acceptance by the authority having jurisdiction of documentation that establishes the qualiﬁ cation of a material, system, component, procedure, or person to fulﬁ ll the requirements of this standard for the intended use. ATTACHMENTS: Means by which nonstruc-tural components or supports of nonstructural compo-nents are secured or connected to the seismic force-resisting system of the structure. Such attach-ments include anchor bolts, welded connections, and mechanical fasteners. BASE: The level at which the horizontal seismic ground motions are considered to be imparted to the structure. BASE SHEAR: Total design lateral force or shear at the base. BOUNDARY ELEMENTS: Diaphragm and shear wall boundary members to which the diaphragm transfers forces. Boundary members include chords and drag struts at diaphragm and shear wall perim-eters, interior openings, discontinuities, and reentrant corners. BOUNDARY MEMBERS: Portions along wall and diaphragm edges strengthened by longitudinal and transverse reinforcement. Boundary members include chords and drag struts at diaphragm and shear wall perimeters, interior openings, discontinuities, and reentrant corners. BUILDING: Any structure whose intended use includes shelter of human occupants. CANTILEVERED COLUMN SYSTEM: A seismic force-resisting system in which lateral forces are resisted entirely by columns acting as cantilevers from the base. CHARACTERISTIC EARTHQUAKE: An earthquake assessed for an active fault having a magnitude equal to the best estimate of the maximum magnitude capable of occurring on the fault, but not less than the largest magnitude that has occurred historically on the fault. COMPONENT: A part of an architectural, electrical, or mechanical system. Component, Nonstructural: A part of an architectural, mechanical, or electrical system within or without a building or nonbuilding structure. Component, Flexible: Nonstructural component having a fundamental period greater than 0.06 s. Component, Rigid: Nonstructural component having a fundamental period less than or equal to 0.06 s. CONCRETE, PLAIN: Concrete that is either unreinforced or contains less reinforcement than the minimum amount speciﬁ ed in ACI 318 for reinforced concrete. CONCRETE, REINFORCED: Concrete reinforced with no less reinforcement than the minimum amount required by ACI 318 prestressed or nonprestressed, and designed on the assumption that the two materials act together in resisting forces. CONSTRUCTION DOCUMENTS: The written, graphic, electronic, and pictorial documents describing the design, locations, and physical charac-teristics of the project required to verify compliance with this standard. COUPLING BEAM: A beam that is used to connect adjacent concrete wall elements to make them act together as a unit to resist lateral loads. DEFORMABILITY: The ratio of the ultimate deformation to the limit deformation. High-Deformability Element: An element whose deformability is not less than 3.5 where subjected to four fully reversed cycles at the limit deformation. Limited-Deformability Element: An element that is neither a low-deformability nor a high-deformability element. Low-Deformability Element: An element whose deformability is 1.5 or less. DEFORMATION: Limit Deformation: Two times the initial deformation that occurs at a load equal to 40 percent of the maximum strength. Ultimate Deformation: The deformation at which failure occurs and that shall be deemed to occur if the sustainable load reduces to 80 percent or less of the maximum strength. DESIGNATED SEISMIC SYSTEMS: Those nonstructural components that require design in accordance with Chapter 13 and for which the component importance factor, Ip, is greater than 1.0. DESIGN EARTHQUAKE: The earthquake effects that are two-thirds of the corresponding Maximum Considered Earthquake (MCER) effects. MINIMUM DESIGN LOADS 59 DESIGN EARTHQUAKE GROUND MOTION: The earthquake ground motions that are two-thirds of the corresponding MCER ground motions. DIAPHRAGM: Roof, ﬂ oor, or other membrane or bracing system acting to transfer the lateral forces to the vertical resisting elements. DIAPHRAGM BOUNDARY: A location where shear is transferred into or out of the diaphragm element. Transfer is either to a boundary element or to another force-resisting element. DIAPHRAGM CHORD: A diaphragm bound-ary element perpendicular to the applied load that is assumed to take axial stresses due to the diaphragm moment. DRAG STRUT (COLLECTOR, TIE, DIA-PHRAGM STRUT): A diaphragm or shear wall boundary element parallel to the applied load that collects and transfers diaphragm shear forces to the vertical force-resisting elements or distributes forces within the diaphragm or shear wall. ENCLOSURE: An interior space surrounded by walls. EQUIPMENT SUPPORT: Those structural members or assemblies of members or manufactured elements, including braces, frames, legs, lugs, snuggers, hangers, or saddles that transmit gravity loads and operating loads between the equipment and the structure. FLEXIBLE CONNECTIONS: Those connec-tions between equipment components that permit rotational and/or translational movement without degradation of performance. Examples include universal joints, bellows expansion joints, and ﬂ exible metal hose. FRAME: Braced Frame: An essentially vertical truss, or its equivalent, of the concentric or eccentric type that is provided in a building frame system or dual system to resist seismic forces. Concentrically Braced Frame (CBF): A braced frame in which the members are subjected primarily to axial forces. CBFs are categorized as ordinary concentrically braced frames (OCBFs) or special concentrically braced frames (SCBFs). Eccentrically Braced Frame (EBF): A diagonally braced frame in which at least one end of each brace frames into a beam a short distance from a beam-column or from another diagonal brace. Moment Frame: A frame in which members and joints resist lateral forces by ﬂ exure as well as along the axis of the members. Moment frames are categorized as intermediate moment frames (IMF), ordinary moment frames (OMF), and special moment frames (SMF). Structural System: Building Frame System: A structural system with an essentially complete space frame providing support for vertical loads. Seismic force resistance is provided by shear walls or braced frames. Dual System: A structural system with an essentially complete space frame providing support for vertical loads. Seismic force resistance is provided by moment-resisting frames and shear walls or braced frames as prescribed in Section 12.2.5.1. Shear Wall-Frame Interactive System: A structural system that uses combinations of ordinary reinforced concrete shear walls and ordinary reinforced concrete moment frames designed to resist lateral forces in proportion to their rigidities considering interaction between shear walls and frames on all levels. Space Frame System: A 3-D structural system composed of interconnected members, other than bearing walls, that is capable of support-ing vertical loads and, where designed for such an application, is capable of providing resis-tance to seismic forces. FRICTION CLIP: A device that relies on friction to resist applied loads in one or more direc-tions to anchor a nonstructural component. Friction is provided mechanically and is not due to gravity loads. GLAZED CURTAIN WALL: A nonbearing wall that extends beyond the edges of building ﬂ oor slabs, and includes a glazing material installed in the curtain wall framing. GLAZED STOREFRONT: A nonbearing wall that is installed between ﬂ oor slabs, typically includ-ing entrances, and includes a glazing material installed in the storefront framing. GRADE PLANE: A horizontal reference plane representing the average of ﬁ nished ground level adjoining the structure at all exterior walls. Where the ﬁ nished ground level slopes away from the exterior walls, the grade plane is established by the lowest points within the area between the structure and the property line or, where the property line is more than 6 ft (1,829 mm) from the structure, between the structure and points 6 ft (1,829 mm) from the structure. CHAPTER 11 SEISMIC DESIGN CRITERIA 60 INSPECTION, SPECIAL: The observation of the work by a special inspector to determine compli-ance with the approved construction documents and these standards in accordance with the quality assurance plan. Continuous Special Inspection: The full-time observation of the work by a special inspector who is present in the area where work is being performed. Periodic Special Inspection: The part-time or intermittent observation of the work by a special inspector who is present in the area where work has been or is being performed. INSPECTOR, SPECIAL (who shall be identi-ﬁ ed as the owner’s inspector): A person approved by the authority having jurisdiction to perform special inspection. INVERTED PENDULUM-TYPE STRUC-TURES: Structures in which more than 50 percent of the structure’s mass is concentrated at the top of a slender, cantilevered structure and in which stability of the mass at the top of the structure relies on rotational restraint to the top of the cantilevered element. JOINT: The geometric volume common to intersecting members. LIGHT-FRAME CONSTRUCTION: A method of construction where the structural assemblies (e.g., walls, ﬂ oors, ceilings, and roofs) are primarily formed by a system of repetitive wood or cold-formed steel framing members or subassemblies of these members (e.g., trusses). LONGITUDINAL REINFORCEMENT RATIO: Area of longitudinal reinforcement divided by the cross-sectional area of the concrete. MAXIMUM CONSIDERED EARTHQUAKE (MCE) GROUND MOTION: The most severe earthquake effects considered by this standard more speciﬁ cally deﬁ ned in the following two terms. MAXIMUM CONSIDERED EARTHQUAKE GEOMETRIC MEAN (MCEG) PEAK GROUND ACCELERATION: The most severe earthquake effects considered by this standard determined for geometric mean peak ground acceleration and without adjustment for targeted risk. The MCEG peak ground acceleration adjusted for site effects (PGAM) is used in this standard for evaluation of liquefaction, lateral spreading, seismic settlements, and other soil related issues. In this standard, general procedures for determining PGAM are provided in Section 11.8.3; site-speciﬁ c procedures are provided in Section 21.5. RISK-TARGETED MAXIMUM CONSID-ERED EARTHQUAKE (MCER) GROUND MOTION RESPONSE ACCELERATION: The most severe earthquake effects considered by this standard determined for the orientation that results in the largest maximum response to horizontal ground motions and with adjustment for targeted risk. In this standard, general procedures for determining the MCER Ground Motion values are provided in Section 11.4.3; site-speciﬁ c procedures are provided in Sections 21.1 and 21.2. MECHANICALLY ANCHORED TANKS OR VESSELS: Tanks or vessels provided with mechani-cal anchors to resist overturning moments. NONBUILDING STRUCTURE: A structure, other than a building, constructed of a type included in Chapter 15 and within the limits of Section 15.1.1. NONBUILDING STRUCTURE SIMILAR TO A BUILDING: A nonbuilding structure that is designed and constructed in a manner similar to buildings, will respond to strong ground motion in a fashion similar to buildings, and has a basic lateral and vertical seismic force-resisting system conforming to one of the types indicated in Tables 12.2-1 or 15.4-1. ORTHOGONAL: To be in two horizontal directions, at 90° to each other. OWNER: Any person, agent, ﬁ rm, or corporation having a legal or equitable interest in the property. PARTITION: A nonstructural interior wall that spans horizontally or vertically from support to support. The supports may be the basic building frame, subsidiary structural members, or other portions of the partition system. P-DELTA EFFECT: The secondary effect on shears and moments of structural members due to the action of the vertical loads induced by horizontal displacement of the structure resulting from various loading conditions. PILE: Deep foundation element, which includes piers, caissons, and piles. PILE CAP: Foundation elements to which piles are connected including grade beams and mats. REGISTERED DESIGN PROFESSIONAL: An architect or engineer, registered or licensed to practice professional architecture or engineering, as deﬁ ned by the statutory requirements of the profes-sional registrations laws of the state in which the project is to be constructed. SEISMIC DESIGN CATEGORY: A classiﬁ ca-tion assigned to a structure based on its Risk Category and the severity of the design earthquake ground motion at the site as deﬁ ned in Section 11.4. MINIMUM DESIGN LOADS 61 SEISMIC FORCE-RESISTING SYSTEM: That part of the structural system that has been considered in the design to provide the required resistance to the seismic forces prescribed herein. SEISMIC FORCES: The assumed forces prescribed herein, related to the response of the structure to earthquake motions, to be used in the design of the structure and its components. SELF-ANCHORED TANKS OR VESSELS: Tanks or vessels that are stable under design overturn-ing moment without the need for mechanical anchors to resist uplift. SHEAR PANEL: A ﬂ oor, roof, or wall element sheathed to act as a shear wall or diaphragm. SITE CLASS: A classiﬁ cation assigned to a site based on the types of soils present and their engineer-ing properties as deﬁ ned in Chapter 20. STORAGE RACKS: Include industrial pallet racks, moveable shelf racks, and stacker racks made of cold-formed or hot-rolled structural members. Does not include other types of racks such as drive-in and drive-through racks, cantilever racks, portable racks, or racks made of materials other than steel. STORY: The portion of a structure between the tops of two successive ﬂ oor surfaces and, for the topmost story, from the top of the ﬂ oor surface to the top of the roof surface. STORY ABOVE GRADE PLANE: A story in which the ﬂ oor or roof surface at the top of the story is more than 6 ft (1,828 mm) above grade plane or is more than 12 ft (3,658 mm) above the ﬁ nished ground level at any point on the perimeter of the structure. STORY DRIFT: The horizontal deﬂ ection at the top of the story relative to the bottom of the story as determined in Section 12.8.6. STORY DRIFT RATIO: The story drift, as determined in Section 12.8.6, divided by the story height, hsx. STORY SHEAR: The summation of design lateral seismic forces at levels above the story under consideration. STRENGTH: Design Strength: Nominal strength multiplied by a strength reduction factor, ϕ. Nominal Strength: Strength of a member or cross-section calculated in accordance with the requirements and assumptions of the strength design methods of this standard (or the reference documents) before application of any strength-reduction factors. Required Strength: Strength of a member, cross-section, or connection required to resist factored loads or related internal moments and forces in such combinations as stipulated by this standard. STRUCTURAL HEIGHT: The vertical distance from the base to the highest level of the seismic force-resisting system of the structure. For pitched or sloped roofs, the structural height is from the base to the average height of the roof. STRUCTURAL OBSERVATIONS: The visual observations to determine that the seismic force-resisting system is constructed in general conformance with the construction documents. STRUCTURE: That which is built or con-structed and limited to buildings and nonbuilding structures as deﬁ ned herein. SUBDIAPHRAGM: A portion of a diaphragm used to transfer wall anchorage forces to diaphragm cross ties. SUPPORTS: Those members, assemblies of members, or manufactured elements, including braces, frames, legs, lugs, snubbers, hangers, saddles, or struts, and associated fasteners that transmit loads between nonstructural components and their attach-ments to the structure. TESTING AGENCY: A company or corporation that provides testing and/or inspection services. VENEERS: Facings or ornamentation of brick, concrete, stone, tile, or similar materials attached to a backing. WALL: A component that has a slope of 60° or greater with the horizontal plane used to enclose or divide space. Bearing Wall: Any wall meeting either of the following classiﬁ cations: 1. Any metal or wood stud wall that supports more than 100 lb/linear ft (1,459 N/m) of vertical load in addition to its own weight. 2. Any concrete or masonry wall that supports more than 200 lb/linear ft (2,919 N/m) of vertical load in addition to its own weight. Light Frame Wall: A wall with wood or steel studs. Light Frame Wood Shear Wall: A wall constructed with wood studs and sheathed with material rated for shear resistance. Nonbearing Wall: Any wall that is not a bearing wall. Nonstructural Wall: All walls other than bearing walls or shear walls. Shear Wall (Vertical Diaphragm): A wall, bearing or nonbearing, designed to resist lateral CHAPTER 11 SEISMIC DESIGN CRITERIA 62 forces acting in the plane of the wall (some-times referred to as a “vertical diaphragm”). Structural Wall: Walls that meet the deﬁ nition for bearing walls or shear walls. WALL SYSTEM, BEARING: A structural system with bearing walls providing support for all or major portions of the vertical loads. Shear walls or braced frames provide seismic force resistance. WOOD STRUCTURAL PANEL: A wood-based panel product that meets the requirements of DOC PS1 or DOC PS2 and is bonded with a water-proof adhesive. Included under this designation are plywood, oriented strand board, and composite panels. 11.3 SYMBOLS The unit dimensions used with the items covered by the symbols shall be consistent throughout except where speciﬁ cally noted. Symbols presented in this section apply only to the seismic requirements in this standard as indicated. Ach = cross-sectional area (in.2 or mm2) of a structural member measured out-to-out of transverse reinforcement A0 = area of the load-carrying foundation (ft2 or m2) Ash = total cross-sectional area of hoop rein-forcement (in.2 or mm2), including supplementary cross-ties, having a spacing of sh and crossing a section with a core dimension of hc Avd = required area of leg (in.2 or mm2) of diagonal reinforcement Ax = torsional ampliﬁ cation factor (Section 12.8.4.3) ai = the acceleration at level i obtained from a modal analysis (Section 13.3.1) ap = the ampliﬁ cation factor related to the response of a system or component as affected by the type of seismic attach-ment, determined in Section 13.3.1 bp = the width of the rectangular glass panel Cd = deﬂ ection ampliﬁ cation factor as given in Tables 12.2-1, 15.4-1, or 15.4-2 CR = site-speciﬁ c risk coefﬁ cient at any period; see Section 21.2.1.1 CRS = mapped value of the risk coefﬁ cient at short periods as given by Fig. 22-17 CR1 = mapped value of the risk coefﬁ cient at a period of 1 s as given by Fig. 22-18 Cs = seismic response coefﬁ cient determined in Section 12.8.1.1 and 19.3.1 (dimensionless) CT = building period coefﬁ cient in Section 12.8.2.1 Cvx = vertical distribution factor as determined in Section 12.8.3 c = distance from the neutral axis of a ﬂ exural member to the ﬁ ber of maximum compressive strain (in. or mm) D = the effect of dead load Dclear = relative horizontal (drift) displacement, measured over the height of the glass panel under consideration, which causes initial glass-to-frame contact. For rectan-gular glass panels within a rectangular wall frame, Dclear is set forth in Section 13.5.9.1 DpI = seismic relative displacement; see Section 13.3.2 Ds = the total depth of stratum in Eq. 19.2-12 (ft or m) dC = The total thickness of cohesive soil layers in the top 100 ft (30 m); see Section 20.4.3 (ft or m) di = The thickness of any soil or rock layer i (between 0 and 100 ft [30 m]); see Section 20.4.1 (ft or m) dS = The total thickness of cohesionless soil layers in the top 100 ft (30 m); see Section 20.4.2 (ft or m) E = effect of horizontal and vertical earth-quake-induced forces (Section 12.4) Fa = short-period site coefﬁ cient (at 0.2 s-period); see Section 11.4.3 Fi, Fn, Fx = portion of the seismic base shear, V, induced at Level i, n, or x, respectively, as determined in Section 12.8.3 Fp = the seismic force acting on a component of a structure as determined in Sections 12.11.1 and 13.3.1 FPGA = site coefﬁ cient for PGA; see Section 11.8.3 Fv = long-period site coefﬁ cient (at 1.0 s-period); see Section 11.4.3 fc′ = speciﬁ ed compressive strength of concrete used in design fs′ = ultimate tensile strength (psi or MPa) of the bolt, stud, or insert leg wires. For ASTM A307 bolts or A108 studs, it is permitted to be assumed to be 60,000 psi (415 MPa) fy = speciﬁ ed yield strength of reinforcement (psi or MPa) fyh = speciﬁ ed yield strength of the special lateral reinforcement (psi or kPa) MINIMUM DESIGN LOADS 63 G = γυs 2/g = the average shear modulus for the soils beneath the foundation at large strain levels (psf or Pa) G0 = γυs0 2 /g = the average shear modulus for the soils beneath the foundation at small strain levels (psf or Pa) g = acceleration due to gravity H = thickness of soil h = height of a shear wall measured as the maximum clear height from top of foundation to bottom of diaphragm framing above, or the maximum clear height from top of diaphragm to bottom of diaphragm framing above h = average roof height of structure with respect to the base; see Chapter 13 h _ = effective height of the building as determined in Section 19.2.1.1 or 19.3.1 (ft or m) hc = core dimension of a component measured to the outside of the special lateral reinforcement (in. or mm) hi, hx = the height above the base to Level i or x, respectively hn = structural height as deﬁ ned in Section 11.2 hp = the height of the rectangular glass panel hsx = the story height below Level x = (hx – hx–1) Ie = the importance factor as prescribed in Section 11.5.1 I0 = the static moment of inertia of the load-carrying foundation; see Section 19.2.1.1 (in.4 or mm4) Ip = the component importance factor as prescribed in Section 13.3.1 i = the building level referred to by the subscript i; i = 1 designates the ﬁ rst level above the base Kp = the stiffness of the component or attach-ment, Section 13.6.2 Ky = the lateral stiffness of the foundation as deﬁ ned in Section 19.2.1.1 (lb/in. or N/m) Kθ = the rocking stiffness of the foundation as deﬁ ned in Section 19.2.1.1 (ft-lb/degree or N-m/rad) KL/r = the lateral slenderness ratio of a compres-sion member measured in terms of its effective length, KL, and the least radius of gyration of the member cross section, r k = distribution exponent given in Section 12.8.3 k _ = stiffness of the building as determined in Section 19.2.1.1 (lb/ft or N/m) ka = coefﬁ cient deﬁ ned in Sections 12.11.2 and 12.14.7.5 L = overall length of the building (ft or m) at the base in the direction being analyzed L0 = overall length of the side of the founda-tion in the direction being analyzed, Section 19.2.1.2 (ft or m) M0, M01 = the overturning moment at the founda-tion–soil interface as determined in Sections 19.2.3 and 19.3.2 (ft-lb or N-m) Mt = torsional moment resulting from eccen-tricity between the locations of center of mass and the center of rigidity (Section 12.8.4.1) Mta = accidental torsional moment as deter-mined in Section 12.8.4.2 m = a subscript denoting the mode of vibra-tion under consideration; that is, m = 1 for the fundamental mode N = standard penetration resistance, ASTM D-1586 N = number of stories above the base (Section 12.8.2.1) N _ = average ﬁ eld standard penetration resistance for the top 100 ft (30 m); see Sections 20.3.3 and 20.4.2 N _ ch = average standard penetration resistance for cohesionless soil layers for the top 100 ft (30 m); see Sections 20.3.3 and 20.4.2 Ni = standard penetration resistance of any soil or rock layer i (between 0 and 100 ft [30 m]); see Section 20.4.2 n = designation for the level that is uppermost in the main portion of the building PGA = mapped MCEG peak ground acceleration shown in Figs. 22-6 through 22-10 PGAM = MCEG peak ground acceleration adjusted for Site Class effects; see Section 11.8.3 Px = total unfactored vertical design load at and above level x, for use in Section 12.8.7 PI = plasticity index, ASTM D4318 QE = effect of horizontal seismic (earthquake-induced) forces R = response modiﬁ cation coefﬁ cient as given in Tables 12.2-1, 12.14-1, 15.4-1, or 15.4-2 Rp = component response modiﬁ cation factor as deﬁ ned in Section 13.3.1 r = a characteristic length of the foundation as deﬁ ned in Section 19.2.1.2 ra = characteristic foundation length as deﬁ ned by Eq. 19.2-7 (ft or m) CHAPTER 11 SEISMIC DESIGN CRITERIA 64 rm = characteristic foundation length as deﬁ ned by Eq. 19.2-8 (ft or m) SS = mapped MCER, 5 percent damped, spectral response acceleration parameter at short periods as deﬁ ned in Section 11.4.1 S1 = mapped MCER, 5 percent damped, spectral response acceleration parameter at a period of 1 s as deﬁ ned in Section 11.4.1 SaM = the site-speciﬁ c MCER spectral response acceleration parameter at any period SDS = design, 5 percent damped, spectral response acceleration parameter at short periods as deﬁ ned in Section 11.4.4 SD1 = design, 5 percent damped, spectral response acceleration parameter at a period of 1 s as deﬁ ned in Section 11.4.4 SMS = the MCER, 5 percent damped, spectral response acceleration parameter at short periods adjusted for site class effects as deﬁ ned in Section 11.4.3 SM1 = the MCER, 5 percent damped, spectral response acceleration parameter at a period of 1 s adjusted for site class effects as deﬁ ned in Section 11.4.3 su = undrained shear strength; see Section 20.4.3 s _ u = average undrained shear strength in top 100 ft (30 m); see Sections 20.3.3 and 20.4.3, ASTM D2166 or ASTM D2850 sui = undrained shear strength of any cohesive soil layer i (between 0 and 100 ft [30 m]); see Section 20.4.3 sh = spacing of special lateral reinforcement (in. or mm) T = the fundamental period of the building T ˜, T ˜1 = the effective fundamental period(s) of the building as determined in Sections 19.2.1.1 and 19.3.1 T a = approximate fundamental period of the building as determined in Section 12.8.2 TL = long-period transition period as deﬁ ned in Section 11.4.5 Tp = fundamental period of the component and its attachment, Section 13.6.2 T0 = 0.2SD1/SDS TS = SD1/SDS T4 = net tension in steel cable due to dead load, prestress, live load, and seismic load (Section 14.1.7) V = total design lateral force or shear at the base Vt = design value of the seismic base shear as determined in Section 12.9.4 Vx = seismic design shear in story x as deter-mined in Section 12.8.4 or 12.9.4 V ˜ = reduced base shear accounting for the effects of soil structure interaction as determined in Section 19.3.1 V ˜1 = portion of the reduced base shear, V ˜, contributed by the fundamental mode, Section 19.3 (kip or kN) ΔV = reduction in V as determined in Section 19.3.1 (kip or kN) ΔV1 = reduction in V1 as determined in Section 19.3.1 (kip or kN) vs = shear wave velocity at small shear strains (greater than 10–3 percent strain); see Section 19.2.1 (ft/s or m/s) v _ s = average shear wave velocity at small shear strains in top 100 ft (30 m); see Sections 20.3.3 and 20.4.1 vsi = the shear wave velocity of any soil or rock layer i (between 0 and 100 ft [30 m]); see Section 20.4.1 vso = average shear wave velocity for the soils beneath the foundation at small strain levels, Section 19.2.1.1 (ft/s or m/s) W = effective seismic weight of the building as deﬁ ned in Section 12.7.2. For calcula-tion of seismic-isolated building period, W is the total effective seismic weight of the building as deﬁ ned in Sections 19.2 and 19.3 (kip or kN) W _ = effective seismic weight of the building as deﬁ ned in Sections 19.2 and 19.3 (kip or kN) Wc = gravity load of a component of the building Wp = component operating weight (lb or N) w = moisture content (in percent), ASTM D2216 wi, wn, wx = portion of W that is located at or assigned to Level i, n, or x, respectively x = level under consideration, 1 designates the ﬁ rst level above the base z = height in structure of point of attachment of component with respect to the base; see Section 13.3.1 β = ratio of shear demand to shear capacity for the story between Level x and x – 1 β _ = fraction of critical damping for the coupled structure-foundation system, determined in Section 19.2.1 MINIMUM DESIGN LOADS 65 β0 = foundation damping factor as speciﬁ ed in Section 19.2.1.2 γ = average unit weight of soil (lb/ft3 or N/m3) Δ = design story drift as determined in Section 12.8.6 Δfallout = the relative seismic displacement (drift) at which glass fallout from the curtain wall, storefront, or partition occurs Δa = allowable story drift as speciﬁ ed in Section 12.12.1 δmax = maximum displacement at Level x, considering torsion, Section 12.8.4.3 δM = maximum inelastic response displace-ment, considering torsion, Section 12.12.3 δMT = total separation distance between adjacent structures on the same property, Section 12.12.3 δavg = the average of the displacements at the extreme points of the structure at Level x, Section 12.8.4.3 δx = deﬂ ection of Level x at the center of the mass at and above Level x, Eq. 12.8-15 δxe = deﬂ ection of Level x at the center of the mass at and above Level x determined by an elastic analysis, Section 12.8-6 δxm = modal deﬂ ection of Level x at the center of the mass at and above Level x as determined by Section 19.3.2 δ _ x, δ _ x1 = deﬂ ection of Level x at the center of the mass at and above Level x, Eqs. 19.2-13 and 19.3-3 (in. or mm) θ = stability coefﬁ cient for P-delta effects as determined in Section 12.8.7 ρ = a redundancy factor based on the extent of structural redundancy present in a building as deﬁ ned in Section 12.3.4 ρs = spiral reinforcement ratio for precast, prestressed piles in Section 14.2.3.2.6 λ = time effect factor Ω0 = overstrength factor as deﬁ ned in Tables 12.2-1, 15.4-1, and 15.4-2 11.4 SEISMIC GROUND MOTION VALUES 11.4.1 Mapped Acceleration Parameters The parameters SS and S1 shall be determined from the 0.2 and 1 s spectral response accelerations shown on Figs. 22-1, 22-3, 22-5, and 22-6 for SS and Figs. 22-2, 22-4, 22-5, and 22-6 for S1. Where S1 is less than or equal to 0.04 and SS is less than or equal to 0.15, the structure is permitted to be assigned to Seismic Design Category A and is only required to comply with Section 11.7. User Note: Electronic values of mapped acceleration parameters, and other seismic design parameters, are provided at the USGS Web site at or through the SEI Web site at 11.4.2 Site Class Based on the site soil properties, the site shall be classiﬁ ed as Site Class A, B, C, D, E, or F in accor-dance with Chapter 20. Where the soil properties are not known in sufﬁ cient detail to determine the site class, Site Class D shall be used unless the authority having jurisdiction or geotechnical data determines Site Class E or F soils are present at the site. 11.4.3 Site Coefﬁ cients and Risk-Targeted Maximum Considered Earthquake (MCER) Spectral Response Acceleration Parameters The MCER spectral response acceleration parameter for short periods (SMS) and at 1 s (SM1), adjusted for Site Class effects, shall be determined by Eqs. 11.4-1 and 11.4-2, respectively. SMS = FaSS (11.4-1) SM1 = FvS1 (11.4-2) where SS = the mapped MCER spectral response acceleration parameter at short periods as determined in accordance with Section 11.4.1, and S1 = the mapped MCER spectral response acceleration parameter at a period of 1 s as determined in accordance with Section 11.4.1 where site coefﬁ cients Fa and Fv are deﬁ ned in Tables 11.4-1 and 11.4-2, respectively. Where the simpliﬁ ed design procedure of Section 12.14 is used, the value of Fa shall be determined in accordance with Section 12.14.8.1, and the values for Fv, SMS, and SM1 need not be determined. 11.4.4 Design Spectral Acceleration Parameters Design earthquake spectral response acceleration parameter at short period, SDS, and at 1 s period, SD1, shall be determined from Eqs. 11.4-3 and 11.4-4, respectively. Where the alternate simpliﬁ ed design procedure of Section 12.14 is used, the value of SDS shall be determined in accordance with Section 12.14.8.1, and the value for SD1 need not be determined. S S DS MS = 2 3 (11.4-3) S S D M 1 1 2 3 = (11.4-4) CHAPTER 11 SEISMIC DESIGN CRITERIA 66 11.4.5 Design Response Spectrum Where a design response spectrum is required by this standard and site-speciﬁ c ground motion proce-dures are not used, the design response spectrum curve shall be developed as indicated in Fig. 11.4-1 and as follows: 1.01 0 0 Period, T (sec) Spectral Response Acceleration,Sa (g) SDS SD1 SD1 Sa T = TL T 2 SD1⋅TL Sa = T0 TS FIGURE 11.4-1 Design Response Spectrum. Table 11.4-1 Site Coefﬁ cient, Fa Site Class Mapped Risk-Targeted Maximum Considered Earthquake (MCER) Spectral Response Acceleration Parameter at Short Period SS ≤ 0.25 SS = 0.5 SS = 0.75 SS = 1.0 SS ≥ 1.25 A 0.8 0.8 0.8 0.8 0.8 B 1.0 1.0 1.0 1.0 1.0 C 1.2 1.2 1.1 1.0 1.0 D 1.6 1.4 1.2 1.1 1.0 E 2.5 1.7 1.2 0.9 0.9 F See Section 11.4.7 Note: Use straight-line interpolation for intermediate values of SS. Table 11.4-2 Site Coefﬁ cient, Fv Site Class Mapped Risk-Targeted Maximum Considered Earthquake (MCER) Spectral Response Acceleration Parameter at 1-s Period S1 ≤ 0.1 S1 = 0.2 S1 = 0.3 S1 = 0.4 S1 ≥ 0.5 A 0.8 0.8 0.8 0.8 0.8 B 1.0 1.0 1.0 1.0 1.0 C 1.7 1.6 1.5 1.4 1.3 D 2.4 2.0 1.8 1.6 1.5 E 3.5 3.2 2.8 2.4 2.4 F See Section 11.4.7 Note: Use straight-line interpolation for intermediate values of S1. 1. For periods less than T0, the design spectral response acceleration, Sa, shall be taken as given by Eq. 11.4-5: S S T T a DS = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 4 0 6 0 . . (11.4-5) 2. For periods greater than or equal to T0 and less than or equal to TS, the design spectral response acceleration, Sa, shall be taken equal to SDS. 3. For periods greater than TS, and less than or equal to TL, the design spectral response acceleration, Sa, shall be taken as given by Eq. 11.4-6: S S T a D = 1 (11.4-6) 4. For periods greater than TL, Sa shall be taken as given by Eq. 11.4-7: S S T T a D L = 1 2 (11.4-7) where SDS = the design spectral response acceleration parameter at short periods MINIMUM DESIGN LOADS 67 SD1 = the design spectral response acceleration parameter at 1-s period T = the fundamental period of the structure, s T0 = 0.2 S S D DS 1 TS = S S D DS 1 and TL = long-period transition period (s) shown in Figs. 22-12 through 22-16. 11.4.6 Risk-Targeted Maximum Considered (MCER) Response Spectrum Where an MCER response spectrum is required, it shall be determined by multiplying the design response spectrum by 1.5. 11.4.7 Site-Speciﬁ c Ground Motion Procedures The site-speciﬁ c ground motion procedures set forth in Chapter 21 are permitted to be used to determine ground motions for any structure. A site response analysis shall be performed in accordance with Section 21.1 for structures on Site Class F sites, unless the exception to Section 20.3.1 is applicable. For seismically isolated structures and for structures with damping systems on sites with S1 greater than or equal to 0.6, a ground motion hazard analysis shall be performed in accordance with Section 21.2. 11.5 IMPORTANCE FACTOR AND RISK CATEGORY 11.5.1 Importance Factor An importance factor, IC, shall be assigned to each structure in accordance with Table 1.5-2. 11.5.2 Protected Access for Risk Category IV Where operational access to a Risk Category IV structure is required through an adjacent structure, the adjacent structure shall conform to the requirements for Risk Category IV structures. Where operational access is less than 10 ft from an interior lot line or another structure on the same lot, protection from potential falling debris from adjacent structures shall be provided by the owner of the Risk Category IV structure. 11.6 SEISMIC DESIGN CATEGORY Structures shall be assigned a Seismic Design Category in accordance with this section. Risk Category I, II, or III structures located where the mapped spectral response acceleration parameter at 1-s period, S1, is greater than or equal to 0.75 shall be assigned to Seismic Design Category E. Risk Category IV structures located where the mapped spectral response acceleration parameter at 1-s period, S1, is greater than or equal to 0.75 shall be assigned to Seismic Design Category F. All other structures shall be assigned to a Seismic Design Category based on their Risk Category and the design spectral response acceleration parameters, SDS and SD1, determined in accordance with Section 11.4.4. Each building and structure shall be assigned to the more severe Seismic Design Category in accordance with Table 11.6-1 or 11.6-2, irrespective of the fundamen-tal period of vibration of the structure, T. Where S1 is less than 0.75, the Seismic Design Category is permitted to be determined from Table 11.6-1 alone where all of the following apply: 1. In each of the two orthogonal directions, the approximate fundamental period of the structure, T a, determined in accordance with Section 12.8.2.1 is less than 0.8T s, where T s is determined in accordance with Section 11.4.5. 2. In each of two orthogonal directions, the funda-mental period of the structure used to calculate the story drift is less than T s. 3. Eq. 12.8-2 is used to determine the seismic response coefﬁ cient Cs. Table 11.6-1 Seismic Design Category Based on Short Period Response Acceleration Parameter Value of SDS Risk Category I or II or III IV SDS < 0.167 A A 0.167 ≤ SDS < 0.33 B C 0.33 ≤ SDS < 0.50 C D 0.50 ≤ SDS D D Table 11.6-2 Seismic Design Category Based on 1-S Period Response Acceleration Parameter Value of SD1 Risk Category I or II or III IV SD1 < 0.067 A A 0.067 ≤ SD1 < 0.133 B C 0.133 ≤ SD1 < 0.20 C D 0.20 ≤ SD1 D D CHAPTER 11 SEISMIC DESIGN CRITERIA 68 Table 11.8-1 Site Coefﬁ cient FPGA Site Class Mapped Maximum Considered Geometric Mean (MCEG) Peak Ground Acceleration, PGA PGA ≤ 0.1 PGA = 0.2 PGA = 0.3 PGA = 0.4 PGA ≥ 0.5 A 0.8 0.8 0.8 0.8 0.8 B 1.0 1.0 1.0 1.0 1.0 C 1.2 1.2 1.1 1.0 1.0 D 1.6 1.4 1.2 1.1 1.0 E 2.5 1.7 1.2 0.9 0.9 F See Section 11.4.7 Note: Use straight-line interpolation for intermediate values of PGA. 4. The diaphragms are rigid as deﬁ ned in Section 12.3.1 or for diaphragms that are ﬂ exible, the distance between vertical elements of the seismic force-resisting system does not exceed 40 ft. Where the alternate simpliﬁ ed design procedure of Section 12.14 is used, the Seismic Design Category is permitted to be determined from Table 11.6-1 alone, using the value of SDS determined in Section 12.14.8.1. 11.7 DESIGN REQUIREMENTS FOR SEISMIC DESIGN CATEGORY A Buildings and other structures assigned to Seismic Design Category A need only comply with the requirements of Section 1.4. Nonstructural compo-nents in SDC A are exempt from seismic design requirements. In addition, tanks assigned to Risk Category IV shall satisfy the freeboard requirement in Section 15.7.6.1.2. 11.8 GEOLOGIC HAZARDS AND GEOTECHNICAL INVESTIGATION 11.8.1 Site Limitation for Seismic Design Categories E and F A structure assigned to Seismic Design Category E or F shall not be located where there is a known potential for an active fault to cause rupture of the ground surface at the structure. 11.8.2 Geotechnical Investigation Report Requirements for Seismic Design Categories C through F A geotechnical investigation report shall be provided for a structure assigned to Seismic Design Category C, D, E, or F in accordance with this section. An investigation shall be conducted and a report shall be submitted that includes an evaluation of the following potential geologic and seismic hazards: a. Slope instability, b. Liquefaction, c. Total and differential settlement, and d. Surface displacement due to faulting or seismically induced lateral spreading or lateral ﬂ ow. The report shall contain recommendations for foundation designs or other measures to mitigate the effects of the previously mentioned hazards. EXCEPTION: Where approved by the authority having jurisdiction, a site-speciﬁ c geotechnical report is not required where prior evaluations of nearby sites with similar soil conditions provide direction relative to the proposed construction. 11.8.3 Additional Geotechnical Investigation Report Requirements for Seismic Design Categories D through F The geotechnical investigation report for a structure assigned to Seismic Design Category D, E, or F shall include all of the following, as applicable: 1. The determination of dynamic seismic lateral earth pressures on basement and retaining walls due to design earthquake ground motions. 2. The potential for liquefaction and soil strength loss evaluated for site peak ground acceleration, earthquake magnitude, and source characteristics consistent with the MCEG peak ground accelera-tion. Peak ground acceleration shall be determined based on either (1) a site-speciﬁ c study taking into account soil ampliﬁ cation effects as speciﬁ ed in MINIMUM DESIGN LOADS 69 Section 11.4.7 or (2) the peak ground acceleration PGAM, from Eq. 11.8-1. PGAM = FPGA PGA (Eq. 11.8-1) where PGAM = MCEG peak ground acceleration adjusted for Site Class effects. PGA = Mapped MCEG peak ground acceleration shown in Figs. 22-6 through 22-10. FPGA = Site coefﬁ cient from Table 11.8-1. 3. Assessment of potential consequences of liquefac-tion and soil strength loss, including, but not limited to, estimation of total and differential settlement, lateral soil movement, lateral soil loads on foundations, reduction in foundation soil-bearing capacity and lateral soil reaction, soil downdrag and reduction in axial and lateral soil reaction for pile foundations, increases in soil lateral pressures on retaining walls, and ﬂ otation of buried structures. 4. Discussion of mitigation measures such as, but not limited to, selection of appropriate foundation type and depths, selection of appropriate structural systems to accommodate anticipated displacements and forces, ground stabilization, or any combina-tion of these measures and how they shall be considered in the design of the structure. 71 Chapter 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 12.1.3 Continuous Load Path and Interconnection A continuous load path, or paths, with adequate strength and stiffness shall be provided to transfer all forces from the point of application to the ﬁ nal point of resistance. All parts of the structure between separation joints shall be interconnected to form a continuous path to the seismic force-resisting system, and the connections shall be capable of transmitting the seismic force (Fp) induced by the parts being connected. Any smaller portion of the structure shall be tied to the remainder of the structure with elements having a design strength capable of transmit-ting a seismic force of 0.133 times the short period design spectral response acceleration parameter, SDS, times the weight of the smaller portion or 5 percent of the portion’s weight, whichever is greater. This connection force does not apply to the overall design of the seismic force-resisting system. Connection design forces need not exceed the maximum forces that the structural system can deliver to the connection. 12.1.4 Connection to Supports A positive connection for resisting a horizontal force acting parallel to the member shall be provided for each beam, girder, or truss either directly to its supporting elements, or to slabs designed to act as diaphragms. Where the connection is through a diaphragm, then the member’s supporting element must also be connected to the diaphragm. The connection shall have a minimum design strength of 5 percent of the dead plus live load reaction. 12.1.5 Foundation Design The foundation shall be designed to resist the forces developed and accommodate the movements imparted to the structure by the design ground motions. The dynamic nature of the forces, the expected ground motion, the design basis for strength and energy dissipation capacity of the structure, and the dynamic properties of the soil shall be included in the determination of the foundation design criteria. The design and construction of foundations shall comply with Section 12.13. 12.1 STRUCTURAL DESIGN BASIS 12.1.1 Basic Requirements The seismic analysis and design procedures to be used in the design of building structures and their members shall be as prescribed in this section. The building structure shall include complete lateral and vertical force-resisting systems capable of providing adequate strength, stiffness, and energy dissipation capacity to withstand the design ground motions within the prescribed limits of deformation and strength demand. The design ground motions shall be assumed to occur along any horizontal direction of a building structure. The adequacy of the structural systems shall be demonstrated through the construc-tion of a mathematical model and evaluation of this model for the effects of design ground motions. The design seismic forces, and their distribution over the height of the building structure, shall be established in accordance with one of the applicable procedures indicated in Section 12.6 and the corresponding internal forces and deformations in the members of the structure shall be determined. An approved alternative procedure shall not be used to establish the seismic forces and their distribution unless the corresponding internal forces and deformations in the members are determined using a model consistent with the procedure adopted. EXCEPTION: As an alternative, the simpliﬁ ed design procedures of Section 12.14 is permitted to be used in lieu of the requirements of Sections 12.1 through 12.12, subject to all of the limitations contained in Section 12.14. 12.1.2 Member Design, Connection Design, and Deformation Limit Individual members, including those not part of the seismic force–resisting system, shall be provided with adequate strength to resist the shears, axial forces, and moments determined in accordance with this standard, and connections shall develop the strength of the connected members or the forces indicated in Section 12.1.1. The deformation of the structure shall not exceed the prescribed limits where the structure is subjected to the design seismic forces. CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 72 12.1.6 Material Design and Detailing Requirements Structural elements including foundation elements shall conform to the material design and detailing requirements set forth in Chapter 14. 12.2 STRUCTURAL SYSTEM SELECTION 12.2.1 Selection and Limitations The basic lateral and vertical seismic force-resist-ing system shall conform to one of the types indicated in Table 12.2-1 or a combination of systems as permitted in Sections 12.2.2, 12.2.3, and 12.2.4. Each type is subdivided by the types of vertical elements used to resist lateral seismic forces. The structural system used shall be in accordance with the structural system limitations and the limits on structural height, hn, contained in Table 12.2-1. The appropriate response modiﬁ cation coefﬁ cient, R, overstrength factor, Ω0, and the deﬂ ection ampliﬁ cation factor, Cd, indicated in Table 12.2-1 shall be used in determining the base shear, element design forces, and design story drift. Each selected seismic force-resisting system shall be designed and detailed in accordance with the speciﬁ c requirements for the system as set forth in the applicable reference document listed in Table 12.2-1 and the additional requirements set forth in Chapter 14. Seismic force-resisting systems not contained in Table 12.2-1 are permitted provided analytical and test data are submitted to the authority having jurisdiction for approval that establish their dynamic characteristics and demonstrate their lateral force resistance and energy dissipation capacity to be equivalent to the structural systems listed in Table 12.2-1 for equivalent values of response modiﬁ cation coefﬁ cient, R, overstrength factor, Ω0, and deﬂ ection ampliﬁ cation factor, Cd. 12.2.2 Combinations of Framing Systems in Different Directions Different seismic force-resisting systems are permitted to be used to resist seismic forces along each of the two orthogonal axes of the structure. Where different systems are used, the respective R, Cd, and Ω0 coefﬁ cients shall apply to each system, including the structural system limitations contained in Table 12.2-1. 12.2.3 Combinations of Framing Systems in the Same Direction Where different seismic force-resisting systems are used in combination to resist seismic forces in the same direction, other than those combinations considered as dual systems, the most stringent applicable structural system limitations contained in Table 12.2-1 shall apply and the design shall comply with the requirements of this section. 12.2.3.1 R, Cd, and Ω0 Values for Vertical Combinations Where a structure has a vertical combination in the same direction, the following requirements shall apply: 1. Where the lower system has a lower Response Modiﬁ cation Coefﬁ cient, R, the design coefﬁ cients (R, Ω0, and Cd) for the upper system are permitted to be used to calculate the forces and drifts of the upper system. For the design of the lower system, the design coefﬁ cients (R, Ω0, and Cd) for the lower system shall be used. Forces transferred from the upper system to the lower system shall be increased by multiplying by the ratio of the higher response modiﬁ cation coefﬁ cient to the lower response modiﬁ cation coefﬁ cient. 2. Where the upper system has a lower Response Modiﬁ cation Coefﬁ cient, the Design Coefﬁ cients (R, Ω0, and Cd) for the upper system shall be used for both systems. EXCEPTIONS: 1. Rooftop structures not exceeding two stories in height and 10 percent of the total structure weight. 2. Other supported structural systems with a weight equal to or less than 10 percent of the weight of the structure. 3. Detached one- and two-family dwellings of light-frame construction. 12.2.3.2 Two Stage Analysis Procedure A two-stage equivalent lateral force procedure is permitted to be used for structures having a ﬂ exible upper portion above a rigid lower portion, provided the design of the structure complies with all of the following: a. The stiffness of the lower portion shall be at least 10 times the stiffness of the upper portion. b. The period of the entire structure shall not be greater than 1.1 times the period of the upper portion considered as a separate structure supported at the transition from the upper to the lower portion. c. The upper portion shall be designed as a separate structure using the appropriate values of R and ρ. MINIMUM DESIGN LOADS 73 Table 12.2-1 Design Coefﬁ cients and Factors for Seismic Force-Resisting Systems Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Overstrength Factor, Ω0 g Deﬂ ection Ampliﬁ cation Factor, Cd b Structural System Limitations Including Structural Height, hn (ft) Limitsc Seismic Design Category B C Dd Ed Fe A. BEARING WALL SYSTEMS 1. Special reinforced concrete shear wallsl, m 14.2 5 2½ 5 NL NL 160 160 100 2. Ordinary reinforced concrete shear wallsl 14.2 4 2½ 4 NL NL NP NP NP 3. Detailed plain concrete shear wallsl 14.2 2 2½ 2 NL NP NP NP NP 4. Ordinary plain concrete shear wallsl 14.2 1½ 2½ 1½ NL NP NP NP NP 5. Intermediate precast shear walls l 14.2 4 2½ 4 NL NL 40k 40k 40k 6. Ordinary precast shear wallsl 14.2 3 2½ 3 NL NP NP NP NP 7. Special reinforced masonry shear walls 14.4 5 2½ 3½ NL NL 160 160 100 8. Intermediate reinforced masonry shear walls 14.4 3½ 2½ 2¼ NL NL NP NP NP 9. Ordinary reinforced masonry shear walls 14.4 2 2½ 1¾ NL 160 NP NP NP 10. Detailed plain masonry shear walls 14.4 2 2½ 1¾ NL NP NP NP NP 11. Ordinary plain masonry shear walls 14.4 1½ 2½ 1¼ NL NP NP NP NP 12. Prestressed masonry shear walls 14.4 1½ 2½ 1¾ NL NP NP NP NP 13. Ordinary reinforced AAC masonry shear walls 14.4 2 2½ 2 NL 35 NP NP NP 14. Ordinary plain AAC masonry shear walls 14.4 1½ 2½ 1½ NL NP NP NP NP 15. Light-frame (wood) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.1 and 14.5 6½ 3 4 NL NL 65 65 65 16. Light-frame (cold-formed steel) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.1 6½ 3 4 NL NL 65 65 65 17. Light-frame walls with shear panels of all other materials 14.1 and 14.5 2 2½ 2 NL NL 35 NP NP 18. Light-frame (cold-formed steel) wall systems using ﬂ at strap bracing 14.1 4 2 3½ NL NL 65 65 65 B. BUILDING FRAME SYSTEMS 1. Steel eccentrically braced frames 14.1 8 2 4 NL NL 160 160 100 2. Steel special concentrically braced frames 14.1 6 2 5 NL NL 160 160 100 3. Steel ordinary concentrically braced frames 14.1 3¼ 2 3¼ NL NL 35j 35j NPj Continued CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 74 Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Overstrength Factor, Ω0 g Deﬂ ection Ampliﬁ cation Factor, Cd b Structural System Limitations Including Structural Height, hn (ft) Limitsc Seismic Design Category B C Dd Ed Fe 4. Special reinforced concrete shear wallsl,m 14.2 6 2½ 5 NL NL 160 160 100 5. Ordinary reinforced concrete shear wallsl 14.2 5 2½ 4½ NL NL NP NP NP 6. Detailed plain concrete shear wallsl 14.2 and 14.2.2.8 2 2½ 2 NL NP NP NP NP 7. Ordinary plain concrete shear wallsl 14.2 1½ 2½ 1½ NL NP NP NP NP 8. Intermediate precast shear wallsl 14.2 5 2½ 4½ NL NL 40k 40k 40k 9. Ordinary precast shear wallsl 14.2 4 2½ 4 NL NP NP NP NP 10. Steel and concrete composite eccentrically braced frames 14.3 8 2 ½ 4 NL NL 160 160 100 11. Steel and concrete composite special concentrically braced frames 14.3 5 2 4½ NL NL 160 160 100 12. Steel and concrete composite ordinary braced frames 14.3 3 2 3 NL NL NP NP NP 13. Steel and concrete composite plate shear walls 14.3 6½ 2½ 5½ NL NL 160 160 100 14. Steel and concrete composite special shear walls 14.3 6 2½ 5 NL NL 160 160 100 15. Steel and concrete composite ordinary shear walls 14.3 5 2½ 4½ NL NL NP NP NP 16. Special reinforced masonry shear walls 14.4 5½ 2½ 4 NL NL 160 160 100 17. Intermediate reinforced masonry shear walls 14.4 4 2½ 4 NL NL NP NP NP 18. Ordinary reinforced masonry shear walls 14.4 2 2½ 2 NL 160 NP NP NP 19. Detailed plain masonry shear walls 14.4 2 2½ 2 NL NP NP NP NP 20. Ordinary plain masonry shear walls 14.4 1½ 2½ 1¼ NL NP NP NP NP 21. Prestressed masonry shear walls 14.4 1½ 2½ 1¾ NL NP NP NP NP 22. Light-frame (wood) walls sheathed with wood structural panels rated for shear resistance 14.5 7 2½ 4½ NL NL 65 65 65 23. Light-frame (cold-formed steel) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.1 7 2½ 4½ NL NL 65 65 65 24. Light-frame walls with shear panels of all other materials 14.1and 14.5 2½ 2½ 2½ NL NL 35 NP NP 25. Steel buckling-restrained braced frames 14.1 8 2½ 5 NL NL 160 160 100 26. Steel special plate shear walls 14.1 7 2 6 NL NL 160 160 100 Table 12.2-1 (Continued) MINIMUM DESIGN LOADS 75 Continued Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Overstrength Factor, Ω0 g Deﬂ ection Ampliﬁ cation Factor, Cd b Structural System Limitations Including Structural Height, hn (ft) Limitsc Seismic Design Category B C Dd Ed Fe C. MOMENT-RESISTING FRAME SYSTEMS 1. Steel special moment frames 14.1 and 12.2.5.5 8 3 5½ NL NL NL NL NL 2. Steel special truss moment frames 14.1 7 3 5½ NL NL 160 100 NP 3. Steel intermediate moment frames 12.2.5.7 and 14.1 4½ 3 4 NL NL 35h NPh NPh 4. Steel ordinary moment frames 12.2.5.6 and 14.1 3½ 3 3 NL NL NPi NPi NPi 5. Special reinforced concrete moment framesn 12.2.5.5 and 14.2 8 3 5½ NL NL NL NL NL 6. Intermediate reinforced concrete moment frames 14.2 5 3 4½ NL NL NP NP NP 7. Ordinary reinforced concrete moment frames 14.2 3 3 2½ NL NP NP NP NP 8. Steel and concrete composite special moment frames 12.2.5.5 and 14.3 8 3 5½ NL NL NL NL NL 9. Steel and concrete composite intermediate moment frames 14.3 5 3 4½ NL NL NP NP NP 10. Steel and concrete composite partially restrained moment frames 14.3 6 3 5½ 160 160 100 NP NP 11. Steel and concrete composite ordinary moment frames 14.3 3 3 2½ NL NP NP NP NP 12. Cold-formed steel—special bolted moment framep 14.1 3½ 3o 3½ 35 35 35 35 35 D. DUAL SYSTEMS WITH SPECIAL MOMENT FRAMES CAPABLE OF RESISTING AT LEAST 25% OF PRESCRIBED SEISMIC FORCES 12.2.5.1 1. Steel eccentrically braced frames 14.1 8 2½ 4 NL NL NL NL NL 2. Steel special concentrically braced frames 14.1 7 2½ 5½ NL NL NL NL NL 3. Special reinforced concrete shear wallsl 14.2 7 2½ 5½ NL NL NL NL NL 4. Ordinary reinforced concrete shear wallsl 14.2 6 2½ 5 NL NL NP NP NP 5. Steel and concrete composite eccentrically braced frames 14.3 8 2½ 4 NL NL NL NL NL 6. Steel and concrete composite special concentrically braced frames 14.3 6 2½ 5 NL NL NL NL NL Table 12.2-1 (Continued) CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 76 Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Overstrength Factor, Ω0 g Deﬂ ection Ampliﬁ cation Factor, Cd b Structural System Limitations Including Structural Height, hn (ft) Limitsc Seismic Design Category B C Dd Ed Fe 7. Steel and concrete composite plate shear walls 14.3 7½ 2½ 6 NL NL NL NL NL 8. Steel and concrete composite special shear walls 14.3 7 2½ 6 NL NL NL NL NL 9. Steel and concrete composite ordinary shear walls 14.3 6 2½ 5 NL NL NP NP NP 10. Special reinforced masonry shear walls 14.4 5½ 3 5 NL NL NL NL NL 11. Intermediate reinforced masonry shear walls 14.4 4 3 3½ NL NL NP NP NP 12. Steel buckling-restrained braced frames 14.1 8 2½ 5 NL NL NL NL NL 13. Steel special plate shear walls 14.1 8 2½ 6½ NL NL NL NL NL E. DUAL SYSTEMS WITH INTERMEDIATE MOMENT FRAMES CAPABLE OF RESISTING AT LEAST 25% OF PRESCRIBED SEISMIC FORCES 12.2.5.1 1. Steel special concentrically braced framesf 14.1 6 2½ 5 NL NL 35 NP NP 2. Special reinforced concrete shear wallsl 14.2 6½ 2½ 5 NL NL 160 100 100 3. Ordinary reinforced masonry shear walls 14.4 3 3 2½ NL 160 NP NP NP 4. Intermediate reinforced masonry shear walls 14.4 3½ 3 3 NL NL NP NP NP 5. Steel and concrete composite special concentrically braced frames 14.3 5½ 2½ 4½ NL NL 160 100 NP 6. Steel and concrete composite ordinary braced frames 14.3 3½ 2½ 3 NL NL NP NP NP 7. Steel and concrete composite ordinary shear walls 14.3 5 3 4½ NL NL NP NP NP 8. Ordinary reinforced concrete shear wallsl 14.2 5½ 2½ 4½ NL NL NP NP NP F. SHEAR WALL-FRAME INTERACTIVE SYSTEM WITH ORDINARY REINFORCED CONCRETE MOMENT FRAMES AND ORDINARY REINFORCED CONCRETE SHEAR WALLSl 12.2.5.8 and 14.2 4½ 2½ 4 NL NP NP NP NP Table 12.2-1 (Continued) MINIMUM DESIGN LOADS 77 Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Overstrength Factor, Ω0 g Deﬂ ection Ampliﬁ cation Factor, Cd b Structural System Limitations Including Structural Height, hn (ft) Limitsc Seismic Design Category B C Dd Ed Fe G. CANTILEVERED COLUMN SYSTEMS DETAILED TO CONFORM TO THE REQUIREMENTS FOR: 12.2.5.2 1. Steel special cantilever column systems 14.1 2½ 1¼ 2½ 35 35 35 35 35 2. Steel ordinary cantilever column systems 14.1 1¼ 1¼ 1¼ 35 35 NPi NPi NPi 3. Special reinforced concrete moment framesn 12.2.5.5 and 14.2 2½ 1¼ 2½ 35 35 35 35 35 4. Intermediate reinforced concrete moment frames 14.2 1½ 1¼ 1½ 35 35 NP NP NP 5. Ordinary reinforced concrete moment frames 14.2 1 1¼ 1 35 NP NP NP NP 6. Timber frames 14.5 1½ 1½ 1½ 35 35 35 NP NP H. STEEL SYSTEMS NOT SPECIFICALLY DETAILED FOR SEISMIC RESISTANCE, EXCLUDING CANTILEVER COLUMN SYSTEMS 14.1 3 3 3 NL NL NP NP NP aResponse modiﬁ cation coefﬁ cient, R, for use throughout the standard. Note R reduces forces to a strength level, not an allowable stress level. bDeﬂ ection ampliﬁ cation factor, Cd, for use in Sections 12.8.6, 12.8.7, and 12.9.2. cNL = Not Limited and NP = Not Permitted. For metric units use 30.5 m for 100 ft and use 48.8 m for 160 ft. dSee Section 12.2.5.4 for a description of seismic force-resisting systems limited to buildings with a structural height, hn, of 240 ft (73.2 m) or less. eSee Section 12.2.5.4 for seismic force-resisting systems limited to buildings with a structural height, hn, of 160 ft (48.8 m) or less. fOrdinary moment frame is permitted to be used in lieu of intermediate moment frame for Seismic Design Categories B or C. gWhere the tabulated value of the overstrength factor, Ω0, is greater than or equal to 2½, Ωo is permitted to be reduced by subtracting the value of 1/2 for structures with ﬂ exible diaphragms. hSee Section 12.2.5.7 for limitations in structures assigned to Seismic Design Categories D, E, or F. i See Section 12.2.5.6 for limitations in structures assigned to Seismic Design Categories D, E, or F. jSteel ordinary concentrically braced frames are permitted in single-story buildings up to a structural height, hn, of 60 ft (18.3 m) where the dead load of the roof does not exceed 20 psf (0.96 kN/m2) and in penthouse structures. kAn increase in structural height, hn, to 45 ft (13.7 m) is permitted for single story storage warehouse facilities. lIn Section 2.2 of ACI 318. A shear wall is deﬁ ned as a structural wall. mIn Section 2.2 of ACI 318. The deﬁ nition of “special structural wall” includes precast and cast-in-place construction. nIn Section 2.2 of ACI 318. The deﬁ nition of “special moment frame” includes precast and cast-in-place construction. oAlternately, the seismic load effect with overstrength, Emh, is permitted to be based on the expected strength determined in accordance with AISI S110. pCold-formed steel – special bolted moment frames shall be limited to one-story in height in accordance with AISI S110. Table 12.2-1 (Continued) CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 78 d. The lower portion shall be designed as a separate structure using the appropriate values of R and ρ. The reactions from the upper portion shall be those determined from the analysis of the upper portion ampliﬁ ed by the ratio of the R/ρ of the upper portion over R/ρ of the lower portion. This ratio shall not be less than 1.0. e. The upper portion is analyzed with the equivalent lateral force or modal response spectrum proce-dure, and the lower portion is analyzed with the equivalent lateral force procedure. 12.2.3.3 R, Cd, and Ω0 Values for Horizontal Combinations The value of the response modiﬁ cation coefﬁ -cient, R, used for design in the direction under consideration shall not be greater than the least value of R for any of the systems utilized in that direction. The deﬂ ection ampliﬁ cation factor, Cd, and the overstrength factor, Ω0, shall be consistent with R required in that direction. EXCEPTION: Resisting elements are permitted to be designed using the least value of R for the different structural systems found in each independent line of resistance if the following three conditions are met: (1) Risk Category I or II building, (2) two stories or less above grade plane, and (3) use of light-frame construction or ﬂ exible diaphragms. The value of R used for design of diaphragms in such structures shall not be greater than the least value of R for any of the systems utilized in that same direction. 12.2.4 Combination Framing Detailing Requirements Structural members common to different framing systems used to resist seismic forces in any direction shall be designed using the detailing requirements of Chapter 12 required by the highest response modiﬁ cation coefﬁ cient, R, of the connected framing systems. 12.2.5 System Speciﬁ c Requirements The structural framing system shall also comply with the following system speciﬁ c requirements of this section. 12.2.5.1 Dual System For a dual system, the moment frames shall be capable of resisting at least 25 percent of the design seismic forces. The total seismic force resistance is to be provided by the combination of the moment frames and the shear walls or braced frames in proportion to their rigidities. 12.2.5.2 Cantilever Column Systems Cantilever column systems are permitted as indicated in Table 12.2-1 and as follows. The required axial strength of individual cantilever column ele-ments, considering only the load combinations that include seismic load effects, shall not exceed 15 percent of the available axial strength, including slenderness effects. Foundation and other elements used to provide overturning resistance at the base of cantilever column elements shall be designed to resist the seismic load effects including overstrength factor of Section 12.4.3. 12.2.5.3 Inverted Pendulum-Type Structures Regardless of the structural system selected, inverted pendulums as deﬁ ned in Section 11.2, shall comply with this section. Supporting columns or piers of inverted pendulum-type structures shall be designed for the bending moment calculated at the base determined using the procedures given in Section 12.8 and varying uniformly to a moment at the top equal to one-half the calculated bending moment at the base. 12.2.5.4 Increased Structural Height Limit for Steel Eccentrically Braced Frames, Steel Special Concentrically Braced Frames, Steel Buckling-restrained Braced Frames, Steel Special Plate Shear Walls and Special Reinforced Concrete Shear Walls The limits on structural height, hn, in Table 12.2-1 are permitted to be increased from 160 ft (50 m) to 240 ft (75 m) for structures assigned to Seismic Design Categories D or E and from 100 ft (30 m) to 160 ft (50 m) for structures assigned to Seismic Design Category F provided the seismic force-resisting systems are limited to steel eccentrically braced frames, steel special concentrically braced frames, steel buckling-restrained braced frames, steel special plate shear walls, or special reinforced concrete cast-in-place shear walls and both of the following requirements are met: 1. The structure shall not have an extreme torsional irregularity as deﬁ ned in Table 12.2-1 (horizontal structural irregularity Type 1b). 2. The steel eccentrically braced frames, steel special concentrically braced frames, steel buckling-restrained braced frames, steel special plate shear walls or special reinforced cast-in-place concrete shear walls in any one plane shall resist no more than 60 percent of the total seismic forces in each direction, neglecting accidental torsional effects. MINIMUM DESIGN LOADS 79 12.2.5.5 Special Moment Frames in Structures Assigned to Seismic Design Categories D through F For structures assigned to Seismic Design Categories D, E, or F, a special moment frame that is used but not required by Table 12.2-1 shall not be discontinued and supported by a more rigid system with a lower response modiﬁ cation coefﬁ cient, R, unless the requirements of Sections 12.3.3.2 and 12.3.3.4 are met. Where a special moment frame is required by Table 12.2-1, the frame shall be continu-ous to the base. 12.2.5.6 Steel Ordinary Moment Frames 12.2.5.6.1 Seismic Design Category D or E. a. Single-story steel ordinary moment frames in structures assigned to Seismic Design Category D or E are permitted up to a structural height, hn, of 65 ft (20 m) where the dead load supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior walls more than 35 ft (10.6 m) above the base tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). EXCEPTION: Single-story structures with steel ordinary moment frames whose purpose is to enclose equipment or machinery and whose occupants are engaged in maintenance or monitoring of that equipment, machinery, or their associated processes shall be permitted to be of unlimited height where the sum of the dead and equipment loads supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior wall system including exterior columns more than 35 ft (10.6 m) above the base shall not exceed 20 psf (0.96 kN/m2). For determining compliance with the exterior wall or roof load limits, the weight of equipment or machinery, including cranes, not self-supporting for all loads shall be assumed fully tributary to the area of the adjacent exterior wall or roof not to exceed 600 ft2 (55.8 m2) regardless of their height above the base of the structure. b. Steel ordinary moment frames in structures assigned to Seismic Design Category D or E not meeting the limitations set forth in Section 12.2.5.6.1.a are permitted within light-frame construction up to a structural height, hn, of 35 ft (10.6 m) where neither the roof dead load nor the dead load of any ﬂ oor above the base supported by and tributary to the moment frames exceeds 35 psf (1.68 kN/m2). In addition, the dead load of the exterior walls tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). 12.2.5.6.2 Seismic Design Category F. Single-story steel ordinary moment frames in structures assigned to Seismic Design Category F are permitted up to a structural height, hn, of 65 ft (20 m) where the dead load supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior walls tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). 12.2.5.7 Steel Intermediate Moment Frames 12.2.5.7.1 Seismic Design Category D a. Single-story steel intermediate moment frames in structures assigned to Seismic Design Category D are permitted up to a structural height, hn, of 65 ft (20 m) where the dead load supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior walls more than 35 ft (10.6 m) above the base tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). EXCEPTION: Single-story structures with steel intermediate moment frames whose purpose is to enclose equipment or machinery and whose occupants are engaged in maintenance or monitoring of that equipment, machinery, or their associated processes shall be permitted to be of unlimited height where the sum of the dead and equipment loads supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior wall system including exterior columns more than 35 ft (10.6 m) above the base shall not exceed 20 psf (0.96 kN/m2). For determining compliance with the exterior wall or roof load limits, the weight of equipment or machinery, including cranes, not self-supporting for all loads shall be assumed fully tributary to the area of the adjacent exterior wall or roof not to exceed 600 ft2 (55.8 m2) regardless of their height above the base of the structure. b. Steel intermediate moment frames in structures assigned to Seismic Design Category D not meeting the limitations set forth in Section 12.2.5.7.1.a are permitted up to a structural height, hn, of 35 ft (10.6 m). CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 80 12.2.5.7.2 Seismic Design Category E. a. Single-story steel intermediate moment frames in structures assigned to Seismic Design Category E are permitted up to a structural height, hn, of 65 ft (20 m) where the dead load supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior walls more than 35 ft (10.6 m) above the base tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). EXCEPTION: Single-story structures with steel intermediate moment frames whose purpose is to enclose equipment or machinery and whose occupants are engaged in maintenance or monitoring of that equipment, machinery, or their associated processes shall be permitted to be of unlimited height where the sum of the dead and equipment loads supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior wall system including exterior columns more than 35 ft (10.6 m) above the base shall not exceed 20 psf (0.96 kN/m2). For determining compliance with the exterior wall or roof load limits, the weight of equipment or machinery, including cranes, not self-supporting for all loads shall be assumed fully tributary to the area of the adjacent exterior wall or roof not to exceed 600 ft2 (55.8 m2) regardless of their height above the base of the structure. b. Steel intermediate moment frames in structures assigned to Seismic Design Category E not meeting the limitations set forth in Section 12.2.5.7.2.a are permitted up to a structural height, hn, of 35 ft (10.6 m) where neither the roof dead load nor the dead load of any ﬂ oor above the base supported by and tributary to the moment frames exceeds 35 psf (1.68 kN/m2). In addition, the dead load of the exterior walls tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). 12.2.5.7.3 Seismic Design Category F. a. Single-story steel intermediate moment frames in structures assigned to Seismic Design Category F are permitted up to a structural height, hn, of 65 ft (20 m) where the dead load supported by and tributary to the roof does not exceed 20 psf (0.96 kN/m2). In addition, the dead load of the exterior walls tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). b. Steel intermediate moment frames in structures assigned to Seismic Design Category F not meeting the limitations set forth in Section 12.2.5.7.3.a are permitted within light-frame construction up to a structural height, hn, of 35 ft (10.6 m) where neither the roof dead load nor the dead load of any ﬂ oor above the base supported by and tributary to the moment frames exceeds 35 psf (1.68 kN/m2). In addition, the dead load of the exterior walls tributary to the moment frames shall not exceed 20 psf (0.96 kN/m2). 12.2.5.8 Shear Wall-Frame Interactive Systems The shear strength of the shear walls of the shear wall-frame interactive system shall be at least 75 percent of the design story shear at each story. The frames of the shear wall-frame interactive system shall be capable of resisting at least 25 percent of the design story shear in every story. 12.3 DIAPHRAGM FLEXIBILITY, CONFIGURATION IRREGULARITIES, AND REDUNDANCY 12.3.1 Diaphragm Flexibility The structural analysis shall consider the relative stiffnesses of diaphragms and the vertical elements of the seismic force-resisting system. Unless a dia-phragm can be idealized as either ﬂ exible or rigid in accordance with Sections 12.3.1.1, 12.3.1.2, or 12.3.1.3, the structural analysis shall explicitly include consideration of the stiffness of the diaphragm (i.e., semirigid modeling assumption). 12.3.1.1 Flexible Diaphragm Condition Diaphragms constructed of untopped steel decking or wood structural panels are permitted to be idealized as ﬂ exible if any of the following conditions exist: a. In structures where the vertical elements are steel braced frames, steel and concrete composite braced frames or concrete, masonry, steel, or steel and concrete composite shear walls. b. In one- and two-family dwellings. c. In structures of light-frame construction where all of the following conditions are met: 1. Topping of concrete or similar materials is not placed over wood structural panel diaphragms except for nonstructural topping no greater than 1 1/2 in. (38 mm) thick. 2. Each line of vertical elements of the seismic force-resisting system complies with the allowable story drift of Table 12.12-1. MINIMUM DESIGN LOADS 81 12.3.1.2 Rigid Diaphragm Condition Diaphragms of concrete slabs or concrete ﬁ lled metal deck with span-to-depth ratios of 3 or less in structures that have no horizontal irregularities are permitted to be idealized as rigid. 12.3.1.3 Calculated Flexible Diaphragm Condition Diaphragms not satisfying the conditions of Sections 12.3.1.1 or 12.3.1.2 are permitted to be idealized as ﬂ exible where the computed maximum in-plane deﬂ ection of the diaphragm under lateral load is more than two times the average story drift of adjoining vertical elements of the seismic force-resisting system of the associated story under equiva-lent tributary lateral load as shown in Fig. 12.3-1. The loadings used for this calculation shall be those prescribed by Section 12.8. 12.3.2 Irregular and Regular Classiﬁ cation Structures shall be classiﬁ ed as having a struc-tural irregularity based upon the criteria in this section. Such classiﬁ cation shall be based on their structural conﬁ gurations. 12.3.2.1 Horizontal Irregularity Structures having one or more of the irregularity types listed in Table 12.3-1 shall be designated as having a horizontal structural irregularity. Such structures assigned to the seismic design categories listed in Table 12.3-1 shall comply with the require-ments in the sections referenced in that table. 12.3.2.2 Vertical Irregularity Structures having one or more of the irregularity types listed in Table 12.3-2 shall be designated as having a vertical structural irregularity. Such struc-tures assigned to the seismic design categories listed in Table 12.3-2 shall comply with the requirements in the sections referenced in that table. EXCEPTIONS: 1. Vertical structural irregularities of Types 1a, 1b, and 2 in Table 12.3-2 do not apply where no story drift ratio under design lateral seismic force is greater than 130 percent of the story drift ratio of the next story above. Torsional effects need not be considered in the calculation of story drifts. The story drift ratio relationship for the top two stories of the structure are not required to be evaluated. 2. Vertical structural irregularities of Types 1a, 1b, and 2 in Table 12.3-2 are not required to be considered for one-story buildings in any seismic design category or for two-story buildings assigned to Seismic Design Categories B, C, or D. 12.3.3 Limitations and Additional Requirements for Systems with Structural Irregularities 12.3.3.1 Prohibited Horizontal and Vertical Irregularities for Seismic Design Categories D through F Structures assigned to Seismic Design Category E or F having horizontal irregularity Type 1b of Table 12.3-1 or vertical irregularities Type 1b, 5a, or 5b of Table 12.3-2 shall not be permitted. Structures assigned to Seismic Design Category D having vertical irregularity Type 5b of Table 12.3-2 shall not be permitted. 12.3.3.2 Extreme Weak Stories Structures with a vertical irregularity Type 5b as deﬁ ned in Table 12.3-2, shall not be over two stories or 30 ft (9 m) in structural height, hn. MAXIMUM DIAPHRAGM (ADVE) AVERAGE DRIFT OF VERTICAL ELEMENT Note: Diaphragm is flexible if MDD > 2(ADVE). DEFLECTION (MDD) SEISMIC LOADING S De FIGURE 12.3-1 Flexible Diaphragm CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 82 Table 12.3-1 Horizontal Structural Irregularities Type Description Reference Section Seismic Design Category Application 1a. Torsional Irregularity: Torsional irregularity is deﬁ ned to exist where the maximum story drift, computed including accidental torsion with Ax = 1.0, at one end of the structure transverse to an axis is more than 1.2 times the average of the story drifts at the two ends of the structure. Torsional irregularity requirements in the reference sections apply only to structures in which the diaphragms are rigid or semirigid. 12.3.3.4 12.7.3 12.8.4.3 12.12.1 Table 12.6-1 Section 16.2.2 D, E, and F B, C, D, E, and F C, D, E, and F C, D, E, and F D, E, and F B, C, D, E, and F 1b. Extreme Torsional Irregularity: Extreme torsional irregularity is deﬁ ned to exist where the maximum story drift, computed including accidental torsion with Ax = 1.0, at one end of the structure transverse to an axis is more than 1.4 times the average of the story drifts at the two ends of the structure. Extreme torsional irregularity requirements in the reference sections apply only to structures in which the diaphragms are rigid or semirigid. 12.3.3.1 12.3.3.4 12.7.3 12.8.4.3 12.12.1 Table 12.6-1 Section 16.2.2 E and F D B, C, and D C and D C and D D B, C, and D 2. Reentrant Corner Irregularity: Reentrant corner irregularity is deﬁ ned to exist where both plan projections of the structure beyond a reentrant corner are greater than 15% of the plan dimension of the structure in the given direction. 12.3.3.4 Table 12.6-1 D, E, and F D, E, and F 3. Diaphragm Discontinuity Irregularity: Diaphragm discontinuity irregularity is deﬁ ned to exist where there is a diaphragm with an abrupt discontinuity or variation in stiffness, including one having a cutout or open area greater than 50% of the gross enclosed diaphragm area, or a change in effective diaphragm stiffness of more than 50% from one story to the next. 12.3.3.4 Table 12.6-1 D, E, and F D, E, and F 4. Out-of-Plane Offset Irregularity: Out-of-plane offset irregularity is deﬁ ned to exist where there is a discontinuity in a lateral force-resistance path, such as an out-of-plane offset of at least one of the vertical elements. 12.3.3.3 12.3.3.4 12.7.3 Table 12.6-1 Section 16.2.2 B, C, D, E, and F D, E, and F B, C, D, E, and F D, E, and F B, C, D, E, and F 5. Nonparallel System Irregularity: Nonparallel system irregularity is deﬁ ned to exist where vertical lateral force-resisting elements are not parallel to the major orthogonal axes of the seismic force-resisting system. 12.5.3 12.7.3 Table 12.6-1 Section 16.2.2 C, D, E, and F B, C, D, E, and F D, E, and F B, C, D, E, and F EXCEPTION: The limit does not apply where the “weak” story is capable of resisting a total seismic force equal to Ω0 times the design force prescribed in Section 12.8. 12.3.3.3 Elements Supporting Discontinuous Walls or Frames Columns, beams, trusses, or slabs supporting discontinuous walls or frames of structures having horizontal irregularity Type 4 of Table 12.3-1 or vertical irregularity Type 4 of Table 12.3-2 shall be designed to resist the seismic load effects including overstrength factor of Section 12.4.3. The connections of such discontinuous elements to the supporting members shall be adequate to transmit the forces for which the discon-tinuous elements were required to be designed. 12.3.3.4 Increase in Forces Due to Irregularities for Seismic Design Categories D through F For structures assigned to Seismic Design Category D, E, or F and having a horizontal structural irregularity of Type 1a, 1b, 2, 3, or 4 in Table 12.3-1 or a vertical structural irregularity of Type 4 in Table 12.3-2, the design forces determined from Section 12.10.1.1 shall be increased 25 percent for the following elements of the seismic force-resisting system: 1. Connections of diaphragms to vertical elements and to collectors. 2. Collectors and their connections, including connections to vertical elements, of the seismic force-resisting system. MINIMUM DESIGN LOADS 83 EXCEPTION: Forces calculated using the seismic load effects including overstrength factor of Section 12.4.3 need not be increased. 12.3.4 Redundancy A redundancy factor, ρ, shall be assigned to the seismic force-resisting system in each of two orthogo-nal directions for all structures in accordance with this section. 12.3.4.1 Conditions Where Value of ρ is 1.0 The value of ρ is permitted to equal 1.0 for the following: 1. Structures assigned to Seismic Design Category B or C. 2. Drift calculation and P-delta effects. 3. Design of nonstructural components. 4. Design of nonbuilding structures that are not similar to buildings. 5. Design of collector elements, splices, and their connections for which the seismic load effects including overstrength factor of Section 12.4.3 are used. 6. Design of members or connections where the seismic load effects including overstrength factor of Section 12.4.3 are required for design. 7. Diaphragm loads determined using Eq. 12.10-1. 8. Structures with damping systems designed in accordance with Chapter 18. 9. Design of structural walls for out-of-plane forces, including their anchorage. Table 12.3-2 Vertical Structural Irregularities Type Description Reference Section Seismic Design Category Application 1a. Stiffness-Soft Story Irregularity: Stiffness-soft story irregularity is deﬁ ned to exist where there is a story in which the lateral stiffness is less than 70% of that in the story above or less than 80% of the average stiffness of the three stories above. Table 12.6-1 D, E, and F 1b. Stiffness-Extreme Soft Story Irregularity: Stiffness-extreme soft story irregularity is deﬁ ned to exist where there is a story in which the lateral stiffness is less than 60% of that in the story above or less than 70% of the average stiffness of the three stories above. 12.3.3.1 Table 12.6-1 E and F D, E, and F 2. Weight (Mass) Irregularity: Weight (mass) irregularity is deﬁ ned to exist where the effective mass of any story is more than 150% of the effective mass of an adjacent story. A roof that is lighter than the ﬂ oor below need not be considered. Table 12.6-1 D, E, and F 3. Vertical Geometric Irregularity: Vertical geometric irregularity is deﬁ ned to exist where the horizontal dimension of the seismic force-resisting system in any story is more than 130% of that in an adjacent story. Table 12.6-1 D, E, and F 4. In-Plane Discontinuity in Vertical Lateral Force-Resisting Element Irregularity: In-plane discontinuity in vertical lateral force-resisting elements irregularity is deﬁ ned to exist where there is an in-plane offset of a vertical seismic force-resisting element resulting in overturning demands on a supporting beam, column, truss, or slab. 12.3.3.3 12.3.3.4 Table 12.6-1 B, C, D, E, and F D, E, and F D, E, and F 5a. Discontinuity in Lateral Strength–Weak Story Irregularity: Discontinuity in lateral strength–weak story irregularity is deﬁ ned to exist where the story lateral strength is less than 80% of that in the story above. The story lateral strength is the total lateral strength of all seismic-resisting elements sharing the story shear for the direction under consideration. 12.3.3.1 Table 12.6-1 E and F D, E, and F 5b. Discontinuity in Lateral Strength–Extreme Weak Story Irregularity: Discontinuity in lateral strength–extreme weak story irregularity is deﬁ ned to exist where the story lateral strength is less than 65% of that in the story above. The story strength is the total strength of all seismic-resisting elements sharing the story shear for the direction under consideration. 12.3.3.1 12.3.3.2 Table 12.6-1 D, E, and F B and C D, E, and F CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 84 12.3.4.2 Redundancy Factor, ρ, for Seismic Design Categories D through F For structures assigned to Seismic Design Category D, E, or F, ρ shall equal 1.3 unless one of the following two conditions is met, whereby ρ is permitted to be taken as 1.0: a. Each story resisting more than 35 percent of the base shear in the direction of interest shall comply with Table 12.3-3. b. Structures that are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force-resisting perimeter framing on each side of the structure in each orthogonal direction at each story resisting more than 35 percent of the base shear. The number of bays for a shear wall shall be calculated as the length of shear wall divided by the story height or two times the length of shear wall divided by the story height, hsx, for light-frame construction. 12.4 SEISMIC LOAD EFFECTS AND COMBINATIONS 12.4.1 Applicability All members of the structure, including those not part of the seismic force-resisting system, shall be designed using the seismic load effects of Section 12.4 unless otherwise exempted by this standard. Seismic load effects are the axial, shear, and ﬂ exural member forces resulting from application of horizon-tal and vertical seismic forces as set forth in Section 12.4.2. Where speciﬁ cally required, seismic load effects shall be modiﬁ ed to account for overstrength, as set forth in Section 12.4.3. 12.4.2 Seismic Load Effect The seismic load effect, E, shall be determined in accordance with the following: 1. For use in load combination 5 in Section 2.3.2 or load combinations 5 and 6 in Section 2.4.1, E shall be determined in accordance with Eq. 12.4-1 as follows: E = Eh + Ev (12.4-1) 2. For use in load combination 7 in Section 2.3.2 or load combination 8 in Section 2.4.1, E shall be determined in accordance with Eq. 12.4-2 as follows: E = Eh – Ev (12.4-2) where E = seismic load effect Eh = effect of horizontal seismic forces as deﬁ ned in Section 12.4.2.1 Ev = effect of vertical seismic forces as deﬁ ned in Section 12.4.2.2 12.4.2.1 Horizontal Seismic Load Effect The horizontal seismic load effect, Eh, shall be determined in accordance with Eq. 12.4-3 as follows: Eh = ρQE (12.4-3) Table 12.3-3 Requirements for Each Story Resisting More than 35% of the Base Shear Lateral Force-Resisting Element Requirement Braced frames Removal of an individual brace, or connection thereto, would not result in more than a 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity Type 1b). Moment frames Loss of moment resistance at the beam-to-column connections at both ends of a single beam would not result in more than a 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity Type 1b). Shear walls or wall piers with a height-to-length ratio greater than 1.0 Removal of a shear wall or wall pier with a height-to-length ratio greater than 1.0 within any story, or collector connections thereto, would not result in more than a 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity Type 1b). The shear wall and wall pier height-to-length ratios are determined as shown in Figure 12.3-2. Cantilever columns Loss of moment resistance at the base connections of any single cantilever column would not result in more than a 33% reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity Type 1b). Other No requirements MINIMUM DESIGN LOADS 85 Shear wall height-to-length-ratio = hwall/Lwall Wall pier height-to-length-ratio = hwp/Lwp hwall = height of shear wall hwp = height of wall pier Lwall = height of shear wall Lwp = height of wall pier Story Level Story Level hwP LwP hwall Lwall FIGURE 12.3-2 Shear Wall and Wall Pier Height-To-Length Ratio Determination CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 86 where QE = effects of horizontal seismic forces from V or Fp. Where required by Section 12.5.3 or 12.5.4, such effects shall result from application of horizontal forces simultaneously in two direc-tions at right angles to each other ρ = redundancy factor, as deﬁ ned in Section 12.3.4 12.4.2.2 Vertical Seismic Load Effect The vertical seismic load effect, Ev, shall be determined in accordance with Eq. 12.4-4 as follows: Ev = 0.2SDSD (12.4-4) where SDS = design spectral response acceleration parameter at short periods obtained from Section 11.4.4 D = effect of dead load EXCEPTIONS: The vertical seismic load effect, Ev, is permitted to be taken as zero for either of the following conditions: 1. In Eqs. 12.4-1, 12.4-2, 12.4-5, and 12.4-6 where SDS is equal to or less than 0.125. 2. In Eq. 12.4-2 where determining demands on the soil–structure interface of foundations. 12.4.2.3 Seismic Load Combinations Where the prescribed seismic load effect, E, deﬁ ned in Section 12.4.2 is combined with the effects of other loads as set forth in Chapter 2, the following seismic load combinations for structures not subject to ﬂ ood or atmospheric ice loads shall be used in lieu of the seismic load combinations in either Section 2.3.2 or 2.4.1: Basic Combinations for Strength Design (see Sections 2.3.2 and 2.2 for notation). 5. (1.2 + 0.2SDS)D + ρQE + L + 0.2S 6. (0.9 – 0.2SDS)D + ρQE + 1.6H NOTES: 1. The load factor on L in combination 5 is permitted to equal 0.5 for all occupancies in which Lo in Table 4-1 is less than or equal to 100 psf (4.79 kN/m2), with the exception of garages or areas occupied as places of public assembly. 2. The load factor on H shall be set equal to zero in combination 7 if the structural action due to H counteracts that due to E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall not be included in H but shall be included in the design resistance. Basic Combinations for Allowable Stress Design (see Sections 2.4.1 and 2.2 for notation). 5. (1.0 + 0.14SDS)D + H + F + 0.7ρQE 6. (1.0 + 0.10SDS)D + H + F + 0.525ρQE + 0.75L + 0.75(Lr or S or R) 8. (0.6 – 0.14SDS)D + 0.7ρQE + H 12.4.3 Seismic Load Effect Including Overstrength Factor Where speciﬁ cally required, conditions requiring overstrength factor applications shall be determined in accordance with the following: 1. For use in load combination 5 in Section 2.3.2 or load combinations 5 and 6 in Section 2.4.1, E shall be taken equal to Em as determined in accordance with Eq. 12.4-5 as follows: Em = Emh + Ev (12.4-5) 2. For use in load combination 7 in Section 2.3.2 or load combination 8 in Section 2.4.1, E shall be taken equal to Em as determined in accordance with Eq. 12.4-6 as follows: Em = Emh – Ev (12.4-6) where Em = seismic load effect including overstrength factor Emh = effect of horizontal seismic forces including overstrength factor as deﬁ ned in Section 12.4.3.1 Ev = vertical seismic load effect as deﬁ ned in Section 12.4.2.2 12.4.3.1 Horizontal Seismic Load Effect with Overstrength Factor The horizontal seismic load effect with over-strength factor, Emh, shall be determined in accordance with Eq. 12.4-7 as follows: Emh = ΩoQE (12.4-7) where QE = effects of horizontal seismic forces from V, Fpx, or Fp as speciﬁ ed in Sections 12.8.1, 12.10, or 13.3.1. Where required by Section 12.5.3 or 12.5.4, such effects shall result from application of horizontal forces simultaneously in two directions at right angles to each other. Ωo = overstrength factor EXCEPTION: The value of Emh need not exceed the maximum force that can develop in the element as determined by a rational, plastic mechanism analysis MINIMUM DESIGN LOADS 87 or nonlinear response analysis utilizing realistic expected values of material strengths. 12.4.3.2 Load Combinations with Overstrength Factor Where the seismic load effect with overstrength factor, Em, deﬁ ned in Section 12.4.3, is combined with the effects of other loads as set forth in Chapter 2, the following seismic load combination for structures not subject to ﬂ ood or atmospheric ice loads shall be used in lieu of the seismic load combinations in either Section 2.3.2 or 2.4.1: Basic Combinations for Strength Design with Overstrength Factor (see Sections 2.3.2 and 2.2 for notation). 5. (1.2 + 0.2SDS)D + ΩoQE + L + 0.2S 7. (0.9 – 0.2SDS)D + ΩoQE + 1.6H NOTES: 1. The load factor on L in combination 5 is permitted to equal 0.5 for all occupancies in which Lo in Table 4-1 is less than or equal to 100 psf (4.79 kN/ m2), with the exception of garages or areas occupied as places of public assembly. 2. The load factor on H shall be set equal to zero in combination 7 if the structural action due to H counteracts that due to E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall not be included in H but shall be included in the design resistance. Basic Combinations for Allowable Stress Design with Overstrength Factor (see Sections 2.4.1 and 2.2 for notation). 5. (1.0 + 0.14SDS)D + H + F + 0.7ΩoQE 6. (1.0 + 0.105SDS)D + H + F + 0.525ΩoQE + 0.75L + 0.75(Lr or S or R) 8. (0.6 – 0.14SDS)D + 0.7ΩoQE + H 12.4.3.3 Allowable Stress Increase for Load Combinations with Overstrength Where allowable stress design methodologies are used with the seismic load effect deﬁ ned in Section 12.4.3 applied in load combinations 5, 6, or 8 of Section 2.4.1, allowable stresses are permitted to be determined using an allowable stress increase of 1.2. This increase shall not be combined with increases in allowable stresses or load combination reductions otherwise permitted by this standard or the material reference document except for increases due to adjustment factors in accordance with AF&PA NDS. 12.4.4 Minimum Upward Force for Horizontal Cantilevers for Seismic Design Categories D through F In structures assigned to Seismic Design Category D, E, or F, horizontal cantilever structural members shall be designed for a minimum net upward force of 0.2 times the dead load in addition to the applicable load combinations of Section 12.4. 12.5 DIRECTION OF LOADING 12.5.1 Direction of Loading Criteria The directions of application of seismic forces used in the design shall be those which will produce the most critical load effects. It is permitted to satisfy this requirement using the procedures of Section 12.5.2 for Seismic Design Category B, Section 12.5.3 for Seismic Design Category C, and Section 12.5.4 for Seismic Design Categories D, E, and F. 12.5.2 Seismic Design Category B For structures assigned to Seismic Design Category B, the design seismic forces are permitted to be applied independently in each of two orthogonal directions and orthogonal interaction effects are permitted to be neglected. 12.5.3 Seismic Design Category C Loading applied to structures assigned to Seismic Design Category C shall, as a minimum, conform to the requirements of Section 12.5.2 for Seismic Design Category B and the requirements of this section. Structures that have horizontal structural irregularity Type 5 in Table 12.3-1 shall use one of the following procedures: a. Orthogonal Combination Procedure. The structure shall be analyzed using the equivalent lateral force analysis procedure of Section 12.8, the modal response spectrum analysis procedure of Section 12.9, or the linear response history procedure of Section 16.1, as permitted under Section 12.6, with the loading applied indepen-dently in any two orthogonal directions. The requirement of Section 12.5.1 is deemed satisﬁ ed if members and their foundations are designed for 100 percent of the forces for one direction plus 30 percent of the forces for the perpendicular direc-tion. The combination requiring the maximum component strength shall be used. b. Simultaneous Application of Orthogonal Ground Motion. The structure shall be analyzed CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 88 using the linear response history procedure of Section 16.1 or the nonlinear response history procedure of Section 16.2, as permitted by Section 12.6, with orthogonal pairs of ground motion acceleration histories applied simultaneously. 12.5.4 Seismic Design Categories D through F Structures assigned to Seismic Design Category D, E, or F shall, as a minimum, conform to the requirements of Section 12.5.3. In addition, any column or wall that forms part of two or more intersecting seismic force-resisting systems and is subjected to axial load due to seismic forces acting along either principal plan axis equaling or exceeding 20 percent of the axial design strength of the column or wall shall be designed for the most critical load effect due to application of seismic forces in any direction. Either of the procedures of Section 12.5.3 a or b are permitted to be used to satisfy this require-ment. Except as required by Section 12.7.3, 2-D analyses are permitted for structures with ﬂ exible diaphragms. 12.6 ANALYSIS PROCEDURE SELECTION The structural analysis required by Chapter 12 shall consist of one of the types permitted in Table 12.6-1, based on the structure’s seismic design category, structural system, dynamic properties, and regularity, or with the approval of the authority having jurisdic-tion, an alternative generally accepted procedure is permitted to be used. The analysis procedure selected shall be completed in accordance with the require-ments of the corresponding section referenced in Table 12.6-1. 12.7 MODELING CRITERIA 12.7.1 Foundation Modeling For purposes of determining seismic loads, it is permitted to consider the structure to be ﬁ xed at the base. Alternatively, where foundation ﬂ exibility is considered, it shall be in accordance with Section 12.13.3 or Chapter 19. 12.7.2 Effective Seismic Weight The effective seismic weight, W, of a structure shall include the dead load, as deﬁ ned in Section 3.1, above the base and other loads above the base as listed below: 1. In areas used for storage, a minimum of 25 percent of the ﬂ oor live load shall be included. EXCEPTIONS: a. Where the inclusion of storage loads adds no more than 5% to the effective seismic weight at that level, it need not be included in the effective seismic weight. b. Floor live load in public garages and open parking structures need not be included. Table 12.6-1 Permitted Analytical Procedures Seismic Design Category Structural Characteristics Equivalent Lateral Force Analysis, Section 12.8a Modal Response Spectrum Analysis, Section 12.9a Seismic Response History Procedures, Chapter 16a B, C All structures P P P D, E, F Risk Category I or II buildings not exceeding 2 stories above the base P P P Structures of light frame construction P P P Structures with no structural irregularities and not exceeding 160 ft in structural height P P P Structures exceeding 160 ft in structural height with no structural irregularities and with T < 3.5T s P P P Structures not exceeding 160 ft in structural height and having only horizontal irregularities of Type 2, 3, 4, or 5 in Table 12.3-1 or vertical irregularities of Type 4, 5a, or 5b in Table 12.3-2 P P P All other structures NP P P aP: Permitted; NP: Not Permitted; T s = SD1/SDS. MINIMUM DESIGN LOADS 89 2. Where provision for partitions is required by Section 4.2.2 in the ﬂ oor load design, the actual partition weight or a minimum weight of 10 psf (0.48 kN/m2) of ﬂ oor area, whichever is greater. 3. Total operating weight of permanent equipment. 4. Where the ﬂ at roof snow load, Pf, exceeds 30 psf (1.44 kN/m2), 20 percent of the uniform design snow load, regardless of actual roof slope. 5. Weight of landscaping and other materials at roof gardens and similar areas. 12.7.3 Structural Modeling A mathematical model of the structure shall be constructed for the purpose of determining member forces and structure displacements resulting from applied loads and any imposed displacements or P-delta effects. The model shall include the stiffness and strength of elements that are signiﬁ cant to the distribution of forces and deformations in the structure and represent the spatial distribution of mass and stiffness throughout the structure. In addition, the model shall comply with the following: a. Stiffness properties of concrete and masonry elements shall consider the effects of cracked sections. b. For steel moment frame systems, the contribution of panel zone deformations to overall story drift shall be included. Structures that have horizontal structural irregu-larity Type 1a, 1b, 4, or 5 of Table 12.3-1 shall be analyzed using a 3-D representation. Where a 3-D model is used, a minimum of three dynamic degrees of freedom consisting of translation in two orthogonal plan directions and rotation about the vertical axis shall be included at each level of the structure. Where the diaphragms have not been classiﬁ ed as rigid or ﬂ exible in accordance with Section 12.3.1, the model shall include representation of the diaphragm’s stiffness characteristics and such additional dynamic degrees of freedom as are required to account for the participation of the diaphragm in the structure’s dynamic response. EXCEPTION: Analysis using a 3-D representation is not required for structures with ﬂ exible diaphragms that have Type 4 horizontal structural irregularities. 12.7.4 Interaction Effects Moment-resisting frames that are enclosed or adjoined by elements that are more rigid and not considered to be part of the seismic force-resisting system shall be designed so that the action or failure of those elements will not impair the vertical load and seismic force-resisting capability of the frame. The design shall provide for the effect of these rigid elements on the structural system at structural deformations corresponding to the design story drift (Δ) as determined in Section 12.8.6. In addition, the effects of these elements shall be considered where determining whether a structure has one or more of the irregularities deﬁ ned in Section 12.3.2. 12.8 EQUIVALENT LATERAL FORCE PROCEDURE 12.8.1 Seismic Base Shear The seismic base shear, V, in a given direction shall be determined in accordance with the following equation: V = CsW (12.8-1) where Cs = the seismic response coefﬁ cient determined in accordance with Section 12.8.1.1 W = the effective seismic weight per Section 12.7.2 12.8.1.1 Calculation of Seismic Response Coefﬁ cient The seismic response coefﬁ cient, Cs, shall be determined in accordance with Eq. 12.8-2. C S R I s DS e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (12.8-2) where SDS = the design spectral response acceleration parameter in the short period range as deter-mined from Section 11.4.4 or 11.4.7 R = the response modiﬁ cation factor in Table 12.2-1 Ie = the importance factor determined in accordance with Section 11.5.1 The value of Cs computed in accordance with Eq. 12.8-2 need not exceed the following: C S T R I s D e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 for T ≤ TL (12.8-3) C S T T R I s D L e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 for T > TL (12.8-4) CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 90 Cs shall not be less than Cs = 0.044SDSIe ≥ 0.01 (12.8-5) In addition, for structures located where S1 is equal to or greater than 0.6g, Cs shall not be less than Cs = 0.5S1/(R/Ie) (12.8-6) where Ie and R are as deﬁ ned in Section 12.8.1.1 and SD1 = the design spectral response acceleration parameter at a period of 1.0 s, as determined from Section 11.4.4 or 11.4.7 T = the fundamental period of the structure(s) determined in Section 12.8.2 TL = long-period transition period(s) determined in Section 11.4.5 S1 = the mapped maximum considered earthquake spectral response acceleration parameter determined in accordance with Section 11.4.1 or 11.4.7 12.8.1.2 Soil Structure Interaction Reduction A soil structure interaction reduction is permitted where determined using Chapter 19 or other generally accepted procedures approved by the authority having jurisdiction. 12.8.1.3 Maximum Ss Value in Determination of Cs For regular structures ﬁ ve stories or less above the base as deﬁ ned in Section 11.2 and with a period, T, of 0.5 s or less, Cs is permitted to be calculated using a value of 1.5 for SS. 12.8.2 Period Determination The fundamental period of the structure, T, in the direction under consideration shall be established using the structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis. The fundamental period, T, shall not exceed the product of the coefﬁ cient for upper limit on calculated period (Cu) from Table 12.8-1 and the approximate fundamental period, T a, determined in accordance with Section 12.8.2.1. As an alternative to performing an analysis to determine the fundamental period, T, it is permitted to use the approximate building period, T a, calculated in accor-dance with Section 12.8.2.1, directly. 12.8.2.1 Approximate Fundamental Period The approximate fundamental period (T a), in s, shall be determined from the following equation: T a = Cthn x (12.8-7) where hn is the structural height as deﬁ ned in Section 11.2 and the coefﬁ cients Ct and x are determined from Table 12.8-2. Alternatively, it is permitted to determine the approximate fundamental period (T a), in s, from the following equation for structures not exceeding 12 stories above the base as deﬁ ned in Section 11.2 where the seismic force-resisting system consists Table 12.8-1 Coefﬁ cient for Upper Limit on Calculated Period Design Spectral Response Acceleration Parameter at 1 s, SD1 Coefﬁ cient Cu ≥ 0.4 1.4 0.3 1.4 0.2 1.5 0.15 1.6 ≤ 0.1 1.7 Table 12.8-2 Values of Approximate Period Parameters Ct and x Structure Type Ct x Moment-resisting frame systems in which the frames resist 100% of the required seismic force and are not enclosed or adjoined by components that are more rigid and will prevent the frames from deﬂ ecting where subjected to seismic forces: Steel moment-resisting frames 0.028 (0.0724)a 0.8 Concrete moment-resisting frames 0.016 (0.0466)a 0.9 Steel eccentrically braced frames in accordance with Table 12.2-1 lines B1 or D1 0.03 (0.0731)a 0.75 Steel buckling-restrained braced frames 0.03 (0.0731)a 0.75 All other structural systems 0.02 (0.0488)a 0.75 aMetric equivalents are shown in parentheses. MINIMUM DESIGN LOADS 91 entirely of concrete or steel moment resisting frames and the average story height is at least 10 ft (3 m): T a = 0.1N (12.8-8) where N = number of stories above the base. The approximate fundamental period, T a, in s for masonry or concrete shear wall structures is permitted to be determined from Eq. 12.8-9 as follows: T C h a w n = 0 0019 . (12.8-9) where Cw is calculated from Eq. 12.8-10 as follows: C A h h A h D w B n i i x i i i = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ∑ 100 1 0 83 1 2 2 . (12.8-10) where AB = area of base of structure, ft2 Ai = web area of shear wall i in ft2 Di = length of shear wall i in ft hi = height of shear wall i in ft x = number of shear walls in the building effective in resisting lateral forces in the direction under consideration 12.8.3 Vertical Distribution of Seismic Forces The lateral seismic force (Fx) (kip or kN) induced at any level shall be determined from the following equations: Fx = CvxV (12.8-11) and C w h w h vx x x k i i k i n = = ∑ 1 (12.8-12) where Cvx = vertical distribution factor V = total design lateral force or shear at the base of the structure (kip or kN) wi and wx = the portion of the total effective seismic weight of the structure (W) located or assigned to Level i or x hi and hx = the height (ft or m) from the base to Level i or x k = an exponent related to the structure period as follows: for structures having a period of 0.5 s or less, k = 1 for structures having a period of 2.5 s or more, k = 2 for structures having a period between 0.5 and 2.5 s, k shall be 2 or shall be determined by linear interpolation between 1 and 2 12.8.4 Horizontal Distribution of Forces The seismic design story shear in any story (Vx) (kip or kN) shall be determined from the following equation: V F x i i x n = = ∑ (12.8-13) where Fi = the portion of the seismic base shear (V) (kip or kN) induced at Level i. The seismic design story shear (Vx) (kip or kN) shall be distributed to the various vertical elements of the seismic force-resisting system in the story under consideration based on the relative lateral stiffness of the vertical resisting elements and the diaphragm. 12.8.4.1 Inherent Torsion For diaphragms that are not ﬂ exible, the distribu-tion of lateral forces at each level shall consider the effect of the inherent torsional moment, Mt, resulting from eccentricity between the locations of the center of mass and the center of rigidity. For ﬂ exible diaphragms, the distribution of forces to the vertical elements shall account for the position and distribu-tion of the masses supported. 12.8.4.2 Accidental Torsion Where diaphragms are not ﬂ exible, the design shall include the inherent torsional moment (Mt) resulting from the location of the structure masses plus the accidental torsional moments (Mta) caused by assumed displacement of the center of mass each way from its actual location by a distance equal to 5 percent of the dimension of the structure perpendicu-lar to the direction of the applied forces. Where earthquake forces are applied concurrently in two orthogonal directions, the required 5 percent displacement of the center of mass need not be applied in both of the orthogonal directions at the same time, but shall be applied in the direction that produces the greater effect. 12.8.4.3 Ampliﬁ cation of Accidental Torsional Moment Structures assigned to Seismic Design Category C, D, E, or F, where Type 1a or 1b torsional irregu-larity exists as deﬁ ned in Table 12.3-1 shall have the effects accounted for by multiplying Mta at each level by a torsional ampliﬁ cation factor (Ax) as illustrated in CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 92 Fig. 12.8-1 and determined from the following equation: Ax = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ δ δ max . 1 2 2 avg (12.8-14) where δmax = the maximum displacement at Level x com-puted assuming Ax = 1 (in. or mm) δavg = the average of the displacements at the extreme points of the structure at Level x computed assuming Ax = 1 (in. or mm) The torsional ampliﬁ cation factor (Ax) shall not be less than 1 and is not required to exceed 3.0. The more severe loading for each element shall be considered for design. 12.8.5 Overturning The structure shall be designed to resist overturn-ing effects caused by the seismic forces determined in Section 12.8.3. 12.8.6 Story Drift Determination The design story drift (Δ) shall be computed as the difference of the deﬂ ections at the centers of mass at the top and bottom of the story under consideration. See Fig. 12.8-2. Where centers of mass do not align vertically, it is permitted to compute the deﬂ ection at the bottom of the story based on the vertical projec-tion of the center of mass at the top of the story. Where allowable stress design is used, Δ shall be computed using the strength level seismic forces speciﬁ ed in Section 12.8 without reduction for allowable stress design. For structures assigned to Seismic Design Category C, D, E, or F having horizontal irregularity Type 1a or 1b of Table 12.3-1, the design story drift, Δ, shall be computed as the largest difference of the deﬂ ections of vertically aligned points at the top and bottom of the story under consideration along any of the edges of the structure. The deﬂ ection at Level x (δx) (in. or mm) used to compute the design story drift, Δ, shall be determined in accordance with the following equation: δ δ x d xe e C I = (12.8-15) where Cd = the deﬂ ection ampliﬁ cation factor in Table 12.2-1 δxe = the deﬂ ection at the location required by this section determined by an elastic analysis Ie = the importance factor determined in accordance with Section 11.5.1 12.8.6.1 Minimum Base Shear for Computing Drift The elastic analysis of the seismic force-resisting system for computing drift shall be made using the prescribed seismic design forces of Section 12.8. EXCEPTION: Eq. 12.8-5 need not be considered for computing drift. δ δB 2 x avg A ; 2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = + = δ δ δA 2 avg max x avg 1.2δ A ; 2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = + = δ δB δA δ FIGURE 12.8-1 Torsional Ampliﬁ cation Factor, Ax MINIMUM DESIGN LOADS 93 12.8.6.2 Period for Computing Drift For determining compliance with the story drift limits of Section 12.12.1, it is permitted to determine the elastic drifts, (δxe), using seismic design forces based on the computed fundamental period of the structure without the upper limit (CuT a) speciﬁ ed in Section 12.8.2. 12.8.7 P-Delta Effects P-delta effects on story shears and moments, the resulting member forces and moments, and the story drifts induced by these effects are not required to be considered where the stability coefﬁ cient (θ) as determined by the following equation is equal to or less than 0.10: θ = Δ P I V h C x e x sx d (12.8-16) where Px = the total vertical design load at and above Level x (kip or kN); where computing Px, no individual load factor need exceed 1.0 Δ = the design story drift as deﬁ ned in Section 12.8.6 occurring simultaneously with Vx (in. or mm) Ie = the importance factor determined in accordance with Section 11.5.1 Vx = the seismic shear force acting between Levels x and x – 1 (kip or kN) hsx = the story height below Level x (in. or mm) Cd = the deﬂ ection ampliﬁ cation factor in Table 12.2-1 The stability coefﬁ cient (θ) shall not exceed θmax determined as follows: θ β max . . = ≤ 0 5 0 25 Cd (12.8-17) where β is the ratio of shear demand to shear capacity for the story between Levels x and x – 1. This ratio is permitted to be conservatively taken as 1.0. Where the stability coefﬁ cient (θ) is greater than 0.10 but less than or equal to θmax, the incremental factor related to P-delta effects on displacements and member forces shall be determined by rational analysis. Alternatively, it is permitted to multiply displacements and member forces by 1.0/(1 – θ). Where θ is greater than θmax, the structure is potentially unstable and shall be redesigned. Where the P-delta effect is included in an automated analysis, Eq. 12.8-17 shall still be satisﬁ ed, however, the value of θ computed from Eq. 12.8-16 using the results of the P-delta analysis is permitted to be divided by (1 + θ) before checking Eq. 12.8-17. L2 L1 Story Level 2 F2 = strength-level design earthquake force δ δ δ δe2 = elastic displacement computed under strength-level design earthquake forces δ δ δ δ2 = Cd δe2/IE = amplified displacement Δ Δ Δ Δ2 = δ (δ (δ (δe2 - δ δ δ δe1) Cd /IE ≤ ≤ ≤ ≤Δ Δ Δ Δa (Table 12.12-1) Story Level 1 F1 = strength-level design earthquake force δ δ δ δe1 = elastic displacement computed under strength-level design earthquake forces δ δ δ δ1 = Cd δ δ δ δe1/IE = amplified displacement Δ Δ Δ Δ1 = δ δ δ1 ≤ ≤ ≤ ≤Δ Δ Δ Δa (Table 12.12-1) Δ Δ Δ Δi = Story Drift Δ Δ Δ Δi/Li = Story Drift Ratio δ δ δ δ2 = Total Displacement e FIGURE 12.8-2 Story Drift Determination CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 94 12.9 MODAL RESPONSE SPECTRUM ANALYSIS 12.9.1 Number of Modes An analysis shall be conducted to determine the natural modes of vibration for the structure. The analysis shall include a sufﬁ cient number of modes to obtain a combined modal mass participation of at least 90 percent of the actual mass in each of the orthogonal horizontal directions of response consid-ered by the model. 12.9.2 Modal Response Parameters The value for each force-related design parameter of interest, including story drifts, support forces, and individual member forces for each mode of response shall be computed using the properties of each mode and the response spectra deﬁ ned in either Section 11.4.5 or 21.2 divided by the quantity R/Ie. The value for displacement and drift quantities shall be multi-plied by the quantity Cd/Ie. 12.9.3 Combined Response Parameters The value for each parameter of interest calcu-lated for the various modes shall be combined using the square root of the sum of the squares (SRSS) method, the complete quadratic combination (CQC) method, the complete quadratic combination method as modiﬁ ed by ASCE 4 (CQC-4), or an approved equivalent approach. The CQC or the CQC-4 method shall be used for each of the modal values where closely spaced modes have signiﬁ cant cross-correlation of translational and torsional response. 12.9.4 Scaling Design Values of Combined Response A base shear (V) shall be calculated in each of the two orthogonal horizontal directions using the calculated fundamental period of the structure T in each direction and the procedures of Section 12.8. 12.9.4.1 Scaling of Forces Where the calculated fundamental period exceeds CuT a in a given direction, CuT a shall be used in lieu of T in that direction. Where the combined response for the modal base shear (Vt) is less than 85 percent of the calculated base shear (V) using the equivalent lateral force procedure, the forces shall be multiplied by 0.85 V − Vt : where V = the equivalent lateral force procedure base shear, calculated in accordance with this section and Section 12.8 Vt = the base shear from the required modal combination 12.9.4.2 Scaling of Drifts Where the combined response for the modal base shear (Vt) is less than 0.85CsW, and where Cs is determined in accordance with Eq. 12.8-6, drifts shall be multiplied by 0 85 . C W V s t 12.9.5 Horizontal Shear Distribution The distribution of horizontal shear shall be in accordance with Section 12.8.4 except that ampliﬁ ca-tion of torsion in accordance with Section 12.8.4.3 is not required where accidental torsion effects are included in the dynamic analysis model. 12.9.6 P-Delta Effects The P-delta effects shall be determined in accordance with Section 12.8.7. The base shear used to determine the story shears and the story drifts shall be determined in accordance with Section 12.8.6. 12.9.7 Soil Structure Interaction Reduction A soil structure interaction reduction is permitted where determined using Chapter 19 or other generally accepted procedures approved by the authority having jurisdiction. 12.10 DIAPHRAGMS, CHORDS, AND COLLECTORS 12.10.1 Diaphragm Design Diaphragms shall be designed for both the shear and bending stresses resulting from design forces. At diaphragm discontinuities, such as openings and reentrant corners, the design shall assure that the dissipation or transfer of edge (chord) forces com-bined with other forces in the diaphragm is within shear and tension capacity of the diaphragm. 12.10.1.1 Diaphragm Design Forces Floor and roof diaphragms shall be designed to resist design seismic forces from the structural analysis, but shall not be less than that determined in accordance with Eq. 12.10-1 as follows: F F w w px i i x n i i x n px = = = ∑ ∑ (12.10-1) MINIMUM DESIGN LOADS 95 where Fpx = the diaphragm design force Fi = the design force applied to Level i wi = the weight tributary to Level i wpx = the weight tributary to the diaphragm at Level x The force determined from Eq. 12.10-1 shall not be less than Fpx = 0.2SDSIewpx (12.10-2) The force determined from Eq. 12.10-1 need not exceed Fpx = 0.4SDSIewpx (12.10-3) Where the diaphragm is required to transfer design seismic force from the vertical resisting elements above the diaphragm to other vertical resisting elements below the diaphragm due to offsets in the placement of the elements or to changes in relative lateral stiffness in the vertical elements, these forces shall be added to those determined from Eq. 12.10-1. The redundancy factor, ρ, applies to the design of diaphragms in structures assigned to Seismic Design Category D, E, or F. For inertial forces calculated in accordance with Eq. 12.10-1, the redundancy factor shall equal 1.0. For transfer forces, the redundancy factor, ρ, shall be the same as that used for the structure. For structures having horizontal or vertical structural irregularities of the types indicated in Section 12.3.3.4, the requirements of that section shall also apply. 12.10.2 Collector Elements Collector elements shall be provided that are capable of transferring the seismic forces originating in other portions of the structure to the element providing the resistance to those forces. 12.10.2.1 Collector Elements Requiring Load Combinations with Overstrength Factor for Seismic Design Categories C through F In structures assigned to Seismic Design Category C, D, E, or F, collector elements (see Fig. 12.10-1) and their connections including connections to vertical elements shall be designed to resist the maximum of the following: 1. Forces calculated using the seismic load effects including overstrength factor of Section 12.4.3 with seismic forces determined by the Equivalent Lateral Force procedure of Section 12.8 or the Modal Response Spectrum Analysis procedure of Section 12.9. 2. Forces calculated using the seismic load effects including overstrength factor of Section 12.4.3 with seismic forces determined by Equation 12.10-1. 3. Forces calculated using the load combinations of Section 12.4.2.3 with seismic forces determined by Equation 12.10-2. Transfer forces as described in Section 12.10.1.1 shall be considered. EXCEPTIONS: 1. The forces calculated above need not exceed those calculated using the load combinations of Section 12.4.2.3 with seismic forces determined by Equation 12.10-3. 2. In structures or portions thereof braced entirely by light-frame shear walls, collector elements and their connections including connections to vertical elements need only be designed to resist forces using the load combinations of Section 12.4.2.3 with seismic forces determined in accordance with Section 12.10.1.1. FULL LENGTH SHEAR WALL (NO COLLECTOR REQUIRED) COLLECTOR ELEMENT TO TRANSFER FORCE BETWEEN DIAPHRAGM AND SHEAR WALL SHEAR WALL AT STAIRWELL FIGURE 12.10-1 Collectors CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 96 12.11 STRUCTURAL WALLS AND THEIR ANCHORAGE 12.11.1 Design for Out-of-Plane Forces Structural walls and their anchorage shall be designed for a force normal to the surface equal to Fp = 0.4SDSIe times the weight of the structural wall with a minimum force of 10 percent of the weight of the structural wall. Interconnection of structural wall elements and connections to supporting framing systems shall have sufﬁ cient ductility, rotational capacity, or sufﬁ cient strength to resist shrinkage, thermal changes, and differential foundation settle-ment when combined with seismic forces. 12.11.2 Anchorage of Structural Walls and Transfer of Design Forces into Diaphragms. 12.11.2.1 Wall Anchorage Forces The anchorage of structural walls to supporting construction shall provide a direct connection capable of resisting the following: Fp = 0.4SDSkaIeWp (12.11-1) Fp shall not be taken less than 0.2kaIeWp. k L a f = + 1 0 100 . (12.11-2) ka need not be taken larger than 2.0. where Fp = the design force in the individual anchors SDS = the design spectral response acceleration parameter at short periods per Section 11.4.4 Ie = the importance factor determined in accordance with Section 11.5.1 ka = ampliﬁ cation factor for diaphragm ﬂ exibility Lf = the span, in feet, of a ﬂ exible diaphragm that provides the lateral support for the wall; the span is measured between vertical elements that provide lateral support to the diaphragm in the direction considered; use zero for rigid diaphragms Wp = the weight of the wall tributary to the anchor Where the anchorage is not located at the roof and all diaphragms are not ﬂ exible, the value from Eq. 12.11-1 is permitted to be multiplied by the factor (1 + 2z/h)/3, where z is the height of the anchor above the base of the structure and h is the height of the roof above the base. Structural walls shall be designed to resist bending between anchors where the anchor spacing exceeds 4 ft (1,219 mm). 12.11.2.2 Additional Requirements for Diaphragms in Structures Assigned to Seismic Design Categories C through F 12.11.2.2.1 Transfer of Anchorage Forces into Diaphragm Diaphragms shall be provided with continuous ties or struts between diaphragm chords to distribute these anchorage forces into the diaphragms. Diaphragm connections shall be positive, mechanical, or welded. Added chords are permitted to be used to form subdiaphragms to transmit the anchorage forces to the main continuous cross-ties. The maximum length-to-width ratio of the structural subdiaphragm shall be 2.5 to 1. Connections and anchorages capable of resisting the prescribed forces shall be provided between the diaphragm and the attached components. Connections shall extend into the diaphragm a sufﬁ cient distance to develop the force transferred into the diaphragm. 12.11.2.2.2 Steel Elements of Structural Wall Anchor-age System The strength design forces for steel elements of the structural wall anchorage system, with the exception of anchor bolts and reinforcing steel, shall be increased by 1.4 times the forces otherwise required by this section. 12.11.2.2.3 Wood Diaphragms In wood diaphragms, the continuous ties shall be in addition to the dia-phragm sheathing. Anchorage shall not be accom-plished by use of toenails or nails subject to withdrawal nor shall wood ledgers or framing be used in cross-grain bending or cross-grain tension. The diaphragm sheathing shall not be considered effective as providing the ties or struts required by this section. 12.11.2.2.4 Metal Deck Diaphragms In metal deck diaphragms, the metal deck shall not be used as the continuous ties required by this section in the direc-tion perpendicular to the deck span. 12.11.2.2.5 Embedded Straps Diaphragm to structural wall anchorage using embedded straps shall be attached to, or hooked around, the reinforcing steel or otherwise terminated so as to effectively transfer forces to the reinforcing steel. 12.11.2.2.6 Eccentrically Loaded Anchorage System Where elements of the wall anchorage system are loaded eccentrically or are not perpendicular to the wall, the system shall be designed to resist all components of the forces induced by the eccentricity. MINIMUM DESIGN LOADS 97 12.11.2.2.7 Walls with Pilasters Where pilasters are present in the wall, the anchorage force at the pilas-ters shall be calculated considering the additional load transferred from the wall panels to the pilasters. However, the minimum anchorage force at a ﬂ oor or roof shall not be reduced. 12.12 DRIFT AND DEFORMATION 12.12.1 Story Drift Limit The design story drift (Δ) as determined in Sections 12.8.6, 12.9.2, or 16.1, shall not exceed the allowable story drift (Δa) as obtained from Table 12.12-1 for any story. 12.12.1.1 Moment Frames in Structures Assigned to Seismic Design Categories D through F For seismic force-resisting systems comprised solely of moment frames in structures assigned to Seismic Design Categories D, E, or F, the design story drift (Δ) shall not exceed Δa/ρ for any story. ρ shall be determined in accordance with Section 12.3.4.2. 12.12.2 Diaphragm Deﬂ ection The deﬂ ection in the plane of the diaphragm, as determined by engineering analysis, shall not exceed the permissible deﬂ ection of the attached elements. Permissible deﬂ ection shall be that deﬂ ection that will permit the attached element to maintain its structural integrity under the individual loading and continue to support the prescribed loads. 12.12.3 Structural Separation All portions of the structure shall be designed and constructed to act as an integral unit in resisting seismic forces unless separated structurally by a distance sufﬁ cient to avoid damaging contact as set forth in this section. Separations shall allow for the maximum inelastic response displacement (δM). δM shall be determined at critical locations with consideration for translational and torsional displacements of the structure including torsional ampliﬁ cations, where applicable, using the following equation: δ δ M d e C I = max (12.12-1) Where δmax = maximum elastic displacement at the critical location. Adjacent structures on the same property shall be separated by at least δMT, determined as follows: δ δ δ MT M M = ( ) +( ) 1 2 2 2 (12.12-2) where δM1 and δM2 are the maximum inelastic response displacements of the adjacent structures at their adjacent edges. Where a structure adjoins a property line not common to a public way, the structure shall be set back from the property line by at least the displace-ment δM of that structure. EXCEPTION: Smaller separations or property line setbacks are permitted where justiﬁ ed by rational analysis based on inelastic response to design ground motions. Table 12.12-1 Allowable Story Drift, Δa a,b Structure Risk Category I or II III IV Structures, other than masonry shear wall structures, 4 stories or less above the base as deﬁ ned in Section 11.2, with interior walls, partitions, ceilings, and exterior wall systems that have been designed to accommodate the story drifts. 0.025hsx c 0.020hsx 0.015hsx Masonry cantilever shear wall structuresd 0.010hsx 0.010hsx 0.010hsx Other masonry shear wall structures 0.007hsx 0.007hsx 0.007hsx All other structures 0.020hsx 0.015hsx 0.010hsx ahsx is the story height below Level x. b For seismic force-resisting systems comprised solely of moment frames in Seismic Design Categories D, E, and F, the allowable story drift shall comply with the requirements of Section 12.12.1.1. c There shall be no drift limit for single-story structures with interior walls, partitions, ceilings, and exterior wall systems that have been designed to accommodate the story drifts. The structure separation requirement of Section 12.12.3 is not waived. dStructures in which the basic structural system consists of masonry shear walls designed as vertical elements cantilevered from their base or foundation support which are so constructed that moment transfer between shear walls (coupling) is negligible. CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 98 12.12.4 Members Spanning between Structures Gravity connections or supports for members spanning between structures or seismically separate portions of structures shall be designed for the maximum anticipated relative displacements. These displacements shall be calculated: 1. Using the deﬂ ection calculated at the locations of support, per Eq. 12.8-15 multiplied by 1.5R/Cd, and 2. Considering additional deﬂ ection due to diaphragm rotation including the torsional ampliﬁ cation factor calculated per Section 12.8.4.3 where either structure is torsionally irregular, and 3. Considering diaphragm deformations, and 4. Assuming the two structures are moving in opposite directions and using the absolute sum of the displacements. 12.12.5 Deformation Compatibility for Seismic Design Categories D through F For structures assigned to Seismic Design Category D, E, or F, every structural component not included in the seismic force-resisting system in the direction under consideration shall be designed to be adequate for the gravity load effects and the seismic forces resulting from displacement due to the design story drift (Δ) as determined in accordance with Section 12.8.6 (see also Section 12.12.1). EXCEPTION: Reinforced concrete frame members not designed as part of the seismic force-resisting system shall comply with Section 21.11 of ACI 318. Where determining the moments and shears induced in components that are not included in the seismic force-resisting system in the direction under consideration, the stiffening effects of adjoining rigid structural and nonstructural elements shall be consid-ered and a rational value of member and restraint stiffness shall be used. 12.13 FOUNDATION DESIGN 12.13.1 Design Basis The design basis for foundations shall be as set forth in Section 12.1.5. 12.13.2 Materials of Construction Materials used for the design and construction of foundations shall comply with the requirements of Chapter 14. Design and detailing of steel piles shall comply with Section 14.1.7 Design and detailing of concrete piles shall comply with Section 14.2.3. 12.13.3 Foundation Load-Deformation Characteristics Where foundation ﬂ exibility is included for the linear analysis procedures in Chapters 12 and 16, the load-deformation characteristics of the foundation–soil system (foundation stiffness) shall be modeled in accordance with the requirements of this section. The linear load-deformation behavior of foundations shall be represented by an equivalent linear stiffness using soil properties that are compatible with the soil strain levels associated with the design earthquake motion. The strain-compatible shear modulus, G, and the associated strain-compatible shear wave velocity, vS, needed for the evaluation of equivalent linear stiffness shall be determined using the criteria in Section 19.2.1.1 or based on a site-speciﬁ c study. A 50 percent increase and decrease in stiffness shall be incorporated in dynamic analyses unless smaller variations can be justiﬁ ed based on ﬁ eld measurements of dynamic soil proper-ties or direct measurements of dynamic foundation stiffness. The largest values of response shall be used in design. 12.13.4 Reduction of Foundation Overturning Overturning effects at the soil–foundation interface are permitted to be reduced by 25 percent for foundations of structures that satisfy both of the following conditions: a. The structure is designed in accordance with the Equivalent Lateral Force Analysis as set forth in Section 12.8. b. The structure is not an inverted pendulum or cantilevered column type structure. Overturning effects at the soil–foundation interface are permitted to be reduced by 10 percent for foundations of structures designed in accordance with the modal analysis requirements of Section 12.9. 12.13.5 Requirements for Structures Assigned to Seismic Design Category C In addition to the requirements of Section 11.8.2, the following foundation design requirements shall apply to structures assigned to Seismic Design Category C. 12.13.5.1 Pole-Type Structures Where construction employing posts or poles as columns embedded in earth or embedded in concrete footings in the earth is used to resist lateral loads, the depth of embedment required for posts or poles to resist seismic forces shall be determined by means of MINIMUM DESIGN LOADS 99 the design criteria established in the foundation investigation report. 12.13.5.2 Foundation Ties Individual pile caps, drilled piers, or caissons shall be interconnected by ties. All ties shall have a design strength in tension or compression at least equal to a force equal to 10 percent of SDS times the larger pile cap or column factored dead plus factored live load unless it is demonstrated that equivalent restraint will be provided by reinforced concrete beams within slabs on grade or reinforced concrete slabs on grade or conﬁ nement by competent rock, hard cohesive soils, very dense granular soils, or other approved means. 12.13.5.3 Pile Anchorage Requirements In addition to the requirements of Section 14.2.3.1, anchorage of piles shall comply with this section. Where required for resistance to uplift forces, anchorage of steel pipe (round HSS sections), concrete-ﬁ lled steel pipe or H piles to the pile cap shall be made by means other than concrete bond to the bare steel section. EXCEPTION: Anchorage of concrete-ﬁ lled steel pipe piles is permitted to be accomplished using deformed bars developed into the concrete portion of the pile. 12.13.6 Requirements for Structures Assigned to Seismic Design Categories D through F In addition to the requirements of Sections 11.8.2, 11.8.3, 14.1.8, and 14.2.3.2, the following foundation design requirements shall apply to structures assigned to Seismic Design Category D, E, or F. Design and construction of concrete foundation elements shall conform to the requirements of ACI 318, Section 21.8, except as modiﬁ ed by the requirements of this section. EXCEPTION: Detached one- and two-family dwellings of light-frame construction not exceeding two stories above grade plane need only comply with the requirements for Sections 11.8.2, 11.8.3 (Items 2 through 4), 12.13.2, and 12.13.5. 12.13.6.1 Pole-Type Structures Where construction employing posts or poles as columns embedded in earth or embedded in concrete footings in the earth is used to resist lateral loads, the depth of embedment required for posts or poles to resist seismic forces shall be determined by means of the design criteria established in the foundation investigation report. 12.13.6.2 Foundation Ties Individual pile caps, drilled piers, or caissons shall be interconnected by ties. In addition, individual spread footings founded on soil deﬁ ned in Chapter 20 as Site Class E or F shall be interconnected by ties. All ties shall have a design strength in tension or compression at least equal to a force equal to 10 percent of SDS times the larger pile cap or column factored dead plus factored live load unless it is demonstrated that equivalent restraint will be provided by reinforced concrete beams within slabs on grade or reinforced concrete slabs on grade or conﬁ nement by competent rock, hard cohesive soils, very dense granular soils, or other approved means. 12.13.6.3 General Pile Design Requirement Piling shall be designed and constructed to withstand deformations from earthquake ground motions and structure response. Deformations shall include both free-ﬁ eld soil strains (without the structure) and deformations induced by lateral pile resistance to structure seismic forces, all as modiﬁ ed by soil–pile interaction. 12.13.6.4 Batter Piles Batter piles and their connections shall be capable of resisting forces and moments from the load combinations with overstrength factor of Section 12.4.3.2 or 12.14.3.2.2. Where vertical and batter piles act jointly to resist foundation forces as a group, these forces shall be distributed to the individual piles in accordance with their relative horizontal and vertical rigidities and the geometric distribution of the piles within the group. 12.13.6.5 Pile Anchorage Requirements In addition to the requirements of Section 12.13.5.3, anchorage of piles shall comply with this section. Design of anchorage of piles into the pile cap shall consider the combined effect of axial forces due to uplift and bending moments due to ﬁ xity to the pile cap. For piles required to resist uplift forces or provide rotational restraint, anchorage into the pile cap shall comply with the following: 1. In the case of uplift, the anchorage shall be capable of developing the least of the nominal tensile strength of the longitudinal reinforcement in a concrete pile, the nominal tensile strength of a steel pile, and 1.3 times the pile pullout resistance, or shall be designed to resist the axial tension force resulting from the seismic load effects including overstrength factor of Section 12.4.3 or 12.14.3.2. CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 100 The pile pullout resistance shall be taken as the ultimate frictional or adhesive force that can be developed between the soil and the pile plus the pile and pile cap weight. 2. In the case of rotational restraint, the anchorage shall be designed to resist the axial and shear forces and moments resulting from the seismic load effects including overstrength factor of Section 12.4.3 or 12.14.3.2 or shall be capable of developing the full axial, bending, and shear nominal strength of the pile. 12.13.6.6 Splices of Pile Segments Splices of pile segments shall develop the nominal strength of the pile section. EXCEPTION: Splices designed to resist the axial and shear forces and moments from the seismic load effects including overstrength factor of Section 12.4.3 or 12.14.3.2. 12.13.6.7 Pile Soil Interaction Pile moments, shears, and lateral deﬂ ections used for design shall be established considering the interac-tion of the shaft and soil. Where the ratio of the depth of embedment of the pile to the pile diameter or width is less than or equal to 6, the pile is permitted to be assumed to be ﬂ exurally rigid with respect to the soil. 12.13.6.8 Pile Group Effects Pile group effects from soil on lateral pile nominal strength shall be included where pile center-to-center spacing in the direction of lateral force is less than eight pile diameters or widths. Pile group effects on vertical nominal strength shall be included where pile center-to-center spacing is less than three pile diameters or widths. 12.14 SIMPLIFIED ALTERNATIVE STRUCTURAL DESIGN CRITERIA FOR SIMPLE BEARING WALL OR BUILDING FRAME SYSTEMS 12.14.1 General 12.14.1.1 Simpliﬁ ed Design Procedure The procedures of this section are permitted to be used in lieu of other analytical procedures in Chapter 12 for the analysis and design of simple buildings with bearing wall or building frame systems, subject to all of the limitations listed in this section. Where these procedures are used, the seismic design category shall be determined from Table 11.6-1 using the value of SDS from Section 12.14.8.1. The simpliﬁ ed design procedure is permitted to be used if the following limitations are met: 1. The structure shall qualify for Risk Category I or II in accordance with Table 1.5-1. 2. The site class, deﬁ ned in Chapter 20, shall not be class E or F. 3. The structure shall not exceed three stories above grade plane. 4. The seismic force-resisting system shall be either a bearing wall system or building frame system, as indicated in Table 12.14-1. 5. The structure shall have at least two lines of lateral resistance in each of two major axis directions. 6. At least one line of resistance shall be provided on each side of the center of mass in each direction. 7. For structures with ﬂ exible diaphragms, over-hangs beyond the outside line of shear walls or braced frames shall satisfy the following: a ≤ d/5 (12.14-1) where a = the distance perpendicular to the forces being considered from the extreme edge of the diaphragm to the line of vertical resistance closest to that edge d = the depth of the diaphragm parallel to the forces being considered at the line of vertical resistance closest to the edge 8. For buildings with a diaphragm that is not ﬂ exible, the distance between the center of rigidity and the center of mass parallel to each major axis shall not exceed 15 percent of the greatest width of the diaphragm parallel to that axis. In addition, the following two equations shall be satisﬁ ed: k d k d e b b k i i i m j j j n i i m 1 1 2 1 2 2 2 1 1 1 1 2 1 1 2 5 0 05 = = = ∑ ∑ ∑ + ≥ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . (Eq. 12.14-2A) k d k d e b b k i i i m j j j n j j m 1 1 2 1 2 2 2 1 2 2 2 2 1 1 2 5 0 05 = = = ∑ ∑ ∑ + ≥ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . (Eq. 12.14-2B) where (see Fig. 12.14-1) k1i = the lateral load stiffness of wall i or braced frame i parallel to major axis 1 k2j = the lateral load stiffness of wall j or braced frame j parallel to major axis 2 MINIMUM DESIGN LOADS 101 Table 12.14-1 Design Coefﬁ cients and Factors for Seismic Force-Resisting Systems for Simpliﬁ ed Design Procedure Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Limitationsb Seismic Design Category B C D, E A. BEARING WALL SYSTEMS 1. Special reinforced concrete shear walls 14.2 5 P P P 2. Ordinary reinforced concrete shear walls 14.2 4 P P NP 3. Detailed plain concrete shear walls 14.2 2 P NP NP 4. Ordinary plain concrete shear walls 14.2 1½ P NP NP 5. Intermediate precast shear walls 14.2 4 P P 40c 6. Ordinary precast shear walls 14.2 3 P NP NP 7. Special reinforced masonry shear walls 14.4 5 P P P 8. Intermediate reinforced masonry shear walls 14.4 3½ P P NP 9. Ordinary reinforced masonry shear walls 14.4 2 P NP NP 10. Detailed plain masonry shear walls 14.4 2 P NP NP 11. Ordinary plain masonry shear walls 14.4 1½ P NP NP 12. Prestressed masonry shear walls 14.4 1½ P NP NP 13. Light-frame (wood) walls sheathed with wood structural panels rated for shear resistance 14.5 6½ P P P 14. Light-frame (cold-formed steel) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.1 6½ P P P 15. Light-frame walls with shear panels of all other materials 14.1 and 14.5 2 P P NPd 16. Light-frame (cold-formed steel) wall systems using ﬂ at strap bracing 14.1 and 14.5 4 P P P B. BUILDING FRAME SYSTEMS 1. Steel eccentrically braced frames 14.1 8 P P P 2. Steel special concentrically braced frames 14.1 6 P P P 3. Steel ordinary concentrically braced frames 14.1 3¼ P P P 4. Special reinforced concrete shear walls 14.2 6 P P P 5. Ordinary reinforced concrete shear walls 14.2 5 P P NP 6. Detailed plain concrete shear walls 14.2 and 14.2.2.8 2 P NP NP 7. Ordinary plain concrete shear walls 14.2 1½ P NP NP 8. Intermediate precast shear walls 14.2 5 P P 40c 9. Ordinary precast shear walls 14.2 4 P NP NP 10. Steel and concrete composite eccentrically braced frames 14.3 8 P P P 11. Steel and concrete composite special concentrically braced frames 14.3 5 P P P 12. Steel and concrete composite ordinary braced frames 14.3 3 P P NP 13. Steel and concrete composite plate shear walls 14.3 6½ P P P 14. Steel and concrete composite special shear walls 14.3 6 P P P 15. Steel and concrete composite ordinary shear walls 14.3 5 P P NP 16. Special reinforced masonry shear walls 14.4 5½ P P P Continued CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 102 d1i = the distance from the wall i or braced frame i to the center of rigidity, perpendicular to major axis 1 d2j = the distance from the wall j or braced frame j to the center of rigidity, perpendicular to major axis 2 e1 = the distance perpendicular to major axis 1 between the center of rigidity and the center of mass b1 = the width of the diaphragm perpendicular to major axis 1 e2 = the distance perpendicular to major axis 2 between the center of rigidity and the center of mass b2 = the width of the diaphragm perpendicular to major axis 2 m = the number of walls and braced frames resisting lateral force in direction 1 n = the number of walls and braced frames resisting lateral force in direction 2 Eq. 12.14-2 A and B need not be checked where a structure fulﬁ lls all the following limitations: 1. The arrangement of walls or braced frames is symmetric about each major axis direction. 2. The distance between the two most separated lines of walls or braced frames is at least 90 percent of the dimension of the structure perpendicular to that axis direction. 3. The stiffness along each of the lines considered for item 2 above is at least 33 percent of the total stiffness in that axis direction. 9. Lines of resistance of the seismic force-resisting system shall be oriented at angles of no more than 15° from alignment with the major orthogonal horizontal axes of the building. 10. The simpliﬁ ed design procedure shall be used for each major orthogonal horizontal axis direction of the building. 11. System irregularities caused by in-plane or out-of-plane offsets of lateral force-resisting elements shall not be permitted. EXCEPTION: Out-of-plane and in-plane offsets of shear walls are permitted in two-story buildings of light-frame construction provided that the framing supporting the upper wall is designed for seismic force effects from overturning of the wall ampliﬁ ed by a factor of 2.5. 12. The lateral load resistance of any story shall not be less than 80 percent of the story above. Seismic Force-Resisting System ASCE 7 Section Where Detailing Requirements Are Speciﬁ ed Response Modiﬁ cation Coefﬁ cient, Ra Limitationsb Seismic Design Category B C D, E 17. Intermediate reinforced masonry shear walls 14.4 4 P P NP 18. Ordinary reinforced masonry shear walls 14.4 2 P NP NP 19. Detailed plain masonry shear walls 14.4 2 P NP NP 20. Ordinary plain masonry shear walls 14.4 1½ P NP NP 21. Prestressed masonry shear walls 14.4 1½ P NP NP 22. Light-frame (wood) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.5 7 P P P 23. Light-frame (cold-formed steel) walls sheathed with wood structural panels rated for shear resistance or steel sheets 14.1 7 P P P 24. Light-frame walls with shear panels of all other materials 14.1and 14.5 2½ P P NPd 25. Steel buckling-restrained braced frames 14.1 8 P P P 26. Steel special plate shear walls 14.1 7 P P P aResponse modiﬁ cation coefﬁ cient, R, for use throughout the standard. bP = permitted; NP = not permitted. cLight-frame walls with shear panels of all other materials are not permitted in Seismic Design Category E. dLight-frame walls with shear panels of all other materials are permitted up to 35 ft (10.6 m) in structural height, hn, in Seismic Design Category D and are not permitted in Seismic Design Category E. Table 12.14-1 (Continued) MINIMUM DESIGN LOADS 103 12.14.1.2 Reference Documents The reference documents listed in Chapter 23 shall be used as indicated in Section 12.14. 12.14.1.3 Deﬁ nitions The deﬁ nitions listed in Section 11.2 shall be used in addition to the following: PRINCIPAL ORTHOGONAL HORIZON-TAL DIRECTIONS: The orthogonal directions that overlay the majority of lateral force-resisting elements. 12.14.1.4 Notation D = The effect of dead load E = The effect of horizontal and vertical earthquake-induced forces Fa = Acceleration-based site coefﬁ cient, see Section 12.14.8.1 Fi = The portion of the seismic base shear, V, induced at Level i Fp = The seismic design force applicable to a particular structural component Fx = See Section 12.14.8.2 hi = The height above the base to Level i hx = The height above the base to Level x Level i = The building level referred to by the subscript i; i = 1 designates the ﬁ rst level above the base Level n = The level that is uppermost in the main portion of the building Level x = See “Level i” QE = The effect of horizontal seismic forces R = The response modiﬁ cation coefﬁ cient as given in Table 12.14-1 SDS = See Section 12.14.8.1 SS = See Section 11.4.1 Y-axis d11 d12 = d13 k23 Axis 2 b1 k22 e1 Center of Rigidity Center of Mass k11 k13 k12 Wall 3 Wall 2 Wall 1 Axis 1 d23 d22 b2 X - axis FIGURE 12.14-1 Notation Used in Torsion Check for Nonﬂ exible Diaphragms CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 104 V = The total design shear at the base of the structure in the direction of interest, as determined using the procedure of 12.14.8.1 Vx = The seismic design shear in Story x. See Section 12.14.8.3 W = See Section 12.14.8.1 Wc = Weight of wall Wp = Weight of structural component wi = The portion of the effective seismic weight, W, located at or assigned to Level i wx = See Section 12.14.8.2 12.14.2 Design Basis The structure shall include complete lateral and vertical force-resisting systems with adequate strength to resist the design seismic forces, speciﬁ ed in this section, in combination with other loads. Design seismic forces shall be distributed to the various elements of the structure and their connections using a linear elastic analysis in accordance with the procedures of Section 12.14.8. The members of the seismic force-resisting system and their connections shall be detailed to conform with the applicable requirements for the selected structural system as indicated in Section 12.14.4.1. A continuous load path, or paths, with adequate strength and stiffness shall be provided to transfer all forces from the point of application to the ﬁ nal point of resistance. The foundation shall be designed to accommodate the forces developed. 12.14.3 Seismic Load Effects and Combinations All members of the structure, including those not part of the seismic force-resisting system, shall be designed using the seismic load effects of Section 12.14.3 unless otherwise exempted by this standard. Seismic load effects are the axial, shear, and ﬂ exural member forces resulting from application of horizon-tal and vertical seismic forces as set forth in Section 12.14.3.1. Where speciﬁ cally required, seismic load effects shall be modiﬁ ed to account for overstrength, as set forth in Section 12.14.3.2. 12.14.3.1 Seismic Load Effect The seismic load effect, E, shall be determined in accordance with the following: 1. For use in load combination 5 in Section 2.3.2 or load combinations 5 and 6 in Section 2.4.1, E shall be determined in accordance with Eq. 12.14-3 as follows: E = Eh + Ev (12.14-3) 2. For use in load combination 7 in Section 2.3.2 or load combination 8 in Section 2.4.1, E shall be determined in accordance with Eq. 12.14-4 as follows: E = Eh – Ev (12.14-4) where E = seismic load effect Eh = effect of horizontal seismic forces as deﬁ ned in Section 12.14.3.1.1 Ev = effect of vertical seismic forces as deﬁ ned in Section 12.14.3.1.2 12.14.3.1.1 Horizontal Seismic Load Effect The horizontal seismic load effect, Eh, shall be determined in accordance with Eq. 12.14-5 as follows: Eh = QE (12.14-5) where QE = effects of horizontal seismic forces from V or Fp as speciﬁ ed in Sections 12.14.7.5, 12.14.8.1, and 13.3.1. 12.14.3.1.2 Vertical Seismic Load Effect The vertical seismic load effect, Ev, shall be determined in accor-dance with Eq. 12.14-6 as follows: Ev = 0.2SDSD (12.14-6) where SDS = design spectral response acceleration parameter at short periods obtained from Section 11.4.4 D = effect of dead load EXCEPTION: The vertical seismic load effect, Ev, is permitted to be taken as zero for either of the following conditions: 1. In Eqs. 12.4-3, 12.4-4, 12.4-7, and 12.14-8 where SDS is equal to or less than 0.125. 2. In Eq. 12.14-4 where determining demands on the soil–structure interface of foundations. 12.14.3.1.3 Seismic Load Combinations Where the prescribed seismic load effect, E, deﬁ ned in Section 12.14.3.1 is combined with the effects of other loads as set forth in Chapter 2, the following seismic load combinations for structures not subject to ﬂ ood or atmospheric ice loads shall be used in lieu of the seismic load combinations in Sections 2.3.2 or 2.4.1: Basic Combinations for Strength Design (see Sections 2.3.2 and 2.2 for notation). MINIMUM DESIGN LOADS 105 5. (1.2 + 0.2SDS)D + QE + L + 0.2S 7. (0.9 – 0.2SDS)D + QE + 1.6H NOTES: 1. The load factor on L in combination 5 is permitted to equal 0.5 for all occupancies in which Lo in Table 4-1 is less than or equal to 100 psf (4.79 kN/m2), with the exception of garages or areas occupied as places of public assembly. 2. The load factor on H shall be set equal to zero in combination 7 if the structural action due to H counteracts that due to E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall not be included in H but shall be included in the design resistance. Basic Combinations for Allowable Stress Design (see Sections 2.4.1 and 2.2 for notation). 5. (1.0 + 0.14SDS)D + H + F + 0.7QE 6. (1.0 + 0.105SDS)D + H + F + 0.525QE + 0.75L + 0.75(Lr or S or R) 8. (0.6 – 0.14SDS)D + 0.7QE + H 12.14.3.2 Seismic Load Effect Including a 2.5 Overstrength Factor Where speciﬁ cally required, conditions requiring overstrength factor applications shall be determined in accordance with the following: 1. For use in load combination 5 in Section 2.3.2 or load combinations 5 and 6 in Section 2.4.1, E shall be taken equal to Em as determined in accordance with Eq. 12.14-7 as follows: Em = Emh + Ev (12.14-7) 2. For use in load combination 7 in Section 2.3.2 or load combination 8 in Section 2.4.1, E shall be taken equal to Em as determined in accordance with Eq. 12.14-8 as follows: Em = Emh – Ev (12.14-8) where Em = seismic load effect including overstrength factor Emh = effect of horizontal seismic forces including overstrength factor as deﬁ ned in Section 12.14.3.2.1 Ev = vertical seismic load effect as deﬁ ned in Section 12.14.3.1.2 12.14.3.2.1 Horizontal Seismic Load Effect with a 2.5 Overstrength Factor The horizontal seismic load effect with overstrength factor, Emh, shall be deter-mined in accordance with Eq. 12.14-9 as follows: Emh = 2.5QE (12.14-9) where QE = effects of horizontal seismic forces from V or Fp as speciﬁ ed in Sections 12.14.7.5, 12.14.8.1, and 13.3.1 EXCEPTION: The value of Emh need not exceed the maximum force that can develop in the element as determined by a rational, plastic mechanism analysis or nonlinear response analysis utilizing realistic expected values of material strengths. 12.14.3.2.2 Load Combinations with Overstrength Factor Where the seismic load effect with over-strength factor, Em, deﬁ ned in Section 12.14.3.2, is combined with the effects of other loads as set forth in Chapter 2, the following seismic load combinations for structures not subject to ﬂ ood or atmospheric ice loads shall be used in lieu of the seismic load combi-nations in Section 2.3.2 or 2.4.1: Basic Combinations for Strength Design with Overstrength Factor (see Sections 2.3.2 and 2.2 for notation). 5. (1.2 + 0.2SDS)D + 2.5QE + L + 0.2S 7. (0.9 – 0.2SDS)D + 2.5QE + 1.6H NOTES: 1. The load factor on L in combination 5 is permitted to equal 0.5 for all occupancies in which Lo in Table 4-1 is less than or equal to 100 psf (4.79 kN/ m2), with the exception of garages or areas occupied as places of public assembly. 2. The load factor on H shall be set equal to zero in combination 7 if the structural action due to H counteracts that due to E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall not be included in H, but shall be included in the design resistance. Basic Combinations for Allowable Stress Design with Overstrength Factor (see Sections 2.4.1 and 2.2 for notation). 5. (1.0 + 0.14SDS)D + H + F + 1.75QE 6. (1.0 + 0.105SDS)D + H + F + 1.313QE + 0.75L + 0.75(Lr or S or R) 8. (0.6 – 0.14SDS)D + 1.75QE + H 12.14.3.2.3 Allowable Stress Increase for Load Combinations with Overstrength Where allowable stress design methodologies are used with the seismic load effect deﬁ ned in Section 12.14.3.2 applied in load combinations 5, 6, or 8 of Section 2.4.1, allowable CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 106 stresses are permitted to be determined using an allowable stress increase of 1.2. This increase shall not be combined with increases in allowable stresses or load combination reductions otherwise permitted by this standard or the material reference document except that combination with the duration of load increases permitted in AF&PA NDS is permitted. 12.14.4 Seismic Force-Resisting System 12.14.4.1 Selection and Limitations The basic lateral and vertical seismic force-resist-ing system shall conform to one of the types indicated in Table 12.14-1 and shall conform to all of the detailing requirements referenced in the table. The appropriate response modiﬁ cation coefﬁ cient, R, indicated in Table 12.14-1 shall be used in determin-ing the base shear and element design forces as set forth in the seismic requirements of this standard. Special framing and detailing requirements are indicated in Section 12.14.7 and in Sections 14.1, 14.2, 14.3, 14.4, and 14.5 for structures assigned to the various seismic design categories. 12.14.4.2 Combinations of Framing Systems 12.14.4.2.1 Horizontal Combinations Different seismic force-resisting systems are permitted to be used in each of the two principal orthogonal building directions. Where a combination of different structural systems is utilized to resist lateral forces in the same direction, the value of R used for design in that direction shall not be greater than the least value of R for any of the systems utilized in that direction. EXCEPTION: For buildings of light-frame construction or having ﬂ exible diaphragms and that are two stories or less above grade plane, resisting elements are permitted to be designed using the least value of R of the different seismic force-resisting systems found in each independent line of framing. The value of R used for design of diaphragms in such structures shall not be greater than the least value for any of the systems utilized in that same direction. 12.14.4.2.2 Vertical Combinations Different seismic force-resisting systems are permitted to be used in different stories. The value of R used in a given direction shall not be greater than the least value of any of the systems used in that direction. 12.14.4.2.3 Combination Framing Detailing Require-ments The detailing requirements of Section 12.14.7 required by the higher response modiﬁ cation coefﬁ -cient, R, shall be used for structural members common to systems having different response modiﬁ cation coefﬁ cients. 12.14.5 Diaphragm Flexibility Diaphragms constructed of steel decking (untopped), wood structural panels, or similar panel-ized construction are permitted to be considered ﬂ exible. 12.14.6 Application of Loading The effects of the combination of loads shall be considered as prescribed in Section 12.14.3. The design seismic forces are permitted to be applied separately in each orthogonal direction and the combi-nation of effects from the two directions need not be considered. Reversal of load shall be considered. 12.14.7 Design and Detailing Requirements The design and detailing of the members of the seismic force-resisting system shall comply with the requirements of this section. The foundation shall be designed to resist the forces developed and accommodate the movements imparted to the structure by the design ground motions. The dynamic nature of the forces, the expected ground motion, the design basis for strength and energy dissipation capacity of the structure, and the dynamic properties of the soil shall be included in the determination of the foundation design criteria. The design and construction of foundations shall comply with Section 12.13. Structural elements including foundation elements shall conform to the material design and detailing requirements set forth in Chapter 14. 12.14.7.1 Connections All parts of the structure between separation joints shall be interconnected, and the connection shall be capable of transmitting the seismic force, Fp, induced by the parts being connected. Any smaller portion of the structure shall be tied to the remainder of the structure with elements having a strength of 0.20 times the short period design spectral response acceleration coefﬁ cient, SDS, times the weight of the smaller portion or 5 percent of the portion’s weight, whichever is greater. A positive connection for resisting a horizontal force acting parallel to the member shall be provided for each beam, girder, or truss either directly to its supporting elements, or to slabs designed to act as diaphragms. Where the connection is through a diaphragm, then the member’s supporting element MINIMUM DESIGN LOADS 107 must also be connected to the diaphragm. The connection shall have minimum design strength of 5 percent of the dead plus live load reaction. 12.14.7.2 Openings or Reentrant Building Corners Except where as otherwise speciﬁ cally provided for in this standard, openings in shear walls, dia-phragms, or other plate-type elements, shall be provided with reinforcement at the edges of the openings or reentrant corners designed to transfer the stresses into the structure. The edge reinforcement shall extend into the body of the wall or diaphragm a distance sufﬁ cient to develop the force in the reinforcement. EXCEPTION: Shear walls of wood structural panels are permitted where designed in accordance with AF&PA SDPWS for perforated shear walls or AISI S213 for Type II shear walls. 12.14.7.3 Collector Elements Collector elements shall be provided with adequate strength to transfer the seismic forces originating in other portions of the structure to the element providing the resistance to those forces (see Fig. 12.10-1). Collector elements, splices, and their connections to resisting elements shall be designed to resist the forces deﬁ ned in Section 12.14.3.2. EXCEPTION: In structures, or portions thereof, braced entirely by light-frame shear walls, collector elements, splices, and connections to resisting elements are permitted to be designed to resist forces in accordance with Section 12.14.7.4. 12.14.7.4 Diaphragms Floor and roof diaphragms shall be designed to resist the design seismic forces at each level, Fx, calculated in accordance with Section 12.14.8.2. Where the diaphragm is required to transfer design seismic forces from the vertical-resisting elements above the diaphragm to other vertical-resisting elements below the diaphragm due to changes in relative lateral stiffness in the vertical elements, the transferred portion of the seismic shear force at that level, Vx, shall be added to the diaphragm design force. Diaphragms shall provide for both the shear and bending stresses resulting from these forces. Diaphragms shall have ties or struts to distribute the wall anchorage forces into the diaphragm. Diaphragm connections shall be positive, mechanical, or welded type connections. 12.14.7.5 Anchorage of Structural Walls Structural walls shall be anchored to all ﬂ oors, roofs, and members that provide out-of-plane lateral support for the wall or that are supported by the wall. The anchorage shall provide a positive direct connec-tion between the wall and ﬂ oor, roof, or supporting member with the strength to resist the out-of-plane force given by Eq. 12.14-10: Fp =0.4kaSDSWp (12.14-10) Fp shall not be taken less than 0.2ka Wp. k L a f = + 1 0 100 . (12.14-11) ka need not be taken larger than 2.0 where Fp = the design force in the individual anchors ka = ampliﬁ cation factor for diaphragm ﬂ exibility Lf = the span, in feet, of a ﬂ exible diaphragm that provides the lateral support for the wall; the span is measured between vertical elements that provide lateral support to the diaphragm in the direction considered; use zero for rigid diaphragms SDS = the design spectral response acceleration at short periods per Section 12.14.8.1 Wp = the weight of the wall tributary to the anchor 12.14.7.5.1 Transfer of Anchorage Forces into Diaphragms Diaphragms shall be provided with continuous ties or struts between diaphragm chords to distribute these anchorage forces into the diaphragms. Added chords are permitted to be used to form subdiaphragms to transmit the anchorage forces to the main continuous cross-ties. The maximum length-to-width ratio of the structural subdiaphragm shall be 2.5 to 1. Connections and anchorages capable of resisting the prescribed forces shall be provided between the diaphragm and the attached components. Connections shall extend into the diaphragm a sufﬁ cient distance to develop the force transferred into the diaphragm. 12.14.7.5.2 Wood Diaphragms In wood diaphragms, the continuous ties shall be in addition to the dia-phragm sheathing. Anchorage shall not be accom-plished by use of toenails or nails subject to withdrawal nor shall wood ledgers or framing be used in cross-grain bending or cross-grain tension. The diaphragm sheathing shall not be considered effective as providing the ties or struts required by this section. 12.14.7.5.3 Metal Deck Diaphragms In metal deck diaphragms, the metal deck shall not be used as the continuous ties required by this section in the direc-tion perpendicular to the deck span. 12.14.7.5.4 Embedded Straps Diaphragm to wall anchorage using embedded straps shall be attached to CHAPTER 12 SEISMIC DESIGN REQUIREMENTS FOR BUILDING STRUCTURES 108 or hooked around the reinforcing steel or otherwise terminated so as to effectively transfer forces to the reinforcing steel. 12.14.7.6 Bearing Walls and Shear Walls Exterior and interior bearing walls and shear walls and their anchorage shall be designed for a force equal to 40 percent of the short period design spectral response acceleration SDS times the weight of wall, Wc, normal to the surface, with a minimum force of 10 percent of the weight of the wall. Inter-connection of wall elements and connections to supporting framing systems shall have sufﬁ cient ductility, rotational capacity, or sufﬁ cient strength to resist shrinkage, thermal changes, and differential foundation settlement where combined with seismic forces. 12.14.7.7 Anchorage of Nonstructural Systems Where required by Chapter 13, all portions or components of the structure shall be anchored for the seismic force, Fp, prescribed therein. 12.14.8 Simpliﬁ ed Lateral Force Analysis Procedure An equivalent lateral force analysis shall consist of the application of equivalent static lateral forces to a linear mathematical model of the structure. The lateral forces applied in each direction shall sum to a total seismic base shear given by Section 12.14.8.1 and shall be distributed vertically in accordance with Section 12.14.8.2. For purposes of analysis, the structure shall be considered ﬁ xed at the base. 12.14.8.1 Seismic Base Shear The seismic base shear, V, in a given direction shall be determined in accordance with Eq. 12.14-11: V FS R W DS = (12.14-11) where S F S DS a s = 2 3 where Fa is permitted to be taken as 1.0 for rock sites, 1.4 for soil sites, or determined in accordance with Section 11.4.3. For the purpose of this section, sites are permitted to be considered to be rock if there is no more than 10 ft (3 m) of soil between the rock surface and the bottom of spread footing or mat foundation. In calculating SDS, Ss shall be in accor-dance with Section 11.4.1, but need not be taken larger than 1.5. F = 1.0 for buildings that are one story above grade plane F = 1.1 for buildings that are two stories above grade plane F = 1.2 for buildings that are three stories above grade plane R = the response modiﬁ cation factor from Table 12.14-1 W = effective seismic weight of the structure that includes the dead load, as deﬁ ned in Section 3.1, above grade plane and other loads above grade plane as listed in the following text: 1. In areas used for storage, a minimum of 25 percent of the ﬂ oor live load shall be included. EXCEPTIONS: a. Where the inclusion of storage loads adds no more than 5% to the effective seismic weight at that level, it need not be included in the effective seismic weight. b. Floor live load in public garages and open parking structures need not be included. 2. Where provision for partitions is required by Section 4.2.2 in the ﬂ oor load design, the actual partition weight, or a minimum weight of 10 psf (0.48 kN/m2) of ﬂ oor area, whichever is greater. 3. Total operating weight of permanent equipment. 4. Where the ﬂ at roof snow load, Pf, exceeds 30 psf (1.44 kN/m2), 20 percent of the uniform design snow load, regardless of actual roof slope. 5. Weight of landscaping and other materials at roof gardens and similar areas. 12.14.8.2 Vertical Distribution The forces at each level shall be calculated using the following equation: F w W V x x = (12.14-12) where wx = the portion of the effective seismic weight of the structure, W, at level x. 12.14.8.3 Horizontal Shear Distribution The seismic design story shear in any story, Vx (kip or kN), shall be determined from the following equation: V F x i i x n = = ∑ (12.14-13) MINIMUM DESIGN LOADS 109 where Fi = the portion of the seismic base shear, V (kip or kN) induced at Level i. 12.14.8.3.1 Flexible Diaphragm Structures The seismic design story shear in stories of structures with ﬂ exible diaphragms, as deﬁ ned in Section 12.14.5, shall be distributed to the vertical elements of the seismic force-resisting system using tributary area rules. Two-dimensional analysis is permitted where diaphragms are ﬂ exible. 12.14.8.3.2 Structures with Diaphragms That Are Not Flexible For structures with diaphragms that are not ﬂ exible, as deﬁ ned in Section 12.14.5, the seismic design story shear, Vx (kip or kN), shall be distributed to the various vertical elements of the seismic force-resisting system in the story under consideration based on the relative lateral stiffnesses of the vertical elements and the diaphragm. 12.14.8.3.2.1 Torsion The design of structures with diaphragms that are not ﬂ exible shall include the torsional moment, Mt (kip-ft or KN-m) resulting from eccentricity between the locations of center of mass and the center of rigidity. 12.14.8.4 Overturning The structure shall be designed to resist overturn-ing effects caused by the seismic forces determined in Section 12.14.8.2. The foundations of structures shall be designed for not less than 75 percent of the foundation overturning design moment, Mf (kip-ft or kN-m) at the foundation–soil interface. 12.14.8.5 Drift Limits and Building Separation Structural drift need not be calculated. Where a drift value is needed for use in material standards, to determine structural separations between buildings or from property lines, for design of cladding, or for other design requirements, it shall be taken as 1 percent of structural height, hn, unless computed to be less. All portions of the structure shall be designed to act as an integral unit in resisting seismic forces unless separated structurally by a distance sufﬁ cient to avoid damaging contact under the total deﬂ ection. 111 Chapter 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 13.1.4 Exemptions The following nonstructural components are exempt from the requirements of this section: 1. Furniture (except storage cabinets as noted in Table 13.5-1). 2. Temporary or movable equipment. 3. Architectural components in Seismic Design Category B other than parapets supported by bearing walls or shear walls provided that the component importance factor, Ip, is equal to 1.0. 4. Mechanical and electrical components in Seismic Design Category B. 5. Mechanical and electrical components in Seismic Design Category C provided that the component importance factor, Ip, is equal to 1.0. 6. Mechanical and electrical components in Seismic Design Categories D, E, or F where all of the following apply: a. The component importance factor, Ip, is equal to 1.0; b. The component is positively attached to the structure; c. Flexible connections are provided between the component and associated ductwork, piping, and conduit; and either i. The component weighs 400 lb (1,780 N) or less and has a center of mass located 4 ft (1.22 m) or less above the adjacent ﬂ oor level; or ii. The component weighs 20 lb (89 N) or less or, in the case of a distributed system, 5 lb/ft (73 N/m) or less. 13.1.5 Application of Nonstructural Component Requirements to Nonbuilding Structures Nonbuilding structures (including storage racks and tanks) that are supported by other structures shall be designed in accordance with Chapter 15. Where Section 15.3 requires that seismic forces be determined in accordance with Chapter 13 and values for Rp are not provided in Table 13.5-1 or 13.6-1, Rp shall be taken as equal to the value of R listed in Section 15. The value of ap shall be deter-mined in accordance with footnote a of Table 13.5-1 or 13.6-1. 13.1 GENERAL 13.1.1 Scope This chapter establishes minimum design criteria for nonstructural components that are permanently attached to structures and for their supports and attachments. Where the weight of a nonstructural component is greater than or equal to 25 percent of the effective seismic weight, W, of the structure as deﬁ ned in Section 12.7.2, the component shall be classiﬁ ed as a nonbuilding structure and shall be designed in accordance with Section 15.3.2. 13.1.2 Seismic Design Category For the purposes of this chapter, nonstructural components shall be assigned to the same seismic design category as the structure that they occupy or to which they are attached. 13.1.3 Component Importance Factor All components shall be assigned a component importance factor as indicated in this section. The component importance factor, Ip, shall be taken as 1.5 if any of the following conditions apply: 1. The component is required to function for life-safety purposes after an earthquake, including ﬁ re protection sprinkler systems and egress stairways. 2. The component conveys, supports, or otherwise contains toxic, highly toxic, or explosive sub-stances where the quantity of the material exceeds a threshold quantity established by the authority having jurisdiction and is sufﬁ cient to pose a threat to the public if released. 3. The component is in or attached to a Risk Cat-egory IV structure and it is needed for continued operation of the facility or its failure could impair the continued operation of the facility. 4. The component conveys, supports, or otherwise contains hazardous substances and is attached to a structure or portion thereof classiﬁ ed by the authority having jurisdiction as a hazardous occupancy. All other components shall be assigned a component importance factor, Ip, equal to 1.0. CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 112 13.1.6 Reference Documents Where a reference document provides a basis for the earthquake-resistant design of a particular type of nonstructural component, that document is permitted to be used, subject to the approval of the authority having jurisdiction and the following conditions: a. The design earthquake forces shall not be less than those determined in accordance with Section 13.3.1. b. Each nonstructural component’s seismic interac-tions with all other connected components and with the supporting structure shall be accounted for in the design. The component shall accommodate drifts, deﬂ ections, and relative displacements determined in accordance with the applicable seismic requirements of this standard. c. Nonstructural component anchorage requirements shall not be less than those speciﬁ ed in Section 13.4. 13.1.7 Reference Documents Using Allowable Stress Design Where a reference document provides a basis for the earthquake-resistant design of a particular type of component, and the same reference document deﬁ nes acceptance criteria in terms of allowable stresses rather than strengths, that reference document is permitted to be used. The allowable stress load combination shall consider dead, live, operating, and earthquake loads in addition to those in the reference document. The earthquake loads determined in accordance with Section 13.3.1 shall be multiplied by a factor of 0.7. The allowable stress design load combinations of Section 2.4 need not be used. The component shall also accommodate the relative displacements speciﬁ ed in Section 13.3.2. 13.2 GENERAL DESIGN REQUIREMENTS 13.2.1 Applicable Requirements for Architectural, Mechanical, and Electrical Components, Supports, and Attachments Architectural, mechanical, and electrical compo-nents, supports, and attachments shall comply with the sections referenced in Table 13.2-1. These requirements shall be satisﬁ ed by one of the following methods: 1. Project-speciﬁ c design and documentation submit-ted for approval to the authority having jurisdiction after review and acceptance by a registered design professional. 2. Submittal of the manufacturer’s certiﬁ cation that the component is seismically qualiﬁ ed by at least one of the following: a. Analysis, or b. Testing in accordance with the alternative set forth in Section 13.2.5, or c. Experience data in accordance with the alterna-tive set forth in Section 13.2.6. 13.2.2 Special Certiﬁ cation Requirements for Designated Seismic Systems Certiﬁ cations shall be provided for designated seismic systems assigned to Seismic Design Catego-ries C through F as follows: 1. Active mechanical and electrical equipment that must remain operable following the design earth-quake ground motion shall be certiﬁ ed by the manufacturer as operable whereby active parts or energized components shall be certiﬁ ed exclusively on the basis of approved shake table testing in accordance with Section 13.2.5 or experience data in accordance with Section 13.2.6 unless it can be Table 13.2-1 Applicable Requirements for Architectural, Mechanical, and Electrical Components: Supports and Attachments Nonstructural Element (i.e., Component, Support, Attachment) General Design Requirements (Section 13.2) Force and Displacement Requirements (Section 13.3) Attachment Requirements (Section 13.4) Architectural Component Requirements (Section 13.5) Mechanical and Electrical Component Requirements (Section 13.6) Architectural components and supports and attachments for architectural components X X X X Mechanical and electrical components with Ip > 1 X X X X Supports and attachments for mechanical and electrical components X X X X MINIMUM DESIGN LOADS 113 shown that the component is inherently rugged by comparison with similar seismically qualiﬁ ed components. Evidence demonstrating compliance with this requirement shall be submitted for approval to the authority having jurisdiction after review and acceptance by a registered design professional. 2. Components with hazardous substances and assigned a component importance factor, Ip, of 1.5 in accordance with Section 13.1.3 shall be certiﬁ ed by the manufacturer as maintaining containment following the design earthquake ground motion by (1) analysis, (2) approved shake table testing in accordance with Section 13.2.5, or (3) experience data in accordance with Section 13.2.6. Evidence demonstrating compliance with this requirement shall be submitted for approval to the authority having jurisdiction after review and acceptance by a registered design professional. 13.2.3 Consequential Damage The functional and physical interrelationship of components, their supports, and their effect on each other shall be considered so that the failure of an essential or nonessential architectural, mechanical, or electrical component shall not cause the failure of an essential architectural, mechanical, or electrical component. 13.2.4 Flexibility The design and evaluation of components, their supports, and their attachments shall consider their ﬂ exibility as well as their strength. 13.2.5 Testing Alternative for Seismic Capacity Determination As an alternative to the analytical requirements of Sections 13.2 through 13.6, testing shall be deemed as an acceptable method to determine the seismic capacity of components and their supports and attachments. Seismic qualiﬁ cation by testing based upon a nationally recognized testing standard proce-dure, such as ICC-ES AC 156, acceptable to the authority having jurisdiction shall be deemed to satisfy the design and evaluation requirements provided that the substantiated seismic capacities equal or exceed the seismic demands determined in accordance with Sections 13.3.1 and 13.3.2. 13.2.6 Experience Data Alternative for Seismic Capacity Determination As an alternative to the analytical requirements of Sections 13.2 through 13.6, use of experience data shall be deemed as an acceptable method to determine the seismic capacity of components and their supports and attachments. Seismic qualiﬁ cation by experience data based upon nationally recognized procedures acceptable to the authority having jurisdic-tion shall be deemed to satisfy the design and evalua-tion requirements provided that the substantiated seismic capacities equal or exceed the seismic demands determined in accordance with Sections 13.3.1 and 13.3.2. 13.2.7 Construction Documents Where design of nonstructural components or their supports and attachments is required by Table 13.2-1, such design shall be shown in construction documents prepared by a registered design profes-sional for use by the owner, authorities having jurisdiction, contractors, and inspectors. Such docu-ments shall include a quality assurance plan if required by Appendix 11A. 13.3 SEISMIC DEMANDS ON NONSTRUCTURAL COMPONENTS 13.3.1 Seismic Design Force The horizontal seismic design force (Fp) shall be applied at the component’s center of gravity and distributed relative to the component’s mass distribu-tion and shall be determined in accordance with Eq. 13.3-1: F a S W R I z h P p DS p p p = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 4 1 2 . (13.3-1) Fp is not required to be taken as greater than Fp = 1.6SDSIpWp (13.3-2) and Fp shall not be taken as less than Fp = 0.3SDSIpWp (13.3-3) where Fp = seismic design force SDS = spectral acceleration, short period, as determined from Section 11.4.4 ap = component ampliﬁ cation factor that varies from 1.00 to 2.50 (select appropriate value from Table 13.5-1 or 13.6-1) Ip = component importance factor that varies from 1.00 to 1.50 (see Section 13.1.3) Wp = component operating weight CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 114 Rp = component response modiﬁ cation factor that varies from 1.00 to 12 (select appropriate value from Table 13.5-1 or 13.6-1) z = height in structure of point of attachment of component with respect to the base. For items at or below the base, z shall be taken as 0. The value of z/h need not exceed 1.0 h = average roof height of structure with respect to the base The force (Fp) shall be applied independently in at least two orthogonal horizontal directions in combination with service loads associated with the component, as appropriate. For vertically cantilevered systems, however, the force Fp shall be assumed to act in any horizontal direction. In addition, the component shall be designed for a concurrent vertical force ±0.2SDSWp. The redundancy factor, ρ, is permit-ted to be taken equal to 1 and the overstrength factor, Ω0, does not apply. EXCEPTION: The concurrent vertical seismic force need not be considered for lay-in access ﬂ oor panels and lay-in ceiling panels. Where nonseismic loads on nonstructural components exceed Fp, such loads shall govern the strength design, but the detailing requirements and limitations prescribed in this chapter shall apply. In lieu of the forces determined in accordance with Eq. 13.3-1, accelerations at any level are permitted to be determined by the modal analysis procedures of Section 12.9 with R = 1.0. Seismic forces shall be in accordance with Eq. 13.3-4: F a a W R I A p i p p p p x = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (13.3-4) where ai is the acceleration at level i obtained from the modal analysis and where Ax is the torsional ampliﬁ cation factor determined by Eq.12.8-14. Upper and lower limits of Fp determined by Eqs. 13.3-2 and 13.3-3 shall apply. 13.3.2 Seismic Relative Displacements The effects of seismic relative displacements shall be considered in combination with displacements caused by other loads as appropriate. Seismic relative displacements, DpI, shall be determined in accordance with with Eq. 13.3-5 as: DpI = DpIe (13.3-5) where Ie = the importance factor in Section 11.5.1 Dp = displacement determined in accordance with the equations set forth in Sections 13.3.2.1 and 13.3.2.2. 13.3.2.1 Displacements within Structures For two connection points on the same Structure A or the same structural system, one at a height hx and the other at a height hy, Dp shall be determined as Dp = ΔxA – ΔyA (13.3-6) Alternatively, Dp is permitted to be determined using modal procedures described in Section 12.9, using the difference in story deﬂ ections calculated for each mode and then combined using appropriate modal combination procedures. Dp is not required to be taken as greater than D h h h p x y aA sx = − ( )Δ (13.3-7) 13.3.2.2 Displacements between Structures For two connection points on separate Structures A and B or separate structural systems, one at a height hx and the other at a height hy, Dp shall be determined as Dp = \|δxA\| + \|δyB\| (13.3-8) Dp is not required to be taken as greater than D h h h h p x aA sx y aB sx = Δ + Δ (13.3-9) where Dp = relative seismic displacement that the compo-nent must be designed to accommodate δxA = deﬂ ection at building Level x of Structure A, determined in accordance with Eq. (12.8-15) δyA = deﬂ ection at building Level y of Structure A, determined in accordance with Eq. (12.8-15). δyB = deﬂ ection at building Level y of Structure B, determined in accordance with Eq. (12.8-15). hx = height of Level x to which upper connection point is attached hy = height of Level y to which lower connection point is attached ΔaA = allowable story drift for Structure A as deﬁ ned in Table 12.12-1 ΔaB = allowable story drift for Structure B as deﬁ ned in Table 12.12-1 hsx = story height used in the deﬁ nition of the allowable drift Δa in Table12.12-1. Note that Δa/hsx = the drift index. MINIMUM DESIGN LOADS 115 The effects of seismic relative displacements shall be considered in combination with displacements caused by other loads as appropriate. 13.4 NONSTRUCTURAL COMPONENT ANCHORAGE Nonstructural components and their supports shall be attached (or anchored) to the structure in accordance with the requirements of this section and the attach-ment shall satisfy the requirements for the parent material as set forth elsewhere in this standard. Component attachments shall be bolted, welded, or otherwise positively fastened without consideration of frictional resistance produced by the effects of gravity. A continuous load path of sufﬁ cient strength and stiffness between the component and the support-ing structure shall be provided. Local elements of the structure including connections shall be designed and constructed for the component forces where they control the design of the elements or their connections. The component forces shall be those determined in Section 13.3.1, except that modiﬁ ca-tions to Fp and Rp due to anchorage conditions need not be considered. The design documents shall include sufﬁ cient information relating to the attach-ments to verify compliance with the requirements of this section. 13.4.1 Design Force in the Attachment The force in the attachment shall be determined based on the prescribed forces and displacements for the component as determined in Sections 13.3.1 and 13.3.2, except that Rp shall not be taken as larger than 6. 13.4.2 Anchors in Concrete or Masonry. 13.4.2.1 Anchors in Concrete Anchors in concrete shall be designed in accor-dance with Appendix D of ACI 318. 13.4.2.2 Anchors in Masonry Anchors in masonry shall be designed in accor-dance with TMS 402/ACI 503/ASCE 5. Anchors shall be designed to be governed by the tensile or shear strength of a ductile steel element. EXCEPTION: Anchors shall be permitted to be designed so that the attachment that the anchor is connecting to the structure undergoes ductile yielding at a load level corresponding to anchor forces not greater than their design strength, or the minimum design strength of the anchors shall be at least 2.5 times the factored forces transmitted by the component. 13.4.2.3 Post-Installed Anchors in Concrete and Masonry Post-installed anchors in concrete shall be prequaliﬁ ed for seismic applications in accordance with ACI 355.2 or other approved qualiﬁ cation procedures. Post-installed anchors in masonry shall be prequaliﬁ ed for seismic applications in accordance with approved qualiﬁ cation procedures. 13.4.3 Installation Conditions Determination of forces in attachments shall take into account the expected conditions of installation including eccentricities and prying effects. 13.4.4 Multiple Attachments Determination of force distribution of multiple attachments at one location shall take into account the stiffness and ductility of the component, component supports, attachments, and structure and the ability to redistribute loads to other attachments in the group. Designs of anchorage in concrete in accordance with Appendix D of ACI 318 shall be considered to satisfy this requirement. 13.4.5 Power Actuated Fasteners Power actuated fasteners in concrete or steel shall not be used for sustained tension loads or for brace applications in Seismic Design Categories D, E, or F unless approved for seismic loading. Power actuated fasteners in masonry are not permitted unless approved for seismic loading. EXCEPTION: Power actuated fasteners in concrete used for support of acoustical tile or lay-in panel suspended ceiling applications and distributed systems where the service load on any individual fastener does not exceed 90 lb (400 N). Power actuated fasteners in steel where the service load on any individual fastener does not exceed 250 lb (1,112 N). 13.4.6 Friction Clips Friction clips in Seismic Design Categories D, E, or F shall not be used for supporting sustained loads in addition to resisting seismic forces. C-type beam and large ﬂ ange clamps are permitted for hangers provided they are equipped with restraining straps equivalent to those speciﬁ ed in NFPA 13, Section 9.3.7. Lock nuts or equivalent shall be provided to prevent loosening of threaded connections. CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 116 13.5 ARCHITECTURAL COMPONENTS 13.5.1 General Architectural components, and their supports and attachments, shall satisfy the requirements of this section. Appropriate coefﬁ cients shall be selected from Table 13.5-1. EXCEPTION: Components supported by chains or otherwise suspended from the structure are not required to satisfy the seismic force and relative displacement requirements provided they meet all of the following criteria: 1. The design load for such items shall be equal to 1.4 times the operating weight acting down with a simultaneous horizontal load equal to 1.4 times the operating weight. The horizontal load shall be applied in the direction that results in the most critical loading for design. 2. Seismic interaction effects shall be considered in accordance with Section 13.2.3. 3. The connection to the structure shall allow a 360° range of motion in the horizontal plane. 13.5.2 Forces and Displacements All architectural components, and their supports and attachments, shall be designed for the seismic forces deﬁ ned in Section 13.3.1. Architectural components that could pose a life-safety hazard shall be designed to accommodate the seismic relative displacement requirements of Section 13.3.2. Architectural components shall be designed considering vertical deﬂ ection due to joint rotation of cantilever structural members. 13.5.3 Exterior Nonstructural Wall Elements and Connections Exterior nonstructural wall panels or elements that are attached to or enclose the structure shall be designed to accommodate the seismic relative dis-placements deﬁ ned in Section 13.3.2 and movements due to temperature changes. Such elements shall be supported by means of positive and direct structural supports or by mechanical connections and fasteners in accordance with the following requirements: a. Connections and panel joints shall allow for the story drift caused by relative seismic displacements (Dp) determined in Section 13.3.2, or 0.5 in. (13 mm), whichever is greatest. b. Connections to permit movement in the plane of the panel for story drift shall be sliding connections using slotted or oversize holes, connections that permit movement by bending of steel, or other connections that provide equivalent sliding or ductile capacity. c. The connecting member itself shall have sufﬁ cient ductility and rotation capacity to preclude fracture of the concrete or brittle failures at or near welds. d. All fasteners in the connecting system such as bolts, inserts, welds, and dowels and the body of the connectors shall be designed for the force (Fp) determined by Section 13.3.1 with values of Rp and ap taken from Table 13.5-1 applied at the center of mass of the panel. e. Where anchorage is achieved using ﬂ at straps embedded in concrete or masonry, such straps shall be attached to or hooked around reinforcing steel or otherwise terminated so as to effectively transfer forces to the reinforcing steel or to assure that pullout of anchorage is not the initial failure mechanism. 13.5.4 Glass Glass in glazed curtain walls and storefronts shall be designed and installed in accordance with Section 13.5.9. 13.5.5 Out-of-Plane Bending Transverse or out-of-plane bending or deforma-tion of a component or system that is subjected to forces as determined in Section 13.5.2 shall not exceed the deﬂ ection capability of the component or system. 13.5.6 Suspended Ceilings Suspended ceilings shall be in accordance with this section. EXCEPTIONS: 1. Suspended ceilings with areas less than or equal to 144 ft2 (13.4 m2) that are surrounded by walls or sofﬁ ts that are laterally braced to the structure above are exempt from the requirements of this section. 2. Suspended ceilings constructed of screw- or nail-attached gypsum board on one level that are surrounded by and connected to walls or sofﬁ ts that are laterally braced to the structure above are exempt from the requirements of this section. 13.5.6.1 Seismic Forces The weight of the ceiling, Wp, shall include the ceiling grid; ceiling tiles or panels; light ﬁ xtures if attached to, clipped to, or laterally supported by the ceiling grid; and other components that are laterally supported by the ceiling. Wp shall be taken as not less than 4 psf (192 N/m2). MINIMUM DESIGN LOADS 117 The seismic force, Fp, shall be transmitted through the ceiling attachments to the building structural elements or the ceiling–structure boundary. 13.5.6.2 Industry Standard Construction for Acousti-cal Tile or Lay-in Panel Ceilings Unless designed in accordance with Section 13.5.6.3, or seismically qualiﬁ ed in accordance with Table 13.5-1 Coefﬁ cients for Architectural Components Architectural Component ap a Rp b Interior nonstructural walls and partitionsb Plain (unreinforced) masonry walls 1.0 1.5 All other walls and partitions 1.0 2.5 Cantilever elements (Unbraced or braced to structural frame below its center of mass) Parapets and cantilever interior nonstructural walls 2.5 2.5 Chimneys where laterally braced or supported by the structural frame 2.5 2.5 Cantilever elements (Braced to structural frame above its center of mass) Parapets 1.0 2.5 Chimneys 1.0 2.5 Exterior nonstructural wallsb 1.0b 2.5 Exterior nonstructural wall elements and connectionsb Wall element 1.0 2.5 Body of wall panel connections 1.0 2.5 Fasteners of the connecting system 1.25 1.0 Veneer Limited deformability elements and attachments 1.0 2.5 Low deformability elements and attachments 1.0 1.5 Penthouses (except where framed by an extension of the building frame) 2.5 3.5 Ceilings All 1.0 2.5 Cabinets Permanent ﬂ oor-supported storage cabinets over 6 ft (1,829 mm) tall, including contents Permanent ﬂ oor-supported library shelving, book stacks, and bookshelves over 6 ft (1,829 mm) tall, including contents 1.0 1.0 2.5 2.5 Laboratory equipment 1.0 2.5 Access ﬂ oors Special access ﬂ oors (designed in accordance with Section 13.5.7.2) 1.0 2.5 All other 1.0 1.5 Appendages and ornamentations 2.5 2.5 Signs and billboards 2.5 3.0 Other rigid components High deformability elements and attachments 1.0 3.5 Limited deformability elements and attachments 1.0 2.5 Low deformability materials and attachments 1.0 1.5 Other ﬂ exible components High deformability elements and attachments 2.5 3.5 Limited deformability elements and attachments 2.5 2.5 Low deformability materials and attachments 2.5 1.5 Egress stairways not part of the building structure 1.0 2.5 aA lower value for ap shall not be used unless justiﬁ ed by detailed dynamic analysis. The value for ap shall not be less than 1.00. The value of ap = 1 is for rigid components and rigidly attached components. The value of ap = 2.5 is for ﬂ exible components and ﬂ exibly attached components. b Where ﬂ exible diaphragms provide lateral support for concrete or masonry walls and partitions, the design forces for anchorage to the diaphragm shall be as speciﬁ ed in Section 12.11.2. CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 118 Section 13.2.5 or 13.2.6, acoustical tile or lay-in panel ceilings shall be designed and constructed in accor-dance with this section. 13.5.6.2.1 Seismic Design Category C Acoustical tile or lay-in panel ceilings in structures assigned to Seismic Design Category C shall be designed and installed in accordance with ASTM C635, ASTM C636, and ASTM E580, Section 4—Seismic Design Category C. 13.5.6.2.2 Seismic Design Categories D through F Acoustical tile or lay-in panel ceilings in Seismic Design Categories D, E, and F shall be designed and installed in accordance with ASTM C635, ASTM C636, and ASTM E580, Section 5—Seismic Design Categories D, E, and F as modiﬁ ed by this section. Acoustical tile or lay-in panel ceilings shall also comply with the following: a. The width of the perimeter supporting closure angle or channel shall be not less than 2.0 in. (50 mm). Where perimeter supporting clips are used, they shall be qualiﬁ ed in accordance with approved test criteria. In each orthogonal horizontal direc-tion, one end of the ceiling grid shall be attached to the closure angle or channel. The other end in each horizontal direction shall have a 0.75 in. (19 mm) clearance from the wall and shall rest upon and be free to slide on a closure angle or channel. b. For ceiling areas exceeding 2,500 ft2 (232 m2), a seismic separation joint or full height partition that breaks the ceiling up into areas not exceeding 2,500 ft2 (232 m2), each with a ratio of the long to short dimension less than or equal to 4, shall be provided unless structural analyses are performed of the ceiling bracing system for the prescribed seismic forces that demonstrate ceiling penetrations and closure angles or channels provide sufﬁ cient clearance to accommodate the anticipated lateral displacement. Each area shall be provided with closure angles or channels in accordance with Section 13.5.6.2.2.a and horizontal restraints or bracing. 13.5.6.3 Integral Construction As an alternate to providing large clearances around sprinkler system penetrations through ceilings, the sprinkler system and ceiling grid are permitted to be designed and tied together as an integral unit. Such a design shall consider the mass and ﬂ exibility of all elements involved, including the ceiling, sprinkler system, light ﬁ xtures, and mechanical (HVAC) appurtenances. Such design shall be performed by a registered design professional. 13.5.7 Access Floors 13.5.7.1 General The weight of the access ﬂ oor, Wp, shall include the weight of the ﬂ oor system, 100 percent of the weight of all equipment fastened to the ﬂ oor, and 25 percent of the weight of all equipment supported by but not fastened to the ﬂ oor. The seismic force, Fp, shall be transmitted from the top surface of the access ﬂ oor to the supporting structure. Overturning effects of equipment fastened to the access ﬂ oor panels also shall be considered. The ability of “slip on” heads for pedestals shall be evaluated for suitability to transfer overturning effects of equipment. Where checking individual pedestals for overturn-ing effects, the maximum concurrent axial load shall not exceed the portion of Wp assigned to the pedestal under consideration. 13.5.7.2 Special Access Floors Access ﬂ oors shall be considered to be “special access ﬂ oors” if they are designed to comply with the following considerations: 1. Connections transmitting seismic loads consist of mechanical fasteners, anchors satisfying the requirements of Appendix D of ACI 318, welding, or bearing. Design load capacities comply with recognized design codes and/or certiﬁ ed test results. 2. Seismic loads are not transmitted by friction, power actuated fasteners, adhesives, or by friction produced solely by the effects of gravity. 3. The design analysis of the bracing system includes the destabilizing effects of individual members buckling in compression. 4. Bracing and pedestals are of structural or mechani-cal shapes produced to ASTM speciﬁ cations that specify minimum mechanical properties. Electrical tubing shall not be used. 5. Floor stringers that are designed to carry axial seismic loads and that are mechanically fastened to the supporting pedestals are used. 13.5.8 Partitions 13.5.8.1 General Partitions that are tied to the ceiling and all partitions greater than 6 ft (1.8 m) in height shall be MINIMUM DESIGN LOADS 119 laterally braced to the building structure. Such bracing shall be independent of any ceiling lateral force bracing. Bracing shall be spaced to limit horizontal deﬂ ection at the partition head to be compatible with ceiling deﬂ ection requirements as determined in Section 13.5.6 for suspended ceilings and elsewhere in this section for other systems. EXCEPTION: Partitions that meet all of the following conditions: 1. The partition height does not exceed 9 ft (2,740 mm). 2. The linear weight of the partition does not exceed the product of 10 lb (0.479 kN) times the height (ft or m) of the partition. 3. The partition horizontal seismic load does not exceed 5 psf (0.24 kN/m2). 13.5.8.2 Glass Glass in glazed partitions shall be designed and installed in accordance with Section 13.5.9. 13.5.9 Glass in Glazed Curtain Walls, Glazed Storefronts, and Glazed Partitions 13.5.9.1 General Glass in glazed curtain walls, glazed storefronts, and glazed partitions shall meet the relative displace-ment requirement of Eq. 13.5-1: Δfallout ≥ 1.25IeDp (13.5-1) or 0.5 in. (13 mm), whichever is greater where: Δfallout = the relative seismic displacement (drift) at which glass fallout from the curtain wall, storefront wall, or partition occurs (Section 13.5.9.2) Dp = the relative seismic displacement that the component must be designed to accommodate (Section 13.3.2.1). Dp shall be applied over the height of the glass component under consideration Ie = the importance factor determined in accor-dance with Section 11.5.1 EXCEPTION: 1. Glass with sufﬁ cient clearances from its frame such that physical contact between the glass and frame will not occur at the design drift, as demon-strated by Eq. 13.5-2, need not comply with this requirement: Dclear ≥ 1.25Dp (13.5-2) where Dclear = relative horizontal (drift) displacement, measured over the height of the glass panel under consideration, which causes initial glass-to-frame contact. For rectangular glass panels within a rectangular wall frame Dclear = 2 1 1 2 1 c h c b c p p + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ where hp = the height of the rectangular glass panel bp = the width of the rectangular glass panel c1 = the average of the clearances (gaps) on both sides between the vertical glass edges and the frame c2 = the average of the clearances (gaps) top and bottom between the horizontal glass edges and the frame 2. Fully tempered monolithic glass in Risk Categories I, II, and III located no more than 10 ft (3 m) above a walking surface need not comply with this requirement. 3. Annealed or heat-strengthened laminated glass in single thickness with interlayer no less than 0.030 in. (0.76 mm) that is captured mechanically in a wall system glazing pocket, and whose perimeter is secured to the frame by a wet glazed gunable curing elastomeric sealant perimeter bead of 0.5 in. (13 mm) minimum glass contact width, or other approved anchorage system need not comply with this requirement. 13.5.9.2 Seismic Drift Limits for Glass Components Δfallout, the drift causing glass fallout from the curtain wall, storefront, or partition shall be deter-mined in accordance with AAMA 501.6 or by engineering analysis. 13.6 MECHANICAL AND ELECTRICAL COMPONENTS 13.6.1 General Mechanical and electrical components and their supports shall satisfy the requirements of this section. The attachment of mechanical and electrical compo-nents and their supports to the structure shall meet the requirements of Section 13.4. Appropriate coefﬁ cients shall be selected from Table 13.6-1. EXCEPTION: Light ﬁ xtures, lighted signs, and ceiling fans not connected to ducts or piping, which are supported by chains or otherwise suspended from the structure, are not required to satisfy the seismic CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 120 Table 13.6-1 Seismic Coefﬁ cients for Mechanical and Electrical Components Mechanical and Electrical Components ap a Rp b Air-side HVAC, fans, air handlers, air conditioning units, cabinet heaters, air distribution boxes, and other mechanical components constructed of sheet metal framing 2.5 6.0 Wet-side HVAC, boilers, furnaces, atmospheric tanks and bins, chillers, water heaters, heat exchangers, evaporators, air separators, manufacturing or process equipment, and other mechanical components constructed of high-deformability materials 1.0 2.5 Engines, turbines, pumps, compressors, and pressure vessels not supported on skirts and not within the scope of Chapter 15 1.0 2.5 Skirt-supported pressure vessels not within the scope of Chapter 15 2.5 2.5 Elevator and escalator components 1.0 2.5 Generators, batteries, inverters, motors, transformers, and other electrical components constructed of high deformability materials 1.0 2.5 Motor control centers, panel boards, switch gear, instrumentation cabinets, and other components constructed of sheet metal framing 2.5 6.0 Communication equipment, computers, instrumentation, and controls 1.0 2.5 Roof-mounted stacks, cooling and electrical towers laterally braced below their center of mass 2.5 3.0 Roof-mounted stacks, cooling and electrical towers laterally braced above their center of mass 1.0 2.5 Lighting ﬁ xtures 1.0 1.5 Other mechanical or electrical components 1.0 1.5 Vibration Isolated Components and Systemsb Components and systems isolated using neoprene elements and neoprene isolated ﬂ oors with built-in or separate elastomeric snubbing devices or resilient perimeter stops 2.5 2.5 Spring isolated components and systems and vibration isolated ﬂ oors closely restrained using built-in or separate elastomeric snubbing devices or resilient perimeter stops 2.5 2.0 Internally isolated components and systems 2.5 2.0 Suspended vibration isolated equipment including in-line duct devices and suspended internally isolated components 2.5 2.5 Distribution Systems Piping in accordance with ASME B31, including in-line components with joints made by welding or brazing 2.5 12.0 Piping in accordance with ASME B31, including in-line components, constructed of high or limited deformability materials, with joints made by threading, bonding, compression couplings, or grooved couplings 2.5 6.0 Piping and tubing not in accordance with ASME B31, including in-line components, constructed of high-deformability materials, with joints made by welding or brazing 2.5 9.0 Piping and tubing not in accordance with ASME B31, including in-line components, constructed of high- or limited-deformability materials, with joints made by threading, bonding, compression couplings, or grooved couplings 2.5 4.5 Piping and tubing constructed of low-deformability materials, such as cast iron, glass, and nonductile plastics 2.5 3.0 Ductwork, including in-line components, constructed of high-deformability materials, with joints made by welding or brazing 2.5 9.0 Ductwork, including in-line components, constructed of high- or limited-deformability materials with joints made by means other than welding or brazing 2.5 6.0 Ductwork, including in-line components, constructed of low-deformability materials, such as cast iron, glass, and nonductile plastics 2.5 3.0 MINIMUM DESIGN LOADS 121 force and relative displacement requirements provided they meet all of the following criteria: 1. The design load for such items shall be equal to 1.4 times the operating weight acting down with a simultaneous horizontal load equal to 1.4 times the operating weight. The horizontal load shall be applied in the direction that results in the most critical loading for the design. 2. Seismic interaction effects shall be considered in accordance with Section 13.2.3. 3. The connection to the structure shall allow a 360° range of motion in the horizontal plane. Where design of mechanical and electrical components for seismic effects is required, consider-ation shall be given to the dynamic effects of the components, their contents, and where appropriate, their supports and attachments. In such cases, the interaction between the components and the support-ing structures, including other mechanical and electrical components, shall also be considered. 13.6.2 Component Period The fundamental period of the nonstructural component (including its supports and attachment to the structure), Tp, shall be determined by the follow-ing equation provided that the component, supports, and attachment can be reasonably represented analytically by a simple spring and mass single degree-of-freedom system: T W K g P p p = 2π (13.6-1) where Tp = component fundamental period Wp = component operating weight g = gravitational acceleration Kp = combined stiffness of the component, supports and attachments, determined in terms of load per unit deﬂ ection at the center of gravity of the component Alternatively, the fundamental period of the component, Tp, in seconds is permitted to be deter-mined from experimental test data or by a properly substantiated analysis. 13.6.3 Mechanical Components HVAC ductwork shall meet the requirements of Section 13.6.7. Piping systems shall meet the require-ments of Section 13.6.8. Boilers and vessels shall meet the requirements of Section 13.6.9. Elevators shall meet the requirements of Section 13.6.10. All other mechanical components shall meet the require-ments of Section 13.6.11. Mechanical components with Ip greater than 1.0 shall be designed for the seismic forces and relative displacements deﬁ ned in Sections 13.3.1 and 13.3.2 and shall satisfy the following additional requirements: 1. Provision shall be made to eliminate seismic impact for components vulnerable to impact, for components constructed of nonductile materials, and in cases where material ductility will be reduced due to service conditions (e.g., low temperature applications). 2. The possibility of loads imposed on components by attached utility or service lines, due to differential movement of support points on separate structures, shall be evaluated. 3. Where piping or HVAC ductwork components are attached to structures that could displace relative to one another and for isolated structures where such components cross the isolation interface, the Distribution Systems Electrical conduit and cable trays 2.5 6.0 Bus ducts 1.0 2.5 Plumbing 1.0 2.5 Manufacturing or process conveyors (nonpersonnel) 2.5 3.0 aA lower value for ap is permitted where justiﬁ ed by detailed dynamic analyses. The value for ap shall not be less than 1.0. The value of ap equal to 1.0 is for rigid components and rigidly attached components. The value of ap equal to 2.5 is for ﬂ exible components and ﬂ exibly attached components. bComponents mounted on vibration isolators shall have a bumper restraint or snubber in each horizontal direction. The design force shall be taken as 2Fp if the nominal clearance (air gap) between the equipment support frame and restraint is greater than 0.25 in. (6 mm). If the nominal clearance speciﬁ ed on the construction documents is not greater than 0.25 in. (6 mm), the design force is permitted to be taken as Fp. Table 13.6-1 (Continued) CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 122 components shall be designed to accommodate the seismic relative displacements deﬁ ned in Section 13.3.2. 13.6.4 Electrical Components Electrical components with Ip greater than 1.0 shall be designed for the seismic forces and relative displacements deﬁ ned in Sections 13.3.1 and 13.3.2 and shall satisfy the following additional requirements: 1. Provision shall be made to eliminate seismic impact between components. 2. Loads imposed on the components by attached utility or service lines that are attached to separate structures shall be evaluated. 3. Batteries on racks shall have wrap-around restraints to ensure that the batteries will not fall from the racks. Spacers shall be used between restraints and cells to prevent damage to cases. Racks shall be evaluated for sufﬁ cient lateral load capacity. 4. Internal coils of dry type transformers shall be positively attached to their supporting substructure within the transformer enclosure. 5. Electrical control panels, computer equipment, and other items with slide-out components shall have a latching mechanism to hold the components in place. 6. Electrical cabinet design shall comply with the applicable National Electrical Manufacturers Association (NEMA) standards. Cutouts in the lower shear panel that have not been made by the manufacturer and reduce signiﬁ cantly the strength of the cabinet shall be speciﬁ cally evaluated. 7. The attachments for additional external items weighing more than 100 lb (445 N) shall be speciﬁ cally evaluated if not provided by the manufacturer. 8. Where conduit, cable trays, or similar electrical distribution components are attached to structures that could displace relative to one another and for isolated structures where such components cross the isolation interface, the components shall be designed to accommodate the seismic relative displacements deﬁ ned in Section 13.3.2. 13.6.5 Component Supports Mechanical and electrical component supports (including those with Ip = 1.0) and the means by which they are attached to the component shall be designed for the forces and displacements determined in Sections 13.3.1 and 13.3.2. Such supports include structural members, braces, frames, skirts, legs, saddles, pedestals, cables, guys, stays, snubbers, and tethers, as well as elements forged or cast as a part of the mechanical or electrical component. 13.6.5.1 Design Basis If standard supports, for example, ASME B31, NFPA 13, or MSS SP-58, or proprietary supports are used, they shall be designed by either load rating (i.e., testing) or for the calculated seismic forces. In addition, the stiffness of the support, where appropri-ate, shall be designed such that the seismic load path for the component performs its intended function. 13.6.5.2 Design for Relative Displacement Component supports shall be designed to accom-modate the seismic relative displacements between points of support determined in accordance with Section 13.3.2. 13.6.5.3 Support Attachment to Component The means by which supports are attached to the component, except where integral (i.e., cast or forged), shall be designed to accommodate both the forces and displacements determined in accordance with Sections 13.3.1 and 13.3.2. If the value of Ip = 1.5 for the component, the local region of the support attachment point to the component shall be evaluated for the effect of the load transfer on the component wall. 13.6.5.4 Material Detailing Requirements The materials comprising supports and the means of attachment to the component shall be constructed of materials suitable for the application, including the effects of service conditions, for example, low temperature applications. Materials shall be in conformance with a nationally recognized standard. 13.6.5.5 Additional Requirements The following additional requirements shall apply to mechanical and electrical component supports: 1. Seismic supports shall be constructed so that support engagement is maintained. 2. Reinforcement (e.g., stiffeners or Belleville washers) shall be provided at bolted connections through sheet metal equipment housings as required to transfer the equipment seismic loads speciﬁ ed in this section from the equipment to the structure. Where equipment has been certiﬁ ed per Section 13.2.2, 13.2.5, or 13.2.6, anchor bolts or other fasteners and associated hardware as included in the certiﬁ cation shall be installed in conformance MINIMUM DESIGN LOADS 123 with the manufacturer’s instructions. For those cases where no certiﬁ cation exists or where instructions for such reinforcement are not pro-vided, reinforcement methods shall be as speciﬁ ed by a registered design professional or as approved by the authority having jurisdiction. 3. Where weak-axis bending of cold-formed steel supports is relied on for the seismic load path, such supports shall be speciﬁ cally evaluated. 4. Components mounted on vibration isolators shall have a bumper restraint or snubber in each hori-zontal direction, and vertical restraints shall be provided where required to resist overturning. Isolator housings and restraints shall be constructed of ductile materials. (See additional design force requirements in footnote b to Table 13.6-1.) A viscoelastic pad or similar material of appropriate thickness shall be used between the bumper and components to limit the impact load. 5. Where post-installed mechanical anchors are used for non-vibration isolated mechanical equipment rated over 10 hp (7.45 kW), they shall be qualiﬁ ed in accordance with ACI 355.2. 6. For piping, boilers, and pressure vessels, attachments to concrete shall be suitable for cyclic loads. 7. For mechanical equipment, drilled and grouted-in-place anchors for tensile load applications shall use either expansive cement or expansive epoxy grout. 13.6.5.6 Conduit, Cable Tray, and Other Electrical Distribution Systems (Raceways) Raceways shall be designed for seismic forces and seismic relative displacements as required in Section 13.3. Conduit greater than 2.5 in. (64 mm) trade size and attached to panels, cabinets, or other equipment subject to seismic relative displacement, Dp, shall be provided with ﬂ exible connections or designed for seismic forces and seismic relative displacements as required in Section 13.3. EXCEPTIONS: 1. Design for the seismic forces and relative displace-ments of Section 13.3 shall not be required for raceways where either: a. Trapeze assemblies are used to support race-ways and the total weight of the raceway supported by trapeze assemblies is less than 10 lb/ft (146 N/m), or b. The raceway is supported by hangers and each hanger in the raceway run is 12 in. (305 mm) or less in length from the raceway support point to the supporting structure. Where rod hangers are used, they shall be equipped with swivels to prevent inelastic bending in the rod. 2. Design for the seismic forces and relative displace-ments of Section 13.3 shall not be required for conduit, regardless of the value of Ip, where the conduit is less than 2.5 in. (64 mm) trade size. 13.6.6 Utility and Service Lines At the interface of adjacent structures or portions of the same structure that may move independently, utility lines shall be provided with adequate ﬂ exibility to accommodate the anticipated differential movement between the portions that move independently. Differential displacement calculations shall be determined in accordance with Section 13.3.2. The possible interruption of utility service shall be considered in relation to designated seismic systems in Risk Category IV as deﬁ ned in Table 1.5-1. Speciﬁ c attention shall be given to the vulnerability of underground utilities and utility interfaces between the structure and the ground where Site Class E or F soil is present, and where the seismic coefﬁ cient SDS at the underground utility or at the base of the structure is equal to or greater than 0.33. 13.6.7 Ductwork HVAC and other ductwork shall be designed for seismic forces and seismic relative displacements as required in Section 13.3. Design for the displacements across seismic joints shall be required for ductwork with Ip = 1.5 without consideration of the exceptions below. EXCEPTIONS: The following exceptions pertain to ductwork not designed to carry toxic, highly toxic, or ﬂ ammable gases or used for smoke control: 1. Design for the seismic forces and relative displace-ments of Section 13.3 shall not be required for ductwork where either: a. Trapeze assemblies are used to support duct-work and the total weight of the ductwork supported by trapeze assemblies is less than 10 lb/ft (146 N/m); or b. The ductwork is supported by hangers and each hanger in the duct run is 12 in. (305 mm) or less in length from the duct support point to the supporting structure. Where rod hangers are used, they shall be equipped with swivels to prevent inelastic bending in the rod. 2. Design for the seismic forces and relative displace-ments of Section 13.3 shall not be required where provisions are made to avoid impact with larger CHAPTER 13 SEISMIC DESIGN REQUIREMENTS FOR NONSTRUCTURAL COMPONENTS 124 ducts or mechanical components or to protect the ducts in the event of such impact; and HVAC ducts have a cross-sectional area of less than 6 ft2 (0.557 m2), or weigh 17 lb/ft (248 N/m) or less. HVAC duct systems fabricated and installed in accordance with standards approved by the authority having jurisdiction shall be deemed to meet the lateral bracing requirements of this section. Components that are installed in-line with the duct system and have an operating weight greater than 75 lb (334 N), such as fans, heat exchangers, and humidiﬁ ers, shall be supported and laterally braced independent of the duct system and such braces shall meet the force requirements of Section 13.3.1. Appurtenances such as dampers, louvers, and diffus-ers shall be positively attached with mechanical fasteners. Unbraced piping attached to in-line equip-ment shall be provided with adequate ﬂ exibility to accommodate the seismic relative displacements of Section 13.3.2. 13.6.8 Piping Systems Unless otherwise noted in this section, piping systems shall be designed for the seismic forces and seismic relative displacements of Section 13.3. ASME pressure piping systems shall satisfy the requirements of Section 13.6.8.1. Fire protection sprinkler piping shall satisfy the requirements of Section 13.6.8.2. Elevator system piping shall satisfy the requirements of Section 13.6.10. Where other applicable material standards or recognized design bases are not used, piping design including consideration of service loads shall be based on the following allowable stresses: a. For piping constructed with ductile materials (e.g., steel, aluminum, or copper), 90 percent of the minimum speciﬁ ed yield strength. b. For threaded connections in piping constructed with ductile materials, 70 percent of the minimum speciﬁ ed yield strength. c. For piping constructed with nonductile materials (e.g., cast iron or ceramics), 10 percent of the material minimum speciﬁ ed tensile strength. d. For threaded connections in piping constructed with nonductile materials, 8 percent of the material minimum speciﬁ ed tensile strength. Piping not detailed to accommodate the seismic relative displacements at connections to other compo-nents shall be provided with connections having sufﬁ cient ﬂ exibility to avoid failure of the connection between the components. 13.6.8.1 ASME Pressure Piping Systems Pressure piping systems, including their supports, designed and constructed in accordance with ASME B31 shall be deemed to meet the force, displacement, and other requirements of this section. In lieu of speciﬁ c force and displacement requirements provided in ASME B31, the force and displacement require-ments of Section 13.3 shall be used. Materials meeting the toughness requirements of ASME B31 shall be considered high-deformability materials. 13.6.8.2 Fire Protection Sprinkler Piping Systems Fire protection sprinkler piping, pipe hangers, and bracing designed and constructed in accordance with NFPA 13 shall be deemed to meet the force and displacement requirements of this section. The exceptions of Section 13.6.8.3 shall not apply. 13.6.8.3 Exceptions Design of piping systems and attachments for the seismic forces and relative displacements of Section 13.3 shall not be required where one of the following conditions apply: 1. Trapeze assemblies are used to support piping whereby no single pipe exceeds the limits set forth in 3a, 3b, or 3c below and the total weight of the piping supported by the trapeze assemblies is less than 10 lb/ft (146 N/m). 2. The piping is supported by hangers and each hanger in the piping run is 12 in. (305 mm) or less in length from the top of the pipe to the supporting structure. Where pipes are supported on a trapeze, the trapeze shall be supported by hangers having a length of 12 in. (305 mm) or less. Where rod hangers are used, they shall be equipped with swivels, eye nuts, or other devices to prevent bending in the rod. 3. Piping having an Rp in Table 13.6-1 of 4.5 or greater is used and provisions are made to avoid impact with other structural or nonstructural components or to protect the piping in the event of such impact and where the following size require-ments are satisﬁ ed: a. For Seismic Design Category C where Ip is greater than 1.0, the nominal pipe size shall be 2 in. (50 mm) or less. b. For Seismic Design Categories D, E, or F and values of Ip are greater than 1.0, the nominal pipe size shall be 1 in. (25 mm) or less. c. For Seismic Design Categories D, E, or F where Ip = 1.0, the nominal pipe size shall be 3 in. (80 mm) or less. MINIMUM DESIGN LOADS 125 13.6.9 Boilers and Pressure Vessels Boilers or pressure vessels designed and con-structed in accordance with ASME BPVC shall be deemed to meet the force, displacement, and other requirements of this section. In lieu of the speciﬁ c force and displacement requirements provided in the ASME BPVC, the force and displacement requirements of Sections 13.3.1 and 13.3.2 shall be used. Materials meeting the toughness requirements of ASME BPVC shall be considered high-deformability materials. Other boilers and pressure vessels designated as having an Ip = 1.5, but not designed and constructed in accordance with the requirements of ASME BPVC, shall comply with the requirements of Section 13.6.11. 13.6.10 Elevator and Escalator Design Requirements Elevators and escalators designed in accordance with the seismic requirements of ASME A17.1 shall be deemed to meet the seismic force requirements of this section, except as modiﬁ ed in the following text. The exceptions of Section 13.6.8.3 shall not apply to elevator piping. 13.6.10.1 Escalators, Elevators, and Hoistway Structural System Escalators, elevators, and hoistway structural systems shall be designed to meet the force and dis-placement requirements of Sections 13.3.1 and 13.3.2. 13.6.10.2 Elevator Equipment and Controller Supports and Attachments Elevator equipment and controller supports and attachments shall be designed to meet the force and displacement requirements of Sections 13.3.1 and 13.3.2. 13.6.10.3 Seismic Controls for Elevators Elevators operating with a speed of 150 ft/min (46 m/min) or greater shall be provided with seismic switches. Seismic switches shall provide an electric signal indicating that structural motions are of such a magnitude that the operation of the elevators may be impaired. Seismic switches in accordance with Section 8.4.10.1.2 of ASME A17.1 shall be deemed to meet the requirements of this section. EXCEPTION: In cases where seismic switches cannot be located near a column in accordance with ASME A17.1, they shall have two horizontal axes of sensitivity and have a trigger level set to 20 percent of the acceleration of gravity where located at or near the base of the structure and 50 percent of the acceleration of gravity in all other locations. Upon activation of the seismic switch, elevator operations shall conform to requirements of ASME A17.1, except as noted in the following text. In facilities where the loss of the use of an elevator is a life-safety issue, the elevator shall only be used after the seismic switch has triggered provided that: 1. The elevator shall operate no faster than the service speed. 2. Before the elevator is occupied, it is operated from top to bottom and back to top to verify that it is operable. 13.6.10.4 Retainer Plates Retainer plates are required at the top and bottom of the car and counterweight. 13.6.11 Other Mechanical and Electrical Components Mechanical and electrical components, including conveyor systems, not designed and constructed in accordance with the reference documents in Chapter 23 shall meet the following: 1. Components, their supports and attachments shall comply with the requirements of Sections 13.4, 13.6.3, 13.6.4, and 13.6.5. 2. For mechanical components with hazardous substances and assigned a component importance factor, Ip, of 1.5 in accordance with Section 13.1.3, and for boilers and pressure vessels not designed in accordance with ASME BPVC, the design strength for seismic loads in combination with other service loads and appropriate environmental effects shall be based on the following material properties: a. For mechanical components constructed with ductile materials (e.g., steel, aluminum, or copper), 90 percent of the minimum speciﬁ ed yield strength. b. For threaded connections in components constructed with ductile materials, 70 percent of the minimum speciﬁ ed yield strength. c. For mechanical components constructed with nonductile materials (e.g., plastic, cast iron, or ceramics), 10 percent of the material minimum speciﬁ ed tensile strength. d. For threaded connections in components constructed with nonductile materials, 8 percent of the material minimum speciﬁ ed tensile strength. 127 Chapter 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 14.1.2.2 Seismic Requirements for Structural Steel Structures The design of structural steel structures to resist seismic forces shall be in accordance with the provi-sions of Section 14.1.2.2.1 or 14.1.2.2.2, as applicable. 14.1.2.2.1 Seismic Design Categories B and C Structural steel structures assigned to Seismic Design Category B or C shall be of any construction permit-ted by the applicable reference documents in Section 14.1.1. Where a response modiﬁ cation coefﬁ cient, R, in accordance with Table 12.2-1 is used for the design of structural steel structures assigned to Seismic Design Category B or C, the structures shall be designed and detailed in accordance with the require-ments of AISC 341. EXCEPTION: The response modiﬁ cation coefﬁ cient, R, designated for “Steel systems not speciﬁ cally detailed for seismic resistance, excluding cantilever column systems” in Table 12.2-1 shall be permitted for systems designed and detailed in accordance with AISC 360 and need not be designed and detailed in accordance with AISC 341. 14.1.2.2.2 Seismic Design Categories D through F Structural steel structures assigned to Seismic Design Category D, E, or F shall be designed and detailed in accordance with AISC 341, except as permitted in Table 15.4-1. 14.1.3 Cold-Formed Steel 14.1.3.1 General The design of cold-formed carbon or low-alloy steel structural members shall be in accordance with the requirements of AISI S100 and the design of cold-formed stainless steel structural members shall be in accordance with the requirements of ASCE 8. Where required, the seismic design of cold-formed steel structures shall be in accordance with the additional provisions of Section 14.1.3.2. 14.1.3.2 Seismic Requirements for Cold-Formed Steel Structures Where a response modiﬁ cation coefﬁ cient, R, in accordance with Table 12.2-1 is used for the design of 14.0 SCOPE Structural elements including foundation elements shall conform to the material design and detailing requirements set forth in this chapter or as otherwise speciﬁ ed for non-building structures in Tables 15.4-1 and 15.4-2. 14.1 STEEL Structures, including foundations, constructed of steel to resist seismic loads shall be designed and detailed in accordance with this standard including the reference documents and additional requirements provided in this section. 14.1.1 Reference Documents The design, construction, and quality of steel members that resist seismic forces shall conform to the applicable requirements, as amended herein, of the following: 1. AISC 360 2. AISC 341 3. AISI S100 4. AISI S110 5. AISI S230 6. AISI S213 7. ASCE 19 8. ASCE 8 9. SJI-K-1.1 10. SJI-LH/DLH-1.1 11. SJI-JG-1.1 12. SJI-CJ-1.0 14.1.2 Structural Steel 14.1.2.1 General The design of structural steel for buildings and structures shall be in accordance with AISC 360. Where required, the seismic design of structural steel structures shall be in accordance with the additional provisions of Section 14.1.2.2. CHAPTER 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 128 cold-formed steel structures, the structures shall be designed and detailed in accordance with the require-ments of AISI S100, ASCE 8, and AISI S110 as modiﬁ ed in Section 14.1.3.3. 14.1.3.3 Modiﬁ cations to AISI S110 The text of AISI S110 shall be modiﬁ ed as indicated in Sections 14.1.3.3.1 through 14.1.3.3.5. Italics are used for text within Sections 14.1.3.3.1 through 14.1.3.3.5 to indicate requirements that differ from AISI S110. 14.1.3.3.1 AISI S110, Section D1 Modify Section D1 to read as follows: D1 Cold-Formed Steel Special Bolted Moment Frames (CFS-SBMF) Cold-formed steel–special bolted moment frame (CFS-SBMF) systems shall withstand signiﬁ cant inelastic deformations through friction and bearing at their bolted connections. Beams, columns, and connections shall satisfy the requirements in this section. CFS-SBMF systems shall be limited to one-story structures, no greater than 35 feet in height, without column splices and satisfying the requirements in this section. The CFS-SBMF shall engage all columns supporting the roof or ﬂ oor above. The single size beam and single size column with the same bolted moment connection detail shall be used for each frame. The frame shall be supported on a level ﬂ oor or foundation. 14.1.3.3.2 AISI S110, Section D1.1.1 Modify Section D1.1.1 to read as follows: D1.1.1 Connection Limitations Beam-to-column connections in CFS-SBMF systems shall be bolted connections with snug-tight high-strength bolts. The bolt spacing and edge distance shall be in accordance with the limits of AISI S100, Section E3. The 8-bolt conﬁ guration shown in Table D1-1 shall be used. The faying surfaces of the beam and column in the bolted moment connection region shall be free of lubricants or debris. 14.1.3.3.3 AISI S110, Section D1.2.1 Modify Section D1.2.1 and add new Section D1.2.1.1 to read as follows: D1.2.1 Beam Limitations In addition to the requirements of Section D1.2.3, beams in CFS-SBMF systems shall be ASTM A653 galvanized 55 ksi (374 MPa) yield stress cold-formed steel C-section members with lips, and designed in accordance with Chapter C of AISI S100. The beams shall have a minimum design thickness of 0.105 in. (2.67 mm). The beam depth shall be not less than 12 in. (305 mm) or greater than 20 in. (508 mm). The ﬂ at depth-to-thickness ratio of the web shall not exceed 6.18 E Fy / . D1.2.1.1 Single-Channel Beam Limitations When single-channel beams are used, torsional effects shall be accounted for in the design. 14.1.3.3.4 AISI S110, Section D1.2.2 Modify Section D1.2.2 to read as follows: D1.2.2 Column Limitations In addition to the requirements of D1.2.3, columns in CFS-SBMF systems shall be ASTM A500 Grade B cold-formed steel hollow structural section (HSS) members painted with a standard industrial ﬁ nished surface, and designed in accordance with Chapter C of AISI S100. The column depth shall be not less than 8 in. (203 mm) or greater than 12 in. (305 mm). The ﬂ at depth-to-thickness ratio shall not exceed 1.40 E Fy / . 14.1.3.3.5 AISI S110, Section D1.3 Delete text in Section D1.3 to read as follows: D1.3 Design Story Drift Where the applicable building code does not contain design coefﬁ cients for CSF-SBMF systems, the provisions of Appendix 1 shall apply. For structures having a period less than TS, as deﬁ ned in the applicable building code, alternate methods of computing Δ shall be permitted, provided such alternate methods are acceptable to the authority having jurisdiction. 14.1.4 Cold-Formed Steel Light-Frame Construction 14.1.4.1 General Cold-formed steel light-frame construction shall be designed in accordance with AISI S100, Section D4. Where required, the seismic design of cold-formed steel light-frame construction shall be in accordance with the additional provisions of Section 14.1.4.2. 14.1.4.2 Seismic Requirements for Cold-Formed Steel Light-Frame Construction Where a response modiﬁ cation coefﬁ cient, R, in accordance with Table 12.2-1 is used for the design of cold-formed steel light-frame construction, the MINIMUM DESIGN LOADS 129 structures shall be designed and detailed in accor-dance with the requirements of AISI S213. 14.1.4.3 Prescriptive Cold-Formed Steel Light-Frame Construction Cold-formed steel light-frame construction for one- and two-family dwellings is permitted to be designed and constructed in accordance with the requirements of AISI S230 subject to the limitations therein. 14.1.5 Steel Deck Diaphragms Steel deck diaphragms shall be made from materials conforming to the requirements of AISI S100 or ASCE 8. Nominal strengths shall be deter-mined in accordance with approved analytical procedures or with test procedures prepared by a registered design professional experienced in testing of cold-formed steel assemblies and approved by the authority having jurisdiction. The required strength of diaphragms, including bracing members that form part of the diaphragm, shall be determined in accordance with Section 12.10.1. The steel deck installation for the building, including fasteners, shall comply with the test assembly arrangement. Quality standards established for the nominal strength test shall be the minimum standards required for the steel deck installation, including fasteners. 14.1.6 Steel Cables The design strength of steel cables shall be determined by the requirements of ASCE 19 except as modiﬁ ed by this chapter. ASCE 19, Section 3.1.2(d), shall be modiﬁ ed by substituting 1.5(T4) where T4 is the net tension in cable due to dead load, prestress, live load, and seismic load. A load factor of 1.1 shall be applied to the prestress force to be added to the load combination of Section 3.1.2 of ASCE 19. 14.1.7 Additional Detailing Requirements for Steel Piles in Seismic Design Categories D through F In addition to the foundation requirements set forth in Sections 12.1.5 and 12.13, design and detailing of H-piles shall conform to the requirements of AISC 341, and the connection between the pile cap and steel piles or unﬁ lled steel pipe piles in structures assigned to Seismic Design Category D, E, or F shall be designed for a tensile force not less than 10 percent of the pile compression capacity. EXCEPTION: Connection tensile capacity need not exceed the strength required to resist seismic load effects including overstrength factor of Section 12.4.3.2 or Section 12.14.2.2.2. Connections need not be provided where the foundation or supported structure does not rely on the tensile capacity of the piles for stability under the design seismic forces. 14.2 CONCRETE Structures, including foundations, constructed of concrete to resist seismic loads shall be designed and detailed in accordance with this standard including the reference documents and additional requirements provided in this section. 14.2.1 Reference Documents The quality and testing of concrete materials and the design and construction of structural concrete members that resist seismic forces shall conform to the requirements of ACI 318, except as modiﬁ ed in Section 14.2.2. 14.2.2 Modiﬁ cations to ACI 318 The text of ACI 318 shall be modiﬁ ed as indi-cated in Sections 14.2.2.1 through 14.2.2.9. Italics are used for text within Sections 14.2.2.1 through 14.2.2.9 to indicate requirements that differ from ACI 318. 14.2.2.1 Deﬁ nitions Add the following deﬁ nitions to Section 2.2. DETAILED PLAIN CONCRETE STRUCTURAL WALL: A wall complying with the requirements of Chapter 22. ORDINARY PRECAST STRUCTURAL WALL: A precast wall complying with the requirements of Chapters 1 through 18. WALL PIER: A wall segment with a horizontal length-to-thickness ratio of at least 2.5, but not exceeding 6, whose clear height is at least two times its horizontal length. 14.2.2.2 ACI 318, Section 7.10 Modify Section 7.10 by revising Section 7.10.5.6 to read as follows: 7.10.5.6 Where anchor bolts are placed in the top of columns or pedestals, the bolts shall be enclosed by lateral reinforcement that also surrounds at least four vertical bars of the column or pedestal. The lateral reinforcement shall be distributed within 5 in. of the top of the column or pedestal, and shall consist of at least two No. 4 or three No. 3 bars. In structures assigned to Seismic Design Categories C, D, E, or F, the ties shall have a hook on each free end that complies with 7.1.4. CHAPTER 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 130 14.2.2.3 Scope Modify Section 21.1.1.3 to read as follows: 21.1.1.3 All members shall satisfy requirements of Chapters 1 to 19 and 22. Structures assigned to SDC B, C, D, E, or F also shall satisfy 21.1.1.4 through 21.1.1.8, as applicable, except as modiﬁ ed by the requirements of Chapters 14 and 15 of this standard. 14.2.2.4 Intermediate Precast Structural Walls Modify Section 21.4 by renumbering Section 21.4.3 to Section 21.4.4 and adding new Sections 21.4.3, 21.4.5, and 21.4.6 to read as follows: 21.4.3 Connections that are designed to yield shall be capable of maintaining 80 percent of their design strength at the deformation induced by design displacement, or shall use type 2 mechanical splices. 21.4.4 Elements of the connection that are not designed to yield shall develop at least 1.5 Sy. 21.4.5 Wall piers in structures assigned to SDC D, E, or F shall comply with Section 14.2.2.4 of this standard. 21.4.6 Wall piers not designed as part of a moment frame in SDC C shall have transverse reinforcement designed to resist the shear forces determined from Section 21.3.3. Spacing of transverse reinforcement shall not exceed 8 in. Transverse reinforcement shall be extended beyond the pier clear height for at least 12 in. EXCEPTIONS: The preceding requirement need not apply in the following situations: 1. Wall piers that satisfy Section 21,13. 2. Wall piers along a wall line within a story where other shear wall segments provide lateral support to the wall piers and such segments have a total stiffness of at least six times the sum of the stiffnesses of all the wall piers. Wall segments with a horizontal length-to-thickness ratio less than 2.5 shall be designed as columns. 14.2.2.5 Wall Piers and Wall Segments Modify Section 21.9 by adding a new Section 21.9.10 to read as follows: 21.9.10 Wall Piers and Wall Segments. 21.9.10.1 Wall piers not designed as a part of a special moment-resisting frame shall have transverse reinforcement designed to satisfy the requirements in Section 21.9.10.2. EXCEPTIONS: 1. Wall piers that satisfy Section 21.13. 2. Wall piers along a wall line within a story where other shear wall segments provide lateral support to the wall piers, and such segments have a total stiffness of at least six times the sum of the in-plane stiffnesses of all the wall piers. 21.9.10.2 Transverse reinforcement with seismic hooks at both ends shall be designed to resist the shear forces determined from Section 21.6.5.1. Spacing of transverse reinforcement shall not exceed 6 in. (152 mm). Transverse reinforcement shall be extended beyond the pier clear height for at least 12 in. (304 mm). 21.9.10.3 Wall segments with a horizontal length-to-thickness ratio less than 2.5 shall be designed as columns. 14.2.2.6 Special Precast Structural Walls Modify Section 21.10.2 to read as follows: 21.10.2 Special structural walls constructed using precast concrete shall satisfy all requirements of Section 21.9 in addition to Section 21.4 as modiﬁ ed by Section 14.2.2. 14.2.2.7 Foundations Modify Section 21.12.1.1 to read as follows: 21.12.1.1 Foundations resisting earthquake-induced forces or transferring earthquake-induced forces between structure and ground in structures assigned to SDC D, E, or F shall comply with requirements of Section 21.12 and other applicable code provisions unless modiﬁ ed by Sections 12.1.5, 12.13, or 14.2 of ASCE 7. 14.2.2.8 Detailed Plain Concrete Shear Walls Modify Section 22.6 by adding a new Section 22.6.7 to read 22.6.7 Detailed Plain Concrete Shear Walls. 22.6.7.1 Detailed plain concrete shear walls are walls conforming to the requirements for ordinary plain concrete shear walls and Section 22.6.7.2. 22.6.7.2 Reinforcement shall be provided as follows: a. Vertical reinforcement of at least 0.20 in.2 (129 mm2) in cross-sectional area shall be provided continuously from support to support at each corner, at each side of each opening, and at the ends of walls. The continuous vertical bar required beside an opening is permitted to substitute for the No. 5 bar required by Section 22.6.6.5. b. Horizontal reinforcement at least 0.20 in.2 (129 mm2) in cross-sectional area shall be provided: 1. Continuously at structurally connected roof and ﬂ oor levels and at the top of walls. MINIMUM DESIGN LOADS 131 2. At the bottom of load-bearing walls or in the top of foundations where doweled to the wall. 3. At a maximum spacing of 120 in. (3,048 mm). Reinforcement at the top and bottom of openings, where used in determining the maximum spacing speciﬁ ed in Item 3 in the preceding text, shall be continuous in the wall. 14.2.2.9 Strength Requirements for Anchors Modify Section D.4 by adding a new exception at the end of Section D.4.2.2 to read as follows: EXCEPTION: If Nb is determined using Eq. D-7, the concrete breakout strength of Section D.4.2 shall be considered satisﬁ ed by the design procedure of Sections D.5.2 and D.6.2 without the need for testing regardless of anchor bolt diameter and tensile embedment. 14.2.3 Additional Detailing Requirements for Concrete Piles In addition to the foundation requirements set forth in Sections 12.1.5 and 12.13 of this standard and in Section 21.12 of ACI 318, design, detailing, and construction of concrete piles shall conform to the requirements of this section. 14.2.3.1 Concrete Pile Requirements for Seismic Design Category C Concrete piles in structures assigned to Seismic Design Category C shall comply with the require-ments of this section. 14.2.3.1.1 Anchorage of Piles All concrete piles and concrete-ﬁ lled pipe piles shall be connected to the pile cap by embedding the pile reinforcement in the pile cap for a distance equal to the development length as speciﬁ ed in ACI 318 as modiﬁ ed by Section 14.2.2 of this standard or by the use of ﬁ eld-placed dowels anchored in the concrete pile. For deformed bars, the development length is the full development length for compression or tension, in the case of uplift, without reduction in length for excess area. Hoops, spirals, and ties shall be terminated with seismic hooks as deﬁ ned in Section 2.2 of ACI 318. Where a minimum length for reinforcement or the extent of closely spaced conﬁ nement reinforce-ment is speciﬁ ed at the top of the pile, provisions shall be made so that those speciﬁ ed lengths or extents are maintained after pile cutoff. 14.2.3.1.2 Reinforcement for Uncased Concrete Piles (SDC C) Reinforcement shall be provided where required by analysis. For uncased cast-in-place drilled or augered concrete piles, a minimum of four longitu-dinal bars, with a minimum longitudinal reinforce-ment ratio of 0.0025, and transverse reinforcement, as deﬁ ned below, shall be provided throughout the minimum reinforced length of the pile as deﬁ ned below starting at the top of the pile. The longitudinal reinforcement shall extend beyond the minimum reinforced length of the pile by the tension develop-ment length. Transverse reinforcement shall consist of closed ties (or equivalent spirals) with a minimum 3/8 in. (9 mm) diameter. Spacing of transverse reinforcing shall not exceed 6 in. (150 mm) or 8 longitudinal-bar diameters within a distance of three times the pile diameter from the bottom of the pile cap. Spacing of transverse reinforcing shall not exceed 16 longitudi-nal-bar diameters throughout the remainder of the minimum reinforced length. The minimum reinforced length of the pile shall be taken as the greater of 1. One-third of the pile length. 2. A distance of 10 ft (3 m). 3. Three times the pile diameter. 4. The ﬂ exural length of the pile, which shall be taken as the length from the bottom of the pile cap to a point where the concrete section cracking moment multiplied by a resistance factor 0.4 exceeds the required factored moment at that point. 14.2.3.1.3 Reinforcement for Metal-Cased Concrete Piles (SDC C) Reinforcement requirements are the same as for uncased concrete piles. EXCEPTION: Spiral-welded metal casing of a thickness not less than No. 14 gauge can be considered as providing concrete conﬁ nement equivalent to the closed ties or equivalent spirals required in an uncased concrete pile, provided that the metal casing is adequately protected against possible deleterious action due to soil constituents, changing water levels, or other factors indicated by boring records of site conditions. 14.2.3.1.4 Reinforcement for Concrete-Filled Pipe Piles (SDC C) Minimum reinforcement 0.01 times the cross-sectional area of the pile concrete shall be provided in the top of the pile with a length equal to two times the required cap embedment anchorage into the pile cap. 14.2.3.1.5 Reinforcement for Precast Nonprestressed Concrete Piles (SDC C) A minimum longitudinal steel reinforcement ratio of 0.01 shall be provided for CHAPTER 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 132 precast nonprestressed concrete piles. The longitudinal reinforcing shall be conﬁ ned with closed ties or equivalent spirals of a minimum 3/8 in. (10 mm) diameter. Transverse conﬁ nement reinforcing shall be provided at a maximum spacing of eight times the diameter of the smallest longitudinal bar, but not to exceed 6 in. (152 mm), within three pile diameters of the bottom of the pile cap. Outside of the conﬁ nement region, closed ties or equivalent spirals shall be provided at a 16 longitudinal-bar-diameter maximum spacing, but not greater than 8 in. (200 mm). Rein-forcement shall be full length. 14.2.3.1.6 Reinforcement for Precast Prestressed Piles (SDC C) For the upper 20 ft (6 m) of precast pre-stressed piles, the minimum volumetric ratio of spiral reinforcement shall not be less than 0.007 or the amount required by the following equation: ρs c yh f f = ′ 0 12 . (14.2-1) where ρs = volumetric ratio (vol. spiral/vol. core) fc′ = speciﬁ ed compressive strength of concrete, psi (MPa) fyh = speciﬁ ed yield strength of spiral reinforcement, which shall not be taken greater than 85,000 psi (586 MPa) A minimum of one-half of the volumetric ratio of spiral reinforcement required by Eq. 14.2-1 shall be provided for the remaining length of the pile. 14.2.3.2 Concrete Pile Requirements for Seismic Design Categories D through F Concrete piles in structures assigned to Seismic Design Category D, E, or F shall comply with the requirements of this section. 14.2.3.2.1 Site Class E or F Soil Where concrete piles are used in Site Class E or F, they shall have trans-verse reinforcement in accordance with Sections 21.6.4.2 through 21.6.4.4 of ACI 318 within seven pile diameters of the pile cap and of the interfaces between strata that are hard or stiff and strata that are liqueﬁ able or are composed of soft to medium stiff clay. 14.2.3.2.2 Nonapplicable ACI 318 Sections for Grade Beam and Piles Section 21.12.3.3 of ACI 318 need not apply to grade beams designed to resist the seismic load effects including overstrength factor of Section 12.4.3 or 12.14.3.2. Section 21.12.4.4(a) of ACI 318 need not apply to concrete piles. Section 21.12.4.4(b) of ACI 318 need not apply to precast, prestressed concrete piles. 14.2.3.2.3 Reinforcement for Uncased Concrete Piles (SDC D through F) Reinforcement shall be provided where required by analysis. For uncased cast-in-place drilled or augered concrete piles, a minimum of four longitudinal bars with a minimum longitudinal reinforcement ratio of 0.005 and transverse conﬁ ne-ment reinforcement in accordance with Sections 21.6.4.2 through 21.6.4.4 of ACI 318 shall be pro-vided throughout the minimum reinforced length of the pile as deﬁ ned below starting at the top of the pile. The longitudinal reinforcement shall extend beyond the minimum reinforced length of the pile by the tension development length. The minimum reinforced length of the pile shall be taken as the greater of 1. One-half of the pile length. 2. A distance of 10 ft (3 m). 3. Three times the pile diameter. 4. The ﬂ exural length of the pile, which shall be taken as the length from the bottom of the pile cap to a point where the concrete section cracking moment multiplied by a resistance factor 0.4 exceeds the required factored moment at that point. In addition, for piles located in Site Classes E or F, longitudinal reinforcement and transverse conﬁ ne-ment reinforcement, as described above, shall extend the full length of the pile. Where transverse reinforcing is required, trans-verse reinforcing ties shall be a minimum of No. 3 bars for up to 20-in.-diameter (500 mm) piles and No. 4 bars for piles of larger diameter. In Site Classes A through D, longitudinal reinforcement and transverse conﬁ nement reinforce-ment, as deﬁ ned above, shall also extend a minimum of seven times the pile diameter above and below the interfaces of soft to medium stiff clay or liqueﬁ able strata except that transverse reinforcing not located within the minimum reinforced length shall be permitted to use a transverse spiral reinforcement ratio of not less than one-half of that required in Section 21.6.4.4(a) of ACI 318. Spacing of transverse rein-forcing not located within the minimum reinforced length is permitted to be increased, but shall not exceed the least of the following: 1. 12 longitudinal bar diameters. 2. One-half the pile diameter. 3. 12 in. (300 mm). MINIMUM DESIGN LOADS 133 14.2.3.2.4 Reinforcement for Metal-Cased Concrete Piles (SDC D through F) Reinforcement requirements are the same as for uncased concrete piles. EXCEPTION: Spiral-welded metal casing of a thickness not less than No. 14 gauge can be considered as providing concrete conﬁ nement equivalent to the closed ties or equivalent spirals required in an uncased concrete pile, provided that the metal casing is adequately protected against possible deleterious action due to soil constituents, changing water levels, or other factors indicated by boring records of site conditions. 14.2.3.2.5 Reinforcement for Precast Concrete Piles (SDC D through F) Transverse conﬁ nement reinforce-ment consisting of closed ties or equivalent spirals shall be provided in accordance with Sections 21.6.4.2 through 21.6.4.4 of ACI 318 for the full length of the pile. EXCEPTION: In other than Site Classes E or F, the speciﬁ ed transverse conﬁ nement reinforcement shall be provided within three pile diameters below the bottom of the pile cap, but it is permitted to use a transverse reinforcing ratio of not less than one-half of that required in Section 21.6.4.4(a) of ACI 318 throughout the remainder of the pile length. 14.2.3.2.6 Reinforcement for Precast Prestressed Piles (SDC D through F) In addition to the requirements for Seismic Design Category C, the following requirements shall be met: 1. Requirements of ACI 318, Chapter 21, need not apply. 2. Where the total pile length in the soil is 35 ft (10,668 mm) or less, the ductile pile region shall be taken as the entire length of the pile. Where the pile length exceeds 35 ft (10,668 mm), the ductile pile region shall be taken as the greater of 35 ft (10,668 mm) or the distance from the underside of the pile cap to the point of zero curvature plus three times the least pile dimension. 3. In the ductile pile region, the center to center spacing of the spirals or hoop reinforcement shall not exceed one-ﬁ fth of the least pile dimension, six times the diameter of the longitudinal strand, or 8 in. (203 mm), whichever is smaller. 4. Spiral reinforcement shall be spliced by lapping one full turn, by welding, or by the use of a mechanical connector. Where spiral reinforcement is lap spliced, the ends of the spiral shall terminate in a seismic hook in accordance with ACI 318, except that the bend shall be not less than 135°. Welded splices and mechanical connectors shall comply with Section 12.14.3 of ACI 318. 5. Where the transverse reinforcement consists of spirals or circular hoops, the volumetric ratio of spiral transverse reinforcement in the ductile pile region shall comply with ρs c yh g ch c g f f A A P f A = ′ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ′ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0 25 1 0 0 5 1 4 . . . . but not less than ρs c yh c g f f P f A = ′ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ′ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0 12 0 5 1 4 . . . and ρs need not exceed 0.021 where ρs = volumetric ratio (vol. of spiral/vol. of core) fc′ ≤ 6,000 psi (41.4 MPa) fyh = yield strength of spiral reinforcement ≤ 85 ksi (586 MPa) Ag = pile cross-sectional area, in.2 (mm2) Ach = core area deﬁ ned by spiral outside diameter, in.2 (mm2) P = axial load on pile resulting from the load combination 1.2D + 0.5L + 1.0E, lb (kN) This required amount of spiral reinforcement is permitted to be obtained by providing an inner and outer spiral. 6. Where transverse reinforcement consists of rectangular hoops and cross ties, the total cross-sectional area of lateral transverse reinforcement in the ductile region with spacing, s, and perpendicu-lar to dimension, hc, shall conform to A sh f f A A P f A sh c c yh g ch c g = ′ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ′ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0 3 1 0 0 5 1 4 . . . . but not less than A sh f f P f A sh c c yh c g = ′ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ′ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0 12 0 5 1 4 . . . where s = spacing of transverse reinforcement measured along length of pile, in. (mm) hc = cross-sectional dimension of pile core mea-sured center to center of hoop reinforcement, in. (mm) fyh ≤ 70 ksi (483 MPa) The hoops and cross ties shall be equivalent to deformed bars not less than No. 3 in size. Rectan-gular hoop ends shall terminate at a corner with seismic hooks. CHAPTER 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 134 7. Outside of the ductile pile region, the spiral or hoop reinforcement with a volumetric ratio not less than one-half of that required for transverse conﬁ nement reinforcement shall be provided. 14.3 COMPOSITE STEEL AND CONCRETE STRUCTURES Structures, including foundations, constructed of composite steel and concrete to resist seismic loads shall be designed and detailed in accordance with this standard, including the reference documents and additional requirements provided in this section. 14.3.1 Reference Documents The design, construction, and quality of compos-ite steel and concrete members that resist seismic forces shall conform to the applicable requirements of the following: 1. AISC 341 2. AISC 360 3. ACI 318, excluding Chapter 22 14.3.2 General Systems of structural steel acting compositely with reinforced concrete shall be designed in accor-dance with AISC 360 and ACI 318, excluding Chapter 22. Where required, the seismic design of composite steel and concrete systems shall be in accordance with the additional provisions of Section 14.3.3. 14.3.3 Seismic Requirements for Composite Steel and Concrete Structures Where a response modiﬁ cation coefﬁ cient, R, in accordance with Table 12.2-1 is used for the design of systems of structural steel acting compositely with reinforced concrete, the structures shall be designed and detailed in accordance with the requirements of AISC 341. 14.3.4 Metal-Cased Concrete Piles Metal-cased concrete piles shall be designed and detailed in accordance with Section 14.2.3.2.4. 14.4 MASONRY Structures, including foundations, constructed of masonry to resist seismic loads shall be designed and detailed in accordance with this standard, including the references and additional requirements provided in this section. 14.4.1 Reference Documents The design, construction, and quality assurance of masonry members that resist seismic forces shall conform to the requirements of TMS 402/ACI 530/ ASCE 5 and TMS 602/ACI 530.1/ASCE 6, except as modiﬁ ed by Section 14.4. 14.4.2 R factors To qualify for the response modiﬁ cation coefﬁ -cients, R, set forth in this standard, the requirements of TMS 402/ACI 530/ASCE 5 and TMS 602/ACI 530.1/ASCE 6, as amended in subsequent sections, shall be satisﬁ ed. Intermediate and special reinforced masonry shear walls designed in accordance with Section 2.3 of TMS 402/ACI 530/ASCE 5 shall also comply with the additional requirements contained in Section 14.4.4. 14.4.3 Modiﬁ cations to Chapter 1 of TMS 402/ACI 530/ASCE 5 14.4.3.1 Separation Joints Add the following new Section 1.19.3 to TMS 402/ACI 530/ASCE 5: 1.19.3 Separation Joints. Where concrete abuts structural masonry and the joint between the materials is not designed as a separation joint, the concrete shall be roughened so that the average height of aggregate exposure is 1/8 in. (3 mm) and shall be bonded to the masonry in accordance with these requirements as if it were masonry. Vertical joints not intended to act as separation joints shall be crossed by horizontal reinforcement as required by Section 1.9.4.2. 14.4.4 Modiﬁ cations to Chapter 2 of TMS 402/ACI 530/ASCE 5 14.4.4.1 Stress Increase If the increase in stress given in Section 2.1.2.3 of TMS 402/ACI 530/ASCE 5 is used, the restriction on load reduction in Section 2.4.1 of this standard shall be observed. 14.4.4.2 Reinforcement Requirements and Details 14.4.4.2.1 Reinforcing Bar Size Limitations Reinforc-ing bars used in masonry shall not be larger than No. 9 (M#29). The nominal bar diameter shall not exceed MINIMUM DESIGN LOADS 135 one-eighth of the nominal member thickness and shall not exceed one-quarter of the least clear dimension of the cell, course, or collar joint in which it is placed. The area of reinforcing bars placed in a cell or in a course of hollow unit construction shall not exceed 4 percent of the cell area. 14.4.4.2.2 Splices Lap splices shall not be used in plastic hinge zones of special reinforced masonry shear walls. The length of the plastic hinge zone shall be taken as at least 0.15 times the distance between the point of zero moment and the point of maximum moment. Reinforcement splices shall comply with TMS 402/ACI 530/ASCE 5 except paragraphs 2.1.9.7.2 and 2.1.9.7.3 shall be modiﬁ ed as follows: 2.1.9.7.2 Welded Splices: A welded splice shall be capable of developing in tension at least 125 percent of the speciﬁ ed yield strength, fy, of the bar. Welded splices shall only be permitted for ASTM A706 steel reinforcement. Welded splices shall not be permitted in plastic hinge zones of intermediate or special reinforced walls of masonry. 2.1.9.7.3 Mechanical Connections: Mechanical splices shall be classiﬁ ed as Type 1 or Type 2 according to Section 21.1.6.1 of ACI 318. Type 1 mechanical splices shall not be used within a plastic hinge zone or within a beam-wall joint of intermediate or special reinforced masonry shear wall system. Type 2 mechanical splices shall be permitted in any location within a member. 14.4.5 Modiﬁ cations to Chapter 3 of TMS 402/ACI 530/ASCE 5 14.4.5.1 Anchoring to Masonry Add the following as the ﬁ rst paragraph in Section 3.1.6 to TMS 402/ACI 530/ASCE 5: 3.1.6 Anchor Bolts Embedded in Grout. Anchorage assemblies connecting masonry elements that are part of the seismic force-resisting system to diaphragms and chords shall be designed so that the strength of the anchor is governed by steel tensile or shear yielding. Alternatively, the anchorage assembly is permitted to be designed so that it is governed by masonry breakout or anchor pullout provided that the anchorage assembly is designed to resist not less than 2.5 times the factored forces transmitted by the assembly. 14.4.5.2 Splices in Reinforcement Replace Sections 3.3.3.4(b) and 3.3.3.4(c) of TMS 402/ACI 530/ASCE 5 with the following: (b) A welded splice shall be capable of developing in tension at least 125 percent of the speciﬁ ed yield strength, fy, of the bar. Welded splices shall only be permitted for ASTM A706 steel reinforcement. Welded splices shall not be permitted in plastic hinge zones of intermediate or special reinforced walls of masonry. (c) Mechanical splices shall be classiﬁ ed as Type 1 or Type 2 according to Section 21.1.6.1 of ACI 318. Type 1 mechanical splices shall not be used within a plastic hinge zone or within a beam-column joint of intermediate or special reinforced masonry shear walls. Type 2 mechani-cal splices are permitted in any location within a member. Add the following new Section 3.3.3.4.1 to TMS 402/ACI 530/ASCE 5: 3.3.3.4.1 Lap splices shall not be used in plastic hinge zones of special reinforced masonry shear walls. The length of the plastic hinge zone shall be taken as at least 0.15 times the distance between the point of zero moment and the point of maximum moment. 14.4.5.3 Coupling Beams Add the following new Section 3.3.4.2.6 to TMS 402/ACI 530/ASCE 5: 3.3.4.2.6 Coupling Beams. Structural members that provide coupling between shear walls shall be designed to reach their moment or shear nominal strength before either shear wall reaches its moment or shear nominal strength. Analysis of coupled shear walls shall comply with accepted principles of mechanics. The design shear strength, φVn, of the coupling beams shall satisfy the following criterion: φV M M L V n c g ≥ + ( ) + 1 25 1 4 1 2 . . where M1 and M2 = nominal moment strength at the ends of the beam Lc = length of the beam between the shear walls Vg = unfactored shear force due to gravity loads The calculation of the nominal ﬂ exural moment shall include the reinforcement in reinforced concrete roof and ﬂ oor systems. The width of the reinforced concrete used for calculations of reinforcement shall be six times the ﬂ oor or roof slab thickness. CHAPTER 14 MATERIAL SPECIFIC SEISMIC DESIGN AND DETAILING REQUIREMENTS 136 14.4.5.4 Deep Flexural Members Add the following new Section 3.3.4.2.7 to TMS 402/ACI 530/ASCE 5: 3.3.4.2.7 Deep Flexural Member Detailing. Flexural members with overall-depth-to-clear-span ratio greater than 2/5 for continuous spans or 4/5 for simple spans shall be detailed in accordance with this section. 3.3.4.2.7.1 Minimum ﬂ exural tension reinforcement shall conform to Section 3.3.4.3.2. 3.3.4.2.7.2 Uniformly distributed horizontal and vertical reinforcement shall be provided throughout the length and depth of deep ﬂ exural members such that the reinforcement ratios in both directions are at least 0.001. Distributed ﬂ exural reinforcement is to be included in the determination of the actual reinforcement ratios. 14.4.5.5 Walls with Factored Axial Stress Greater Than 0.05 fm ′ Add the following exception following the second paragraph of Section 3.3.5.3 of TMS 402/ACI 530/ ASCE 5: EXCEPTION: A nominal thickness of 4 in. (102 mm) is permitted where load-bearing reinforced hollow clay unit masonry walls satisfy all of the following conditions. 1. The maximum unsupported height-to-thickness or length-to-thickness ratios do not exceed 27. 2. The net area unit strength exceeds 8,000 psi (55 MPa). 3. Units are laid in running bond. 4. Bar sizes do not exceed No. 4 (13 mm). 5. There are no more than two bars or one splice in a cell. 6. Joints are not raked. 14.4.5.6 Shear Keys Add the following new Section 3.3.6.6 to TMS 402/ACI 530/ASCE 5: 3.3.6.11 Shear Keys. The surface of concrete upon which a special reinforced masonry shear wall is constructed shall have a minimum surface roughness of 1/8 in. (3 mm). Shear keys are required where the calculated tensile strain in vertical reinforcement from in-plane loads exceeds the yield strain under load combinations that include seismic forces based on an R factor equal to 1.5. Shear keys that satisfy the following requirements shall be placed at the interface between the wall and the foundation. 1. The width of the keys shall be at least equal to the width of the grout space. 2. The depth of the keys shall be at least 1.5 in. (38 mm). 3. The length of the key shall be at least 6 in. (152 mm). 4. The spacing between keys shall be at least equal to the length of the key. 5. The cumulative length of all keys at each end of the shear wall shall be at least 10 percent of the length of the shear wall (20 percent total). 6. At least 6 in. (150 mm) of a shear key shall be placed within 16 in. (406 mm) of each end of the wall. 7. Each key and the grout space above each key in the ﬁ rst course of masonry shall be grouted solid. 14.4.6 Modiﬁ cations to Chapter 6 of TMS 402/ACI 530/ASCE 5 14.4.6.1 Corrugated Sheet Metal Anchors Add Section 6.2.2.10.1 to TMS 402/ACI 530/ ASCE 5 as follows: 6.2.2.10.1 Provide continuous single wire joint reinforcement of wire size W1.7 (MW11) at a maximum spacing of 18 in. (457 mm) on center vertically. Mechanically attach anchors to the joint reinforcement with clips or hooks. Corrugated sheet metal anchors shall not be used. 14.4.7 Modiﬁ cations to TMS 602/ACI 530.1/ASCE 6 14.4.7.1 Construction Procedures Add the following new Article 3.5 I to TMS 602/ ACI 530.1/ASCE 6: 3.5 I. Construction procedures or admixtures shall be used to facilitate placement and control shrinkage of grout. 14.5 WOOD Structures, including foundations, constructed of wood to resist seismic loads shall be designed and detailed in accordance with this standard including the references and additional requirements provided in this section. 14.5.1 Reference Documents The quality, testing, design, and construction of members and their fastenings in wood systems that resist seismic forces shall conform to the requirements of the applicable following reference documents,: 1. AF&PA NDS 2. AF&PA SDPWS MINIMUM DESIGN LOADS 137 14.5.2 Framing All wood columns and posts shall be framed to provide full end bearing. Alternatively, column and post end connections shall be designed to resist the full compressive loads, neglecting all end-bearing capacity. Continuity of wall top plates or provision for transfer of induced axial load forces shall be provided. Where offsets occur in the wall line, portions of the shear wall on each side of the offset shall be considered as separate shear walls unless provisions for force transfer around the offset are provided. 139 Chapter 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES selected in accordance with Section 12.6. Nonbuilding structures that are not similar to buildings shall be designed using either the equivalent lateral force procedure in accordance with Section 12.8, the modal analysis procedure in accordance with Section 12.9, the linear response history analysis procedure in accordance with Section 16.1, the nonlinear response history analysis procedure in accordance with Section 16.2, or the procedure prescribed in the speciﬁ c reference document. 15.2 REFERENCE DOCUMENTS Reference documents referred to in Chapter 15 are listed in Chapter 23 and have seismic requirements based on the same force and displacement levels used in this standard or have seismic requirements that are speciﬁ cally modiﬁ ed by Chapter 15. 15.3 NONBUILDING STRUCTURES SUPPORTED BY OTHER STRUCTURES Where nonbuilding structures identiﬁ ed in Table 15.4-2 are supported by other structures, and the nonbuilding structures are not part of the primary seismic force-resisting system, one of the following methods shall be used. 15.3.1 Less Than 25 percent Combined Weight Condition For the condition where the weight of the nonbuilding structure is less than 25 percent of the combined effective seismic weights of the nonbuilding structure and supporting structure, the design seismic forces of the nonbuilding structure shall be determined in accordance with Chapter 13 where the values of Rp and ap shall be determined in accordance to Section 13.1.5. The supporting structure shall be designed in accordance with the requirements of Chapter 12 or Section 15.5 as appropriate with the weight of the nonbuilding structure considered in the determination of the effective seismic weight, W. 15.1 GENERAL 15.1.1 Nonbuilding Structures Nonbuilding structures include all self-supporting structures that carry gravity loads and that may be required to resist the effects of earthquake, with the exception of building structures speciﬁ cally excluded in Section 11.1.2, and other nonbuilding structures where speciﬁ c seismic provisions have yet to be developed, and therefore, are not set forth in Chapter 15. Nonbuilding structures supported by the earth or supported by other structures shall be designed and detailed to resist the minimum lateral forces speciﬁ ed in this section. Design shall conform to the applicable requirements of other sections as modiﬁ ed by this section. Foundation design shall comply with the requirements of Sections 12.1.5, 12.13, and Chapter 14. 15.1.2 Design The design of nonbuilding structures shall provide sufﬁ cient stiffness, strength, and ductility consistent with the requirements speciﬁ ed herein for buildings to resist the effects of seismic ground motions as represented by these design forces: a. Applicable strength and other design criteria shall be obtained from other portions of the seismic requirements of this standard or its reference documents. b. Where applicable strength and other design criteria are not contained in, or referenced by the seismic requirements of this standard, such criteria shall be obtained from reference documents. Where reference documents deﬁ ne acceptance criteria in terms of allowable stresses as opposed to strength, the design seismic forces shall be obtained from this section and used in combination with other loads as speciﬁ ed in Section 2.4 of this standard and used directly with allowable stresses speciﬁ ed in the reference documents. Detailing shall be in accordance with the reference documents. 15.1.3 Structural Analysis Procedure Selection Structural analysis procedures for nonbuilding structures that are similar to buildings shall be CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 140 15.3.2 Greater Than or Equal to 25 Percent Combined Weight Condition For the condition where the weight of the nonbuilding structure is equal to or greater than 25 percent of the combined effective seismic weights of the nonbuilding structure and supporting structure, an analysis combining the structural characteristics of both the nonbuilding structure and the supporting structures shall be performed to determine the seismic design forces as follows: 1. Where the fundamental period, T, of the nonbuild-ing structure is less than 0.06 s, the nonbuilding structure shall be considered a rigid element with appropriate distribution of its effective seismic weight. The supporting structure shall be designed in accordance with the requirements of Chapter 12 or Section 15.5 as appropriate, and the R value of the combined system is permitted to be taken as the R value of the supporting structural system. The nonbuilding structure and attachments shall be designed for the forces using the procedures of Chapter 13 where the value of Rp shall be taken as equal to the R value of the nonbuilding structure as set forth in Table 15.4-2, and ap shall be taken as 1.0. 2. Where the fundamental period, T, of the nonbuild-ing structure is 0.06 s or greater, the nonbuilding structure and supporting structure shall be modeled together in a combined model with appropriate stiffness and effective seismic weight distributions. The combined structure shall be designed in accordance with Section 15.5 with the R value of the combined system taken as the lesser R value of the nonbuilding structure or the supporting struc-ture. The nonbuilding structure and attachments shall be designed for the forces determined for the nonbuilding structure in the combined analysis. 15.3.3 Architectural, Mechanical, and Electrical Components Architectural, mechanical, and electrical components supported by nonbuilding structures shall be designed in accordance with Chapter 13 of this standard. 15.4 STRUCTURAL DESIGN REQUIREMENTS 15.4.1 Design Basis Nonbuilding structures having speciﬁ c seismic design criteria established in reference documents shall be designed using the standards as amended herein. Where reference documents are not cited herein, nonbuilding structures shall be designed in compliance with Sections 15.5 and 15.6 to resist minimum seismic lateral forces that are not less than the requirements of Section 12.8 with the following additions and exceptions: 1. The seismic force-resisting system shall be selected as follows: a. For nonbuilding structures similar to buildings, a system shall be selected from among the types indicated in Table 12.2-1 or Table 15.4-1 subject to the system limitations and limits on structural height, hn, based on the seismic design category indicated in the table. The appropriate values of R, Ω0, and Cd indicated in the selected table shall be used in determining the base shear, element design forces, and design story drift as indicated in this standard. Design and detailing requirements shall comply with the sections referenced in the selected table. b. For nonbuilding structures not similar to buildings, a system shall be selected from among the types indicated in Table 15.4-2 subject to the system limitations and limits on structural height, hn, based on seismic design category indicated in the table. The appropriate values of R, Ωo, and Cd indicated in Table 15.4-2 shall be used in determining the base shear, element design forces, and design story drift as indicated in this standard. Design and detailing requirements shall comply with the sections referenced in Table 15.4-2. c. Where neither Table 15.4-1 nor Table 15.4-2 contains an appropriate entry, applicable strength and other design criteria shall be obtained from a reference document that is applicable to the speciﬁ c type of nonbuilding structure. Design and detailing requirements shall comply with the reference document. 2. For nonbuilding systems that have an R value provided in Table 15.4-2, the minimum speciﬁ ed value in Eq. 12.8-5 shall be replaced by Cs = 0.044SDSIe (15.4-1) The value of Cs shall not be taken as less than 0.03. And for nonbuilding structures located where S1 ≥ 0.6g, the minimum speciﬁ ed value in Eq. 12.8-6 shall be replaced by Cs = 0.8S1/(R/Ie) (15.4-2) EXCEPTION: Tanks and vessels that are designed to AWWA D100, AWWA D103, API MINIMUM DESIGN LOADS 141 Table 15.4-1 Seismic Coefﬁ cients for Nonbuilding Structures Similar to Buildings Nonbuilding Structure Type Detailing Requirements R Ω0 Cd Structural System and Structural Height, hn, Limits (ft)a B C D E F Steel storage racks 15.5.3 4 2 3.5 NL NL NL NL NL Building frame systems: Steel special concentrically braced frames AISC 341 6 2 5 NL NL 160 160 100 Steel ordinary concentrically braced frame AISC 341 3¼ 2 3¼ NL NL 35b 35b NPb With permitted height increase AISC 341 2½ 2 2½ NL NL 160 160 100 With unlimited height AISC 360 1.5 1 1.5 NL NL NL NL NL Moment-resisting frame systems: Steel special moment frames AISC 341 8 3 5.5 NL NL NL NL NL Special reinforced concrete moment frames 14.2.2.6 & ACI 318, including Chapter 21 8 3 5.5 NL NL NL NL NL Steel intermediate moment frames AISC 341 4.5 3 4 NL NL 35c,d NPc,d NPc,d With permitted height increase AISC 341 2.5 2 2.5 NL NL 160 160 100 With unlimited height AISC 341 1.5 1 1.5 NL NL NL NL NL Intermediate reinforced concrete moment frames ACI 318, including Chapter 21 5 3 4.5 NL NL NP NP NP With permitted height increase ACI 318, including Chapter 21 3 2 2.5 NL NL 50 50 50 With unlimited height ACI 318, including Chapter 21 0.8 1 1 NL NL NL NL NL Steel ordinary moment frames AISC 341 3.5 3 3 NL NL NPc,d NPc,d NPc,d With permitted height increase AISC 341 2.5 2 2.5 NL NL 100 100 NPc,d With unlimited height AISC 360 1 1 1 NL NL NL NL NL Ordinary reinforced concrete moment frames ACI 318, excluding Chapter 21 3 3 2.5 NL NP NP NP NP With permitted height increase ACI 318, excluding Chapter 21 0.8 1 1 NL NL 50 50 50 aNL = no limit and NP = not permitted. bSteel ordinary braced frames are permitted in pipe racks up to 65 ft (20 m). cSteel ordinary moment frames and intermediate moment frames are permitted in pipe racks up to a height of 65 ft (20 m) where the moment joints of ﬁ eld connections are constructed of bolted end plates. dSteel ordinary moment frames and intermediate moment frames are permitted in pipe racks up to a height of 35 ft (11 m). CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 142 Table 15.4-2 Seismic Coefﬁ cients for Nonbuilding Structures not Similar to Buildings Nonbuilding Structure Type Detailing Requirementsc R Ω0 Cd Structural Height, hn, Limits (ft)ad B C D E F Elevated tanks, vessels, bins or hoppers On symmetrically braced legs (not similar to buildings) 15.7.10 3 2b 2.5 NL NL 160 100 100 On unbraced legs or asymmetrically braced legs (not similar buildings) 15.7.10 2 2b 2.5 NL NL 100 60 60 Horizontal, saddle supported welded steel vessels 15.7.14 3 2b 2.5 NL NL NL NL NL Tanks or vessels supported on structural towers similar to buildings 15.5.5 Use values for the appropriate structure type in the categories for building frame systems and moment resisting frame systems listed in Table 12.2-1 or Table 15.4-1. Flat-bottom ground-supported tanks: 15.7 Steel or ﬁ ber-reinforced plastic: Mechanically anchored 3 2b 2.5 NL NL NL NL NL Self-anchored 2.5 2b 2 NL NL NL NL NL Reinforced or prestressed concrete: Reinforced nonsliding base 2 2b 2 NL NL NL NL NL Anchored ﬂ exible base 3.25 2b 2 NL NL NL NL NL Unanchored and unconstrained ﬂ exible base 1.5 1.5b 1.5 NL NL NL NL NL All other 1.5 1.5b 1.5 NL NL NL NL NL Cast-in-place concrete silos having walls continuous to the foundation 15.6.2 3 1.75 3 NL NL NL NL NL All other reinforced masonry structures not similar to buildings detailed as intermediate reinforced masonry shear walls 14.4.1f 3 2 2.5 NL NL 50 50 50 All other reinforced masonry structures not similar to buildings detailed as ordinary reinforced masonry shear walls 14.4.1 2 2.5 1.75 NL 160 NP NP NP All other nonreinforced masonry structures not similar to buildings 14.4.1 1.25 2 1.5 NL NL NP NP NP Concrete chimneys and stacks 15.6.2 and ACI 307 2 1.5 2.0 NL NL NL NL NL MINIMUM DESIGN LOADS 143 Nonbuilding Structure Type Detailing Requirementsc R Ω0 Cd Structural Height, hn, Limits (ft)ad B C D E F All steel and reinforced concrete distributed mass cantilever structures not otherwise covered herein including stacks, chimneys, silos, skirt-supported vertical vessels and single pedestal or skirt supported Welded steel Welded steel with special detailinge Prestressed or reinforced concrete Prestressed or reinforced concrete with special detailing 15.6.2 15.7.10 15.7.10 & 15.7.10.5 a and b 15.7.10 15.7.10 and ACI 318 Chapter 21, Sections 21.2 and 21.7 2 3 2 3 2b 2b 2b 2b 2 2 2 2 NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL NL Trussed towers (freestanding or guyed), guyed stacks, and chimneys 15.6.2 3 2 2.5 NL NL NL NL NL Cooling towers Concrete or steel 3.5 1.75 3 NL NL NL NL NL Wood frames 3.5 3 3 NL NL NL 50 50 Telecommunication towers 15.6.6 Truss: Steel 3 1.5 3 NL NL NL NL NL Pole: Steel 1.5 1.5 1.5 NL NL NL NL NL Wood 1.5 1.5 1.5 NL NL NL NL NL Concrete 1.5 1.5 1.5 NL NL NL NL NL Frame: Steel 3 1.5 1.5 NL NL NL NL NL Wood 1.5 1.5 1.5 NL NL NL NL NL Concrete 2 1.5 1.5 NL NL NL NL NL Amusement structures and monuments 15.6.3 2 2 2 NL NL NL NL NL Inverted pendulum type structures (except elevated tanks, vessels, bins, and hoppers) 12.2.5.3 2 2 2 NL NL NL NL NL Signs and billboards 3.0 1.75 3 NL NL NL NL NL All other self-supporting structures, tanks, or vessels not covered above or by reference standards that are similar to buildings 1.25 2 2.5 NL NL 50 50 50 aNL = no limit and NP = not permitted. bSee Section 15.7.3a for the application of the overstrength factors, Ω0, for tanks and vessels. cIf a section is not indicated in the Detailing Requirements column, no speciﬁ c detailing requirements apply. dFor the purpose of height limit determination, the height of the structure shall be taken as the height to the top of the structural frame making up the primary seismic force-resisting system. eSections 15.7.10.5a and 15.7.10.5b shall be applied for any Risk Category. f Detailed with an essentially complete vertical load carrying frame. Table 15.4-2 (Continued) CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 144 650 Appendix E, and API 620 Appendix L as modiﬁ ed by this standard, and stacks and chimneys that are designed to ACI 307 as modiﬁ ed by this standard, shall be subject to the larger of the minimum base shear value deﬁ ned by the reference document or the value determined by replacing Eq. 12.8-5 with the following: Cs = 0.044SDS Ie (15.4-3) The value of Cs shall not be taken as less than 0.01. and for nonbuilding structures located where S1 ≥ 0.6g, Eq. 12.8-6 shall be replaced by Cs = 0.5S1/(R/Ie) (15.4-4) Minimum base shear requirements need not apply to the convective (sloshing) component of liquid in tanks. 3. The importance factor, Ie, shall be as set forth in Section 15.4.1.1. 4. The vertical distribution of the lateral seismic forces in nonbuilding structures covered by this section shall be determined: a. Using the requirements of Section 12.8.3, or b. Using the procedures of Section 12.9, or c. In accordance with the reference document applicable to the speciﬁ c nonbuilding structure. 5. For nonbuilding structural systems containing liquids, gases, and granular solids supported at the base as deﬁ ned in Section 15.7.1, the minimum seismic design force shall not be less than that required by the reference document for the speciﬁ c system. 6. Where a reference document provides a basis for the earthquake resistant design of a particular type of nonbuilding structure covered by Chapter 15, such a standard shall not be used unless the following limitations are met: a. The seismic ground accelerations, and seismic coefﬁ cients, shall be in conformance with the requirements of Section 11.4. b. The values for total lateral force and total base overturning moment used in design shall not be less than 80 percent of the base shear value and overturning moment, each adjusted for the effects of soil–structure interaction that is obtained using this standard. 7. The base shear is permitted to be reduced in accordance with Section 19.2.1 to account for the effects of soil–structure interaction. In no case shall the reduced base shear be less than 0.7V. 8. Unless otherwise noted in Chapter 15, the effects on the nonbuilding structure due to gravity loads and seismic forces shall be combined in accor-dance with the factored load combinations as presented in Section 2.3. 9. Where speciﬁ cally required by Chapter 15, the design seismic force on nonbuilding structures shall be as deﬁ ned in Section 12.4.3. 15.4.1.1 Importance Factor The importance factor, Ie, and risk category for nonbuilding structures are based on the relative hazard of the contents and the function. The value of Ie shall be the largest value determined by the following: a. Applicable reference document listed in Chapter 23. b. The largest value as selected from Table 1.5-2. c. As speciﬁ ed elsewhere in Chapter 15. 15.4.2 Rigid Nonbuilding Structures Nonbuilding structures that have a fundamental period, T, less than 0.06 s, including their anchorages, shall be designed for the lateral force obtained from the following: V = 0.30SDSWIe (15.4-5) where V = the total design lateral seismic base shear force applied to a nonbuilding structure SDS = the site design response acceleration as deter-mined from Section 11.4.4 W = nonbuilding structure operating weight Ie = the importance factor determined in accordance with Section 15.4.1.1 The force shall be distributed with height in accordance with Section 12.8.3. 15.4.3 Loads The seismic effective weight W for nonbuilding structures shall include the dead load and other loads as deﬁ ned for structures in Section 12.7.2. For purposes of calculating design seismic forces in nonbuilding structures, W also shall include all normal operating contents for items such as tanks, vessels, bins, hoppers, and the contents of piping. W shall include snow and ice loads where these loads consti-tute 25 percent or more of W or where required by the authority having jurisdiction based on local environ-mental characteristics. 15.4.4 Fundamental Period The fundamental period of the nonbuilding structure shall be determined using the structural MINIMUM DESIGN LOADS 145 properties and deformation characteristics of the resisting elements in a properly substantiated analysis as indicated in Section 12.8.2. Alternatively, the fundamental period T is permitted to be computed from the following equation: T f g f i i i n i i i n = = = ∑ ∑ 2 2 1 1 π δ δ (15.4-6) The values of fi represent any lateral force distribution in accordance with the principles of structural mechanics. The elastic deﬂ ections, δi, shall be calculated using the applied lateral forces, fi. Equations 12.8-7, 12.8-8, 12.8-9, and 12.8-10 shall not be used for determining the period of a nonbuild-ing structure. 15.4.5 Drift Limitations The drift limitations of Section 12.12.1 need not apply to nonbuilding structures if a rational analysis indicates they can be exceeded without adversely affecting structural stability or attached or interconnected components and elements such as walkways and piping. P-delta effects shall be consid-ered where critical to the function or stability of the structure. 15.4.6 Materials Requirements The requirements regarding speciﬁ c materials in Chapter 14 shall be applicable unless speciﬁ cally exempted in Chapter 15. 15.4.7 Deﬂ ection Limits and Structure Separation Deﬂ ection limits and structure separation shall be determined in accordance with this standard unless speciﬁ cally amended in Chapter 15. 15.4.8 Site-Speciﬁ c Response Spectra Where required by a reference document or the authority having jurisdiction, speciﬁ c types of nonbuilding structures shall be designed for site-speciﬁ c criteria that account for local seismicity and geology, expected recurrence intervals, and magnitudes of events from known seismic hazards (see Section 11.4.7 of this standard). If a longer recurrence interval is deﬁ ned in the reference docu-ment for the nonbuilding structure, such as liqueﬁ ed natural gas (LNG) tanks (NFPA 59A), the recurrence interval required in the reference document shall be used. 15.4.9 Anchors in Concrete or Masonry 15.4.9.1 Anchors in Concrete Anchors in concrete used for nonbuilding structure anchorage shall be designed in accordance with Appendix D of ACI 318. 15.4.9.2 Anchors in Masonry Anchors in masonry used for nonbuilding structure anchorage shall be designed in accordance with TMS402/ACI 530/ASCE 6. Anchors shall be designed to be governed by the tensile or shear strength of a ductile steel element. EXCEPTION: Anchors shall be permitted to be designed so that the attachment that the anchor is connecting to the structure undergoes ductile yielding at a load level corresponding to anchor forces not greater than their design strength, or the minimum design strength of the anchors shall be at least 2.5 times the factored forces transmitted by the attachment. 15.4.9.3 Post-Installed Anchors in Concrete and Masonry Post-installed anchors in concrete shall be prequaliﬁ ed for seismic applications in accordance with ACI 355.2 or other approved qualiﬁ cation procedures. Post-installed anchors in masonry shall be prequaliﬁ ed for seismic applications in accordance with approved qualiﬁ cation procedures. 15.5 NONBUILDING STRUCTURES SIMILAR TO BUILDINGS 15.5.1 General Nonbuilding structures similar to buildings as deﬁ ned in Section 11.2 shall be designed in accor-dance with this standard as modiﬁ ed by this section and the speciﬁ c reference documents. This general category of nonbuilding structures shall be designed in accordance with the seismic requirements of this standard and the applicable portions of Section 15.4. The combination of load effects, E, shall be deter-mined in accordance with Section 12.4. 15.5.2 Pipe Racks 15.5.2.1 Design Basis In addition to the requirements of Section 15.5.1, pipe racks supported at the base of the structure shall be designed to meet the force requirements of Section 12.8 or 12.9. Displacements of the pipe rack and CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 146 potential for interaction effects (pounding of the piping system) shall be considered using the ampliﬁ ed deﬂ ections obtained from the following equation: δ δ x d xe e C I = (15.5-1) where Cd = deﬂ ection ampliﬁ cation factor in Table 15.4-1 δxe = deﬂ ections determined using the prescribed seismic design forces of this standard Ie = importance factor determined in accordance with Section 15.4.1.1 See Section 13.6.3 for the design of piping systems and their attachments. Friction resulting from gravity loads shall not be considered to provide resistance to seismic forces. 15.5.3 Steel Storage Racks Steel storage racks supported at or below grade shall be designed in accordance with ANSI/RMI MH 16.1 and its force and displacement requirements, except as follows: 15.5.3.1 Modify Section 2.6.2 of ANSI/RMI MH 16.1 as follows: 2.6.2 Minimum Seismic Forces The storage rack shall be designed… Above-Grade Elevation: Storage rack installed at elevations above grade shall be designed, fabricated, and installed in accordance with the following requirements: Storage racks shall meet the force and displacement requirements required of nonbuilding structures supported by other structures, including the force and displacement effects caused by ampliﬁ cations of upper-story motions. In no case shall the value of V be taken as less than the value of Fp determined in accordance with Section 13.3.1 of ASCE/SEI 7, where Rp is taken equal to R, and ap is taken equal to 2.5. 15.5.3.2 Modify Section 7.2.2 of ANSI/RMI MH 16.1 as follows: 7.2.2 Base Plate Design Once the required bearing area has been determined from the allowable bearing stress F’p the minimum thickness of the base plate is determined by rational analysis or by appropriate test using a test load 1.5 times the ASD design load or the factored LRFD load. Design forces that include seismic loads for anchorage of steel storage racks to concrete or masonry shall be determined using load combinations with overstrength provided in Section 12.4.3.2 of ASCE/SEI 7. The overstrength factor shall be taken as 2.0. Anchorage of steel storage racks to concrete shall be in accordance with the requirements of Section 15.4.9 of ASCE/SEI 7. Upon request, information shall be given to the owner or the owner’s agent on the location, size, and pressures under the column base plates of each type of upright frame in the installation. When rational analysis is used to determine base plate thickness and other applicable standards do not apply, the base plate shall be permitted to be designed for the following loading conditions, where applicable: (balance of section unchanged) 15.5.3.3 Modify Section 7.2.4 of ANSI/RMI MH 16.1 as follows: 7.2.4 Shims Shims may be used under the base plate to maintain the plumbness of the storage rack. The shims shall be made of a material that meets or exceeds the design bearing strength (LRFD) or allowable bearing strength (ASD) of the ﬂ oor. The shim size and location under the base plate shall be equal to or greater than the required base plate size and location. In no case shall the total thickness of any set of shims under a base plate exceed six times the diameter of the largest anchor bolt used in that base. Shims that are a total thickness of less than or equal to six times the anchor bolt diameter under bases with less than two anchor bolts shall be interlocked or welded together in a fashion that is capable of transferring all the shear forces at the base. Shims that are a total thickness of less than or equal to two times the anchor bolt diameter need not be interlocked or welded together. Bending in the anchor associated with shims or grout under the base plate shall be taken into account in the design of the anchor bolts. 15.5.3.4 Alternative As an alternative to ANSI MH 16.1 as modiﬁ ed above, steel storage racks shall be permitted to be designed in accordance with the requirements of MINIMUM DESIGN LOADS 147 Sections 15.1, 15.2, 15.3, 15.5.1, and 15.5.3.5 through 15.5.3.8 of this standard. 15.5.3.5 General Requirements Steel storage racks shall satisfy the force require-ments of this section. EXCEPTION: Steel storage racks supported at the base are permitted to be designed as structures with an R of 4, provided that the seismic requirements of this standard are met. Higher values of R are permitted to be used where the detailing requirements of reference documents listed in Section 14.1.1 are met. The importance factor, Ie, for storage racks in structures open to the public, such as warehouse retail stores, shall be taken equal to 1.5. 15.5.3.6 Operating Weight Steel storage racks shall be designed for each of the following conditions of operating weight, W or Wp. a. Weight of the rack plus every storage level loaded to 67 percent of its rated load capacity. b. Weight of the rack plus the highest storage level only loaded to 100 percent of its rated load capacity. The design shall consider the actual height of the center of mass of each storage load component. 15.5.3.7 Vertical Distribution of Seismic Forces For all steel storage racks, the vertical distribution of seismic forces shall be as speciﬁ ed in Section 12.8.3 and in accordance with the following: a. The base shear, V, of the typical structure shall be the base shear of the steel storage rack where loaded in accordance with Section 15.5.3.6. b. The base of the structure shall be the ﬂ oor support-ing the steel storage rack. Each steel storage level of the rack shall be treated as a level of the structure with heights hi and hx measured from the base of the structure. c. The factor k is permitted to be taken as 1.0. 15.5.3.8 Seismic Displacements Steel storage rack installations shall accommodate the seismic displacement of the storage racks and their contents relative to all adjacent or attached components and elements. The assumed total relative displacement for storage racks shall be not less than 5 percent of the structural height above the base, hn, unless a smaller value is justiﬁ ed by test data or analysis in accordance with Section 11.1.4. 15.5.4 Electrical Power Generating Facilities 15.5.4.1 General Electrical power generating facilities are power plants that generate electricity by steam turbines, combustion turbines, diesel generators, or similar turbo machinery. 15.5.4.2 Design Basis In addition to the requirements of Section 15.5.1, electrical power generating facilities shall be designed using this standard and the appropriate factors contained in Section 15.4. 15.5.5 Structural Towers for Tanks and Vessels 15.5.5.1 General In addition to the requirements of Section 15.5.1, structural towers that support tanks and vessels shall be designed to meet the requirements of Section 15.3. In addition, the following special considerations shall be included: a. The distribution of the lateral base shear from the tank or vessel onto the supporting structure shall consider the relative stiffness of the tank and resisting structural elements. b. The distribution of the vertical reactions from the tank or vessel onto the supporting structure shall consider the relative stiffness of the tank and resisting structural elements. Where the tank or vessel is supported on grillage beams, the calcu-lated vertical reaction due to weight and overturn-ing shall be increased at least 20 percent to account for nonuniform support. The grillage beam and vessel attachment shall be designed for this increased design value. c. Seismic displacements of the tank and vessel shall consider the deformation of the support structure where determining P-delta effects or evaluating required clearances to prevent pounding of the tank on the structure. 15.5.6 Piers and Wharves 15.5.6.1 General Piers and wharves are structures located in waterfront areas that project into a body of water or that parallel the shoreline. 15.5.6.2 Design Basis In addition to the requirements of Section 15.5.1, piers and wharves that are accessible to the general CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 148 public, such as cruise ship terminals and piers with retail or commercial ofﬁ ces or restaurants, shall be designed to comply with this standard. Piers and wharves that are not accessible to the general public are beyond the scope of this section. The design shall account for the effects of liquefaction and soil failure collapse mechanisms, as well as consider all applicable marine loading combi-nations, such as mooring, berthing, wave, and current on piers and wharves as required. Structural detailing shall consider the effects of the marine environment. 15.6 GENERAL REQUIREMENTS FOR NONBUILDING STRUCTURES NOT SIMILAR TO BUILDINGS Nonbuilding structures that do not have lateral and vertical seismic force-resisting systems that are similar to buildings shall be designed in accordance with this standard as modiﬁ ed by this section and the speciﬁ c reference documents. Loads and load distribu-tions shall not be less demanding than those deter-mined in this standard. The combination of earthquake load effects, E, shall be determined in accordance with Section 12.4.2. EXCEPTION: The redundancy factor, ρ, per Section 12.3.4 shall be taken as 1. 15.6.1 Earth-Retaining Structures This section applies to all earth-retaining struc-tures assigned to Seismic Design Category D, E, or F. The lateral earth pressures due to earthquake ground motions shall be determined in accordance with Section 11.8.3. The risk category shall be determined by the proximity of the earth-retaining structure to other buildings and structures. If failure of the earth-retain-ing structure would affect the adjacent building or structure, the risk category shall not be less than that of the adjacent building or structure. Earth-retaining walls are permitted to be designed for seismic loads as either yielding or nonyielding walls. Cantilevered reinforced concrete or masonry retaining walls shall be assumed to be yielding walls and shall be designed as simple ﬂ exural wall elements. 15.6.2 Stacks and Chimneys Stacks and chimneys are permitted to be either lined or unlined and shall be constructed from con-crete, steel, or masonry. Steel stacks, concrete stacks, steel chimneys, concrete chimneys, and liners shall be designed to resist seismic lateral forces determined from a substantiated analysis using reference docu-ments. Interaction of the stack or chimney with the liners shall be considered. A minimum separation shall be provided between the liner and chimney equal to Cd times the calculated differential lateral drift. Concrete chimneys and stacks shall be designed in accordance with the requirements of ACI 307 except that (1) the design base shear shall be deter-mined based on Section 15.4.1 of this standard; (2) the seismic coefﬁ cients shall be based on the values provided in Table 15.4-2, and (3) openings shall be detailed as required below. When modal response spectrum analysis is used for design, the procedures of Section 12.9 shall be permitted to be used. For concrete chimneys and stacks assigned to SDC D, E, and F, splices for vertical rebar shall be staggered such that no more than 50% of the bars are spliced at any section and alternate lap splices are staggered by the development length. In addition, where the loss of cross-sectional area is greater than 10%, cross sections in the regions of breachings/ openings shall be designed and detailed for vertical force, shear force, and bending moment demands along the vertical direction, determined for the affected cross section using an overstrength factor of 1.5. The region where the overstrength factor applies shall extend above and below the opening(s) by a distance equal to half of the width of the largest opening in the affected region. Appropriate reinforce-ment development lengths shall be provided beyond the required region of overstrength. The jamb regions around each opening shall be detailed using the column tie requirements in Section 7.10.5 of ACI 318. Such detailing shall extend for a jamb width of a minimum of two times the wall thickness and for a height of the opening height plus twice the wall thickness above and below the opening, but no less than the development length of the longitudinal bars. Where the existence of a footing or base mat precludes the ability to achieve the extension distance below the opening and within the stack, the jamb reinforcing shall be extended and developed into the footing or base mat. The percentage of longitudinal reinforce-ment in jamb regions shall meet the requirements of Section 10.9 of ACI 318 for compression members. 15.6.3 Amusement Structures Amusement structures are permanently ﬁ xed structures constructed primarily for the conveyance and entertainment of people. Amusement structures shall be designed to resist seismic lateral forces determined from a substantiated analysis using reference documents. MINIMUM DESIGN LOADS 149 15.6.4 Special Hydraulic Structures Special hydraulic structures are structures that are contained inside liquid-containing structures. These structures are exposed to liquids on both wall surfaces at the same head elevation under normal operating conditions. Special hydraulic structures are subjected to out-of-plane forces only during an earthquake where the structure is subjected to differential hydrodynamic ﬂ uid forces. Examples of special hydraulic structures include separation walls, bafﬂ e walls, weirs, and other similar structures. 15.6.4.1 Design Basis Special hydraulic structures shall be designed for out-of-phase movement of the ﬂ uid. Unbalanced forces from the motion of the liquid must be applied simultaneously “in front of” and “behind” these elements. Structures subject to hydrodynamic pressures induced by earthquakes shall be designed for rigid body and sloshing liquid forces and their own inertia force. The height of sloshing shall be determined and compared to the freeboard height of the structure. Interior elements, such as bafﬂ es or roof supports, also shall be designed for the effects of unbalanced forces and sloshing. 15.6.5 Secondary Containment Systems Secondary containment systems, such as impoundment dikes and walls, shall meet the require-ments of the applicable standards for tanks and vessels and the authority having jurisdiction. Secondary containment systems shall be designed to withstand the effects of the maximum considered earthquake ground motion where empty and two-thirds of the maximum considered earthquake ground motion where full including all hydrodynamic forces as determined in accordance with the procedures of Section 11.4. Where determined by the risk assess-ment required by Section 1.5.2 or by the authority having jurisdiction that the site may be subject to aftershocks of the same magnitude as the maximum considered motion, secondary containment systems shall be designed to withstand the effects of the maximum considered earthquake ground motion where full including all hydrodynamic forces as determined in accordance with the procedures of Section 11.4. 15.6.5.1 Freeboard Sloshing of the liquid within the secondary containment area shall be considered in determining the height of the impound. Where the primary containment has not been designed with a reduction in the structure category (i.e., no reduction in importance factor Ie) as permitted by Section 1.5.3, no freeboard provision is required. Where the primary containment has been designed for a reduced structure category (i.e., importance factor Ie reduced) as permitted by Section 1.5.3, a minimum freeboard, δs, shall be provided where δs = 0.42DSac (15.6-1) where Sac is the spectral acceleration of the convective component and is determined according to the procedures of Section 15.7.6.1 using 0.5 percent damping. For circular impoundment dikes, D shall be taken as the diameter of the impoundment dike. For rectangular impoundment dikes, D shall be taken as the plan dimension of the impoundment dike, L, for the direction under consideration. 15.6.6 Telecommunication Towers Self-supporting and guyed telecommunication towers shall be designed to resist seismic lateral forces determined from a substantiated analysis using reference documents. 15.7 TANKS AND VESSELS 15.7.1 General This section applies to all tanks, vessels, bins, and silos, and similar containers storing liquids, gases, and granular solids supported at the base (hereafter referred to generically as “tanks and vessels”). Tanks and vessels covered herein include reinforced con-crete, prestressed concrete, steel, aluminum, and ﬁ ber-reinforced plastic materials. Tanks supported on elevated levels in buildings shall be designed in accordance with Section 15.3. 15.7.2 Design Basis Tanks and vessels storing liquids, gases, and granular solids shall be designed in accordance with this standard and shall be designed to meet the requirements of the applicable reference documents listed in Chapter 23. Resistance to seismic forces shall be determined from a substantiated analysis based on the applicable reference documents listed in Chapter 23. a. Damping for the convective (sloshing) force component shall be taken as 0.5 percent. b. Impulsive and convective components shall be combined by the direct sum or the square root of CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 150 the sum of the squares (SRSS) method where the modal periods are separated. If signiﬁ cant modal coupling may occur, the complete quadratic combination (CQC) method shall be used. c. Vertical earthquake forces shall be considered in accordance with the applicable reference document. If the reference document permits the user the option of including or excluding the vertical earthquake force to comply with this standard, it shall be included. For tanks and vessels not covered by a reference document, the forces due to the vertical acceleration shall be deﬁ ned as follows: (1) Hydrodynamic vertical and lateral forces in tank walls: The increase in hydrostatic pres-sures due to the vertical excitation of the contained liquid shall correspond to an effective increase in unit weight, γL, of the stored liquid equal to 0.2SDS γL. (2) Hydrodynamic hoop forces in cylindrical tank walls: In a cylindrical tank wall, the hoop force per unit height, Nh, at height y from the base, associated with the vertical excitation of the contained liquid, shall be computed in accor-dance with Eq. 15.7-1. N S H y D h DS L L i = − ( )⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 2 2 . γ (15.7-1) where Di = inside tank diameter HL = liquid height inside the tank y = distance from base of the tank to height being investigated γL = unit weight of stored liquid (3) Vertical inertia forces in cylindrical and rectangular tank walls: Vertical inertia forces associated with the vertical acceleration of the structure itself shall be taken equal to 0.2SDSW. 15.7.3 Strength and Ductility Structural members that are part of the seismic force-resisting system shall be designed to provide the following: a. Connections to seismic force-resisting elements, excluding anchors (bolts or rods) embedded in concrete, shall be designed to develop Ω0 times the calculated connection design force. For anchors (bolts or rods) embedded in concrete, the design of the anchor embedment shall meet the requirements of Section 15.7.5. Additionally, the connection of the anchors to the tank or vessel shall be designed to develop the lesser of the strength of the anchor in tension as determined by the reference document or Ω0 times the calculated anchor design force. The overstrength requirements of Section 12.4.3, and the Ω0 values tabulated in Table 15.4-2, do not apply to the design of walls, including interior walls, of tanks or vessels. b. Penetrations, manholes, and openings in shell elements shall be designed to maintain the strength and stability of the shell to carry tensile and compressive membrane shell forces. c. Support towers for tanks and vessels with irregular bracing, unbraced panels, asymmetric bracing, or concentrated masses shall be designed using the requirements of Section 12.3.2 for irregular structures. Support towers using chevron or eccentric braced framing shall comply with the seismic requirements of this standard. Support towers using tension-only bracing shall be designed such that the full cross-section of the tension element can yield during overload conditions. d. In support towers for tanks and vessels, compres-sion struts that resist the reaction forces from tension braces shall be designed to resist the lesser of the yield load of the brace, AgFy, or Ωo times the calculated tension load in the brace. e. The vessel stiffness relative to the support system (foundation, support tower, skirt, etc.) shall be considered in determining forces in the vessel, the resisting elements, and the connections. f. For concrete liquid-containing structures, system ductility, and energy dissipation under unfactored loads shall not be allowed to be achieved by inelastic deformations to such a degree as to jeopardize the serviceability of the structure. Stiffness degradation and energy dissipation shall be allowed to be obtained either through limited microcracking, or by means of lateral force resistance mechanisms that dissipate energy without damaging the structure. 15.7.4 Flexibility of Piping Attachments Design of piping systems connected to tanks and vessels shall consider the potential movement of the connection points during earthquakes and provide sufﬁ cient ﬂ exibility to avoid release of the product by failure of the piping system. The piping system and supports shall be designed so as not to impart signiﬁ -cant mechanical loading on the attachment to the tank or vessel shell. Mechanical devices that add ﬂ exibil-ity, such as bellows, expansion joints, and other ﬂ exible apparatus, are permitted to be used where MINIMUM DESIGN LOADS 151 they are designed for seismic displacements and deﬁ ned operating pressure. Unless otherwise calculated, the minimum displacements in Table 15.7-1 shall be assumed. For attachment points located above the support or foundation elevation, the displacements in Table 15.7-1 shall be increased to account for drift of the tank or vessel relative to the base of support. The piping system and tank connection shall also be designed to tolerate Cd times the displacements given in Table 15.7-1 without rupture, although permanent deformations and inelastic behavior in the piping supports and tank shell is permitted. For attachment points located above the support or foundation elevation, the displacements in Table 15.7-1 shall be increased to account for drift of the tank or vessel. The values given in Table 15.7-1 do not include the inﬂ uence of relative movements of the foundation and piping anchorage points due to foundation movements (e.g., settlement, seismic displacements). The effects of the foundation movements shall be included in the piping system design including the determination of the mechanical loading on the tank or vessel, and the total displacement capacity of the mechanical devices intended to add ﬂ exibility. The anchorage ratio, J, for self-anchored tanks shall comply with the criteria shown in Table 15.7-2 and is deﬁ ned as J M D w w rw t a = + ( ) 2 (15.7-2) Table 15.7-1 Minimum Design Displacements for Piping Attachments Condition Displacements (in.) Mechanically Anchored Tanks and Vessels Upward vertical displacement relative to support or foundation 1 (25.4 mm) Downward vertical displacement relative to support or foundation 0.5 (12.7 mm) Range of horizontal displacement (radial and tangential) relative to support or foundation 0.5 (12.7 mm) Self-Anchored Tanks or Vessels (at grade) Upward vertical displacement relative to support or foundation If designed in accordance with a reference document as modiﬁ ed by this standard Anchorage ratio less than or equal to 0.785 (indicates no uplift) 1 (25.4 mm) Anchorage ratio greater than 0.785 (indicates uplift) 4 (101.1 mm) If designed for seismic loads in accordance with this standard but not covered by a reference document For tanks and vessels with a diameter less than 40 ft 8 (202.2 mm) For tanks and vessels with a diameter equal to or greater than 40 ft 12 (0.305 m) Downward vertical displacement relative to support or foundation For tanks with a ringwall/mat foundation 0.5 (12.7 mm) For tanks with a berm foundation 1 (25.4 mm) Range of horizontal displacement (radial and tangential) relative to support or foundation 2 (50.8mm) Table 15.7-2 Anchorage Ratio J Anchorage Ratio Criteria J < 0.785 No uplift under the design seismic overturning moment. The tank is self-anchored. 0.785 < J < 1.54 Tank is uplifting, but the tank is stable for the design load providing the shell compression requirements are satisﬁ ed. The tank is self-anchored. J > 1.54 Tank is not stable and shall be mechanically anchored for the design load. where w W D w t s r = + π (15.7-3) wr = roof load acting on the shell in pounds per foot (N/m) of shell circumference. Only permanent roof loads shall be included. Roof live load shall not be included wa = maximum weight of the tank contents that may be used to resist the shell overturning moment in pounds per foot (N/m) of shell circumfer-ence. Usually consists of an annulus of liquid limited by the bending strength of the tank bottom or annular plate CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 152 Mrw = the overturning moment applied at the bottom of the shell due to the seismic design loads in foot-pounds (N-m) (also known as the “ringwall moment”) D = tank diameter in feet Ws = total weight of tank shell in pounds 15.7.5 Anchorage Tanks and vessels at grade are permitted to be designed without anchorage where they meet the requirements for unanchored tanks in reference documents. Tanks and vessels supported above grade on structural towers or building structures shall be anchored to the supporting structure. The following special detailing requirements shall apply to steel tank and vessel anchor bolts in SDC C, D, E, and F. Anchorage shall be in accordance with Section 15.4.9, whereby the anchor embedment into the concrete shall be designed to develop the steel strength of the anchor in tension. The steel strength of the anchor in tension shall be determined in accor-dance with ACI 318, Appendix D, Eq. D-3. The anchor shall have a minimum gauge length of eight diameters. Post-installed anchors are permitted to be used in accordance with Section 15.4.9.3 provided the anchor embedment into the concrete is designed to develop the steel strength of the anchor in tension. In either case, the load combinations with overstrength of Section 12.4.3 are not to be used to size the anchor bolts for tanks and horizontal and vertical vessels. 15.7.6 Ground-Supported Storage Tanks for Liquids 15.7.6.1 General Ground-supported, ﬂ at bottom tanks storing liquids shall be designed to resist the seismic forces calculated using one of the following procedures: a. The base shear and overturning moment calculated as if the tank and the entire contents are a rigid mass system per Section 15.4.2 of this standard. b. Tanks or vessels storing liquids in Risk Category IV, or with a diameter greater than 20 ft (6.1 m), shall be designed to consider the hydrodynamic pressures of the liquid in determining the equiva-lent lateral forces and lateral force distribution per the applicable reference documents listed in Chapter 23 and the requirements of Section 15.7 of this standard. c. The force and displacement requirements of Section 15.4 of this standard. The design of tanks storing liquids shall consider the impulsive and convective (sloshing) effects and their consequences on the tank, foundation, and attached elements. The impulsive component corresponds to the high-frequency ampliﬁ ed response to the lateral ground motion of the tank roof, the shell, and the portion of the contents that moves in unison with the shell. The convective component corresponds to the low-frequency ampliﬁ ed response of the contents in the fundamental sloshing mode. Damping for the convective component shall be 0.5 percent for the sloshing liquid unless otherwise deﬁ ned by the reference document. The following deﬁ nitions shall apply: Di = inside diameter of tank or vessel HL = design liquid height inside the tank or vessel L = inside length of a rectangular tank, parallel to the direction of the earthquake force being investigated Nh = hydrodynamic hoop force per unit height in the wall of a cylindrical tank or vessel T c = natural period of the ﬁ rst (convective) mode of sloshing Ti = fundamental period of the tank structure and impulsive component of the content Vi = base shear due to impulsive component from weight of tank and contents Vc = base shear due to the convective component of the effective sloshing mass y = distance from base of the tank to level being investigated γL = unit weight of stored liquid The seismic base shear is the combination of the impulsive and convective components: V = Vi + Vc (15.7-4) where V S W R I i ai i e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (15.7-5) V S I W c ac e c = 1 5 . (15.7-6) Sai = the spectral acceleration as a multiplier of gravity including the site impulsive components at period Ti and 5 percent damping For Ti ≤ T s Sai = SDS (15.7-7) For T s < Ti ≤ TL S S T ai D i = 1 (15.7-8) MINIMUM DESIGN LOADS 153 For Ti > TL S S T T ai D L i = 1 2 (15.7-9) NOTES: a. Where a reference document is used in which the spectral acceleration for the tank shell, and the impulsive component of the liquid is independent of Ti, then Sai = SDS. b. Equations 15.7-8 and 15.7-9 shall not be less than the minimum values required in Section 15.4.1 Item 2 multiplied by R/Ie. c. For tanks in Risk Category IV, the value of the importance factor, Ie, used for freeboard determina-tion only shall be taken as 1.0. d. For tanks in Risk Categories I, II, and III, the value of TL used for freeboard determination is permitted to be set equal to 4 s. The value of the importance factor, Ie, used for freeboard determination for tanks in Risk Categories I, II, and III shall be the value determined from Table 1.5-1. e. Impulsive and convective seismic forces for tanks are permitted to be combined using the square root of the sum of the squares (SRSS) method in lieu of the direct sum method shown in Section 15.7.6 and its related subsections. Sac = the spectral acceleration of the sloshing liquid (convective component) based on the sloshing period T c and 0.5 percent damping For T c ≤ TL: S S T S ac D c DS = ≤ 1 5 1 5 1 . . (15.7-10) For T c > TL: S S T T ac D L c = 1 5 1 2 . (15.7-11) EXCEPTION: For T c > 4 s, Sac is permitted be determined by a site-speciﬁ c study using one or more of the following methods: (i) the procedures found in Chapter 21, provided such procedures, which rely on ground-motion attenuation equations for computing response spectra, cover the natural period band containing T c, (ii) ground-motion simulation methods employing seismological models of fault rupture and wave propagation, and (iii) analysis of representative strong-motion accelerogram data with reliable long-period content extending to periods greater than T c. Site-speciﬁ c values of Sac shall be based on one standard deviation determinations. However, in no case shall the value of Sac be taken as less than the value determined in accordance with Eq. 15.7-11 using 50% of the mapped value of TL from Chapter 22. The 80 percent limit on Sa required by Sections 21.3 and 21.4 shall not apply to the determination of site-speciﬁ c values of Sac, which satisfy the requirements of this exception. In determining the value of Sac, the value of TL shall not be less than 4 s where T D g H D c = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 3 68 3 68 π . tanh . (15.7-12) and where D = the tank diameter in ft (m), H = liquid height in ft (m), and g = acceleration due to gravity in consistent units Wi = impulsive weight (impulsive component of liquid, roof and equipment, shell, bottom, and internal elements) Wc = the portion of the liquid weight sloshing 15.7.6.1.1 Distribution of Hydrodynamic and Inertia Forces Unless otherwise required by the appropriate reference document listed in Chapter 23, the method given in ACI 350.3 is permitted to be used to deter-mine the vertical and horizontal distribution of the hydrodynamic and inertia forces on the walls of circular and rectangular tanks. 15.7.6.1.2 Sloshing Sloshing of the stored liquid shall be taken into account in the seismic design of tanks and vessels in accordance with the following requirements: a. The height of the sloshing wave, δs, shall be computed using Eq. 15.7-13 as follows: δs = 0.42DiIeSac (15.7-13) For cylindrical tanks, Di shall be the inside diameter of the tank; for rectangular tanks, the term Di shall be replaced by the longitudinal plan dimension of the tank, L, for the direction under consideration. b. The effects of sloshing shall be accommodated by means of one of the following: 1. A minimum freeboard in accordance with Table 15.7-3. 2. A roof and supporting structure designed to contain the sloshing liquid in accordance with subsection 3 below. 3. For open-top tanks or vessels only, an overﬂ ow spillway around the tank or vessel perimeter. CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 154 c. If the sloshing is restricted because the freeboard is less than the computed sloshing height, then the roof and supporting structure shall be designed for an equivalent hydrostatic head equal to the computed sloshing height less the freeboard. In addition, the design of the tank shall use the conﬁ ned portion of the convective (sloshing) mass as an additional impulsive mass. 15.7.6.1.3 Equipment and Attached Piping Equipment, piping, and walkways or other appurtenances attached to the structure shall be designed to accommodate the displacements imposed by seismic forces. For piping attachments, see Section 15.7.4. 15.7.6.1.4 Internal Elements The attachments of internal equipment and accessories that are attached to the primary liquid or pressure retaining shell or bottom or that provide structural support for major elements (e.g., a column supporting the roof rafters) shall be designed for the lateral loads due to the sloshing liquid in addition to the inertial forces by a substantiated analysis method. 15.7.6.1.5 Sliding Resistance The transfer of the total lateral shear force between the tank or vessel and the subgrade shall be considered: a. For unanchored ﬂ at bottom steel tanks, the overall horizontal seismic shear force is permitted to be resisted by friction between the tank bottom and the foundation or subgrade. Unanchored storage tanks shall be designed such that sliding will not occur where the tank is full of stored product. The maximum calculated seismic base shear, V, shall not exceed V < W tan 30° (15.7-14) W shall be determined using the effective seismic weight of the tank, roof, and contents after reduc-tion for coincident vertical earthquake. Lower values of the friction factor shall be used if the design of the tank bottom to supporting foundation does not justify the friction value above (e.g., leak detection membrane beneath the bottom with a lower friction factor, smooth bottoms, etc.). Alternatively, the friction factor is permitted to be determined by testing in accordance with Section 11.1.4. b. No additional lateral anchorage is required for anchored steel tanks designed in accordance with reference documents. c. The lateral shear transfer behavior for special tank conﬁ gurations (e.g., shovel bottoms, highly crowned tank bottoms, tanks on grillage) can be unique and are beyond the scope of this standard. 15.7.6.1.6 Local Shear Transfer Local transfer of the shear from the roof to the wall and the wall of the tank into the base shall be considered. For cylindrical tanks and vessels, the peak local tangential shear per unit length shall be calculated by v V D max = 2 π (15.7-15) a. Tangential shear in ﬂ at bottom steel tanks shall be transferred through the welded connection to the steel bottom. This transfer mechanism is deemed acceptable for steel tanks designed in accordance with the reference documents where SDS < 1.0g. b. For concrete tanks with a sliding base where the lateral shear is resisted by friction between the tank wall and the base, the friction coefﬁ cient value used for design shall not exceed tan 30°. c. Fixed-base or hinged-base concrete tanks transfer the horizontal seismic base shear shared by membrane (tangential) shear and radial shear into the foundation. For anchored ﬂ exible-base concrete tanks, the majority of the base shear is resisted by membrane (tangential) shear through the anchoring system with only insigniﬁ cant vertical bending in the wall. The connection between the wall and ﬂ oor shall be designed to resist the maximum tangential shear. Table 15.7-3 Minimum Required Freeboard Value of SDS Risk Category I or II III IV SDS < 0.167g a a δs c 0.167g ≤ SDS < 0.33g a a δs c 0.33g ≤ SDS < 0.50g a 0.7δs b δs c SDS ≥ 0.50g a 0.7δs b δs c aNOTE: No minimum freeboard is required. cFreeboard equal to the calculated wave height, δs, is required unless one of the following alternatives is provided: (1) Secondary containment is provided to control the product spill. (2) The roof and supporting structure are designed to contain the sloshing liquid. bA freeboard equal to 0.7δs is required unless one of the following alternatives is provided: (1) Secondary containment is provided to control the product spill. (2) The roof and supporting structure are designed to contain the sloshing liquid. MINIMUM DESIGN LOADS 155 15.7.6.1.7 Pressure Stability For steel tanks, the internal pressure from the stored product stiffens thin cylindrical shell structural elements subjected to membrane compression forces. This stiffening effect is permitted to be considered in resisting seismically induced compressive forces if permitted by the reference document or the authority having jurisdiction. 15.7.6.1.8 Shell Support Steel tanks resting on concrete ring walls or slabs shall have a uniformly supported annulus under the shell. Uniform support shall be provided by one of the following methods: a. Shimming and grouting the annulus. b. Using ﬁ berboard or other suitable padding. c. Using butt-welded bottom or annular plates resting directly on the foundation. d. Using closely spaced shims (without structural grout) provided that the localized bearing loads are considered in the tank wall and foundation to prevent local crippling and spalling. Anchored tanks shall be shimmed and grouted. Local buckling of the steel shell for the peak com-pressive force due to operating loads and seismic overturning shall be considered. 15.7.6.1.9 Repair, Alteration, or Reconstruction Repairs, modiﬁ cations, or reconstruction (i.e., cut down and re-erect) of a tank or vessel shall conform to industry standard practice and this standard. For welded steel tanks storing liquids, see API 653 and the applicable reference document listed in Chapter 23. Tanks that are relocated shall be re-evaluated for the seismic loads for the new site and the requirements of new construction in accordance with the appropriate reference document and this standard. 15.7.7 Water Storage and Water Treatment Tanks and Vessels 15.7.7.1 Welded Steel Welded steel water storage tanks and vessels shall be designed in accordance with the seismic requirements of AWWA D100. 15.7.7.2 Bolted Steel Bolted steel water storage structures shall be designed in accordance with the seismic requirements of AWWA D103 except that the design input forces of AWWA D100 shall be modiﬁ ed in the same manner shown in Section 15.7.7.1 of this standard. 15.7.7.3 Reinforced and Prestressed Concrete Reinforced and prestressed concrete tanks shall be designed in accordance with the seismic require-ments of AWWA D110, AWWA D115, or ACI 350.3 except that the importance factor, Ie, shall be deter-mined according to Section 15.4.1.1, the response modiﬁ cation coefﬁ cient, R, shall be taken from Table 15.4-2, and the design input forces for strength design procedures shall be determined using the procedures of ACI 350.3 except a. Sac shall be substituted for Cc in ACI 350.3 Section 9.4.2 using Eqs. 15.7-10 for T c ≤ TL and 15.7-11. for T c > TL from Section 15.7.6.1; and b. The value of Ct from ACI 350.3 Section 9.4.3 shall be determined using the procedures of Section 15.7.2(c). The values of I, Ri, and b as deﬁ ned in ACI 350.3 shall be taken as 1.0 in the determination of vertical seismic effects. 15.7.8 Petrochemical and Industrial Tanks and Vessels Storing Liquids 15.7.8.1 Welded Steel Welded steel petrochemical and industrial tanks and vessels storing liquids under an internal pressure of less than or equal to 2.5 psig (17.2 kpa g) shall be designed in accordance with the seismic requirements of API 650. Welded steel petrochemical and industrial tanks and vessels storing liquids under an internal pressure of greater than 2.5 psig (17.2 kpa g) and less than or equal to 15 psig (104.4 kpa g) shall be designed in accordance with the seismic requirements of API 620. 15.7.8.2 Bolted Steel Bolted steel tanks used for storage of production liquids. API 12B covers the material, design, and erection requirements for vertical, cylindrical, above-ground bolted tanks in nominal capacities of 100 to 10,000 barrels for production service. Unless required by the authority having jurisdiction, these temporary structures need not be designed for seismic loads. If design for seismic load is required, the loads are permitted to be adjusted for the temporary nature of the anticipated service life. 15.7.8.3 Reinforced and Prestressed Concrete Reinforced concrete tanks for the storage of petrochemical and industrial liquids shall be designed in accordance with the force requirements of Section 15.7.7.3. CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 156 15.7.9 Ground-Supported Storage Tanks for Granular Materials 15.7.9.1 General The intergranular behavior of the material shall be considered in determining effective mass and load paths, including the following behaviors: a. Increased lateral pressure (and the resulting hoop stress) due to loss of the intergranular friction of the material during the seismic shaking. b. Increased hoop stresses generated from temperature changes in the shell after the material has been compacted. c. Intergranular friction, which can transfer seismic shear directly to the foundation. 15.7.9.2 Lateral Force Determination The lateral forces for tanks and vessels storing granular materials at grade shall be determined by the requirements and accelerations for short period structures (i.e., SDS). 15.7.9.3 Force Distribution to Shell and Foundation 15.7.9.3.1 Increased Lateral Pressure The increase in lateral pressure on the tank wall shall be added to the static design lateral pressure but shall not be used in the determination of pressure stability effects on the axial buckling strength of the tank shell. 15.7.9.3.2 Effective Mass A portion of a stored granular mass will act with the shell (the effective mass). The effective mass is related to the physical characteristics of the product, the height-to-diameter (H/D) ratio of the tank, and the intensity of the seismic event. The effective mass shall be used to determine the shear and overturning loads resisted by the tank. 15.7.9.3.3 Effective Density The effective density factor (that part of the total stored mass of product that is accelerated by the seismic event) shall be determined in accordance with ACI 313. 15.7.9.3.4 Lateral Sliding For granular storage tanks that have a steel bottom and are supported such that friction at the bottom to foundation interface can resist lateral shear loads, no additional anchorage to prevent sliding is required. For tanks without steel bottoms (i.e., the material rests directly on the foundation), shear anchorage shall be provided to prevent sliding. 15.7.9.3.5 Combined Anchorage Systems If separate anchorage systems are used to prevent overturning and sliding, the relative stiffness of the systems shall be considered in determining the load distribution. 15.7.9.4 Welded Steel Structures Welded steel granular storage structures shall be designed in accordance with the seismic requirements of this standard. Component allowable stresses and materials shall be per AWWA D100, except the allowable circumferential membrane stresses and material requirements in API 650 shall apply. 15.7.9.5 Bolted Steel Structures Bolted steel granular storage structures shall be designed in accordance with the seismic requirements of this section. Component allowable stresses and materials shall be per AWWA D103. 15.7.9.6 Reinforced Concrete Structures Reinforced concrete structures for the storage of granular materi-als shall be designed in accordance with the seismic force requirements of this standard and the require-ments of ACI 313. 15.7.9.7 Prestressed Concrete Structures Prestressed concrete structures for the storage of granular materials shall be designed in accordance with the seismic force requirements of this standard and the requirements of ACI 313. 15.7.10 Elevated Tanks and Vessels for Liquids and Granular Materials 15.7.10.1 General This section applies to tanks, vessels, bins, and hoppers that are elevated above grade where the supporting tower is an integral part of the structure, or where the primary function of the tower is to support the tank or vessel. Tanks and vessels that are sup-ported within buildings or are incidental to the primary function of the tower are considered mechani-cal equipment and shall be designed in accordance with Chapter 13. Elevated tanks shall be designed for the force and displacement requirements of the applicable reference document or Section 15.4. 15.7.10.2 Effective Mass The design of the supporting tower or pedestal, anchorage, and foundation for seismic overturning shall assume the material stored is a rigid mass acting at the volumetric center of gravity. The effects of MINIMUM DESIGN LOADS 157 ﬂ uid–structure interaction are permitted to be consid-ered in determining the forces, effective period, and mass centroids of the system if the following require-ments are met: a. The sloshing period, T c is greater than 3T where T = natural period of the tank with conﬁ ned liquid (rigid mass) and supporting structure. b. The sloshing mechanism (i.e., the percentage of convective mass and centroid) is determined for the speciﬁ c conﬁ guration of the container by detailed ﬂ uid–structure interaction analysis or testing. Soil–structure interaction is permitted to be included in determining T providing the requirements of Chapter 19 are met. 15.7.10.3 P-Delta Effects The lateral drift of the elevated tank shall be considered as follows: a. The design drift, the elastic lateral displacement of the stored mass center of gravity, shall be increased by the factor Cd for evaluating the additional load in the support structure. b. The base of the tank shall be assumed to be ﬁ xed rotationally and laterally. c. Deﬂ ections due to bending, axial tension, or compression shall be considered. For pedestal tanks with a height-to-diameter ratio less than 5, shear deformations of the pedestal shall be considered. d. The dead load effects of roof-mounted equipment or platforms shall be included in the analysis. e. If constructed within the plumbness tolerances speciﬁ ed by the reference document, initial tilt need not be considered in the P-delta analysis. 15.7.10.4 Transfer of Lateral Forces into Support Tower For post supported tanks and vessels that are cross-braced: a. The bracing shall be installed in such a manner as to provide uniform resistance to the lateral load (e.g., pretensioning or tuning to attain equal sag). b. The additional load in the brace due to the eccentricity between the post to tank attachment and the line of action of the bracing shall be included. c. Eccentricity of compression strut line of action (elements that resist the tensile pull from the bracing rods in the seismic force-resisting systems) with their attachment points shall be considered. d. The connection of the post or leg with the founda-tion shall be designed to resist both the vertical and lateral resultant from the yield load in the bracing assuming the direction of the lateral load is oriented to produce the maximum lateral shear at the post to foundation interface. Where multiple rods are connected to the same location, the anchorage shall be designed to resist the concurrent tensile loads in the braces. 15.7.10.5 Evaluation of Structures Sensitive to Buckling Failure Shell structures that support substantial loads may exhibit a primary mode of failure from localized or general buckling of the support pedestal or skirt due to seismic loads. Such structures may include single pedestal water towers, skirt-supported process vessels, and similar single member towers. Where the struc-tural assessment concludes that buckling of the support is the governing primary mode of failure, structures speciﬁ ed in this standard to be designed to subsections a and b below and those that are assigned as Risk Category IV shall be designed to resist the seismic forces as follows: a. The seismic response coefﬁ cient for this evaluation shall be in accordance with Section 12.8.1.1 of this standard with Ie/R set equal to 1.0. Soil–structure and ﬂ uid–structure interaction is permitted to be utilized in determining the structural response. Vertical or orthogonal combinations need not be considered. b. The resistance of the structure shall be deﬁ ned as the critical buckling resistance of the element, that is, a factor of safety set equal to 1.0. 15.7.10.6 Welded Steel Water Storage Structures Welded steel elevated water storage structures shall be designed and detailed in accordance with the seismic requirements of AWWA D100 with the structural height limits imposed by Table 15.4-2. 15.7.10.7 Concrete Pedestal (Composite) Tanks Concrete pedestal (composite) elevated water storage structures shall be designed in accordance with the requirements of ACI 371R except that the design input forces shall be modiﬁ ed as follows: In Eq. 4-8a of ACI 371R, For T s < T ≤ 2.5 s, replace the term 1 2 2 3 . / C RT v with S T R I D e 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (15.7-24) CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 158 In Eq. 4-8b of ACI 371R, replace the term 2 5 . C R a with S R I DS e ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (15.7-25) In Eq. 4-9 of ACI 371R, replace the term 0.5Ca with 0.2SDS (15.7-26) 15.7.10.7.1 Analysis Procedures The equivalent lateral force procedure is permitted for all concrete pedestal tanks and shall be based on a ﬁ xed-base, single degree-of-freedom model. All mass, including the liquid, shall be considered rigid unless the sloshing mechanism (i.e., the percentage of convective mass and centroid) is determined for the speciﬁ c conﬁ gura-tion of the container by detailed ﬂ uid–structure interaction analysis or testing. Soil–structure interac-tion is permitted to be included. A more rigorous analysis is permitted. 15.7.10.7.2 Structure Period The fundamental period of vibration of the structure shall be established using the uncracked structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis. The period used to calculate the seismic response coefﬁ cient shall not exceed 2.5 s. 15.7.11 Boilers and Pressure Vessels 15.7.11.1 General Attachments to the pressure boundary, supports, and seismic force-resisting anchorage systems for boilers and pressure vessels shall be designed to meet the force and displacement requirements of Section 15.3 or 15.4 and the additional requirements of this section. Boilers and pressure vessels categorized as Risk Categories III or IV shall be designed to meet the force and displacement requirements of Section 15.3 or 15.4. 15.7.11.2 ASME Boilers and Pressure Vessels Boilers or pressure vessels designed and con-structed in accordance with ASME BPVC shall be deemed to meet the requirements of this section provided that the force and displacement requirements of Section 15.3 or 15.4 are used with appropriate scaling of the force and displacement requirements to the working stress design basis. 15.7.11.3 Attachments of Internal Equipment and Refractory Attachments to the pressure boundary for internal and external ancillary components (refractory, cyclones, trays, etc.) shall be designed to resist the seismic forces speciﬁ ed in this standard to safeguard against rupture of the pressure boundary. Alternatively, the element attached is permitted to be designed to fail prior to damaging the pressure boundary provided that the consequences of the failure do not place the pressure boundary in jeopardy. For boilers or vessels containing liquids, the effect of sloshing on the internal equipment shall be considered if the equipment can damage the integrity of the pressure boundary. 15.7.11.4 Coupling of Vessel and Support Structure Where the mass of the operating vessel or vessels supported is greater than 25 percent of the total mass of the combined structure, the structure and vessel designs shall consider the effects of dynamic coupling between each other. Coupling with adjacent, connected structures such as multiple towers shall be considered if the structures are interconnected with elements that will transfer loads from one structure to the other. 15.7.11.5 Effective Mass Fluid–structure interaction (sloshing) shall be considered in determining the effective mass of the stored material providing sufﬁ cient liquid surface exists for sloshing to occur and the T c is greater than 3T. Changes to or variations in material density with pressure and temperature shall be considered. 15.7.11.6 Other Boilers and Pressure Vessels Boilers and pressure vessels designated Risk Category IV, but not designed and constructed in accordance with the requirements of ASME BPVC, shall meet the following requirements: The seismic loads in combination with other service loads and appropriate environmental effects shall not exceed the material strength shown in Table 15.7-4. Consideration shall be made to mitigate seismic impact loads for boiler or vessel elements constructed of nonductile materials or vessels operated in such a way that material ductility is reduced (e.g., low temperature applications). 15.7.11.7 Supports and Attachments for Boilers and Pressure Vessels Attachments to the pressure boundary and support for boilers and pressure vessels shall meet the following requirements: MINIMUM DESIGN LOADS 159 a. Attachments and supports transferring seismic loads shall be constructed of ductile materials suitable for the intended application and environ-mental conditions. b. Anchorage shall be in accordance with Section 15.4.9, whereby the anchor embedment into the concrete is designed to develop the steel strength of the anchor in tension. The steel strength of the anchor in tension shall be determined in accor-dance with ACI 318 Appendix D Eq. D-3. The anchor shall have a minimum gauge length of eight diameters. The load combinations with over-strength of Section 12.4.3 are not to be used to size the anchor bolts for tanks and horizontal and vertical vessels. c. Seismic supports and attachments to structures shall be designed and constructed so that the support or attachment remains ductile throughout the range of reversing seismic lateral loads and displacements. d. Vessel attachments shall consider the potential effect on the vessel and the support for uneven vertical reactions based on variations in relative stiffness of the support members, dissimilar details, nonuniform shimming, or irregular supports. Uneven distribution of lateral forces shall consider the relative distribution of the resisting elements, the behavior of the connection details, and vessel shear distribution. The requirements of Sections 15.4 and 15.7.10.5 shall also be applicable to this section. 15.7.12 Liquid and Gas Spheres 15.7.12.1 General Attachments to the pressure or liquid boundary, supports, and seismic force-resisting anchorage systems for liquid and gas spheres shall be designed to meet the force and displacement requirements of Section 15.3 or 15.4 and the additional requirements of this section. Spheres categorized as Risk Category III or IV shall themselves be designed to meet the force and displacement requirements of Section 15.3 or 15.4. 15.7.12.2 ASME Spheres Spheres designed and constructed in accordance with Section VIII of ASME BPVC shall be deemed to meet the requirements of this section providing the force and displacement requirements of Section 15.3 or 15.4 are used with appropriate scaling of the force and displacement requirements to the working stress design basis. 15.7.12.3 Attachments of Internal Equipment and Refractory Attachments to the pressure or liquid boundary for internal and external ancillary components (refractory, cyclones, trays, etc.) shall be designed to resist the seismic forces speciﬁ ed in this standard to safeguard against rupture of the pressure boundary. Alternatively, the element attached to the sphere could be designed to fail prior to damaging the pressure or liquid boundary providing the conse-quences of the failure does not place the pressure boundary in jeopardy. For spheres containing liquids, the effect of sloshing on the internal equipment shall be considered if the equipment can damage the pressure boundary. 15.7.12.4 Effective Mass Fluid–structure interaction (sloshing) shall be considered in determining the effective mass of the stored material providing sufﬁ cient liquid surface exists for sloshing to occur and the T c is greater than 3T. Changes to or variations in ﬂ uid density shall be considered. Table 15.7-4 Maximum Material Strength Material Minimum Ratio Fu/Fy Max. Material Strength Vessel Material Max. Material Strength Threaded Materiala Ductile (e.g., steel, aluminum, copper) 1.33b 90%d 70%d Semiductile 1.2c 70%d 50%d Nonductile (e.g., cast iron, ceramics, ﬁ berglass) NA 25%e 20%e aThreaded connection to vessel or support system. bMinimum 20% elongation per the ASTM material speciﬁ cation. dBased on material minimum speciﬁ ed yield strength. cMinimum 15% elongation per the ASTM material speciﬁ cation. eBased on material minimum speciﬁ ed tensile strength. CHAPTER 15 SEISMIC DESIGN REQUIREMENTS FOR NONBUILDING STRUCTURES 160 15.7.12.5 Post and Rod Supported For post supported spheres that are cross-braced: a. The requirements of Section 15.7.10.4 shall also be applicable to this section. b. The stiffening effect (reduction in lateral drift) from pretensioning of the bracing shall be consid-ered in determining the natural period. c. The slenderness and local buckling of the posts shall be considered. d. Local buckling of the sphere shell at the post attachment shall be considered. e. For spheres storing liquids, bracing connections shall be designed and constructed to develop the minimum published yield strength of the brace. For spheres storing gas vapors only, bracing connection shall be designed for Ω0 times the maximum design load in the brace. Lateral bracing connec-tions directly attached to the pressure or liquid boundary are prohibited. 15.7.12.6 Skirt Supported For skirt-supported spheres, the following requirements shall apply: a. The requirements of Section 15.7.10.5 shall also apply. b. The local buckling of the skirt under compressive membrane forces due to axial load and bending moments shall be considered. c. Penetration of the skirt support (manholes, piping, etc.) shall be designed and constructed to maintain the strength of the skirt without penetrations. 15.7.13 Refrigerated Gas Liquid Storage Tanks and Vessels 15.7.13.1 General Tanks and facilities for the storage of liqueﬁ ed hydrocarbons and refrigerated liquids shall meet the requirements of this standard. Low-pressure welded steel storage tanks for liqueﬁ ed hydrocarbon gas (e.g., LPG, butane, etc.) and refrigerated liquids (e.g., ammonia) shall be designed in accordance with the requirements of Section 15.7.8 and API 620. 15.7.14 Horizontal, Saddle Supported Vessels for Liquid or Vapor Storage 15.7.14.1 General Horizontal vessels supported on saddles (some-times referred to as “blimps”) shall be designed to meet the force and displacement requirements of Section 15.3 or 15.4. 15.7.14.2 Effective Mass Changes to or variations in material density shall be considered. The design of the supports, saddles, anchorage, and foundation for seismic overturning shall assume the material stored is a rigid mass acting at the volumetric center of gravity. 15.7.14.3 Vessel Design Unless a more rigorous analysis is performed a. Horizontal vessels with a length-to-diameter ratio of 6 or more are permitted to be assumed to be a simply supported beam spanning between the saddles for determining the natural period of vibration and global bending moment. b. For horizontal vessels with a length-to-diameter ratio of less than 6, the effects of “deep beam shear” shall be considered where determining the fundamental period and stress distribution. c. Local bending and buckling of the vessel shell at the saddle supports due to seismic load shall be considered. The stabilizing effects of internal pressure shall not be considered to increase the buckling resistance of the vessel shell. d. If the vessel is a combination of liquid and gas storage, the vessel and supports shall be designed both with and without gas pressure acting (assume piping has ruptured and pressure does not exist). 161 Chapter 16 SEISMIC RESPONSE HISTORY PROCEDURES horizontal ground motion acceleration components that shall be selected and scaled from individual recorded events. Appropriate ground motions shall be selected from events having magnitudes, fault distance, and source mechanisms that are consistent with those that control the maximum considered earthquake. Where the required number of recorded ground motion pairs is not available, appropriate simulated ground motion pairs are permitted to be used to make up the total number required. For each pair of horizontal ground motion components, a square root of the sum of the squares (SRSS) spec-trum shall be constructed by taking the SRSS of the 5 percent-damped response spectra for the scaled components (where an identical scale factor is applied to both components of a pair). Each pair of motions shall be scaled such that in the period range from 0.2T to 1.5T, the average of the SRSS spectra from all horizontal component pairs does not fall below the corresponding ordinate of the response spectrum used in the design, determined in accordance with Section 11.4.5 or 11.4.7. At sites within 3 miles (5 km) of the active fault that controls the hazard, each pair of components shall be rotated to the fault-normal and fault-parallel directions of the causative fault and shall be scaled so that the average of the fault-normal components is not less than the MCER response spectrum for the period range from 0.2T to 1.5T. 16.1.4 Response Parameters For each ground motion analyzed, the individual response parameters shall be multiplied by the following scalar quantities: a. Force response parameters shall be multiplied by Ie/R, where Ie is the importance factor determined in accordance with Section 11.5.1 and R is the Response Modiﬁ cation Coefﬁ cient selected in accordance with Section 12.2.1. b. Drift quantities shall be multiplied by Cd/R, where Cd is the deﬂ ection ampliﬁ cation factor speciﬁ ed in Table 12.2-1. For each ground motion i, where i is the designa-tion assigned to each ground motion, the maximum value of the base shear, Vi, member forces, QEi, scaled as indicated in the preceding text and story drifts, Δi, at each story as deﬁ ned in Section 12.8.6 shall be 16.1 LINEAR RESPONSE HISTORY PROCEDURE Where linear response history procedure is performed the requirements of this chapter shall be satisﬁ ed. 16.1.1 Analysis Requirements A linear response history analysis shall consist of an analysis of a linear mathematical model of the structure to determine its response, through methods of numerical integration, to suites of ground motion acceleration histories compatible with the design response spectrum for the site. The analysis shall be performed in accordance with the requirements of this section. 16.1.2 Modeling Mathematical models shall conform to the requirements of Section 12.7. 16.1.3 Ground Motion A suite of not less than three appropriate ground motions shall be used in the analysis. Ground motion shall conform to the requirements of this section. 16.1.3.1 Two-Dimensional Analysis Where two-dimensional analyses are performed, each ground motion shall consist of a horizontal acceleration history, selected from an actual recorded event. Appropriate acceleration histories shall be obtained from records of events having magnitudes, fault distance, and source mechanisms that are consistent with those that control the maximum considered earthquake. Where the required number of appropriate recorded ground motion records are not available, appropriate simulated ground motion records shall be used to make up the total number required. The ground motions shall be scaled such that the average value of the 5 percent damped response spectra for the suite of motions is not less than the design response spectrum for the site for periods ranging from 0.2T to 1.5T where T is the natural period of the structure in the fundamental mode for the direction of response being analyzed. 16.1.3.2 Three-Dimensional Analysis Where three-dimensional analyses are performed, ground motions shall consist of pairs of appropriate CHAPTER 16 SEISMIC RESPONSE HISTORY PROCEDURES 162 determined. Where the maximum scaled base shear predicted by the analysis, Vi, is less than 85 percent of the value of V determined using the minimum value of Cs set forth in Eq. 12.8-5 or when located where S1 is equal to or greater than 0.6g, the minimum value of Cs set forth in Eq. 12.8-6, the scaled member forces, QEi, shall be additionally multiplied by V Vi where V is the minimum base shear that has been determined using the minimum value of Cs set forth in Eq. 12.8-5, or when located where S1 is equal to or greater than 0.6g, the minimum value of Cs set forth in Eq. 12.8-6. Where the maximum scaled base shear predicted by the analysis, Vi, is less than 0.85CsW, where Cs is from Eq. 12.8-6, drifts shall be multiplied by 0.85C W V s i . If at least seven ground motions are analyzed, the design member forces used in the load combinations of Section 12.4.2.1 and the design story drift used in the evaluation of drift in accordance with Section 12.12.1 are permitted to be taken respectively as the average of the scaled QEi and Δi values determined from the analyses and scaled as indicated in the preceding text. If fewer than seven ground motions are analyzed, the design member forces and the design story drift shall be taken as the maximum value of the scaled QEi and Δi values determined from the analyses. Where this standard requires consideration of the seismic load effects including overstrength factor of Section 12.4.3, the value of Ω0QE need not be taken larger than the maximum of the unscaled value, QEi, obtained from the analyses. 16.1.5 Horizontal Shear Distribution The distribution of horizontal shear shall be in accordance with Section 12.8.4 except that ampliﬁ ca-tion of torsion in accordance with Section 12.8.4.3 is not required where accidental torsion effects are included in the dynamic analysis model. 16.2 NONLINEAR RESPONSE HISTORY PROCEDURE Where nonlinear response history procedure is performed the requirements of Section 16.2 shall be satisﬁ ed. 16.2.1 Analysis Requirements A nonlinear response history analysis shall consist of an analysis of a mathematical model of the structure that directly accounts for the nonlinear hysteretic behavior of the structure’s elements to determine its response through methods of numerical integration to suites of ground motion acceleration histories compatible with the design response spec-trum for the site. The analysis shall be performed in accordance with this section. See Section 12.1.1 for limitations on the use of this procedure. 16.2.2 Modeling A mathematical model of the structure shall be constructed that represents the spatial distribution of mass throughout the structure. The hysteretic behavior of elements shall be modeled consistent with suitable laboratory test data and shall account for all signiﬁ -cant yielding, strength degradation, stiffness degrada-tion, and hysteretic pinching indicated by such test data. Strength of elements shall be based on expected values considering material overstrength, strain hardening, and hysteretic strength degradation. Linear properties, consistent with the requirements of Section 12.7.3, are permitted to be used for those elements demonstrated by the analysis to remain within their linear range of response. The structure shall be assumed to have a ﬁ xed-base, or alternatively, it is permitted to use realistic assumptions with regard to the stiffness and load-carrying characteristics of the foundations consistent with site-speciﬁ c soils data and rational principles of engineering mechanics. For regular structures with independent orthogo-nal seismic force-resisting systems, independent 2-D models are permitted to be constructed to represent each system. For structures having a horizontal structural irregularity of Type 1a, 1b, 4, or 5 of Table 12.3-1 or structures without independent orthogonal systems, a 3-D model incorporating a minimum of three dynamic degrees of freedom consisting of translation in two orthogonal plan directions and torsional rotation about the vertical axis at each level of the structure shall be used. Where the diaphragms are not rigid compared to the vertical elements of the seismic force-resisting system, the model should include representation of the diaphragm’s ﬂ exibility and such additional dynamic degrees of freedom as are required to account for the participation of the diaphragm in the structure’s dynamic response. 16.2.3 Ground Motion and Other Loading Ground motion shall conform to the requirements of Section 16.1.3. The structure shall be analyzed for the effects of these ground motions simultaneously with the effects of dead load in combination with not less than 25 percent of the required live loads. MINIMUM DESIGN LOADS 163 16.2.4 Response Parameters For each ground motion analyzed, individual response parameters consisting of the maximum value of the individual member forces, QEi, member inelastic deformations, ψi, and story drifts, Δi, at each story shall be determined, where i is the designation assigned to each ground motion. If at least seven ground motions are analyzed, the design values of member forces, QE, member inelastic deformations, ψ, and story drift, Δ, are permitted to be taken as the average of the QEi, ψi, and Δi values determined from the analyses. If fewer than seven ground motions are analyzed, the design member forces, QE, design member inelastic deformations, ψ, and the design story drift, Δ, shall be taken as the maximum value of the QEi, ψi, and Δi values deter-mined from the analyses. 16.2.4.1 Member Strength The adequacy of members to resist the combina-tion of load effects of Section 12.4 need not be evaluated. EXCEPTION: Where this standard requires consideration of the seismic load effects including overstrength factor of Section 12.4.3, the maximum value of QEi obtained from the suite of analyses shall be taken in place of the quantity Ω0QE· 16.2.4.2 Member Deformation The adequacy of individual members and their connections to withstand the estimated design deformation values, ψi, as predicted by the analyses shall be evaluated based on laboratory test data for similar elements. The effects of gravity and other loads on member deformation capacity shall be considered in these evaluations. Member deformation shall not exceed two-thirds of a value that results in loss of ability to carry gravity loads or that results in deterioration of member strength to less than the 67 percent of the peak value. 16.2.4.3 Story Drift The design story drift, Δi, obtained from the analyses shall not exceed 125 percent of the drift limit speciﬁ ed in Section 12.12.1. 16.2.5 Design Review A design review of the seismic force-resisting system and the structural analysis shall be performed by an independent team of registered design profes-sionals in the appropriate disciplines and others experienced in seismic analysis methods and the theory and application of nonlinear seismic analysis and structural behavior under extreme cyclic loads. The design review shall include, but need not be limited to, the following: 1. Review of any site-speciﬁ c seismic criteria employed in the analysis including the develop-ment of site-speciﬁ c spectra and ground motion time histories. 2. Review of acceptance criteria used to demonstrate the adequacy of structural elements and systems to withstand the calculated force and deformation demands, together with that laboratory and other data used to substantiate these criteria. 3. Review of the preliminary design including the selection of structural system and the conﬁ guration of structural elements. 4. Review of the ﬁ nal design of the entire structural system and all supporting analyses. 165 Chapter 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES EFFECTIVE STIFFNESS: The value of the lateral force in the isolation system, or an element thereof, divided by the corresponding lateral displacement. ISOLATION INTERFACE: The boundary between the upper portion of the structure, which is isolated, and the lower portion of the structure, which moves rigidly with the ground. ISOLATION SYSTEM: The collection of structural elements that includes all individual isolator units, all structural elements that transfer force between elements of the isolation system, and all connections to other structural elements. The isolation system also includes the wind-restraint system, energy-dissipation devices, and/or the displacement restraint system if such systems and devices are used to meet the design requirements of this chapter. ISOLATOR UNIT: A horizontally ﬂ exible and vertically stiff structural element of the isolation system that permits large lateral deformations under design seismic load. An isolator unit is permitted to be used either as part of, or in addition to, the weight-supporting system of the structure. MAXIMUM DISPLACEMENT: The maximum considered earthquake lateral displacement, excluding additional displacement due to actual and accidental torsion. SCRAGGING: Cyclic loading or working of rubber products, including elastomeric isolators, to effect a reduction in stiffness properties, a portion of which will be recovered over time. WIND-RESTRAINT SYSTEM: The collection of structural elements that provides restraint of the seismic-isolated structure for wind loads. The wind-restraint system is permitted to be either an integral part of isolator units or a separate device. 17.1.3 Notation BD = numerical coefﬁ cient as set forth in Table 17.5-1 for effective damping equal to βD BM = numerical coefﬁ cient as set forth in Table 17.5-1 for effective damping equal to βM b = shortest plan dimension of the structure, in ft (mm) measured perpendicular to d 17.1 GENERAL Every seismically isolated structure and every portion thereof shall be designed and constructed in accor-dance with the requirements of this section and the applicable requirements of this standard. 17.1.1 Variations in Material Properties The analysis of seismically isolated structures, including the substructure, isolators, and superstruc-ture, shall consider variations in seismic isolator material properties over the projected life of the structure including changes due to aging, contamina-tion, environmental exposure, loading rate, scragging, and temperature. 17.1.2 Deﬁ nitions DISPLACEMENT: Design Displacement: The design earthquake lateral displacement, excluding additional displacement due to actual and accidental torsion, required for design of the isolation system. Total Design Displacement: The design earthquake lateral displacement, including additional displacement due to actual and accidental torsion, required for design of the isolation system or an element thereof. Total Maximum Displacement: The maximum considered earthquake lateral displacement, including additional displacement due to actual and accidental torsion, required for veriﬁ cation of the stability of the isolation system or elements thereof, design of structure separa-tions, and vertical load testing of isolator unit prototypes. DISPLACEMENT RESTRAINT SYSTEM: A collection of structural elements that limits lateral displacement of seismically isolated structures due to the maximum considered earthquake. EFFECTIVE DAMPING: The value of equiva-lent viscous damping corresponding to energy dissipated during cyclic response of the isolation system. CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 166 DD = design displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.5-1 D′D = design displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.6-1 DM = maximum displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consider-ation, as prescribed by Eq. 17.5-3 D′M = maximum displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consider-ation, as prescribed by Eq. 17.6-2 DTD = total design displacement, in in. (mm), of an element of the isolation system includ-ing both translational displacement at the center of rigidity and the component of torsional displacement in the direction under consideration, as prescribed by Eq. 17.5-5 DTM = total maximum displacement, in in. (mm), of an element of the isolation system including both translational displacement at the center of rigidity and the component of torsional displacement in the direction under consideration, as prescribed by Eq. 17.5-6 d = longest plan dimension of the structure, in ft (mm) Eloop = energy dissipated in kips-in. (kN-mm), in an isolator unit during a full cycle of reversible load over a test displacement range from Δ+ to Δ–, as measured by the area enclosed by the loop of the force-deﬂ ection curve e = actual eccentricity, in ft (mm), measured in plan between the center of mass of the structure above the isolation interface and the center of rigidity of the isolation system, plus accidental eccentricity, in ft. (mm), taken as 5 percent of the maximum building dimension perpen dicular to the direction of force under consideration F– = minimum negative force in an isolator unit during a single cycle of prototype testing at a displacement amplitude of Δ– F+ = maximum positive force in kips (kN) in an isolator unit during a single cycle of proto type testing at a displacement amplitude of Δ+ Fx = total force distributed over the height of the structure above the isolation interface as prescribed by Eq. 17.5-9 kDmax = maximum effective stiffness, in kips/in. (kN/mm), of the isolation system at the design displacement in the horizontal direction under consideration, as prescribed by Eq. 17.8-3 kDmin = minimum effective stiffness, in kips/in. (kN/mm), of the isolation system at the design displacement in the horizontal direction under consideration, as prescribed by Eq. 17.8-4 kMmax = maximum effective stiffness, in kips/in. (kN/mm), of the isolation system at the maximum displacement in the horizontal direction under consideration, as prescribed by Eq. 17.8-5 kMmin = minimum effective stiffness, in kips/in. (kN/mm), of the isolation system at the maximum displacement in the horizontal direction under consideration, as pre-scribed by Eq. 17.8-6 keff = effective stiffness of an isolator unit, as prescribed by Eq. 17.8-1 L = effect of live load in Chapter 17 TD = effective period, in s, of the seismically isolated structure at the design displace-ment in the direction under consideration, as prescribed by Eq. 17.5-2 TM = effective period, in s, of the seismically isolated structure at the maximum displacement in the direction under consideration, as prescribed by Eq. 17.5-4 Vb = total lateral seismic design force or shear on elements of the isolation system or elements below isolation system, as prescribed by Eq. 17.5-7 Vs = total lateral seismic design force or shear on elements above the isolation system, as prescribed by Eq. 17.5-8 y = distance, in ft (mm), between the center of rigidity of the isolation system rigidity and the element of interest measured perpendicular to the direction of seismic loading under consideration βD = effective damping of the isolation system at the design displacement, as prescribed by Eq. 17.8-7 βM = effective damping of the isolation system at the maximum displacement, as pre-scribed by Eq. 17.8-8 MINIMUM DESIGN LOADS 167 βeff = effective damping of the isolation system, as prescribed by Eq. 17.8-2 Δ+ = maximum positive displacement of an isolator unit during each cycle of prototype testing Δ– = minimum negative displacement of an isolator unit during each cycle of prototype testing ΣED = total energy dissipated, in kips-in. (kN-mm), in the isolation system during a full cycle of response at the design displacement, DD ΣEM = total energy dissipated, in kips-in. (kN-mm), in the isolation system during a full cycle of response at the maximum displacement, DM Σ\|FD +\|max = sum, for all isolator units, of the maximum absolute value of force, in kips (kN), at a positive displacement equal to DD Σ\|FD +\|min = sum, for all isolator units, of the minimum absolute value of force, in kips (kN), at a positive displacement equal to DD Σ\|FD –\|max = sum, for all isolator units, of the maximum absolute value of force, in kips (kN), at a negative displacement equal to DD Σ\|FD –\|min = sum, for all isolator units, of the minimum absolute value of force, in kips (kN), at a negative displacement equal to DD Σ\|FM +\|max = sum, for all isolator units, of the maximum absolute value of force, in kips (kN), at a positive displacement equal to DM Σ\|FM +\|min = sum, for all isolator units, of the minimum absolute value of force, in kips (kN), at a positive displacement equal to DM Σ\|FM –\|max = sum, for all isolator units, of the maximum absolute value of force, in kips (kN), at a negative displacement equal to DM Σ\|FM –\|min = sum, for all isolator units, of the minimum absolute value of force, in kips (kN), at a negative displacement equal to DM 17.2 GENERAL DESIGN REQUIREMENTS 17.2.1 Importance Factor All portions of the structure, including the structure above the isolation system, shall be assigned a risk category in accordance with Table 1.5-1. The importance factor, Ie, shall be taken as 1.0 for a seismically isolated structure, regardless of its risk category assignment. 17.2.2 MCER Spectral Response Acceleration Parameters, SMS and SM1 The MCER spectral response acceleration param-eters SMS and SM1 shall be determined in accordance with Section 11.4.3. 17.2.3 Conﬁ guration Each structure shall be designated as having a structural irregularity based on the structural conﬁ gu-ration above the isolation system. 17.2.4 Isolation System 17.2.4.1 Environmental Conditions In addition to the requirements for vertical and lateral loads induced by wind and earthquake, the isolation system shall provide for other environmental conditions including aging effects, creep, fatigue, operating temperature, and exposure to moisture or damaging substances. 17.2.4.2 Wind Forces Isolated structures shall resist design wind loads at all levels above the isolation interface. At the isolation interface, a wind-restraint system shall be provided to limit lateral displacement in the isolation system to a value equal to that required between ﬂ oors of the structure above the isolation interface in accordance with Section 17.5.6. 17.2.4.3 Fire Resistance Fire resistance for the isolation system shall meet that required for the columns, walls, or other such gravity-bearing elements in the same region of the structure. 17.2.4.4 Lateral Restoring Force The isolation system shall be conﬁ gured to produce a restoring force such that the lateral force at the total design displacement is at least 0.025W greater than the lateral force at 50 percent of the total design displacement. 17.2.4.5 Displacement Restraint The isolation system shall not be conﬁ gured to include a displacement restraint that limits lateral displacement due to the maximum considered earthquake to less than the total maximum displace-ment unless the seismically isolated structure is designed in accordance with the following criteria where more stringent than the requirements of Section 17.2: CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 168 1. Maximum considered earthquake response is calculated in accordance with the dynamic analysis requirements of Section 17.6, explicitly considering the nonlinear characteristics of the isolation system and the structure above the isolation system. 2. The ultimate capacity of the isolation system and structural elements below the isolation system shall exceed the strength and displacement demands of the maximum considered earthquake. 3. The structure above the isolation system is checked for stability and ductility demand of the maximum considered earthquake. 4. The displacement restraint does not become effective at a displacement less than 0.75 times the total design displacement unless it is demonstrated by analysis that earlier engagement does not result in unsatisfactory performance. 17.2.4.6 Vertical-Load Stability Each element of the isolation system shall be designed to be stable under the design vertical load where subjected to a horizontal displacement equal to the total maximum displacement. The design vertical load shall be computed using load combination 5 of Section 2.3.2 for the maximum vertical load and load combination 7 of Section 12.4.2.3 for the minimum vertical load where SDS in these equations is replaced by SMS. The vertical loads that result from application of horizontal seismic forces, QE, shall be based on peak response due to the maximum considered earthquake. 17.2.4.7 Overturning The factor of safety against global structural overturning at the isolation interface shall not be less than 1.0 for required load combinations. All gravity and seismic loading conditions shall be investigated. Seismic forces for overturning calculations shall be based on the maximum considered earthquake, and W shall be used for the vertical restoring force. Local uplift of individual elements shall not be allowed unless the resulting deﬂ ections do not cause overstress or instability of the isolator units or other structure elements. 17.2.4.8 Inspection and Replacement a. Access for inspection and replacement of all components of the isolation system shall be provided. b. A registered design professional shall complete a ﬁ nal series of inspections or observations of structure separation areas and components that cross the isolation interface prior to the issuance of the certiﬁ cate of occupancy for the seismically isolated structure. Such inspections and observa-tions shall indicate that the conditions allow free and unhindered displacement of the structure to maximum design levels and that all components that cross the isolation interface as installed are able to accommodate the stipulated displacements. c. Seismically isolated structures shall have a moni-toring, inspection, and maintenance program for the isolation system established by the registered design professional responsible for the design of the isolation system. d. Remodeling, repair, or retroﬁ tting at the isolation system interface, including that of components that cross the isolation interface, shall be performed under the direction of a registered design professional. 17.2.4.9 Quality Control A quality control testing program for isolator units shall be established by the registered design professional responsible for the structural design. 17.2.5 Structural System 17.2.5.1 Horizontal Distribution of Force A horizontal diaphragm or other structural elements shall provide continuity above the isolation interface and shall have adequate strength and ductility to transmit forces (due to nonuniform ground motion) from one part of the structure to another. 17.2.5.2 Building Separations Minimum separations between the isolated structure and surrounding retaining walls or other ﬁ xed obstructions shall not be less than the total maximum displacement. 17.2.5.3 Nonbuilding Structures Nonbuilding structures shall be designed and constructed in accordance with the requirements of Chapter 15 using design displacements and forces calculated in accordance with Sections 17.5 or 17.6. 17.2.6 Elements of Structures and Nonstructural Components Parts or portions of an isolated structure, perma-nent nonstructural components and the attachments to them, and the attachments for permanent equipment supported by a structure shall be designed to resist seismic forces and displacements as prescribed by this section and the applicable requirements of Chapter 13. MINIMUM DESIGN LOADS 169 17.2.6.1 Components at or above the Isolation Interface Elements of seismically isolated structures and nonstructural components, or portions thereof, that are at or above the isolation interface shall be designed to resist a total lateral seismic force equal to the maximum dynamic response of the element or component under consideration. EXCEPTION: Elements of seismically isolated structures and nonstructural components or portions designed to resist seismic forces and displacements as prescribed in Chapter 12 or 13 as appropriate. 17.2.6.2 Components Crossing the Isolation Interface Elements of seismically isolated structures and nonstructural components, or portions thereof, that cross the isolation interface shall be designed to withstand the total maximum displacement. 17.2.6.3 Components below the Isolation Interface Elements of seismically isolated structures and nonstructural components, or portions thereof, that are below the isolation interface shall be designed and constructed in accordance with the requirements of Section 12.1 and Chapter 13. 17.3 GROUND MOTION FOR ISOLATED SYSTEMS 17.3.1 Design Spectra The site-speciﬁ c ground motion procedures set forth in Chapter 21 are permitted to be used to determine ground motions for any structure. For structures on Site Class F sites, site response analysis shall be performed in accordance with Section 21.1. For seismically isolated structures on sites with S1 greater than or equal to 0.6, a ground motion hazard analysis shall be performed in accordance with Section 21.2. Structures that do not require or use site-speciﬁ c ground motion procedures shall be analyzed using the design spectrum for the design earthquake developed in accordance with Section 11.4.5. A spectrum shall be constructed for the MCER ground motion. The spectrum for MCER ground motions shall not be taken as less than 1.5 times the spectrum for the design earthquake ground motions. 17.3.2 Ground Motion Histories Where response-history procedures are used, ground motions shall consist of pairs of appropriate horizontal ground motion acceleration components developed per Section 16.1.3.2 except that 0.2T and 1.5T shall be replaced by 0.5TD and 1.25TM, respec-tively, where TD and TM are deﬁ ned in Section 17.5.3. 17.4 ANALYSIS PROCEDURE SELECTION Seismically isolated structures except those deﬁ ned in Section 17.4.1 shall be designed using the dynamic procedures of Section 17.6. 17.4.1 Equivalent Lateral Force Procedure The equivalent lateral force procedure of Section 17.5 is permitted to be used for design of a seismi-cally isolated structure provided that 1. The structure is located at a site with S1 less than 0.60g. 2. The structure is located on a Site Class A, B, C, or D. 3. The structure above the isolation interface is less than or equal to four stories or 65 ft (19.8 m) in structural height, hn, measured from the base as deﬁ ned in Section 11.2. 4. The effective period of the isolated structure at the maximum displacement, TM, is less than or equal to 3.0 s. 5. The effective period of the isolated structure at the design displacement, TD, is greater than three times the elastic, ﬁ xed-base period of the structure above the isolation system as determined by Eq. 12.8-7 or 12.8-8. 6. The structure above the isolation system is of regular conﬁ guration. 7. The isolation system meets all of the following criteria: a. The effective stiffness of the isolation system at the design displacement is greater than one-third of the effective stiffness at 20 percent of the design displacement. b. The isolation system is capable of producing a restoring force as speciﬁ ed in Section 17.2.4.4. c. The isolation system does not limit maximum considered earthquake displacement to less than the total maximum displacement. 17.4.2 Dynamic Procedures The dynamic procedures of Section 17.6 are permitted to be used as speciﬁ ed in this section. 17.4.2.1 Response-Spectrum Procedure Response-spectrum analysis shall not be used for design of a seismically isolated structure unless: CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 170 1. The structure is located on a Site Class A, B, C, or D. 2. The isolation system meets the criteria of Item 7 of Section 17.4.1. 17.4.2.2 Response-History Procedure The response-history procedure is permitted for design of any seismically isolated structure and shall be used for design of all seismically isolated struc-tures not meeting the criteria of Section 17.4.2.1. 17.5 EQUIVALENT LATERAL FORCE PROCEDURE 17.5.1 General Where the equivalent lateral force procedure is used to design seismically isolated structures, the requirements of this section shall apply. 17.5.2 Deformation Characteristics of the Isolation System Minimum lateral earthquake design displacements and forces on seismically isolated structures shall be based on the deformation characteristics of the isolation system. The deformation characteristics of the isolation system shall explicitly include the effects of the wind-restraint system if such a system is used to meet the design requirements of this standard. The deformation characteristics of the isolation system shall be based on properly substantiated tests per-formed in accordance with Section 17.8. 17.5.3 Minimum Lateral Displacements 17.5.3.1 Design Displacement The isolation system shall be designed and constructed to withstand minimum lateral earthquake displacements, DD, that act in the direction of each of the main horizontal axes of the structure using Eq. 17.5-1: D gS T B D D D D = 1 2 4π (17.5-1) where g = acceleration due to gravity. The units for g are in./s2 (mm/s2) if the units of the design displace-ment, DD, are in. (mm) SD1 = design 5 percent damped spectral acceleration parameter at 1-s period in units of g-s, as determined in Section 11.4.4 TD = effective period of the seismically isolated structure in seconds, at the design displacement in the direction under consideration, as pre-scribed by Eq. 17.5-2 BD = numerical coefﬁ cient related to the effective damping of the isolation system at the design displacement, βD, as set forth in Table 17.5-1 17.5.3.2 Effective Period at Design Displacement The effective period of the isolated structure at design displacement, TD, shall be determined using the deformational characteristics of the isolation system and Eq. 17.5-2: T W k g D D = 2π min (17.5-2) where W = effective seismic weight of the structure above the isolation interface as deﬁ ned in Section 12.7.2 kDmin = minimum effective stiffness in kips/in. (kN/ mm) of the isolation system at the design displacement in the horizontal direction under consideration, as prescribed by Eq. 17.8-4 g = acceleration due to gravity 17.5.3.3 Maximum Displacement The maximum displacement of the isolation system, DM, in the most critical direction of horizontal response shall be calculated using Eq. 17.5-3: D gS T B M M M M = 1 2 4π (17.5-3) Table 17.5-1 Damping Coefﬁ cient, BD or BM Effective Damping, βD or βM (percentage of critical)a,b BD or BM Factor ≤2 0.8 5 1.0 10 1.2 20 1.5 30 1.7 40 1.9 ≥50 2.0 a The damping coefﬁ cient shall be based on the effective damping of the isolation system determined in accordance with the requirements of Section 17.8.5.2. b The damping coefﬁ cient shall be based on linear interpolation for effective damping values other than those given. MINIMUM DESIGN LOADS 171 where g = acceleration of gravity SM1 = maximum considered earthquake 5 percent damped spectral acceleration parameter at 1-s period, in units of g-s, as determined in Section 11.4.3 TM = effective period, in seconds, of the seismically isolated structure at the maximum displacement in the direction under consideration, as pre-scribed by Eq. 17.5-4 BM = numerical coefﬁ cient related to the effective damping of the isolation system at the maximum displacement, βM, as set forth in Table 17.5-1 17.5.3.4 Effective Period at Maximum Displacement The effective period of the isolated structure at maximum displacement, TM, shall be determined using the deformational characteristics of the isolation system and Eq. 17.5-4: T W k g M M = 2π min (17.5-4) where W = effective seismic weight of the structure above the isolation interface as deﬁ ned in Section 12.7.2 (kip or kN) kMmin = minimum effective stiffness, in kips/in. (kN/mm), of the isolation system at the maximum displacement in the horizontal direction under consideration, as prescribed by Eq. 17.8-6 g = the acceleration of gravity 17.5.3.5 Total Displacement The total design displacement, DTD, and the total maximum displacement, DTM, of elements of the isolation system shall include additional displacement due to actual and accidental torsion calculated from the spatial distribution of the lateral stiffness of the isolation system and the most disadvantageous location of eccentric mass. The total design displacement, DTD, and the total maximum displacement, DTM, of elements of an isolation system with uniform spatial distribution of lateral stiffness shall not be taken as less than that prescribed by Eqs. 17.5-5 and 17.5-6: D D y e b d TD D = + + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 12 2 2 (17.5-5) D D y e b d TM M = + + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 12 2 2 (17.5-6) where DD = design displacement at the center of rigidity of the isolation system in the direction under consideration as prescribed by Eq. 17.5-1 DM = maximum displacement at the center of rigidity of the isolation system in the direction under consideration as prescribed by Eq. 17.5-3 y = the distance between the centers of rigidity of the isolation system and the element of interest measured perpendicular to the direction of seismic loading under consideration e = the actual eccentricity measured in plan between the center of mass of the structure above the isolation interface and the center of rigidity of the isolation system, plus accidental eccentricity, in ft (mm), taken as 5 percent of the longest plan dimension of the structure perpendicular to the direction of force under consideration b = the shortest plan dimension of the structure measured perpendicular to d d = the longest plan dimension of the structure EXCEPTION: The total design displacement, DTD, and the total maximum displacement, DTM, are permitted to be taken as less than the value prescribed by Eqs. 17.5-5 and 17.5-6, respectively, but not less than 1.1 times DD and DM, respectively, provided the isolation system is shown by calculation to be conﬁ gured to resist torsion accordingly. 17.5.4 Minimum Lateral Forces 17.5.4.1 Isolation System and Structural Elements below the Isolation System The isolation system, the foundation, and all structural elements below the isolation system shall be designed and constructed to withstand a minimum lateral seismic force, Vb, using all of the appropriate requirements for a nonisolated structure and as prescribed by Eq. 17.5-7: Vb = kDmaxDD (17.5-7) where kDmax = maximum effective stiffness, in kips/in. (kN/mm), of the isolation system at the design displacement in the horizontal direction under consideration as prescribed by Eq. 17.8-3 DD = design displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.5-1 CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 172 Vb shall not be taken as less than the maximum force in the isolation system at any displacement up to and including the design displacement. 17.5.4.2 Structural Elements above the Isolation System The structure above the isolation system shall be designed and constructed to withstand a minimum shear force, Vs, using all of the appropriate require-ments for a nonisolated structure and as prescribed by Eq. 17.5-8: V k D R s D D I = max (17.5-8) where kDmax = maximum effective stiffness, in kips/in. (kN/mm), of the isolation system at the design displacement in the horizontal direction under consideration DD = design displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.5-1 RI = numerical coefﬁ cient related to the type of seismic force-resisting system above the isolation system The RI factor shall be based on the type of seismic force-resisting system used for the structure above the isolation system and shall be three-eighths of the value of R given in Table 12.2-1, with a maximum value not greater than 2.0 and a minimum value not less than 1.0. 17.5.4.3 Limits on Vs The value of Vs shall not be taken as less than the following: 1. The lateral seismic force required by Section 12.8 for a ﬁ xed-base structure of the same effective seismic weight, W, and a period equal to the isolated period, TD. 2. The base shear corresponding to the factored design wind load. 3. The lateral seismic force required to fully activate the isolation system (e.g., the yield level of a softening system, the ultimate capacity of a sacriﬁ cial wind-restraint system, or the break-away friction level of a sliding system) multiplied by 1.5. 17.5.5 Vertical Distribution of Force The shear force Vs shall be distributed over the height of the structure above the isolation interface using Eq. 17.5-9: F V w h w h x s x x i i i n = = ∑ 1 (17.5-9) where Fx = portion of Vs that is assigned to Level x Vs = total lateral seismic design force or shear on elements above the isolation system as pre-scribed by Eq. 17.5-8 wx = portion of W that is located at or assigned to Level x hx = height above the base of Level x At each level designated as x, the force, Fx, shall be applied over the area of the structure in accordance with the mass distribution at the level. 17.5.6 Drift Limits The maximum story drift of the structure above the isolation system shall not exceed 0.015hsx. The drift shall be calculated by Eq. 12.8-15 with Cd for the isolated structure equal to RI as deﬁ ned in Section 17.5.4.2. 17.6 DYNAMIC ANALYSIS PROCEDURES 17.6.1 General Where dynamic analysis is used to design seismically isolated structures, the requirements of this section shall apply. 17.6.2 Modeling The mathematical models of the isolated structure including the isolation system, the seismic force-resisting system, and other structural elements shall conform to Section 12.7.3 and to the requirements of Sections 17.6.2.1 and 17.6.2.2. 17.6.2.1 Isolation System The isolation system shall be modeled using deformational characteristics developed and veriﬁ ed by test in accordance with the requirements of Section 17.5.2. The isolation system shall be modeled with sufﬁ cient detail to a. Account for the spatial distribution of isolator units. b. Calculate translation, in both horizontal directions, and torsion of the structure above the isolation MINIMUM DESIGN LOADS 173 interface considering the most disadvantageous location of eccentric mass. c. Assess overturning/uplift forces on individual isolator units. d. Account for the effects of vertical load, bilateral load, and/or the rate of loading if the force-deﬂ ection properties of the isolation system are dependent on one or more of these attributes. The total design displacement and total maximum displacement across the isolation system shall be calculated using a model of the isolated structure that incorporates the force-deﬂ ection characteristics of nonlinear elements of the isolation system and the seismic force-resisting system. 17.6.2.2 Isolated Structure The maximum displacement of each ﬂ oor and design forces and displacements in elements of the seismic force-resisting system are permitted to be calculated using a linear elastic model of the isolated structure provided that both of the following condi-tions are met: 1. Stiffness properties assumed for the nonlinear components of the isolation system are based on the maximum effective stiffness of the isolation system; and 2. All elements of the seismic force-resisting system of the structure above the isolation system remain elastic for the design earthquake. Seismic force-resisting systems with elastic elements include, but are not limited to, irregular structural systems designed for a lateral force not less than 100 percent of Vs and regular structural systems designed for a lateral force not less than 80 percent of Vs, where Vs is determined in accordance with Section 17.5.4.2. 17.6.3 Description of Procedures 17.6.3.1 General Response-spectrum and response-history proce-dures shall be performed in accordance with Section 12.9 and Chapter 16, and the requirements of this section. 17.6.3.2 Input Earthquake The design earthquake ground motions shall be used to calculate the total design displacement of the isolation system and the lateral forces and displace-ments in the isolated structure. The maximum considered earthquake shall be used to calculate the total maximum displacement of the isolation system. 17.6.3.3 Response-Spectrum Procedure Response-spectrum analysis shall be performed using a modal damping value for the fundamental mode in the direction of interest not greater than the effective damping of the isolation system or 30 percent of critical, whichever is less. Modal damping values for higher modes shall be selected consistent with those that would be appropriate for response-spectrum analysis of the structure above the isolation system assuming a ﬁ xed base. Response-spectrum analysis used to determine the total design displacement and the total maximum displacement shall include simultaneous excitation of the model by 100 percent of the ground motion in the critical direction and 30 percent of the ground motion in the perpendicular, horizontal direction. The maximum displacement of the isolation system shall be calculated as the vectorial sum of the two orthogo-nal displacements. The design shear at any story shall not be less than the story shear resulting from application of the story forces calculated using Eq. 17.5-9 and a value of Vs equal to the base shear obtained from the response-spectrum analysis in the direction of interest. 17.6.3.4 Response-History Procedure Where a response-history procedure is performed, a suite of not fewer than three pairs of appropriate ground motions shall be used in the analysis; the ground motion pairs shall be selected and scaled in accordance with Section 17.3.2. Each pair of ground motion components shall be applied simultaneously to the model considering the most disadvantageous location of eccentric mass. The maximum displacement of the isolation system shall be calculated from the vectorial sum of the two orthogonal displacements at each time step. The parameters of interest shall be calculated for each ground motion used for the response-history analysis. If seven or more pairs of ground motions are used for the response-history analysis, the average value of the response parameter of interest is permit-ted to be used for design. If fewer than seven pairs of ground motions are used for analysis, the maximum value of the response parameter of interest shall be used for design. CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 174 17.6.4 Minimum Lateral Displacements and Forces 17.6.4.1 Isolation System and Structural Elements below the Isolation System The isolation system, foundation, and all struc-tural elements below the isolation system shall be designed using all of the appropriate requirements for a nonisolated structure and the forces obtained from the dynamic analysis without reduction, but the design lateral force shall not be taken as less than 90 percent of Vb determined in accordance as prescribed by Eq. 17.5-7. The total design displacement of the isolation system shall not be taken as less than 90 percent of DTD as speciﬁ ed by Section 17.5.3.5. The total maximum displacement of the isolation system shall not be taken as less than 80 percent of DTM as prescribed by Section 17.5.3.5. The limits on displacements speciﬁ ed by this section shall be evaluated using values of DTD and DTM determined in accordance with Section 17.5.5 except that DD ′ is permitted to be used in lieu of DD and DM ′ is permitted to be used in lieu of DM as prescribed in Eqs. 17.6-1 and 17.6-2: ′ = +( ) D D T T D D D 1 2 / (17.6-1) ′ = +( ) D D T T M M M 1 2 / (17.6-2) where DD = design displacement, in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.5-1 DM = maximum displacement in in. (mm), at the center of rigidity of the isolation system in the direction under consideration, as prescribed by Eq. 17.5-3 T = elastic, ﬁ xed-base period of the structure above the isolation system as determined by Section 12.8.2 TD = effective period of seismically isolated structure in s, at the design displacement in the direction under consideration, as prescribed by Eq. 17.5-2 TM = effective period, in s, of the seismically isolated structure, at the maximum displacement in the direction under consideration, as prescribed by Eq. 17.5-4 17.6.4.2 Structural Elements above the Isolation System Subject to the procedure-speciﬁ c limits of this section, structural elements above the isolation system shall be designed using the appropriate requirements for a nonisolated structure and the forces obtained from the dynamic analysis reduced by a factor of RI as determined in accordance with Section 17.5.4.2. The design lateral shear force on the structure above the isolation system, if regular in conﬁ guration, shall not be taken as less than 80 percent of Vs, or less than the limits speciﬁ ed by Section 17.5.4.3. EXCEPTION: The lateral shear force on the structure above the isolation system, if regular in conﬁ guration, is permitted to be taken as less than 80 percent, but shall not be less than 60 percent of Vs, where the response-history procedure is used for analysis of the seismically isolated structure. The design lateral shear force on the structure above the isolation system, if irregular in conﬁ gura-tion, shall not be taken as less than Vs or less than the limits speciﬁ ed by Section 17.5.4.3. EXCEPTION: The design lateral shear force on the structure above the isolation system, if irregular in conﬁ guration, is permitted to be taken as less than 100 percent, but shall not be less than 80 percent of Vs, where the response-history procedure is used for analysis of the seismically isolated structure. 17.6.4.3 Scaling of Results Where the factored lateral shear force on struc-tural elements, determined using either response-spectrum or response-history procedure, is less than the minimum values prescribed by Sections 17.6.4.1 and 17.6.4.2, all response parameters, including member forces and moments, shall be adjusted upward proportionally. 17.6.4.4 Drift Limits Maximum story drift corresponding to the design lateral force including displacement due to vertical deformation of the isolation system shall not exceed the following limits: 1. The maximum story drift of the structure above the isolation system calculated by response-spectrum analysis shall not exceed 0.015hsx. 2. The maximum story drift of the structure above the isolation system calculated by response-history analysis based on the force-deﬂ ection characteris-tics of nonlinear elements of the seismic force-resisting system shall not exceed 0.020hsx. Drift shall be calculated using Eq. 12.8-15 with the Cd of the isolated structure equal to RI as deﬁ ned in Section 17.5.4.2. The secondary effects of the maximum consid-ered earthquake lateral displacement of the structure MINIMUM DESIGN LOADS 175 above the isolation system combined with gravity forces shall be investigated if the story drift ratio exceeds 0.010/RI. 17.7 DESIGN REVIEW A design review of the isolation system and related test programs shall be performed by an independent engineering team including persons licensed in the appropriate disciplines and experienced in seismic analysis methods and the theory and application of seismic isolation. Isolation system design review shall include, but not be limited to, the following: 1. Review of site-speciﬁ c seismic criteria including the development of site-speciﬁ c spectra and ground motion histories and all other design criteria developed speciﬁ cally for the project. 2. Review of the preliminary design including the determination of the total design displacement, the total maximum displacement, and the lateral force level. 3. Overview and observation of prototype testing (Section 17.8). 4. Review of the ﬁ nal design of the entire structural system and all supporting analyses. 5. Review of the isolation system quality control testing program (Section 17.2.4.9). 17.8 TESTING 17.8.1 General The deformation characteristics and damping values of the isolation system used in the design and analysis of seismically isolated structures shall be based on tests of a selected sample of the components prior to construction as described in this section. The isolation system components to be tested shall include the wind-restraint system if such a system is used in the design. The tests speciﬁ ed in this section are for estab-lishing and validating the design properties of the isolation system and shall not be considered as satisfying the manufacturing quality control tests of Section 17.2.4.9. 17.8.2 Prototype Tests Prototype tests shall be performed separately on two full-size specimens (or sets of specimens, as appropriate) of each predominant type and size of isolator unit of the isolation system. The test speci-mens shall include the wind-restraint system as well as individual isolator units if such systems are used in the design. Specimens tested shall not be used for construction unless accepted by the registered design professional responsible for the design of the structure and approved by the authority having jurisdiction. 17.8.2.1 Record For each cycle of each test, the force-deﬂ ection and hysteretic behavior of the test specimen shall be recorded. 17.8.2.2 Sequence and Cycles The following sequence of tests shall be per-formed for the prescribed number of cycles at a vertical load equal to the average dead load plus one-half the effects due to live load on all isolator units of a common type and size: 1. Twenty fully reversed cycles of loading at a lateral force corresponding to the wind design force. 2. Three fully reversed cycles of loading at each of the following increments of the total design displacement—0.25DD, 0.5DD, 1.0DD, and 1.0DM where DD and DM are as determined in Sections 17.5.3.1 and 17.5.3.3, respectively, or Section 17.6 as appropriate. 3. Three fully reversed cycles of loading at the total maximum displacement, 1.0DTM. 4. 30SD1/SDSBD, but not less than 10, fully reversed cycles of loading at 1.0 times the total design displacement, 1.0DTD. If an isolator unit is also a vertical-load-carrying element, then item 2 of the sequence of cyclic tests speciﬁ ed in the preceding text shall be performed for two additional vertical load cases speciﬁ ed in Section 17.2.4.6. The load increment due to earthquake overturning, QE, shall be equal to or greater than the peak earthquake vertical force response corresponding to the test displacement being evaluated. In these tests, the combined vertical load shall be taken as the typical or average downward force on all isolator units of a common type and size. 17.8.2.3 Units Dependent on Loading Rates If the force-deﬂ ection properties of the isolator units are dependent on the rate of loading, each set of tests speciﬁ ed in Section 17.8.2.2 shall be performed dynamically at a frequency equal to the inverse of the effective period, TD. If reduced-scale prototype specimens are used to quantify rate-dependent properties of isolators, the CHAPTER 17 SEISMIC DESIGN REQUIREMENTS FOR SEISMICALLY ISOLATED STRUCTURES 176 reduced-scale prototype specimens shall be of the same type and material and be manufactured with the same processes and quality as full-scale prototypes and shall be tested at a frequency that represents full-scale prototype loading rates. The force-deﬂ ection properties of an isolator unit shall be considered to be dependent on the rate of loading if the measured property (effective stiffness or effective damping) at the design displacement when tested at any frequency in the range of 0.1 to 2.0 times the inverse of TD is different from the property when tested at a frequency equal to the inverse of TD by more than 15 percent. 17.8.2.4 Units Dependent on Bilateral Load If the force-deﬂ ection properties of the isolator units are dependent on bilateral load, the tests speciﬁ ed in Sections 17.8.2.2 and 17.8.2.3 shall be augmented to include bilateral load at the following increments of the total design displacement, DTD: 0.25 and 1.0, 0.5 and 1.0, 0.75 and 1.0, and 1.0 and 1.0 If reduced-scale prototype specimens are used to quantify bilateral-load-dependent properties, the reduced-scale specimens shall be of the same type and material and manufactured with the same processes and quality as full-scale prototypes. The force-deﬂ ection properties of an isolator unit shall be considered to be dependent on bilateral load if the effective stiffness where subjected to bilateral loading is different from the effective stiffness where subjected to unilateral loading, by more than 15 percent. 17.8.2.5 Maximum and Minimum Vertical Load Isolator units that carry vertical load shall be statically tested for maximum and minimum down-ward vertical load at the total maximum displacement. In these tests, the combined vertical loads shall be taken as speciﬁ ed in Section 17.2.4.6 on any one isolator of a common type and size. The dead load, D, and live load, L, are speciﬁ ed in Section 12.4. The seismic load E is given by Eqs. 12.4-1 and 12.4-2 where SDS in these equations is replaced by SMS and the vertical loads that result from application of horizontal seismic forces, QE, shall be based on the peak response due to the maximum considered earthquake. 17.8.2.6 Sacriﬁ cial Wind-Restraint Systems If a sacriﬁ cial wind-restraint system is to be utilized, its ultimate capacity shall be established by test. 17.8.2.7 Testing Similar Units Prototype tests are not required if an isolator unit is of similar size and of the same type and material as a prototype isolator unit that has been previously tested using the speciﬁ ed sequence of tests. 17.8.3 Determination of Force-Deﬂ ection Characteristics The force-deﬂ ection characteristics of the isolation system shall be based on the cyclic load tests of prototype isolator speciﬁ ed in Section 17.8.2. As required, the effective stiffness of an isolator unit, keff, shall be calculated for each cycle of loading as prescribed by Eq. 17.8-1: k F F eff = + Δ + Δ + − + − (17.8-1) where F+ and F– are the positive and negative forces, at Δ+ and Δ–, respectively. As required, the effective damping, βeff, of an isolator unit shall be calculated for each cycle of loading by Eq. 17.8-2: β π eff loop eff = Δ + Δ ( ) + − 2 2 E k (17.8-2) where the energy dissipated per cycle of loading, Eloop, and the effective stiffness, keff, shall be based on peak test displacements of Δ+ and Δ–. 17.8.4 Test Specimen Adequacy The performance of the test specimens shall be deemed adequate if the following conditions are satisﬁ ed: 1. The force-deﬂ ection plots for all tests speciﬁ ed in Section 17.8.2 have a positive incremental force-resisting capacity. 2. For each increment of test displacement speciﬁ ed in item 2 of Section 17.8.2.2 and for each vertical load case speciﬁ ed in Section 17.8.2.2, a. For each test specimen, the difference between the effective stiffness at each of the three cycles of test and the average value of effective stiffness is no greater than 15 percent. b. For each cycle of test, the difference between effective stiffness of the two test specimens of a common type and size of the isolator unit and the average effective stiffness is no greater than 15 percent. 3. For each specimen there is no greater than a 20 percent change in the initial effective stiffness over the cycles of test speciﬁ ed in item 4 of Section 17.8.2.2. MINIMUM DESIGN LOADS 177 4. For each specimen there is no greater than a 20 percent decrease in the initial effective damping over the cycles of test speciﬁ ed in item 4 of Section 17.8.2.2. 5. All specimens of vertical-load-carrying elements of the isolation system remain stable where tested in accordance with Section 17.8.2.5. 17.8.5 Design Properties of the Isolation System 17.8.5.1 Maximum and Minimum Effective Stiffness At the design displacement, the maximum and minimum effective stiffness of the isolated system, kDmax and kDmin, shall be based on the cyclic tests of item 2 of Section 17.8.2.2 and calculated using Eqs. 17.8-3 and 17.8-4: k F F D Dmax D max D max D = + + − ∑ ∑ 2 (17.8-3) k F F D D D D D min min min = + + − ∑ ∑ 2 (17.8-4) At the maximum displacement, the maximum and minimum effective stiffness of the isolation system, kMmax and kMmin, shall be based on the cyclic tests of item 3 of Section 17.8.2.2 and calculated using Eqs. 17.8-5 and 17.8-6: k F F D M M M M max max max = + + − ∑ ∑ 2 (17.8-5) k F F D M M M M min min min = + + − ∑ 2 (17.8-6) The maximum effective stiffness of the isolation system, kDmax (or kMmax), shall be based on forces from the cycle of prototype testing at a test displacement equal to DD (or DM) that produces the largest value of effective stiffness. Minimum effective stiffness of the isolation system, kDmin (or kMmin), shall be based on forces from the cycle of prototype testing at a test displacement equal to DD (or DM) that produces the smallest value of effective stiffness. For isolator units that are found by the tests of Sections 17.8.2.2, 17.8.2.3, and 17.8.2.4 to have force-deﬂ ection characteristics that vary with vertical load, rate of loading, or bilateral load, respectively, the values of kDmax and kMmax shall be increased and the values of kDmin and kMmin shall be decreased, as necessary, to bound the effects of measured variation in effective stiffness. 17.8.5.2 Effective Damping At the design displacement, the effective damping of the isolation system, βD, shall be based on the cyclic tests of item 2 of Section 17.8.2.2 and calcu-lated using Eq. 17.8-7: β π D D D D E k D = ∑ 2 2 max (17.8-7) In Eq. 17.8-7, the total energy dissipated per cycle of design displacement response, ΣED, shall be taken as the sum of the energy dissipated per cycle in all isolator units measured at a test displacement equal to DD and shall be based on forces and deﬂ ections from the cycle of prototype testing at test displacement DD that produces the smallest values of effective damping. At the maximum displacement, the effective damping of the isolation system, βM, shall be based on the cyclic tests of item 2 of Section 17.8.2.2 and calculated using Eq. 17.8-8 β π M M M M E k D = ∑ 2 2 max (17.8-8) In Eq. 17.8-8, the total energy dissipated per cycle of design displacement response, ΣEM, shall be taken as the sum of the energy dissipated per cycle in all isolator units measured at a test displacement equal to DM and shall be based on forces and deﬂ ections from the cycle of prototype testing at test displacement DM that produces the smallest value of effective damping. 179 Chapter 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 18.1.3 Notation The following notations apply to the provisions of this chapter: B1D = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βml (m = 1) and period of structure equal to T1D B1E = numerical coefﬁ cient as set forth in Table 18.6-1 for the effective damping equal to βI + βV1 and period equal to T1 B1M = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmM (m = 1) and period of structure equal to T1M BmD = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βml and period of structure equal to Tm BmM = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmM and period of structure equal to Tm BR = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βR and period of structure equal to TR BV + I = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to the sum of viscous damping in the fundamental mode of vibration of the structure in the direction of interest, βVm (m = 1), plus inherent damping, βI, and period of structure equal to T1 CmFD = force coefﬁ cient as set forth in Table 18.7-1 CmFV = force coefﬁ cient as set forth in Table 18.7-2 CS1 = seismic response coefﬁ cient of the funda-mental mode of vibration of the structure in the direction of interest, Section 18.4.2.4 or 18.5.2.4 (m = 1) CSm = seismic response coefﬁ cient of the mth mode of vibration of the structure in the direction of interest, Section 18.4.2.4 (m = 1) or Section 18.4.2.6 (m > 1) CSR = seismic response coefﬁ cient of the residual mode of vibration of the structure in the direction of interest, Section 18.5.2.8 D1D = fundamental mode design displacement at the center rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.2 18.1 GENERAL 18.1.1 Scope Every structure with a damping system and every portion thereof shall be designed and constructed in accordance with the requirements of this standard as modiﬁ ed by this section. Where damping devices are used across the isolation interface of a seismically isolated structure, displacements, velocities, and accelerations shall be determined in accordance with Chapter 17. 18.1.2 Deﬁ nitions The following deﬁ nitions apply to the provisions of Chapter 18: DAMPING DEVICE: A ﬂ exible structural element of the damping system that dissipates energy due to relative motion of each end of the device. Damping devices include all pins, bolts, gusset plates, brace extensions, and other components required to connect damping devices to the other elements of the structure. Damping devices may be classiﬁ ed as either displacement-dependent or velocity-dependent, or a combination thereof, and may be conﬁ gured to act in either a linear or nonlinear manner. DAMPING SYSTEM: The collection of structural elements that includes all the individual damping devices, all structural elements or bracing required to transfer forces from damping devices to the base of the structure, and the structural elements required to transfer forces from damping devices to the seismic force-resisting system. DISPLACEMENT-DEPENDENT DAMPING DEVICE: The force response of a displacement-dependent damping device is primarily a function of the relative displacement between each end of the device. The response is substantially independent of the relative velocity between each of the devices and/ or the excitation frequency. VELOCITY-DEPENDENT DAMPING DEVICE: The force-displacement relation for a velocity-dependent damping device is primarily a function of the relative velocity between each end of the device and could also be a function of the relative displacement between each end of the device. CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 180 D1M = fundamental mode maximum displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.5 DmD = design displacement at the center of rigidity of the roof level of the structure due to the mth mode of vibration in the direction under consideration, Section 18.4.3.2 DmM = maximum displacement at the center of rigidity of the roof level of the structure due to the mth mode of vibration in the direction under consideration, Section 18.4.3.5 DRD = residual mode design displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.2 DRM = residual mode maximum displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.5 DY = displacement at the center of rigidity of the roof level of the structure at the effective yield point of the seismic force-resisting system, Section 18.6.3 fi = lateral force at Level i of the structure distributed approximately in accordance with Section 12.8.3, Section 18.5.2.3 Fi1 = inertial force at Level i (or mass point i) in the fundamental mode of vibration of the structure in the direction of interest, Section 18.5.2.9 Fim = inertial force at Level i (or mass point i) in the mth mode of vibration of the structure in the direction of interest, Section 18.4.2.7 FiR = inertial force at Level i (or mass point i) in the residual mode of vibration of the struc-ture in the direction of interest, Section 18.5.2.9 hr = height of the structure above the base to the roof level, Section 18.5.2.3 qH = hysteresis loop adjustment factor as deter-mined in Section 18.6.2.2.1 QDSD = force in an element of the damping system required to resist design seismic forces of displacement-dependent damping devices, Section 18.7.2.5 QmDSV = forces in an element of the damping system required to resist design seismic forces of velocity-dependent damping devices due to the mth mode of vibration of the structure in the direction of interest, Section 18.7.2.5 QmSFRS = force in an element of the damping system equal to the design seismic force of the mth mode of vibration of the structure in the direction of interest, Section 18.7.2.5 T1 = the fundamental period of the structure in the direction under consideration T1D = effective period, in seconds, of the funda-mental mode of vibration of the structure at the design displacement in the direction under consideration, as prescribed by Section 18.4.2.5 or 18.5.2.5 T1M = effective period, in seconds, of the funda-mental mode of vibration of the structure at the maximum displacement in the direction under consideration, as prescribed by Section 18.4.2.5 or 18.5.2.5 TR = period, in seconds, of the residual mode of vibration of the structure in the direction under consideration, Section 18.5.2.7 Vm = design value of the seismic base shear of the mth mode of vibration of the structure in the direction of interest, Section 18.4.2.2 Vmin = minimum allowable value of base shear permitted for design of the seismic force-resisting system of the structure in the direction of interest, Section 18.2.2.1 VR = design value of the seismic base shear of the residual mode of vibration of the structure in a given direction, as determined in Section 18.5.2.6 W _ 1 = effective fundamental mode seismic weight determined in accordance with Eq. 18.4-2b for m = 1 W _ R = effective residual mode seismic weight determined in accordance with Eq. 18.5-13 α = velocity exponent relating damping device force to damping device velocity βmD = total effective damping of the mth mode of vibration of the structure in the direction of interest at the design displacement, Section 18.6.2 βmM = total effective damping of the mth mode of vibration of the structure in the direction of interest at the maximum displacement, Section 18.6.2 βHD = component of effective damping of the structure in the direction of interest due to post-yield hysteretic behavior of the seismic force-resisting system and elements of the damping system at effective ductility demand μD, Section 18.6.2.2 βHM = component of effective damping of the structure in the direction of interest due to post-yield hysteretic behavior of the seismic MINIMUM DESIGN LOADS 181 force-resisting system and elements of the damping system at effective ductility demand, μM, Section 18.6.2.2 βI = component of effective damping of the structure due to the inherent dissipation of energy by elements of the structure, at or just below the effective yield displacement of the seismic force-resisting system, Section 18.6.2.1 βR = total effective damping in the residual mode of vibration of the structure in the direction of interest, calculated in accordance with Section 18.6.2 (using μD = 1.0 and μM = 1.0) βVm = component of effective damping of the mth mode of vibration of the structure in the direction of interest due to viscous dissipa-tion of energy by the damping system, at or just below the effective yield displacement of the seismic force-resisting system, Section 18.6.2.3 δi = elastic deﬂ ection of Level i of the structure due to applied lateral force, fi, Section 18.5.2.3 δi1D = fundamental mode design deﬂ ection of Level i at the center of rigidity of the structure in the direction under consideration, Section 18.5.3.1 δiD = total design deﬂ ection of Level i at the center of rigidity of the structure in the direction under consideration, Section 18.5.3 δiM = total maximum deﬂ ection of Level i at the center of rigidity of the structure in the direction under consideration, Section 18.5.3 δiRD = residual mode design deﬂ ection of Level i at the center of rigidity of the structure in the direction under consideration, Section 18.5.3.1 δim = deﬂ ection of Level i in the mth mode of vibration at the center of rigidity of the structure in the direction under consideration, Section 18.6.2.3 Δ1D = design story drift due to the fundamental mode of vibration of the structure in the direction of interest, Section 18.5.3.3 ΔD = total design story drift of the structure in the direction of interest, Section 18.5.3.3 ΔM = total maximum story drift of the structure in the direction of interest, Section 18.5.3 ΔmD = design story drift due to the mth mode of vibration of the structure in the direction of interest, Section 18.4.3.3 ΔRD = design story drift due to the residual mode of vibration of the structure in the direction of interest, Section 18.5.3.3 μ = effective ductility demand on the seismic force-resisting system in the direction of interest μD = effective ductility demand on the seismic force-resisting system in the direction of interest due to the design earthquake ground motions, Section 18.6.3 μM = effective ductility demand on the seismic force-resisting system in the direction of interest due to the maximum considered earthquake ground motions, Section 18.6.3 μmax = maximum allowable effective ductility demand on the seismic force-resisting system due to the design earthquake ground motions, Section 18.6.4 φi1 = displacement amplitude at Level i of the fundamental mode of vibration of the structure in the direction of interest, normal-ized to unity at the roof level, Section 18.5.2.3 φiR = displacement amplitude at Level i of the residual mode of vibration of the structure in the direction of interest normalized to unity at the roof level, Section 18.5.2.7 Γ1 = participation factor of the fundamental mode of vibration of the structure in the direction of interest, Section 18.4.2.3 or 18.5.2.3 (m = 1) Γm = participation factor in the mth mode of vibration of the structure in the direction of interest, Section 18.4.2.3 ΓR = participation factor of the residual mode of vibration of the structure in the direction of interest, Section 18.5.2.7 ∇1D = design story velocity due to the fundamental mode of vibration of the structure in the direction of interest, Section 18.5.3.4 ∇D = total design story velocity of the structure in the direction of interest, Section 18.4.3.4 ∇M = total maximum story velocity of the structure in the direction of interest, Section 18.5.3 ∇mD = design story velocity due to the mth mode of vibration of the structure in the direction of interest, Section 18.4.3.4 CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 182 18.2 GENERAL DESIGN REQUIREMENTS 18.2.1 Seismic Design Category A Seismic Design Category A structures with a damping system shall be designed using the design spectral response acceleration determined in accor-dance with Section 11.4.4 and the analysis methods and design requirements for Seismic Design Category B structures. 18.2.2 System Requirements Design of the structure shall consider the basic requirements for the seismic force-resisting system and the damping system as deﬁ ned in the following sections. The seismic force-resisting system shall have the required strength to meet the forces deﬁ ned in Section 18.2.2.1. The combination of the seismic force-resisting system and the damping system is permitted to be used to meet the drift requirement. 18.2.2.1 Seismic Force-Resisting System Structures that contain a damping system are required to have a seismic force-resisting system that, in each lateral direction, conforms to one of the types indicated in Table 12.2-1. The design of the seismic force-resisting system in each direction shall satisfy the requirements of Section 18.7 and the following: 1. The seismic base shear used for design of the seismic force-resisting system shall not be less than Vmin, where Vmin is determined as the greater of the values computed using Eqs. 18.2-1 and 18.2-2: V V BV I min = + (18.2-1) Vmin = 0.75V (18.2-2) where V = seismic base shear in the direction of interest, determined in accordance with Section 12.8 BV + I = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to the sum of viscous damping in the fundamental mode of vibration of the structure in the direction of interest, βVm (m = 1), plus inherent damping, βI, and period of structure equal to T1 EXCEPTION: The seismic base shear used for design of the seismic force-resisting system shall not be taken as less than 1.0V, if either of the following conditions apply: a. In the direction of interest, the damping system has less than two damping devices on each ﬂ oor level, conﬁ gured to resist torsion. b. The seismic force-resisting system has horizontal irregularity Type 1b (Table 12.3-1) or vertical irregularity Type 1b (Table 12.3-2). 2. Minimum strength requirements for elements of the seismic force-resisting system that are also elements of the damping system or are otherwise required to resist forces from damping devices shall meet the additional requirements of Section 18.7.2. 18.2.2.2 Damping System Elements of the damping system shall be designed to remain elastic for design loads including unreduced seismic forces of damping devices as required in Section 18.7.2.1, unless it is shown by analysis or test that inelastic response of elements would not adversely affect damping system function and inelastic response is limited in accordance with the requirements of Section 18.7.2.6. 18.2.3 Ground Motion 18.2.3.1 Design Spectra Spectra for the design earthquake ground motions and maximum considered earthquake ground motions developed in accordance with Section 17.3.1 shall be used for the design and analysis of a structure with a damping system. Site-speciﬁ c design spectra shall be developed and used for design of a structure with a damping system if either of the following conditions apply: 1. The structure is located on a Class F site. 2. The structure is located at a site with S1 greater than or equal to 0.6. 18.2.3.2 Ground Motion Histories Ground motion histories for the design earthquake and the maximum considered earthquake developed in accordance with Section 17.3.2 shall be used for design and analysis of all structures with a damping system if either of the following conditions apply: 1. The structure is located at a site with S1 greater than or equal to 0.6. 2. The damping system is explicitly modeled and analyzed using the response-history analysis method. MINIMUM DESIGN LOADS 183 18.2.4 Procedure Selection A structure with a damping system shall be designed using linear procedures, nonlinear proce-dures, or a combination of linear and nonlinear procedures, as permitted in this section. Regardless of the analysis method used, the peak dynamic response of the structure and elements of the damping system shall be conﬁ rmed by using the nonlinear response-history procedure if the structure is located at a site with S1 greater than or equal to 0.6. 18.2.4.1 Nonlinear Procedures The nonlinear procedures of Section 18.3 are permitted to be used for design of all structures with damping systems. 18.2.4.2 Response-Spectrum Procedure The response-spectrum procedure of Section 18.4 is permitted to be used for design of a structure with a damping system provided that 1. In the direction of interest, the damping system has at least two damping devices in each story, conﬁ gured to resist torsion. 2. The total effective damping of the fundamental mode, βmD (m = 1), of the structure in the direction of interest is not greater than 35 percent of critical. 18.2.4.3 Equivalent Lateral Force Procedure The equivalent lateral force procedure of Section 18.5 is permitted to be used for design of a structure with a damping system provided that 1. In the direction of interest, the damping system has at least two damping devices in each story, conﬁ gured to resist torsion. 2. The total effective damping of the fundamental mode, βmD (m = 1), of the structure in the direction of interest is not greater than 35 percent of critical. 3. The seismic force-resisting system does not have horizontal irregularity Type 1a or 1b (Table 12.3-1) or vertical irregularity Type 1a, 1b, 2, or 3 (Table 12.3-2). 4. Floor diaphragms are rigid as deﬁ ned in Section 12.3.1. 5. The height of the structure above the base does not exceed 100 ft (30 m). 18.2.5 Damping System 18.2.5.1 Device Design The design, construction, and installation of damping devices shall be based on response to maximum considered earthquake ground motions and consideration of the following: 1. Low-cycle, large-displacement degradation due to seismic loads. 2. High-cycle, small-displacement degradation due to wind, thermal, or other cyclic loads. 3. Forces or displacements due to gravity loads. 4. Adhesion of device parts due to corrosion or abrasion, biodegradation, moisture, or chemical exposure. 5. Exposure to environmental conditions, including, but not limited to, temperature, humidity, moisture, radiation (e.g., ultraviolet light), and reactive or corrosive substances (e.g., salt water). Damping devices subject to failure by low-cycle fatigue shall resist wind forces without slip, move-ment, or inelastic cycling. The design of damping devices shall incorporate the range of thermal conditions, device wear, manu-facturing tolerances, and other effects that cause device properties to vary during the design life of the device. 18.2.5.2 Multiaxis Movement Connection points of damping devices shall provide sufﬁ cient articulation to accommodate simultaneous longitudinal, lateral, and vertical displacements of the damping system. 18.2.5.3 Inspection and Periodic Testing Means of access for inspection and removal of all damping devices shall be provided. The registered design professional responsible for design of the structure shall establish an appropriate inspection and testing schedule for each type of damping device to ensure that the devices respond in a dependable manner throughout their design life. The degree of inspection and testing shall reﬂ ect the established in-service history of the damping devices and the likelihood of change in properties over the design life of the devices. 18.2.5.4 Quality Control As part of the quality assurance plan developed in accordance with Section 11A.1.2, the registered design professional responsible for the structural design shall establish a quality control plan for the manufacture of damping devices. As a minimum, this plan shall include the testing requirements of Section 18.9.2. CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 184 18.3 NONLINEAR PROCEDURES The stiffness and damping properties of the damping devices used in the models shall be based on or veriﬁ ed by testing of the damping devices as speciﬁ ed in Section 18.9. The nonlinear force-deﬂ ection characteristics of damping devices shall be modeled, as required, to explicitly account for device depen-dence on frequency, amplitude, and duration of seismic loading. 18.3.1 Nonlinear Response-History Procedure A nonlinear response-history analysis shall utilize a mathematical model of the structure and the damping system as provided in Section 16.2.2 and this section. The model shall directly account for the nonlinear hysteretic behavior of elements of the structure and the damping devices to determine its response. The analysis shall be performed in accordance with Section 16.2 together with the requirements of this section. Inherent damping of the structure shall not be taken as greater than 5 percent of critical unless test data consistent with levels of deformation at or just below the effective yield displacement of the seismic force-resisting system support higher values. If the calculated force in an element of the seismic force-resisting system does not exceed 1.5 times its nominal strength, that element is permitted to be modeled as linear. 18.3.1.1 Damping Device Modeling Mathematical models of displacement-dependent damping devices shall include the hysteretic behavior of the devices consistent with test data and accounting for all signiﬁ cant changes in strength, stiffness, and hysteretic loop shape. Mathematical models of velocity-dependent damping devices shall include the velocity coefﬁ cient consistent with test data. If this coefﬁ cient changes with time and/or tempera-ture, such behavior shall be modeled explicitly. The elements of damping devices connecting damper units to the structure shall be included in the model. EXCEPTION: If the properties of the damping devices are expected to change during the duration of the time history analysis, the dynamic response is permitted to be enveloped by the upper and lower limits of device properties. All these limit cases for variable device properties must satisfy the same conditions as if the time-dependent behavior of the devices were explicitly modeled. 18.3.1.2 Response Parameters In addition to the response parameters given in Section 16.2.4, for each ground motion used for response-history analysis, individual response param-eters consisting of the maximum value of the discrete damping device forces, displacements, and velocities, in the case of velocity-dependent devices, shall be determined. If at least seven pairs of ground motions are used for response-history analysis, the design values of the damping device forces, displacements, and velocities are permitted to be taken as the average of the values determined by the analyses. If less than seven pairs of ground motions are used for response-history analysis, the design damping device forces, displacements, and velocities shall be taken as the maximum value determined by the analyses. A minimum of three pairs of ground motions shall be used. 18.3.2 Nonlinear Static Procedure The nonlinear modeling described in Section 16.2.2 and the lateral loads described in Section 16.2 shall be applied to the seismic force-resisting system. The resulting force-displacement curve shall be used in lieu of the assumed effective yield displacement, DY, of Eq. 18.6-10 to calculate the effective ductility demand due to the design earthquake ground motions, μD, and due to the maximum considered earthquake ground motions, μM, in Eqs. 18.6-8 and 18.6-9, respectively. The value of (R/Cd) shall be taken as 1.0 in Eqs. 18.4-4, 18.4-5, 18.4-8, and 18.4-9 for the response-spectrum procedure, and in Eqs. 18.5-6, 18.5-7, and 18.5-15 for the equivalent lateral force procedure. 18.4 RESPONSE-SPECTRUM PROCEDURE Where the response-spectrum procedure is used to analyze a structure with a damping system, the requirements of this section shall apply. 18.4.1 Modeling A mathematical model of the seismic force-resist-ing system and damping system shall be constructed that represents the spatial distribution of mass, stiffness, and damping throughout the structure. The model and analysis shall comply with the require-ments of Section 12.9 for the seismic force-resisting system and to the requirements of this section for the MINIMUM DESIGN LOADS 185 damping system. The stiffness and damping properties of the damping devices used in the models shall be based on or veriﬁ ed by testing of the damping devices as speciﬁ ed in Section 18.9. The elastic stiffness of elements of the damping system other than damping devices shall be explicitly modeled. Stiffness of damping devices shall be modeled depending on damping device type as follows: 1. Displacement-dependent damping devices: Displacement-dependent damping devices shall be modeled with an effective stiffness that represents damping device force at the response displacement of interest (e.g., design story drift). Alternatively, the stiffness of hysteretic and friction damping devices is permitted to be excluded from response spectrum analysis provided design forces in displacement-dependent damping devices, QDSD, are applied to the model as external loads (Section 18.7.2.5). 2. Velocity-dependent damping devices: Velocity-dependent damping devices that have a stiffness component (e.g., viscoelastic damping devices) shall be modeled with an effective stiffness corresponding to the amplitude and frequency of interest. 18.4.2 Seismic Force-Resisting System 18.4.2.1 Seismic Base Shear The seismic base shear, V, of the structure in a given direction shall be determined as the combina-tion of modal components, Vm, subject to the limits of Eq. 18.4-1: V ≥ Vmin (18.4-1) The seismic base shear, V, of the structure shall be determined by the sum of the square root method (SRSS) or complete quadratic combination of modal base shear components, Vm. 18.4.2.2 Modal Base Shear Modal base shear of the mth mode of vibration, Vm, of the structure in the direction of interest shall be determined in accordance with Eqs. 18.4-2: Vm = CsmW _ (18.4-2a) W w w m i im i n i im i n = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = ∑ ∑ φ φ 1 2 2 1 (18.4-2b) where Csm = seismic response coefﬁ cient of the mth mode of vibration of the structure in the direction of interest as determined from Section 18.4.2.4 (m = 1) or Section 18.4.2.6 (m > 1) W _ m = effective seismic weight of the mth mode of vibration of the structure 18.4.2.3 Modal Participation Factor The modal participation factor of the mth mode of vibration, Γm, of the structure in the direction of interest shall be determined in accordance with Eq. 18.4-3: Γm m i im i n W w = = ∑ φ 1 (18.4-3) where φim = displacement amplitude at the ith level of the structure in the mth mode of vibration in the direction of interest, normalized to unity at the roof level. 18.4.2.4 Fundamental Mode Seismic Response Coefﬁ cient The fundamental mode (m = 1) seismic response coefﬁ cient, CS1, in the direction of interest shall be determined in accordance with Eqs. 18.4-4 and 18.4-5: For T1D < TS, C R C S B S d DS D 1 0 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟Ω (18.4-4) For T1D ≥ TS, C R C S T B S d D D D 1 1 1 0 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) Ω (18.4-5) 18.4.2.5 Effective Fundamental Mode Period Determination The effective fundamental mode (m = 1) period at the design earthquake ground motion, T1D, and at the MCER ground motion, T1M, shall be based on either explicit consideration of the post-yield force deﬂ ection characteristics of the structure or determined in accordance with Eqs. 18.4-6 and 18.4-7: T T D D 1 1 = μ (18.4-6) T T M M 1 1 = μ (18.4-7) CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 186 18.4.2.6 Higher Mode Seismic Response Coefﬁ cient Higher mode (m > 1) seismic response coefﬁ -cient, CSm, of the mth mode of vibration (m > 1) of the structure in the direction of interest shall be deter-mined in accordance with Eqs. 18.4-8 and 18.4-9: For Tm < TS, C R C S B Sm d D mD = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0 Ω (18.4-8) For Tm ≥ TS, C R C S T B Sm d D m mD = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) 1 0 Ω (18.4-9) where Tm = period, in seconds, of the mth mode of vibration of the structure in the direction under consideration BmD = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmD and period of the structure equal to Tm 18.4.2.7 Design Lateral Force Design lateral force at Level i due to the mth mode of vibration, Fim, of the structure in the direction of interest shall be determined in accordance with Eq. 18.4-10: F w W V im i im m m m = φ Γ (18.4-10) Design forces in elements of the seismic force-resisting system shall be determined by the SRSS or complete quadratic combination of modal design forces. 18.4.3 Damping System Design forces in damping devices and other elements of the damping system shall be determined on the basis of the ﬂ oor deﬂ ection, story drift, and story velocity response parameters described in the following sections. Displacements and velocities used to determine maximum forces in damping devices at each story shall account for the angle of orientation of each device from the horizontal and consider the effects of increased response due to torsion required for design of the seismic force-resisting system. Floor deﬂ ections at Level i, δiD and δiM, story drifts, ΔD and ΔM, and story velocities, ∇D and ∇M, shall be calculated for both the design earthquake ground motions and the maximum considered earthquake ground motions, respectively, in accor-dance with this section. 18.4.3.1 Design Earthquake Floor Deﬂ ection The deﬂ ection of structure due to the design earthquake ground motions at Level i in the mth mode of vibration, δimD, of the structure in the direction of interest shall be determined in accordance with Eq. 18.4-11: δimD = DmDφim (18.4-11) The total design deﬂ ection at each ﬂ oor of the structure shall be calculated by the SRSS or complete quadratic combination of modal design earthquake deﬂ ections. 18.4.3.2 Design Earthquake Roof Displacement Fundamental (m = 1) and higher mode (m > 1) roof displacements due to the design earthquake ground motions, D1D and DmD, of the structure in the direction of interest shall be determined in accordance with Eqs. 18.4-12 and to 18.4-13: For m = 1, D g S T B g S T B T T D DS D D DS E D S 1 2 1 1 2 1 2 1 1 2 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < π Γ π Γ , (18.4-12a) D g S T B g S T B T T D D D D D E D S 1 2 1 1 1 1 2 1 1 1 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥ π Γ π Γ , (18.4-12b) For m > 1, D g S T B g S T B mD m D m mD m DS m mD = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 4 2 1 2 2 π Γ π Γ (18.4-13) 18.4.3.3 Design Earthquake Story Drift Design story drift in the fundamental mode, Δ1D, and higher modes, ΔmD (m > 1), of the structure in the direction of interest shall be calculated in accordance with Section 12.8.6 using modal roof displacements of Section 18.4.3.2. Total design story drift, ΔD, shall be determined by the SRSS or complete quadratic combination of modal design earthquake drifts. 18.4.3.4 Design Earthquake Story Velocity Design story velocity in the fundamental mode, ∇1D, and higher modes, ∇mD (m > 1), of the structure in the direction of interest shall be calculated in accordance with Eqs. 18.4-14 and 18.4-15: For m = 1, ∇1D = 2π Δ1 1 D D T (18.4-14) For m > 1, ∇mD = 2π ΔmD m T (18.4-15) MINIMUM DESIGN LOADS 187 Total design story velocity, ΔD, shall be determined by the SRSS or complete quadratic combination of modal design velocities. 18.4.3.5 Maximum Considered Earthquake Response Total modal maximum ﬂ oor deﬂ ection at Level i, design story drift values, and design story velocity values shall be based on Sections 18.4.3.1, 18.4.3.3, and 18.4.3.4, respectively, except design roof displacement shall be replaced by maximum roof displacement. Maximum roof displacement of the structure in the direction of interest shall be calculated in accordance with Eqs. 18.4-16 and to 18.4-17: For m = 1, D g S T B g S T B T T M MS M M MS E M S 1 2 1 1 2 1 2 1 1 2 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < π π Γ Γ , (18.4-16a) D g S T B g S T B T T M M M M M E M S 1 2 1 1 1 1 2 1 1 1 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥ π π Γ Γ , (18.4-16b) For m >1, D g S T B g S T B mM m M m mM m MS m mM = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 4 2 1 2 2 π π Γ Γ (18.4-17) where BmM = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmM and period of the structure equal to Tm 18.5 EQUIVALENT LATERAL FORCE PROCEDURE Where the equivalent lateral force procedure is used to design structures with a damping system, the requirements of this section shall apply. 18.5.1 Modeling Elements of the seismic force-resisting system shall be modeled in a manner consistent with the requirements of Section 12.8. For purposes of analysis, the structure shall be considered to be ﬁ xed at the base. Elements of the damping system shall be modeled as required to determine design forces transferred from damping devices to both the ground and the seismic force-resisting system. The effective stiffness of velocity-dependent damping devices shall be modeled. Damping devices need not be explicitly modeled provided effective damping is calculated in accor-dance with the procedures of Section 18.6 and used to modify response as required in Sections 18.5.2 and 18.5.3. The stiffness and damping properties of the damping devices used in the models shall be based on or veriﬁ ed by testing of the damping devices as speciﬁ ed in Section 18.9. 18.5.2 Seismic Force-Resisting System 18.5.2.1 Seismic Base Shear The seismic base shear, V, of the seismic force-resist-ing system in a given direction shall be determined as the combination of the two modal components, V1 and VR, in accordance with Eq. 18.5-1: V V V V R = + ≥ 1 2 2 min (18.5-1) where V1 = design value of the seismic base shear of the fundamental mode in a given direction of response, as determined in Section 18.5.2.2 VR = design value of the seismic base shear of the residual mode in a given direction, as deter-mined in Section 18.5.2.6 Vmin = minimum allowable value of base shear permitted for design of the seismic force-resisting system of the structure in direction of the interest, as determined in Section 18.2.2.1 18.5.2.2 Fundamental Mode Base Shear The fundamental mode base shear, V1, shall be determined in accordance with Eq. 18.5-2: V1 = CS1W _ 1 (18.5-2) where CS1 = the fundamental mode seismic response coef-ﬁ cient, as determined in Section 18.5.2.4 W _ 1 = the effective fundamental mode seismic weight including portions of the live load as deﬁ ned by Eq. 18.4-2b for m = 1 18.5.2.3 Fundamental Mode Properties The fundamental mode shape, φi1, and participa-tion factor, Γ1, shall be determined by either dynamic analysis using the elastic structural properties and deformational characteristics of the resisting elements or using Eqs. 18.5-3 and 18.5-4: φ i i r h h 1 = (18.5-3) CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 188 Γ1 1 1 = = ∑ W wi il i n φ (18.5-4) where hi = the height above the base to Level i hr = the height of the structure above the base to the roof level wi = the portion of the total effective seismic weight, W, located at or assigned to Level i The fundamental period, T1, shall be determined either by dynamic analysis using the elastic structural properties and deformational characteristics of the resisting elements, or using Eq. 18.5-5 as follows: T w g f i i n i i i n 1 1 2 1 1 2 = = = ∑ ∑ π δ δ (18.5-5) where fi = lateral force at Level i of the structure distributed in accordance with Section 12.8.3 δi = elastic deﬂ ection at Level i of the structure due to applied lateral forces fi 18.5.2.4 Fundamental Mode Seismic Response Coefﬁ cient The fundamental mode seismic response coefﬁ cient, CS1, shall be determined using Eq. 18.5-6 or 18.5-7: For T1D < TS, C R C S B S d D D 1 1 0 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟Ω (18.5-6) For T1D ≥ TS, C R C S T B S d D D D 1 1 1 0 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) Ω (18.5-7) where SDS = the design spectral response acceleration parameter in the short period range SD1 = the design spectral response acceleration parameter at a period of 1 s B1D = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmD (m = 1) and period of the structure equal to T1D 18.5.2.5 Effective Fundamental Mode Period Determination The effective fundamental mode period at the design earthquake, T1D, and at the maximum consid-ered earthquake, T1M, shall be based on explicit consideration of the post-yield force deﬂ ection characteristics of the structure or shall be calculated using Eqs. 18.5-8 and 18.5-9: T T D D 1 1 = μ (18.5-8) T T M M 1 1 = μ (18.5-9) 18.5.2.6 Residual Mode Base Shear Residual mode base shear, VR, shall be deter-mined in accordance with Eq. 18.5-10: VR = CSRW _ R (18.5-10) where CSR = the residual mode seismic response coefﬁ cient as determined in Section 18.5.2.8 W _ R = the effective residual mode effective weight of the structure determined using Eq. 18.5-13 18.5.2.7 Residual Mode Properties Residual mode shape, φiR, participation factor, ΓR, effective residual mode seismic weight of the structure, W _ R, and effective period, TR, shall be determined using Eqs. 18.5-11 through 18.5-14: φ φ iR i = − − 1 1 1 1 1 Γ Γ (18.5-11) ΓR = 1 – Γ1 (18.5-12) W _ R = W – W _ 1 (18.5-13) TR = 0.4T1 (18.5-14) 18.5.2.8 Residual Mode Seismic Response Coefﬁ cient The residual mode seismic response coefﬁ cient, CSR, shall be determined in accordance with Eq. 18.5-15: C R C S B SR d DS R = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟Ω0 (18.5-15) where BR = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βR, and period of the structure equal to TR 18.5.2.9 Design Lateral Force The design lateral force in elements of the seismic force-resisting system at Level i due to fundamental mode response, Fi1, and residual mode MINIMUM DESIGN LOADS 189 response, FiR, of the structure in the direction of interest shall be determined in accordance with Eqs. 18.5-16 and 18.5-17: F w W V i i i 1 1 1 1 = φ 1 Γ (18.5-16) F w W V iR i iR R R R = φ Γ (18.5-17) Design forces in elements of the seismic force-resisting system shall be determined by taking the SRSS of the forces due to fundamental and residual modes. 18.5.3 Damping System Design forces in damping devices and other elements of the damping system shall be determined on the basis of the ﬂ oor deﬂ ection, story drift, and story velocity response parameters described in the following sections. Displacements and velocities used to determine maximum forces in damping devices at each story shall account for the angle of orientation of each device from the horizontal and consider the effects of increased response due to torsion required for design of the seismic force-resisting system. Floor deﬂ ections at Level i, δiD and δiM, story drifts, ΔD and ΔM, and story velocities, ∇D and ∇M, shall be calculated for both the design earthquake ground motions and the maximum considered earthquake ground motions, respectively, in accordance with the following sections. 18.5.3.1 Design Earthquake Floor Deﬂ ection The total design deﬂ ection at each ﬂ oor of the structure in the direction of interest shall be calculated as the SRSS of the fundamental and residual mode ﬂ oor deﬂ ections. The fundamental and residual mode deﬂ ections due to the design earthquake ground motions, δi1D and δiRD, at the center of rigidity of Level i of the structure in the direction of interest shall be determined using Eqs. 18.5-18 and 18.5-19: δi1D = D1Dφi1 (18.5-18) δiRD = DRDφiR (18.5-19) where D1D = fundamental mode design displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.2 DRD = residual mode design displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.5.3.2 18.5.3.2 Design Earthquake Roof Displacement Fundamental and residual mode displacements due to the design earthquake ground motions, D1D and D1R, at the center of rigidity of the roof level of the structure in the direction of interest shall be determined using Eqs. 18.5-20 and 18.5-21: D g S T B g S T B T T D DS D D DS D D S 1 2 1 1 2 1 2 1 1 2 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < π π Γ Γ , (18.5-20a) D g S T B g S T B T T D D D D D E D S 1 2 1 1 1 1 2 1 1 1 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥ π π Γ Γ , (18.5-20b) D g S T B g S T B RD R D R R R DS R R = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 4 2 1 2 2 π π Γ Γ (18.5-21) 18.5.3.3 Design Earthquake Story Drift Design story drifts, ΔD, in the direction of interest shall be calculated using Eq. 18.5-22: Δ = Δ + Δ D D RD 1 2 2 (18.5-22) where Δ1D = design story drift due to the fundamental mode of vibration of the structure in the direction of interest ΔRD = design story drift due to the residual mode of vibration of the structure in the direction of interest Modal design story drifts, Δ1D and ΔRD, shall be determined as the difference of the deﬂ ections at the top and bottom of the story under consideration using the ﬂ oor deﬂ ections of Section 18.5.3.1. 18.5.3.4 Design Earthquake Story Velocity Design story velocities, ∇D, in the direction of interest shall be calculated in accordance with Eqs. 18.5-23 through 18.5-25: ∇ = ∇ + ∇ D D RD 1 2 2 (18.5-23) ∇ = Δ 1 1 1 D D D T 2π (18.5-24) CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 190 ∇ = Δ RD RD R T 2π (18.5-25) where ∇1D = design story velocity due to the fundamental mode of vibration of the structure in the direction of interest ∇RD = design story velocity due to the residual mode of vibration of the structure in the direction of interest 18.5.3.5 Maximum Considered Earthquake Response Total and modal maximum ﬂ oor deﬂ ections at Level i, design story drifts, and design story velocities shall be based on the equations in Sections 18.5.3.1, 18.5.3.3, and 18.5.3.4, respectively, except that design roof displacements shall be replaced by maximum roof displacements. Maximum roof displacements shall be calculated in accordance with Eqs. 18.5-26 and 18.5-27: D g S T B g S T B T T M MS M M MS E M S 1 2 1 1 2 1 2 1 1 2 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < π π Γ Γ , (18.5-26a) D g S T B g S T B T T M M M M M E M S 1 2 1 1 1 1 2 1 1 1 1 1 4 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≥ π π Γ Γ , (18.5-26b) D g S T B g S T B RM R M R R R MS R R = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 4 2 1 2 2 π π Γ Γ (18.5-27) where SM1 = the MCER, 5 percent damped, spectral response acceleration parameter at a period of 1 s adjusted for site class effects as deﬁ ned in Section 11.4.3 SMS = the MCER, 5 percent damped, spectral response acceleration parameter at short periods adjusted for site class effects as deﬁ ned in Section 11.4.3 B1M = numerical coefﬁ cient as set forth in Table 18.6-1 for effective damping equal to βmM (m = 1) and period of structure equal to T1M 18.6 DAMPED RESPONSE MODIFICATION As required in Sections 18.4 and 18.5, response of the structure shall be modiﬁ ed for the effects of the damping system. 18.6.1 Damping Coefﬁ cient Where the period of the structure is greater than or equal to T0, the damping coefﬁ cient shall be as prescribed in Table 18.6-1. Where the period of the structure is less than T0, the damping coefﬁ cient shall be linearly interpolated between a value of 1.0 at a 0-second period for all values of effective damping and the value at period T0 as indicated in Table 18.6-1. 18.6.2 Effective Damping The effective damping at the design displace-ment, βmD, and at the maximum displacement, βmM, of the mth mode of vibration of the structure in the direction under consideration shall be calculated using Eqs. 18.6-1 and 18.6-2: β β β μ β mD I Vm D HD = + + (18.6-1) β β β μ β mM I Vm M HM = + + (18.6-2) where βHD = component of effective damping of the structure in the direction of interest due to post-yield hysteretic behavior of the seismic force-resisting system and elements of the damping system at effective ductility demand, μD βHM = component of effective damping of the struc-ture in the direction of interest due to post-yield hysteretic behavior of the seismic force-resist-ing system and elements of the damping system at effective ductility demand, μM βI = component of effective damping of the struc-ture due to the inherent dissipation of energy Table 18.6-1 Damping Coefﬁ cient, BV+I, B1D, BR, B1M, BmD, BmM (Where Period of the Structure ≥ T0) Effective Damping, β (percentage of critical) Bv+I, B1D, BR, B1M, BmD, BmM (where period of the structure ≥ T0) ≤2 0.8 5 1.0 10 1.2 20 1.5 30 1.8 40 2.1 50 2.4 60 2.7 70 3.0 80 3.3 90 3.6 ≥100 4.0 MINIMUM DESIGN LOADS 191 by elements of the structure, at or just below the effective yield displacement of the seismic force-resisting system βVm = component of effective damping of the mth mode of vibration of the structure in the direction of interest due to viscous dissipation of energy by the damping system, at or just below the effective yield displacement of the seismic force-resisting system μD = effective ductility demand on the seismic force-resisting system in the direction of interest due to the design earthquake ground motions μM = effective ductility demand on the seismic force-resisting system in the direction of interest due to the maximum considered earthquake ground motions Unless analysis or test data supports other values, the effective ductility demand of higher modes of vibration in the direction of interest shall be taken as 1.0. 18.6.2.1 Inherent Damping Inherent damping, βI, shall be based on the material type, conﬁ guration, and behavior of the structure and nonstructural components responding dynamically at or just below yield of the seismic force-resisting system. Unless analysis or test data supports other values, inherent damping shall be taken as not greater than 5 percent of critical for all modes of vibration. 18.6.2.2 Hysteretic Damping Hysteretic damping of the seismic force-resisting system and elements of the damping system shall be based either on test or analysis or shall be calculated using Eqs. 18.6-3 and 18.6-4: β β μ HD H I D q = − ( ) − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 64 1 1 . (18.6-3) β β μ HM H I M q = − ( ) − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 64 1 1 . (18.6-4) where qH = hysteresis loop adjustment factor, as deﬁ ned in Section 18.6.2.2.1 μD = effective ductility demand on the seismic force-resisting system in the direction of interest due to the design earthquake ground motions μM = effective ductility demand on the seismic force-resisting system in the direction of interest due to the maximum considered earthquake ground motions Unless analysis or test data supports other values, the hysteretic damping of higher modes of vibration in the direction of interest shall be taken as zero. 18.6.2.2.1 Hysteresis Loop Adjustment Factor The calculation of hysteretic damping of the seismic force-resisting system and elements of the damping system shall consider pinching and other effects that reduce the area of the hysteresis loop during repeated cycles of earthquake demand. Unless analysis or test data support other values, the fraction of full hyster-etic loop area of the seismic force-resisting system used for design shall be taken as equal to the factor, qH, calculated using Eq. 18.6-5: q T T H S = 0 67 1 . (18.6-5) where TS = period deﬁ ned by the ratio, SD1/SDS T1 = period of the fundamental mode of vibration of the structure in the direction of the interest The value of qH shall not be taken as greater than 1.0 and need not be taken as less than 0.5. 18.6.2.3 Viscous Damping Viscous damping of the mth mode of vibration of the structure, βVm, shall be calculated using Eqs. 18.6-6 and 18.6-7: β π Vm mj j m W W = ∑ 4 (18.6-6) W F m im im j = ∑ 1 2 δ (18.6-7) where Wmj = work done by jth damping device in one complete cycle of dynamic response corre-sponding to the mth mode of vibration of the structure in the direction of interest at modal displacements, δim Wm = maximum strain energy in the mth mode of vibration of the structure in the direction of interest at modal displacements, δim Fim = mth mode inertial force at Level i δim = deﬂ ection of Level i in the mth mode of vibration at the center of rigidity of the struc-ture in the direction under consideration Viscous modal damping of displacement-dependent damping devices shall be based on a CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 192 response amplitude equal to the effective yield displacement of the structure. The calculation of the work done by individual damping devices shall consider orientation and participation of each device with respect to the mode of vibration of interest. The work done by individual damping devices shall be reduced as required to account for the ﬂ exibility of elements, including pins, bolts, gusset plates, brace extensions, and other components that connect damping devices to other elements of the structure. 18.6.3 Effective Ductility Demand The effective ductility demand on the seismic force-resisting system due to the design earthquake ground motions, μD, and due to the maximum considered earthquake ground motions, μM, shall be calculated using Eqs. 18.6-8, 18.6-9, and 18.6-10: μD D Y D D = ≥ 1 1 0 . (18.6-8) μM M Y D D = ≥ 1 1 0 . (18.6-9) D g C R C T Y d S = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 2 0 1 1 1 2 π Ω Γ (18.6-10) where D1D = fundamental mode design displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.4.3.2 or 18.5.3.2 D1M = fundamental mode maximum displacement at the center of rigidity of the roof level of the structure in the direction under consideration, Section 18.4.3.5 or 18.5.3.5 DY = displacement at the center of rigidity of the roof level of the structure at the effective yield point of the seismic force-resisting system R = response modiﬁ cation coefﬁ cient from Table 12.2-1 Cd = deﬂ ection ampliﬁ cation factor from Table 12.2-1 Ω0 = overstrength factor from Table 12.2-1 Γ1 = participation factor of the fundamental mode of vibration of the structure in the direction of interest, Section 18.4.2.3 or 18.5.2.3 (m = 1) CS1 = seismic response coefﬁ cient of the fundamental mode of vibration of the structure in the direction of interest, Section 18.4.2.4 or 18.5.2.4 (m = 1) T1 = period of the fundamental mode of vibration of the structure in the direction of interest The design ductility demand, μD, shall not exceed the maximum value of effective ductility demand, μmax, given in Section 18.6.4. 18.6.4 Maximum Effective Ductility Demand For determination of the hysteresis loop adjust-ment factor, hysteretic damping, and other parameters, the maximum value of effective ductility demand, μmax, shall be calculated using Eqs. 18.6-11 and 18.6-12: For T1D ≤ TS, μmax = 0.5[(R/(Ω0Ie))2 + 1] (18.6-11) For T1 ≥ TS, μmax = R/(Ω0Ie) (18.6-12) where Ie = the importance factor determined in accordance with Section 11.5.1 T1D = effective period of the fundamental mode of vibration of the structure at the design displace-ment in the direction under consideration For T1 < TS < T1D, μmax shall be determined by linear interpolation between the values of Eqs. 18.6-11 and 18.6-12. 18.7 SEISMIC LOAD CONDITIONS AND ACCEPTANCE CRITERIA For the nonlinear procedures of Section 18.3, the seismic force-resisting system, damping system, loading conditions, and acceptance criteria for response parameters of interest shall conform with Section 18.7.1. Design forces and displacements determined in accordance with the response-spectrum procedure of Section 18.4 or the equivalent lateral force procedure of Section 18.5 shall be checked using the strength design criteria of this standard and the seismic loading conditions of Section 18.7.1 and 18.7.2. 18.7.1 Nonlinear Procedures Where nonlinear procedures are used in analysis, the seismic force-resisting system, damping system, seismic loading conditions, and acceptance criteria shall conform to the following subsections. 18.7.1.1 Seismic Force-Resisting System The seismic force-resisting system shall satisfy the strength requirements of Section 12.2.1 using the seismic base shear, Vmin, as given by Section 18.2.2.1. The story drift shall be determined using the design earthquake ground motions. MINIMUM DESIGN LOADS 193 18.7.1.2 Damping Systems The damping devices and their connections shall be sized to resist the forces, displacements, and velocities from the maximum considered earthquake ground motions. 18.7.1.3 Combination of Load Effects The effects on the damping system due to gravity loads and seismic forces shall be combined in accor-dance with Section 12.4 using the effect of horizontal seismic forces, QE, determined in accordance with the analysis. The redundancy factor, ρ, shall be taken equal to 1.0 in all cases, and the seismic load effect with overstrength factor of Section 12.4.3 need not apply to the design of the damping system. 18.7.1.4 Acceptance Criteria for the Response Parameters of Interest The damping system components shall be evaluated using the strength design criteria of this standard using the seismic forces and seismic loading conditions determined from the nonlinear procedures and φ = 1.0. The members of the seismic force-resist-ing system need not be evaluated where using the nonlinear procedure forces. 18.7.2 Response-Spectrum and Equivalent Lateral Force Procedures Where response-spectrum or equivalent lateral force procedures are used in analysis, the seismic force-resisting system, damping system, seismic loading conditions, and acceptance criteria shall conform to the following subsections. 18.7.2.1 Seismic Force-Resisting System The seismic force-resisting system shall satisfy the requirements of Section 12.2.1 using seismic base shear and design forces determined in accordance with Section 18.4.2 or 18.5.2. The design story drift, ΔD, as determined in either Section 18.4.3.3 or 18.5.3.3 shall not exceed (R/Cd) times the allowable story drift, as obtained from Table 12.12-1, considering the effects of torsion as required in Section 12.12.1. 18.7.2.2 Damping System The damping system shall satisfy the require-ments of Section 12.2.1 for seismic design forces and seismic loading conditions determined in accordance with this section. 18.7.2.3 Combination of Load Effects The effects on the damping system and its components due to gravity loads and seismic forces shall be combined in accordance with Section 12.4 using the effect of horizontal seismic forces, QE, determined in accordance with Section 18.7.2.5. The redundancy factor, ρ, shall be taken equal to 1.0 in all cases, and the seismic load effect with overstrength factor of Section 12.4.3 need not apply to the design of the damping system. 18.7.2.4 Modal Damping System Design Forces Modal damping system design forces shall be calculated on the basis of the type of damping devices and the modal design story displacements and velocities determined in accordance with either Section 18.4.3 or 18.5.3. Modal design story displacements and velocities shall be increased as required to envelop the total design story displacements and velocities determined in accordance with Section 18.3 where peak response is required to be conﬁ rmed by response-history analysis. 1. Displacement-dependent damping devices: Design seismic force in displacement-dependent damping devices shall be based on the maximum force in the device at displacements up to and including the design story drift, ΔD. 2. Velocity-dependent damping devices: Design seismic force in each mode of vibration in veloc-ity-dependent damping devices shall be based on the maximum force in the device at velocities up to and including the design story velocity for the mode of interest. Displacements and velocities used to determine design forces in damping devices at each story shall account for the angle of orientation of the damping device from the horizontal and consider the effects of increased ﬂ oor response due to torsional motions. 18.7.2.5 Seismic Load Conditions and Combination of Modal Responses Seismic design force, QE, in each element of the damping system shall be taken as the maximum force of the following three loading conditions: 1. Stage of maximum displacement: Seismic design force at the stage of maximum displacement shall be calculated in accordance with Eq. 18.7-1: Q Q Q E mSFRS m DSD = ( ) ± ∑ Ω0 2 (18.7-1) CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 194 where QmSFRS = force in an element of the damping system equal to the design seismic force of the mth mode of vibration of the structure in the direction of interest QDSD = force in an element of the damping system required to resist design seismic forces of displacement-dependent damping devices Seismic forces in elements of the damping system, QDSD, shall be calculated by imposing design forces of displacement-dependent damping devices on the damping system as pseudostatic forces. Design seismic forces of displacement-dependent damping devices shall be applied in both positive and negative directions at peak displacement of the structure. 2. Stage of maximum velocity: Seismic design force at the stage of maximum velocity shall be calcu-lated in accordance with Eq. 18.7-2: Q Q E mDSV m = ( ) ∑ 2 (18.7-2) where QmDSV = force in an element of the damping system required to resist design seismic forces of velocity-dependent damping devices due to the mth mode of vibration of the structure in the direction of interest Modal seismic design forces in elements of the damping system, QmDSV, shall be calculated by imposing modal design forces of velocity-dependent devices on the nondeformed damping system as pseudostatic forces. Modal seismic design forces shall be applied in directions consis-tent with the deformed shape of the mode of interest. Horizontal restraint forces shall be applied at each ﬂ oor Level i of the nondeformed damping system concurrent with the design forces in velocity-dependent damping devices such that the horizontal displacement at each level of the structure is zero. At each ﬂ oor Level i, restraint forces shall be proportional to and applied at the location of each mass point. 3. Stage of maximum acceleration: Seismic design force at the stage of maximum acceleration shall be calculated in accordance with Eq. 18.7-3: Q C Q C Q Q E mFD 0 mSFRS mFV mDSV m DSD = + ( ) ± ∑ Ω 2 (18.7-3) The force coefﬁ cients, CmFD and CmFV, shall be determined from Tables 18.7-1 and 18.7-2, respectively, using values of effective damping determined in accordance with the following requirements: For fundamental-mode response (m = 1) in the direction of interest, the coefﬁ cients, C1FD and C1FV, shall be based on the velocity exponent, α, that Table 18.7-1 Force Coefﬁ cient, CmFD a,b Effective Damping μ ≤ 1.0 CmFD = 1.0c α ≤ 0.25 α = 0.5 α = 0.75 α ≥ 1.0 ≤0.05 1.00 1.00 1.00 1.00 μ ≥ 1.0 0.1 1.00 1.00 1.00 1.00 μ ≥ 1.0 0.2 1.00 0.95 0.94 0.93 μ ≥ 1.1 0.3 1.00 0.92 0.88 0.86 μ ≥ 1.2 0.4 1.00 0.88 0.81 0.78 μ ≥ 1.3 0.5 1.00 0.84 0.73 0.71 μ ≥ 1.4 0.6 1.00 0.79 0.64 0.64 μ ≥ 1.6 0.7 1.00 0.75 0.55 0.58 μ ≥ 1.7 0.8 1.00 0.70 0.50 0.53 μ ≥ 1.9 0.9 1.00 0.66 0.50 0.50 μ ≥ 2.1 ≥1.0 1.00 0.62 0.50 0.50 μ ≥ 2.2 aUnless analysis or test data support other values, the force coefﬁ cient CmFD for viscoelastic systems shall be taken as 1.0. bInterpolation shall be used for intermediate values of velocity exponent, α, and ductility demand, μ. cCmFD shall be taken as equal to 1.0 for values of ductility demand, μ, greater than or equal to the values shown. MINIMUM DESIGN LOADS 195 relates device force to damping device velocity. The effective fundamental-mode damping shall be taken as equal to the total effective damping of the fundamental mode less the hysteretic component of damping (β1D – βHD or β1M – βHM) at the response level of interest (μ = μD or μ = μM). For higher-mode (m > 1) or residual-mode response in the direction of interest, the coefﬁ -cients, CmFD and CmFV, shall be based on a value of α equal to 1.0. The effective modal damping shall be taken as equal to the total effective damping of the mode of interest (βmD or βmM). For determina-tion of the coefﬁ cient CmFD, the ductility demand shall be taken as equal to that of the fundamental mode (μ = μD or μ = μM). 18.7.2.6 Inelastic Response Limits Elements of the damping system are permitted to exceed strength limits for design loads provided it is shown by analysis or test that 1. Inelastic response does not adversely affect damping system function. 2. Element forces calculated in accordance with Section 18.7.2.5, using a value of Ω0 taken as equal to 1.0, do not exceed the strength required to satisfy the load combinations of Section 12.4. 18.8 DESIGN REVIEW A design review of the damping system and related test programs shall be performed by an independent team of registered design professionals in the appro-priate disciplines and others experienced in seismic analysis methods and the theory and application of energy dissipation systems. The design review shall include, but need not be limited to, the following: 1. Review of site-speciﬁ c seismic criteria including the development of the site-speciﬁ c spectra and ground motion histories and all other project-speciﬁ c design criteria. 2. Review of the preliminary design of the seismic force-resisting system and the damping system, including design parameters of damping devices. 3. Review of the ﬁ nal design of the seismic force-resisting system and the damping system and all supporting analyses. 4. Review of damping device test requirements, device manufacturing quality control and assur-ance, and scheduled maintenance and inspection requirements. 18.9 TESTING The force-velocity displacement and damping proper-ties used for the design of the damping system shall be based on the prototype tests speciﬁ ed in this section. The fabrication and quality control procedures used for all prototype and production damping devices shall be identical. 18.9.1 Prototype Tests The following tests shall be performed separately on two full-size damping devices of each type and size used in the design, in the order listed as follows. Representative sizes of each type of device are permitted to be used for prototype testing, provided both of the following conditions are met: 1. Fabrication and quality control procedures are identical for each type and size of device used in the structure. 2. Prototype testing of representative sizes is accepted by the registered design professional responsible for design of the structure. Test specimens shall not be used for construction, unless they are accepted by the registered design professional responsible for design of the structure and meet the requirements for prototype and produc-tion tests. Table 18.7-2 Force Coefﬁ cient, CmFV a,b Effective Damping α ≤ 0.25 α = 0.5 α = 0.75 α ≥ 1.0 ≤0.05 1.00 0.35 0.20 0.10 0.1 1.00 0.44 0.31 0.20 0.2 1.00 0.56 0.46 0.37 0.3 1.00 0.64 0.58 0.51 0.4 1.00 0.70 0.69 0.62 0.5 1.00 0.75 0.77 0.71 0.6 1.00 0.80 0.84 0.77 0.7 1.00 0.83 0.90 0.81 0.8 1.00 0.90 0.94 0.90 0.9 1.00 1.00 1.00 1.00 ≥1.0 1.00 1.00 1.00 1.00 aUnless analysis or test data support other values, the force coefﬁ cient CmFD for viscoelastic systems shall be taken as 1.0. bInterpolation shall be used for intermediate values of velocity exponent, α. CHAPTER 18 SEISMIC DESIGN REQUIREMENTS FOR STRUCTURES WITH DAMPING SYSTEMS 196 18.9.1.1 Data Recording The force-deﬂ ection relationship for each cycle of each test shall be recorded. 18.9.1.2 Sequence and Cycles of Testing For the following test sequences, each damping device shall be subjected to gravity load effects and thermal environments representative of the installed condition. For seismic testing, the displacement in the devices calculated for the maximum considered earthquake ground motions, termed herein as the maximum device displacement, shall be used. 1. Each damping device shall be subjected to the number of cycles expected in the design wind-storm, but not less than 2,000 continuous fully reversed cycles of wind load. Wind load shall be at amplitudes expected in the design windstorm and shall be applied at a frequency equal to the inverse of the fundamental period of the structure (f1 = 1/T1). EXCEPTION: Damping devices need not be subjected to these tests if they are not subject to wind-induced forces or displacements or if the design wind force is less than the device yield or slip force. 2. Each damping device shall be loaded with ﬁ ve fully reversed, sinusoidal cycles at the maximum earthquake device displacement at a frequency equal to 1/T1M as calculated in Section 18.4.2.5. Where the damping device characteristics vary with operating temperature, these tests shall be conducted at a minimum of three temperatures (minimum, ambient, and maximum) that bracket the range of operating temperatures. EXCEPTION: Damping devices are permitted to be tested by alternative methods provided all of the following conditions are met: a. Alternative methods of testing are equivalent to the cyclic testing requirements of this section. b. Alternative methods capture the dependence of the damping device response on ambient temperature, frequency of loading, and tempera-ture rise during testing. c. Alternative methods are accepted by the registered design professional responsible for the design of the structure. 3. If the force-deformation properties of the damping device at any displacement less than or equal to the maximum device displacement change by more than 15 percent for changes in testing frequency from 1/T1M to 2.5/T1, then the preceding tests shall also be performed at frequencies equal to 1/T1 and 2.5/T1. If reduced-scale prototypes are used to qualify the rate-dependent properties of damping devices, the reduced-scale prototypes should be of the same type and materials, and manufactured with the same processes and quality control procedures, as full-scale prototypes, and tested at a similitude-scaled frequency that represents the full-scale loading rates. 18.9.1.3 Testing Similar Devices Damping devices need not be prototype tested provided that both of the following conditions are met: 1. All pertinent testing and other damping device data are made available to and are accepted by the registered design professional responsible for the design of the structure. 2. The registered design professional substantiates the similarity of the damping device to previously tested devices. 18.9.1.4 Determination of Force-Velocity-Displacement Characteristics The force-velocity-displacement characteristics of a damping device shall be based on the cyclic load and displacement tests of prototype devices speciﬁ ed in the preceding text. Effective stiffness of a damping device shall be calculated for each cycle of deforma-tion using Eq. 17.8-1. 18.9.1.5 Device Adequacy The performance of a prototype damping device shall be deemed adequate if all of the conditions listed below are satisﬁ ed. The 15 percent limits speciﬁ ed in the following text are permitted to be increased by the registered design professional responsible for the design of the structure provided that the increased limit has been demonstrated by analysis not to have a deleterious effect on the response of the structure. 18.9.1.5.1 Displacement-Dependent Damping Devices The performance of the prototype displacement-dependent damping devices shall be deemed adequate if the following conditions, based on tests speciﬁ ed in Section 18.9.1.2, are satisﬁ ed: 1. For Test 1, no signs of damage including leakage, yielding, or breakage. 2. For Tests 2 and 3, the maximum force and minimum force at zero displacement for a damping device for any one cycle does not differ by more MINIMUM DESIGN LOADS 197 than 15 percent from the average maximum and minimum forces at zero displacement as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 3. For Tests 2 and 3, the maximum force and minimum force at maximum device displacement for a damping device for any one cycle does not differ by more than 15 percent from the average maximum and minimum forces at the maximum device displacement as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 4. For Tests 2 and 3, the area of hysteresis loop (Eloop) of a damping device for any one cycle does not differ by more than 15 percent from the average area of the hysteresis loop as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 5. The average maximum and minimum forces at zero displacement and maximum displacement, and the average area of the hysteresis loop (Eloop), calculated for each test in the sequence of Tests 2 and 3, shall not differ by more than 15 percent from the target values speciﬁ ed by the registered design professional responsible for the design of the structure. 18.9.1.5.2 Velocity-Dependent Damping Devices The performance of the prototype velocity-dependent damping devices shall be deemed adequate if the following conditions, based on tests speciﬁ ed in Section 18.9.1.2, are satisﬁ ed: 1. For Test 1, no signs of damage including leakage, yielding, or breakage. 2. For velocity-dependent damping devices with stiffness, the effective stiffness of a damping device in any one cycle of Tests 2 and 3 does not differ by more than 15 percent from the average effective stiffness as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 3. For Tests 2 and 3, the maximum force and minimum force at zero displacement for a damping device for any one cycle does not differ by more than 15 percent from the average maximum and minimum forces at zero displacement as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 4. For Tests 2 and 3, the area of hysteresis loop (Eloop) of a damping device for any one cycle does not differ by more than 15 percent from the average area of the hysteresis loop as calculated from all cycles in that test at a speciﬁ c frequency and temperature. 5. The average maximum and minimum forces at zero displacement, effective stiffness (for damping devices with stiffness only), and average area of the hysteresis loop (Eloop) calculated for each test in the sequence of Tests 2 and 3, does not differ by more than 15 percent from the target values speciﬁ ed by the registered design professional responsible for the design of the structure. 18.9.2 Production Testing Prior to installation in a building, damping devices shall be tested to ensure that their force-velocity-displacement characteristics fall within the limits set by the registered design professional responsible for the design of the structure. The scope and frequency of the production-testing program shall be determined by the registered design professional responsible for the design of the structure. 199 Chapter 19 SOIL–STRUCTURE INTERACTION FOR SEISMIC DESIGN β ˜ = the fraction of critical damping for the structure-foundation system determined in Section 19.2.1.2 W _ = the effective seismic weight of the structure, which shall be taken as 0.7W, except for struc-tures where the effective seismic weight is concentrated at a single level, it shall be taken as equal to W 19.2.1.1 Effective Building Period The effective period (T ˜ ) shall be determined as follows: T T k K K h K y y = + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 2 θ (19.2-3) where T = the fundamental period of the structure as determined in Section 12.8.2 k _ = the stiffness of the structure where ﬁ xed at the base, deﬁ ned by the following: k W gT = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 2 2 π (19.2-4) where h _ = the effective height of the structure, which shall be taken as 0.7 times the structural height (hn), except for structures where the gravity load is effectively concentrated at a single level, the effective height of the structure shall be taken as the height to that level Ky = the lateral stiffness of the foundation deﬁ ned as the horizontal force at the level of the foundation necessary to produce a unit deﬂ ection at that level, the force and the deﬂ ection being mea-sured in the direction in which the structure is analyzed Kθ = the rocking stiffness of the foundation deﬁ ned as the moment necessary to produce a unit average rotation of the foundation, the moment and rotation being measured in the direction in which the structure is analyzed g = the acceleration of gravity The foundation stiffnesses (Ky and Kθ) shall be computed by established principles of foundation mechanics using soil properties that are compatible 19.1 GENERAL If the option to incorporate the effects of soil–struc-ture interaction is exercised, the requirements of this section are permitted to be used in the determination of the design earthquake forces and the corresponding displacements of the structure if the model used for structural response analysis does not directly incorpo-rate the effects of foundation ﬂ exibility (i.e., the model corresponds to a ﬁ xed-based condition with no foundation springs). The provisions in this section shall not be used if a ﬂ exible-base foundation is included in the structural response model. The provisions for use with the equivalent lateral force procedure are given in Section 19.2, and those for use with the modal analysis procedure are given in Section 19.3. 19.2 EQUIVALENT LATERAL FORCE PROCEDURE The following requirements are supplementary to those presented in Section 12.8. 19.2.1 Base Shear To account for the effects of soil–structure interaction, the base shear (V) determined from Eq. 12.8-1 shall be reduced to V ˜ = V – ΔV (19.2-1) The reduction (ΔV) shall be computed as follows and shall not exceed 0.3V: Δ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≤ V C C W V s s 0 05 0 3 0 4 . . . β (19.2-2) where Cs = the seismic design coefﬁ cient computed from Eqs. 12.8-2, 12.8-3, and through 12.8-4 using the fundamental natural period of the ﬁ xed-base structure (T or T a) as speciﬁ ed in Section 12.8.2 C ˜ = the value of Cs computed from Eqs. 12.8-2, 12.8-3, and through 12.8-4 using the fundamen-tal natural period of the ﬂ exibly supported structure (T ˜ ) deﬁ ned in Section 19.2.1.1 CHAPTER 19 SOIL–STRUCTURE INTERACTION FOR SEISMIC DESIGN 200 with the soil strain levels associated with the design earthquake motion. The average shear modulus (G) for the soils beneath the foundation at large strain levels and the associated shear wave velocity (vs) needed in these computations shall be determined from Table 19.2-1 where vso = the average shear wave velocity for the soils beneath the foundation at small strain levels (10–3 percent or less) Go = γv2 so/g = the average shear modulus for the soils beneath the foundation at small strain levels γ = the average unit weight of the soils Alternatively, for structures supported on mat foundations that rest at or near the ground surface or are embedded in such a way that the side wall contact with the soil is not considered to remain effective during the design ground motion, the effective period of the structure is permitted to be determined from T T r h v T r h r a s a m = + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 25 1 1 12 2 2 2 3 α αθ . (19.2-5) where α = the relative weight density of the structure and the soil deﬁ ned by α γ = W A h o (19.2-6) ra and rm = characteristic foundation lengths deﬁ ned by r A a o = π (19.2-7) and r I m o = 4 4 π (19.2-8) where Ao = the area of the load-carrying foundation Io = the static moment of inertia of the load-carrying foundation about a horizontal centroidal axis normal to the direction in which the structure is analyzed αθ = dynamic foundation stiffness modiﬁ er for rocking as determined from Table 19.2-2 vs = shear wave velocity T = fundamental period as determined in Section 12.8.2 19.2.1.2 Effective Damping The effective damping factor for the structure-foundation system (β ˜ ) shall be computed as follows: β β = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ o T T 0 05 3 . (19.2-9) where βo = the foundation damping factor as speciﬁ ed in Fig. 19.2-1 For values of SDS 2 5 . between 0.10 and 0.20 the values of βo shall be determined by linear interpola-tion vbetween the solid lines and the dashed lines of Fig. 19.2-1. The quantity r in Fig. 19.2-1 is a characteristic foundation length that shall be determined as follows: For h L0 0 5 ≤. , r = ra (19.2-10) For h L0 1 ≥, r = rm (19.2-11) Table 19.2-1 Values of G/Go and vs/vso Site Class Value of vs/vso Value of G/Go SDS/2.5 SDS/2.5 ≤0.1 0.4 ≥0.8 ≤0.1 0.4 ≥0.8 A 1.00 1.00 1.00 1.00 1.00 1.00 B 1.00 0.97 0.95 1.00 0.95 0.90 C 0.97 0.87 0.77 0.95 0.75 0.60 D 0.95 0.71 0.32 0.90 0.50 0.10 E 0.77 0.22 a 0.60 0.05 a F a a a a a a Note: Use straight-line interpolation for intermediate values of SDS/2.5. aShould be evaluated from site speciﬁ c analysis Table 19.2-2 Values of αθ rm/vsT αθ <0.05 1.0 0.15 0.85 0.35 0.7 0.5 0.6 MINIMUM DESIGN LOADS 201 where Lo = the overall length of the side of the foundation in the direction being analyzed ra and rm = characteristic foundation lengths deﬁ ned in Eqs. 19.2-7 and 19.2-8, respectively For intermediate values of h L0 , the value of r shall be determined by linear interpolation. EXCEPTION: For structures supported on point-bearing piles and in all other cases where the foundation soil consists of a soft stratum of reasonably uniform properties underlain by a much stiffer, rock-like deposit with an abrupt increase in stiffness, the factor βo in Eq. 19.2-9 shall be replaced by βo ′ if 4 1 D v T s s < where Ds is the total depth of the stratum. βo ′ shall be determined as follows: ′ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ β β o s s o D v T 4 2 (19.2-12) The value of β ˜ computed from Eq. 19.2-9, both with or without the adjustment represented by Eq. 19.2-12, shall in no case be taken as less than β ˜ = 0.05 or greater than β ˜ = 0.20. 19.2.2 Vertical Distribution of Seismic Forces The distribution over the height of the structure of the reduced total seismic force (V ˜ ) shall be considered to be the same as for the structure without interaction. 19.2.3 Other Effects The modiﬁ ed story shears, overturning moments, and torsional effects about a vertical axis shall be determined as for structures without interaction using the reduced lateral forces. The modiﬁ ed deﬂ ections (δ ˜ ) shall be determined as follows: δ δ θ x o x x V V M h K = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (19.2-13) where Mo = the overturning moment at the base using the unmodiﬁ ed seismic forces and not including the reduction permitted in the design of the foundation hx = the height above the base to the level under consideration δx = the deﬂ ections of the ﬁ xed-base structure as determined in Section 12.8.6 using the unmodi-ﬁ ed seismic forces The modiﬁ ed story drifts and P-delta effects shall be evaluated in accordance with the provisions of Sections 12.8.6 and 12.8.7 using the modiﬁ ed story shears and deﬂ ections determined in this section. 19.3 MODAL ANALYSIS PROCEDURE The following provisions are supplementary to those presented in Section 12.9. 19.3.1 Modal Base Shears To account for the effects of soil–structure interaction, the base shear corresponding to the fundamental mode of vibration (V1) shall be reduced to V ˜1 = V1 – ΔV1 (19.3-1) The reduction (ΔV1) shall be computed in accordance with Eq. 19.2-2 with W _ taken as equal to the effective seismic weight of the fundamental period of vibration, W _ , and Cs computed in accordance with Eq. 12.8-1, except that SDS shall be replaced by design spectral response acceleration of the design response spectra at the fundamental period of the ﬁ xed-base structure (T1). The period T ˜ shall be determined from Eq. 19.2-3 or from Eq. 19.2-5 where applicable, taking T = T1, evaluating k _ from Eq. 19.2-4 with W _ = W _ 1, and computing h _ as follows: FIGURE 19.2-1 Foundation Damping Factor CHAPTER 19 SOIL–STRUCTURE INTERACTION FOR SEISMIC DESIGN 202 h w h w i i i i n i i i n = = = ∑ ∑ ϕ ϕ 1 1 1 1 (19.3-2) where wi = the portion of the total gravity load of the structure at Level i ϕi1 = the displacement amplitude at the ith level of the structure when vibrating in its fundamental mode hi = the height above the base to Level i The preceding designated values of W _ , h _ , T, and T ˜ also shall be used to evaluate the factor α from Eq. 19.2-6 and the factor βo from Fig. 19.2-1. No reduc-tion shall be made in the shear components contrib-uted by the higher modes of vibration. The reduced base shear (V ˜1) shall in no case be taken less than 0.7V1. 19.3.2 Other Modal Effects The modiﬁ ed modal seismic forces, story shears, and overturning moments shall be determined as for structures without interaction using the modiﬁ ed base shear (V ˜1) instead of V1. The modiﬁ ed modal deﬂ ec-tions (δ ˜ xm) shall be determined as follows: δ δ θ x o x x V V M h K 1 1 1 1 1 = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (19.3-3) and δ ˜ xm = δxm for m = 2, 3, . . . (19.3-4) where Mo1 = the overturning base moment for the fundamen-tal mode of the ﬁ xed-base structure using the unmodiﬁ ed modal base shear V1 δxm = the modal deﬂ ections at Level x of the ﬁ xed-base structure using the unmodiﬁ ed modal shears, Vm The modiﬁ ed modal drift in a story (Δ ˜ m) shall be computed as the difference of the deﬂ ections (δ ˜ xm) at the top and bottom of the story under consideration. 19.3.3 Design Values The design values of the modiﬁ ed shears, moments, deﬂ ections, and story drifts shall be determined as for structures without interaction by taking the square root of the sum of the squares (SRSS) of the respective modal contributions. In the design of the foundation, it is permitted to reduce the overturning moment at the foundation–soil interface determined in this manner by 10 percent as for structures without interaction. The effects of torsion about a vertical axis shall be evaluated in accordance with the provisions of Section 12.8.4, and the P-delta effects shall be evaluated in accordance with the provisions of Section 12.8.7 using the story shears and drifts determined in Section 19.3.2. 203 Chapter 20 SITE CLASSIFICATION PROCEDURE FOR SEISMIC DESIGN accelerations for liqueﬁ able soils. Rather, a site class is permitted to be determined in accordance with Section 20.3 and the corresponding values of Fa and Fv determined from Tables 11.4-1 and 11.4-2. 2. Peats and/or highly organic clays [H > 10 ft (3 m)] of peat and/or highly organic clay where H = thickness of soil. 3. Very high plasticity clays [H > 25 ft (7.6 m) with PI > 75]. 4. Very thick soft/medium stiff clays [H > 120 ft (37 m)] with su < 1,000 psf (50 kPa). 20.3.2 Soft Clay Site Class E Where a site does not qualify under the criteria for Site Class F and there is a total thickness of soft clay greater than 10 ft (3 m) where a soft clay layer is deﬁ ned by su < 500 psf (25 kPa), w ≥ 40 percent, and PI > 20, it shall be classiﬁ ed as Site Class E. 20.3.3 Site Classes C, D, and E The existence of Site Class C, D, and E soils shall be classiﬁ ed by using one of the following three methods with v _ s, N _ , and s _ u computed in all cases as speciﬁ ed in Section 20.4: 1. v _ s for the top 100 ft (30 m) (v _ s method). 2. N _ for the top 100 ft (30 m) (N _ method). 3. N _ ch for cohesionless soil layers (PI < 20) in the top 100 ft (30 m) and s _ u for cohesive soil layers (PI > 20) in the top 100 ft (30 m) (s _ u method). Where the N _ ch and s _ u criteria differ, the site shall be assigned to the category with the softer soil. 20.3.4 Shear Wave Velocity for Site Class B The shear wave velocity for rock, Site Class B, shall be either measured on site or estimated by a geotechnical engineer, engineering geologist, or seismologist for competent rock with moderate fracturing and weathering. Softer and more highly fractured and weathered rock shall either be measured on site for shear wave velocity or classiﬁ ed as Site Class C. 20.3.5 Shear Wave Velocity for Site Class A The hard rock, Site Class A, category shall be supported by shear wave velocity measurement either 20.1 SITE CLASSIFICATION The site soil shall be classiﬁ ed in accordance with Table 20.3-1 and Section 20.3 based on the upper 100 ft (30 m) of the site proﬁ le. Where site-speciﬁ c data are not available to a depth of 100 ft (30 m), appropri-ate soil properties are permitted to be estimated by the registered design professional preparing the soil investigation report based on known geologic condi-tions. Where the soil properties are not known in sufﬁ cient detail to determine the site class, Site Class D shall be used unless the authority having jurisdic-tion or geotechnical data determine Site Class E or F soils are present at the site. Site Classes A and B shall not be assigned to a site if there is more than 10 ft (10.1 m) of soil between the rock surface and the bottom of the spread footing or mat foundation. 20.2 SITE RESPONSE ANALYSIS FOR SITE CLASS F SOIL A site response analysis in accordance with Section 21.1 shall be provided for Site Class F soils, unless the exception to Section 20.3.1 is applicable. 20.3 SITE CLASS DEFINITIONS Site class types shall be assigned in accordance with the deﬁ nitions provided in Table 20.3-1 and this section. 20.3.1 Site Class F Where any of the following conditions is satis-ﬁ ed, the site shall be classiﬁ ed as Site Class F and a site response analysis in accordance with Section 21.1 shall be performed. 1. Soils vulnerable to potential failure or collapse under seismic loading, such as liqueﬁ able soils, quick and highly sensitive clays, and collapsible weakly cemented soils. EXCEPTION: For structures having fundamental periods of vibration equal to or less than 0.5 s, site response analysis is not required to determine spectral CHAPTER 20 SITE CLASSIFICATION PROCEDURE FOR SEISMIC DESIGN 204 on site or on proﬁ les of the same rock type in the same formation with an equal or greater degree of weathering and fracturing. Where hard rock condi-tions are known to be continuous to a depth of 100 ft (30 m), surﬁ cial shear wave velocity measurements are permitted to be extrapolated to assess v _ s. 20.4 DEFINITIONS OF SITE CLASS PARAMETERS The deﬁ nitions presented in this section shall apply to the upper 100 ft (30 m) of the site proﬁ le. Proﬁ les containing distinct soil and rock layers shall be subdivided into those layers designated by a number that ranges from 1 to n at the bottom where there are a total of n distinct layers in the upper 100 ft (30 m). Where some of the n layers are cohesive and others are not, k is the number of cohesive layers and m is the number of cohesionless layers. The symbol i refers to any one of the layers between 1 and n. 20.4.1 v _ s, Average Shear Wave Velocity v _ s shall be determined in accordance with the following formula: v d d v s i i n i si i n = = = ∑ ∑ 1 1 (20.4-1) where di = the thickness of any layer between 0 and 100 ft (30 m) vsi = the shear wave velocity in ft/s (m/s) di i n = ∑ 1 = 100 ft (30 m) 20.4.2 N _ , Average Field Standard Penetration Resistance and N _ ch, Average Standard Penetration Resistance for Cohesionless Soil Layers N _ and N _ ch shall be determined in accordance with the following formulas: N d d N i i n i i i n = = = ∑ ∑ 1 1 (20.4-2) where Ni and di in Eq. 20.4-2 are for cohesionless soil, cohesive soil, and rock layers. N d d N ch s i i i m = = ∑ 1 (20.4-3) where Ni and di in Eq. 20.4-3 are for cohesionless soil layers only and d d i i m s = ∑ = 1 where ds is the total thickness of cohesionless soil layers in the top 100 ft (30 m). Ni is the standard penetration resistance (ASTM D1586) not to exceed 100 blows/ft (305 blows/m) as directly measured in the ﬁ eld without corrections. Where refusal is met for a rock layer, Ni shall be taken as 100 blows/ft (305 blows/m). 20.4.3 s _ u, Average Undrained Shear Strength s _ u shall be determined in accordance with the following formula: Table 20.3-1 Site Classiﬁ cation Site Class v _ s N _ or N _ ch s _ u A. Hard rock >5,000 ft/s NA NA B. Rock 2,500 to 5,000 ft/s NA NA C. Very dense soil and soft rock 1,200 to 2,500 ft/s >50 >2,000 psf D. Stiff soil 600 to 1,200 ft/s 15 to 50 1,000 to 2,000 psf E. Soft clay soil <600 ft/s <15 <1,000 psf Any proﬁ le with more than 10 ft of soil having the following characteristics: —Plasticity index PI > 20, —Moisture content w ≥ 40%, —Undrained shear strength s _ u < 500 psf F. Soils requiring site response analysis in accordance with Section 21.1 See Section 20.3.1 For SI: 1 ft/s = 0.3048 m/s; 1 lb/ft2 = 0.0479 kN/m2. MINIMUM DESIGN LOADS 205 s d d s u c i ui i k = = ∑ 1 (20.4-4) where di i k = ∑ 1 = dc dc = the total thickness of cohesive soil layers in the top 100 ft (30 m) PI = the plasticity index as determined in accor-dance with ASTM D4318 w = the moisture content in percent as determined in accordance with ASTM D2216 sui = the undrained shear strength in psf (kPa), not to exceed 5,000 psf (240 kPa) as determined in accordance with ASTM D2166 or ASTM D2850 207 Chapter 21 SITE-SPECIFIC GROUND MOTION PROCEDURES FOR SEISMIC DESIGN site coefﬁ cients in Section 11.4.3 consistent with the classiﬁ cation of the soils at the proﬁ le base. 21.1.3 Site Response Analysis and Computed Results Base ground motion time histories shall be input to the soil proﬁ le as outcropping motions. Using appropriate computational techniques that treat nonlinear soil properties in a nonlinear or equivalent-linear manner, the response of the soil proﬁ le shall be determined and surface ground motion time histories shall be calculated. Ratios of 5 percent damped response spectra of surface ground motions to input base ground motions shall be calculated. The recom-mended surface MCER ground motion response spectrum shall not be lower than the MCER response spectrum of the base motion multiplied by the average surface-to-base response spectral ratios (calculated period by period) obtained from the site response analyses. The recommended surface ground motions that result from the analysis shall reﬂ ect consideration of sensitivity of response to uncertainty in soil properties, depth of soil model, and input motions. 21.2 RISK-TARGETED MAXIMUM CONSIDERED EARTHQUAKE (MCER) GROUND MOTION HAZARD ANALYSIS The requirements of Section 21.2 shall be satisﬁ ed where a ground motion hazard analysis is performed or required by Section 11.4.7. The ground motion hazard analysis shall account for the regional tectonic setting, geology, and seismicity, the expected recur-rence rates and maximum magnitudes of earthquakes on known faults and source zones, the characteristics of ground motion attenuation, near source effects, if any, on ground motions, and the effects of subsurface site conditions on ground motions. The characteristics of subsurface site conditions shall be considered either using attenuation relations that represent regional and local geology or in accordance with Section 21.1. The analysis shall incorporate current seismic interpreta-tions, including uncertainties for models and param-eter values for seismic sources and ground motions. The analysis shall be documented in a report. 21.1 SITE RESPONSE ANALYSIS The requirements of Section 21.1 shall be satisﬁ ed where site response analysis is performed or required by Section 11.4.7. The analysis shall be documented in a report. 21.1.1 Base Ground Motions A MCER response spectrum shall be developed for bedrock, using the procedure of Sections 11.4.6 or 21.2. Unless a site-speciﬁ c ground motion hazard analysis described in Section 21.2 is carried out, the MCER rock response spectrum shall be developed using the procedure of Section 11.4.6 assuming Site Class B. If bedrock consists of Site Class A, the spectrum shall be adjusted using the site coefﬁ cients in Section 11.4.3 unless other site coefﬁ cients can be justiﬁ ed. At least ﬁ ve recorded or simulated horizontal ground motion acceleration time histories shall be selected from events having magnitudes and fault distances that are consistent with those that control the MCER ground motion. Each selected time history shall be scaled so that its response spectrum is, on average, approximately at the level of the MCER rock response spectrum over the period range of signiﬁ -cance to structural response. 21.1.2 Site Condition Modeling A site response model based on low-strain shear wave velocities, nonlinear or equivalent linear shear stress–strain relationships, and unit weights shall be developed. Low-strain shear wave velocities shall be determined from ﬁ eld measurements at the site or from measurements from similar soils in the site vicinity. Nonlinear or equivalent linear shear stress– strain relationships and unit weights shall be selected on the basis of laboratory tests or published relation-ships for similar soils. The uncertainties in soil properties shall be estimated. Where very deep soil proﬁ les make the development of a soil model to bedrock impractical, the model is permitted to be terminated where the soil stiffness is at least as great as the values used to deﬁ ne Site Class D in Chapter 20. In such cases, the MCER response spectrum and acceleration time histories of the base motion devel-oped in Section 21.1.1 shall be adjusted upward using CHAPTER 21 SITE-SPECIFIC GROUND MOTION PROCEDURES FOR SEISMIC DESIGN 208 21.2.1 Probabilistic (MCER) Ground Motions The probabilistic spectral response accelerations shall be taken as the spectral response accelerations in the direction of maximum horizontal response represented by a 5 percent damped acceleration response spectrum that is expected to achieve a 1 percent probability of collapse within a 50-year period. For the purpose of this standard, ordinates of the probabilistic ground motion response spectrum shall be determined by either Method 1 of Section 21.2.1.1 or Method 2 of Section 21.2.1.2. 21.2.1.1 Method 1 At each spectral response period for which the acceleration is computed, ordinates of the probabilistic ground motion response spectrum shall be determined as the product of the risk coefﬁ cient, CR, and the spectral response acceleration from a 5 percent damped acceleration response spectrum having a 2 percent probability of exceedance within a 50-year period. The value of the risk coefﬁ cient, CR, shall be determined using values of CRS and CR1 from Figs. 22-3 and 22-4, respectively. At spectral response periods less than or equal to 0.2 s, CR shall be taken as equal to CRS. At spectral response periods greater than or equal to 1.0 s, CR shall be taken as equal to CR1. At response spectral periods greater than 0.2 s and less than 1.0 s, CR shall be based on linear interpolation of CRS and CR1. 21.2.1.2 Method 2 At each spectral response period for which the acceleration is computed, ordinates of the probabilistic ground motion response spectrum shall be determined from iterative integration of a site-speciﬁ c hazard curve with a lognormal probability density function representing the collapse fragility (i.e., probability of collapse as a function of spectral response accelera-tion). The ordinate of the probabilistic ground motion response spectrum at each period shall achieve a 1 percent probability of collapse within a 50-year period for a collapse fragility having (i) a 10 percent prob-ability of collapse at said ordinate of the probabilistic ground motion response spectrum and (ii) a logarith-mic standard deviation value of 0.6. 21.2.2 Deterministic (MCER) Ground Motions The deterministic spectral response acceleration at each period shall be calculated as an 84th-percentile 5 percent damped spectral response acceleration in the direction of maximum horizontal response computed at that period. The largest such acceleration calculated for the characteristic earthquakes on all known active faults within the region shall be used. For the purposes of this standard, the ordinates of the deterministic ground motion response spectrum shall not be taken as lower than the corresponding ordinates of the response spectrum determined in accordance with Fig. 21.2-1, where Fa and Fv are determined using Tables 11.4-1 and 11.4-2, respectively, with the value of SS taken as 1.5 and the value of S1 taken as 0.6. 21.2.3 Site-Speciﬁ c MCER The site-speciﬁ c MCER spectral response acceleration at any period, SaM, shall be taken as the lesser of the spectral response accelerations from the probabilistic ground motions of Section 21.2.1 and the deterministic ground motions of Section 21.2.2. 21.3 DESIGN RESPONSE SPECTRUM The design spectral response acceleration at any period shall be determined from Eq. 21.3-1: S S a aM = 2 3 (21.3-1) where SaM is the MCER spectral response acceleration obtained from Section 21.1 or 21.2. The design spectral response acceleration at any period shall not be taken as less than 80 percent of Sa determined in accordance with Section 11.4.5. For sites classiﬁ ed as Site Class F requiring site response analysis in accordance with Section 11.4.7, the design spectral response acceleration at any period shall not be taken as less than 80 percent of Sa determined for Site Class E in accordance with Section 11.4.5. 21.4 DESIGN ACCELERATION PARAMETERS Where the site-speciﬁ c procedure is used to determine the design ground motion in accordance with Section 21.3, the parameter SDS shall be taken as the spectral acceleration, Sa, obtained from the site-speciﬁ c spectra at a period of 0.2 s, except that it shall not be taken as less than 90 percent of the peak spectral acceleration, Sa, at any period larger than 0.2 s. The parameter SD1 shall be taken as the greater of the spectral accelera-tion, Sa, at a period of 1 s or two times the spectral acceleration, Sa, at a period of 2 s. The parameters SMS and SM1 shall be taken as 1.5 times SDS and SD1, respectively. The values so obtained shall not be less than 80 percent of the values determined in MINIMUM DESIGN LOADS 209 accordance with Section 11.4.3 for SMS and SM1 and Section 11.4.4 for SDS and SD1. For use with the Equivalent Lateral Force Procedure, the site-speciﬁ c spectral acceleration, Sa, at T shall be permitted to replace SD1/T in Eq. 12.8-3 and SD1TL/T2 in Eq. 12.8-4. The parameter SDS calcu-lated per this section shall be permitted to be used in Eqs. 12.8-2, 12.8-5, 15.4-1, and 15.4-3. The mapped value of S1 shall be used in Eqs. 12.8-6, 15.4-2, and 15.4-4. 21.5 MAXIMUM CONSIDERED EARTHQUAKE GEOMETRIC MEAN (MCEG) PEAK GROUND ACCELERATION 21.5.1 Probabilistic MCEG Peak Ground Acceleration The probabilistic geometric mean peak ground acceleration shall be taken as the geometric mean peak ground acceleration with a 2 percent probability of exceedance within a 50-year period. 21.5.2 Deterministic MCEG Peak Ground Acceleration The deterministic geometric mean peak ground acceleration shall be calculated as the largest 84th-percentile geometric mean peak ground acceleration for characteristic earthquakes on all known active faults within the site region. The deterministic geometric mean peak ground acceleration shall not be taken as lower than 0.5 FPGA, where FPGA is deter-mined using Table 11.8-1 with the value of PGA taken as 0.5 g. 21.5.3 Site-Speciﬁ c MCEG Peak Ground Acceleration The site-speciﬁ c MCEG peak ground acceleration, PGAM, shall be taken as the lesser of the probabilistic geometric mean peak ground acceleration of Section 21.5.1 and the deterministic geometric mean peak ground acceleration of Section 21.5.2. The site-speciﬁ c MCEG peak ground acceleration shall not be taken as less than 80 percent of PGAM determined from Eq. 11.8-1. FIGURE 21.2-1 Deterministic Lower Limit on MCER Response Spectrum
6272	https://www.chegg.com/homework-help/questions-and-answers/319-many-ways-n-distinguishable-balls-arranged-n-distinguishable-boxes-box-empty-b-exactly-q92218369	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: 3.19 In how many ways can n distinguishable balls be arranged in n distinguishable boxes so that a) no box is empty? b) exactly one box is empty? c) at least one box is empty? Please explain in detial This AI-generated tip is based on Chegg's full solution. Sign up to see more! To determine the number of ways distinguishable balls can be arranged in distinguishable boxes such that no box is empty, consider that each ball must be placed in a unique box, making use of the counting principle. Sol. (a) As we have to arranged n balls in to n boxes. Therefore one ball is placed in each box. Also ball and boxes are distinguishable. So first ball is placed by n ways As first ball is placed in a box so second ball cannot be placed in same box S… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
6273	http://www.360doc.com/content/19/0618/18/1862743_843358807.shtml	34个奥数公式技巧，让孩子轻松搞定小学奥数搜索我的图书馆查看信箱系统消息官方通知设置开始对话有 11 人和你对话，查看忽略历史对话记录通知设置发文章发文工具撰写网文摘手文档视频思维导图随笔相册原创同步助手其他工具图片转文字文件清理AI助手留言交流搜索分享 QQ空间QQ好友新浪微博微信生成长图转Word打印朗读全屏修改转藏+1 × 微信扫一扫关注查看更多精彩文章 × 微信扫一扫将文章发送给好友 34个奥数公式技巧，让孩子轻松搞定小学奥数品茗堂 2019-06-18 来源\|114阅读\|1 转藏大中小转藏全屏朗读打印转Word生成长图分享 QQ空间QQ好友新浪微博微信展开全文 34个小学奥数必考公式 1、和差倍问题： 2、年龄问题的三个基本特征： ①两个人的年龄差是不变的； ②两个人的年龄是同时增加或者同时减少的； ③两个人的年龄的倍数是发生变化的； 3、归一问题的基本特点：问题中有一个不变的量，一般是那个'单一量'，题目一般用'照这样的速度'……等词语来表示。关键问题：根据题目中的条件确定并求出单一量； 4、植树问题： 5、鸡兔同笼问题：基本概念：鸡兔同笼问题又称为置换问题、假设问题，就是把假设错的那部分置换出来；基本思路： ①假设，即假设某种现象存在（甲和乙一样或者乙和甲一样）： ②假设后，发生了和题目条件不同的差，找出这个差是多少； ③每个事物造成的差是固定的，从而找出出现这个差的原因； ④再根据这两个差作适当的调整，消去出现的差。基本公式： ①把所有鸡假设成兔子：鸡数＝（兔脚数×总头数－总脚数）÷（兔脚数－鸡脚数） ②把所有兔子假设成鸡：兔数＝（总脚数一鸡脚数×总头数）÷（兔脚数一鸡脚数）关键问题：找出总量的差与单位量的差。 6、盈亏问题：基本概念：一定量的对象，按照某种标准分组，产生一种结果：按照另一种标准分组，又产生一种结果，由于分组的标准不同，造成结果的差异，由它们的关系求对象分组的组数或对象的总量。基本思路：先将两种分配方案进行比较，分析由于标准的差异造成结果的变化，根据这个关系求出参加分配的总份数，然后根据题意求出对象的总量。基本题型： ①一次有余数，另一次不足；基本公式：总份数＝（余数＋不足数）÷两次每份数的差 ②当两次都有余数；基本公式：总份数＝（较大余数一较小余数）÷两次每份数的差 ③当两次都不足；基本公式：总份数＝（较大不足数一较小不足数）÷两次每份数的差基本特点：对象总量和总的组数是不变的。关键问题：确定对象总量和总的组数。 7、牛吃草问题：基本思路：假设每头牛吃草的速度为'1'份，根据两次不同的吃法，求出其中的总草量的差；再找出造成这种差异的原因，即可确定草的生长速度和总草量。基本特点：原草量和新草生长速度是不变的；关键问题：确定两个不变的量。基本公式：生长量=（较长时间×长时间牛头数-较短时间×短时间牛头数）÷（长时间-短时间）；总草量=较长时间×长时间牛头数-较长时间×生长量； 8、周期循环与数表规律：周期现象：事物在运动变化的过程中，某些特征有规律循环出现。周期：我们把连续两次出现所经过的时间叫周期。关键问题：确定循环周期。闰年：一年有366天； ①年份能被4整除；②如果年份能被100整除，则年份必须能被400整除；平年：一年有365天。 ①年份不能被4整除；②如果年份能被100整除，但不能被400整除； 9、平均数：基本公式： ①平均数=总数量÷总份数总数量=平均数×总份数总份数=总数量÷平均数 ②平均数=基准数＋每一个数与基准数差的和÷总份数基本算法： ①求出总数量以及总份数，利用基本公式①进行计算. ②基准数法：根据给出的数之间的关系，确定一个基准数；一般选与所有数比较接近的数或者中间数为基准数；以基准数为标准，求所有给出数与基准数的差；再求出所有差的和；再求出这些差的平均数；最后求这个差的平均数和基准数的和，就是所求的平均数，具体关系见基本公式② 10、抽屉原理：抽屉原则一：如果把（n+1）个物体放在n个抽屉里，那么必有一个抽屉中至少放有2个物体。例：把4个物体放在3个抽屉里，也就是把4分解成三个整数的和，那么就有以下四种情况： ①4=4+0+0 ②4=3+1+0 ③4=2+2+0 ④4=2+1+1 观察上面四种放物体的方式，我们会发现一个共同特点：总有那么一个抽屉里有2个或多于2个物体，也就是说必有一个抽屉中至少放有2个物体。抽屉原则二：如果把n个物体放在m个抽屉里，其中n>m，那么必有一个抽屉至少有: ①k=[n/m ]+1个物体：当n不能被m整除时。 ②k=n/m个物体：当n能被m整除时。理解知识点： [X]表示不超过X的最大整数。例[4.351]=4；[0.321]=0；[2.9999]=2；关键问题：构造物体和抽屉。也就是找到代表物体和抽屉的量，而后依据抽屉原则进行运算。 11、定义新运算：基本概念：定义一种新的运算符号，这个新的运算符号包含有多种基本（混合）运算。基本思路：严格按照新定义的运算规则，把已知的数代入，转化为加减乘除的运算，然后按照基本运算过程、规律进行运算。关键问题：正确理解定义的运算符号的意义。注意事项： ①新的运算不一定符合运算规律，特别注意运算顺序。 ②每个新定义的运算符号只能在本题中使用。 12、数列求和：等差数列：在一列数中，任意相邻两个数的差是一定的，这样的一列数，就叫做等差数列。基本概念：首项：等差数列的第一个数，一般用a1表示；项数：等差数列的所有数的个数，一般用n表示；公差：数列中任意相邻两个数的差，一般用d表示；通项：表示数列中每一个数的公式，一般用an表示；数列的和：这一数列全部数字的和，一般用Sn表示．基本思路：等差数列中涉及五个量：a1 ,an, d, n,sn,,通项公式中涉及四个量，如果己知其中三个，就可求出第四个；求和公式中涉及四个量，如果己知其中三个，就可以求这第四个。基本公式：通项公式：an = a1+（n－1）d；通项＝首项＋（项数一1)×公差；数列和公式：sn,= (a1+ an)×n÷2；数列和＝（首项＋末项）×项数÷2；项数公式：n= (an+ a1)÷d＋1；项数=（末项-首项）÷公差＋1；公差公式：d =（an－a1））÷（n－1）；公差=（末项－首项）÷（项数－1）；关键问题：确定已知量和未知量，确定使用的公式； 13、二进制及其应用：十进制：用0～9十个数字表示，逢10进1；不同数位上的数字表示不同的含义，十位上的2表示20，百位上的2表示200。所以234=200+30+4=2×102+3×10+4。 =An×10n-1+An-1×10n-2+An-2×10n-3+An-3×10n-4+An-4×10n-5+An-6×10n-7+……+A3×102+A2×101+A1×100 注意：N0=１；N１=N（其中N是任意自然数）二进制：用0～1两个数字表示，逢2进1；不同数位上的数字表示不同的含义。（2）= An×2n-1+An-1×2n-2+An-2×2n-3+An-3×2n-4+An-4×2n-5+An-6×2n-7 +……+A3×22+A2×21+A1×20 注意：An不是0就是1。十进制化成二进制： ①根据二进制满2进1的特点，用2连续去除这个数，直到商为0，然后把每次所得的余数按自下而上依次写出即可。 ②先找出不大于该数的2的n次方，再求它们的差，再找不大于这个差的2的n次方，依此方法一直找到差为0，按照二进制展开式特点即可写出。 14、加法乘法原理和几何计数：加法原理：如果完成一件任务有n类方法，在第一类方法中有m1种不同方法，在第二类方法中有m2种不同方法……，在第n类方法中有mn种不同方法，那么完成这件任务共有：m1+ m2....... +mn种不同的方法。关键问题：确定工作的分类方法。基本特征：每一种方法都可完成任务。乘法原理：如果完成一件任务需要分成n个步骤进行，做第1步有m1种方法，不管第1步用哪一种方法，第2步总有m2种方法……不管前面n-1步用哪种方法，第n步总有mn种方法，那么完成这件任务共有：m1×m2.......×mn种不同的方法。关键问题：确定工作的完成步骤。基本特征：每一步只能完成任务的一部分。直线：一点在直线或空间沿一定方向或相反方向运动，形成的轨迹。直线特点：没有端点，没有长度。线段：直线上任意两点间的距离。这两点叫端点。线段特点：有两个端点，有长度。射线：把直线的一端无限延长。射线特点：只有一个端点；没有长度。 ①数线段规律：总数＝1+2+3+…+（点数一1）； ②数角规律=1+2+3+…+（射线数一1）； ③数长方形规律：个数=长的线段数×宽的线段数： ④数长方形规律：个数=1×1+2×2+3×3+…+行数×列数 15、质数与合数：质数：一个数除了1和它本身之外，没有别的约数，这个数叫做质数，也叫做素数。合数：一个数除了1和它本身之外，还有别的约数，这个数叫做合数。质因数：如果某个质数是某个数的约数，那么这个质数叫做这个数的质因数。分解质因数：把一个数用质数相乘的形式表示出来，叫做分解质因数。通常用短除法分解质因数。任何一个合数分解质因数的结果是唯一的。分解质因数的标准表示形式： N= ，其中a1、a2、a3……an都是合数N的质因数，且a1<a2<a3<……<an。求约数个数的公式： P=(r1+1)×(r2+1)×(r3+1)×……×(rn+1) 互质数：如果两个数的最大公约数是1，这两个数叫做互质数。 16、约数与倍数：约数和倍数：若整数a能够被b整除，a叫做b的倍数，b就叫做a的约数。公约数：几个数公有的约数，叫做这几个数的公约数；其中最大的一个，叫做这几个数的最大公约数。最大公约数的性质： 1、几个数都除以它们的最大公约数，所得的几个商是互质数。 2、几个数的最大公约数都是这几个数的约数。 3、几个数的公约数，都是这几个数的最大公约数的约数。 4、几个数都乘以一个自然数m，所得的积的最大公约数等于这几个数的最大公约数乘以m。例如：12的约数有1、2、3、4、6、12； 18的约数有：1、2、3、6、9、18；那么12和18的公约数有：1、2、3、6；那么12和18最大的公约数是：6，记作（12，18）=6；求最大公约数基本方法： 1、分解质因数法：先分解质因数，然后把相同的因数连乘起来。 2、短除法：先找公有的约数，然后相乘。 3、辗转相除法：每一次都用除数和余数相除，能够整除的那个余数，就是所求的最大公约数。公倍数：几个数公有的倍数，叫做这几个数的公倍数；其中最小的一个，叫做这几个数的最小公倍数。 12的倍数有：12、24、36、48……； 18的倍数有：18、36、54、72……；那么12和18的公倍数有：36、72、108……；那么12和18最小的公倍数是36，记作[12，18]=36；最小公倍数的性质： 1、两个数的任意公倍数都是它们最小公倍数的倍数。 2、两个数最大公约数与最小公倍数的乘积等于这两个数的乘积。求最小公倍数基本方法：1、短除法求最小公倍数；2、分解质因数的方法 17、数的整除：基本概念和符号： 1、整除：如果一个整数a，除以一个自然数b，得到一个整数商c，而且没有余数，那么叫做a能被b整除或b能整除a，记作b\|a。 2、常用符号：整除符号'\|'，不能整除符号' '；因为符号'∵'，所以的符号'∴'；整除判断方法： 1.能被2、5整除：末位上的数字能被2、5整除。 2.能被4、25整除：末两位的数字所组成的数能被4、25整除。 3.能被8、125整除：末三位的数字所组成的数能被8、125整除。 4.能被3、9整除：各个数位上数字的和能被3、9整除。 5.能被7整除： ①末三位上数字所组成的数与末三位以前的数字所组成数之差能被7整除。 ②逐次去掉最后一位数字并减去末位数字的2倍后能被7整除。 6.能被11整除： ①末三位上数字所组成的数与末三位以前的数字所组成的数之差能被11整除。 ②奇数位上的数字和与偶数位数的数字和的差能被11整除。 ③逐次去掉最后一位数字并减去末位数字后能被11整除。 7.能被13整除： ①末三位上数字所组成的数与末三位以前的数字所组成的数之差能被13整除。 ②逐次去掉最后一位数字并减去末位数字的9倍后能被13整除。整除的性质： 1.如果a、b能被c整除，那么（a+b）与（a-b）也能被c整除。 2.如果a能被b整除，c是整数，那么a乘以c也能被b整除。 3.如果a能被b整除，b又能被c整除，那么a也能被c整除。 4.如果a能被b、c整除，那么a也能被b和c的最小公倍数整除。 18、余数及其应用：基本概念：对任意自然数a、b、q、r，如果使得a÷b=q……r，且0<r<b,那么r叫做a除以b的余数，q叫做a除以b的不完全商。余数的性质： ①余数小于除数。 ②若a、b除以c的余数相同，则c\|a-b或c\|b-a。 ③a与b的和除以c的余数等于a除以c的余数加上b除以c的余数的和除以c的余数。 ④a与b的积除以c的余数等于a除以c的余数与b除以c的余数的积除以c的余数。 19、余数、同余与周期：同余的定义： ①若两个整数a、b除以m的余数相同，则称a、b对于模m同余。 ②已知三个整数a、b、m，如果m\|a-b，就称a、b对于模m同余，记作a≡b(mod m)，读作a同余于b模m。同余的性质： ①自身性：a≡a(mod m)； ②对称性：若a≡b(mod m)，则b≡a(mod m)； ③传递性：若a≡b(mod m)，b≡c(mod m)，则a≡ c(mod m)； ④和差性：若a≡b(mod m)，c≡d(mod m)，则a+c≡b+d(mod m)，a-c≡b-d(mod m)； ⑤相乘性：若a≡ b(mod m)，c≡d(mod m)，则a×c≡ b×d(mod m)； ⑥乘方性：若a≡b(mod m)，则an≡bn(mod m)； ⑦同倍性:若a≡ b(mod m)，整数c，则a×c≡ b×c(mod m×c)；关于乘方的预备知识： ①若A=a×b，则MA=Ma×b=（Ma）b ②若B=c+d则MB=Mc+d=Mc×Md 被3、9、11除后的余数特征： ①一个自然数M，n表示M的各个数位上数字的和，则M≡n(mod 9)或（mod 3）； ②一个自然数M，X表示M的各个奇数位上数字的和，Y表示M的各个偶数数位上数字的和，则M≡Y-X或M≡11-（X-Y）(mod 11)；费尔马小定理：如果p是质数（素数），a是自然数，且a不能被p整除，则ap-1≡1(mod p)。 20、分数与百分数的应用：基本概念与性质：分数：把单位'1'平均分成几份，表示这样的一份或几份的数。分数的性质：分数的分子和分母同时乘以或除以相同的数（0除外），分数的大小不变。分数单位：把单位'1'平均分成几份，表示这样一份的数。百分数：表示一个数是另一个数百分之几的数。常用方法： ①逆向思维方法：从题目提供条件的反方向（或结果）进行思考。 ②对应思维方法：找出题目中具体的量与它所占的率的直接对应关系。 ③转化思维方法：把一类应用题转化成另一类应用题进行解答。最常见的是转换成比例和转换成倍数关系；把不同的标准（在分数中一般指的是一倍量）下的分率转化成同一条件下的分率。常见的处理方法是确定不同的标准为一倍量。 ④假设思维方法：为了解题的方便，可以把题目中不相等的量假设成相等或者假设某种情况成立，计算出相应的结果，然后再进行调整，求出最后结果。 ⑤量不变思维方法：在变化的各个量当中，总有一个量是不变的，不论其他量如何变化，而这个量是始终固定不变的。有以下三种情况：A、分量发生变化，总量不变。B、总量发生变化，但其中有的分量不变。C、总量和分量都发生变化，但分量之间的差量不变化。 ⑥替换思维方法：用一种量代替另一种量，从而使数量关系单一化、量率关系明朗化。 ⑦同倍率法：总量和分量之间按照同分率变化的规律进行处理。 ⑧浓度配比法：一般应用于总量和分量都发生变化的状况。 21、分数大小的比较：基本方法： ①通分分子法：使所有分数的分子相同，根据同分子分数大小和分母的关系比较。 ②通分分母法：使所有分数的分母相同，根据同分母分数大小和分子的关系比较。 ③基准数法：确定一个标准，使所有的分数都和它进行比较。 ④分子和分母大小比较法：当分子和分母的差一定时，分子或分母越大的分数值越大。 ⑤倍率比较法：当比较两个分子或分母同时变化时分数的大小，除了运用以上方法外，可以用同倍率的变化关系比较分数的大小。（具体运用见同倍率变化规律） ⑥转化比较方法：把所有分数转化成小数（求出分数的值）后进行比较。 ⑦倍数比较法：用一个数除以另一个数，结果得数和1进行比较。 ⑧大小比较法：用一个分数减去另一个分数，得出的数和0比较。 ⑨倒数比较法：利用倒数比较大小，然后确定原数的大小。 ⑩基准数比较法：确定一个基准数，每一个数与基准数比较。 22、分数拆分：将一个分数单位分解成两个分数之和的公式： 23、完全平方数：完全平方数特征： 1.末位数字只能是：0、1、4、5、6、9；反之不成立。 2.除以3余0或余1；反之不成立。 3.除以4余0或余1；反之不成立。 4.约数个数为奇数；反之成立。 5.奇数的平方的十位数字为偶数；反之不成立。 6.奇数平方个位数字是奇数；偶数平方个位数字是偶数。 7.两个相临整数的平方之间不可能再有平方数。平方差公式： X2-Y2=（X-Y）（X+Y）完全平方和公式：（X+Y）2=X2+2XY+Y2 完全平方差公式：（X-Y）2=X2-2XY+Y2 24、比和比例：比：两个数相除又叫两个数的比。比号前面的数叫比的前项，比号后面的数叫比的后项。比值：比的前项除以后项的商，叫做比值。比的性质：比的前项和后项同时乘以或除以相同的数（零除外），比值不变。比例：表示两个比相等的式子叫做比例。a:b=c:d或比例的性质：两个外项积等于两个内项积(交叉相乘)，ad=bc。正比例：若A扩大或缩小几倍，B也扩大或缩小几倍（AB的商不变时），则A与B成正比。反比例：若A扩大或缩小几倍，B也缩小或扩大几倍（AB的积不变时），则A与B成反比。比例尺：图上距离与实际距离的比叫做比例尺。按比例分配：把几个数按一定比例分成几份，叫按比例分配。 25、综合行程：基本概念：行程问题是研究物体运动的，它研究的是物体速度、时间、路程三者之间的关系. 基本公式：路程=速度×时间；路程÷时间=速度；路程÷速度=时间关键问题：确定运动过程中的位置和方向。相遇问题：速度和×相遇时间=相遇路程（请写出其他公式）追及问题：追及时间＝路程差÷速度差（写出其他公式）流水问题：顺水行程=（船速+水速）×顺水时间逆水行程=（船速-水速）×逆水时间顺水速度=船速+水速逆水速度=船速-水速静水速度=（顺水速度+逆水速度）÷2 水速=（顺水速度-逆水速度）÷2 流水问题：关键是确定物体所运动的速度，参照以上公式。过桥问题：关键是确定物体所运动的路程，参照以上公式。主要方法：画线段图法基本题型：已知路程（相遇路程、追及路程）、时间（相遇时间、追及时间）、速度（速度和、速度差）中任意两个量，求第三个量。 26、工程问题：基本公式： ①工作总量=工作效率×工作时间 ②工作效率=工作总量÷工作时间 ③工作时间=工作总量÷工作效率基本思路： ①假设工作总量为'1'（和总工作量无关）； ②假设一个方便的数为工作总量（一般是它们完成工作总量所用时间的最小公倍数），利用上述三个基本关系，可以简单地表示出工作效率及工作时间. 关键问题：确定工作量、工作时间、工作效率间的两两对应关系。 27、逻辑推理：条件分析—假设法：假设可能情况中的一种成立，然后按照这个假设去判断，如果有与题设条件矛盾的情况，说明该假设情况是不成立的，那么与他的相反情况是成立的。例如，假设a是偶数成立，在判断过程中出现了矛盾，那么a一定是奇数。条件分析—列表法：当题设条件比较多，需要多次假设才能完成时，就需要进行列表来辅助分析。列表法就是把题设的条件全部表示在一个长方形表格中，表格的行、列分别表示不同的对象与情况，观察表格内的题设情况，运用逻辑规律进行判断。条件分析—图表法：当两个对象之间只有两种关系时，就可用连线表示两个对象之间的关系，有连线则表示'是，有'等肯定的状态，没有连线则表示否定的状态。例如A和B两人之间有认识或不认识两种状态，有连线表示认识，没有表示不认识。逻辑计算：在推理的过程中除了要进行条件分析的推理之外，还要进行相应的计算，根据计算的结果为推理提供一个新的判断筛选条件。简单归纳与推理：根据题目提供的特征和数据，分析其中存在的规律和方法，并从特殊情况推广到一般情况，并递推出相关的关系式，从而得到问题的解决。 28、几何面积：基本思路：在一些面积的计算上，不能直接运用公式的情况下，一般需要对图形进行割补，平移、旋转、翻折、分解、变形、重叠等，使不规则的图形变为规则的图形进行计算；另外需要掌握和记忆一些常规的面积规律。常用方法： 1.连辅助线方法 2.利用等底等高的两个三角形面积相等。 3.大胆假设（有些点的设置题目中说的是任意点，解题时可把任意点设置在特殊位置上）。 4.利用特殊规律 ①等腰直角三角形，已知任意一条边都可求出面积。（斜边的平方除以4等于等腰直角三角形的面积） ②梯形对角线连线后，两腰部分面积相等。 ③圆的面积占外接正方形面积的78.5%。 29、时钟问题—快慢表问题：基本思路： 1、按照行程问题中的思维方法解题； 2、不同的表当成速度不同的运动物体； 3、路程的单位是分格（表一周为60分格）； 4、时间是标准表所经过的时间； 5、合理利用行程问题中的比例关系； 30、时钟问题—钟面追及：基本思路：封闭曲线上的追及问题。关键问题： ①确定分针与时针的初始位置； ②确定分针与时针的路程差；基本方法： ①分格方法：时钟的钟面圆周被均匀分成60小格，每小格我们称为1分格。分针每小时走60分格，即一周；而时针只走5分格，故分针每分钟走1分格，时针每分钟走1／12分格。 ②度数方法：从角度观点看，钟面圆周一周是360°，分针每分钟转 360/60度，即6°，时针每分钟转360/12X60度，即1/2度。 31、浓度与配比：经验总结：在配比的过程中存在这样的一个反比例关系，进行混合的两种溶液的重量和他们浓度的变化成反比。溶质：溶解在其它物质里的物质（例如糖、盐、酒精等）叫溶质。溶剂：溶解其它物质的物质（例如水、汽油等）叫溶剂。溶液：溶质和溶剂混合成的液体（例如盐水、糖水等）叫溶液。基本公式：溶液重量=溶质重量+溶剂重量；溶质重量=溶液重量×浓度；浓度= 溶质/溶液×100%=溶质/（溶剂+溶质）×100% 经验总结：在配比的过程中存在这样的一个反比例关系，进行混合的两种溶液的重量和他们浓度的变化成反比。 32、经济问题：利润的百分数=（卖价-成本）÷成本×100%；卖价=成本×（1+利润的百分数）；成本=卖价÷（1+利润的百分数）；商品的定价按照期望的利润来确定；定价=成本×（1+期望利润的百分数）；本金：储蓄的金额；利率：利息和本金的比；利息=本金×利率×期数；含税价格=不含税价格×（1+增值税税率）； 33、不定方程：一次不定方程：含有两个未知数的一个方程，叫做二元一次方程，由于它的解不唯一，所以也叫做二元一次不定方程；常规方法：观察法、试验法、枚举法；多元不定方程：含有三个未知数的方程叫三元一次方程，它的解也不唯一；多元不定方程解法：根据已知条件确定一个未知数的值，或者消去一个未知数，这样就把三元一次方程变成二元一次不定方程，按照二元一次不定方程解即可；涉及知识点：列方程、数的整除、大小比较；解不定方程的步骤： 1、列方程；2、消元；3、写出表达式；4、确定范围；5、确定特征；6、确定答案；技巧总结： A、写出表达式的技巧：用特征不明显的未知数表示特征明显的未知数，同时考虑用范围小的未知数表示范围大的未知数； B、消元技巧：消掉范围大的未知数； 34、循环小数：把循环小数的小数部分化成分数的规则： ①纯循环小数小数部分化成分数：将一个循环节的数字组成的数作为分子，分母的各位都是9，9的个数与循环节的位数相同，最后能约分的再约分。 ②混循环小数小数部分化成分数：分子是第二个循环节以前的小数部分的数字组成的数与不循环部分的数字所组成的数之差，分母的头几位数字是9，9的个数与一个循环节的位数相同，末几位是0，0的个数与不循环部分的位数相同。分数转化成循环小数的判断方法： ①一个最简分数，如果分母中既含有质因数2和5，又含有2和5以外的质因数，那么这个分数化成的小数必定是混循环小数。 ②一个最简分数，如果分母中只含有2和5以外的质因数，那么这个分数化成的小数必定是纯循环小数。本站是提供个人知识管理的网络存储空间，所有内容均由用户发布，不代表本站观点。请注意甄别内容中的联系方式、诱导购买等信息，谨防诈骗。如发现有害或侵权内容，请点击一键举报。转藏分享 QQ空间QQ好友新浪微博微信献花（0） +1 来自：品茗堂>《数学》举报/认领上一篇：小学数学：36个奥数专题汇总，建议要掌握，教你超越80%的对手！下一篇： 6道简单的小学数学题，题题是深坑，大学生也能错一半猜你喜欢 0 条评论发表请遵守用户评论公约查看更多评论类似文章更多小学奥数很简单，就这30个知识点 max-width: 100%;box-sizing: border-box !important;">ortant;">1.ortant;">2ortant;">3ortant;">4ortant;">5ortant;">6ortant;">7... 小升初加油站\|2017小升初数学常见知识点 ②对称性：若a≡b(mod m)，则b≡a(mod m);③传递性：若a≡b(mod m)，b≡c(mod m)，则a≡ c(mod m);④和差性：若a≡b(mod m)，c≡d(mod m)，则a+c≡b+d(mod m)，a-c≡b-d(mod m);⑤相乘性：若a≡ b(mod ... 5-3-1约数与倍数_题库学生版如果为、的最大公约数，且，，那么互质，所以、的最小公倍数为，所以最大公约数与最小公倍数有如下一些基本关系：如果你写出12的所有约数，1和12除外，你会发现最大的约数是最小约数的3倍．现有一个整... 小学奥数必须掌握的30个知识模块汇总（详细版）已知条件几个数的和与差几个数的和与倍数几个数的差与倍数。基本公式：总份数＝（较大余数一较小余数）÷两次每份数的差。3、几个数的公约数，都是这几个数的最大公约数的约数。4、几个数都乘以一... 小学奥数知识5-4-3 约数与倍数（三）.学生版 (1)约数:在正整数范围内约数又叫因数,整数a能被整数b整除，a叫做b的倍数，b就叫做a的约数；【例 17】设A共有9个不同的约数，B共有6个不同的约数，C共有8个不同的约数，这三个数中的任何两个都不整除... 数论综合一个整数的约数个数与约数和的计算方法,两数的最大公约数与最小公倍数之间的关系,分数的最小公倍数.涉及一个整数的约数,以及若干整数最大公约数与最小公倍数的问题,其中分解质因数发挥着重要作用。【分... 2017年备战小升初：数学考点归纳分析最小公倍数：几个数公有的倍数，叫做这几个数的公倍数。3(或9)的倍数的特征：各个数位上的数之和是3(或9)的倍数。4(或25)的倍数的特征：末2位是4(或25)的倍数。8(或125)的倍数的特征：末3位是8(或125)... 最大公约数算法最大公约数算法1、查找约数法．先分别找出每个数的所有约数，再从两个数的约数中找出公有的约数，其中最大的一个就是最大公约数．例如，求 12 和 30 的最大公约数． 12 的约数有：1、2、3、4、6、... 164418e8tve6ytzq806tqx.doc 所以，两个数的最小公倍数，可由这两个数的乘积除以这两个数的最大公约数来求得。若要求三个或三个以上的数的最小公倍数，可以先求其中两个数的最小公倍数，再求这个最小公倍数与第三个数的最小公倍数... 个图VIP年卡，限时优惠价168元>>x 品茗堂关注对话 TA的最新馆藏美味家常下饭菜，米饭炫了一碗又一碗【新学期必备】科学课班级教学常规完全版趣谈小学科学教育小学科学课堂管理系列课程——第1课《小学科学课堂管理基础与规则建立》 [转]补肾化瘀通淋汤治疗慢性前列腺炎42例 [转]换了3次吸顶灯我才懂，这5款灯千万不能买！血的教训告诉你原因喜欢该文的人也喜欢更多混社会，秘而不宣的10个潜规则阅72 摄影：飞禽的世界-丹顶鹤（3）阅96 13岁少女的生死噩梦！2000年永吉县“3·7”恶性强奸案侦破始...阅217 六安猪头肉的商业制作配方阅80 北大最受欢迎的哲学课，你为什么会痛苦？阅40 热门阅读换一换「敏感资源」分享几个看片网站！都是精品网站哦阅883469 最全超高清的人体穴位图阅119822 基坑支护专项方案（专家论证修改版- 根据建办质〔2021〕48号编制...阅1722 最完美的借款合同（民间借贷个人最新版本）阅26371 统编版小学语文教材文言文编排简析阅8567 复制打印文章发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文分享文章QQ空间QQ好友新浪微博微信 AI解释复制打印文章 adsbygoogle.js 发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文 AI助手阅读时有疑惑？点击向AI助手提问吧联系客服在线客服： 360doc小助手2 客服QQ： 1732698931 联系电话：4000-999-276 客服工作时间9:00-18:00，晚上非工作时间，请在QQ留言，第二天客服上班后会立即联系您。
6274	https://mathworld.wolfram.com/Exponentiation.html	Exponentiation -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Algebra Algebraic Operations General Algebraic Operations Calculus and Analysis Special Functions Powers History and Terminology Terminology Exponentiation Exponentiation is the process of taking a quantity (the base) to the power of another quantity (the exponent). This operation most commonly denoted . In TeX, the Wolfram Language, and many other computer languages, exponentiation is denoted with a caret, i.e., as b^e. However, in FORTRAN, it is denoted be (Calderbank 1989, p.29). See also Antilogarithm, Asterisk, Base, Caret, Exponent, Logarithm, Power Explore with Wolfram\|Alpha More things to try: powers .142857... express 4.8675 through pi and e References Calderbank, V.J. Programming in FORTRAN, 3rd ed. London: Chapman and Hall, 1989. Referenced on Wolfram\|Alpha Exponentiation Cite this as: Weisstein, Eric W. "Exponentiation." From MathWorld--A Wolfram Resource. Subject classifications Algebra Algebraic Operations General Algebraic Operations Calculus and Analysis Special Functions Powers History and Terminology Terminology About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
6275	https://www.englishgrammar.org/adverbs-exercise-8/	Adverbs Exercise Home of English Grammar Grammar Guide Home Exercises Rules Test Yourself Tools Grammar Checker Word Counter Top Social Media Posts Writing Guides Contact You are here: Home/Adverbs/ Adverbs Exercise Adverbs Exercise September 27, 2020 - Fill in the blanks with an appropriate adverb. 1. She worked ………………….. and passed the examination. hard hardly Correct!Wrong! Hard is the adverb for this meaning. 2. We ………………….. see a lion. rarely scarcely hardly Correct!Wrong! Rarely means seldom. Scarcely and hardly have different meanings. 3. He narrated the incident in ……………………… detail details Correct!Wrong! In detail means thoroughly. 4. He often comes …………………… to school. late lately Correct!Wrong! Late is the opposite of early. Lately means recently. 5. I really feel ……………………… about it. badly bad Correct!Wrong! Badly means poorly. 6. He is …………………….. rich. very much so Correct!Wrong! 7. The journey was ………………….. uncomfortable. rather fairly fair Correct!Wrong! Fairly means reasonably or satisfactorily. It is not normally used with negative ideas. 8. The lecture was …………………. boring fairly rather Correct!Wrong! Fairly is not normally used with negative ideas. 9. Hardly ……………………. had any rest for weeks. have I I have Correct!Wrong! When a negative word comes at the beginning of the sentence, we use the inverted word order. 10. He is getting …………………….. day by day. well better best Correct!Wrong! We use the comparative form when an idea of progress is implied. 11. He plays hockey very …………………… well good Correct!Wrong! Good is an adjective; well is an adverb. 12. He is not ……………………. to go to school. enough old old enough Correct!Wrong! When enough is an adverb, it goes after the adjective or adverb modified by it. Play Again! Answers She worked hard and passed the examination. We rarely see a lion. He narrated the incident in detail. He often comes late to school. I really feel bad about it. He is very rich. The journey was rather uncomfortable. The lecture was rather boring Hardly have I had any rest for weeks. He is getting better day by day. He plays hockey very well. He is not old enough to go to school. Share Post Send Mail 2,485,429 Facebook 726,224 X (formerly Twitter) 2026 Grammar Guide "120 Mistakes You Must Avoid in 2026" Free Weekly Exercises & Infographics Download Now (PDF) » Grammar Checker Hint → Bookmark GrammarCheck for future use. Latest Exercises Modal Verbs ExerciseSeptember 20, 2025 Gap Filling Tenses Grammar ExerciseSeptember 14, 2025 Modal Verbs Exercise for ESL StudentsSeptember 13, 2025 Modal Auxiliary Verbs Exercise for ESL StudentsSeptember 11, 2025 ESL Grammar ExerciseSeptember 10, 2025 100 High-Impact Verbs for Action-TakersSeptember 9, 2025 Do vs. HaveSeptember 6, 2025 Gap Filling Grammar Exercise for BeginnersSeptember 5, 2025 Gap Filling Grammar Exercise for ESL StudentsSeptember 3, 2025 Copyright © 2025 · EnglishGrammar.org Disclaimer · Privacy Policy · Sitemap
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Δ Converting 0.3 as a Fraction Updated: August 18, 2023 \| Category:Advanced Math, Percentages Math can sometimes feel like a giant puzzle. Think of decimals like 0.3 and fractions as different pieces of the same puzzle. Today, we will learn how to turn the decimal 0.3 into a fraction. Stick with me, and we’ll make this super easy to understand! Step-By-Step Guide to Convert 0.3 as a Fraction Step 1: Write down the decimal & divide by 1. 0.3/1 Step 2: Multiply the numerator (top number) and the denominator (bottom number) by a power of 10, making the numerator a whole number. In the case of 0.3, we multiply by 10 because the decimal extends to the tenth place. 0.3×10/1×10=3/10 Step 3: Simplify the fraction if necessary. In this case, 3/10 is already the simplest, so no further simplification is needed. Therefore, 0.3 as a fraction is 3/10. 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6277	https://artofproblemsolving.com/articles/files/SatoNT.pdf?srsltid=AfmBOooebN9hMYy5wAX6yHMKOCZ3tH_IWjedJ2MauTwStENJkXLugX5w	Number Theory Naoki Sato sato@artofproblemsolving.com 0 Preface This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material that an IMO student should be familiar with. This text is meant to be a reference, and not a replacement but rather a supplement to a number theory textbook; several are given at the back. Proofs are given when appropriate, or when they illustrate some insight or important idea. The problems are culled from various sources, many from actual contests and olympiads, and in general are very difficult. The author welcomes any corrections or suggestions. 1 Divisibility For integers a and b, we say that a divides b, or that a is a divisor (or factor ) of b, or that b is a multiple of a, if there exists an integer c such that b = ca , and we denote this by a \| b. Otherwise, a does not divide b, and we denote this by a - b. A positive integer p is a prime if the only divisors of p are 1 and p. If pk \| a and pk+1 - a where p is a prime, i.e. pk is the highest power of p dividing a, then we denote this by pk‖a.Useful Facts • If a, b > 0, and a \| b, then a ≤ b. • If a \| b1, a \| b2, . . . , a \| bn, then for any integers c1, c2, . . . , cn, a \| n ∑ i=1 bici. Theorem 1.1 . The Division Algorithm . For any positive integer a and integer b, there exist unique integers q and r such that b = qa + r and 0 ≤ r < a , with r = 0 iff a \| b.1Theorem 1.2 . The Fundamental Theorem of Arithmetic . Every integer greater than 1 can be written uniquely in the form pe1 1 pe2 2 · · · pek k , where the pi are distinct primes and the ei are positive integers. Theorem 1.3 . (Euclid) There exist an infinite number of primes. Proof . Suppose that there are a finite number of primes, say p1, p2, . . . , pn. Let N = p1p2 · · · pn + 1. By the fundamental theorem of arithmetic, N is divisible by some prime p. This prime p must be among the pi, since by assumption these are all the primes, but N is seen not to be divisible by any of the pi, contradiction. Example 1.1 . Let x and y be integers. Prove that 2 x + 3 y is divisible by 17 iff 9 x + 5 y is divisible by 17. Solution . 17 \| (2 x + 3 y) ⇒ 17 \| [13(2 x + 3 y)], or 17 \| (26 x + 39 y) ⇒ 17 \| (9 x + 5 y), and conversely, 17 \| (9 x + 5 y) ⇒ 17 \| [4(9 x + 5 y)], or 17 \| (36 x + 20 y) ⇒ 17 \| (2 x + 3 y). Example 1.2 . Find all positive integers d such that d divides both n2 +1 and ( n + 1) 2 + 1 for some integer n. Solution . Let d \| (n2 + 1) and d \| [( n + 1) 2 + 1], or d \| (n2 + 2 n + 2). Then d \| [( n2 + 2 n + 2) − (n2 + 1)], or d \| (2 n + 1) ⇒ d \| (4 n2 + 4 n + 1), so d \| [4( n2+2 n+2) −(4 n2+4 n+1)], or d \| (4 n+7). Then d \| [(4 n+7) −2(2 n+1)], or d \| 5, so d can only be 1 or 5. Taking n = 2 shows that both of these values are achieved. Example 1.3 . Suppose that a1, a2, . . . , a2n are distinct integers such that the equation (x − a1)( x − a2) · · · (x − a2n) − (−1) n(n!) 2 = 0 has an integer solution r. Show that r = a1 + a2 + · · · + a2n 2n . (1984 IMO Short List) Solution . Clearly, r 6 = ai for all i, and the r − ai are 2 n distinct integers, so \|(r − a1)( r − a2) · · · (r − a2n)\| ≥ \| (1)(2) · · · (n)( −1)( −2) · · · (−n)\| = ( n!) 2, 2with equality iff {r − a1, r − a2, . . . , r − a2n} = {1, 2, . . . , n, −1, −2, . . . , −n}. Therefore, this must be the case, so (r − a1) + ( r − a2) + · · · + ( r − a2n)= 2 nr − (a1 + a2 + · · · + a2n)= 1 + 2 + · · · + n + ( −1) + ( −2) + · · · + ( −n) = 0 ⇒ r = a1 + a2 + · · · + a2n 2n . Example 1.4 . Let 0 < a 1 < a 2 < · · · < a mn +1 be mn + 1 integers. Prove that you can select either m + 1 of them no one of which divides any other, or n + 1 of them each dividing the following one. (1966 Putnam Mathematical Competition) Solution . For each i, 1 ≤ i ≤ mn + 1, let ni be the length of the longest sequence starting with ai and each dividing the following one, among the integers ai, ai+1 , . . . , amn +1 . If some ni is greater than n then the problem is solved. Otherwise, by the pigeonhole principle, there are at least m + 1 values of ni that are equal. Then, the integers ai corresponding to these ni cannot divide each other. Useful Facts • Bertrand’s Postulate . For every positive integer n, there exists a prime p such that n ≤ p ≤ 2n. • Gauss’s Lemma . If a polynomial with integer coefficients factors into two polynomials with rational coefficients, then it factors into two poly-nomials with integer coefficients. Problems 1. Let a and b be positive integers such that a \| b2, b2 \| a3, a3 \| b4, b4 \| a5,. . . . Prove that a = b.2. Let a, b, and c denote three distinct integers, and let P denote a poly-nomial having all integral coefficients. Show that it is impossible that P (a) = b, P (b) = c, and P (c) = a.(1974 USAMO) 33. Show that if a and b are positive integers, then ( a + 12 )n + ( b + 12 )n is an integer for only finitely many positive integers n.(A Problem Seminar , D.J. Newman) 4. For a positive integer n, let r(n) denote the sum of the remainders when n is divided by 1, 2, . . . , n respectively. Prove that r(k) = r(k − 1) for infinitely many positive integers k.(1981 K¨ ursch´ ak Competition) 5. Prove that for all positive integers n,0 < n ∑ k=1 g(k) k − 2n 3 < 23, where g(k) denotes the greatest odd divisor of k.(1973 Austrian Mathematics Olympiad) 6. Let d be a positive integer, and let S be the set of all positive integers of the form x2 + dy 2, where x and y are non-negative integers. (a) Prove that if a ∈ S and b ∈ S, then ab ∈ S.(b) Prove that if a ∈ S and p ∈ S, such that p is a prime and p \| a,then a/p ∈ S.(c) Assume that the equation x2 + dy 2 = p has a solution in non-negative integers x and y, where p is a given prime. Show that if d ≥ 2, then the solution is unique, and if d = 1, then there are exactly two solutions. 2 GCD and LCM The greatest common divisor of two positive integers a and b is the great-est positive integer that divides both a and b, which we denote by gcd( a, b ), and similarly, the lowest common multiple of a and b is the least positive 4integer that is a multiple of both a and b, which we denote by lcm( a, b ). We say that a and b are relatively prime if gcd( a, b ) = 1. For integers a1, a2,. . . , an, gcd( a1, a 2, . . . , a n) is the greatest positive integer that divides all of a1, a2, . . . , an, and lcm( a1, a 2, . . . , a n) is defined similarly. Useful Facts • For all a, b, gcd( a, b ) · lcm( a, b ) = ab . • For all a, b, and m, gcd( ma, mb ) = m gcd( a, b ) and lcm( ma, mb ) = mlcm( a, b ). • If d \| gcd( a, b ), then gcd (ad , bd ) = gcd( a, b ) d . In particular, if d = gcd( a, b ), then gcd( a/d, b/d ) = 1; that is, a/d and b/d are relatively prime. • If a \| bc and gcd( a, c ) = 1, then a \| b. • For positive integers a and b, if d is a positive integer such that d \| a, d \| b, and for any d′, d′ \| a and d′ \| b implies that d′ \| d, then d =gcd( a, b ). This is merely the assertion that any common divisor of a and b divides gcd( a, b ). • If a1a2 · · · an is a perfect kth power and the ai are pairwise relatively prime, then each ai is a perfect kth power. • Any two consecutive integers are relatively prime. Example 2.1 . Show that for any positive integer N , there exists a multiple of N that consists only of 1s and 0s. Furthermore, show that if N is relatively prime to 10, then there exists a multiple that consists only of 1s. Solution . Consider the N + 1 integers 1, 11, 111, . . . , 111...1 ( N + 1 1s). When divided by N , they leave N + 1 remainders. By the pigeonhole princi-ple, two of these remainders are equal, so the difference in the corresponding integers, an integer of the form 111...000, is divisible by N . If N is relatively prime to 10, then we may divide out all powers of 10, to obtain an integer of the form 111...1 that remains divisible by N .5Theorem 2.1 . For any positive integers a and b, there exist integers x and y such that ax + by = gcd( a, b ). Furthermore, as x and y vary over all integers, ax + by attains all multiples and only multiples of gcd( a, b ). Proof . Let S be the set of all integers of the form ax +by , and let d be the least positive element of S. By the division algorithm, there exist integers q and r such that a = qd + r, 0 ≤ r < d . Then r = a − qd = a − q(ax + by ) = (1 − qx )a − (qy )b, so r is also in S. But r < d , so r = 0 ⇒ d \| a, and similarly, d \| b, so d \| gcd( a, b ). However, gcd( a, b ) divides all elements of S,so in particular gcd( a, b ) \| d ⇒ d = gcd( a, b ). The second part of the theorem follows. Corollary 2.2 . The positive integers a and b are relatively prime iff there exist integers x and y such that ax + by = 1. Corollary 2.3 . For any positive integers a1, a2, . . . , an, there exist integers x1, x2, . . . , xn, such that a1x1+a2x2+· · · +anxn = gcd( a1, a 2, . . . , a n). Corollary 2.4 . Let a and b be positive integers, and let n be an integer. Then the equation ax + by = n has a solution in integers x and y iff gcd( a, b ) \| n. If this is the case, then all solutions are of the form (x, y ) = ( x0 + t · bd , y 0 − t · ad ) , where d = gcd( a, b ), ( x0, y 0) is a specific solution of ax + by = n, and t is an integer. Proof . The first part follows from Theorem 2.1. For the second part, as stated, let d = gcd( a, b ), and let ( x0, y 0) be a specific solution of ax + by = n,so that ax 0 + by 0 = n. If ax + by = n, then ax + by − ax 0 − by 0 = a(x − x0) + b(y − y0) = 0, or a(x − x0) = b(y0 − y), and hence (x − x0) · ad = ( y0 − y) · bd. Since a/d and b/d are relatively prime, b/d must divide x − x0, and a/d must divide y0 − y. Let x − x0 = tb/d and y0 − y = ta/d . This gives the solutions described above. 6Example 2.2 . Prove that the fraction 21 n + 4 14 n + 3 is irreducible for every positive integer n. (1959 IMO) Solution . For all n, 3(14 n + 3) − 2(21 n + 4) = 1, so the numerator and denominator are relatively prime. Example 2.3 . For all positive integers n, let Tn = 2 2n Show that if m 6 = n, then Tm and Tn are relatively prime. Solution . We have that Tn − 2 = 2 2n − 1 = 2 2n−1·2 − 1= ( Tn−1 − 1) 2 − 1 = T 2 n−1 − 2Tn−1 = Tn−1(Tn−1 − 2) = Tn−1Tn−2(Tn−2 − 2) = · · · = Tn−1Tn−2 · · · T1T0(T0 − 2) = Tn−1Tn−2 · · · T1T0, for all n. Therefore, any common divisor of Tm and Tn must divide 2. But each Tn is odd, so Tm and Tn are relatively prime. Remark . It immediately follows from this result that there are an infinite number of primes. The Euclidean Algorithm . By recursive use of the division algorithm, we may find the gcd of two positive integers a and b without factoring either, and the x and y in Theorem 2.1 (and so, a specific solution in Corollary 2.4). For example, for a = 329 and b = 182, we compute 329 = 1 · 182 + 147 , 182 = 1 · 147 + 35 , 147 = 4 · 35 + 7 , 35 = 5 · 7, and stop when there is no remainder. The last dividend is the gcd, so in our example, gcd(329,182) = 7. Now, working through the above equations 7backwards, 7 = 147 − 4 · 35 = 147 − 4 · (182 − 1 · 147) = 5 · 147 − 4 · 182 = 5 · (329 − 182) − 4 · 182 = 5 · 329 − 9 · 182 . Remark . The Euclidean algorithm also works for polynomials. Example 2.4 . Let n be a positive integer, and let S be a subset of n + 1 elements of the set {1, 2, . . . , 2n}. Show that (a) There exist two elements of S that are relatively prime, and (b) There exist two elements of S, one of which divides the other. Solution . (a) There must be two elements of S that are consecutive, and thus, relatively prime. (b) Consider the greatest odd factor of each of the n + 1 elements in S. Each is among the n odd integers 1, 3, . . . , 2 n − 1. By the pigeon-hole principle, two must have the same greatest odd factor, so they differ (multiplication-wise) by a power of 2, and so one divides the other. Example 2.5 . The positive integers a1, a2, . . . , an are such that each is less than 1000, and lcm( ai, a j ) > 1000 for all i, j, i 6 = j. Show that n ∑ i=1 1 ai < 2. (1951 Russian Mathematics Olympiad) Solution . If 1000 m+1 < a ≤ 1000 m , then the m multiples a, 2 a, . . . , ma do not exceed 1000. Let k1 the number of ai in the interval ( 1000 2 , 1000], k2 in (1000 3 , 1000 2 ], etc. Then there are k1 + 2 k2 + 3 k3 + · · · integers, no greater than 1000, that are multiples of at least one of the ai. But the multiples are distinct, so k1 + 2 k2 + 3 k3 + · · · < 1000 ⇒ 2k1 + 3 k2 + 4 k3 + · · · = ( k1 + 2 k2 + 3 k3 + · · · ) + ( k1 + k2 + k3 + · · · ) < 1000 + n< 2000 . 8Therefore, n ∑ i=1 1 ai ≤ k1 21000 + k2 31000 + k3 41000 + · · · = 2k1 + 3 k2 + 4 k3 + · · · 1000 < 2. Note: It can be shown that n ≤ 500 as follows: Consider the greatest odd divisor of a1, a2, . . . , a1000 . Each must be distinct; otherwise, two differ, multiplication-wise, by a power of 2, which means one divides the other, contradiction. Also, there are only 500 odd numbers between 1 and 1000, from which the result follows. It also then follows that n ∑ i=1 1 ai < 32. Useful Facts • Dirichlet’s Theorem . If a and b are relatively prime positive integers, then the arithmetic sequence a, a + b, a + 2 b, . . . , contains infinitely many primes. Problems 1. The symbols ( a, b, . . . , g ) and [ a, b, . . . , g ] denote the greatest common divisor and lowest common multiple, respectively of the positive inte-gers a, b, . . . , g. Prove that [a, b, c ]2 [a, b ][ a, c ][ b, c ] = (a, b, c )2 (a, b )( a, c )( b, c ). (1972 USAMO) 2. Show that gcd( am − 1, a n − 1) = agcd( m,n ) − 1 for all positive integers a > 1, m, n.93. Let a, b, and c be positive integers. Show that lcm( a, b, c ) = abc · gcd( a, b, c )gcd( a, b ) · gcd( a, c ) · gcd( b, c ). Express gcd( a, b, c ) in terms of abc , lcm( a, b, c ), lcm( a, b ), lcm( a, c ), and lcm( b, c ). Generalize. 4. Let a, b be odd positive integers. Define the sequence ( fn) by putting f1 = a, f2 = b, and by letting fn for n ≥ 3 be the greatest odd divisor of fn−1 + fn−2. Show that fn is constant for n sufficiently large and determine the eventual value as a function of a and b.(1993 USAMO) 5. Let n ≥ a1 > a 2 > · · · > a k be positive integers such that lcm( ai, a j ) ≤ n for all i, j. Prove that ia i ≤ n for i = 1, 2, . . . , k. 3 Arithmetic Functions There are several important arithmetic functions, of which three are pre-sented here. If the prime factorization of n > 1 is pe1 1 pe2 2 · · · pek k , then the number of positive integers less than n, relatively prime to n, is φ(n) = ( 1 − 1 p1 ) ( 1 − 1 p2 ) · · · ( 1 − 1 pk ) n = pe1−11 pe2−12 · · · pek −1 k (p1 − 1)( p2 − 1) · · · (pk − 1) , the number of divisors of n is τ (n) = ( e1 + 1)( e2 + 1) · · · (ek + 1) , and the sum of the divisors of n is σ(n) = ( pe1 1 pe1−11 + · · · + 1)( pe2 2 pe2−12 + · · · + 1) · · · (pek k pek −1 k · · · + 1) = (pe1+1 1 − 1 p1 − 1 ) ( pe2+1 2 − 1 p2 − 1 ) · · · (pek +1 k − 1 pk − 1 ) . Also, φ(1), τ (1), and σ(1) are defined to be 1. We say that a function f is multiplicative if f (mn ) = f (m)f (n) for all relatively prime positive 10 integers m and n, and f (1) = 1 (otherwise, f (1) = 0, which implies that f (n) = 0 for all n). Theorem 3.1 . The functions φ, τ , and σ are multiplicative. Hence, by taking the prime factorization and evaluating at each prime power, the formula above are found easily. Example 3.1 . Find the number of solutions in ordered pairs of positive integers ( x, y ) of the equation 1 x + 1 y = 1 n, where n is a positive integer. Solution . From the given, 1 x + 1 y = 1 n ⇔ xy = nx + ny ⇔ (x − n)( y − n) = n2. If n = 1, then we immediately deduce the unique solution (2,2). For n ≥ 2, let n = pe1 1 pe2 2 · · · pek k be the prime factorization of n. Since x, y > n ,there is a 1-1 correspondence between the solutions in ( x, y ) and the factors of n2, so the number of solutions is τ (n2) = (2 e1 + 1)(2 e2 + 1) · · · (2 ek + 1) . Example 3.2 . Let n be a positive integer. Prove that ∑ d\|n φ(d) = n. Solution . For a divisor d of n, let Sd be the set of all a, 1 ≤ a ≤ n, such that gcd( a, n ) = n/d . Then Sd consists of all elements of the form b · n/d ,where 0 ≤ b ≤ d, and gcd( b, d ) = 1, so Sd contains φ(d) elements. Also, it is clear that each integer between 1 and n belongs to a unique Sd. The result then follows from summing over all divisors d of n.Problems 1. Let n be a positive integer. Prove that n ∑ k=1 τ (k) = n ∑ k=1 ⌊nk ⌋ . 11 2. Let n be a positive integer. Prove that ∑ d\|n τ 3(d) = ∑ d\|n τ (d)  2 . Prove that if σ(N ) = 2 N + 1, then N is the square of an odd integer. (1976 Putnam Mathematical Competition) 4 Modular Arithmetic For a positive integer m and integers a and b, we say that a is congruent to b modulo m if m \| (a − b), and we denote this by a ≡ b modulo m, or more commonly a ≡ b (mod m). Otherwise, a is not congruent to b modulo m,and we denote this by a 6 ≡ b (mod m) (although this notation is not used often). In the above notation, m is called the modulus , and we consider the integers modulo m. Theorem 4.1 . If a ≡ b and c ≡ d (mod m), then a + c ≡ b + d (mod m)and ac ≡ bd (mod m). Proof . If a ≡ b and c ≡ d (mod m), then there exist integers k and l such that a = b + km and c = d + lm . Hence, a + c = b + d + ( k + l)m, so a + c ≡ b + d (mod m). Also, ac = bd + dkm + blm + klm 2 = bd + ( dk + bl + klm )m, so ac ≡ bd (mod m). Useful Facts • For all integers n, n2 ≡ { 01 } (mod 4) { if n is even , if n is odd . • For all integers n, n2 ≡  041  (mod 8)  if n ≡ 0 (mod 4) , if n ≡ 2 (mod 4) , if n ≡ 1 (mod 2) . 12 • If f is a polynomial with integer coefficients and a ≡ b (mod m), then f (a) ≡ f (b) (mod m). • If f is a polynomial with integer coefficients of degree n, not identically zero, and p is a prime, then the congruence f (x) ≡ 0 (mod p)has at most n solutions modulo p, counting multiplicity. Example 4.1 . Prove that the only solution in rational numbers of the equation x3 + 3 y3 + 9 z3 − 9xyz = 0 is x = y = z = 0. (1983 K¨ ursch´ ak Competition) Solution . Suppose that the equation has a solution in rationals, with at least one non-zero variable. Since the equation is homogeneous, we may obtain a solution in integers ( x0, y 0, z 0) by multiplying the equation by the cube of the lowest common multiple of the denominators. Taking the equa-tion modulo 3, we obtain x30 ≡ 0 (mod 3). Therefore, x0 must be divisible by 3, say x0 = 3 x1. Substituting, 27 x31 + 3 y30 + 9 z30 − 27 x1y0z0 = 0 ⇒ y30 + 3 z30 + 9 x31 − 9x1y0z0 = 0 . Therefore, another solution is ( y0, z 0, x 1). We may then apply this reduction recursively, to obtain y0 = 3 y1, z0 = 3 z1, and another solution ( x1, y 1, z 1). Hence, we may divide powers of 3 out of our integer solution an arbitrary number of times, contradiction. Example 4.2 . Does one of the first 10 8 + 1 Fibonacci numbers terminate with 4 zeroes? Solution . The answer is yes. Consider the sequence of pairs ( Fk, F k+1 )modulo 10 4. Since there are only a finite number of different possible pairs (10 8 to be exact), and each pair is dependent only on the previous one, this sequence is eventually periodic. Also, by the Fibonacci relation, one can find the previous pair to a given pair, so this sequence is immediately periodic. But F0 ≡ 0 (mod 10 4), so within 10 8 terms, another Fibonacci number divisible by 10 4 must appear. 13 In fact, a computer check shows that 10 4 \| F7500 , and ( Fn) modulo 10 4 has period 15000, which is much smaller than the upper bound of 10 8.If ax ≡ 1 (mod m), then we say that x is the inverse of a modulo m,denoted by a−1, and it is unique modulo m. Theorem 4.2 . The inverse of a modulo m exists and is unique iff a is relatively prime to m. Proof . If ax ≡ 1 (mod m), then ax = 1+ km for some k ⇒ ax −km = 1. By Corollary 2.2, a and m are relatively prime. Now, if gcd( a, m ) = 1, then by Corollary 2.2, there exist integers x and y such that ax + my = 1 ⇒ ax =1 − my ⇒ ax ≡ 1 (mod m). The inverse x is unique modulo m, since if x′ is also an inverse, then ax ≡ ax ′ ≡ 1 ⇒ xax ≡ xax ′ ≡ x ≡ x′. Corollary 4.3 . If p is a prime, then the inverse of a modulo p exists and is unique iff p does not divide a. Corollary 4.4 . If ak ≡ bk (mod m) and k is relatively prime to m, then a ≡ b (mod m). Proof . Multiplying both sides by k−1, which exists by Theorem 4.2, yields the result. We say that a set {a1, a 2, . . . , a m} is a complete residue system modulo m if for all i, 0 ≤ i ≤ m−1, there exists a unique j such that aj ≡ i (mod m). Example 4.3 . Find all positive integers n such that there exist complete residue systems {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} modulo n for which {a1 + b1, a 2 + b2, . . . , a n + bn} is also a complete residue system. Solution . The answer is all odd n. First we prove necessity. For any complete residue system {a1, a 2, . . . , a n} modulo n, we have that a1 + a2 + · · · + an ≡ n(n + 1) /2 (mod n). So, if all three sets are complete residue systems, then a1 +a2 +· · · +an +b1 +b2 +· · · +bn ≡ n2 +n ≡ 0 (mod n)and a1 + b1 + a2 + b2 + · · · + an + bn ≡ n(n + 1) /2 (mod n), so n(n + 1) /2 ≡ 0(mod n). The quantity n(n + 1) /2 is divisible by n iff ( n + 1) /2 is an integer, which implies that n is odd. Now assume that n is odd. Let ai = bi = i for all i. Then ai + bi = 2 i for all i, and n is relatively prime to 2, so by Corollary 4.4, {2, 4, . . . , 2n} is a complete residue system modulo n. Theorem 4.5 . Euler’s Theorem . If a is relatively prime to m, then aφ(m) ≡ 1 (mod m). 14 Proof . Let a1, a2, . . . , aφ(m) be the positive integers less than m that are relatively prime to m. Consider the integers aa 1, aa 2, . . . , aa φ(m). We claim that they are a permutation of the original φ(m) integers ai, modulo m. For each i, aa i is also relatively prime to m, so aa i ≡ ak for some k. Since aa i ≡ aa j ⇔ ai ≡ aj (mod m), each ai gets taken to a different ak under multiplication by a, so indeed they are permuted. Hence, a1a2 · · · aφ(m) ≡ (aa 1)( aa 2) · · · (aa φ(m)) ≡ aφ(m)a1a2 · · · aφ(m) ⇒ 1 ≡ aφ(m) (mod m). Remark . This gives an explicit formula for the inverse of a modulo m: a−1 ≡ aφ(m)−2 (mod m). Alternatively, one can use the Euclidean algorithm to find a−1 ≡ x as in the proof of Theorem 4.2. Corollary 4.6 . Fermat’s Little Theorem (FLT) . If p is a prime, and p does not divide a, then ap−1 ≡ 1 (mod p). Example 4.4 . Show that if a and b are relatively prime positive integers, then there exist integers m and n such that am + bn ≡ 1 (mod ab ). Solution . Let S = am + bn, where m = φ(b) and n = φ(a). Then by Euler’s Theorem, S ≡ bφ(a) ≡ 1 (mod a), or S − 1 ≡ 0 (mod a), and S ≡ aφ(b) ≡ 1 (mod b), or S − 1 ≡ 0 (mod b). Therefore, S − 1 ≡ 0, or S ≡ 1(mod ab ). Example 4.5 . For all positive integers i, let Si be the sum of the products of 1, 2, . . . , p − 1 taken i at a time, where p is an odd prime. Show that S1 ≡ S2 ≡ · · · ≡ Sp−2 ≡ 0 (mod p). Solution . First, observe that (x − 1)( x − 2) · · · (x − (p − 1)) = xp−1 − S1xp−2 + S2xp−3 − · · · − Sp−2x + Sp−1. This polynomial vanishes for x = 1, 2, . . . , p − 1. But by Fermat’s Little Theorem, so does xp−1 − 1 modulo p. Taking the difference of these two polynomials, we obtain another polynomial of degree p − 2 with p − 1 roots modulo p, so it must be the zero polynomial, and the result follows from comparing coefficients. 15 Remark . We immediately have that ( p − 1)! ≡ Sp−1 ≡ − 1 (mod p), which is Wilson’s Theorem. Also, xp − x ≡ 0 (mod p) for all x, yet we cannot compare coefficients here. Why not? Theorem 4.7 . If p is a prime and n is an integer such that p \| (4 n2 + 1), then p ≡ 1 (mod 4). Proof . Clearly, p cannot be 2, so we need only show that p 6 ≡ 3 (mod 4). Suppose p = 4 k + 3 for some k. Let y = 2 n, so by Fermat’s Little Theorem, yp−1 ≡ 1 (mod p), since p does not divide n. But, y2 + 1 ≡ 0, so yp−1 ≡ y4k+2 ≡ (y2)2k+1 ≡ (−1) 2k+1 ≡ − 1 (mod p), contradiction. Therefore, p ≡ 1 (mod 4). Remark . The same proof can be used to show that if p is a prime and p \| (n2 + 1), then p = 2 or p ≡ 1 (mod 4). Example 4.6 . Show that there are an infinite number of primes of the form 4 k + 1 and of the form 4 k + 3. Solution . Suppose that there are a finite number of primes of the form 4k + 1, say p1, p2, . . . , pn. Let N = 4( p1p2 · · · pn)2 + 1. By Theorem 4.7, N is only divisible by primes of the form 4 k + 1, but clearly N is not divisible by any of these primes, contradiction. Similarly, suppose that there are a finite number of primes of the form 4k + 3, say q1, q2, . . . , qm. Let M = 4 q1q2 · · · qm − 1. Then M ≡ 3 (mod 4), so M must be divisible by a prime of the form 4 k + 3, but M is not divisible by any of these primes, contradiction. Example 4.7 . Show that if n is an integer greater than 1, then n does not divide 2 n − 1. Solution . Let p be the least prime divisor of n. Then gcd( n, p − 1) = 1, and by Corollary 2.2, there exist integers x and y such that nx +( p−1) y = 1. If p \| (2 n − 1), then 2 ≡ 2nx +( p−1) y ≡ (2 n)x(2 p−1)y ≡ 1 (mod p) by Fermat’s Little Theorem, contradiction. Therefore, p - (2 n − 1) ⇒ n - (2 n − 1). Theorem 4.8 . Wilson’s Theorem . If p is a prime, then ( p − 1)! ≡ − 1(mod p). (See also Example 4.5.) Proof . Consider the congruence x2 ≡ 1 (mod p). Then x2 − 1 ≡ (x − 1)( x + 1) ≡ 0, so the only solutions are x ≡ 1 and −1. Therefore, for each i,2 ≤ i ≤ p − 2, there exists a unique inverse j 6 = i of i, 2 ≤ j ≤ p − 2, modulo 16 p. Hence, when we group in pairs of inverses, (p − 1)! ≡ 1 · 2 · · · (p − 2) · (p − 1) ≡ 1 · 1 · · · 1 · (p − 1) ≡ − 1 (mod p). Example 4.8 . Let {a1, a 2, . . . , a 101 } and {b1, b 2, . . . , b 101 } be complete residue systems modulo 101. Can {a1b1, a 2b2, . . . , a 101 b101 } be a complete residue system modulo 101? Solution . The answer is no. Suppose that {a1b1, a 2b2, . . . , a 101 b101 } is a complete residue system modulo 101. Without loss of generality, assume that a101 ≡ 0 (mod 101). Then b101 ≡ 0 (mod 101), because if any other bj was congruent to 0 modulo 101, then aj bj ≡ a101 b101 ≡ 0 (mod 101), contradiction. By Wilson’s Theorem, a1a2 · · · a100 ≡ b1b2 · · · b100 ≡ 100! ≡−1 (mod 101), so a1b1a2b2 · · · a100 b100 ≡ 1 (mod 101). But a101 b101 ≡ 0(mod 101), so a1b1a2b2 · · · a100 b100 ≡ 100! ≡ − 1 (mod 101), contradiction. Theorem 4.9 . If p is a prime, then the congruence x2 + 1 ≡ 0 (mod p)has a solution iff p = 2 or p ≡ 1 (mod 4). (Compare to Theorem 7.1) Proof . If p = 2, then x = 1 is a solution. If p ≡ 3 (mod 4), then by the remark to Theorem 4.7, no solutions exist. Finally, if p = 4 k + 1, then let x = 1 · 2 · · · (2 k). Then x2 ≡ 1 · 2 · · · (2 k) · (2 k) · · · 2 · 1 ≡ 1 · 2 · · · (2 k) · (−2k) · · · (−2) · (−1) (multiplying by 2 k −1s) ≡ 1 · 2 · · · (2 k) · (p − 2k) · · · (p − 2) · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p). Theorem 4.10 . Let p be a prime such that p ≡ 1 (mod 4). Then there exist positive integers x and y such that p = x2 + y2. Proof . By Theorem 4.9, there exists an integer a such that a2 ≡ − 1(mod p). Consider the set of integers of the form ax − y, where x and y are integers, 0 ≤ x, y < √p. The number of possible pairs ( x, y ) is then (b√pc + 1) 2 > (√p)2 = p, so by pigeonhole principle, there exist integers 0 ≤ x1, x 2, y 1, y 2 < √p, such that ax 1−y1 ≡ ax 2−y2 (mod p). Let x = x1−x2 and y = y1 − y2. At least one of x and y is non-zero, and ax ≡ y ⇒ a2x2 ≡ 17 −x2 ≡ y2 ⇒ x2 + y2 ≡ 0 (mod p). Thus, x2 + y2 is a multiple of p, and 0 < x 2 + y2 < (√p)2 + ( √p)2 = 2 p, so x2 + y2 = p. Theorem 4.11 . Let n be a positive integer. Then there exist integers x and y such that n = x2 + y2 iff each prime factor of n of the form 4 k + 3 appears an even number of times. Theorem 4.12 . The Chinese Remainder Theorem (CRT) . If a1, a2, . . . , ak are integers, and m1, m2, . . . , mk are pairwise relatively prime integers, then the system of congruences x ≡ a1 (mod m1),x ≡ a2 (mod m2), ... x ≡ ak (mod mk)has a unique solution modulo m1m2 · · · mk. Proof . Let m = m1m2 · · · mk, and consider m/m 1. This is relatively prime to m1, so there exists an integer t1 such that t1 · m/m 1 ≡ 1 (mod m1). Accordingly, let s1 = t1 · m/m 1. Then s1 ≡ 1 (mod m1) and s1 ≡ 0(mod mj ), j 6 = 1. Similarly, for all i, there exists an si such that si ≡ 1(mod mi) and si ≡ 0 (mod mj ), j 6 = i. Then, x = a1s1 + a2s2 + · · · + aksk is a solution to the above system. To see uniqueness, let x′ be another solution. Then x − x′ ≡ 0 (mod mi) for all i ⇒ x − x′ ≡ 0 (mod m1m2 · · · mk). Remark . The proof shows explicitly how to find the solution x. Example 4.9 . For a positive integer n, find the number of solutions of the congruence x2 ≡ 1 (mod n). Solution . Let the prime factorization of n be 2 epe1 1 pe2 2 · · · pek k . By CRT, x2 ≡ 1 (mod n) ⇔ x2 ≡ 1 (mod pei i ) for all i, and x2 ≡ 1 (mod 2 e). We consider these cases separately. We have that x2 ≡ 1 (mod pei i ) ⇔ x2 − 1 = ( x − 1)( x + 1) ≡ 0 (mod pei i ). But pi cannot divide both x − 1 and x + 1, so it divides one of them; that is, x ≡ ± 1 (mod pei i ). Hence, there are two solutions. Now, if ( x − 1)( x + 1) ≡ 0 (mod 2 e), 2 can divide both x − 1 and x + 1, but 4 cannot divide both. For e = 1 and e = 2, it is easily checked that there are 1 and 2 solutions respectively. For e ≥ 3, since there is at most one factor 18 of 2 in one of x − 1 and x + 1, there must be at least e − 1 in the other, for their product to be divisible by 2 e. Hence, the only possibilities are x − 1 or x + 1 ≡ 0, 2 e−1 (mod 2 e), which lead to the four solutions x ≡ 1, 2 e−1 − 1, 2e−1 + 1, and 2 e − 1. Now that we know how many solutions each prime power factor con-tributes, the number of solutions modulo n is simply the product of these, by CRT. The following table gives the answer: e Number of solutions 0, 1 2k 2 2k+1 ≥ 3 2k+2 Theorem 4.11 . Let m be a positive integer, let a and b be integers, and let k = gcd( a, m ). Then the congruence ax ≡ b (mod m) has k solutions or no solutions according as k \| b or k - b.Problems 1. Prove that for each positive integer n there exist n consecutive positive integers, none of which is an integral power of a prime. (1989 IMO) 2. For an odd positive integer n > 1, let S be the set of integers x,1 ≤ x ≤ n, such that both x and x + 1 are relatively prime to n. Show that ∏ x∈S x ≡ 1 (mod n). Find all positive integer solutions to 3 x + 4 y = 5 z .(1991 IMO Short List) 4. Let n be a positive integer such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is divisible by 24. (1969 Putnam Mathematical Competition) 5. (Wolstenholme’s Theorem) Prove that if 1 + 12 + 13 + · · · + 1 p − 119 is expressed as a fraction, where p ≥ 5 is a prime, then p2 divides the numerator. 6. Let a be the greatest positive root of the equation x3 − 3x2 + 1 = 0. Show that ba1788 c and ba1988 c are both divisible by 17. (1988 IMO Short List) 7. Let {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} be complete residue systems modulo n, such that {a1b1, a 2b2, . . . , a nbn} is also a complete residue system modulo n. Show that n = 1 or 2. 8. Let m, n be positive integers. Show that 4 mn − m − n can never be a square. (1984 IMO Proposal) 5 Binomial Coefficients For non-negative integers n and k, k ≤ n, the binomial coefficient (nk ) is defined as n! k!( n − k)! , and has several important properties. By convention, (nk ) = 0 if k > n .In the following results, for polynomials f and g with integer coefficients, we say that f ≡ g (mod m) if m divides every coefficient in f − g. Theorem 5.1 . If p is a prime, then the number of factors of p in n! is ⌊np ⌋ + ⌊ np2 ⌋ + ⌊ np3 ⌋ · · · . It is also n − sn p − 1 , where sn is the sum of the digits of n when expressed in base p. Theorem 5.2 . If p is a prime, then (pi ) ≡ 0 (mod p)20 for 1 ≤ i ≤ p − 1. Corollary 5.3 . (1 + x)p ≡ 1 + xp (mod p). Lemma 5.4 . For all real numbers x and y, bx + yc ≥ b xc + byc. Proof . x ≥ b xc ⇒ x + y ≥ b xc + byc ∈ Z, so bx + yc ≥ b xc + byc. Theorem 5.5 . If p is a prime, then (pk i ) ≡ 0 (mod p)for 1 ≤ i ≤ pk − 1. Proof . By Lemma 5.4, k ∑ j=1 (⌊ ipj ⌋ + ⌊pk − ipj ⌋) ≤ k ∑ j=1 ⌊pk pj ⌋ , where the LHS and RHS are the number of factors of p in i!( pk − i)! and pk! respectively. But, ⌊ ipk ⌋ = ⌊pk −ipk ⌋ = 0 and ⌊pk pk ⌋ = 1, so the inequality is strict, and at least one factor of p divides (pk i ). Corollary 5.6 . (1 + x)pk ≡ 1 + xpk (mod p). Example 5.1 . Let n be a positive integer. Show that the product of n consecutive positive integers is divisible by n!. Solution . If the consecutive integers are m, m + 1, . . . , m + n − 1, then m(m + 1) · · · (m + n − 1) n! = (m + n − 1 n ) . Example 5.2 . Let n be a positive integer. Show that (n + 1) lcm (( n 0 ) , (n 1 ) , . . . , (nn )) = lcm(1 , 2, . . . , n + 1) . (AMM E2686) Solution . Let p be a prime ≤ n + 1 and let α (respectively β) be the highest power of p in the LHS (respectively RHS) of the above equality. Choose r so that pr ≤ n + 1 < p r+1 . Then clearly β = r. We claim that if pr ≤ m < p r+1 , then pr+1 - (mk ) for 0 ≤ k ≤ m. (∗)21 Indeed, the number of factors of p in (mk ) is γ = r ∑ s=1 (⌊ mps ⌋ − ⌊ kps ⌋ − ⌊m − kps ⌋) . Since each summand in this sum is 0 or 1, we have γ ≤ r; that is, () holds. For 0 ≤ k ≤ n, let ak = ( n + 1) (nk ) = ( n − k + 1) (n + 1 k ) = ( k + 1) (n + 1 k + 1 ) . By ( ∗), pr+1 does not divide any of the integers (nk ), (n+1 k ), or (n+1 k+1 ). Thus, pr+1 can divide ak only if p divides each of the integers n + 1, n − k + 1, and k + 1. This implies that p divides ( n + 1) − (n − k + 1) − (k + 1) = −1, contradiction. Therefore, pr+1 - ak. On the other hand, for k = pr − 1, we have that k ≤ n and ak = ( k + 1) (n+1 k+1 ) is divisible by pr. Therefore, β = r = α. Theorem 5.7 . Lucas’s Theorem . Let m and n be non-negative integers, and p a prime. Let m = mkpk + mk−1pk−1 + · · · + m1p + m0, and n = nkpk + nk−1pk−1 + · · · + n1p + n0 be the base p expansions of m and n respectively. Then (mn ) ≡ (mk nk )( mk−1 nk−1 ) · · · (m1 n1 )( m0 n0 ) (mod p). Proof . By Corollary 5.6, (1 + x)m ≡ (1 + x)mk pk +mk−1pk−1+··· +m1p+m0 ≡ (1 + x)pk mk (1 + x)pk−1mk−1 · · · (1 + x)pm 1 (1 + x)m0 ≡ (1 + xpk )mk (1 + xpk−1 )mk−1 · · · (1 + xp)m1 (1 + x)m0 (mod p). By base p expansion, the coefficient of xn on both sides is (mn ) ≡ (mk nk )( mk−1 nk−1 ) · · · (m1 n1 )( m0 n0 ) (mod p). 22 Corollary 5.8 . Let n be a positive integer. Let A(n) denote the number of factors of 2 in n!, and let B(n) denote the number of 1s in the binary expansion of n. Then the number of odd entries in the nth row of Pascal’s Triangle, or equivalently the number of odd coefficients in the expansion of (1 + x)n, is 2 B(n). Furthermore, A(n) + B(n) = n for all n.Useful Facts • For a polynomial f with integer coefficients and prime p,[f (x)] pn ≡ f (xpn ) (mod p). Problems 1. Let a and b be non-negative integers, and p a prime. Show that (pa pb ) ≡ (ab ) (mod p). Let an be the last non-zero digit in the decimal representation of the number n!. Is the sequence a1, a2, a3, . . . eventually periodic? (1991 IMO Short List) 3. Find all positive integers n such that 2 n \| (3 n − 1). 4. Find the greatest integer k for which 1991 k divides 1990 1991 1992 1992 1991 1990 . (1991 IMO Short List) 5. For a positive integer n, let a(n) and b(n) denote the number of binomial coefficients in the nth row of Pascal’s Triangle that are congruent to 1 and 2 modulo 3 respectively. Prove that a(n) − b(n) is always a power of 2. 6. Let n be a positive integer. Prove that if the number of factors of 2 in n! is n − 1, then n is a power of 2. 23 7. For a positive integer n, let Cn = 1 n + 1 (2nn ) , and Sn = C1 + C2 + · · · + Cn.Prove that Sn ≡ 1 (mod 3) if and only if there exists a 2 in the base 3 expansion of n + 1. 6 Order of an Element We know that if a is relatively prime to m, then there exists a positive integer n such that an ≡ 1 (mod m). Let d be the smallest such n. Then we say that d is the order of a modulo m, denoted by ord m(a), or simply ord( a) if the modulus m is understood. Theorem 6.1 . If a is relatively prime to m, then an ≡ 1 (mod m) iff ord( a) \| n. Furthermore, an0 ≡ an1 iff ord( a) \| (n0 − n1). Proof . Let d = ord( a). It is clear that d \| n ⇒ an ≡ 1 (mod m). On the other hand, if an ≡ 1 (mod m), then by the division algorithm, there exist integers q and r such that n = qd + r, 0 ≤ r < d . Then an ≡ aqd +r ≡ (ad)qar ≡ ar ≡ 1 (mod m). But r < d , so r = 0 ⇒ d \| n. The second part of the theorem follows. Remark . In particular, by Euler’s Theorem, ord( a) \| φ(m). Example 6.1 . Show that the order of 2 modulo 101 is 100. Solution . Let d = ord(2). Then d \| φ(101), or d \| 100. If d < 100, then d divides 100/2 or 100/5; that is, d is missing at least one prime factor. However, 250 ≡ 1024 5 ≡ 14 5 ≡ 196 · 196 · 14 ≡ (−6) · (−6) · 14 ≡ − 1 (mod 101) , and 220 ≡ 1024 2 ≡ 14 2 ≡ − 6 (mod 101) , so d = 100. Example 6.2 . Prove that if p is a prime, then every prime divisor of 2p − 1 is greater than p.24 Solution . Let q \| (2 p − 1), where q is a prime. Then 2 p ≡ 1 (mod q), so ord(2) \| p. But ord(2) 6 = 1, so ord(2) = p. And by Fermat’s Little Theorem, ord(2) \| (q − 1) ⇒ p ≤ q − 1 ⇒ q > p .In fact, for p > 2, q must be of the form 2 kp + 1. From the above, ord(2) \| (q − 1), or p \| (q − 1) ⇒ q = mp + 1. Since q must be odd, m must be even. Example 6.3 . Let p be a prime that is relatively prime to 10, and let n be an integer, 0 < n < p . Let d be the order of 10 modulo p.(a) Show that the length of the period of the decimal expansion of n/p is d.(b) Prove that if d is even, then the period of the decimal expansion of n/p can be divided into two halves, whose sum is 10 d/ 2 − 1. For example, 1/7 = 0 .142857, so d = 6, and 142 + 857 = 999 = 10 3 − 1. Solution . (a) Let m be the length of the period, and let n/p =0.a 1a2 . . . a m. Then 10 mnp = a1a2 . . . a m.a 1a2 . . . a m ⇒ (10 m − 1) np = a1a2 . . . a m, an integer. Since n and p are relatively prime, p must divide 10 m − 1, so d divides m. Conversely, p divides 10 d −1, so (10 d −1) n/p is an integer, with at most d digits. If we divide this integer by 10 d − 1, then we obtain a rational number, whose decimal expansion has period at most d. Therefore, m = d.(b) Let d = 2 k, so n/p = 0 .a 1a2 . . . a kak+1 . . . a 2k. Now p divides 10 d −1 = 10 2k − 1 = (10 k − 1)(10 k + 1). However, p cannot divide 10 k − 1 (since the order of 10 is 2 k), so p divides 10 k + 1. Hence, 10 knp = a1a2 . . . a k.a k+1 . . . a 2k ⇒ (10 k + 1) np = a1a2 . . . a k + 0 .a 1a2 . . . a k + 0 .a k+1 . . . a 2k is an integer. This can occur iff a1a2 . . . a k +ak+1 . . . a 2k is a number consisting only of 9s, and hence, equal to 10 k − 1. Problems 25 1. Prove that for all positive integers a > 1 and n, n \| φ(an − 1). 2. Prove that if p is a prime, then pp−1 has a prime factor that is congruent to 1 modulo p.3. For any integer a, set na = 101 a−100 ·2a. Show that for 0 ≤ a, b, c, d ≤ 99 , n a + nb ≡ nc + nd (mod 10100) implies {a, b } = {c, d }.(1994 Putnam Mathematical Competition) 4. Show that if 3 ≤ d ≤ 2n+1 , then d - (a2n 1) for all positive integers a. 7 Quadratic Residues Let m be an integer greater than 1, and a an integer relatively prime to m. If x2 ≡ a (mod m) has a solution, then we say that a is a quadratic residue of m. Otherwise, we say that a is a quadratic non-residue . Now, let p be an odd prime. Then the Legendre symbol (ap ) is assigned the value of 1 if a is a quadratic residue of p. Otherwise, it is assigned the value of −1. Theorem 7.1 . Let p be an odd prime, and a and b be integers relatively prime to p. Then (a) (ap ) ≡ a(p−1) /2 (mod p), and (b) (ap )( bp ) = (ab p ) . Proof . If the congruence x2 ≡ a (mod p) has a solution, then a(p−1) /2 ≡ xp−1 ≡ 1 (mod p), by Fermat’s Little Theorem. If the congruence x2 ≡ a (mod p) has no solutions, then for each i, 1 ≤ i ≤ p − 1, there is a unique j 6 = i, 1 ≤ j ≤ p − 1, such that ij ≡ a. Therefore, all the integers from 1 to p − 1 can be arranged into ( p − 1) /2 such pairs. Taking their product, a(p−1) /2 ≡ 1 · 2 · · · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p), 26 by Wilson’s Theorem. Part (b) now follows from part (a). Remark . Part (a) is known as Euler’s criterion. Example 7.1 . Show that if p is an odd prime, then (1 p ) + (2 p ) · · · + (p − 1 p ) = 0 . Solution . Note that 1 2, 2 2, . . . , (( p − 1) /2) 2 are distinct modulo p,and that (( p + 1) /2) 2, . . . , ( p − 1) 2 represent the same residues, simply in reverse. Hence, there are exactly ( p−1) /2 quadratic residues, leaving ( p−1) /2quadratic non-residues. Therefore, the given sum contains ( p − 1) /2 1s and (p − 1) /2 −1s. Theorem 7.2 . Gauss’s Lemma . Let p be an odd prime and let a be relatively prime to p. Consider the least non-negative residues of a, 2 a, . . . , (( p − 1) /2) a modulo p. If n is the number of these residues that are greater than p/ 2, then (ap ) = ( −1) n. Theorem 7.3 . If p is an odd prime, then (−1 p ) = ( −1) (p−1) /2; that is, (−1 p ) = { 1 if p ≡ 1 (mod 4) , −1 if p ≡ 3 (mod 4) . Proof . This follows from Theorem 4.9 (and Theorem 7.1). Theorem 7.4 . If p is an odd prime, then (2 p ) = ( −1) (p2−1) /8; that is, (2 p ) = { 1 if p ≡ 1 or 7 (mod 8) , −1 if p ≡ 3 or 5 (mod 8) . 27 Proof . If p ≡ 1 or 5 (mod 8), then 2(p−1) /2 (p − 12 ) ! ≡ 2 · 4 · 6 · · · (p − 1) ≡ 2 · 4 · 6 · · · (p − 12 ) · ( −p − 32 ) · · · (−5) · (−3) · (−1) ≡ (−1) (p−1) /4 (p − 12 ) ! ⇒ 2(p−1) /2 ≡ (−1) (p−1) /4 (mod p). By Theorem 7.1, (2 p ) = ( −1) (p−1) /4. Hence, (2 p ) = 1 or −1 according as p ≡ 1 or 5 (mod 8). Similarly, if p ≡ 3 or 7 (mod 8), then 2(p−1) /2 (p − 12 ) ! ≡ 2 · 4 · 6 · · · (p − 32 ) · ( −p − 12 ) · · · (−5) · (−3) · (−1) ≡ (−1) (p+1) /4 (p − 12 ) ! ⇒ 2(p−1) /2 ≡ (−1) (p+1) /4 (mod p). Hence, (2 p ) = 1 or −1 according as p ≡ 7 or 3 (mod 8). Example 7.2 . Prove that if n is an odd positive integer, then every prime divisor of 2 n − 1 is of the form 8 k ± 1. (Compare to Example 6.2) Solution . Let p \| (2 n − 1), where p is prime. Let n = 2 m + 1. Then 2n ≡ 22m+1 ≡ 2(2 m)2 ≡ 1 (mod p) ⇒ (2 p ) = 1 ⇒ p is of the form 8 k ± 1. Theorem 7.5 . The Law of Quadratic Reciprocity . For distinct odd primes p and q, (pq ) ( qp ) = ( −1) p−12 · q−12 . Example 7.3 . For which primes p > 3 does the congruence x2 ≡ − 3(mod p) have a solution? Solution . We seek p for which (−3 p ) = (−1 p ) ( 3 p ) = 1. By quadratic reciprocity, (3 p ) ( p 3 ) = ( −1) (p−1) /2 = (−1 p ) , 28 by Theorem 7.3. Thus, in general, (−3 p ) = (−1 p ) ( 3 p ) = (p 3 ) ( −1 p )2 = (p 3 ) . And, ( p 3 ) = 1 iff p ≡ 1 (mod 3). Since p 6 ≡ 4 (mod 6), we have that x2 ≡ − 3(mod p) has a solution iff p ≡ 1 (mod 6). Example 7.4 . Show that if p = 2 n + 1, n ≥ 2, is prime, then 3 (p−1) /2 + 1 is divisible by p. Solution . We must have that n is even, say 2 k, for otherwise p ≡ 0(mod 3). By Theorem 7.1, (3 p ) ≡ 3(p−1) /2 (mod p). However, p ≡ 1 (mod 4), and p ≡ 4k + 1 ≡ 2 (mod 3) ⇒ (p 3 ) = −1, and by quadratic reciprocity, (3 p ) ( p 3 ) = ( −1) (p−1) /2 = 1 , so (3 p ) = −1 ⇒ 3(p−1) /2 + 1 ≡ 0 (mod p). Useful Facts • (a) If p is a prime and p ≡ 1 or 3 (mod 8), then there exist positive integers x and y such that p = x2 + 2 y2.(b) If p is a prime and p ≡ 1 (mod 6), then there exist positive integers x and y such that p = x2 + 3 y2.Problems 1. Show that if p > 3 is a prime, then the sum of the quadratic residues among the integers 1, 2, . . . , p − 1 is divisible by p.2. Let Fn denote the nth Fibonacci number. Prove that if p > 5 is a prime, then Fp ≡ (p 5 ) (mod p). 29 3. Show that 16 is a perfect 8 th power modulo p for any prime p.4. Let a, b, and c be positive integers that are pairwise relatively prime, and that satisfy a2 − ab + b2 = c2. Show that every prime factor of c is of the form 6 k + 1. 5. Let p be an odd prime and let ζ be a primitive pth root of unity; that is, ζ is a complex number such that ζp = 1 and ζk 6 = 1 for 1 ≤ k ≤ p − 1. Let Ap and Bp denote the set of quadratic residues and non-residues modulo p, respectively. Finally, let α = ∑ k∈Ap ζk and β = ∑ k∈Bp ζk.For example, for p = 7, α = ζ + ζ2 + ζ4 and β = ζ3 + ζ5 + ζ6. Show that α and β are the roots of x2 + x +1 − (−1 p ) p 4 = 0 . 8 Primitive Roots If the order of g modulo m is φ(m), then we say that g is a primitive root modulo m, or simply of m. Example 8.1 . Show that 2 is a primitive root modulo 3 n for all n ≥ 1. Solution . The statement is easily verified for n = 1, so assume the result is true for some n = k; that is, 2 φ(3 k ) ≡ 22·3k−1 ≡ 1 (mod 3 k). Now, let d be the order of 2 modulo 3 k+1 . Then 2 d ≡ 1 (mod 3 k+1 ) ⇒ 2d ≡ 1 (mod 3 k), so 2 · 3k−1 \| d. However, d \| φ(3 k+1 ), or d \| 2 · 3k. We deduce that d is either 2 · 3k−1 or 2 · 3k. Now we require the following lemma: Lemma. 2 2·3n−1 ≡ 1 + 3 n (mod 3 n+1 ), for all n ≥ 1. This is true for n = 1, so assume it is true for some n = k. Then by assumption, 22·3k−1 = 1 + 3 k + 3 k+1 m for some integer m ⇒ 22·3k = 1 + 3 k+1 + 3 k+2 M for some integer M (obtained by cubing) ⇒ 22·3k ≡ 1 + 3 k+1 (mod 3 k+2 ). By induction, the lemma is proved. Therefore, 2 2·3k−1 ≡ 1 + 3 k 6 ≡ 1 (mod 3 k+1 ), so the order of 2 modulo 3 k+1 is 2 · 3k, and again by induction, the result follows. 30 Corollary 8.2 . If 2 n ≡ − 1 (mod 3 k), then 3 k−1 \| n. Proof . The given implies 2 2n ≡ 1 (mod 3 k) ⇒ φ(3 k) \| 2n, or 3 k−1 \| n. Theorem 8.3 . If m has a primitive root, then it has φ(φ(m)) (distinct) primitive roots modulo m. Theorem 8.4 . The positive integer m has a primitive root iff m is one of 2, 4, pk, or 2 pk, where p is an odd prime. Theorem 8.5 . If g is a primitive root of m, then gn ≡ 1 (mod m) iff φ(m) \| n. Furthermore, gn0 ≡ gn1 iff φ(m) \| (n0 − n1). Proof . This follows directly from Theorem 6.1. Theorem 8.6 . If g is a primitive root of m, then the powers 1, g, g2,. . . , gφ(m)−1 represent each integer relatively prime to m uniquely modulo m.In particular, if m > 2, then gφ(m)/2 ≡ − 1 modulo m. Proof . Clearly, each power gi is relatively prime to m, and there are φ(m) integers relatively prime to m. Also, if gi ≡ gj (mod m), then gi−j ≡ 1 ⇒ φ(m) \| (i − j) by Theorem 8.6, so each of the powers are distinct modulo m. Hence, each integer relatively prime to m is some power gi modulo m. Furthermore, there is a unique i, 0 ≤ i ≤ φ(m) − 1, such that gi ≡ − 1 ⇒ g2i ≡ 1 ⇒ 2i = φ(m), or i = φ(m)/2. Proposition 8.7 . Let m be a positive integer. Then the only solutions to the congruence x2 ≡ 1 (mod m) are x ≡ ± 1 (mod m) iff m has a primitive root. Proof . This follows from Example 4.9. Example 8.2 . For a positive integer m, let S be the set of positive integers less than m that are relatively prime to m, and let P be the product of the elements in S. Show that P ≡ ± 1 (mod m), with P ≡ − 1 (mod m)iff m has a primitive root. Solution . We use a similar strategy as in the proof of Wilson’s Theorem. The result is clear for m = 2, so assume that m ≥ 3. We partition S as follows: Let A be the elements of S that are solutions to the congruence x2 ≡ 1 (mod m), and let B be the remaining elements. The elements in B can be arranged into pairs, by pairing each with its distinct multiplicative inverse. Hence, the product of the elements in B is 1 modulo m.The elements in A may also be arranged into pairs, by pairing each with 31 its distinct additive inverse, i.e. x and m − x. These must be distinct, because otherwise, x = m/ 2, which is not relatively prime to m. Note that their product is x(m − x) ≡ mx − x2 ≡ − 1 (mod m). Now if m has a primitive root, then by Proposition 8.7, A consists of only the two elements 1 and −1, so P ≡ − 1 (mod m). Otherwise, by Example 4.9, the number of elements of A is a power of two that is at least 4, so the number of such pairs in A is even, and P ≡ 1 (mod m). Remark . For m prime, this simply becomes Wilson’s Theorem. Theorem 8.8 .(1) If g is a primitive root of p, p a prime, then g or g + p is a primitive root of p2, according as gp−1 6 ≡ 1 (mod p2) or gp−1 ≡ 1 (mod p2). (2) If g is a primitive root of pk, where k ≥ 2 and p is prime, then g is a primitive root of pk+1 .By Theorem 8.6, given a primitive root g of m, for each a relatively prime to m, there exists a unique integer i modulo φ(m) such that gi ≡ a (mod m). This i is called the index of a with respect to the base g, denoted by ind g(a)(i is dependent on g, so it must be specified). Indices have striking similarity to logarithms, as seen in the following properties: (1) ind g(1) ≡ 0 (mod φ(m)), ind g(g) ≡ 1 (mod φ(m)), (2) a ≡ b (mod m) ⇒ ind g(a) ≡ ind g(b) (mod φ(m)), (3) ind g(ab ) ≡ ind g(a) + ind g(b) (mod φ(m)), (4) ind g(ak) ≡ k ind g(a) (mod φ(m)). Theorem 8.9 . If p is a prime and a is not divisible by p, then the con-gruence xn ≡ a (mod p) has gcd( n, p − 1) solutions or no solutions according as a(p−1) / gcd( n,p −1) ≡ 1 (mod p) or a(p−1) / gcd( n,p −1) 6 ≡ 1 (mod p). Proof . Let g be a primitive root of p, and let i be the index of a with respect to g. Also, any solution x must be relatively prime to p, so let u be the index of x. Then the congruence xn ≡ a becomes gnu ≡ gi (mod p) ⇔ nu ≡ i (mod p − 1). Let k = gcd( n, p − 1). Since g is a primitive root of p, k \| i ⇔ gi(p−1) /k ≡ a(p−1) /k ≡ 1. The result now follows from Theorem 4.11. 32 Remark . Taking p to be an odd prime and n = 2, we deduce Euler’s criterion. Example 8.3 Let n ≥ 2 be an integer and p = 2 n + 1. Show that if 3(p−1) /2 + 1 ≡ 0 (mod p), then p is a prime. (The converse to Example 7.4.) Solution . From 3 (p−1) /2 ≡ 32n−1 ≡ − 1 (mod p), we obtain 3 2n ≡ 1(mod p), so the order of 3 is 2 n = p−1, but the order also divides φ(p) ≥ p−1. Therefore, φ(p) = p − 1, and p is a prime. Example 8.4 . Prove that if n = 3 k−1, then 2 n ≡ − 1 (mod 3 k). (A partial converse to Corollary 8.2.) Solution . By Example 8.1, 2 is a primitive root of 3 k. Therefore, 2 has order φ(3 k) = 2 · 3k−1 = 2 n ⇒ 22n ≡ 1 ⇒ (2 n − 1)(2 n + 1) ≡ 0 (mod 3 k). However, 2 n − 1 ≡ (−1) 3k−1 − 1 ≡ 1 6 ≡ 0 (mod 3), so 2 n + 1 ≡ 0 (mod 3 k). Example 8.5 . Find all positive integers n > 1 such that 2n + 1 n2 is an integer. (1990 IMO) Solution . Clearly, n must be odd. Now assume that 3 k‖n; that is, 3 k is the highest power of 3 dividing n. Then 3 2k \| n2 \| (2 n + 1) ⇒ 2n ≡ − 1(mod 3 2k) ⇒ 32k−1 \| n, by Corollary 8.2 ⇒ 2k − 1 ≤ k ⇒ k ≤ 1, showing that n has at most one factor of 3. We observe that n = 3 is a solution. Suppose that n has a prime factor greater than 3; let p be the least such prime. Then p \| (2 n +1) ⇒ 2n ≡ − 1 (mod p). Let d be the order of 2 modulo p. Since 2 2n ≡ 1, d \| 2n. If d is odd, then d \| n ⇒ 2n ≡ 1, contradiction, so d is even, say d = 2 d1. Then 2 d1 \| 2n ⇒ d1 \| n. Also, d \| (p − 1), or 2d1 \| (p − 1) ⇒ d1 ≤ (p − 1) /2 < p . But d1 \| n, so d1 = 1 or d1 = 3. If d1 = 1, then d = 2, and 2 2 ≡ 1 (mod p), contradiction. If d1 = 3, then d = 6, and 26 ≡ 1 (mod p), or p \| 63 ⇒ p = 7. However, the order of 2 modulo 7 is 3, which is odd, again contradiction. Therefore, no such p can exist, and the only solution is n = 3. Useful Facts • All prime divisors of the Fermat number 2 2n 1, n > 1, are of the form 2n+2 k + 1. 33 Problems 1. Let p be an odd prime. Prove that 1i + 2 i + · · · + ( p − 1) i ≡ 0 (mod p)for all i, 0 ≤ i ≤ p − 2. 2. Show that if p is an odd prime, then the congruence x4 ≡ − 1 (mod p)has a solution iff p ≡ 1 (mod 8). 3. Show that if a and n are positive integers with a odd, then a2n ≡ 1(mod 2 n+2 ). 4. The number 142857 has the remarkable property that multiplying it by 1, 2, 3, 4, 5, and 6 cyclically permutes the digits. What are other numbers that have this property? Hint: Compute 142857 × 7. 9 Dirichlet Series Despite the intimidating name, Dirichlet series are easy to work with, and can provide quick proofs to certain number-theoretic identities, such as Example 3.2. Let α be a function taking the positive integers to the integers. Then we say that f (s) = ∞ ∑ n=1 α(n) ns = α(1) + α(2) 2s + α(3) 3s + · · · is the Dirichlet series generating function (Dsgf) of the function α,which we denote by f (s) ↔ α(n). Like general generating functions, these generating functions are used to provide information about their correspond-ing number-theoretic functions, primarily through manipulation of the gen-erating functions. Let 1 denote the function which is 1 for all positive integers; that is, 1( n) = 1 for all n. Let δ1(n) be the function defined by δ1(n) = { 1 if n = 1 , 0 if n > 1. It is easy to check that 1 and δ1 are multiplicative. 34 Now, let α and β be functions taking the positive integers to the integers. The convolution of α and β, denoted α ∗ β, is defined by (α ∗ β)( n) = ∑ d\|n α(d)β(n/d ). Note that convolution is symmetric; that is, α ∗ β = β ∗ α. Theorem 9.1 . Let f (s) ↔ α(n) and g(s) ↔ β(n). Then ( f · g)( s) ↔ (α ∗ β)( n). We now do three examples. The Dsgf of 1( n) is the well-known Riemann Zeta function ζ(s): ζ(s) = ∞ ∑ n=1 1 ns = 1 + 12s + 13s + · · · , so ζ(s) ↔ 1( n). This function will play a prominent role in this theory. What makes this theory nice to work with is that we may work with these functions at a purely formal level; no knowledge of the analytic properties of ζ(s) or indeed of any other Dsgf is required. By Theorem 9.1, the number-theoretic function corresponding to ζ2(s) is ∑ d\|n 1( d)1( n/d ) = ∑ d\|n 1 = τ (n). Hence, ζ2(s) ↔ τ (n). Finally, it is clear that 1 ↔ δ1(n). If α is a multiplicative function, then we can compute the Dsgf corre-sponding to α using the following theorem. Theorem 9.2 . Let α be a multiplicative function. Then ∞ ∑ n=1 α(n) ns = ∏ p ∞ ∑ k=0 α(pk) pks = ∏ p [1 + α(p)p−s + α(p2)p−2s + α(p3)p−3s + · · · ], where the product on the right is taken over all prime numbers. 35 As before, if we take α = 1, then we obtain ζ(s) = ∏ p (1 + p−s + p−2s + p−3s + · · · )= ∏ p ( 11 − p−s ) = 1 ∏ p (1 − p−s), an identity that will be useful. We say that a positive integer n > 1 is square-free if n contains no repeated prime factors; that is, p2 - n for all primes p. With this in mind, we define the M¨ obius function μ as follows: μ(n) =  1 if n = 1 , 0 if n is not square-free, and (−1) k if n is square-free and has k prime factors . It is easy to check that μ is multiplicative. By Theorem 9.2, the corresponding Dsgf is given by ∏ p (1 − p−s) = 1 ζ(s). Hence, 1 /ζ (s) ↔ μ(n), and this property makes the the seemingly mysterious function μ very important, as seen in the following theorem. Theorem 9.3 . (M¨ obius Inversion Formula) Let α and β be functions such that β(n) = ∑ d\|n α(d). Then α(n) = ∑ d\|n μ(n/d )β(d). Proof . Let f (s) ↔ α(n) and g(s) ↔ β(n). The condition is equivalent to β = α ∗ 1, or g(s) = f (s)ζ(s), and the conclusion is equivalent to α = β ∗ μ,or f (s) = g(s)/ζ (s). Theorem 9.4 . Let f (s) ↔ α(n). Then for any integer k, f (s − k) ↔ nkα(n). 36 For more on Dirichlet series, and generating functions in general, see H. Wilf, Generatingfunctionology .Problems 1. Let α, β, and γ be functions taking the positive integers to the integers. (a) Prove that α ∗ δ1 = α.(b) Prove that ( α ∗ β) ∗ γ = α ∗ (β ∗ γ). (c) Prove that if α and β are multiplicative, then so is α ∗ β.2. Prove that the following relations hold: ζ(s − 1) ζ(s) ↔ φ(n),ζ(s)ζ(s − 1) ↔ σ(n),ζ(s) ζ(2 s) ↔ \| μ(n)\|. Let the prime factorization of a positive integer n > 1 be pe1 1 pe2 2 · · · pek k .Define the functions λ and θ by λ(n) = ( −1) e1+e2+··· +ek and θ(n) = 2 k.Set λ(1) = θ(1) = 1. Show that λ and θ are multiplicative, and that ζ(2 s) ζ(s) ↔ λ(n) and ζ2(s) ζ(2 s) ↔ θ(n). For all positive integers n, let f (n) = n ∑ m=1 n gcd( m, n ). (a) Show that f (n) = ∑ d\|n dφ (d). (b) Let n = pe1 1 pe2 2 · · · pek k 1 be the prime factorization of n. Show that f (n) = (p2e1+1 1 + 1 p1 + 1 ) ( p2e2+1 2 + 1 p2 + 1 ) · · · (p2ek +1 1 + 1 pk + 1 ) . Verify Example 3.2 in one calculation. 37 6. Let id denote the identity function; that is, id( n) = n for all n. Verify each of the following identities in one calculation: (a) φ ∗ τ = σ.(b) μ ∗ 1 = δ1.(c) μ ∗ id = φ.(d) φ ∗ σ = id · τ .(e) σ ∗ id = 1 ∗ (id · τ ). 7. Let a1, a2, . . . , be the sequence of positive integers satisfying ∑ d\|n ad = 2 n for all n. Hence, a1 = 2, a2 = 2 2 − 2 = 2, a3 = 2 3 − 2 = 6, a4 =24 − 2 − 2 = 12, and so on. Show that for all n, n \| an.Hint: Don’t use the Dsgf of ( an)∞ 1 ; use the M¨ obius Inversion Formula. Bigger Hint: Consider the function f : [0 , 1] → [0 , 1] defined by f (x) = {2x}, where {x} = x − b xc is the fractional part of x. Find how the formula in the problem relates to the function f (n) = f ◦ f ◦ · · · ◦ f ︸︷︷︸ n .8. For all non-negative integers k, let σk be the function defined by σk(n) = ∑ d\|n dk. Thus, σ0 = τ and σ1 = σ. Prove that ζ(s)ζ(s − k) ↔ σk(n). 10 Miscellaneous Topics 10.1 Pell’s Equations Pell’s equations (or Fermat’s equations, as they are rightly called) are diophantine equations of the form x2 − dy 2 = N , where d is a positive non-square integer. There always exist an infinite number of solutions when N = 1, which we characterize. 38 Theorem 10.1.1 . If ( a, b ) is the lowest positive integer solution of x2 − dy 2 = 1, then all positive integer solutions are of the form (xn, y n) = ( (a + b√d)n + ( a − b√d)n 2 , (a + b√d)n − (a − b√d)n 2√d ) . We will not give a proof here, but we will verify that every pair indicated by the formula is a solution. The pair ( xn, y n) satisfy the equations xn + yn √d = ( a + b√d)n, and xn − yn √d = ( a − b√d)n. Therefore, x2 n − dy 2 n = ( xn + yn √d)( xn − yn √d)= ( a + b√d)n(a − b√d)n = ( a2 − db 2)n = 1 , since ( a, b ) is a solution. Remark . The sequences ( xn), ( yn) satisfy the recurrence relations xn =2ax n−1 − xn−2, yn = 2 ay n−1 − yn−2.For x2 − dy 2 = −1, the situation is similar. If ( a, b ) is the least positive solution, then the ( xn, y n) as above for n odd are the solutions of x2 − dy 2 = −1, and the ( xn, y n) for n even are the solutions of x2 − dy 2 = 1. Example 10.1.1 Find all solutions in pairs of positive integers ( x, y ) to the equation x2 − 2y2 = 1. Solution . We find that the lowest positive integer solution is (3,2), so all positive integer solutions are given by (xn, y n) = ( (3 + 2 √2) n + (3 − 2√2) n 2 , (3 + 2 √2) n − (3 − 2√2) n 2√2 ) . The first few solutions are (3,2), (17,12), and (99,70). 39 Example 10.1.2 . Prove that the equation x2 −dy 2 = −1 has no solution in integers if d ≡ 3 (mod 4). Solution . It is apparent that d must have a prime factor of the form 4 k +3, say q. Then x2 ≡ − 1 (mod q), which by Theorem 4.9 is a contradiction. Problems 1. In the sequence 12, 53, 11 8 , 27 19 , . . . , the denominator of the nth term ( n > 1) is the sum of the numerator and the denominator of the ( n − 1) th term. The numerator of the nth term is the sum of the denominators of the nth and ( n−1) th term. Find the limit of this sequence. (1979 Atlantic Region Mathematics League) 2. Let x0 = 0, x1 = 1, xn+1 = 4 xn − xn−1, and y0 = 1, y1 = 2, yn+1 =4yn − yn−1. Show for all n ≥ 0 that y2 n = 3 x2 n (1988 Canadian Mathematical Olympiad) 3. The polynomials P , Q are such that deg P = n, deg Q = m, have the same leading coefficient, and P 2(x) = ( x2 − 1) Q2(x) + 1. Show that P ′(x) = nQ (x). (1978 Swedish Mathematical Olympiad, Final Round) 10.2 Farey Sequences The nth Farey sequence is the sequence of all reduced rationals in [0,1], with both numerator and denominator no greater than n, in increasing order. Thus, the first 5 Farey sequences are: 0/1, 1/1, 0/1, 1/2, 1/1, 0/1, 1/3, 1/2, 2/3, 1/1, 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1, 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1. Properties of Farey sequences include the following: 40 (1) If a/b and c/d are consecutive fractions in the same sequence, in that order, then ad − bc = 1. (2) If a/b , c/d , and e/f are consecutive fractions in the same sequence, in that order, then a + eb + f = cd. (3) If a/b and c/d are consecutive fractions in the same sequence, then among all fractions between the two, ( a + c)/(b + d) (reduced) is the unique fraction with the smallest denominator. For proofs of these and other interesting properties, see Ross Honsberger, “Farey Sequences”, Ingenuity in Mathematics .Problems 1. Let a1, a2, . . . , am be the denominators of the fractions in the nth Farey sequence, in that order. Prove that 1 a1a2 1 a2a3 · · · + 1 am−1am = 1 . 10.3 Continued Fractions Let a0, a1, . . . , an be real numbers, all positive, except possibly a0. Then let 〈a0, a 1, . . . , a n〉 denote the continued fraction a0 + 1 a1 + · · · + 1 an−1 + 1 an . If each ai is an integer, then we say that the continued fraction is simple .Define sequences ( pk) and ( qk) as follows: p−1 = 0 , p0 = a0, and pk = akpk−1 + pk−2,q−1 = 0 , q0 = 1 , and qk = akpk−1 + qk−2, for k ≥ 1. Theorem 10.3.1 . For all x > 0 and k ≥ 1, 〈a0, a 1, . . . , a k−1, x 〉 = xp k−1 + pk−2 xq k−1 + qk−2 . 41 In particular, 〈a0, a 1, . . . , a k〉 = pk qk . Theorem 10.3.2 . For all k ≥ 0, (1) pkqk−1 − pk−1qk = ( −1) k−1,(2) pkqk−2 − pk−2qk = ( −1) kak.Define ck to be the kth convergence 〈a0, a 1, . . . , a k〉 = pk/q k. Theorem 10.3.3 . c0 < c 2 < c 4 < · · · < c 5 < c 3 < c 1.For a nice connection between continued fractions, linear diophantine equations, and Pell’s equations, see Andy Liu, “Continued Fractions and Diophantine Equations”, Volume 3, Issue 2, Mathematical Mayhem .Problems 1. Let a = 〈1, 2, . . . , 99 〉 and b = 〈1, 2, . . . , 99 , 100 〉. Prove that \|a − b\| < 199!100! . (1990 Tournament of Towns) 2. Evaluate 8 √√√√√2207 − 12207 − 12207 − · · · . Express your answer in the form a+b√cd , where a, b, c, d are integers. (1995 Putnam) 10.4 The Postage Stamp Problem Let a and b be relatively prime positive integers greater than 1. Consider the set of integers of the form ax + by , where x and y are non-negative integers. The following are true: (1) The greatest integer that cannot be written in the given form is ( a − 1)( b − 1) − 1 = ab − a − b.42 (2) There are 12 (a − 1)( b − 1) positive integers that cannot be written in the given form. (3) For all integers t, 0 ≤ t ≤ ab − a − b, t can be written in the given form iff ab − a − b − t cannot be. (If you have not seen or attempted this enticing problem, it is strongly suggested you have a try before reading the full solution.) Before presenting the solution, it will be instructive to look at an example. Take a = 12 and b = 5. The first few non-negative integers, in rows of 12, with integers that cannot be written in the given form in bold, are shown: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 With this arrangement, one observation should become immediately ap-parent, namely that bold numbers in each column end when they reach a multiple of 5. It should be clear that when reading down a column, once one hits an integer that can be written in the given form, then all successive integers can be as well, since we are adding 12 for each row we go down. It will turn out that this one observation is the key to the solution. Proof . Define a grapefruit to be an integer that may be written in the given form. For each i, 0 ≤ i ≤ a − 1, let mi be the least non-negative integer such that b \| (i + am i). It is obvious that for k ≥ mi, i + ak is a grapefruit. We claim that for 0 ≤ k ≤ mi − 1, i + ak is not a grapefruit. It is sufficient to show that i + a(mi − 1) is not a grapefruit, if mi ≥ 1. Let i + am i = bn i, ni ≥ 0. Since i + a(mi − b) = b(ni − a), mi must be strictly less than b; otherwise, we can find a smaller mi. Then i + a(mi − b) ≤ a−1−a = −1, so ni < a , or ni ≤ a−1. Suppose that ax +by = i+a(mi −1) = bn i − a, for some non-negative integers x and y. Then a(x + 1) = b(ni − y), so ni − y is positive. Since a and b are relatively prime, a divides ni − y.However, ni ≤ a − 1 ⇒ ni − y ≤ a − 1, contradiction. Therefore, the greatest non-grapefruit is of the form bn i − a, ni ≤ a − 1. The above argument also shows that all positive integers of this form are also non-grapefruits. Hence, the greatest non-grapefruit is b(a−1) −a = ab −a−b,proving (1). 43 Now, note that there are mi non-grapefruits in column i. The above tells us the first grapefruit appearing in column i is nib. Since 0, b, 2 b, . . . , ( a−1) b appear in different columns (because a and b are relatively prime), and there are a columns, we conclude that as i varies from 0 to a − 1, ni takes on 0, 1, . . . , a − 1, each exactly once. Therefore, summing over i, 0 ≤ i ≤ a − 1, ∑ i (i + am i) = ∑ i i + ∑ i am i = a(a − 1) 2 + a ∑ i mi = ∑ i bn i = a(a − 1) b 2 ⇒ a ∑ i mi = a(a − 1)( b − 1) 2 ⇒ ∑ i mi = (a − 1)( b − 1) 2 , proving (2). Finally, suppose that ax 1 + by 1 = t, and ax 2 + by 2 = ab − a − b − t, for some non-negative integers x1, x2, y1, and y2. Then a(x1 + x2) + b(y1 + y2) = ab − a − b, contradicting (1). So, if we consider the pairs ( t, ab − a − b − t), 0 ≤ t ≤ (a − 1)( b − 1) /2 − 1, at most one element in each pair can be written in the given form. However, we have shown that exactly ( a − 1)( b − 1) /2 integers cannot be written in the given form, which is the number of pairs. Therefore, exactly one element of each pair can be written in the given form, proving (3). Remark . There is a much shorter proof using Corollary 2.4. Can you find it? For me, this type of problem epitomizes problem solving in number theory, and generally mathematics, in many ways. If I merely presented the proof by itself, it would look artificial and unmotivated. However, by looking at a specific example, and finding a pattern, we were able to use that pattern as a springboard and extend it into a full proof. The algebra in the proof is really nothing more than a translation of observed patterns into formal notation. (Mathematics could be described as simply the study of pattern.) Note also that we used nothing more than very elementary results, showing how powerful basic concepts can be. It may have been messy, but one should never be afraid to get one’s hands dirty; indeed, the deeper you go, the 44 more you will understand the importance of these concepts and the subtle relationships between them. By trying to see an idea through to the end, one can sometimes feel the proof almost working out by itself. The moral of the story is: A simple idea can go a long way. For more insights on the postage stamp problem, see Ross Honsberger, “A Putnam Paper Problem”, Mathematical Gems II .Problems 1. Let a, b, and c be positive integers, no two of which have a common divisor greater than 1. Show that 2 abc − ab − bc − ca is the largest integer that cannot be expressed in the form xab + yca + zab , where x, y, and z are non-negative integers. (1983 IMO) References A. Adler & J. Coury, The Theory of Numbers , Jones and Bartlett I. Niven & H. Zuckerman, An Introduction to the Theory of Numbers ,John Wiley & Sons c© First Version October 1995 c© Second Version January 1996 c© Third Version April 1999 c© Fourth Version May 2000 Thanks to Ather Gattami for an improvement to the proof of the Postage Stamp Problem. This document was typeset under L ATEX, and may be freely distributed provided the contents are unaltered and this copyright notice is not removed. Any comments or corrections are always welcomed. It may not be sold for profit or incorporated in commercial documents without the express permis-sion of the copyright holder. So there. 45
6278	https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Ag	Atomic Weights and Isotopic Compositions for Silver Atomic Weights and Isotopic Compositions for Silver \| Isotope \| \| Relative Atomic Mass \| \| Isotopic Composition \| Standard Atomic Weight \| Notes \| --- --- --- \| \| \| \| \| \| \| \| \| \| \| 47 \| Ag \| 107 \| 106.905 0916(26) \| 0.518 39(8) \| 107.8682(2) \| g \| \| 109 \| 108.904 7553(14) \| 0.481 61(8) \| \| \| \| \|
6279	https://math.stackexchange.com/questions/1975942/prove-that-max-ab-cd-leq-max-a-c-max-b-d	real numbers - Prove that $\max{a+b,c+d} \leq \max{a,c} + \max{b,d}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prove that max{a+b,c+d}≤max{a,c}+max{b,d}max{a+b,c+d}≤max{a,c}+max{b,d} [closed] Ask Question Asked 8 years, 11 months ago Modified8 years, 11 months ago Viewed 8k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. Closed. This question needs details or clarity. It is not currently accepting answers. Want to improve this question? As written, this question is lacking some of the information it needs to be answered. If the author adds details in comments, consider editing them into the question. Once there's sufficient detail to answer, vote to reopen the question. Closed 8 years ago. Improve this question I've got the number max{a,b}max{a,b} and a,b∈R a,b∈R max{a,b}=a,a≥b max{a,b}=a,a≥b or max{a,b}=b,a<b max{a,b}=b,a<b We can see that max{a,b}≤c max{a,b}≤c only if a≤c a≤c and b≤c b≤c Now, if a,b,c,d∈R a,b,c,d∈R , prove that max{a+b,c+d}≤max{a,c}+max{b,d}max{a+b,c+d}≤max{a,c}+max{b,d} I am really stuck here, I need this for my University. real-numbers Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 19, 2016 at 16:58 Em. 16.2k 7 7 gold badges 28 28 silver badges 40 40 bronze badges asked Oct 19, 2016 at 16:51 George K.George K. 129 2 2 silver badges 7 7 bronze badges 2 I've got the number max{a,b} and a,b∈R max{a,b}=a, a≥b or max{a,b}=b, a<b We can see that max{a,b}≤c only if a≤c and b≤c Now, if a,b,c,d ∈ R , prove that max{a+b,c+d} ≤ max{a,c} + max{b,d} I am really stuck here, I need this for my University.George K. –George K. 2016-10-19 16:52:21 +00:00 Commented Oct 19, 2016 at 16:52 Formatting tips here.Em. –Em. 2016-10-19 16:59:02 +00:00 Commented Oct 19, 2016 at 16:59 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. Notice that a≤max{a,c}a≤max{a,c} and b≤max{b,d}b≤max{b,d} which means that a+b≤max{a,c}+max{b,d}a+b≤max{a,c}+max{b,d} Can you finish from here? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2016 at 17:05 EnnarEnnar 24.5k 3 3 gold badges 41 41 silver badges 70 70 bronze badges 2 Can you please complete this proof?agdhruv –agdhruv 2018-02-14 20:21:25 +00:00 Commented Feb 14, 2018 at 20:21 3 @agdhruv, do the same for c+d c+d. Then you have a+b a+b and c+d c+d are both bounded by max{a,c}+max{b,d}max{a,c}+max{b,d}, so max{a+b,c+d}≤max{a,c}+max{b,d}max{a+b,c+d}≤max{a,c}+max{b,d}.Ennar –Ennar 2018-02-15 10:01:43 +00:00 Commented Feb 15, 2018 at 10:01 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. An alternative is to consider max{x,y}=x+y+\|x−y\|2 max{x,y}=x+y+\|x−y\|2 and use the triangle inequality. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2016 at 18:41 Dean C WillsDean C Wills 444 2 2 silver badges 11 11 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-numbers See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 0How to deal with the following MinMax inequality involving graph structure? 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6280	https://www.stlouisfed.org/open-vault/2023/july/real-values-how-they-are-used	Skip to Main Explore Our Apps Tools and resources to find and use economic data worldwide. U.S. Financial, economic, and banking history. Vintages of economic data from specific dates in history. View banking market concentrations and perform HHI analysis. What Are Real Values, and How Are They Used? July 12, 2023 By Cole Schnell SHARE THIS PAGE: Link Copied Taking a quick glance at the menu below, you might mistakenly think the prices are dollar amounts. Five dollars for a cup of coffee seems reasonable today, if a bit expensive. However, in the winter of 1868, you could buy a cup of coffee for 5 cents, rather than $5, at Hardy’s Delmonico Lunch in Boston. This 1868 Hardy’s Delmonico Lunch menu was retrieved from the Library of Congress. Five cents for a cup of coffee may seem foreign today because the overall price level for goods and services—not just coffee—has increased. There has been inflation. In other words, 5 cents buys less today than it did in the past. Or, as an economist would say, the purchasing power of 5 cents has decreased. Hardy’s is no longer in business. However, a Starbucks that sells coffee just a few blocks away from the former Hardy’s location is. The smallest cup of coffee at this Starbucks cost $3.35 as of June 2023, a dollar amount that is 67 times larger than what Hardy’s was charging in 1868. So, was coffee at Hardy’s that cheap? We can use real values to answer that question by measuring the cost of coffee in terms of other goods or labor over time. What Are Real Values? “The real price of everything, what everything really costs to the man who wants to acquire it, is the toil and trouble of acquiring it,” Adam Smith wrote in his 1776 book “The Wealth of Nations.” Nominal values represent the money used in a transaction. For example, the nominal price of coffee in 1868 was 5 cents. Real values adjust for the effects of inflation. How Do You Calculate Real Values? To calculate the real value of a good, like coffee, you must understand how the price has changed relative to a market basket of goods and services, which represents the items purchased by an average consumer. (Note that while real values compare the cost of an item relative to the cost of a market basket, comparing real values over long periods becomes less precise due to changes in the basket.) If the prices of all goods and services in the basket were to rise at the same rate, the real value of any one good would be unchanged. Real values tell us when the prices of goods change at different rates than the overall basket’s. Price indexes can help us with this. Indexes of consumer prices are the price levels of market baskets of goods and services relative to the price level at a “base” date.We can set the price of the basket to $1 at a base date just by mentally varying the size of the basket. For example, if we constructed a basket that happened to have a price of $50, a basket with one-fiftieth the quantity of goods would have a price of $1. When overall prices increase, so do price indexes. By dividing the nominal value by the value of a price index and multiplying by 100, we can convert to a price in terms of units of the consumption basket. The following formula can be used to calculate real values: Real value = Nominal value divided by Price index × 100 Because the consumption basket cost $1 at the base date, the formula for real value also tells us the price of good in terms of money at the base date. The table below shows two popular indexes of consumer prices and their base dates. \| Source \| Price Index \| Base Date \| Unit after Conversion \| --- --- \| \| Bureau of Labor Statistics (BLS) \| Consumer price index (CPI) \| 1982-84 \| 1982-84 dollars \| \| Bureau of Economic Analysis (BEA) \| Personal consumption expenditures price index (PCEPI) \| 2012 \| 2012 dollars \| In the graph from St. Louis Fed online economic database FRED below, the blue line is the real price of one pound of coffee in terms of the CPI market basket of U.S. goods and services in 1982-84. Though the red line shows an increase in the nominal price over the last 43 years, the fact that the blue line decreased over that period shows that coffee prices haven’t risen as fast as those for the market basket of all goods and services. How Do You Find a Value in Today’s Dollars? What if you want to know the history of the real price of coffee—or some other price—in today’s dollars instead of in 2012 or 1982-84 dollars? To do this, we can adjust the formula by multiplying by the value of today’s price index (instead of 100), as the Page One Economics essay, “Adjusting for Inflation,” explained using the CPI. Current value = Original value ×Current CPI divided by Past CPI Using this new equation, we can update the blue line in the FRED graph using May 2023 as the base date instead of 1982-84. Notice the red line is not changed from the previous graph; and the real and nominal lines meet at our new base date. How Much Labor for a Cup of Coffee? Real prices are used as a proxy for the labor required to acquire a good or service. So, instead of using real values, we can determine, as an example, the amount of minimum wage labor required to obtain a cup of coffee in 2023 vs. in 1980. (Note that the comparison isn’t exact because many minimum wage workers are eligible for the earned income tax credit and that also contributes to their income.) The table shows data obtained through FRED on the federal minimum wage and on the price of coffee grounds for each of the two years. Divide the cost of grounds for a cup in a year by the minimum wage for that year and you can get an idea of how much time it would take to earn enough money for the coffee. In this case, the 2023 worker comes out 16 seconds ahead. \| January 1980 \| May 2023 \| --- \| \| Price per pound (453.6 grams) of grounds \| $3.21 \| $6.09 \| \| Price per cup (assuming 45 cups at 10.08 grams each) \| $0.07 \| $0.14 \| \| Federal minimum wage \| $3.10 \| $7.25 \| \| Labor at minimum wage for one cup of coffee \| 1 minute 23 seconds \| 1 minute 7 seconds \| \| SOURCES: U.S. Bureau of Labor Statistics, U.S. Department of Labor and author’s calculations. \| \| NOTE: The table shows rounded values, but the calculations are based on values that are not rounded. \| In What Cases Are “Real” Values Used? Here are a few of many ways economists, banks, the government—and now you—can use real values, plus some helpful resources. Real GDP Gross domestic product (GDP) is often used as a measure of the size of an economy. Comparing real GDP—rather than nominal GDP—over time ensures that you are observing changes in economic output —i.e., the real value of goods and services produced—rather than in inflation. Check out the state of the economy in terms of real GDP (and other real variables) using the St. Louis Fed’s Macro Snapshot. Real Interest Rates Nominal interest rates are the percentage of a loan’s amount that the lender charges the borrower for the loan each year. Real interest rates adjust nominal interest rates for inflation to show the loss of purchasing power required to borrow money. (Or, the borrower can actually gain purchasing power if the nominal interest rate is less than the inflation rate.) Learn more about real interest rates from an Economic Lowdown Podcast episode. Real Wages Wages have a certain amount of “stickiness,” which means that it takes time for them to change even when the economic environment changes. When nominal wages are left unchanged, inflation causes real wages to decrease. Check out an On the Economy blog post on real wage growth in 2022. Inflation-Indexed Bonds In 1997, the U.S. Treasury Department issued inflation-indexed bonds for the first time. For these bonds, the nominal interest payments are adjusted to maintain the same real value. Learn more about bonds using St. Louis Fed’s educational resources. We can set the price of the basket to $1 at a base date just by mentally varying the size of the basket. For example, if we constructed a basket that happened to have a price of $50, a basket with one-fiftieth the quantity of goods would have a price of $1. ABOUT THE AUTHOR Cole Schnell Cole Schnell served as an intern with the St. Louis Fed’s communications team. Cole Schnell Cole Schnell served as an intern with the St. Louis Fed’s communications team. Related Topics Inflation Subscribe to Open Vault This blog explains everyday economics and the Fed, while also spotlighting St. Louis Fed people and programs. Views expressed are not necessarily those of the St. Louis Fed or Federal Reserve System. Email Us Media questions
6281	https://pubmed.ncbi.nlm.nih.gov/8555482/	Internalization of bound fibrinogen modulates platelet aggregation - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Internalization of bound fibrinogen modulates platelet aggregation J D Wencel-Drake1,C Boudignon-Proudhon,M G Dieter,A B Criss,L V Parise Affiliations Expand Affiliation 1 Department of Medical Laboratory Sciences, University of Illinois at Chicago 60612, USA. PMID: 8555482 Item in Clipboard Internalization of bound fibrinogen modulates platelet aggregation J D Wencel-Drake et al. Blood.1996. Show details Display options Display options Format Blood Actions Search in PubMed Search in NLM Catalog Add to Search . 1996 Jan 15;87(2):602-12. Authors J D Wencel-Drake1,C Boudignon-Proudhon,M G Dieter,A B Criss,L V Parise Affiliation 1 Department of Medical Laboratory Sciences, University of Illinois at Chicago 60612, USA. PMID: 8555482 Item in Clipboard Cite Display options Display options Format Abstract In agonist-stimulated platelets, the integrin alpha IIb beta 3 (glycoprotein IIb-IIIa) is converted from an inactive to an active fibrinogen receptor, thereby mediating platelet aggregation. With time after agonist addition, at least two events occur: fibrinogen becomes irreversibly bound to the platelet and, when stirring is delayed, platelets lose the ability to aggregate despite the presence of maximally bound fibrinogen. Because we previously identified an actively internalized pool of alpha IIb, beta 3 in platelets, we explored the possibility that both of these events might result from the internalization of fibrinogen bound to active alpha IIb beta 3. Under conditions of irreversible fibrinogen binding, fluorescence microscopy showed that biotinylated fibrinogen is rapidly internalized by activated platelets to a surface-inaccessible, intracellular pool. Flow cytometric analysis showed that the observed loss in accessibility to extracellular probes immediately precedes a loss in ability to the platelets to aggregate. Moreover, prevention of irreversible fibrinogen binding results in a prevention of internalization and a retention of aggregation capacity. Thus, the internalization of fibrinogen from the activated platelet surface appears to contribute not only to the irreversible phase of fibrinogen binding, but also to the downregulation of platelet adhesiveness. Fibrinogen internalization is therefore likely to represent a fundamental regulatory mechanism that modulates platelet function. PubMed Disclaimer Similar articles Incomplete inhibition of platelet aggregation and glycoprotein IIb-IIIa receptor blockade by abciximab: importance of internal pool of glycoprotein IIb-IIIa receptors.Gawaz M, Ruf A, Pogatsa-Murray G, Dickfeld T, Rüdiger S, Taubitz W, Fischer J, Müller I, Meier D, Patscheke H, Schömig A.Gawaz M, et al.Thromb Haemost. 2000 Jun;83(6):915-22.Thromb Haemost. 2000.PMID: 10896249 Mapping the binding domains of the alpha(IIb) subunit. A study performed on the activated form of the platelet integrin alpha(IIb)beta(3).Biris N, Abatzis M, Mitsios JV, Sakarellos-Daitsiotis M, Sakarellos C, Tsoukatos D, Tselepis AD, Michalis L, Sideris D, Konidou G, Soteriadou K, Tsikaris V.Biris N, et al.Eur J Biochem. 2003 Sep;270(18):3760-7. doi: 10.1046/j.1432-1033.2003.03762.x.Eur J Biochem. 2003.PMID: 12950259 Fibrinogen fragments and platelet dysfunction in uremia.Kozek-Langenecker SA, Masaki T, Mohammad H, Green W, Mohammad SF, Cheung AK.Kozek-Langenecker SA, et al.Kidney Int. 1999 Jul;56(1):299-305. doi: 10.1046/j.1523-1755.1999.00518.x.Kidney Int. 1999.PMID: 10411706 Glycoprotein IIb-IIIa-liposomes bind fibrinogen but do not undergo fibrinogen-mediated aggregation.Sloan SM, Brown EB, Liu Q, Frojmovic MM.Sloan SM, et al.Platelets. 2000 Mar;11(2):99-110. doi: 10.1080/09537100075715.Platelets. 2000.PMID: 10938888 Glycoprotein IIb-IIIa content and platelet aggregation in healthy volunteers and patients with acute coronary syndrome.Yakushkin VV, Zyuryaev IT, Khaspekova SG, Sirotkina OV, Ruda MY, Mazurov AV.Yakushkin VV, et al.Platelets. 2011;22(4):243-51. doi: 10.3109/09537104.2010.547959. Epub 2011 Feb 17.Platelets. 2011.PMID: 21329420 Review. See all similar articles Cited by Platelet Toll-Like-Receptor-2 and -4 Mediate Different Immune-Related Responses to Bacterial Ligands.Niklaus M, Klingler P, Weber K, Koessler A, Kuhn S, Boeck M, Kobsar A, Koessler J.Niklaus M, et al.TH Open. 2022 Jul 11;6(3):e156-e167. doi: 10.1055/a-1827-7365. eCollection 2022 Jul.TH Open. 2022.PMID: 36046205 Free PMC article. Clustering induces a lateral redistribution of alpha 2 beta 1 integrin from membrane rafts to caveolae and subsequent protein kinase C-dependent internalization.Upla P, Marjomäki V, Kankaanpää P, Ivaska J, Hyypiä T, Van Der Goot FG, Heino J.Upla P, et al.Mol Biol Cell. 2004 Feb;15(2):625-36. doi: 10.1091/mbc.e03-08-0588. Epub 2003 Dec 2.Mol Biol Cell. 2004.PMID: 14657242 Free PMC article. Platelet hyperreactivity explains the bleeding abnormality and macrothrombocytopenia in a murine model of sitosterolemia.Kanaji T, Kanaji S, Montgomery RR, Patel SB, Newman PJ.Kanaji T, et al.Blood. 2013 Oct 10;122(15):2732-42. doi: 10.1182/blood-2013-06-510461. Epub 2013 Aug 7.Blood. 2013.PMID: 23926302 Free PMC article. Arf6 controls platelet spreading and clot retraction via integrin αIIbβ3 trafficking.Huang Y, Joshi S, Xiang B, Kanaho Y, Li Z, Bouchard BA, Moncman CL, Whiteheart SW.Huang Y, et al.Blood. 2016 Mar 17;127(11):1459-67. doi: 10.1182/blood-2015-05-648550. Epub 2016 Jan 6.Blood. 2016.PMID: 26738539 Free PMC article. Sulfatides partition disabled-2 in response to platelet activation.Drahos KE, Welsh JD, Finkielstein CV, Capelluto DG.Drahos KE, et al.PLoS One. 2009 Nov 24;4(11):e8007. doi: 10.1371/journal.pone.0008007.PLoS One. 2009.PMID: 19956625 Free PMC article. See all "Cited by" articles Publication types Research Support, Non-U.S. Gov't Actions Search in PubMed Search in MeSH Add to Search Research Support, U.S. Gov't, P.H.S. Actions Search in PubMed Search in MeSH Add to Search MeSH terms Adenosine Diphosphate / pharmacology Actions Search in PubMed Search in MeSH Add to Search Amino Acid Sequence Actions Search in PubMed Search in MeSH Add to Search Cell Compartmentation Actions Search in PubMed Search in MeSH Add to Search Endocytosis Actions Search in PubMed Search in MeSH Add to Search Fibrinogen / metabolism Actions Search in PubMed Search in MeSH Add to Search Flow Cytometry Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Microscopy, Fluorescence Actions Search in PubMed Search in MeSH Add to Search Molecular Sequence Data Actions Search in PubMed Search in MeSH Add to Search Peptide Fragments / pharmacology Actions Search in PubMed Search in MeSH Add to Search Platelet Aggregation / drug effects Actions Search in PubMed Search in MeSH Add to Search Platelet Aggregation / physiology Actions Search in PubMed Search in MeSH Add to Search Platelet Glycoprotein GPIIb-IIIa Complex / metabolism Actions Search in PubMed Search in MeSH Add to Search Protein Binding Actions Search in PubMed Search in MeSH Add to Search Zinc / pharmacology Actions Search in PubMed Search in MeSH Add to Search Substances Peptide Fragments Actions Search in PubMed Search in MeSH Add to Search Platelet Glycoprotein GPIIb-IIIa Complex Actions Search in PubMed Search in MeSH Add to Search thrombin receptor peptide (42-55) Actions Search in PubMed Search in MeSH Add to Search Adenosine Diphosphate Actions Search in PubMed Search in MeSH Add to Search Fibrinogen Actions Search in PubMed Search in MeSH Add to Search Zinc Actions Search in PubMed Search in MeSH Add to Search Related information PubChem Compound (MeSH Keyword) Grants and funding HL38405/HL/NHLBI NIH HHS/United States HL45100/HL/NHLBI NIH HHS/United States [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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6282	https://pubmed.ncbi.nlm.nih.gov/821345/	Malignant external otitis: a severe form of otitis in diabetic patients - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? 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Malignant external otitis: a severe form of otitis in diabetic patients D A Zaky,D W Bentley,K Lowy,R F Betts,R G Douglas Jr PMID: 821345 DOI: 10.1016/0002-9343(76)90181-9 Item in Clipboard Case Reports Malignant external otitis: a severe form of otitis in diabetic patients D A Zaky et al. Am J Med.1976 Aug. Show details Display options Display options Format Am J Med Actions Search in PubMed Search in NLM Catalog Add to Search . 1976 Aug;61(2):298-302. doi: 10.1016/0002-9343(76)90181-9. Authors D A Zaky,D W Bentley,K Lowy,R F Betts,R G Douglas Jr PMID: 821345 DOI: 10.1016/0002-9343(76)90181-9 Item in Clipboard Cite Display options Display options Format Abstract Two cases of malignant external otitis are presented and the literature is reviewed. The disease seems to occur exclusively in elderly diabetic patients. Diagnosis is mostly a clinical one, and requires a high index of suspicion. The characteristic clinical manifestations are pain and severe tenderness of the tissues around the ear and mastoid, persistent drainage and the presence of granulation tissue at the junction of the osseus and cartilagenous portions of the external ear. Roentgenographic findings are not helpful in the early stages. The pathogenesis of this disease depends on the presence of clefts in the cartilage forming the floor of the external auditory canal at its junction with the osseus portion through which infection can spread from the external ear to the deep soft tissues. Serious and often fatal complications may ensue. The most common and earliest symptom to appear if facial nerve palsy. Pseudomonas aeruginosa has been isolated uniformly, in pure or mixed cultures. This entity, therefore, should be borne in mind whenever an elderly diabetic patient presents with external otitis not amenable to the usual methods of therapy. Ps. aeruginosa should be strongly suspected, and its isolation should prompt vigorous systemic treatment with gentamicin and carbenicillin before extensive necrosis of cartilage and bone takes place. Any delay in diagnosis and management will lead to a serious and often fatal complications. PubMed Disclaimer Similar articles Pathogenesis and treatment of facial paralysis due to malignant external otitis.Chandler JR.Chandler JR.Ann Otol Rhinol Laryngol. 1972 Oct;81(5):648-58. doi: 10.1177/000348947208100505.Ann Otol Rhinol Laryngol. 1972.PMID: 4631220 No abstract available. Malignant external otitis: further considerations.Chandler JR.Chandler JR.Ann Otol Rhinol Laryngol. 1977 Jul-Aug;86(4 Pt 1):417-28. doi: 10.1177/000348947708600401.Ann Otol Rhinol Laryngol. 1977.PMID: 407826 Neurological sequelae of malignant external otitis.Faden A.Faden A.Arch Neurol. 1975 Mar;32(3):204-5. doi: 10.1001/archneur.1975.00490450084012.Arch Neurol. 1975.PMID: 164167 Diagnosis and treatment of necrotising otitis externa and diabetic foot osteomyelitis - similarities and differences.Peled C, Kraus M, Kaplan D.Peled C, et al.J Laryngol Otol. 2018 Sep;132(9):775-779. doi: 10.1017/S002221511800138X. Epub 2018 Aug 28.J Laryngol Otol. 2018.PMID: 30149824 Review. Malignant otitis externa in HIV and AIDS.Hern JD, Almeyda J, Thomas DM, Main J, Patel KS.Hern JD, et al.J Laryngol Otol. 1996 Aug;110(8):770-5. doi: 10.1017/s0022215100134929.J Laryngol Otol. 1996.PMID: 8869614 Review. See all similar articles Cited by The effect of diabetes mellitus on chronic rhinosinusitis and sinus surgery outcome.Zhang Z, Adappa ND, Lautenbach E, Chiu AG, Doghramji L, Howland TJ, Cohen NA, Palmer JN.Zhang Z, et al.Int Forum Allergy Rhinol. 2014 Apr;4(4):315-20. doi: 10.1002/alr.21269. Epub 2014 Jan 10.Int Forum Allergy Rhinol. 2014.PMID: 24415555 Free PMC article. Malignant pseudomonas external otitis.Edwards JE, Combs WA, Guze LB.Edwards JE, et al.West J Med. 1979 Sep;131(3):247-50.West J Med. 1979.PMID: 18748487 Free PMC article.No abstract available. Infection and diabetes mellitus.[No authors listed][No authors listed]West J Med. 1979 Jun;130(6):515-21.West J Med. 1979.PMID: 516690 Free PMC article.No abstract available. [Clinical and progressive profile of skin and soft tissue lesions in diabetics in 2017 at the dressing room of the Marc Sankale Center in Dakar].Mané DI, Demba D, Djiby S, Assane NM, Limane BA, Marie KC, Anna S, Maimouna NM.Mané DI, et al.Pan Afr Med J. 2019 Apr 29;32:209. doi: 10.11604/pamj.2019.32.209.18524. eCollection 2019.Pan Afr Med J. 2019.PMID: 31312321 Free PMC article.French. Diabetes mellitus and infection.Larkin JG, Frier BM, Ireland JT.Larkin JG, et al.Postgrad Med J. 1985 Mar;61(713):233-7. doi: 10.1136/pgmj.61.713.233.Postgrad Med J. 1985.PMID: 3885204 Free PMC article.Review.No abstract available. See all "Cited by" articles Publication types Case Reports Actions Search in PubMed Search in MeSH Add to Search MeSH terms Age Factors Actions Search in PubMed Search in MeSH Add to Search Aged Actions Search in PubMed Search in MeSH Add to Search Carbenicillin / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Diabetes Complications Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Middle Aged Actions Search in PubMed Search in MeSH Add to Search Otitis Externa / complications Actions Search in PubMed Search in MeSH Add to Search Otitis Externa / diagnosis Actions Search in PubMed Search in MeSH Add to Search Otitis Externa / etiology Actions Search in PubMed Search in MeSH Add to Search Pseudomonas Infections / drug therapy Actions Search in PubMed Search in MeSH Add to Search Pseudomonas aeruginosa / isolation & purification Actions Search in PubMed Search in MeSH Add to Search Substances Carbenicillin Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound (MeSH Keyword) [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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6283	https://thejns.org/view/journals/j-neurosurg/68/1/article-p62.xml	Intolerance to enteral feeding in the brain-injured patient in: Journal of Neurosurgery Volume 68 Issue 1 (1988) Journals Jump to Content This site uses cookies, tags, and tracking settings to store information that help give you the very best browsing experience. Dismiss this warning My StuffSign inSign upSubmitSubscribe HomeCOVID-19Journals Publish Before Print Collections For AuthorsFor LibrariansAbout Us Search Intolerance to enteral feeding in the brain-injured patient Jane A. Norton Jane A. Norton Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Jane A. Norton in Current site Google Scholar PubMedClose B.S.N. , Linda G. Ott Linda G. Ott Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Linda G. Ott in Current site Google Scholar PubMedClose M.S. , Craig McClain Craig McClain Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Craig McClain in Current site Google Scholar PubMedClose M.D. , Linas Adams Linas Adams Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Linas Adams in Current site Google Scholar PubMedClose M.D. , Robert J. Dempsey Robert J. Dempsey Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Robert J. Dempsey in Current site Google Scholar PubMedClose M.D. , Dennis Haack Dennis Haack Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Dennis Haack in Current site Google Scholar PubMedClose Ph.D. , Phillip A. Tibbs Phillip A. Tibbs Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by Phillip A. Tibbs in Current site Google Scholar PubMedClose M.D. , and A. Byron Young A. Byron Young Divisions of Neurosurgery and of Digestive Diseases and Nutrition, University of Kentucky Medical Center and Veterans Administration Hospital, Lexington, Kentucky Search for other papers by A. Byron Young in Current site Google Scholar PubMedClose M.D. View More View Less Page Range: 62–66 Volume/Issue: Volume 68: Issue 1 DOI link: Restricted access Purchase Now USD$45.00 JNS + Pediatrics - 1 year subscription bundle (Individuals Only) USD$536.00 JNS + Pediatrics + Spine - 1 year subscription bundle (Individuals Only) USD$636.00 Click here for subscription options Purchase Now USD$45.00 ### JNS + Pediatrics - 1 year subscription bundle (Individuals Only) USD$536.00 ### JNS + Pediatrics + Spine - 1 year subscription bundle (Individuals Only) USD$636.00 USD$45.00 USD$536.00 USD$636.00 Print or Print + OnlineSign in ABSTRACT ✓ Calorie and protein supplementation improves nutritional status. This support may improve outcome and decrease morbidity and mortality in acutely brain-injured patients. Investigators have observed a poor tolerance to enteral feedings after brain injury and have noted that this persists for approximately 14 days postinjury. This delay has been attributed to increased gastric residuals, prolonged paralytic ileus, abdominal distention, aspiration pneumonitis, and diarrhea. In the present investigation, 23 brain-injured patients with an admission 24-hour peak Glasgow Coma Scale (GCS) score between 4 and 10 were studied for 18 days from hospital admission. The mean duration from injury to initiation of full-strength, full-rate enteral feeding was 11.5 days. Seven of the 23 patients tolerated enteral feedings within the first 7 days following hospital admission (mean 4.3 days), four patients tolerated feedings between 7 and 10 days postadmission (mean 9 days), and 12 patients did not tolerate feedings until after 10 days postinjury (mean 15.9 days). There was a marginally significant relationship between low GCS scores on admission and length of days to enteral feeding tolerance (p = 0.07). A significant inverse relationship was observed between daily peak intracranial pressure (ICP) and time to tolerance of feedings (p = 0.02). There was no significant relationship between feeding tolerance and days to return of bowel sounds (p = 0.12). Serum albumin levels decreased during the investigation (mean ± standard error to the mean: 3.2 ± 0.12 gm/dl on Day 1; 2.7 ± 0.23 gm/dl on Day 16; normal = 3.5 to 5.0 gm/dl), whereas the percentage of patients tolerating feedings increased over the course of the study. The authors conclude that patients with acute severe brain injury do not adequately tolerate feedings via the enteral route in the early postinjury period. Tolerance of enteral feeding is inversely related to increased ICP and severity of brain injury. It is suggested that parenteral nutritional support is required following brain injury until enteral nutrition can be tolerated. Save Cite Email this contentShare Link Copy this link, or click below to email it to a friend Email this contentor copy the link directly: The link was not copied. Your current browser may not support copying via this button. Link copied successfully Copy link Collapse Expand Volume 68: Issue 1 (Jan 1988) in Journal of Neurosurgery Search within Journal... 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Byron Young, M.D., Division of Neurosurgery, University of Kentucky Medical Center, 800 Rose Street, Lexington, Kentucky 40536-0084. Keywords: enteral feeding; nutritional support; head injury Page Count: 5 Headings Clinical Material and Methods Patient Population Enteral Feeding Regimen Albumin Analysis Statistical Analysis Results Discussion References Export References - [x] 1. Andrassay RJ: Serum albumin levels correlate with tolerance to enteral feedings.Pharm Pract News, Nov 1987, pp 1, 37–39 Andrassay RJ: Serum albumin levels correlate with tolerance to enteral feedings. Pharm Pract News, Nov 1987, pp 1, 37–39 PubMed Andrassay RJ: Serum albumin levels correlate with tolerance to enteral feedings.Pharm Pract News, Nov 1987, pp 1, 37–39 Andrassay RJ: Serum albumin levels correlate with tolerance to enteral feedings. Pharm Pract News, Nov 1987, pp 1, 37–39)\| false Search Google Scholar Export Citation [x] 2. Benawra R, , Kucera P, &Mangurten HH: Nasojejunal feedings in infants with increased intracranial pressure.Clin Pediatr 19:409–411, 1980 Benawra R, Kucera P, Mangurten HH: Nasojejunal feedings in infants with increased intracranial pressure. Clin Pediatr 19:409–411, 1980 PubMed Benawra R, Kucera P, Mangurten HH: Nasojejunal feedings in infants with increased intracranial pressure.Clin Pediatr 19:409–411, 1980 Benawra R, Kucera P, Mangurten HH: Nasojejunal feedings in infants with increased intracranial pressure. Clin Pediatr 19:409–411, 1980)\| false Search Google Scholar Export Citation [x] 3. Bettice D: The cause of vomiting of tube feedings by neurosurgical patients.J Neurosurg Nurs 3:93–112, 1971 Bettice D: The cause of vomiting of tube feedings by neurosurgical patients. 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JPEN 9:121, 1985 (Abstract) PubMed Hunt D, Rowlands B, Allen S: The inadequacy of enteral nutritional support in head injury patients during the early postinjury period.JPEN 9:121, 1985(Abstract)Hunt D, Rowlands B, Allen S: The inadequacy of enteral nutritional support in head injury patients during the early postinjury period. JPEN 9:121, 1985 (Abstract))\| false Search Google Scholar Export Citation [x] 8. Kahrilas PJ, , Dodds WJ, &Dent J, et al: Effect of sleep, spontaneous gastroesophageal (GE) reflux, and a meal on UES pressure.Gastroenterology 90:1481, 1986(Abstract)Kahrilas PJ, Dodds WJ, Dent J, et al: Effect of sleep, spontaneous gastroesophageal (GE) reflux, and a meal on UES pressure. Gastroenterology 90:1481, 1986 (Abstract) PubMed%20reflux,%20and%20a%20meal%20on%20UES%20pressure. "Navigate to this content via PubMed. Opens a new browser window.") Kahrilas PJ, Dodds WJ, Dent J, et al: Effect of sleep, spontaneous gastroesophageal (GE) reflux, and a meal on UES pressure.Gastroenterology 90:1481, 1986(Abstract)Kahrilas PJ, Dodds WJ, Dent J, et al: Effect of sleep, spontaneous gastroesophageal (GE) reflux, and a meal on UES pressure. Gastroenterology 90:1481, 1986 (Abstract))\| false Search Google Scholar Export Citation [x] 9. Kiver KF, , Hays DP, &Fortin DF, et al: Pre-and postpyloric enteral feeding: analysis of safety and complications.JPEN 8:679–691, 1984 Kiver KF, Hays DP, Fortin DF, et al: Pre-and postpyloric enteral feeding: analysis of safety and complications. JPEN 8:679–691, 1984 PubMed Kiver KF, Hays DP, Fortin DF, et al: Pre-and postpyloric enteral feeding: analysis of safety and complications.JPEN 8:679–691, 1984 Kiver KF, Hays DP, Fortin DF, et al: Pre-and postpyloric enteral feeding: analysis of safety and complications. JPEN 8:679–691, 1984)\| false Search Google Scholar Export Citation [x] 10. 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Gastroenterology 86:1592–1610, 1984 PubMed Minami H, McCallum RW: The physiology and pathophysiology of gastric emptying in humans.Gastroenterology 86:1592–1610, 1984 Minami H, McCallum RW: The physiology and pathophysiology of gastric emptying in humans. Gastroenterology 86:1592–1610, 1984)\| false Search Google Scholar Export Citation [x] 12. Olivares L, , Segovia A, &Revuelta R: Tube feeding and lethal aspiration in neurological patients: a review of 720 autopsy cases.Stroke 5:654–657, 1974 Olivares L, Segovia A, Revuelta R: Tube feeding and lethal aspiration in neurological patients: a review of 720 autopsy cases. Stroke 5:654–657, 1974 PubMed Olivares L, Segovia A, Revuelta R: Tube feeding and lethal aspiration in neurological patients: a review of 720 autopsy cases.Stroke 5:654–657, 1974 Olivares L, Segovia A, Revuelta R: Tube feeding and lethal aspiration in neurological patients: a review of 720 autopsy cases. 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Rapp RP, , Young B, &Twyman D, et al: The favorable effect of early parenteral feedings on survival in head-injured patients.J Neurosurg 58:906–912, 1983 Rapp RP, Young B, Twyman D, et al: The favorable effect of early parenteral feedings on survival in head-injured patients. J Neurosurg 58:906–912, 1983 PubMed Rapp RP, Young B, Twyman D, et al: The favorable effect of early parenteral feedings on survival in head-injured patients.J Neurosurg 58:906–912, 1983 Rapp RP, Young B, Twyman D, et al: The favorable effect of early parenteral feedings on survival in head-injured patients. J Neurosurg 58:906–912, 1983)\| false Search Google Scholar Export Citation [x] 15. Rombeau JL, , Twomey PL, &McLean GK, et al: Experience with a new gastrostomy-jejunal feeding tube.Surgery 93:574–578, 1983 Rombeau JL, Twomey PL, McLean GK, et al: Experience with a new gastrostomy-jejunal feeding tube. Surgery 93:574–578, 1983 PubMed Rombeau JL, Twomey PL, McLean GK, et al: Experience with a new gastrostomy-jejunal feeding tube.Surgery 93:574–578, 1983 Rombeau JL, Twomey PL, McLean GK, et al: Experience with a new gastrostomy-jejunal feeding tube. Surgery 93:574–578, 1983)\| false Search Google Scholar Export Citation [x] 16. Stanghellini V, , Malagelada JR, &Zinsmeister AR, et al: Stress-induced gastroduodenal motor disturbances in humans: possible humoral mechanisms.Gastroenterology 85:83–91, 1983 Stanghellini V, Malagelada JR, Zinsmeister AR, et al: Stress-induced gastroduodenal motor disturbances in humans: possible humoral mechanisms. Gastroenterology 85:83–91, 1983 PubMed Stanghellini V, Malagelada JR, Zinsmeister AR, et al: Stress-induced gastroduodenal motor disturbances in humans: possible humoral mechanisms.Gastroenterology 85:83–91, 1983 Stanghellini V, Malagelada JR, Zinsmeister AR, et al: Stress-induced gastroduodenal motor disturbances in humans: possible humoral mechanisms. Gastroenterology 85:83–91, 1983)\| false Search Google Scholar Export Citation [x] 17. Teasdale G, &Jennett B: Assessment of coma and impaired consciousness. A practical scale.Lancet 2:81–84, 1974 Teasdale G, Jennett B: Assessment of coma and impaired consciousness. A practical scale. Lancet 2:81–84, 1974 PubMed Teasdale G, Jennett B: Assessment of coma and impaired consciousness. A practical scale.Lancet 2:81–84, 1974 Teasdale G, Jennett B: Assessment of coma and impaired consciousness. A practical scale. Lancet 2:81–84, 1974)\| false Search Google Scholar Export Citation [x] 18. Thompson DG, , Richelson E, &Malagelada JR: Perturbation of upper gastrointestinal function by cold stress.Gut 24:277–283, 1983 Thompson DG, Richelson E, Malagelada JR: Perturbation of upper gastrointestinal function by cold stress. Gut 24:277–283, 1983 PubMed Thompson DG, Richelson E, Malagelada JR: Perturbation of upper gastrointestinal function by cold stress.Gut 24:277–283, 1983 Thompson DG, Richelson E, Malagelada JR: Perturbation of upper gastrointestinal function by cold stress. Gut 24:277–283, 1983)\| false Search Google Scholar Export Citation [x] 19. Vane DW, , Shiffler M, &Grosfeld JL, et al: Reduced lower esophageal sphincter (LES) pressure after acute and chronic brain injury.J Pediatr Surg 17:960–964, 1982 Vane DW, Shiffler M, Grosfeld JL, et al: Reduced lower esophageal sphincter (LES) pressure after acute and chronic brain injury. J Pediatr Surg 17:960–964, 1982 PubMed%20pressure%20after%20acute%20and%20chronic%20brain%20injury. "Navigate to this content via PubMed. Opens a new browser window.") Vane DW, Shiffler M, Grosfeld JL, et al: Reduced lower esophageal sphincter (LES) pressure after acute and chronic brain injury.J Pediatr Surg 17:960–964, 1982 Vane DW, Shiffler M, Grosfeld JL, et al: Reduced lower esophageal sphincter (LES) pressure after acute and chronic brain injury. J Pediatr Surg 17:960–964, 1982)\| false Search Google Scholar Export Citation [x] 20. Whatley K, , Turner WN, &Dey M, et al: When does metoclopramide facilitate transpyloric intubation?JPEN 8:679–691, 1984 Whatley K, Turner WN, Dey M, et al: When does metoclopramide facilitate transpyloric intubation? JPEN 8:679–691, 1984 PubMed Whatley K, Turner WN, Dey M, et al: When does metoclopramide facilitate transpyloric intubation?JPEN 8:679–691, 1984 Whatley K, Turner WN, Dey M, et al: When does metoclopramide facilitate transpyloric intubation? JPEN 8:679–691, 1984)\| false Search Google Scholar Export Citation [x] 21. Wingate DL: The brain-gut link.Viewpoints on Digestive Diseases 17 (5):17–20, 1985 Wingate DL: The brain-gut link. Viewpoints on Digestive Diseases 17(5):17–20, 1985 PubMed Wingate DL: The brain-gut link.Viewpoints on Digestive Diseases 17 (5):17–20, 1985 Wingate DL: The brain-gut link. Viewpoints on Digestive Diseases 17(5):17–20, 1985)\| false Search Google Scholar Export Citation [x] 22. Wolf S: The stomach's link to the brain.Fed Proc 44:2889–2893, 1985 Wolf S: The stomach's link to the brain. Fed Proc 44:2889–2893, 1985 PubMed Wolf S: The stomach's link to the brain.Fed Proc 44:2889–2893, 1985 Wolf S: The stomach's link to the brain. Fed Proc 44:2889–2893, 1985)\| false Search Google Scholar Export Citation [x] 23. Young B, , Ott L, &Norton J, et al: Metabolic and nutritional sequelae in the non-steroid treated head injury patient.Neurosurgery 17:784–791, 1985 Young B, Ott L, Norton J, et al: Metabolic and nutritional sequelae in the non-steroid treated head injury patient. Neurosurgery 17:784–791, 1985 PubMed Young B, Ott L, Norton J, et al: Metabolic and nutritional sequelae in the non-steroid treated head injury patient.Neurosurgery 17:784–791, 1985 Young B, Ott L, Norton J, et al: Metabolic and nutritional sequelae in the non-steroid treated head injury patient. Neurosurgery 17:784–791, 1985)\| false Search Google Scholar Export Citation [x] 24. Young B, , Ott L, &Twyman D, et al: The effect of nutritional support on outcome from severe head injury.J Neurosurg 67:668–676, 1987 Young B, Ott L, Twyman D, et al: The effect of nutritional support on outcome from severe head injury. J Neurosurg 67:668–676, 1987 PubMed Young B, Ott L, Twyman D, et al: The effect of nutritional support on outcome from severe head injury.J Neurosurg 67:668–676, 1987 Young B, Ott L, Twyman D, et al: The effect of nutritional support on outcome from severe head injury. J Neurosurg 67:668–676, 1987)\| false Search Google Scholar Export Citation Cited By Publications Citing This Document You are looking at 1-1 of 1 citations Urwin, Susan C and Menon, David K, 2004, "Comparative Tolerability of Sedative Agents in Head-Injured Adults" Drug Safety Vol. 27, No. 2, pp 107, 0114-5916 Crossref Metrics Metrics \| \| All Time \| Past Year \| Past 30 Days \| --- --- \| \| Abstract Views \| 4003 \| 1083 \| 9 \| \| Full Text Views \| 410 \| 17 \| 2 \| \| PDF Downloads \| 258 \| 26 \| 2 \| \| EPUB Downloads \| 0 \| 0 \| 0 \| PubMed PubMed Citation Articles by Jane A. Norton Articles by Linda G. Ott Articles by Craig McClain Articles by Linas Adams Articles by Robert J. Dempsey Articles by Dennis Haack Articles by Phillip A. Tibbs Articles by A. Byron Young Similar articles in PubMed Google Scholar Article by Jane A. Norton Article by Linda G. Ott Article by Craig McClain Article by Linas Adams Article by Robert J. Dempsey Article by Dennis Haack Article by Phillip A. Tibbs Article by A. Byron Young Similar articles in Google Scholar © Copyright 1944-2025 American Association of Neurological Surgeons. All rights reserved, including those for text and data mining, AI training, and similar technologies. PermissionsLegalNoticesFeedback Editorial BoardsCMEFor AdvertisersJob SeekersJNSPG Shop FacebookTwitterInstagram Powered by PubFactory Powered by PubFactory [2600:1900:0:2107::1a00] 2600:1900:0:2107::1a00 Sign in to annotate Close Edit Character limit 500/500 Delete Cancel Save @! 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6284	https://proofwiki.org/wiki/Altitude,_Median_and_Perpendicular_Bisector_Coincide_iff_Triangle_is_Isosceles	Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles From ProofWiki Jump to navigation Jump to search 1 Theorem 2 Proof 2.1 Necessary Condition 2.2 Converse Statement Theorem Let $\triangle ABC$ be a triangle. Then: : : the altitude from $AB$ to $C$ : the median from $AB$ to $C$ : the perpendicular bisector of $AB$ : are all the same straight line if and only if: : $\triangle ABC$ is isosceles where $AB$ is the base. Proof Necessary Condition Let $\triangle ABC$ be an isosceles triangle whose base is $AB$. Let $D$ be the midpoint of $AB$. By definition of isosceles triangle, $AC = BC$. We have $AD = DB$ by construction, and $CD$ is common. So by Triangle Side-Side-Side Congruence: : $\triangle ACD = \triangle BCD$ From Two Angles on Straight Line make Two Right Angles: : $\angle ADC + \angle BDC$ equals two right angles and as $\angle ADC = \angle BDC$ they are each both right angles. Thus $CD$ is the perpendicular bisector of $AB$. By definition: : As $D$ is the midpoint of $AB$, $CD$ is the median from $AB$ to $C$. : As $CD$ is perpendicular to $AB$ and passes through $C$, which is a vertex of $\triangle ABC$, $CD$ is the altitude from $AB$ to $C$. Thus all three lines coincide. $\Box$ Converse Statement Let $AC \ne BC$ in $\triangle ABC$. Let $CD$ be the altitude from $AB$ to $C$. From Triangle with Two Equal Angles is Isosceles, it follows that $\angle CAB \ne \angle CBA$. Thus $\triangle ACD \ne \triangle BCD$. Let $E$ be the midpoint of $AB$. Let $EF$ be the perpendicular bisector of $AB$. By definition, $CE$ is the median from $AB$ to $C$. Aiming for a contradiction, suppose $D = E$. Then: : $AD = BD$ : $AC = BC$ : $CD$ is common. So by Triangle Side-Side-Side Congruence: : $\triangle ACD = \triangle BCD$ But we have that $\angle CAB \ne \angle CBA$. So by Proof by Contradiction it follows that $D \ne E$. It follows that $CD$, $CE$ and $EF$ are all different lines. $\blacksquare$ Retrieved from " Categories: Proofs by Contradiction Proven Results Isosceles Triangles Perpendicular Bisectors Medians of Triangles Navigation menu
6285	https://www.education.ky.gov/curriculum/conpro/Math/Documents/4_KDE_Number_and_Operations_Base_Ten_Multi-Digit_Multiplication_Strategies_Grade_4.pdf	1 Designed and revised by the Kentucky Department of Education Field-tested by Kentucky Mathematics Leadership Network Teachers Rights and Usage Agreement: If you encounter errors or other issues with this file, please contact the KDE math team at: kdemath@education.ky.gov Revised 2019 Number and Operations Base Ten: Multi-Digit Multiplication Strategies Grade 4 Formative Assessment Lesson 2 Multi-Digit Multiplication Strategies 4th Grade Mathematical goals This concept-based lesson is intended to help you assess how well students are able to use a variety of strategies to multiply. In particular, this unit aims to identify and help students who have difficulties with:  Representing multiplication in multiple ways. Kentucky Academic Standards This lesson asks students to select and apply mathematical content from within the grade, including the content standards: Grade 4 Number and Operations in Base Ten Cluster: Use place value understanding and properties of operations to perform multi-digit arithmetic. This lesson involves a range of mathematical practices from the standards, with emphasis on: MP2. Reason abstractly and quantitatively. MP7. Look for and make use of structure. MP8. Look for and make use of repeated reasoning. Introduction This lesson is structured in the following way:  A day or two before the lesson, students work individually on an assessment task that is designed to reveal their current understandings and difficulties. You then review their work and create questions for students to answer in order to improve their solutions.  A whole class introduction provides students with guidance on how to engage with the content of the task.  Students work with a partner on a collaborative discussion task using number lines to show evidence of their thinking. Throughout their work, students justify and explain their decisions to their peers and teacher(s).  In a final whole class discussion, students synthesize and reflect on the learning to make connections within the content of the lesson.  Finally, students revisit their original work or a similar task, and try to improve their individual responses. This Formative Assessment Lesson is designed to be part of an instructional unit. This task should be implemented approximately two-thirds of the way through the instructional unit. The results of this task should be used to inform the instruction that will take place for the remainder of your unit. 3 Materials required Each individual student will need two copies of the worksheet Multiplication Strategies & Representations. • Each pair of students will need a packet of Card Set A - D copied in color cut up before the lesson. {Note: you may want to make color copies, and laminate these for use in multiple classes over multiple years.} Time needed Approximately fifteen minutes for the assessment task, a one-hour lesson, and 15 minutes for the students to review their work for changes. All timings are approximate. Exact timings will depend on the needs of the class. Before the lesson Assessment task: Have the students do this task in class a day or more before the formative assessment lesson. This will give you an opportunity to assess the work and to find out the kinds of difficulties students have with it. Then you will be able to target your help more effectively in the follow-up lesson. Give each student a copy of Multiplication Strategies & Representations. Introduce the task briefly and help the class to understand the problem and its context. Frame the task: Spend fifteen minutes on your own, answering these questions. Don’t worry if you can’t figure it out. There will be a lesson on this material [tomorrow] that will help you improve your work. Your goal is to be able to answer these questions with confidence by the end of that lesson. It is important that students answer the question without assistance, as far as possible. If students are struggling to get started, ask them questions that help them understand what is required, but do not do the task for them. 4 Assessing students’ responses Collect students’ responses to the task. Make some notes on what their work reveals about their current levels of understanding and their different problem solving approaches. The purpose of this is to inform you of the issues that will arise during the lesson, so that you may prepare questions carefully. We suggest that you do not score students’ work. The research shows this is counterproductive, as it encourages students to compare scores, and distracts their attention from how they may improve their mathematics. Instead, help students to make further progress by asking questions that focus attention on aspects of their work. Some suggestions for these are given on the next page. These have been drawn from common difficulties anticipated. We suggest that you write your own lists of questions, based on your own students’ work, using the ideas below. You may choose to write questions on each student’s work. If you do not have time to do this, select a few questions that will be of help to the majority of students. These can be written/displayed on the board at the beginning of the lesson. It is also suggested that you plan student pairings based on their work on this initial task - pairing students homogeneously (common understandings). Common Issues Suggested questions and prompts Student doesn’t match the cards correctly because he or she doesn’t have a conceptual understanding of multiplication.  If you are multiplying 27x4, what does the 2 represent? the 7?  What would happen if you multiplied 20 x 4 and 7 x 4? Could you use those answers and to get the answer to 27x4? Student doesn’t understand Distributive Property.  How can the number(s) we are multiplying be broken apart?  What could you do with those numbers to solve this problem? Student doesn’t understand the area model for multiplication.  In the problem 27x14 let’s look at the number 27. How many 10s are in 27? How many ones? How could you model 27? Now let’s look at 14? How many tens? ones? How could you model 14?  Is there a way to take those two models and fit them on a rectangle to discover 27x14 without doing any calculations? Student doesn’t understand the base ten method for multiplication.  How does the area of the base ten model compare to the numbers in the partial products method?  How can use the units along each side of the model to get the final product?  5 Suggested lesson outline Whole Class Introduction (10 minutes) Teacher says: Today we are going to do some more work on multiplication strategies. Think about the strategies we have used for multiplication and use your white boards to show at least two different ways to solve other than the standard algorithm: Share your strategy with a partner. Explain your thinking. Showcase the area model, partial products, base ten representation, distributive property, etc. Now try this problem: During the next month, you have entered a video game challenge. There are twenty-one players that will each play seventeen hours during the challenge. How many total hours will be spent playing the video game challenge? Follow the same directions as above. Collaborative Activity: matching Card Sets Models A, B, C, D, and E. (30 minutes) Strategically partner students based on pre assessment data. Partner students with others who display similar errors/misconceptions on the pre-assessment task. While this may seem counterintuitive, this will allow each student to more confidently share their thinking. This may result in partnering students who were very successful together, those who did fairly well together, and those who did not do very well together. Give each pair Card Sets A, B, and C – Word problem, base ten model, and area model. 14 x 14 6 Introduce the collaborative activity carefully: Teacher says: I want you to work as partners. Take turns to match a word problem card with both a base ten model and area model card. Each time you do this, explain your thinking clearly and carefully. If your partner disagrees with the placement of a card, then challenge him/her. It is important that you both understand the math for all the placements. There is a lot of work to do today, and it doesn't matter if you don't all finish. The important thing is to learn something new, so take your time. As the teacher, your tasks during the partner work are to make a note of student approaches to the task, and to support student problem solving You can then use this information to focus a whole-class discussion towards the end of the lesson. In particular, notice any common mistakes. Make a note of student approaches to the task Try not to make suggestions that move students towards a particular approach to this task. Instead, ask questions to help students clarify their thinking. Encourage students to use each other as a resource for learning. Students will correct their own errors once the Partial Product cards are added. For students struggling to get started: There is more than one way to tackle this task. Can you think what one of them might be? [focusing on either the base ten strategy or the area model.] What is the story problem asking you to find? What information is there? Is there any information missing? What do you already know? How can you calculate products with the base ten model? With the area model? If one student has placed a particular card, challenge their partner to provide an explanation. 7 Maria placed this base ten card with this area model. Martin, why has Maria placed it here? If you find students have difficulty articulating their decisions, then you may want to use the questions from the Common Issues table to support your questioning. If the whole class is struggling on the same issue, then you may want to write a couple of questions on the board and organize a whole class discussion. Placing Card Sets D and E: Distributive Property and Partial Products As students finish placing the word problems, base ten models, and area model cards, hand out Card Sets D, E: Distributive Property and Partial Products These provide students with different ways of interpreting the situation. Do not collect the card sets they have been using. An important part of this task is for students to make connections between all the different representations of multiplication problems. As you monitor the work, listen to the discussion and help students to look for patterns and generalizations. Pairs should have 8 different clusters of cards with 6 cards in each. The original cards show the correct matches on each row of the table as they are originally arranged. Sharing work (10 minutes) When students get as far as they can with matching cards, ask one student from each pair to visit another pair's work. Students remaining at their desk should explain their reasoning for the matched cards on their own desk. Teacher says: If you are staying at your desk, be ready to explain the reasons for your pair's matches. If you are visiting another pair, make note of your card placements on a piece of paper. Go to another pair's desk and check to see which matches are different from your own. If there are differences, ask for an explanation. If you still don't agree, explain your own thinking. When you return to your own desk, you need to consider, as a pair, whether to make any changes to your work. Students may now want to make changes. 8 Whole Class Discussion: comparing different approaches (10 minutes) Conduct a whole-class discussion about what has been learned and highlight misconceptions and strategies you want to be revealed. Select students or pairs who demonstrated strategies and misconceptions you want to share with the class. Be intentional about the order of student sharing from least complex to most complex thinking. As each pair shares, highlight the connections between strategies. The discussion offers students the opportunity to learn from each other and for you to address some of the common misconceptions observed in the initial task and during the collaborative activity. Students should be expected to use this time to compare their solutions, discuss misconceptions and eventually evaluating their own responses based on correct answers. As part of the culture in your mathematics classroom students need to feel safe to share their solution strategies and ask questions of the teacher and each other. Conclude the lesson by discussing and generalizing what has been learned. The generalization involves first extending what has been learned to new examples, and then examining some of the conclusions the students come up with.  Were there certain problems that were more difficult?  Which strategy do you find most useful?  Which strategy do you find most difficult?  How can one strategy help you understand another strategy? Try to avoid making evaluative comments yourself. Instead, encourage students to respond to other students’ explanations. Improve individual solutions to the assessment task (10 minutes) Return to the students their original assessment, Multiplication Strategies & Representations as well as a second blank copy of the task. Teacher says: Look at your original responses and think about what you have learned this lesson. Using what you have learned, try to improve your work. If you have not added questions to individual pieces of work then write your list of questions on the board. Students should select from this list only the questions appropriate to their own work. If you find you are running out of time, then you could set this task in the next lesson, or for homework. 9 Solutions Assessment Task: Multiplication Strategies & Representations Question 1: 28 x 17 = 476 Students show work, but strategies may vary. Question 2: Each of the responses from Julie, Pete, Lisa, & Fred are all correct. Julie used the distributive property, Pete used clustering with the distributive property & some partial products, Lisa used a box method that shows the partial products in the area model, and Fred drew a base-10 block area model representation. Be sure that the responses have correct interpretations of each model, but answers may vary in the way each is described. Question 3: Student should state which person’s strategy most closely matched their own work in question #1. They should then use a different strategy to solve 39 x 14 = 546 correctly. These materials were adapted from Everyday Mathematics, Uncovering Student Misconceptions in Mathematics, and the National Library of Virtual Manipulatives. When teaching these multiplication and division strategies, Teaching Student Centered Mathematics by Van de Walle will be a useful resource. This lesson format was designed from the Classroom Challenge Lessons intended for students in grades 6 through 12 from the Math Assessment Project. 10 Multiplication Strategies & Representations Task Name_____ 1.) Multiply 28 by 17 and show your work: 2.) Julie, Pete, Lisa, & Fred each multipled 28 by 17. Below each method indicate if the work is correct and then explain whether that method makes sense mathematically or not. Julie Pete Lisa Fred (20 + 8) x (10 +7) 20x10 +8 x10+20x7+8x7 Check one: _correct incorrect Explain why: Check one: _correct incorrect Explain why: Check one: _correct _incorrect Explain why: Check one: _correct _incorrect Explain why: 3.) Which method most closely matches how you solved the original problem? __ Choose a different method than what you used in #1 to multiply 39 by 14. Show your work below: Student Materials 11 Problem Card Set A Base Ten Card Set B Area Model Card Set C Distributive Property Card Set D Partial Products Card Set E Each pack of baseball cards has fifteen cards. How many cards are in twenty-two packs? 15 X 22 10 20 100 200 How many eggs are in twelve dozen? 12 X 12 4 20 20 100 The boy scouts traveled a distance of twenty-three feet in their boat. The girl scouts traveled eleven times farther than the boy scouts? How far did the girls travel? 11 X 23 3 30 20 200 12 Problem Card Set A Base Ten Card Set B Area Model Card Set C Distributive Property Card Set D Partial Products Card Set E The tree house Scott is building needs twenty-six boards and each board needs seventeen nails. How many nails does Scott need to buy? 26 X 17 42 140 60 200 An opossum sleeps an average of nineteen hours per day. How many hours does it sleep in a 2-week time period? 14 X 19 36 90 40 100 Cam bought thirteen different colored folders and each had twenty-eight dots. How many total dots are on her folders? 13 X 28 24 80 60 200 13 Problem Card Set A Base Ten Card Set B Area Model Card Set C Distributive Property Card Set D Partial Products Card Set E Bags of Reese’s cups have twenty-one individually wrapped peanut butter cups. How many cups are in twenty-five bags? 21 X 25 5 100 20 400 The zoo has fifteen monkeys who eat fifteen bananas each day. How many bananas do they need each day for the monkeys? 15 X 15 25 50 50 100
6286	https://www.khanacademy.org/math/ka-math-class-11/x0419e5b3b578592a:probability-ncert-new/x0419e5b3b578592a:algebra-of-events/v/mutually-exclusive-and-exhaustive-events	Mutually exclusive and exhaustive events (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content KA Math Class 11 Course: KA Math Class 11>Unit 14 Lesson 2: Algebra of events Algebra of events Worked example - Algebra of events Mutually exclusive events Exhaustive events Mutually exclusive and exhaustive events Mutually exclusive and exhaustive events Algebra of probabilities of events Math> KA Math Class 11> Probability> Algebra of events © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Mutually exclusive and exhaustive events Google Classroom Microsoft Teams About About this video In this worked example, we'll tackle a hairy problem on the topic - Mutually exclusive and Exhaustive events. We roll two dice and list down three events A, B, and C. We then test these events and their complements against each other to find pairs that are mutually exclusive, exhaustive, and both.Created by Ashish Gupta. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? 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6287	https://www.mathplanet.com/education/algebra-1/linear-inequalities/linear-inequalities-in-two-variables	Linear inequalities in two variables - Mathplanet Mathplanet A free service from Mattecentrum Other tools Mathplanet Matteboken Matteboken (Arabic) Webmatematik (Danish) MenuAlgebra 1/Linear inequalities/ Linear inequalities in two variables Do excercisesShow all 3 exercises Linear inequalities with two variables I Linear inequalities with two variables II Linear inequalities with two variables III The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are substituted into the inequality. Example Is (1, 2) a solution to the inequality 2 x+3 y>1 2 x+3 y>1 2⋅1+3⋅2>?1 2⋅1+3⋅2>?1 2+5>?1 2+5>?1 7>1 7>1 The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥. The half-plane that is a solution to the inequality is usually shaded. Example Graph the inequality y≥−x+1 y≥−x+1 Video lesson Graph the linear inequality y≥2 x−3 y≥2 x−3 You need to accept marketing cookies to play the video. 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6288	https://www.chemspider.com/Chemical-Structure.35840.html	hexadiene \| C6H10 Accessed: Mon, 29 Sep 2025 01:18:43 GMT Simple Structure Advanced ChemSpider Webinars hexadiene Download .mol Cite this record hexadiene Molecular formula:C 6 H 10 Average mass:82.146 Monoisotopic mass:82.078250 ChemSpider ID:35840 1 of 1 defined double bonds 1 of 1 defined double bonds 1/1 Spectra available Spectra available Download .mol Cite this record Structural identifiers Names Properties Spectra Vendors More Names and synonyms Verified (3E)-1,3-Hexadien [German] [IUPAC name – generated by ACD/Name] (3E)-1,3-Hexadiene [IUPAC name – generated by ACD/Name] (3E)-1,3-Hexadiène [French] [IUPAC name – generated by ACD/Name] (3E)-Hexa-1,3-diene (E)-1,3-Hexadiene 1,3-Hexadiene [IUPAC name – generated by ACD/Name][IUPAC index name – generated by ACD/Name] 1,3-Hexadiene, (3E)- [IUPAC index name – generated by ACD/Name] 1,3-Hexadiene, (E)- 1,3-Hexadiene, mixture of cis and trans 209-759-4 [EINECS] 592-48-3 [RN] cis,trans-hexa-1,3-diene hexadiene Hexadiene (DOT) MFCD00009385 [MDL number] trans-1,3-Hexadiene Unverified (E,Z)-1,3-Hexadiene 1,3-Hexadiene (cis & trans) 1,3-Hexadiene (mixture of cis and trans) 1,3-HEXADIENE (TRANS) 1,3-Hexadiene cis + trans 1,3-Hexadiene(c,t) 1,3-Hexadiene, (3Z)- [ACD/Index Name] 1,3-Hexadiene,c&t 1,trans-3-hexadiene 14596-92-0 [RN] 20237-34-7 [RN] 42296-74-2 [RN] Hexa-1,3-diene n-Hexadien trans,trans-hexadiene Advertisement Spotlight Advertisement About us Data sources Blog Help Promotion on ChemSpider About us Campaigning & outreach News & events Awards & funding Advertise Journals, books & databases Locations & contacts Terms & conditions Membership Careers Help & legal Privacy policy Home Campaigning & outreach News & events Awards & funding Privacy policy About us Terms & conditions Journals, books & databases Locations & contacts Advertise Terms & conditions Membership & professional community Teaching & learning Careers Help & legal © Royal Society of Chemistry 2025 Registered charity number 207890 This website collects cookies to deliver a better user experience. 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6289	https://mching.people.amherst.edu/Work/composition-product.pdf	A NOTE ON THE COMPOSITION PRODUCT OF SYMMETRIC SEQUENCES MICHAEL CHING Abstract. We consider the composition product of symmetric sequences in the case where the underlying symmetric monoidal structure does not commute with coproducts. Even though this composition product is not a monoidal structure on symmetric sequences, it has enough structure, namely that of a ‘normal oplax’ monoidal product, to be able to deﬁne monoids (which are then operads on the underlying category) and make a bar construction. The main beneﬁt of this work is in the dual setting, where it allows us to deﬁne a cobar construction for cooperads. Introduction The category of symmetric sequences, in a closed symmetric monoidal category (C, ∧, S), is well known to have a monoidal product (called the composition product) whose monoids are precisely the operads in C. This result is due to Smirnov . This perspective allows for standard techniques in the theory of monoids, in particular bar constructions, to be applied to operads. This result depends, however, on the assumption that the monoidal structure ∧on C is closed, or more speciﬁcally, on the fact that ∧commutes with colimits. This condition is necessary in order that the composition product be associative. On the other hand, operads can be deﬁned in any symmetric monoidal category without the closed condition. It is therefore natural to wonder to what extent operads can still be treated as monoids when the composition product is not strictly associative. In particular, can we form a bar construction for operads in this wider context? In this paper, we show that the composition product is a ‘normal oplax monoidal product’ in the sense of Day-Street and that this slightly weaker structure allows, nonetheless, for sensible deﬁnitions of monoids and of bar constructions over them. In particular, it follows that operads in any symmetric monoidal category have a natural bar construction. The main motivation for this paper is, in fact, cooperads. A cooperad in C can be consid-ered as an operad in Cop (with the canonical symmetric monoidal structure on the opposite category). Although most interesting symmetric monoidal categories are closed symmetric monoidal (for example, based compactly-generated topological spaces, or any of the stan-dard models for stable homotopy theory), their opposite categories are less often so. This is reﬂected in the fact that the monoidal product does not generally commute with ﬁnite products in these topological examples. Our results therefore are necessary to deﬁne cobar constructions for cooperads in these categories. Such cobar constructions were described previously by the author in , in which the theory presented here was outlined vaguely. The present paper is intended to ﬁll in the gaps in that presentation and to demonstrate the usefulness of considering normal oplax monoidal structures, and their monoids, in homotopy theory. Here is an outline of the paper. In §1 we describe what is meant by a normal oplax monoidal structure. In §2 we show that the composition product forms part of a normal 2 MICHAEL CHING oplax monoidal structure on the category of symmetric sequences. In §3 we describe monoids for normal oplax monoidal structures, and note that an operad is precisely such an monoid for the composition product. Finally, in §4 we describe the simplicial bar construction in this setting. Acknowledgements This note is a clariﬁcation of parts of my thesis . Thanks to my advisor Haynes Miller for making me think more about this. Thanks also to various referees for bringing the notion of normal oplax monoidal structure to my attention and for other useful comments. 1. Normal oplax monoidal structures In this section we deﬁne the notion of a normal oplax monoidal structure on a category. This structure appears in as an example of a lax monoid (in the 2-cell dual of the 2-category of categories). It is a weakening of the notion of monoidal structure in the sense that the unitivity and associativity morphisms are not required to be isomorphisms. To deal with this change, we need explicit functors to stand for the higher iterates of the usual monoidal structure. These functors are then related by a set of associativity morphisms that satisfy an appropriate set of conditions analogous to the usual axioms for a monoidal structure. Deﬁnition 1.1. Let E be a category. A normal oplax monoidal structure on E consists of the following data: • an object I in E (which we refer to as the unit of the structure); • for each integer n ≥2, a functor µn : En →E; • for each triple of integers (n, l, r) with 0 ≤l ≤l + r ≤n, a natural transformation αn,l,r : µn(X1, . . . , Xn) →µn−r+1(X1, . . . , Xl, µr(Xl+1, . . . , Xl+r), Xl+r+1, . . . , Xn). To make sense of αn,l,r when r = 0, 1, n, we take µ1 : E →E to be the identity functor, and µ0 : E0 →E to be the functor whose only value is the unit I. For example the map αn,l,0 takes the form µn(X1, . . . , Xn) →µn+1(X1, . . . , Xl, I, Xl+1, . . . , Xn). The above data must satisfy the following conditions: (1) for any values of n and l, the maps αn,l,1 and αn,0,n are the identity morphism on µn(X1, . . . , Xn); (2) for any values of n, l, r, k, s that make sense, the diagram µn(X1,...,Xn) µn−r+1(X1,...,µr(Xl+1,...,Xl+r),...,Xn) µn−s+1(X1,...,µs(Xk+1,...,Xk+s),...,Xn) µn−r−s+2(X1,...,µr(Xl+1,...,Xl+r),...,µs(Xk+1,...,Xk+s),...,Xn) / αn,l,r αn,k,s αn−r+1,k−r+1,s / αn−s+1,l,r commutes; COMPOSITION PRODUCT 3 (3) for any values of n, l, r, k, s that make sense, the diagram µn(X1,...,Xn) µn−r+1(X1,...,µr(Xl+1,...,Xl+r),...,Xn) µn−s+1(X1,...,µs(Xk+1,...,Xk+s),...,Xn) µn−r+1(X1,...,µr−s+1(Xl+1,...,µs(Xk+1,...,Xk+s),...,Xl+r),...,Xn) / αn,l,r αn,k,s µn−r+1(X1,...,Xl,αr,k−l,s,Xl+r+1,...,Xn) / αn−s+1,l,r−s+1 commutes. A normal oplax monoidal category consists of a category E together with a normal oplax monoidal structure on E. Example 1.2. Let E be a category with a monoidal structure given by the bifunctor ⊗: E2 →E with unit object I. Set µn(X1, . . . , Xn) := (. . . (X1 ⊗X2) ⊗. . . ) ⊗Xn where µ0 takes the value I. The unit isomorphisms for the monoidal structure determine morphisms αn,l,0 and the associativity isomorphisms determine αn,l,r for r ≥1, as in Deﬁni-tion 1.1. The necessary conditions follow from the axioms for a monoidal structure. Thus a monoidal structure determines a normal oplax monoidal structure. Conversely, suppose we have a normal oplax monoidal structure on E in which all the associativity morphisms αn,l,r are isomorphisms. Then µ2 is a monoidal structure on E with associativity isomorphism α3,1,2 ◦α−1 3,0,2 : µ2(µ2(X, Y ), Z) →µ3(X, Y, Z) →µ2(X, µ2(Y, Z)) and unit isomorphisms α1,0,0 : X →µ2(I, X) and α1,1,0 : X →µ2(X, I). More generally, any non-symmetric cooperad Q in a braided monoidal category E deter-mines a normal oplax monoidal structure given by µn(X1, . . . , Xn) = Q(n) ⊗X1 ⊗· · · ⊗Xn. This is dual to Example 2.6 of . Remark 1.3. Day and Street deﬁne a normal oplax monoidal structure with an associa-tivity morphism of the form β(n1,...,nr) : µn(X1, . . . , Xn) →µr(µn1(X1, . . . , Xr1), . . . , µnr(Xr1+···+rn−1+1, . . . , Xn)) for each ordered partition n = n1 + · · · + nr. These determine our structure maps αn,l,r by considering partitions in which all but one of the values ni is 1 and recalling that µ1 is the identity functor. On the other hand we can reconstruct β(n1,...,nr) from the αn,l,r as the composite αr−1+nr,r−1,nr ◦αr−2+nr−1+nr,r−2,nr−1 ◦· · · ◦α1+n2+···+nr,1,n2 ◦αn,0,n1. Via this correspondence, the conditions (1), (2) and (3) of Deﬁnition 1.1 are equivalent to the unit and associativity conditions satisﬁed by the β(n1,...,nr). In Day and Street’s original context, the adjective ‘normal’ refers to the restriction that µ1 is the identity, and we obtain an ‘oplax monoidal structure’ by replacing this restriction 4 MICHAEL CHING with a natural transformation η : 1E →µ1. In this case µ1 is a comonad on E via the maps β(1) and η. There is also a dual notion of ‘lax monoidal structure’ on a category E. The only diﬀerence is that the directions of the associativity morphisms are reversed, that is, a (normal) lax monoidal structure on E is the same as a (normal) oplax monoidal structure on Eop. Remark 1.4. From a category E with normal oplax monoidal structure we can construct, in a natural way, a multicategory (see, for example, [4, 2.1.1]) with the same class of objects as E. For X1, . . . , Xn, Y ∈E, we deﬁne the set of multi-maps from (X1, . . . , Xn) to Y by HomE(X1, . . . , Xn; Y ) := HomE(µn(X1, . . . , Xn), Y ). In this way, we can view a normal oplax monoidal structure as a special kind of multicategory: namely, one in which the functors Y 7→HomE(X1, . . . , Xn; Y ) are represented, in a natural way, by single objects of E. 2. Composition of symmetric sequences Our main example of a normal oplax monoidal structure comes from the composition product on symmetric sequences. Deﬁnition 2.1. A symmetric sequence in a category C is a functor A : FinSet →C from the category FinSet, whose objects are ﬁnite sets and whose morphisms are bijections, to C. Denote the category of all symmetric sequences in C by CΣ (in which morphisms are natural transformations). There is an equivalent way to deﬁne a symmetric sequence which explains the terminology. Let FinSet0 be the category whose objects are the ﬁnite sets r := {1, . . . , r} for r ≥0 (with 0 the empty set), and whose morphisms are bijections. The category FinSet0 is a skeletal subcategory of FinSet, so any symmetric sequence A : FinSet →C is determined, up to canonical isomorphism, by its restriction A : FinSet0 →C. This restriction consists of the sequence A(0), A(1), A(2), . . . of objects in C, together with an action of the symmetric group Σr on A(r), hence the name ‘symmetric sequence’. Remark 2.2. For the remainder of this section, we ﬁx a symmetric monoidal structure ∧ with unit object S on the category C. The symmetry and associativity isomorphisms for the monoidal product ∧allow us to write expressions such as X ∧Y ∧Z and ^ α∈A Xα without caring about parentheses or ordering of the factors. We also assume throughout this paper that the underlying category C has all colimits (and, when we refer to the opposite category Cop, we assume that C has all limits). Remark 2.3. In this paper we focus on the composition product for symmetric sequences and our application is to symmetric operads. It seems likely that there are corresponding results for non-symmetric sequences in an arbitrary braided (not necessarily symmetric) monoidal category C. This would provide a simplicial bar construction for non-symmetric operads in the case that the monoidal structure on C is not closed. COMPOSITION PRODUCT 5 Deﬁnition 2.4. For a ﬁnite set J, we deﬁne a category J/FinSet0 as follows. The class of objects of J/FinSet0 consist of all functions f : J →I for some I ∈FinSet0, and the set of morphisms from (f : J →I) to (f ′ : J →I′) is the set of bijections σ : I →I′ such that f ′ = σ ◦f. Deﬁnition 2.5. Let A and B be two symmetric sequences in C. For each ﬁnite set J, we deﬁne a functor (A, B) : J/FinSet0 →C on objects by (A, B)(f : J →I) := A(I) ∧ ^ i∈I B(f −1(i)). For a morphism σ : I →I′ in J/FinSet0 we deﬁne (A, B)(σ) := A(I) ∧ ^ i∈I B(f −1(i)) →A(I′) ∧ ^ i′∈I′ B(f −1(σ−1(i′))) by combining map A(σ) with the permutation of the smash product identifying the term corresponding to i ∈I with the term corresponding to σ(i) ∈I′. Deﬁnition 2.6. For symmetric sequences A, B, we deﬁne a symmetric sequence A ◦B by (A ◦B)(J) := colim f∈J/FinSet0(A, B)(f). A bijection θ : J →J′ determines a map (A, B)(f) →(A, B)(f ◦θ−1) that is the map A(I) ∧ ^ i∈I B(f −1(i)) →A(I) ∧ ^ i∈I B(θ(f −1(i))) via the identity on A(I) and the action of the symmetric sequence B on the bijections f −1(i) ∼ = θ(f −1(i)) given by restricting θ. We thus obtain induced maps (A ◦B)(θ) := (A ◦B)(J) →(A ◦B)(J′) that make A ◦B into a symmetric sequence in C. Remark 2.7. The deﬁnition of the composition product in Deﬁnition 2.6 is isomorphic to more familiar descriptions involving a coproduct over partitions of J. Note that each element of J/FinSet0 amounts to a partition of J. (Note that here partitions are allowed to include multiple copies of the empty set. So even the empty set itself has inﬁnitely many partitions, each corresponding to a function ∅→r.) Morphisms in J/FinSet0 connect only those partitions with the same number of pieces. The colimit (A ◦B)(J) can therefore be written as (A ◦B)(J) ∼ = ∞ a r=0 " a J=J1⨿···⨿Jr A(r) ∧B(J1) ∧. . . ∧B(Jr) # Σr where Σr acts by permuting the sets J1, . . . , Jr and on A(r) in the usual way. Deﬁnition 2.8. The unit symmetric sequence I in the symmetric monoidal category (C, ∧, S) is given by I(J) = ( S if \|J\| = 1; 0 otherwise; where 0 is a ﬁxed initial object in C. The map I(J) →I(J′) induced by a bijection J →J′ is the identity morphism on S or 0 as appropriate. 6 MICHAEL CHING Proposition 2.9. Let C be a closed symmetric monoidal category. Then the composition product determines a monoidal structure on the category CΣ of symmetric sequences in C with unit object I. Proof. See §I.1.8 of . □ The importance in Proposition 2.9 of the hypothesis that the monoidal structure be closed is that it implies that the monoidal product ∧has a right adjoint in each variable and hence commutes with colimits. This condition is needed to show that the composition product is associative. We now drop this assumption. Our goal then is to show that the composition product is now part of a normal oplax monoidal structure on CΣ. To do this we need to deﬁne the higher composition products of three or more symmetric sequences. Deﬁnition 2.10. For a ﬁnite set J and integer n ≥2, we denote by J/FinSet0[n] the category whose objects are sequences of functions of ﬁnite sets J / fn−1In−1 / fn−2. . . / f1 I1 with I1, . . . , In−1 ∈FinSet0, and whose morphisms are commutative diagrams In−1 In−2 . . . I1 J I′ n−1 I′ n−2 . . . I′ 1 σn−1 / fn−2 / fn−3 σn−2 / f1 σ1 : fn−1 $ f′ n−1 / f′ n−2 / f′ n−3 / f′ 1 in which the vertical maps σ1, . . . , σn−1 are bijections. Notice that J/FinSet0 is the category J/FinSet0 of Deﬁnition 2.4. We also deﬁne J/FinSet0 to be the terminal category (with one object and one morphism). Deﬁnition 2.11. Now let A1, . . . , An be symmetric sequences in the symmetric monoidal category C. We deﬁne a functor (A1, . . . , An) : J/FinSet0[n] →C as follows. For the sequence f : J / fn−1 In−1 / fn−2 . . . / f1 I1 we set (A1, . . . , An)(f) := A1(I1) ∧ ^ i∈I1 A2(f −1 1 (i)) ∧. . . ∧ ^ i∈In−1 An(f −1 n−1(i)). For a morphism σ : f →f ′ in J/FinSet0[n], the bijections σk determine an isomorphism (A1, . . . , An)(f) →(A1, . . . , An)(f ′). Deﬁnition 2.12. Fix n ≥1 and take symmetric sequences A1, . . . , An in C. The higher composition product of A1, . . . , An is the symmetric sequence (A1 ◦· · · ◦An) deﬁned by (A1 ◦· · · ◦An)(J) := colim f∈J/FinSet0n(f). For n = 2 this reduces to the ordinary composition product of Deﬁnition 2.6 and when n = 1 it produces the identity functor on the category of symmetric sequences. To deﬁne COMPOSITION PRODUCT 7 also the case n = 0, we deﬁne the composition product of the empty collection to be the unit symmetric sequence I of Deﬁnition 2.8. Remark 2.13. As for the binary composition product, there is a formulation of the higher products in terms of partitions. Objects of the groupoid J/FinSet0[n] are in one-to-one correspondence with sequences of nested partitions of J of length n −1. If the pieces of the kth partition are labelled Jk,1, . . . , Jk,rk we have (A1 ◦· · · ◦An)(J) ∼ = a r1,...,rn−1 "a A1(r1) ∧ r1 ^ i=1 A2(J1,i) ∧. . . ∧ rn−1 ^ i=1 An(Jn−1,i) # Σr1×···×Σrn−1 . Remark 2.14. If ∧commutes with colimits in C, the higher composition product (A1 ◦· · ·◦ An) is isomorphic to any of the possible ways of iterating the binary product, for example, (A1 ◦A2 ◦A3) ∼ = (A1 ◦A2) ◦A3 ∼ = A1 ◦(A2 ◦A3), and this observation produces the associativity isomorphisms for the composition product in this case. Without this assumption, there are still maps from (A1 ◦A2 ◦A3) to each of these iterated composition products. We construct these next and then show that they form a normal oplax monoidal structure on CΣ. Deﬁnition 2.15. Let A1, . . . , An be symmetric sequences in the symmetric monoidal cate-gory C. Our aim is to construct natural maps of symmetric sequences: αn,l,r : (A1 ◦· · · ◦An) →(A1 ◦· · · ◦Al ◦(Al+1 ◦· · · ◦Al+r) ◦Al+r+1 ◦· · · ◦An). For each ﬁnite set J, we need to give a map colim f∈J/FinSet0n(f) → colim g∈J/FinSet0n−r+1(g). We do so by picking, for each f ∈J/FinSet0[n], an element g ∈J/FinSet0[n −r + 1] and giving maps αn,l,r(f) : (A1, . . . , An)(f) →(A1, . . . , Al, (Al+1 ◦· · · ◦Al+r), Al+r+1, . . . , An)(g) that commute with the maps induced by a morphism σ : f →f ′ in J/FinSet0[n]. So consider a sequence f : J / fn−1 In−1 / fn−2 . . . / f1 I1 and take g ∈FinSet0[n −r + 1] to be sequence g : J / fn−1 . . . / fl+r Il+r / fl+r−1...flIl / fl−1 . . . / f1 I1. Comparing the objects involved in the required map αn,l,r(f), we see that it is suﬃcient to produce a map   ^ i∈Il Al+1(f −1 l (i)) ! ∧. . . ∧   ^ i∈Il+r−1 Al+r(f −1 l+r−1(i))    → ^ i∈Il (Al+1◦· · ·◦Al+r)((fl+r−1 . . . fl)−1(i)). But the left-hand side here is isomorphic to ^ i∈Il (Al+1, . . . , Al+r)(f[i]) where f[i] is the sequence (fl+r−1 . . . fl)−1(i) / fl+r−1 (fl+r−2 . . . fl)−1(i) / fl+r−2 . . . / fl+1 (fl)−1(i) 8 MICHAEL CHING and so the required map αn,l,r(f) is formed by taking the ∧-product over i ∈Il of the canonical maps (Al+1, . . . , Al+r)(f[i]) →(Al+1 ◦· · · ◦Al+r)((fl+r−1 . . . fl)−1(i)). These αn,l,r(f) commute with the maps induced by a morphism σ : f →f ′ in J/FinSet0[n] so deﬁne the natural transformation αn,l,r. Theorem 2.16. Let C be a symmetric monoidal category with all colimits. Then the functors µn(A1, . . . , An) := (A1 ◦· · · ◦An), (with µ1(A) = A and µ0 = I, the unit symmetric sequence), and the maps αn,l,r of Deﬁnition 2.15, form a normal oplax monoidal structure on the category CΣ of symmetric sequences in C. Proof. We check axioms (1)-(3) of Deﬁnition 1.1. For (1), we ﬁrst look at Deﬁnition 2.15 for αn,l,1 and note that g = f and αn,l,1(f) is the identity map on (A1, . . . , An)(f). Thus αn,l,1 is the identity on (A1 ◦· · · ◦An). For αn,0,n we notice that g the unique object in J/FinSet0, and that αn,l,r(f) is the canonical map (A1, . . . , An)(f) →(A1 ◦· · · ◦An)(J). It follows that αn,0,n is the identity. For each of (2) and (3), we have to compare two maps originating in the colimit colim f∈J/FinSet0n(f) so it is suﬃcient to check that for each f ∈J/FinSet0[n], the components associated to f by these two maps are equal. So consider a sequence f : J / fn−1 In−1 / fn−2 . . . / f1 I1. For (2) each of the relevant composites is built from maps (A1, . . . , An)(f) →(A1, . . . , (Al+1 ◦· · · ◦Al+r), . . . , (Ak+1 ◦· · · ◦Ak+s), . . . , An)(h) where h is the sequence h : J / fn−1 . . . / fk+s Ik+s / fk+s−1...fkIk / fk−1 . . . / fl+r Il+r / fl+r−1...flIl / fl−1 . . . / f1 I1 and each of these is built from the ∧-product of canonical maps of the form (Al+1, . . . , Al+r)(f[i]) →(Al+1 ◦· · · ◦Al+r)((fl+r−1 . . . fl)−1(i)) for i ∈Il and (Ak+1, . . . , Ak+s)(f[i′]) →(Ak+1 ◦· · · ◦Ak+s)((fk+s−1 . . . fk)−1(i′)) for i′ ∈Ik. For (3) each composite is built from maps (A1, . . . , An)(f) →(A1, . . . , (Al+1 ◦· · · ◦(Ak+1 ◦· · · ◦Ak+s) ◦· · · ◦Al+r), . . . , An)(g) where g is the sequence g : J / fn−1 . . . / fl+r Il+r / fl+r−1...flIl / fl−1 . . . / f1 I1. Each of these is built from the ∧-product of maps (Al+1, . . . , Al+r)(f[i]) →(Al+1 ◦· · · ◦(Ak+1 ◦· · · ◦Ak+s) ◦· · · ◦Al+r)(h[i]) COMPOSITION PRODUCT 9 for i ∈Il where h[i] is the sequence (fl+r−1 . . . fl)−1(i) / fl+r−1 . . . (fk+s−1 . . . fl)−1(i) / fk+s−1...fk(fk−1 . . . fl)−1(i) . . . / fl+1 (fl)−1(i) and each of these is built from the ∧-product of the canonical maps (Ak+1, . . . , Ak+s)(h[i][i′]) →(Ak+1 ◦· · · ◦Ak+s)((fk+s−1 . . . fk)−1(i′)) for i′ ∈Ik. □ 3. Monoids in normal oplax monoidal categories We now return to the general context and say what is meant by a monoid in an arbitrary normal oplax monoidal structure on a category E. We saw in Remark 1.4 that such a structure determines a multicategory with the same class of objects. A monoid in E is then precisely a monoid in that multicategory in the sense of Leinster [4, 2.1.11]. This entails the following. Deﬁnition 3.1. Let E be a normal oplax monoidal category as in Deﬁnition 1.1. A monoid with respect to this structure consists of an object M ∈E, morphisms mn : µn(M, . . . , M) →M, for n ≥0, such that m1 is the identity morphism on M, and such that for each triple (n, l, r) with 0 ≤l ≤l + r ≤n the diagram µn(M, . . . , M) µn−r+1(M, . . . , µr(M, . . . , M), . . . , M) µn−r+1(M, . . . , M) M $ mn / αn,l,r mr mn−r+1 commutes. Example 3.2. If the normal oplax monoidal structure on E comes from an actual monoidal structure as in Example 1.2, then a monoid in the sense of Deﬁnition 3.1 is the same as a monoid for the monoidal structure in the usual sense. Remark 3.3. A monoid in a multicategory M corresponds uniquely to a functor of mul-ticategories M : ∗→M where ∗denotes the terminal multicategory, with a single object and a single multi-map of each possible type (see [4, 2.1.11]). A monoid M in the normal oplax monoidal category E can similarly be described as a functor ∗→E of normal oplax monoidal categories. We now show that a monoid M in a normal oplax monoidal category is determined by the structure maps m2 : µ2(M, M) →M and m0 : I →M. This is a consequence of the corresponding statement for monoids in a multicategory but we are unable to ﬁnd such a result explicitly in the literature. 10 MICHAEL CHING Proposition 3.4. Let E be a normal oplax monoidal category an M and object in E. Suppose given maps m2 : µ2(M, M) →M; m0 : I →M such that the following diagrams commute (1) µ2(µ2(M, M), M) µ2(M, M) µ3(M, M, M) M µ2(M, µ2(M, M)) µ2(M, M) / µ2(m2,M) m2 4 α3,0,2 α3,1,2 / µ2(M,m2) 4 m2 (2) M µ2(I, M) µ2(M, M) M / α1,0,0 1M / µ2(m0,M) m2 (3) M µ2(M, I) µ2(M, M) M / α1,1,0 1M / µ2(M,m0) m2 Then there is a unique monoid in E based on the object M that includes the maps m2 and m0. Proof. The axioms for a monoid in Deﬁnition 3.1 imply that for n ≥3, mn is uniquely determined by m2. This gives us the uniqueness part. For the existence claim, we take m1 to be the identity on M, and deﬁne m3 to be the composed map µ3(M, M, M) →M in diagram (1) above. Our goal now is to show that it is possible to deﬁne mn for n > 3 in such a way that all diagrams of the form (D(n, l, r)) µn(M, . . . , M) µn−r+1(M, . . . , µr(M, . . . , M), . . . , M) µn−r+1(M, . . . , M) M $ mn / αn,l,r mr mn−r+1 COMPOSITION PRODUCT 11 commute. Notice that the cases r = 1 and r = n are trivially satisﬁed (since m1 is the identity). We focus ﬁrst on the case 2 ≤r ≤n −1 and save r = 0 for later. Suppose that n ≥4 and we have deﬁned mp : µp(M, . . . , M) →M for p < n such that diagrams D(p, l, r), for p < n and 2 ≤r ≤p−1, commute. This is true for n = 4 by diagram (1) and our choice of m3. For integers l, r satisfying 2 ≤r ≤n −1 and r + l ≤n, let φl,r : µn(M, . . . , M) →M be the composite appearing in diagram D(n, l, r) (and to which we would like mn to be equal). We claim that φl,r = φk,s for any two choices of (l, r) and (k, s). (We assume l ≤k.) To see this we write Cl,r := {l + 1, . . . , l + r} to refer to the sequence of copies of M that are combined to form µr(M, . . . , M) in the dia-gram D(n, l, r). The comparison between φl,r and φk,s depends on the relationship between the sequences Cl,r and Ck,s. When the sequences Cl,r and Ck,s are either disjoint or one is contained in the other, we consider the following diagram whose vertical composites are φl,r and φk,s: µn(M, . . . , M) µn−r+1(. . . , µr(M, . . . , M), . . . ) µn−s+1(. . . , µs(M, . . . , M), . . . ) µn−r+1(M, . . . , M) µn−s+1(M, . . . , M) M u αn,l,r ) αn,k,s mr ms ) mn−r+1 u mn−s+1 This diagram can be ﬁlled in with either axiom (2) or (3) from Deﬁnition 1.1 together with naturality squares and copies of diagrams D(p, l, r) for p < n. Thus φl,r = φk,s. Now suppose that the two sequences have intersection of length at least two or union of length at most n −1. Then, using the previous comparison of sequences for which one is contained in the other we have φl,r = φk,l+r−k = φk,s. Finally, suppose the intersection is length one and the union is length n. Since n ≥4, we must have one of r, s equal to at least 3 in which case we either have φl,r = φl,r−1 = φk,s or φl,r = φk,s−1 = φk,s 12 MICHAEL CHING by the previous comparisons. The conclusion is that φl,r = φk,s for all possibilities and so we deﬁne mn to be this common map. Recursively, this deﬁnes mn for all n ≥0 such that all diagrams D(n, l, r) for r ≥1 commute. It remains to show that with these choices each diagram D(n, l, 0) commutes. This takes the form µn(M, . . . , M) µn+1(M, . . . , M, I, M, . . . , M) µn+1(M, . . . , M) M " mn / αn,l,0 m0 mn+1 When l ̸= n, we can expand out the term mn+1 using the diagram D(n + 1, l, 2) to get µn(M, . . . , M, . . . , M) µn(M, . . . , µ2(I, M), . . . , M) µn(M, . . . , µ2(M, M), . . . , M) µn(M, . . . , M, . . . , M) M / α1,0,0 $ m0 m2 mn The diagonal map is the identity on µn(M, . . . , M) by diagram (2) and so the overall com-posite is mn as required. Finally, if l = n, we use a similar argument based on diagram (3) instead. □ Example 3.5. Let (C, ∧, S) be a symmetric monoidal category. A monoid in the normal oplax monoidal category CΣ of Theorem 2.16 is precisely an operad in C. This generalizes the result of Smirnov to the case that the symmetric monoidal structure on C is not closed. We can also talk about actions of a monoid in a normal oplax monoidal category. Again these are examples of corresponding notions for multicategories but these notions do not seem to be explicitly described in the literature (though are no doubt well known). We spell out the details. Deﬁnition 3.6. Let M be a monoid in the normal oplax monoidal category E with structure maps mn. A left M-module consists of an object L ∈E and, for each n ≥1 a morphism ln : µn(M, . . . , M, L) →L COMPOSITION PRODUCT 13 (where l1 : L →L is the identity morphism) such that each of the following diagrams commutes (1) for l + r < n µn(M, . . . , M, L) µn−r+1(M, . . . , µr(M, . . . , M), . . . , M, L) µn−r+1(M, . . . , M, L) L & ln / αn,l,r mr ln−r+1 (2) for l + r = n µn(M, . . . , M, L) µn−r+1(M, . . . , M, µr(M, . . . , M, L)) µn−r+1(M, . . . , M, L) L $ ln / αn,l,r lr ln−r+1 The corresponding notion of right M-module consists of an object R ∈E with rn : µn(R, M, . . . , M) →R satisfying similar conditions. Remark 3.7. The structure of a left (or right) M-module is uniquely determined by the map l2 : µ2(M, L) →L (respectively, r2) subject to conditions similar to diagrams (1) and (2) (respectively, (1) and (3)) of Proposition 3.4 with the ﬁnal copy of M replaced by L (respectively, the ﬁrst copy replaced by R). The combination of a monoid M and left M-module L in E can be described as a map of multicategories F : (m ▶l) →E where (m ▶l) is the multicategory with two objects and exactly one multi-map of each of the types (m, . . . , m) →m and (m, . . . , m, l) →l. There is a similar characterization of right modules. Example 3.8. If M is a monoid in the normal oplax monoidal category CΣ of Theorem 2.16 (that is, M is an operad in C), then a left or right M-module is a left or right module over the operad M in the usual sense of, for example, [5, 3.26]. 14 MICHAEL CHING 4. Simplicial bar construction on a monoid In this section we show that we can form a simplicial bar construction for a monoid and its modules in any normal oplax monoidal category. This generalizes the usual construction for a monoid in a monoidal category. Deﬁnition 4.1. Let M be a monoid with respect to a normal oplax monoidal structure on a category E. Let L be a left M-module and R a right M-module. For n ≥0, we deﬁne Bn(R, M, L) := µn+2(R, M, . . . , M \| {z } n , L) Now let θ : [n] →[m] be an order-preserving map (where [n] denotes the totally-ordered set {0, . . . , n}). Then we deﬁne θ∗: Bm(R, M, L) →Bn(R, M, L) to be the composite µm+2(R, M, . . . , M, L) µn+2(µθ0(R, . . . ), . . . , µθj(M, . . . , M), . . . , µθn+1(. . . , L)) µn+2(R, M, . . . , M, L) / α + θ∗ (rθ0,...,lθn+1) where θi :=      θ(0) if i = 0; θ(i) −θ(i −1) if 1 ≤i ≤n; m −θ(n) if i = n + 1. The map marked α is a composite of the associativity maps of the normal oplax monoidal structure on E, and that marked (rθ0, . . . , lθn+1) is built from the monoid and module action maps for R, M and L. Theorem 4.2. Let E be a normal oplax monoidal category with monoid M, left M-module L and right M-module R. Deﬁnition 4.1 then determines a simplicial object B•(R, M, L) in E which we refer to as the simplicial bar construction for R, M, L. Proof. We must show that for θ : [n] →[m] and ζ : [m] →[l] we have (ζθ)∗= θ∗ζ∗. This amounts to the commutativity of the following diagram (notation condensed to ﬁt): µl+2(R,...,L) µn+2(µ(ζθ)0(R,... ),...,µ(ζθ)n+1(...,L)) µm+2(µζ0(R,...,),...,...,µζm+1(...,L)) µn+2(µθ0(µζ0(R,... ),... ),...,µθn+1(...,µζm+1(...,L))) µn+2(R,...,L) µm+2(R,...,L) µn+2(µθ0(R,... ),...,µθn+1(...,L)) / α α + (r(ζθ)0,...,l(ζθ)n+1) α / α (rζ0,...,lζm+1) (rζ0,...,lζm+1) / α 3 (rθ0,...,lθn+1) in which maps marked α are composites of the associativity maps for the normal oplax monoidal structure, and those marked (r, . . . , l) involve the module and monoid structure COMPOSITION PRODUCT 15 maps for L, R and M. The top-left square commutes by combining copies of the axioms of Deﬁnition 1.1, the bottom-left is a naturality square, and the quadrilateral on the right commutes by combining copies of the axioms of Deﬁnitions 3.6 and 3.1. □ Example 4.3. When the normal oplax monoidal structure on E arises from an actual monoidal product ⊗on E as in Example 1.2, the bar construction B•(R, M, L) is given by Bn(R, M, L) = R ⊗M ⊗· · · ⊗M \| {z } n ⊗L with the usual simplicial structure. Example 4.4. Now let CΣ be the category of symmetric sequences on a symmetric monoidal category C, with the normal oplax monoidal structure given by Theorem 2.16. Then the sim-plicial bar construction described in this section is the ‘standard’ simplicial bar construction on an operad P with respect to a left P-module L and right P-module R. See, for example [2, 7.9]. If we replace C with its opposite category Cop, together with the canonical symmetric monoidal structure, the simplicial bar construction on an operad in Cop gives us a cosimplicial cobar construction C•(R, Q, L) on a cooperad Q in C with respect to a right Q-comodule R and left Q-comodule L. Explicitly: Cn(R, Q, L)(J) := lim f∈J/FinSet0[n]op(R, Q, . . . , Q \| {z } n , L)(f). References 1. Michael Batanin and Mark Weber, Algebras of higher operads as enriched categories, Applied Categorical Structures 19 (2011), no. 1, 93–135. 2. Michael Ching, Bar constructions for topological operads and the Goodwillie derivatives of the identity, Geom. Topol. 9 (2005), 833–933 (electronic). MR MR2140994 3. Brian Day and Ross Street, Lax monoids, pseudo-operads, and convolution, Diagrammatic morphisms and applications (San Francisco, CA, 2000), Contemp. Math., vol. 318, Amer. Math. Soc., Providence, RI, 2003, pp. 75–96. MR 1973511 (2004c:18011) 4. Tom Leinster, Higher operads, higher categories, London Mathematical Society Lecture Note Series, vol. 298, Cambridge University Press, Cambridge, 2004. MR 2094071 (2005h:18030) 5. Martin Markl, Steve Shnider, and Jim Stasheﬀ, Operads in algebra, topology and physics, Mathematical Surveys and Monographs, vol. 96, American Mathematical Society, Providence, RI, 2002. MR 2003f:18011 6. V. A. Smirnov, On the cochain complex of topological spaces, Math. USSR Sbornik 43 (1982), 133–144. Department of Mathematics, Amherst College, Amherst, MA 01002 E-mail address: mching@amherst.edu
6290	https://jscholarship.library.jhu.edu/bitstream/handle/1774.2/34173/31151006451797.pdf	FROM THE LIBRARY OF FRANK MORLEY, 1860-1937 P i t o i ' E S S O i i OP Mathematics in this University. 1900-1929; Emeritiw, 1029-1937 T H E GIFT O F MRS. MORLEY A TEXT-BOOK E U C L I D ' S E L E M E N T S . A T E X T - B O O K OF EUCLID'S ELEMENTS FOE THE USE OF SCHOOLS PARTS I. and II. CONTAINING BOOKS I.—VI. BY H. S.V'HALL, M.A. FORMERLY SCHOLAR OF CHRIST S COLLESE, CAMBRIDGE ; AND F. H. STEVENS, M.A. FORMERLY SCHOLAR OF QUEEN'S COLLEGE, OXFORD ; MASTERS OP THE MILITARY AND ENGINEERING SIDE, CLIFTON COLLEGE. Uontron: M A C M I L L A N A N D CO. AND NEW TOEK. 1888 [All Bights reserved.] K m . (EambriUgt: PRINTED BY 0. J . CLAY, M.A. & SONS, AT THE UNIVERSITY PRESS. 6IFT OF MRS. FRANK M9RbB^ P E E F A C E . This volume contains the first Six Books of Euclid's Elements, together with Appendices giving the mos im-portant elementary developments o{ Euclidean Geometry. The text has been carefully revised, and special atten-tion given to those points which experience has shewn to present difficulties to beginners. In the course of this revision the Enunciations have been altered as little as possible; and, except in Book V., very few departures have been made from Euclid's proofs: in each case changes have been adopted only where the old text has been generally found a cause of difficulty; and such changes are for the most par in favour of well-recog-nised alternatives. For example, the ambiguity has been removed from the Enunciations of Propositions 18 and 19 of Book I.: the fact that Propositions 8 and 26 establish the complete identical equality of the two triangles considered has been strongly urged; and thus the redundant step has been removed from Proposition 34. In Book II. Simson's ar-rangement of Proposition 13 has been abandoned for a well-known alternative proof. In Book III. Proposition 25 is not given at length, and its place is taken by a simple equivalent. Propositions 35 and 36 have been treated generally, and it has not been thought necessary to do more than call attention in a note to the special eases. Finally, in Book VI. we have adopted an alterna-tive proo{ of Proposition 7, a theorem which has been too much neglected, owing to the cumbrous form in which it has been usually given. These are the chief deviations from the ordinary text as regards method and arrangement of proof: they are points familiar as difficulties to most teachers, and to name them indicates sufficiently, without further enumeration, the general principles which have guided our revision. A few alternative proofs of difficult propositions are given for the convenience of those teachers who care to use them. With regard to Book V. we have established the princi-pal propositions, both from the algebraical and geometrical definitions of ratio and proportion, and w e have endeavoured to bring out clearly the distinction between these two modes of treatment. In compiling the geometrical section of Book V . we have followed the system first advocated by the late Pro-fessor D e Morgan; and here we derived very material assistance from the exposition of the subject given in the text-book of the Association for the Improvement of Geo-metrical Teaching. To this source we are indebted for the improved and more precise wording of definitions (as given on pages 286, 288 to 291), as well as for the order and substance of most of the propositions which appear between pages 297 and 306. But as we have not (except in the points above mentioned) adhered verbally to the text of the Association, we are anxious, while expressing in the fullest manner our obligation to their work, to exempt the PREFACE. V l l .Association from all responsibility for our treatment of the subject. One purpose of the book is to gradually familiarise the student with the use of legitimate symbols and abbrevia-tions; for a geometrical argument may thus be thrown into a form which is not only more readily seized by an advanced reader, but is useful as a.guide to the way in which Euclid's propositions may be handled in written work. O n the other hand, we think it very desirable to defer the intro-duction of symbols until the beginner has learnt that they can only be properly used in Pure Geometry as abbrevia-tions for verbal argument: and we hope thus to prevent the slovenly and inaccurate habits which are very apt to arise from their employment before this principle is fully recognised. Accordingly in Book I. we have used no contractions or symbols of any kind, though we have introduced verbal alterations into the text wherever it appeared that con-ciseness or clearness would be gained. In Book II. abbreviated forms of constantly recurring words are used, and the phrases therefore and is equal to are replaced by the usual symbols, In the Third and following Books, and in additional matter throughout the whole, we have employed all such signs and abbreviations as we believe to add to the clear-ness of the reasoning, care being taken that the symbols chosen are compatible with a rigorous geometrical method, and are recognised by the majority of teachers. It must be understood that our use of symbols, and the removal of unnecessary verbiage and repetition, by no means implies a desire to secure brevity at all hazards. O n the contrary, nothing appears to us more mischievous than an abridgement which is attained by omitting VU1 PREFACE. steps, or condensing two or more steps into one. Such uses spring from the pressure of examinations; but an examination is not, or ought not to be, a mere race; and while we wish to indicate generally in the later books how a geometrical argument may be abbreviated for the pur-poses of written work, we have not thought well to reduce the propositions to the bare skeleton so often presented to an Examiner. Indeed it does not follow that the form most suitable for the page of a text-book is also best adapted to examination purposes; for the object to be attained in each case is entirely different. The text-book should present the argument in the clearest possible manner to the mind of a reader to w h o m it is new: the written proposition need only convey to the Examiner the assurance that the proposition has been thoroughly grasped and remembered by the pupil. From first to last we have kept in mind the undoubted fact that a very small proportion of those who study Ele-mentary Geometry, and study it with profit, are destined to become mathematicians in any real sense; and that to a large majority o{ students, Euclid is intended to serve not so much as a first lesson in mathematical reasoning, as the first, and sometimes the only, model of formal and rigid argument presented in an elementary education. This consideration has determined not only the full treatment of the earlier Books, but the retention of the formal, i f somewhat cumbrous, methods of Euclid in many places where proofs of greater brevity and mathematical elegance are available. W e hope that the additional matter introduced into tho book will provide sufficient exercise for pupils whose study of Euclid is preliminary to a mathematical edu-cation. PREFACE. IX The questions distributed through the text follow very easily from the propositions to which they are attached, and we think that teachers are likely to find in them all that is needed for an average pupil reading the subject for the first time. The Theorems and Examples at the end of each Book contain questions of a slightly more difficult type : they have been very carefully classified and arranged, and brought into close connection with typical examples worked out either partially or in full; and it is hoped that this section o{ the book, on which much thought has been expended, will do something towards removing that extreme want of freedom in solving deductions that is so commonly found even among students who have a good knowledge of the text of Euclid. In the course of our work we have made ourselves acquainted with most modern English books on Euclidean Geometry: among these we have already expressed our special indebtedness to the text-book recently published by the Association for the Improvement of Geometrical Teach-ing; and we must also mention the Edition of Euclid's Ele-ments prepared by M r J. S. Mackay, whose historical notes and frequent references to original authorities have been of the utmost service to us. Our treatment of Maxima and Minima on page 239 is based upon suggestions derived from a discussion of the subject which took place at the annual meeting of the Geometrical Association in January 1887. Of the Riders and Deductions some are original; but the greater part have been drawn from that large store of floating material which has furnished Examination Papers for the last 30 years, and must necessarily form the basis of any elementary collection. Proofs which have been X PREFACE. found in two or more books without acknowledgement have been regarded as common property. As regards figures, in accordance with a usage not uncommon in recent editions of Euclid, w e have made a distinction between given lines and lines of construction. Throughout the book we have italicised those deductions on which w e desired to lay special stress as being in them-selves important geometrical results: this arrangement we think will be useful to teachers who have little time to devote to riders, or who wish to sketch out a suitable course for revision. W e have in conclusion to tender our thanks to many of our friends for the valuable criticism and advice which we received from them as the book was passing through the press, and especially to the Rev. H. C. Watson, o{ Clifton College, who added to these services much kind assistance in the revision of proof-sheets. H. S. HALL, F. H. STEVEN. July, 1888. C O N T E N T S . BOOK I . PACE D e f i n i t i o n s , Postulates, Axioms 1 S e c t i o n I . P r o p o s i t i o n s 1—26 1 1 S e c t i o n I I . Parallels and P a r a l l e l o g r a m s . P r o p o s i t i o n s 27—34 50 S e c t i o n in. The Areas of Parallelograms a n d T r i a n g l e s . P r o p o s i t i o n s 35—48 6 6 Theorems and Examples on Booh I. A n a l y s i s , S y n t h e s i s 87 I . On the I d e n t i c a l Equality o f T r i a n g l e s 90 H. On I n e q u a l i t i e s 93 I I I . On Parallels 95 IV. On Parallelograms 9 6 V. Miscellaneous Theorems and Examples 1 0 0 VI. On the Concurrence o f S t r a i g h t Lines i n a T r i -angle 1 0 2 V I E . On the C o n s t r u c t i o n o f Triangles with g i v e n Parts 1 0 7 VIH. On Areas 1 0 9 IX. On L o c i 1 1 4 X. On the I n t e r s e c t i o n o f Loci 1 1 7 X l l CONTENTS. BOOK IL PAGE Definitions, &o. 120 Propositions 1—14 . 122 Theorems and Examples on Book II. 144 BOOK in. Definitions, &a. 149 Propositions 1—37 153 Note on the Method of Limits as Applied to Tangency 213 Theorems and Examples on Book III. I. On the Centre and Chords op a Circle 215 II. On the Tangent and the Contact of Circles. The Common Tangent t o Two Circles, Problems on Tangency, Orthogonal C i r c l e s . 217 III. On Angles in Segments, and Angles at the Centres and Circumferences of Circles. The Orthocentre of a Triangle, and properties of the Pedal Triangle, Loci, Simson's Line 222 IV. On the Circle in Connection with Bectangles. Further Problems on Tangency 233 V. On Maxima and Minima . 239 VI. Harder Miscellaneous Examples 216 BOOK IV. Definitions, &o. 250 Propositions 1—16 251 Note on Begular Polygons 274 Theorems and Examples on Book IV. I . On the Triangle and its Circles. Circumsoribed, Inscribed, and Escribed Circles, The Nine-points C i r o l e 277 II. Miscellaneous Examples 2S3 CONTENTS. X I I I BOOK V. P A G E I n t r o d u c t o r y 2 8 5 D e f i n i t i o n s 2 8 6 Summary, with A l g e b r a i c a l Proofs, o f the P r i n c i p a l Theorems of Book V. 292 Proofs o f the P r o p o s i t i o n s d e r i v e d from the Geometrical D e f i n i t i o n o f P r o p o r t i o n 297 BOOK VI Definitions 307 P r o p o s i t i o n s 1—D. 308 Theorems and Examples on Book VI. L On Harmonic Section 359 IX On Centres o f S i m i l a r i t y and S i m i l i t u d e 363 HL, On Pole and Polar . 365 IV. On the Badioal Axis o f Two or More C i r c l e s 3 7 1 V. On T r a n s v e r s a l s 3 7 4 VI. Miscellaneous Examples on Book VI. 3 7 7 E U C L I D ' S E L E M E N T S . B O O K I . Definitions. 1. A point is that which has position, but no mag-nitude. 2. A line is that which has length without breadth. The extremities of a line are points, and. the intersection of two lines is a point. 3. A straight line is that which lies evenly between its extreme points. Any portion cut off from a straight line is called a segment of i t . 4. A surface is that which has length and breadth, but no thickness. The boundaries of a surface are lines. 5. A- plane surface is one in which any two points being taken, the straight line between them lies wholly in that surface. A plane surface is frequently referred to simply as a plane. Note. Euclid regards a point merely as a mark of position, and he therefore attaches to i t no idea of size and shape. Similarly he considers that the properties of a line arise only from its length and position, without reference to that minute breadth which every line must really have if actually drawn, even though the most perfect instruments are used. The definition of a surface is to be understood in a similar .way. H. E. 1 10 EUCLID'S ELEMENTS. 6. A plane angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. The point at which the straight lines meet is called the vertex of the angle, and the straight lines themselves the arms of the angle. When several angles are at one point O, any one of them is expressed by three letters, of which the letter that refers to the vertex is put between the other two. Thus if the straight lines OA, OB, O C meet at the point O, the angle contained by the straight lines OA, O B is named the angle A O B or B O A ; and the angle contained by OA, O C is named the angle A O C or COA. Similarly the angle con-tained by OB, O C is referred to as the angle B O C or C O B . But if there be only one angle at a point, i t may be expressed by a single letter, as the angle atO. Of the two straight lines OB, OC shewn in the adjoining figure, we recognize that O C is more in-clined than O B to the straight line O A : this we express by saying that the angle A O C is greater than the angle A O B . Thus an angle must be regarded as having magnitude. It should be observed that the angle AOC is the sum of the angles A O B and B O C ; and that A O B is the difference of the angles A O C and BOC. The beginner is cautioned against supposing that the size of an angle is altered either by increasing or diminishing the length of i t s arms. [Another view of an angle is recognized in many branches of mathematics; and though not employed by Euclid, i t is here given because i t furnishes more clearly than any other a conception of what is meant by the magnitude of an angle. Suppose that the straight line O P in the figure /q is capable of revolution about the point O, like the / . „ hand of a watch, but in the opposite direction; and suppose that in this way it has passed successively from the position O A to the positions ocoupied bv O B and O C . Such a line must have undergone more turning ° " in passing from O A to O C , than in passing from O A to O B ; and consequently tho angle A O C is said to be greater than the angle AOB.] definitions. 7. W h e n a straight line standing on another straight line makes the adjacent angles equal to one another, each of the an-gles is called a right angle; and the straight line which stands on the other is called a perpendicular to it. 8. An obtuse angle is an angle which is greater than one right angle, but less than two right angles. 9. An acute angle is an angle which is less than a right angle. [In the adjoining figure the straight line O B may be supposed to have arrived at i t s present position, from the position occu-pied by OA, by revolution about the point O in either of the two directions indicated by the arrows: thus two straight lines drawn from a point may be considered as forming two angles, (marked ( i ) and ( i i ) in the figure) of which the greater ( i i ) is said to be reflex. If the arms OA, O B are in the same ( i straight line, the angle formed by them B O A on either side is called a straight angle.] 10. Any portion of a plane surface bounded by one or more lines, straight or curved, is called a plane figure. The sum of the bounding lines is called the perimeter of the figure. Two figures are said to be equal in area, when they enclose equal portions of a plane surface. 11. A circle is a plane figure contained by one line, which is called the circum-ference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : this point is called the centre of the circle. A radius of a circle is a straight line drawn from the centre to the circumference. 1—2 4 EUCLID'S ELEMENTS. 12. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 13. A semicircle is the figure bounded by a diameter of a circle and the part of the circumference cut off by the diameter. 14. A segment of a circle is the figure bounded by a straight line and the part of the circumference which it cuts off. 15. Rectilineal figures are those which are bounded by straight lines. 16. A triangle is a plane figure bounded by three straight lines. Any one of the angular points of a triangle may be regarded as i t s vertex; and the opposite side is then called the base. 17. A quadrilateral is a plane figure bounded by four straight lines. The straight line which joins opposite angular points in a quadri-lateral is called a diagonal. 18. A polygon is a plane figure bounded by more than four straight lines. 19. A n equilateral triangle is a triangL whose three sides are equal. 20. A n isosceles triangle is a triangle two of whose sides are equal. 21. A scalene triangle is a triangle which ••y7 has three unequal sides. DEFINITIONS. 22. A right-angled triangle is a triangle which has a right angle. The side opposite to the right angle in a right-angled triangle is called the hypotenuse. 23. A n obtuse-angled triangle is a triangle which has an obtuse angle. 24. A n acute-angled triangle is a triangle which has three acute angles. [It will be seen hereafter (Book I. Proposition 17) that every triangle must have at least two acute angles.] 25. Parallel straight lines are such as, being in the same plane, do not meet, however far they are produced in either direction. 26. A Parallelogram is a four-sided figure which has its opposite sides pa-rallel. 27. A rectangle is a parallelogram which has one of its angles a right angle. 28. A square is a four-sided figure which has all its sides equal and one of its angles a right angle. [ I t may easily be shewn that i f a quadrilateral has all its sides equal and one angle a right angle, then all its angles will be right angles.] 29. A rhombus is a four-sided figure which has all its sides equal, but its angles are not right angles. 30. A trapezium is a four-sided figure F which has two of its sides parallel. / EUCLID'S ELEMENTS. ON THE POSTULATES. In order to effect the constructions necessary to the study of geometry, it must be supposed that certain instruments are available; but it has always been held that such instruments should be as few in number, and as simple in character as possible. For the purposes of the first Six Books a straiglu ruler and a pair of compasses are all that are needed; and in the follow-ing Postulates, or requests, Euclid demands the use of such instruments, and assumes that they suffice, theoretically as well as practically, to cany out the processes mentioned below. Postulates. Let it be granted, 1. That a straight line may be drawn from any one point to any other point. When we draw a straight line from the point A to the point B, we are said to join A B. 2. That a. fittid', that is to say, a terminated straight line m a y be produced to any length in that straight line. .'!. That a circle may be described from any centre, at any distance from that centre, that is, with a radius equal to any finite straight line drawn from the centre. It is important to notice that the Postulates include no means of direct measurement: hence the straight ruler is not supposed to be graduated; and the compasses, in accordance with Euclid's use are not to be employed for transferring distances from one part of a figure to another. o\ THE axioms. The science of Geometry is based upon certain simple state-ments, the truth of which is assumed at the outset to be self-evident. These self-evident truths, called by Euclid Common Xotiom are now known as the Axioms. GENERAL AXIOMS. 7 The necessary characteristics of an Axiom are ( i ) That i t should be self-evident; that is, that i t s truth should be immediately accepted without proof. ( i i ) That i t should be fundamental; that is, that i t s truth should not be derivable from any other truth more simple than i t s e l f . ( i i i ) That i t should supply a basis for the establishment . o f further truths. These characteristics may be summed up in the following definition. Definition. A n Axiom is a self-evident truth, which neither requires nor is capable of proof, but which serves as a founda-tion for future reasoning. Axioms are of two kinds, general and geometrical. General Axioms apply to magnitudes of all kinds. Geometri-cal Axioms refer exclusively to geometrical magnitudes, such as have been already indicated in the definitions. General Axioms. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be taken from equals, the remainders are equal. 4. If equals be added to unequals, the wholes are un-equal, the greater sum being that which includes the greater of the unequals. 5. If equals be taken from unequals, the remainders are unequal, the greater remainder being that which is left from the greater of the unequals. 6. Things which are double of the same thing, or of equal things, are equal to one another. 7. Things which are halves of the same thing, or of equal things, are equal to one another. 9. The whole is greater than its part. To preserve the classification of general and geometrical we have placed Euclid's ninth axiom before the eighth. EUCLID'S ELEMENTS. Geometrical Axioms. 8. Magnitudes which can be made to coincide with one another, are equal. This axiom affords the ultimate test of the equality of two geome-trical magnitudes. It implies that any line, angle, or figure, may be Bupposed to be taken up from its position, and without change in size or form, laid down upon a second line, angle, or figure, for the purpose of comparison. This process is called superposition, and the first magnitude is said to be applied to the other. 10. Two straight lines cannot enclose a space. 11. All right angles are equal. [The statement that all right angles are equal, admits of proof. and is therefore perhaps out of place as an Axiom.] 12. If a straight line meet two straight lines so as to m a k e the interior angles on one side of it together less than two right angles, these straight lines will meet if con-tinually produced on the side on which are the angles which are together less than two right angles. That is to say, if the two straight lines A B and C D are met by the straight line E H at F and G, in such a way that the angles BFG, D G F are together less than two right angles, it is asserted that AB and C D will meet if continually pro-duced in the direction of B and D. [Axiom 12 has been objected to on the double ground that it cannot be considered sclf-evident, and that its truth may be d e d u c e d T Z simpler principles. It is employed for the first tune in the -lth PrrT position of Book I., where a short discussion of the difficult^ will K, found. • u D c The converse of this Axiom is proved in Book I. Proposition 17 l INTRODUCTORY. IXTEODTJCTOBY. Plane Geometry deals with the properties of all lines and figures that m a y be drawn upon a plane surface. Euclid in his first Six Books confines himself to the properties of straight lines, rectilineal figures, and circles. The Definitions indicate the subject-matter of these books: the Postulates and Axioms lay down the fundamental principles which regulate all investigation and argument relating to this subject-matter. Euclid's method of exposition divides the subject into a number of separate discussions, called propositions; each pro-position, though in one sense complete in itself, is derived from results previously obtained, and itself leads up to subsequent propositions. Propositions are of two kinds, Problems and Theorems. A Problem proposes to effect some geometrical construction, such as to draw some particular line, or to construct some re-quired figure. A Theorem proposes to demonstrate some geometrical truth. A Proposition consists'of the following parts: The General Enunciation, the Particular Enunciation, the Construction, and the Demonstration or Proof. (i) The General Enunciation is a preliminary statement, describing in general terms' the purpose of the proposition. In a problem the Enunciation states the construction which it is proposed to effect:- it therefore names first the Data, or things given, secondly the Quaasita, or things required. In a theorem the Enunciation states the property which it is proposed to demonstrate: it names first, the Hypothesis, or the conditions assumed; secondly, the Conclusion, or the asser-tion to be proved. 10 euclid's elements. (ii) The Particular Enunciation repeats in special terms the statement already made, and refers it to a diagram, which enables the reader to follow the reasoning more easily. (iii) The Construction then directs the drawing of such straight lines and circles as may be required to effect the purpose of a problem, or to prove the truth of a theorem. (iv) Lastly, the Demonstration proves that the object pro-posed in a problem has been accomplished, or that the property stated in a theorem is true. Euclid's reasoning is said to be Deductive, because by a con-nected chain of argument it deduces new truths from truths already proved or admitted. The initial letters q.e.e., placed at the end of a problem, stand for Quod erat Faciendum, which was to be done. The letters Q. E. d. are appended to a theorem, and stand for Quod erat Demonstrandum, which was to be proved. A Corollary is a statement the truth of which follows readily from an established proposition; it is therefore appended to the proposition as an inference or deduction, which usually requires no further proof. The following symbols and abbreviations may be employed in writing out the propositions of Book I . , though their use i s not recommended to beginners. .-. for therefore, par1 (or \|;) for parallel, „ is, or are, equal to, par" parallelogram, L „ angle, sq. , , square, it. • - „ right angle, rectil. „ rectilineal, A „ triangle, st. line „ straight line, perp. , , perpendicular, pt. point: and all obvious contractions of words, such as opp., adj., diag., &c, for opposite, adjacent, diagonal, &e. BOOK 1 . PROP. 1. 1 1 SECTION I . Proposition 1. Problem. To describe an equilatercd triangle on a given finite straight line. -A. Let AB be the given straight line. It is required to describe an equilateral triangle on AB. Construction. Prom centre A, with radius AB, describe the circle BCD. Post. 3. From centre B, with radius BA, describe the circle ACE. Post. 3. From the point C at which the circles cut one another, draw the straight lines CA and CB to the points A and B. Post. 1. Then shall ABC be an equilateral triangle. Proof. Because A is the centre of the circle BCD, therefore AC is equal to AB. Def. 11. A n d because B is the centre of the circle ACE, therefore BC is equal to BA. Def. 11. But it has been shewn that AC is equal to AB; therefore AC and BC are each equal to AB. But things which are equal to the same thing are equal to one another. Ax. 1. Therefore AC is equal to BC. Therefore CA, AB, BC are equal to one another. Therefore the triangle ABC is equilateral; and it is described on the given straight line AB. q.e.p. i s e u c l i d ' s elements. Proposition 2. Problem. Frmii, a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw from the point A a straight line equal to BC. Construction. Join A B ; Post. 1. and on AB describe an equilateral triangle DAB. I . 1. From centre B, with radius BC, describe the circle CGH. Punt. 3. Produce DB to meet the circle C G H at G. Post. 2. From centre D, with radius DG, describe the circle GKF. Produce DA to meet the circle G K F at F. Post. 2. Then AF shall be equal to BC. Proof. Because B is the centre of the circle CGH, therefore BC is equal to BG. Def. 11. A n d because D is the centre of the circle GKF, therefore DF is equal to DG ; Def. 11. and DA, DB, parts of them are equal; Def. 19. therefore the remainder AF is equal to the remainder BG. Ax. 3. A n d it has been shown that BC is equal to BG ; therefore AF and BC are each equal to BG. But things which are equal to the same thing are equal to one another. j.r. ]_ Therefore AF is equal to B C ; and it has been drawn from the given point A. c j . e. F. [This Proposition i s rendered necessary by the restriction, t a e i t l v imposed by Euclid, that oompasses shall not be used to -t r a n s f e r diitnnct's.] BOOK I . PROP. 3 . 13 Proposition 3 . Problem. From the greater of two given straight lines to cut off" a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB a part equal to C. Construction. From the point A draw the straight line A D equal t o C ; I . 2. and from centre A, with radius AD, describe the circle DEF, meeting AB at E. Post. 3. Then A E shall be equal to C. Proof. Because A is the centre of the circle DEF, therefore A E is equal to AD. Def. 11. But C is equal to AD. Constr. Therefore A E and C are each equal to AD. Therefore A E is equal to C ; and it has been cut off from the given straight line AB. Q.e.p. exercises. 1. On a given straight line describe an isosceles triangle havin each of the equal sides equal to a given straight line. 2. On a given base describe an isosceles triangle haying each of the equal sides double of the base. 3. In the figure of i . 2, i f AB is equal to BC, shew that D, the vertex of the equilateral triangle, will f a l l on the circumference of the circle C G H . 14 Euclid's elements. Obs. Every triangle has six parts, namely its three sides and three angles. Two triangles are said to be equal in all respects, when they can be made to coincide with one another by superposition (see note on Axiom 8), and in this case each part of the one is equal to a corresponding part of the other. Proposition 4. Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also t / t e angles contained by those sides equal; then shall i h ' -i r bases or third sides be equal, and the triangles shall be equal in area, and their remaining angles shall be equal, each to each, namely those to which the equal sides are opposite : thnt is to say, the triangles shall be equal in all respects. D c E Let ABC, DEF be two triangles, which have the side AB equal to the side DE, the side AC equal to the side DF. and the contained angle BAC equal to the contained anide EOF. Then shall the base BC be equal to the base EF, and the triangle ABC shall be equal to the triangle DEF in area: and the remaining angles shall be equal, each to each, to which the equal sides are opposite, namely the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB along the straight line DE then because AB is equal to DE, H i m therefore the point B must coincide with the point E. BOOK I . PROP. 4. 15 And because AB falls along DE, and the angle BAC is equal to the angle EDF, Hyp. therefore A C must fall along DF. A n d because A C is equal to DF, Hyp. therefore the point C must coincide with the point F. Then B coinciding with E, and C with F, the base BC must coincide with the base EF; for if not, two straight lines would enclose a space; which is impossible. Ax. 10. Thus the base BC coincides with the base EF, and is therefore equal to it. Ax. 8. A n d the triangle A B C coincides with the triangle DEF, and is therefore equal to it in area. Ax. 8. A n d the remaining angles of the one coincide with the re-maining angles of the other, and are therefore equal to them, namely, the angle A B C to the angle DEF, and the angle A C B to the angle DFE. That is, the-triangles are equal in all respects. q.e.d. Note. It follows that two triangles which are equal in their several parts are equal also in area; but i t should be observed that equality of area in two triangles does not necessarily imply equality in their several parts: that is to say, triangles may be equal in area, without being of the same shape. Two triangles which are equal in all respects have identity of fo and magnitude, and are therefore said to be identically equal, or congruent. The following application of Proposition 4 anticipates the chief difficulty of Proposition 5. In the equal sides AB, AC of an isosceles triangle ABC, the points X and Y are taken, so that A X i s equal to AY; and BY and C X are joined. .Shew that BY is equal to CX. In the two triangles XAC, YAB, X A is equal to YA, and A C is equal to AB; Hyp. that is, the two sides XA, A C are equal to the two sides YA, AB, each to each; and the angle at A, which is contained by these sides, is common to both triangles: therefore the triangles are equal in a l l respects; bo that X C is equal to YB. I B E u c l i d ' s elements. Proposition 5 . Theorem. The angles at the bane of an isosceles triangle are equal to one another; and if t l i e equal sides be produced, the angles on the other side of the base shall also be equal to one another. Let ABC be an isosceles triangle, having the side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E : then shall the angle ABC be equal to the angle ACB, and the angle CBD to the angle BCE. Construction. In BD take any point F; and from AE the greater cut off AG equal to AF the less. I . 3. Join FC, GB. Proof. Tnen in the triangles FAC, GAB, { F A is equal to GA, Conttr. and AC is equal to AB, Hyp. also the contained angle at A is common to the two triangles; therefore the triangle FAC is equal to the triangle GAB in a l l respects ; I _ 4 that is, the base FC is equal to the base GB, and the angle ACF is equal to the angle ABG, also the angle AFC i s equal to the angle AGB, Again, because the whole AF i s equal to the whole A G of which the parts AB, AC are equal, Hup. therefore the remainder BF is equal to the remainder CG. BOOK I. PROP. 5. 17 Then in the two triangles BFC, CGB, I B F is equal to CG, Proved. and FC is equal to G B, Proved. also the contained angle BFC is equal to the contained angle CGB, Proved. therefore the triangles BFC, C G B are equal in all respects; so that the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. I. 4. Now it has been shewn that the whole angle ABG is equal to the whole angle ACF, and that parts of these, namely the angles CBG, BCF, are also equal; therefore the remaining angle A B C is equal to the remain-ing angle A C B ; and these are the angles at the base of the triangle ABC. Also it has been shewn that the angle FBC is equal to the angle G C B ; and these are the angles on the other side of the base, q.e.d. Corollary. Hence if a triangle is equilateral it is also equiangular. Definition. Each of two Theorems is said to be the Con-verse of the other, when the hypothesis of each is the conclusion of the other. It will be seen, on comparing the hypotheses and conclusions of Props. 5 and 6, that each proposition i s the converse of the other. Note. Proposition 6 furnishes the first instance of an indirect method of proof, frequently used by Euclid. It consists in shewing that an absurdity must result from supposing the theorem to be otherwise than true. This form of demonstration is known as the Eeductio ad Absurdum, and is most commonly employed in establish-ing the converse of some foregoing theorem. It must not be supposed that the converse of a true theorem is i t s e l f necessarily true: for instance, i t will be seen from Prop. 8, Cor. that i f two triangles have their sides equal, each to each, then their angles will also be equal, each to each; but i t may easily be shewn by means of a figure that the converse of this theorem i s not necessarily true. H. E. 1 8 e u c l i d ' s e l e m e n t s . P r o p o s i t i o n 6 . Theorem. If two angles of a triangle be equal to one another, tlien ths sides also which subtend, or are opposite to, t f c e equal angles, shall be equal to one another. Let ABC be a triangle, having the angle ABC equal to the angle ACB : then shall the side AC be equal to the side AB. Construction. For if AC be not equal to AB, one of them must be greater than the other. If possible, let AB be the greater; and from it cut off BD equal to AC. I . 3. Join DC. Then in the triangles DBC, ACB, DB is equal to AC, Constr. and BC is common to both, \| also the contained angle D B C is equal to the contained angle ACB ; Uyp. therefore the triangle DBC is equal in area to the triangle ACB, 1.1. the part equal to the whole; which is absurd. J . i \ 9 . Therefore AB is not unequal to AC ; that is, AB is equal to AC. Q.E.D. Corollary. Hence if a triangle is equiangular i t is also equilateral. book i . p r o p . 7 . 1 9 Proposition 7. Theorem. On the same base, and on the same side of it, there cannot be two triangles having their sides which are termi-nated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another. If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB, having their sides AC, AD, which are terminated at A, equal to one another, and likewise their sides BC, BD, which are termi-nated at B, equal to one another. Case I. When the vertex of each triangle is without the other triangle. Construction. Join CD. Post. 1. Proof. Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angle ACD is equal to the angle ADC. I. 5. But the whole angle ACD is greater than its part, the angle BCD, therefore also the angle A D C is greater than the angle BCD; s t i l l more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, because BC is equal to BD, Hyp, therefore the angle BDC is equal to the angle BCD: I . 5. but it was shewn to be greater; which is impossible. 2—2 20 EUCLID'S ELEMENTS. Case II. When one of the vertices, as D, is within the other triangle ACB. E/ C/ /F Construction. A s before, join C D ; Post. 1. and produce AC, AD to E and F. Post. 2. Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angles ECD, FDC, on the other side of the base, are equal to one another. I . 5. But the angle ECD is greater than its part, the angle BCD; therefore the angle FDC is also greater than the angle BCD: s t i l l more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, because BC is equal to BD, Hyp. therefore the angle BDC is equal to the angle BCD : I . 5. but i t has been shewn to be greater; which is impossible The case in which the vertex of one triangle is on a side of the other needs no demonstration. Therefore AC cannot be equal to AD, and at the same time, BC equal to BD. Q.E.D. Note. The sides AC, AD are c a l l e d conterminous sides; s i m i l a r l y the sides BC, BD are conterminous. Proposition 8. Theorem. If two triangles have, two sides of tlie one equal to two sides of the other, each to each, and have likewise their bases equal, then the angle which is contained by the tioo s i < l v s or the one shall be equal to t h e , angle which i s contained bn the two sides of the other. book i . prop. 8. 21 A D G Let ABC, DEF be two triangles, having the two sides BA, AC equal to the two sides ED, DF, each to each, namely BA to ED, and AC to DF, and also the base BC equal to the base EF: then shall the angle BAC be equal to the angle EDF. Proof. For if the triangle ABC be applied to the triangle DEF, so that the point B may be on E, and the straight line BC along EF; then because BC is equal to EF, Hyp. therefore the point C must coincide with the point F. Then, BC coinciding with EF, it follows that BA and AC must coincide with ED and DF : for i f not, they would have a different situation, as EG, G F: then, on the same base and on the same side of it there would be two triangles having their conterminous sides equal. But this is impossible. i . 7. Therefore the sides BA, AC coincide with the sides ED, DF. That is, the angle BAC coincides with the angle EDF, and is therefore equal to it. Ax. 8. Note. In this Proposition the three sides of one triangle are given equal respectively to the three sides of the other; and from t h i s i t i s shewn that the two triangles may be made t o coincide with one another. Hence we are led to the following important Corollary. Corollary. If in two triangles the three sides of the one are equal to the three sides of the o i l i e r , each to each, then the triangles are equal in all respects. 22 Euclid's elements. The following proof of Prop. 8 is worthy of attention as it is ind pendent of Prop. 7, which frequently presents d i f f i c u l t y to a beginner. Proposition 8. Alternative Proof. A p Let A B C and DEF be two triangles, which have the sides BA, AC equal respectively to the sides ED, DF, and the base BC equal to the base EF: then shall the angle B A C be equal to the angle EDF. For apply the triangle A B C to the triangle DEF, so that B may f a l l on E, and BC along EF, and so that the point A may be on the side of EF remote from D, then C must f a l l on F, Bince BC i s equal to EF. Let A'EF be the new position of the triangle ABC. If neither DF, FA' nor DE, EA' are in one straight line, join DA'. Cake I. "When DA' intersects EF. Then because ED is equal to EA', therefore the angle EDA' is equal to the angle EA'D. i . 5. Again because FD i s equal to FA', therefore the angle FDA' is equal to the angle FA'D. i . 5. Henoe the whole angle EDF i s equal to the whole angle EA'F; that is, the angle EDF is equal to the angle BAC. Two cases remain which may be dealt with in a similar manner: namely, Case II. When DA' meets EF produced. Case III. When one pair of sides, as DF, FA', are in one straight lino. book i . prop. 9 . 2 3 , Proposition 9. Problem. To bisect a given angle, that is, to divide it into two equal parts. Let BAC be the given angle: it is required to bisect it. Construction. In AB take any point D; and from A C cut off A E equal to AD. I . 3. Join DE; and on DE, on the side remote from A, describe an equi-lateral triangle DEF. I . 1. Join AF. Then shall the straight line AF bisect the angle BAC. Proof. For in the two triangles DAF, EAF, ( D A is equal to EA, Constr. and AF is common to both; and the third side DF is equal to the third side EF; Def. 19. therefore the angle DAF is equal to the angle EAF. I . 8. Therefore the given angle BAC is bisected by the straight line AF. Q.e.f. exercises. 1. If in the above figure the equilateral triangle DFE were de-scribed on the same side of D E as A, what different cases would arise? And under what circumstances would the construction f a i l ? 2. In the same figure, shew that AF also bisects the angle DFE. 3. Divide an angle into four equal parts. 24 e u c l i d ' s elements. Proposition 10. Problem. To bisect a given finite straight line, that is, to divide i t into two equal parts. Let AB be the given straight line : it is required to divide it into two equal parts. Constr. On AB describe an equilateral triangle ABC, I. 1. and bisect the angle A C B by the straight line CD, meeting AB at D. I . 9. Then shall AB be bisected at the point D. Proof. Def. 19. Because For in the triangles ACD, BCD, A C is equal to BC, and C D is common to both; \ also the contained angle A C D is equal to the con-tained angle BCD; Constr. Therefore the triangles are equal in all respects: so that the base AD is equal to the base BD. I. 4. Therefore the straight line AB is bisected at the point D. o, K. F. EXERCISES. 1. Shew that the straight line whioh bisects the vertical angle an isosceles triangle, also bisects the base. 2. On a given base describe an isosceles triangle such that the Bum of i t s equal sides may be equal to a given straight line. book i . prop. 1 1 . 2 5 Proposition 11. Problem. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C the given point in it. It is required to draw from the point C a straight line at right angles to AB. Construction. In A C take any point D, and from CB cut off CE equal to CD. i . 3. O n DE describe the equilateral triangle DFE. I . 1. Join CF. Then shall the straight line CF be at right angles to AB. Proof. For in the triangles DCF, EOF, I D C is equal to EC, Constr. and CF is common to both; and the third side DF is equal to the third side EF: Def. 19. Therefore the angle DGF is equal to the angle ECF: I . 8. and these are adjacent angles. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle; Def. 7. therefore each of the angles DCF, ECF is a right angle. Therefore CF is at right angles to AB, and has been drawn from a point C in it. q.e.f. exercise. In the figure of the above proposition, shew that any point in FC, or FC produced, i s equidistant from D and E. 2 6 E u c l i d ' s e l e m e n t s . Proposition 12. Problem. To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without i t . C Let AB be the given straight line, which may be pro-duced in either direction, and let C be the given point with-out it. It is required to draw from the point C a straight line perpendicular to AB. Construction. O n the side of A B remote from C take any point D; and from centre C, with radius CD, describe the circle FDG, meeting AB at F and G. Post. 3. Bisect FG at H ; I . 10. and join CH. Then shall the straight line C H be perpendicular to AB. Join CF and CG. Proof. Then in the triangles FHC, GHC, I F H is equal to GH, Co;--. BOOK I . PROP. 13. 29 Definitions. (i) The complement of an acute angle is its defect from a right angle, that is, the angle by which it falls short of a right angle. Thus two angles are complementary, when their sum is a right angle. (ii) The supplement of an angle is its defect from two right angles, that is, the angle by which it falls short of two right angles. Thus two angles are supplementary, when their sum is two right angles. Corollary. Angles which are complementary or supple-mentary to the same angle are equal to one another. exercises. 1. If the two exterior angles formed by producing a-side of a tr angle both ways are equal, shew that the triangle is isosceles. 2. The oisectors of the adjacent angles which one straight line makes with another contain a right angle. Note. In the adjoining figure AOB i s a given angle; and one of i t s arms A O i s produced to C: the adjacent angles AOB, B O C are bisected by OX, OY. Then O X and OY are called respect-ively the Internal and external bisectors of the angle AOB. Hence Exercise 2 may be thus enunciated: The internal and external bisectors of an angle are at right ang t o one another. 3. Shew that the angles AOX and COY are complementary. 4. Shew that the angles BOX and COX are supplementary; and also that the angles A O Y and BOY are supplementary. 3 0 E u c l i d ' s e l e m e n t s . P r o p o s i t i o n 14. Theorem. If, at a point in a straight line, two oilier straight hues, on opposite sides of it, make the adjacent angles togetlier equal to two right angles, then i l i e s e two straight lines shall be in one and the same straight line. A t the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles : then BD shall be in the same straight line with BC. Proof For if BD be not in the same straight line with BC, i f possible, let BE be in the same straight line with BC. Then because AB meets the straight line CBE. therefore the adjacent angles CBA, A B E are together equal to two right angles. i. 13. But the angles CBA, ABD are also together equal to two right angles. H'tP-Therefore the angles CBA, ABE are together equal to the angles CBA, ABD. Ax. 11. From each of these equals take the common angle CBA; then the remaining angle ABE is equal to the remaining angle ABD; the part equal to the whole; which is impossible Therefore BE is not in the same straight line with BC. A n d in the same way it may be shewn that no other line but BD can be in the same straight line with BC. Therefore BD is in the same straight line with BC. q.e.1). exercise. ABCD i s a rhombus; and the diagonal AC i s bisected at O. If O i s joined to the angular points B and D ; shew that O B and O D are in one straight line, BOOK I . PROP. 15. 31 06s. When two straight lines intersect at a point, four angles are formed; and any two of these angles which are not adjacent, are said to be vertically opposite to one another. Proposition 15. Theorem. If two straight lines intersect one another, tlien the vertic opposite angles s l i a i l be equal. A Let the two straight lines AB, CD cut one another at the point E: then shall the angle A EC be equal to the angle DEB, and the angle CEB to the angle A ED. Proof. Because AE makes with CD the adjacent angles CEA, AED, therefore these angles are together equal to two right angles. I . 13. Again, because DE makes with AB the adjacent angles AED, DEB, therefore these also are together equal to two right angles. Therefore the angles CEA, AED are together equal to the angles AED, DEB. From each of these equals take the common angle AED; then the remaining angle CEA is equal to the remaining angle DEB. Ax. 3. In a similar way it may be shewn that the angle CEB i s equal to the angle AED. q. e.d. Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles. Corollary 2. Consequently, when any number of straight lines meet at a point, the sum of the angles made by con-secutive lines is equal to four right angles. 3 2 E u c l i d ' s elements. P r o p o s i t i o n 16. Theorem. If one side of a triangle be produced, then Hie exterior angle shall be greater tlian either of the interior opposite angles. Let ABC be a triangle, and let one side BC be produced to D : then shall the exterior angle A C D be greater than either of the interior opposite angles CBA, BAC. Construction. Bisect AC at E: I . 10. Join BE; and produce it to F, making EF equal to BE. I . 3. Join FC. Proof. Then in the triangles AEB, CEF, 1 A E is equal to CE, Constr. and EB to EF ; Conn',: also the angle AEB is equal to the vertically opposite angle CEF; i . 15. therefore the triangle AEB is equal to the triangle CEF in all respects : I. 4. so that the angle BAE is equal to the angle ECF. But the angle ECD is greater than its part, the angle ECF; therefore the angle ECD is greater than the angle BAE; that is, the angle ACD is greater than the angle BAC. In a similar way, i f BC be bisected, and the side AC produced to G, it may be shewn that the angle BCG i s greater than the angle ABC. But the angle BCG is equal to the angle ACD: i . 15. therefore also tho angle ACD is greater than the angle ABC. Q. K. D. book i . prop. 17. 33 Proposition 17. Theorem. A n y two angles of a triangle are togetlier less than two right angles. Let A B C be a triangle: then shall any two of its angles, as ABC, ACB, be together less than two right angles. Construction. Produce the side B C to D. Proof. Then because A C D is an exterior angle of the triangle ABC, therefore it is greater than the interior opposite angle ABC. i . 16. To each of these add the angle A C B : then the angles ACD, A C B are together greater than the angles ABC, ACB. Ax. 4. But the adjacent angles ACD, A C B are together equal to two right angles. I . 13. Therefore the angles ABC, A C B are together less than two right angles. Similarly it m a y be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles. q. e. d. Note. It follows from this Proposition that every triangle must have at least two acute angles: for i f one angle is obtuse, or a right angle, each of the other angles must be less than a right angle. EXERCISES. 1. Enunciate this Proposition so as to shew that i t is the converse of Axiom 12. 2. If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles. 3. Shew how a proof of Proposition 17 may be obtained by joining each vertex in turn to any point in the opposite side. h. e. 3 34 E u c l i d ' s e l e m e n t s . P r o p o s i t i o n 18. Theorem. If one side of a triangle be greater t/ian anot/ter, then the angle opposite to the greater side sliall be greater than the angle opposite to the less. Let A B C be a triangle, in which the side A C is greater than the side AB : then shall the angle ABC be greater than the angle ACB. Construction. From AC, the greater, cut off a part A D equal to AB. L 3. Join BD. Proof. Then in the triangle ABD, because AB is equal to AD, therefore the angle ABD is equal to the angle ADB. I. 5 But the exterior angle A D B of the triangle B D C i s greater than the interior opposite angle DCB, that is, greater than the angle ACB. I. 16. Therefore also the angle ABD is greater than the angle ACB; s t i l l more then is the angle A B O greater than the angle ACB. Q.E.D. Euclid enunciated Proposition 18 as follows: The greater side of every triangle has the greater angle opposite to it. [This form of enunoiation is found to be a common source of d i f f i -culty with beginners, who f a i l to distinguish what i s assumed in i t and what i s to be proved.] [For Exeroises see page 38.] BOOK I . PROP. 19. 35 Proposition 19. Theorem. If one angle of a triangle be greater than another, then the side opposite to the greater angle shall be greater than the side opposite to the less. Let A B C be a triangle in which the angle A B C is greater than the angle A C B : then shall the side A C be greater than the side AB. Proof. For if A C be not greater than AB, it must be either equal to, or less than AB. But A C is not equal to AB, for then the angle A B C would be equal to the angle A C B ; I. 5. but it is not. Hyp. Neither is A C less than AB ; for then the angle A B C would be less than the angle A C B ; 1.18. but it is not: Hyp. Therefore A C is neither equal to, nor less than AB. That is, A C is greater than AB. q.e.d. Note. The mode of demonstration used in this Proposition is known as the Proof by Exhaustion. It i s applicable to cases in which one of certain mutually exclusive suppositions must necessarily be true; and i t consists in shewing the falsity of each of these supposi-tions in turn with one exception: hence the truth of the remaining supposition is inferred. Euclid enunciated Proposition 19 as follows : The greater angle of every triangle is subtended by the greater side, or, has the greater side opposite to it. [For Exercises see page 38.] 3—2 3 6 E u c l i d ' s e l e m e n t s . Proposition 20. Theorem. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle: then shall any two of its sides be together greater than the third side : namely, BA, AC, shall be greater than C D ; AC, CB greater than BA ; and CB, BA greater than AC. Construction. Produce BA to the point D, making AD equal to AC. I . 3. Join DC. Proof. Then in the triangle ADC, because AD is equal to AC, Constr. therefore the angle ACD is equal to the angle ADC. I . 5. But the angle BCD is greater than the angle ACD ; Ax. 9 . therefore also the angle BCD is greater than the angle ADC, that is, than the angle BDC. A n d in the triangle BCD, because the angle BCD is greater than the angle CCC, Pr. therefore the side BD is greater than the side CC. I . 19. But BA and AC are together equal to BD ; therefore BA and AC are together greater than CB. Similarly it may be shewn that AC, CB are together greater than BA : and CB, BA are together greater than AC. q. e. d. [For Exercises see page. oS.] book i . p r o p . 21. 3 7 Proposition 21. Theorem. If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, then these straight lines shall be less than the other two sides of the triangle, but shall contain a greater angle. A VE D/ B C Let ABC be a triangle, and from B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle : then (i) BD and DC shall be together less than BA and AC ; (ii) the angle BDC shall be greater than the angle BAC. Construction. Produce BD to meet AC in E. Proof, (i) In the triangle BAE, the two sides BA, AE are together greater than the third side BE : I . 20. to each of these add EC : then BA, AC are together greater than BE, EC. Ax. 4. Again, in the triangle DEC, the two sides DE, EC are to-gether greater than DC : I . 20. to each of these add BD ; then BE, EC are together greater than BD, DC. But it has been shewn that BA, AC are together greater than BE, EC : s t i l l more then are BA, AC greater than BD, DC. ( i i ) Again, the exterior angle BDC of the triangle DEC is greater than the interior opposite angle DEC ; I . 16. and the exterior angle DEC of the triangle BAE is greater than the interior opposite angle BAE, that is, than the angle BAC; I . 16. s t i l l more then is the angle BDC greater than the angle BAC. Q.E.D. 38 euclid's elements, exercises on Propositions 18 and 19. 1. The hypotenuse is the greatest side of a right-angled triangle. 2. If two angles of a triangle are equal to one another, the sides also, which subtend the equal angles, are equal to one another. Prop. 0. Prove this indirectly by using the result of Prop. 18. 3. BC, the base of an isosceles triangle A B C , is produced to any point D ; shew that A D is greater than either of the equal sides. 4. If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle. 5. In a triangle A B C , if A C is not greater than AB, shew that any straight line drawn through the vertex A and terminated by the base BC, is less than AB. 6. A B C is a triangle, in which O B , O C bisect the angles A B C , A C B respectively: shew that, if A B is greater than AC, then O B is greater than O C . on Proposition 20. 7. The difference of any two sides of a triangle is less than the third side. 8. In a quadrilateral, if two opposite sides which are not parallel are produced to meet one another; shew that the perimeter of the greater of the two triangles so formed is greater than the perimeter of the quadrilateral. 9. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter. 10. The perimeter of a quadrilateral is greater than the snm of its diagonals. 11. Obtain a proof of Proposition 20 by bisecting an angle by a straight line which meets the opposite side. on Proposition 21. 12. In Proposition 21 shew that the angle BDC is greater than the angle B A C by joining AD, and producing it towards the base. 13. The sum of tho distances of any point within a triangle from its angular points is less than tho perimeter of the triangle. BOOK I . PROP. 22. 3 9 Proposition 22. Problem. To describe a triangle having its sides equal to three given straight lines, any two of which are together greater than the third. Let A, B, C be the three given straight lines, of which any two are together greater than the third. It is required to describe a triangle of which the sides shall be equal to A, B, C. Construction. Take a straight line DE terminated at the point D, but unlimited towards E. Make DF equal to A, FG equal to B, and GH equal to C. I . 3. From centre F, with radius FD, describe the circle DLK. From centre G with radius GH, describe the circle MHK, cutting the former circle at K. Join FK, GK. Then shall the triangle KFG have its sides equal to the three straight lines A, B, C. Proof. Because F is the centre of the circle DLK, therefore FK is equal to FD : Def. 11. but FD is equal to A; Constr. therefore also FK is equal to A. Ax.d-Again, because G is the centre of the circle MHK, therefore GK is equal to GH : Def. 11. but GH is equal to C; Constr. therefore also GK is equal to C. Ax. 1. A n d FG is equal to B. Constr. Therefore the triangle KFG has its sides KF, FG, GK equal respectively to the three given lines A, B, C. Q.E.E. 40. Euclid's elements. exercise. On a given base describe a triangle, whose remaining sides sh equal to two given straight lines. Point out how the construction f a i l s , i f any one of the three given lines i s greater than the sum of the other two. Proposition 23. Problem. At a given point in a given straight line, to make an angle equal to a given angle. Let AB be the given straight line, and A the given point in it; and let DCE be the given angle. It is required to draw from A a straight line making with AB an angle equal to the given angle DCE. Construction. In CD, CE take any points D and E : and join DE. From AB cut off AF equal to CD. i . 3. O n A F describe the triangle FAG, having the remaining sides AG, GF equal respectively to CE, ED. i . 22. Then shall the angle FAG be equal to the angle DCE. Proof. For in the triangles FAG, DCE. ( FA is equal to DC, Constr. Because ^ and AG is equal to CE : Comtr. (and the base FG is equal to the base D E : Constr. therefore the angle FAG is equal to the angle DCE. i . 8 . ' That is, AG makes with AB, at the given point A, an an^le equal to the given angle DCE. ^.E-Ff book i . prop. 24j 41 Proposition 24. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one greater than the angle contained by the corresponding sides of the other; then the base of that which has the greater angle shall be greater than the base of the other. Let ABC, DEF be two triangles, in which the two sides BA, AC are equal to the two sides ED, DF, each to each, but the angle BAC greater than the angle EDF : then shall the base BC be greater than the base EF. Of the two sides DE, DF, let DE be that which is not greater than DF. Construction. At the point D, in the straight line ED, and on the same side of it as DF, make the angle EDG equal to the angle BAC. I . 23. Make DG equal to DF or AC; I . 3. and join EG, GF. Proof. Then in the triangles BAC, EDG, ( BA is equal to ED, Hyp. • c and AC is equal to DG, Constr. I also the contained angle BAC is equal to the ( contained angle EDG ; Constr. Therefore the triangle BAC is equal to the triangle EDG in all respects : I . 4. so that the base BC is equal to the base EG. See note on the next page. 42 EUCLID'S ELEMENTS. Again, in the triangle FDG, because D G is equal to DF, therefore the angle DFG is equal to the angle DGF, L 5. but the angle D G F is greater than the angle E G F ; therefore also the angle DFG is greater than the angle E G F ; still more then is the angle EFG greater than the angle EGF. And in the triangle EFG, because the angle EFG is greater than the angle EGF, therefore the side EG is greater than the side E F ; I. 19. but EG was shewn to be equal to B C ; therefore BC is greater than EF. Q.E.D. This condition was inserted by Simson to ensure that, in the complete construction, the point F should f a l l below EG. Without this condition i t would be necessary to consider three case s: for F might f a l l above, or upon, or below EG ; and each figure would require separate proof. W e are however scarcely at liberty to employ Simson's condition without proving that i t f u l f i l s the object for which i t was introduced. This may be done as follows: Let EG, DF, produced i f necessary, intersect at K. Then, since D E is not greater than DF, that is, since D E is not greater than DG, therefore the angle D G E is not greater than the angle DEG. i. 18. But the exterior anglo D K G is greater than the angle D E K : i . 16. therefore the angle D K G is greater than the angle DGK. Hence D G is greater than DK. I , 19, But D G is equal to DF ; therefore DF is greater than DK. So that the point F must f a l l below EG. book I. prop. 24. 43 Or the following method may be adopted. Proposition 24. [Alternative Proof.] In the triangles ABC, DEF, l e t BA be equal to ED, and AC equal to DF, but l e t the angle BAC be greater than the angle EDF: then shall the base B C be greater than the base EF. For apply the triangle DEF to the triangle ABC, so that D may f a l l on A, and D E along AB: then because D E is equal to AB, therefore E must f a l l on B. And because the angle EDF i s less than the angle BAC, therefore DF must f a l l between AB and AC. Let DF occupy the position AG. Case I. K G f a l l s on BC: Then G must be between B and C: therefore BC is greater than BG. But BG i s equal to EF : therefore BC is greater than EF. Case II. If G does not f a l l on BC. Bisect the angle C A G by the straight line AK which meets BC in K. i . 9. Join GK. Then in the triangles GAK, CAK, / G A i s equal to CA, -p ) and AK i s common to both; because F> Let the straight line EF cut the two straight lines AB, CD in G and H: and First, let the exterior angle EG B be equal to the interior opposite angle G H D: then shall AB and C D be parallel. Proof. Because the angle EGB is equal to the angle G H D ; and because the angle EG B is also equal to the vertically op-posite angle AGH; I . 15. therefore the angle A G H is equal to the angle G H D ; but these are alternate angles; therefore AB and C D are parallel. I . Q. E.D. Secondly, let the two interior angles BGH, G H D be to-gether equal to two right angles: then shall AB and C D be parallel. Proof. Because the angles BGH, G H D are together equal to two right angles; Hyp. and because the adjacent angles BGH, A G H are also together equal to two right angles; I . 13. therefore the angles BGH, A G H are together equal to the two angles BGH, GHD. From these equals take the common angle BGH: then the remaining angle A G H is equal to the remaining angle G H D : and these are alternate angles; therefore AB and C D are parallel. i . 27. q. E. D. book I . p r o p . 29. 53 Proposition 29. Theorem. If a straight line fall on two parallel straight lines, then it shall make the alternate angles equal to one another, and t l i e exterior angle equal to the interior opposite angle on the same side; and also the two interior angles on the same side equal to two right angles. F^ Let the straight line EF fall on the parallel straight lines AB, CD: then (i) the alternate angles-AGH, G H D shall be equal to one another; ( i i ) the exterior angle EG B shall be equal to the interior opposite angle G H D ; ( i i i ) the two interior angles BGH, G H D shall be together equal to two right angles. Proof, (i) For if the angle AG H be not equal to the angle GHD, one of them must be greater than the other. If possible, let the angle AGH be greater than the angle G H D ; add to each the angle BG H: then the angles AGH, BGH are together greater than the angles BGH, GHD. But the adjacent angles AGH, BGH are together equal to two right angles; 1.13. therefore the'angles BGH, G H D are together less than two right angles; therefore AB and C D meet towards B and D. Ax. 12. • But they never meet, since they are parallel. Hyp. Therefore the angle AGH is not unequal to the angle GHD: that is, the alternate angles AGH, G H D are equal. (Over) 54 Euclid's elements. (ii) Again, because the angle A G H is equal to the verti-cally opposite angle E G B ; I. 15. and because the angle A G H is equal to the angle G H D ; Proved. therefore the exterior angle EG B is equal to the interior op-posite angle G H D . (iii) Lastly, the angle EGB is equal to the angle GHD; Proved. add to each the angle BG H; then the angles EGB, BGH are together equal to the angles BGH, GHD. But the adjacent angles EGB, BGH are together equal to two right angles: 1.13. therefore also the two interior angles BGH, G H D are to-gether equal to two right angles. Q.E.D. exercises on propositions 27, 2S, 29. 1. Two straight lines AB, CD bisect one another at O: shew t the straight lines joining A C and BD are parallel. [i. 27.] 2. Straight lines which are perpendicular to the same straiaht lin are parallel t o one another. [i. 27 or i . 2S.1 3. If a straight line meet two or more parallel straight lines, and perpendicular to one of them, i t is also perpendicular to all the others. [t 29.] 4. If two straight lines are parallel to two other straight lines, each to each, then the angles containedby the first pair are equal' respectivelii t o the angles contained by the second pair. [i. 2 9 . ' ] . b o o k i . prop. 29. 55 Note on the Twelfth Axiom. It must be admitted that Euclid's twelfth Axiom is satisfactory as the basis of a theory of parallel straight lines. It cannot be regarded as either simple or self-evident, and it therefore falls short of the essential characteristics of an axiom: nor is the difficulty entirely removed by considering it as a cor-rollary to Proposition 17, of which it is the converse. Many substitutes have been proposed; but we need only notice here the system which has met with most general approval. This system rests on the following hypothesis, which is put forward as a fundamental Axiom. Axiom. Two intersecting straight lines cannot be both parallel to a third straight line. This statement is known as Playfair's Axiom; and though i t is not altogether free from objection, i t is recommended as both simpler and more fundamental than that employed by Euclid, and more readily admitted without proof. Propositions 27 and 28 having been proved in the usual way, the first part of Proposition 29 is then given thus. Proposition 29. [Alternative Proof.] If a straight line fall on two parallel straight lines, then it shall make the alternate angles equal. Let the straight line EF meet the two parallel straight lines AB, CD, at G and H: then shall the alternate angles AGH, G H D be equal. For i f the angle AG H is not equal to the angle G H D : " at G in the straight line H G make the angle H G P equal to the angle G H D , and alternate to it. i . 23. Then PG and C D are parallel, i . 27. But AB and C D are parallel: Hyp. therefore the two intersecting straight lines AG, PG are both parallel to C D : which is impossible. Playfair's Axiom. Therefore the angle A G H is not unequal to the angle GHD, that is, the alternate angles AGH, G H D are equal, q.e.d. The second and third parts of the Proposition may then be deduced as in the text; and Euclid's Axiom 12 follows as a Corollary. 5 6 e u c l i d ' s e l e m e n t s . Proposition 30. Theorem. Straight lines which are parallel to the same straight line are parallel to one another. C / ' D P %. Q Let the straight lines AB, C D be each parallel to the straight line P Q : then shall A B and C D be parallel to one another. Construction. Draw any straight line EF cutting AB, CD, and P Q in the points G, H, and K. Proof. Then because AB and PQ are parallel, and EF meets them, therefore the angle AGK is equal to the alternate angle GKQ. 1.29. A n d because C D and P Q are parallel, and EF meets them, therefore the exterior angle G H D is equal to the interior opposite angle HKQ. l 29. Therefore the angle AGH is equal to the angle GHD: and these are alternate angles; therefore A B and C D are parallel. I. 27. Q.E.D. Note. If PQ lies between AB and CD, the Proposition may be established in a similar manner, though in this case i t scarcely needs proof; for i t is inconceivable that two straight lines, which "do not meet an intermediate straight line, should meet one another. The truth of this Proposition may be readily deduced from Playfair's Axiom, of which i t is the converse. For i f AB and C D were not parallel, they would meet when pro-duced. Then there would be two intersecting straight lines both parallel to a third straight line: whioh i s impossible. Therefore AB and C D never meet; that is, they are parallel. book i . p r o p . 3 1 . 57 Proposition 31. Problem. To d r a w a straight line through a given point parallel to a given straight line. B-Let A be the given point, and BC the given straight line. It is required to draw through A a straight line parallel to BC. Construction. In BC take any point D; and join AD. At the point A in DA, make the angle DAE equal to the angle ADC, and alternate to it. I . 23. and produce EA to F. Then shall EF be parallel to BC. Proof. Because the straight line AD, meeting the two straight lines EF, BC, makes the alternate angles EAD, ADC equal; Constr. therefore EF is parallel to B C ; I. 27. and it has been drawn through the given point A. EXERCISES. 1. Any straight line drawn parallel to the base of an isosceles triangle makes equal angles with the sides. 2. If from any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle, the triangle thus formed is isosceles. 3. From a given point draw a straight line that shall make with a given straight line an angle equal to a given angle. 4. From X, a point in the base B C of an isosceles triangle ABC, a straight line is drawn at right angles to the base, cutting A B in Y, and C A produced in Z : shew the triangle A Y Z is isosceles. 5. If the straight line which bisects an exterior angle of a triangle is parallel to the base, shew that the triangle is isosceles. 5 8 e u c l i d ' s elements. Proposition 32. Theorem. If a side of a triangle be produced, then tlie exterior angle shall be equal to t l i e sum of t l i e two interior opposite angles: also the three interior angles of a triangle are together equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D: then (i) the exterior angle A C D shall be equal to the sum of the two interior opposite angles CAB, ABC; (ii) the three interior angles ABC, BCA, CAB shall be together equal to two right angles. Construction. Through C draw C E parallel to BA. I . 31. Proof, (i) Then because BA and C E are parallel, and AC meets them, therefore the angle ACE is equal to the alternate angle CAB. I . 29. Again, because BA and C E are parallel, and BD meets them, therefore the exterior angle ECD i s equal to the interior opposite angle ABC. I . 29. Therefore the whole exterior angle ACD is equal to the sum of the two interior opposite angles CAB, ABC. (ii) Again, since the angle ACD is equal to the sum of the angles CAB, ABC; Proved. to each of these equals add the angle BCA: then the angles BCA, ACD are together equal to the three angles BCA, CAB, ABC. But the adjacent angles BCA, A C D are together equal to two right angles; l . 13. therefore also the angles BCA, CAB, ABC are together equal to two right angles. q. e. d. BOOK I. PROP. 32. 59 From this Proposition we draw the following important inferences. 1. If two triangles have two angles of the one equal to two angles of the other, each to each, then the third angle of the one is equal to the third angle of the other. 2. In any right-angled triangle the two acute angles are com-plementary. 3. In a right-angled isosceles triangle each of the equal angles is half a right angle. 4. If one angle of a triangle is equal to the sum of the other two, the triangle is right-angled. 5. The sum of the angles of any quadrilateral figure is equal to four right angles. 6. Each angle of an equilateral triangle is two-thirds of a right angle. EXERCISES ON PROPOSITION 32 1. Prove that the three angles of a triangle are together equal to two right angles, (i) by drawing through the vertex a straight line parallel to the base; ( i i ) by joining the vertex to any point in the base. 2. If the base of any triangle iB produced both ways, shew that the sum of the two exterior angles diminished by the vertical angle is equal to two right angles. 3. If two straight lines are perpendicular to two other straight lines, each to each, the acute angle between the first pair is equal to the acute angle between the second pair. 4. Every right-angled triangle is divided into two isosceles tri-angles by a straight line drawn from the right angle to the middle point of the hypotenuse. Hence the joining line is equal to half the hypotenuse. 5. Draw a straight line at right angles to a given finite straight line from one of its extremities, without producing. the given straight line. [Let A B be the given straight line. O n A B describe any isosceles triangle A C B . Produce B C to D, making C D equal to BC. Join AD. Then shall A D be perpendicular to AB.] 60 EUCLID'S ELEMENTS. 6. Trisect a right angle. 7. The angle contained by the bisectors, of the angles at the bas of an isosceles triangle is equal to an exterior angle formed by pro-ducing the base. 8. The angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles. The following theorems were added as corollaries to Proposition 32 by Robert Simson. Corollary 1. All the interior angles of any rectilineal figure, toith four right angles, are together equal to twice as m a n y right angles as the figure has sides. B Let A B O D E be any rectilineal figure. Take F, any point within it, and join F to each of the angular points of the figure. Then the figure is divided into as m a n y triangles as it has sides. A n d the three angles of each triangle are together equal to two right angles. I . 32. Hence all the angles of all the triangles are together equal to twice as m a n y right angles as the figure has sides. But all the angles of all the triangles" make up the in-terior angles of the figure, together with the angles at F; and the angles at F are together equal to four riiiht angles: I. 15, Cor. Therefore t i l l the interior angles of the figure, with four right angles, are together equal to twice as m a n y right angles as tho figure has sides. q.'e. d. BOOK I. PROP. 32. 61 Corollary 2. If the sides of a rectilineal figure, which Aos no re-entrant angle, are produced in order, then all the ex-terior angles so formed are together equal to four right angles. For at each angular point of the figure, the interior angle and the exterior angle are together equal to two right angles. " i . 13. Therefore all the interior angles, with all the exterior angles, are together equal to twice as m a n y right angles as the figure has sides. But all the interior angles, with four right angles, are to-gether equal to twice as m a n y right angles as the figure has sides. I. 32, Cor. 1. Therefore all the interior angles, with all the exterior angles, are together equal to all the interior angles, with four right angles. Therefore the exterior angles are together equal to four right angles. q. e. d. EXERCISES ON SIMSON'S COROLLARIES. [A polygon is said to be regular when it has all its sides and all angles equal.] 1. Express in terms of a right angle the magnitude of each angle of ( i ) a regular hexagon, ( i i ) a regular octagon. 2. If one side of a regular hexagon is produced, shew that the ex-terior angle is equal to the angle of an equilateral triangle. 3. Prove Simson's first Corollary by joining one vertex of the rectilineal figure to each of the other vertices. 4. Find the magnitude of each angle of a regular polygon of n sides. 5. If the alternate sides of any polygon be produced to meet, the sum of the included angles, together with eight right angles, will be equal to twice as many right angles as the figure has sides. 62 e u o l i d ' s elements. P r o p o s i t i o n 33. Theorem. The straight lines v)hich join tlie extremities of two equal and parallel straight lines towards t l i e same parts are them-selves equal and parallel. B Let AB and CD be equal and parallel straight lines; and let them be joined towards the same parts by the straight lines AC and BD: then shall AC and BD be equal and paralleL Construction. Join BC. Proof. Then because AB and CD are parallel, and BC meets them, therefore the alternate angles ABC, BCD are equal, i . 29. Now in the triangles ABC, DCB, ( A B is equal to DC, Hyp. and BC is common to both; also the angle ABC is equal to the an^le DCB; Proved. therefore the triangles are equal in all respects; I . 4. so that the base AC is equal to the base DB, and the angle ACB equal to the angle DBC;' but these are alternate angles; therefore AC and BD are parallel: i . 27. and it has been shewn that they are also equal. Q. E. D. Definition. A Parallelogram is a four-sided figure whose opposite sides are parallel. book I . prop. 34. 63 Proposition 34. Theorem. The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects t l i e parallelo-gram. ,B Let ACDB be a parallelogram, of which BC is a diagonal: then shall the opposite sides and angles of the figure be equal to one another; and the diagonal BC shall bisect it. Proof Because AB and C D are parallel, and BC meets them, therefore the alternate angles ABC, DCB are equal. I . 29. Again, because AC and BD are parallel, and BC meets them, therefore the alternate angles ACB, DBC are equal. I . 29. Hence in the triangles ABC, DCB, (the angle ABC is equal to the angle DCB, and the angle ACB is equal to the angle DBC; also the side BC, which is adjacent to the equal angles, is common to both, therefore the two triangles ABC, DCB are equal in all respects; I . 26. so that AB is equal to DC, and AC to DB; and the angle BAC is equal to the angle CDB. Also, because the angle ABC is equal to the angle DCB, and the angle CBD equal to the angle BCA, therefore the whole angle ABD is equal to the whole angle DCA. And since it has been shewn that the triangles ABC, DCB are equal in all respects, therefore the diagonal BC bisects the parallelogram ACDB. Q.E.D. [See note on next page.] 64 EUCLID'S ELEMENTS. Note. To the proof which is here given Euclid added an applica-tion of Proposition 4, with a view to shewing that the triangles A B C , D C B are equal in area, and that therefore the diagonal B C bisects the parallelogram. This equality of area is however sufficiently established by the Btep which depends upon i. 26. [See page 48.] EXERCISES. 1. If one angle of a parallelogram is a right angle, all its angles are right angles. 2. If tlie opposite sides of a quadrilateral are equal, the figure is a parallelogram. 3. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. 4. If a quadrilateral has all its sides equal and one angle a right angle, all its angles are right angles; that is, all the angles of a square are right angles. 5. The diagonals of a parallelogram bisect each other. C. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. 7. If two opposite angles of a parallelogram are bisected by the diagonal which joins them, the figure is equilateral. 8. If the diagonals of a parallelogram are equal, all its angles are right angles. 9. In a parallelogram which is not rectangular the diagonals are unequal. 10. Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides, is bisected at that point. 11. If two parallelograms have two adjacent sides of one equal to two adjacent sides of the other, each to each, and one angle of one equal to one angle of the other, the parallelograms are equal in all respects. 12. Two rectangles are equal -if two adjacent sides of one are equal to two adjacent sides of the other, each to each. 13. In a parallelogram the perpendiculars drawn from one pair of opposite angles to the diagonal whioh joins the other pair are equal. 14. If ABCD is a parallelogram, and X, Y respectively the middle points of tho sides A D , B C ; shew that tlie figure A Y C X is a parallelo-gram. MISCELLANEOUS EXERCISES ON SECTIONS I. AND II. 65 MISCELLANEOUS EXERCISES O N SECTIONS I. A N D II. 1. Shew that the construction in Proposition 2 may generally be performed in eight different ways. Point out the exceptional case. 2. The bisectors of two vertically opposite angles are in the same straight line. 3. In the figure of Proposition 16, if AF is joined, shew ( i ) that A F is equal to B C ; ( i i ) that the triangle A B C is equal to the triangle C F A in all respects. 4. ABC is a triangle right-angled at B, and BC is produced to D: shew that the angle A C D is obtuse. 5. Shew that in any regular polygon of n sides each angle contains 2(«-2) . , , , right angles. 6. The angle contained by the bisectors of the angles at the base of any triangle is equal to the vertical angle together with half the sum of the base angles. 7. The angle contained by the bisectors of two exterior angles of any triangle is equal to half the sum of the two corresponding interior angles. 8. If perpendiculars are drawn to two intersecting straight lines from any point between them, shew that the bisector of the angle between the perpendiculars is parallel to (or coincident with) the bisector of the angle between the given straight lines. 9. If two points P, Q be taken in the equal sides of an isosceles triangle A B C , so that B P is equal to C Q , shew that P Q is parallel to BC. 10. ABC and DEF are two triangles, such that AB, BC are equal and parallel to DE, EF, each to each; shew that A C is equal and parallel to DF. 11. Prove the second Corollary to Prop. 32 by drawing through any angular point lines parallel to all the sides. 12. If two sides of a quadrilateral are parallel, and the remaining two sides equal but not parallel, shew that the opposite angles are supplementary; also that the diagonals are equal. H, E, 5 SECTION III. THE AREAS OF PARALLELOGRAMS AND TRIANGLES. Hitherto when two figures have been said to l>e equal, it has been implied that they are identically equal, that is, equal in all respects. In Section III. of Euclid's first Book, we have t<> consider the equality in area of parallelograms and triangles which are not necessarily equal in all respects. [The ultimate test of equality, as we have already seen, is afforded by Axiom 8, which asserts that magnitudes which may be made to coincide with one another are equal. N o w figures which are not identi-cally equal, cannot be made to coincide without first undergoing some change of form: hence the method of direct superposition is unsuited to the purposes of the present section. W e shall see however from Euclid's proof of Proposition 35, that two figures which are not identically equal, may nevertheless be so related to a third figure, that it is possible to infer the equality of their areas.] Definitions. 1. The Altitude of a parallelogram with reference to a given side as base, is the perpendicular distance between the base and the opposite side. 2. The Altitude of a triangle with reference to a given side as base, is the perpendicular distance of the opposite vertex from the base. book i . prop. 35. 67 Proposition 35. Theorem. Parallelograms on the same base, and between the same parallels, are equal in area. D F Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels BC, AF : then shall the parallelogram ABCD be equal in area to the parallelogram EBCF. Case I. If the sides of the given parallelograms, oppo-site to the common base BC, are terminated at the same point D: then because each of the parallelograms is double of the triangle BDC; i . 34. therefore they are equal to one another. Ax. 6 . Case II. But if the sides-AD, EF, opposite to the base BC, are not terminated at the same point: then because ABCD is a parallelogram, therefore AD is equal to the opposite side BC; l . 34. and for a similar reaison, EF is equal to BC ; therefore AD is equal to EF. Ax. 1. Hence the whole, or remainder, EA is equal to the whole, or remainder, FD. Then in the triangles FDC, EAB, i ' F D is equal to EA, Proved. and DC is equal to the opposite sideAB, I . 34. also the exterior angle FDC is equal to the interior opposite angle EAB, I . 29. therefore the. triangle FDC is equal to the triangle EAB. I . 4. From the whole figure ABCF take the triangle FDC; and 'from the same 'figure take the equal triangle EAB ; then the remainders are equal; Ax. 3. that is, the parallelogram ABCD is equal to the parallelo-Q. e . d. 5—2 68 euclp's ELEMENTS, Proposition 36. Theorem. Parallelograms on equal bases, and between the same parallels, are equal in area. Let ABCD, EFGH be parallelograms on equal bases BC, FG, and between the same parallels AH, BG: then shall the parallelogram ABCD be equal to the paral-lelogram EFGH. Construction. Join BE, CH. Proof. Then because BC is equal to FG; Hyp. and FG is equal to the opposite side EH; i . 34. therefore BC is equal to EH: Ax. 1 . and they are also parallel; Hyp. therefore BE and CH, which join them towards the same parts, are also equal and parallel. I . 33. Therefore EBCH is a parallelogram. Def. 26. N o w the parallelogram ABCD is equal to EBCH ; for they are on the same base BC, and between the same parallels BC, AH. I . 35. Also the parallelogram EFGH is equal to EBCH; for they are on the same base EH, and between the same parallels EH, BG. I . 35. Therefore the parallelogram ABCD is equal to the paral-lelogram EFGH. A.c 1. Q. E. D. From the last two Propositions we infer that: (i) A parallelogram, is equal in area to a rectangle of equa base and equal altitude. ( i i ) Parallelograms on equal bases and of equal altitudes are equal in area. book I . p r o p . 37. ' 6 9 (hi) Of tiooparallelograms of equal altitudes, that is tlie which has the greater base ; and of two parallelograms on equal bases, that is the greater which has the greater altitude. Proposition 37. Theorem. Triangles on the same base, and between the same parol-j, are equal in area. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels BC, AD. Then shall the triangle ABC be equal to the triangle DBC. Construction. Through B draw BE parallel to CA, to meet DA produced in E; I. 31. through C draw CF parallel to BD, to meet AD produced in F. Proof. Then, by construction, each of the figures EBCA, DBCF is a parallelogram. Def. 26. A n d EBCA is equal to DBCF; for they are on, the same base BC, and between the same parallels BC, EF. I. 35. A n d the triangle ABC is half of the parallelogram EBCA, for the diagonal AB bisects it. I . 34. Also the triangle DBC is half of the parallelogram DBCF, for the diagonal D C bisects it. I . 34. But the halves of equal things are equal; Ax. 7. therefore the triangle ABC is equal to the triangle DBC, Q.E.D. [For Exercises see page 73.] 70 e u c l i d ' s elements, P r o p o s i t i o n 38. Theorem. Triangles on equal bases, and between t l i e same parallels, are equal in area. Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD : then shall the triangle ABC be equal to the triangle DEF. Construction. Through B draw BG parallel to CA, to meet DA produced in G; i . 31. through F draw FH parallel to ED, to meet A D produced in H. Proof. Then, by construction, each of the figures GBCA, DEFH is a parallelogram. Def. 2 . 6 . A n d GBCA is equal to DEFH ; for they are on equal bases BC, EF, and between the same parallels BF, GH. I . 36. A n d the triangle ABC is half of the parallelogram GBCA, for the diagonal AB bisects it. I . 34. Also the triangle DEF is half the parallelogram DEFH, for the diagonal DF bisects it. I . 34. But the halves of equal things are equal: Ax. 7. therefore the triangle ABC is equal to the triangle DEF. Q.E.D. From this Proposition w e infer that: (i) Triangles on equal bases and of equal altitude are equal in area. ( i i ) Of two triangles of the same altitude, that is the greater which has the greater base: and of two triangles on the same base. or on equal bases, that is the greater which has the greater altitude. [For Exeroisea see page 73.] BOOK I . PROP. 39. 71 Proposition 39. Theorem. Equal triangles on i l l s same base, and on the same side of it, are between t l i e same parallels. Let the triangles ABC, D B C which stand on the same Irase BC, and on the same side of it, be equal in area : then shall they be between the same parallels ; that is, i f AD be joined, AD shall be parallel to BC. Construction. For i f AD be not parallel to BC, i f possible, through A draw AE parallel to BC, I . 31. meeting BD, or BD produced, in E. Join EC. Proof. N o w the triangle ABC is equal to the triangle EBC, for they are on the same base BC, and between the same parallels BC, AE. I . 37. But the triangle ABC is equal to the triangle DBC; Hyp. therefore also the triangle DBC is equal to the triangle EBC; the whole equal to the part; which is impossible. Therefore AEls not parallel to BC. Similarly it can be shewn that no other straight line through A, except AD, is parallel to BC. Therefore AD is parallel to BC. Q.E.D. From this Proposition it follows that: Equal triangles on the same base have equal altitudes. [For Exercises see page 73.] EUCLID'S ELEMENTS. Proposition 40. Theorem. Equal triangles, on equal bases in the sam.e straight Une, and on t l i e same side of i t , are between the same parallels. Let the triangles ABC, DEF which stand on equal bases BC, EF, in the same straight line BF, and on the same side of it, be equal in area : then shall they be between the same parallels; that is, i f AD be joined, AD shall be parallel to BF. Construction. For i f AD be not parallel to BF, i f possible, through A draw AG parallel to BF, I . 31. meeting ED, or ED produced, in G. Join GF. Proof. N o w the triangle ABC i s equal to the triangle GEF, for they are on equal bases BC, EF, and between the same parallels BF, AG. I . 3S. But the triangle ABC is equal to the triangle DEF: Hyp. therefore also the triangle DEF is equal to the triangle GEF: the whole equal to the part; which is impossible. Therefore AG is not parallel to BF. Similarly i t can be shewn that no other straight line through A, except AD, i s parallel to BF. Therefore AD is parallel to BF. Q.E.D. From this Proposition it follows that: (i) Equal triangles on equal bases have equal altitudes. ( i i ) Equal triangles of equal altitudes have equal bases. EXERCISES ON PROPS. 37—40. 73 EXERCISES ON PROPOSITIONS 3 7 — 4 0 . Definition. Each of the three straight lines which join the angular points of a triangle to the middle points of the opposite sides is called a Median of the triangle. on Prop. 37. 1. If, in the figure of Prop. 37, A C and B D intersect in K, shew that ( i ) the triangles AKB, D K C are equal in area. ( i i ) the quadrilaterals EBKA, F C K D are equal. 2. In the' figure of i. 16, shew that the triangles ABC, FBC are equal in area. 3. On the base of a given triangle construct a second triangle, equal in area to the first, and having its vertex in a given straight line. 4. Describe an isosceles triangle equal in area to a given triangle and standing on the same base. on Peop. 38. 5. A triangle is divided by each of its medians into two parts of equal area. 6. A parallelogram is divided by its diagonals into four triangles of equal area. 7. A B C is a triangle, and its base B C is bisected at X; i f Y be any-point in the median AX, shew that the triangles ABY, A C Y are equal in area. 8. In AC, a diagonal of the parallelogram ABCD, any point X is taken, and XB, X D are drawn: shew that the triangle B A X is equal to the triangle DAX. 9. If two triangles have two sides of one respectively equal to two sides of the other, and the angles contained by those sides supplement-ary, the triangles are equal in area. on Peop. 39. 10. The straight line which joins the middle points of two sides of a triangle is parallel to the third side. 11. If two straight lines AB, C D intersect in O, so that the triangle A O C is equal to the triangle DOB, shew that A D and C B are parallel. on Peop. 40. 12. Deduce Prop. 40 from Prop. 39 by joining AE, AF in the figure of page 72. 7 4 ' e u c l i d ' s elements. Proposition 41. Theorem. If a parallelogram and a triangle be on t l i e same base and between the same parallels, t l i e parallelogram s/udl be double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the. same parallels BC, AE : then shall the parallelogram ABCD be double of the triangle EBC. Construction. Join AC. Proof. Then the triangle ABC is equal to the triangle EBC, for they are on the same base BC, and between tlie same parallels BC, AE. I. 37. But the parallelogram ABCD is double of the triangle ABC, for the diagonal A C bisects the parallelogram. 1.34. Therefore tlie parallelogram ABCD is also double of the triangle EBC. Q.E.D. EXERCISES. 1. ABCD is a parallelogram, and X, Y are the middle points of the sides AD, BC; i f Z i s any point in XY, or XY produced, shew that the triangle A Z B i s one quarter of the parallelogram ABCD. 2. Describe a right-angled isosceles triangle equal to a given 3. If ABCD is a parallelogram, and XY any points in DC and AD respectively: shew that the triangles AXB, BYC are equal in area. 4. ABCD is a parallelogram, and P is any point within it; shew that the sum of the triangles PAB, PCD i s equal to half the paral-lelogram. BOOK I . PR'OP. 42. Proposition 42. Problem. To describe a parallelogram iliat jshall be- equal to a given triangle, a n d have one of its angles equal to a given angle. Let ABC be the given triangle, and D the given angle. It is required to describe a parallelogram equal to ABC, and having one of its angles equal to D. Construction. Bisect BC at E. I . 10. At E in CE, make the angle CEF equal to D; I . 23. through A draw AFG parallel to EC; I . 31. and through C draw CG parallel to EF. Then FECG shall be the parallelogram required. Join AE. Proof. N o w the triangles ABE, AEC are equal, for they are on equal bases BE, EC, and between the same parallels; I . 38. therefore the triangle ABC i s double of the triangle AEC. But FECG is a parallelogram by construction; Def. 26. and it is double of the triangle AEC, for they are on the same base EC, and between the same parallels EC and AG. I . 41. Therefore the parallelogram FECG i s equal to the triangle ABC; and it has one of its angles CEF equal to the given angle D. Q. E. F. EXERCISES. 1. Describe a parallelogram equal to a given square standing on the same base, and having an angle equal to half a right angle. 2. Describe a rhombus equal to a given parallelogram and stand-ing on the same base. W h e n does the construction fail? 76 EUCLID'S ELEMENTS. Definition. If in the diagonal of a parallelogram any point is taken, and straight lines are drawn through it parallel to the sides of the parallelogram; then of the four parallelograms into which the whole figure is divided, the two through which the diagonal passes are called Paral-lelograms about that diagonal, and the other two, which with these make up the whole figure, are called the complements of the parallelograms about the diagonal. Thus in the figure given below, AEKH, KGCF are parallelograms about the diagonal AC; and HKFD, EBGK are the complements of those parallelograms. Note. A parallelogram is often named by two letters only, these being placed at opposite angular points. Proposition 43. Theorem. The complements of tlie parallelograms about tlie diagonal of any parallelogram, are equal to one anotlier. Let ABCD be a parallelogram, and KD, KB the comple-ments of the parallelograms EH, GF about the diagonal AC: then shall the complement BK be equal to the comple-ment KD. Proof. Because EH is a parallelogram, and AK its diagonal, therefore the triangle AEK is equal to the triangle AHK. I . 34. For a similar reason the triangle KGC is equal to the triangle KFC. Hence the triangles AEK, K G C are together equal to the triangles AHK, KFC. BOOK I. PROP. 44.. 77. But the whole triangle ABC is equal to the whole triangle ADC, for A C bisects the parallelogram A B C D ; 'I. 34. therefore the remainder, the complement BK, is equal to the remainder, the complement KD. Q.E.D. EXERCISES. In the figure of Prop. 43, prove that ( i ) The parallelogram ED is equal to the parallelogram BH. ( i i ) If KB, KD are joined, the triangle A K B is equal to the triangle AKD. Proposition 44. Problem. To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle. Let AB be the given straight line, C the given triangle, and D the given angle. It is required to apply to the straight line AB a paral-lelogram equal to the triangle C, and having an angle equal to the angle D. Construction. O n AB produced describe a parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D; I. 22 and I. 42. through A draw A H parallel to BG or EF, to meet FG pro-duced in H. i - 31. Join HB. This step of the construction is effected by first describing on AB produced a triangle whose sides are respectively equal to those of the triangle C (i. 22); and by then making a parallelogram equal to the triangle so drawn, and having an angle equal to D ( i . 42). 78. Euclid's elements. Then because AH and EF are parallel, and HF meets them, therefore the angles AHF, HFE are together equal to two right angles: I . 29. hence the angles BHF, HFE are together less than two right angles; therefore HB and FE will meet i f produced towards B and E. Ax. 12. Produce them to meet at K. Through K draw KL parallel to EA or FH; i . 31. and produce HA, GB to meet KL in the points L and M. Then shall BL be the parallelogram required. Proof. Now FHLK is a parallelogram, Constr. and LB, BF are the complements of the parallelograms about the diagonal HK: therefore LB is equal to BF. I . 43. But the triangle C is equal to BF; Constr. therefore LB is equal to the triangle C. And because the angle GBE is equal to the vertically oppo-site angle ABM, I . 15. and is likewise equal to the angle D; Constr. therefore the angle A B M is equal to the angle D. Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. g.K.F. book i . prop. 45. 79, Proposition 45. Problem, To describe a parallelogram equal to a given rectilineal figure, and having an angle equal t o a given angle. G L K H M Let ABCD be the given rectilineal figure, and E the given angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Suppose the given rectilineal figure to be a quadrilateral. Construction. Join BD. Describe the parallelogram FH equal to the triangle ABD, and having the angle FKH equal to the angle E. I . 42. To G H apply the parallelogram GM, equal to the triangle DBC, and having the angle G H M equal to E. I . 44. Then shall FKML be the parallelogram required. Proof. Because each of the angles GHM, FKH i s equal to E, therefore the angle FKH i s equal to the angle GHM. To each of these equals add the angle GHK; then the angles FKH, GHK are together equal to the angles GHM, GHK. But since FK, GH are parallel, and KH meets them, therefore the angles FKH, GHK are together equal to two right angles : I . 29. therefore also the angles GHM, GHK are together equal to two right angles: therefore KH, H M are in the same straight line. t . 14, 80 Again, because KM, FG are parallel, and HG meets them, therefore the alternate angles M H G , HGF are equal: I . 29 to each of these equals add the angle H G L ; then the angles M H G , HGL are together equal to the angles HGF, HGL. But because HM, GL are parallel, and HG meets them, therefore the angles M H G , H G L are together equal to two right angles: I. 29. therefore also the angles HGF, HGL are together equal to two right angles: therefore FG, GL are in the same straight line. I . 14. And because KF and M L are each parallel to HG, Constr. therefore KF i s parallel to M L ; I . 30. and KM, FL are parallel; Constr. therefore FKML is a parallelogram. Def. 26. And because the parallelogram FH is equal to the triangle ABD, Constr. and the parallelogram G M to the triangle D B C ; Constr. therefore the whole parallelogram FKML is equal to the whole figure ABCD ; and i t has the angle FKM equal to the angle E. By a series of similar steps, a parallelogram may be constructed equal to » rectilineal figure of more than four sides. Q.E.F. book i . p r o p . 4 6 . Proposition 46. Problem. To describe a square on a given straight line. C 8 1 Let AB be the given straight line : it is required to describe a square on AB. Constr. From A draw AC at right angles to AB ; I. 11. and make AD equal to AB. I . 3. Through D draw DE parallel to AB; I . 31. and through B draw BE parallel to AD, meeting DE in E. Then shall ADEB be a square. Proof. For, by construction, ADEB is a parallelogram : therefore AB is equal to DE, and AD to BE. I. 34. But AD is equal to A B ; Constr, therefore the four straight lines AB, AD, DE, EB are equal to one another; that is, the figure ADEB is equilateral. A n d the angle BAD is a right angle. Constr. But a square is a four-sided figure which has all its sides equal, and one angle a right angle. Def. 28. Therefore ADEB is a square; and it is described on AB. Q.E.F. From this proposition we see that a square is an equilateral parallelogram; and since one of i t s angles is a right angle, i t follows from Prop. 29 that each of the remaining angles i s a right angle. Hence we are led to the following Corollary. Corollary. All the angles of a square are right angles. H. E. G 82 EUCLID'S ELEMENTS. Proposition 47. Theorem. In a right-angled triangle tlie square described on tlie hypotenuse is equal to the sum of the squares described on the other two sides. Let ABC be a right-angled triangle, having the angle BAC a right angle : then shall the square described on the hypotenuse BC be equal to the sum of the squares described on BA, AC. Construction. On BC describe the square BDEC; I. 46. and on BA, AC describe the squares BAGF, ACKH. Through A draw AL parallel to BD or CE : I . 31. and join AD, FC. Proof. Then because each of the angles BAC, BAG is a right angle, therefore CA and AG are in the same straight line. I . 14. N o w the angle CBD i s equal to the angle FBA, for each of them is a right angle. Add to each the angle ABC : then the whole angle ABD i s equal to the whole angle FBC. BOOK I . PROP. 47. S3 Then in the triangles ABD, FBC, ( AB is equal to FB, Because \| and BD is equal to BC, [also the angle A B D is equal to the angle FBC ; therefore the triangle A B D is equal to the triangle FBC. 1.4. N o w the parallelogram BL is double of the triangle ABD, for they are on the same base BD, and between the same parallels BD, AL. I . 41. A n d the square G B is double of the triangle FBC, for they are on the same base FB, and between the same parallels FB, GO. I. 41. But doubles of equals are equal: Ax. 6. therefore the parallelogram BL is equal to the square G B. In a similar way, by joining AE, BK, it can be shewn that the parallelogram CL is equal to the square CH. Therefore the whole square BE is equal to the sum of. the squares GB, H C : that is, the square described on the hypotenuse BC is equal to the sum of the squares described on the two sides BA, AC. Q.E.D. Note. It is not necessary to the proof of this Proposition that the three squares should be described external to the triangle ABC; and since each square may be drawn either towards or away from the triangle, i t may be shewn that there are 2x2x2, or eight, possible constructions. EXERCISES. In the figure of this Proposition, shew that ( i ) If BG, C H are joined, these straight lines are parallel; ( i i ) The points F, A, K are in one straight line; ( i i i ) FC and A D are at right angles to one another; ( i v ) If GH, KE, FD are joined, the triangle G A H is equal to the given triangle in a l l respects; and the triangles FBD, K G E are each equal in area to the triangle ABC.( [See Ex. 9, p. 73.] 6-2 8 4 ' EUCLID'S ELEMENTS. 2. O n the sides AB, A C of any triangle ABC, squares ABFG, A C K H are described both toward the triangle, or both on the side remote from it: shew that the straight lines BH and C G are equal. 3. On the sides of any triangle ABC, equilateral triangles BCX, CAY, A B Z are described, all externally, or all towards the triangle: shew that AX, BY, C Z are all equal. 4. The square described on the diagonal of «. given square, is double of the given square. 5. ABC is an equilateral triangle, and AX is the perpendicular drawn from P i to B C : sheio that the square on A X is three times t l i e square on BX. 6. Describe a square equal to the sum of two given Bquares. 7. From the vertex A of a triangle ABC, AX is drawn perpendi-cular to the base: shew that the difference of the squares on the sides A B and AC, is equal to the difference of the squares on B X and CX, the segments of the base. 8. If from any point O within a triangle ABC, perpendiculars O X , OY, O Z are drawn to the sides BC, CA, A B respectively; shew that the sum of the squares on the segments A Z , BX, C Y is equal to the sum of the squares on the segments AY, C X , B Z . Proposition 47. Alternative Proof. C. B /A K / H Let C A B be a right-angled triangle, having the angle ut A a right angle: then shall the square on the hypotenuse B C be equal to the sum ot the squares on BA, AC. book I. prop. 47. 85 On AB describe the square ABFG. i. 46. From FG and G A cut o f f respectively FD and GK, each equal to AC. i . 3. On G K describe the square G K E H : i . 46. then H G and G F are in the same straight line. 1.14. Join CE, ED, DB. It will first be shewn that the figure C E D B i s the square on CB. Now CA is equal to KG ; add to each AK: therefore C K is equal to AG. Similarly D H is equal to G F: hence the four lines BA, CK, DH, BF are all equal. Then in the triangles BAC, CKE, / BA is equal to CK, Proved. • n J and A C i s equal to KE; Constr, j also the contained angle BAC is equal to the contained \ angle CKE, being right angles; therefore the triangles BAC, C K E are equal in all respects. I. 4. Similarly the four triangles BAC, CKE, DHE, BFD maybe shewn to be equal in all respects. Therefore the four straight lines BC, CE, ED, DB are all equal; that is, the figure C E D B is equilateral. Again the angle C B A is equal to the angle DBF; Proved. add to each the angle A B D : then the angle C B D is equal to the angle ABF: therefore the angle C B D is a right angle. Hence the figure C E D B is the square on BC. Def. 28. And E H G K is equal to the square on AC. Constr. Now the square on CEDB is made up of the two triangles BAC, CKE, and the rectilineal figure A K E D B ; therefore the square C E D B is equal to the triangles EHD, DFB together with the same rectilineal figure; but these make up the squares EHGK, AGFB: hence the square C E D B is equal to the sum of the squares EHGK, AGFB: that is, the square on the hypotenuse BC is equal to the sum of the squares on the two sides CA, AB. < 3 . e. d. 06s. The following properties of a square, though not formally enunciated by Euclid, are employed in subsequent proofs. [See I. 48.] (i) The squares on equal straight lines are equal. (ii) Equal squares stand upon equal straight lines. kg euclid's elements. Proposition 48. Theorem. If the square described on one side of a triangle be equal t o the sum of I l i e squares described on the other t v ; o sides, then the angle contained by these two sides s l i a l l be a right angle. Let ABC be a triangle; and let the square described on BC be equal to the sum of the squares described on BA, AC: then shall the angle BAC be a right angle. Construction. From A draw AD at right angles to AC: 1.11. and make AD equal to AB. I . 3. Join DC. Proof. Then, because AD is equal to AB, Constr. therefore the square on AD i s equal to the square on AB. To each of these add the square on CA; then the sum of the squares on CA, AD i s equal to the sum of the squares on CA, AB. But, because the angle DAC is a right angle, Constr. therefore the square on DC is equal to the sum of t l i e squares on CA, AD. I . 47. And, by hypothesis, the square on BC i s equal to the sum of the squares on CA, AB; therefore the square on DC is equal to the square on BC: therefore also the side DC i s equal to the side BC. Then in the triangles DAC, BAC, I D A i s equal to BA, Constr. and AC is common to both; also the third side DC i s equal to the third side BC; Prored. therefore the angle DAC i s equal to the angle BAC. I . 8. But DAC i s a right angle; Constr. therefore also BAC i s a light angle. Q. E.D. THEOREMS A N D EXAMPLES ON BOOK I . INTRODUCTORY. HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES. ANALYSIS. SYNTHESIS. It is commonly found that exercises in Pure Geometry present to a beginner far more difficulty than examples in any other branch of Elementary Mathematics. This seems to be due to the following causes. (i) The main Propositions in the text of Euclid must be not merely understood, but thoroughly digested, before the exercises depending upon them can be successfully attempted. (ii) • The variety of such exercises is practically unlimited; and it is impossible to lay down for their treatment any definite methods, such as the student has "been accustomed to find in the rules of Elementary Arithmetic and Algebra. ( i i i ) The arrangement of Euclid's Propositions, though per-haps the most convincing of all forms of argument, affords in most cases little clue as to the way in which the proof or con-struction was discovered. Euclid's propositions are arranged, synthetically : that is to say, they start from the hypothesis, or data; they next pro-ceed to a construction in accordance with postulates, and pro-blems already solved; then by successive steps based on known theorems, they finally establish the result indicated by the enun-ciation. . Thus Geometrical Synthesis is a building up of known results, in' order to obtain a new result. But as this is not the way in which constructions or proofs are usually discovered, we draw the attention of the student to the following hints. Begin by assuming the result it is desired to establish; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some simpler theorem which is already known to be true, or on. some condition which suggests.the necessary construction. If this attempt is suc-cessful, the steps of the argument m a y in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form. 88 euclid's elements. This unravelling of the conditions of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis: it is the natural way of attack-ing most exercises of a more difficult type, and it is especially adapted to the solution of problems. These directions are so general that they cannot be said to amount to a method: all that can be claimed for Geometrical Analysis is that it furnishes a mode of searching for a suggestion, and its success will necessarily depend on the skill and ingenuity with which it is employed : these m a y be expected to come with experience, but a thorough grasp of the chief Pro-positions of Euclid is essential to attaining them. The practical application of these hints is illustrated by the following examples. 1. Construct an isosceles triangle having given the base, and the sum of one of the equal sides and the perpendicular drawn from the vertex to the base. Let A B be the given base, and K the sum of one side and the perpendicular drawn from the vertex to the base. Analysis. Suppose ABC to be the required triangle. From C draw C X perpendicular to A B : then A B is bisected at X. i. 26. Now i f we produce X C to H, making X H equal to K, i t follows that C H = C A ; and i f A H is joined, we notice that the angle C A H = the angle CHA. i. 5. Now the straight lines X H and A H can be drawn before the position of C is known ; Hence wo have the following construction, which we arrange Bynthctioallr. THEOREMS and examples.on BOOK I . 89 Synthesis. Bisect AB at X : from X draw X H perpendicular to AB, making X H equal to K. . Join AH. At the point A in HA, make the angle HAC equal to the anale A H X ; and join CB. Then A C B shall be the triangle required. First the triangle is isosceles, for AC = BC. i . 4. Again, since the angle HAC = the angle AHC, Constr. . -. HC = AC. 1.6. To each add CX; then the sum of AC, CX = the sum o f HC, CX = HX. That i s , the sum of AC, CX = K. q . c. r . 2. To divide a given straight line so that the square on one pa may be double of the square on the other. D C Let AB be the given straight line. Analysis. Suppose AB to be divided as required at X : that is, suppose the square on A X to be double of the square on X B. Now we remember' that in an isosceles right-angled triangle, the square on the hypotenuse is double of the square, on either of the equal sides. This suggests to us to draw BC perpendicular to AB, and to make BC equal to BX. Join XC. Then the square on X C is double of the square on XB, i . 47. . -. X C = AX. And when we join AC, we notice that the angle X AC = the angle X C A. j. 5. Hence the exterior angle C X B i s double of the angle XAC. ± . 32. But the angle C X B is half of a right angle : i . 32. . • . the angle X A C i s one-fourth of a right angle. This supplies the clue to the following construction:— 90 Euclid's elements. Synthesis. From B draw BD perpendicular to AB ; and from A draw AC, making BAC one-fourth of a right angle. From C, the intersection of A C and BD, draw CX, making the angle A C X equal to the angle BAC. > . 23. Then AB shall be divided as required at X. For since the angle X C A = the angle XAC, . -. XA=XC. 1 . 6 . And because the angle B X C = the sum of the angles BAC, ACX, i . 32. . " . the angle B X C is half a right angle; and the angle at B is a right angle; therefore the angle B C X i s half a right angle; i . 32. therefore the angle B X C = the angle B C X ; . -. BX = BC. Hence the square on X C is double of the square on X B : i . 47. that is, the square on AX is double of the square on XB. q.e.f. I . ON THE IDENTICAL EQUALITY OF TRIANGLES. See Propositions 4, 8, 26. 1. If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles. 2. If the bisector of the vertical angle of a triangle is also per-pendicular to the base, the triangle i s isosceles. 3. If the bisector of the vertical angle of a triangle also bisect the base, the triangle is isosceles. [Produce the bisector, and complete the construction after the manner of i . 16.] 4. If in a triangle a pair of straight lines drawn from the ex-tremities of the base, making equal angles with the sides, are equal, the triangle is isosceles. 5. If in a triangle the perpendiculars drawn from the extremities of the base to the opposite sides are equal, the triangle is isosceles. 6. Two triangles ABC, ABD on the same base AB, and on opposite Bides of i t , are such that A C is equal to AD, and BC i s equal to B D : shew that the line joining the points C and D is perpendicular to AB. 7. If from the extremities of the base of an isosceles triangle per pendiculars aro drawn to the opposite sides, shew that the straight line joining the vertex to the intersection of these perpendiculars bisects the vertical angle. THEOREMS AND EXAMPLES ON BOOK I . 9 1 8. A B C is a triangle in which the vertical angle BAC i s bisected by the straight line A X : from B draw BD perpendicular to AX, and produce i t to meet AC, or A C produced, in E; then shew that BD is equal to D E, 9. In a quadrilateral ABCD, AB is equal to AD, and BC is equal to DC: shew that the diagonal A C bisects each of the angles which i t joins. 10. In a quadrilateral A B C D the opposite sides AD, BC are equal, and also the diagonals AC, BD are equal: i f A C and BD intersect at K, shew that each of the triangles AKB, D K C is isosceles. 11. If one angle of a triangle be equal to the sum of the other t the greatest side i s double of the distance of i t s middle point from the opposite angle. 12. Two right-angled triangles which have their hypotenuses equal and one side of one equal to one side of the other, are identically equal. Let ABC, DEF be two As right-angled at B and E, having A C equal to DF, and AB equal to D E : then shall the As be identically equal. For apply the a A B C to the a DEF, so that A may f a l l on D, and AB along D E ; and so that C may f a l l on the side of DE remote from F . -Let C be the point on which C f a l l s . Then since AB = DE, . -. B must f a l l on E; so that D E C represents the A A B C in i t s new position. Now each of the z s DEF, D E C is a rt. i _ ; . • , EF and E C are in one st. line. Then in the A C D F , because D F = D C , . -. the / DFC=the l DCF. Hence in the two a3 DEF, D E C , ( the L DEF = the / D E C , being rt. Ls; Because^and the L DFE=the / D C E ; ( also the side D E is common to both; , \ the As DEF, D E C are equal in a l l respects; that is, the As DEF, A B C are equal in a l l respects. Hyp. 1.14. 5 . Proved. . . 2G. Q.E.D. 92 Euclid's elements. 13. If two triangles have two sides of the one equal to two sides of the o t l i e r , each to each, and have likewise t l i e angles opposite to one pair of equal sides equal, tlien the angles opposite to the other pair of equal sides are either equal or supplementary, and in t l i e former ease t l i e tri-angles are equal in all respects. Let ABC, DEF be two As, having the side A B equal to the side DE, the side A C equal to the side DF, also the Z A B C equal to the L D E F : then shall the / s ACB, D F E be either equal or supplementary, and in the former case, the As shall be equal in all respects. Apply the A A B C to the A DEF, so that A may fall on D, and A B along D E ; then because A B = D E, Hyp. . -. B will fall on E : and because the Z ABC=the z DEF, Hyp. . -. B C will fall along EF: Then must C fall on F, or in EF, or EF produced. If C falls on F, the A s coincide, and therefore are identically equal: so that the z A C B = the z DFE. But if C falls in EF, or EF produced, as at C: then D E C represents the A A B C in its new position. Then because DF=; AC, Hyp. . -. DF = DC, . -. t h e Z DCF = the z DFC.. t . o . But the Z5 DFC, DFE are supplementary; i . 13, . -. the Z s D C F , D F E are supplementary: that is, the Z 8 ACB, D F E are supplementary. y.K.n. Three cases of this theorem deserve special attention. It has been proved that i f the angles ACB, D F E are not equal, they are supplementary; And we know that of angles whioh are supplementary and un-equal, one must be acute and the other obtuse. THEOREMS A N D EXAMPLES O N BOOK I. 93 Corollaries. Hence if in addition to the hypothesis of this theorem it is given (i) That-the angles ACB, DFE, opposite to the two equal sides AB, D E are both acute, both obtuse, or i f one of them is a right angle, it follows that these angles are equal, and therefore that the triangles are equal in all respects. (ii) That the two given angles are right angles or obtuse angles, it follows that the angles A C B , D F E must be both acute, and therefore equal, by (i): so that the triangles are equal in all respects. (hi) That in each triangle the side opposite the given angle is not less than the other given side; that is, if A C and DF are not less than A B and D E respectively, then the angles A C B , D F E cannot be greater than the angles A B C , DEF, respectively; therefore the angles A C B , DFE, are both acute ; hence, as above, they are equal; and the triangles A B C , D E F are equal in all respects. II. ON INEQUALITIES. See Propositions 16, 17, 18, 19, 20, 21, 24, 25. 1. In a triangle ABC, if AC is not greater than.AB, shew that any straight line drawn through the vertex A, and terminated by the base BC, is less than AB. 2. ABC is a triangle, and the vertical angle BAC is bisected by a straight line which meets the base B C in X ; shew that B A is greater than BX, and C A greater than C X . Hence obtain aproofofi. 20. 3. The • perpendicular is the shortest straight line that can be drawn from a given point to a given straight line ; and of others, that which is nearer to the perpendicular is less than the more remote ; and two, and only two equal straight lines can be drawn from the given point to the given straight line, one on each side of the perpendicular. 4. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter. 5. The sum of the distances of any point within a triangle from its angular points is less than the perimeter of the triangle. 94 Euclid's elements. Ii. The perimeter of a quadrilateral is greater than the sum of diagonals. 7. The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case. 8. In a triangle any two sides are together greater than twice t median which bisects the remaining side. [See Def. p. 73.] [Produce the median, and complete the construction after the manner of i . 16.] 9. In any triangle the sum of the medium is less than the peri-meter. 10. In a triangle an angle is acute, obtuse, or a right angle, according as the median drawn from i t is greater than, less than, or equal to half the opposite side. [See Ex. 4, p. 59.[ 11. The diagonals of a rhombus are unequal. 12. If the vertical angle of a triangle is contained by unequal sides, and if from the vertex the median and the bisector of the angle are drawn, then the median l i e s within t l i e angle contained by the bisector and the longer side. Let ABC be a A, in which AB is greater than AC; l e t AX be the median drawn from A, and AP the bisector of the vertical ZBAC: then shall A X l i e between AP and AB. Produce AX to K, making XK equal to AX. Join KC. Then the A" BXA, C X K may be shewn to be equal in a l l respects; I . 4. hence BA = CK, and the Z BAX = the z CKX. But since BA i s greater than AC, Hyp. : . CK is greater than AC; . -. the z CAK i s greater than the zCKA: i 1 - . that is, the Z C A X is greater than the Z BAX : . -. the Z CAX must be more than half the vert. Z BAC; hence AX l i e s within the angle BAP. q.k.k. 13. If two sides of a triangle are unequal, and if from their po of intersection three straight l i n e s are drawn, namely the bisector of the vertical angle, t l i e median, and the perpendicular to the base, the first i s intermediate in position and magnitude t o the other two. THEOREMS AND EXAMPLES ON BOOK 1. 95 III. ON PARALLELS. See Propositions 27—31. 1. If a straight line meets two parallel straight lines, and the two interior angles on the same side are bisected; shew that the bisectors meet at right angles, [i. 29, i. 32.] 2. The straight lines drawn from any point in the bisector of an angle parallelto the arms of the angle,1 and terminated by them, are equal; and the resulting figure is a rhombus. 3. A B and C D are two straight lines intersecting at D, and the adjacent angles so formed are bisected: if through any point X in D C a straight line Y X Z be drawn parallel to A B and meeting the bisectors in Y and Z, shew that X Y is equal to X Z . 4. If two straight lines are parallel to two other straight lines, each to each; and if the angles contained by each pair are bisected; shew that the bisecting lines are parallel. 5. The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels. 6. A straight line drawn between two parallels and terminated by them, is bisected ; shew that any other straight line passing through the middle point and terminated by the parallels, is also bisected at that point. 7. If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal. Problems. 8. AB and CD are two given straight lines, and X is a given point in A B : find a point Y in A B such that Y X may he equal to the perpendicular distance of Y from CD. 9. -ABC is an isosceles triangle; required to draw a straight line D E parallel to the base BC, and meeting the equal sides in D and E, so that BD, DE, E C may be all equal. 10. A B C is any triangle; required to draw a straight line D E parallel to the base BC; and meeting the other sides in D and E, so that D E m a y be equal to the sum of B D and CE. 11. ABC is any triangle; required to draw a straight line parallel to the base BC, and meeting the other side,s in D and E, so that D E may be equal to the. difference of B D and CE. 96 euclid's elements. IV. ON PARALLELOGRAMS. See Propositions 33, 34, and the deductions from these Props. given on page 64. 1. The straight line drawn through the middle point of a side of triangle parallel t o the base, b i s e c t s the remaining side. Let ABC be a A, and Z the middle point of the side AB. Through Z, Z Y i s drawn par1 to BC; then shall Y be the middle point of AC. Through Z draw Z X par1 to AC. i.31. Then in the A AZYi ZBX, because Z Y and BC are par', .-.the Z A Z Y = the z Z B X ; 1.29. and because Z X and A C are par', . -. the z ZAY = the z BZX; i . 29. also A Z = Z B : Hyp. .: AY = ZX. 1.26. But Z X C Y is a par"1 by construction; . -. Z X = YC. U 34. Hence AY = YC; that is, A C is bisected at Y. q.e.d. 2. The straight line which joins the middle points of two sides triangle, i s parallel t o the third side. Let ABC be a A, and Z, Y the middle points of the sides AB, AC: then shall Z Y be par' to BC. Produce Z Y to V, making YV equal to ZY. Join CV. Then in the A'AYZ.CYV, ( AY = CY, Hyp. Because' andYZ=YV, Constr. I and the z AYZ = the vert. opp. Z CYV; . -. A Z = CV, and the z ZAY = the z VCY; hence CV is par' to AZ. i . 27. But CV i s equal to AZ, that is. to B Z : Hyp. . • . CV is equal and par1 to B Z : . • . ZV is equal and par1 to B C : 1.33. that is, Z Y i s par' to BC. q.bjd. [A second proof of this proposition may be derived from i. 38,3 THEOREMS AND EXAMPLES ON BOOK I. 97 3. The straight line which joins the middle points of two sides of a triangle is equal to half the third side. 4. Shew that tlie three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identi-cally equal. 5. Any straight line drawn from tlie vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle. 6. Given the three middle points of the sides of a triangle, con-struct the triangle. 7. AB, AC are two given straight lines, and P is a given point between them; required to draw through P a straight line termi-nated by AB, A C , and bisected by P. 8. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides A D , B C : shew that B X and D Y trisect the dia-gonal A C . 9. If the middle points of adjacent sides of any quadrilateral be joined, the figure thus formed is a parallelogram. 10. Shew that the straight lines which join the middle points of opposite sides of a quadrilateral, bisect one another. 11. The straight line which joins the middle points of the oblique sides of a trapezium, is parallel to the two parallel sides, and passes through the middle points of the diagonals. 12. The straight line which joins the middle points of tlie oblique sides of a trapezium is equal to half the sum of the parallel sides ; and the portion intercepted between the diagonals is equal to half the difference of the parallel sides. Definition. If from, the extremities of one straight line per-pendiculars are drawn to another, the portion of the latter intercepted between the perpendiculars is said to be the Ortho-gonal Projection of the first line upon the second. B Y Q B x ^ 1 P \ P ^ Y Q Thus in the adjoining figures, if from the extremities of the Btraight line A B the perpendiculars AX, BY are drawn to PQ, then X Y is the orthogonal projection of A B on PQ. H. E. 7 9 8 EUCLID S ELEMENTS. 13. A given straight l i n e AB i s bisected at C; shew t l u i t the pro-jections of AC, C B on any other straight line are equal. Let X Z , Z Y be the projections of AC, C B on any straight line PQ: then X Z and Z Y shall be equal. Through A draw a straight line parallel to PQ, meeting CZ, BY or these lines produced, in H, K. i . 31. Now AX, C Z , BY are parallel, for they are perp. to PQ; i . 28. . -. the figures XH, HY are par"; . -. AH = X Z , and H K = Z Y . i 34. But through C, the middle point of AB, a side of the a ABK, CH has been drawn parallel to the side BK ; . -. C H bisects AK: Ex. 1, p. 96. that is, AH = HK; . -. X Z = ZY. Q.E.D. 14. If three parallel straight lines make equal intercepts on a fourth straight line which meets them, they u r i l l also make equal i n t e r -cepts on any other straight l i n e which meets them. 15. Equal and parallel straight lines hare equal projections on a other straight line. 16. AB is a given straight line bisected at O; and AX, BY are perpendiculars drawn from A and B on any other straight line: shew that O X i s equal to OY. 17. AB is a given straight line bisected at O : and AX. BY and OZ are perpendiculars drawn to any straight line PQ, which does not pass between A and B: shew that O Z i s equal t o half the sum of AX, BY. [OZ is said to be the Arithmetic Mean between AX and BY.] 18. AB i s a given straight line bisected at O: and through A. B and O parallel straight lines are drawn to meet a given straight line PQ in X, Y, Z: shew that O Z is equal to half the sum, or half the difference of A X and BY, according as A and B l i e on the same side or on apposite sides of PQ. THEOREMS AND EXAMPLES ON BOOK I. 99 19. parts. To divide a given finite straight line into any number of equal A [For example, required to divide the straight line A B into five equal parts. From A draw AC, a , straight line of un-limited length, making any angle with AB. In A C take any point P, and mark off successive parts PQ, QR, RS, S T each equal to AP. Join B T ; and through P, Q, R, S draw parallels to B T . It may be shewn by Ex. 14, p. 98, that these parallels divide A B into five equal parts.] 20. If through an angle of a parallelogram any straight line i s drawn, the perpendicular drawn to it from the opposite angle is equal to the sum or difference of the perpendiculars drawn to i t from the two remaining angles, according as the given straight line falls without t l i e parallelogram, or intersects it. [Through the opposite angle draw a straight line parallel to the given straight line, so as to meet the perpendicular from one of the remaining angles, produced if necessary: then apply i. 34, i. 26. Or proceed as in the following example.] 21. From the angular points of parallelogram perpendiculars -are drawn to any straight line which is without the parallelogram: shew that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair. [Draw the diagonals, and from their point of intersection let fall a perpendicular upon the given straight line. See Ex. 17, p. 98.] 22. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal sides is equal to the perpendi-cular drawn from either extremity of the base to the opposite side. [It follows that the sum of the distances of any point in the base of ah isosceles triangle from the equal sides is constant, that is, •the same whatever point in the base is taken.] 23. In the base produced of an isosceles triangle any point is taken: shew that the difference of its distances from the equal sides is constant. 24. The sum of the perpendiculars drawn from any point within an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore constant. 100 EUCLID'S ELEMENTS. Problems. [Problems marked () admit of more than one solution.] 25. Draw a straight hue through a given point, so that the part of i t intercepted between two given parallel straight lineB may be of given length. 26. Draw a straight line parallel to a given straight line, so that the part intercepted between two other given straight lines may be of given length. 27. Draw a straight line equally inclined to two given straight lines that meet, so that the part intercepted between them may be of given length. 28. AB, AC are two given straight lines, and P is a given point without the angle contained by them. It is required to draw through P a straight line to meet the given lines, so that the part intercepted between them may be equal to the part between P and the nearer line. V. MISCELLANEOUS THEOREMS AND EXAMPLES. Chiefly on I. 32. 1. A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that A D is equal to BA ; if D C is drawn, shew that B C D is a right angle. 2. The straight line joining the middle point of the hypotermse of a right-angled triangle to the right angle is equal to half the hypotenuse. 3. From the extremities of the base of a triangle perpendiculars are drawn to the opposite sides (produced i f necessary); shew that the straight lines which join the middle point of the base to the fast of the perpendiculars are equal. 4. In a triangle ABC, A D is drawn perpendicular to B C ; and X, Y, Z are the middle points of the sides BC, CA, A B respective!;); shew that each of t l i e angles Z X Y , Z D Y i s equal to the angle BAC. 5. In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the two triangles thus formed are equiangular to one another. 6. In a right-angled triangle two straight lines are drawn from the right angle, one bisecting the hypotenuse, the other perpendicular to it: shew that they contain an angle equal to the diftrente o/tAe.iioo acute angles of the triangle. [See above, Ex. 2 and Ex. &.] THEOREMS AND EXAMPLES ON BOOK I . 101 7. In a triangle if a perpendicular be drawn from one extremity of t l i e base to the bisector of t l i e vertical angle, ( i ) it will make with either of the sides containing t l i e vertical angle an angle equal to half the sum of the angles at the base; ( i i ) it will make with the base an angle equal t o half the difference of the angles at the base. Let ABC be the given A, and AH the bi-sector of the vertical z BAC. Let CLK meet AH at right angles. ( i ) Then shall each of the zs AKC, ACK be equal to half the sum of the z • ABC, ACB. In the A8AKL, ACL, B f the Z KAL=the Z CAL, Because -l also the z ALK = the z ALC, being rt. [ and AL is common to both AB; . -. the z AKL=the z ACL. i . 26. Again, the Z AKC=the sum of the Z 3 KBC, KCB ; i. 32. that is, the Z A C K = the sum of the z B KBC, KCB. To each add the z ACK, then twice the z ACK = the sum of the z • ABC, ACB, . -. the / ACK=half the sum of the Z» ABC, ACB. (ii) The z KCB shall be equal to half the difference of the Z8 ACB, ABC. As before, the z ACK = the sum of the Z > KBC, KCB, To each of these add the z KCB : then the z ACB = the z KBC together with twice the Z KCB. . -. twice the Z KCB = the difference of the Z»ACB, KBC, that is, the z KCB=half the difference of the Z ' ACB, ABC. Corollary. IfXbe the middle point of the base, and XL be joined, it may be shewn by Ex. 3, p. 97, that X L is half BK; that is, that X L is half the difference of the sides AB, AC. 8. In any triangle the angle contained by the bisector of the vertical angle and the perpendicular from the vertex to the base is equal to half the difference of the angles at the base. [See Ex. 3, p. 59.] 9. In a triangle A B C the side A C is produced to D, and the angles BAC, B C D are bisected by straight lines which meet at F; shew that they contain an angle equal to half the angle at B. 10. If in a right-angled triangle one of the acute angles is doubl of the other, shew that the hypotenuse i s double of the shorter side. 11. If in a diagonal of a parallelogram any two points equidistant from its extremities be joined to the opposite angles, thefigurethus formed will be also a parallelogram. 102 eucxid's elements. 12. ABC is a given equilateral triangle, and in the sides BC, CA, AB the points X, Y, Z are taken respectively, so that BX, C Y and A Z are all equal. AX, BY, C Z are now drawn, intersecting in P, Q, R: Bhew that the triangle P Q R is equilateral. 13. If in the sides AB, BC, CD, DA of a parallelogram ABCD four points P, Q, R, S be taken in order, one in each side, so that AP, BQ, CR, D S are air egual; shew that the figure P Q R S is a parallelo-gram. 14. In the figure of i. 1, if the circles intersect at F, and if C A and C B are produced to meet the circles in P and Q respectively; shew that the points P, F, Q are in the same straight line; and shew also that the triangle C P Q is equilateral. [Problems marked () admit of more than one solution.] 15. Through two given points draw two straight lines forming with a straight line given in position, an equilateral triangle. 16. From a given point it is required to draw to two parallel straight lines two equal straight lines at right angles to one another. 17. Three given straight lines meet at a point; draw another straight line so that the two portions of it intercepted between the given lines may be equal to one another. 18. From a given point draw three straight lines of given lengths, so that their extremities may be in the same straight line, and inter-cept equal distances on that line. [See Fig. to i. 16.] 19. Use the properties of the equilateral triangle to trisect a given finite straight line. 20. In a given triangle inscribe a rhombus, having one of its angles coinciding with an angle of the triangle. VI. ON THE CONCURRENCE OF STRAIGHT LINKS IN A TRIANGLE. Definitions, (i) Three or more straight lines are said to be concurrent when they meet in one point. (ii) Three or more points are said to be collinear when they lie upon one straight line. "We here give some propositions relating to tlie concurrence of certain groups of straight lines drawn in" a triangle: the im-portance of these theorems will he more fully appreciated when the student is familiar with Books nr. and IV. THEOREMS, AND EXAMPLES ON BOOK I . 1 0 3 1. The perpendiculars drawn to the sides of a triangle from their middle points are concurrent. Let ABC be a a, and X, Y, Z the middle points of its sides : then shall the perp8 drawn to the sides from X, Y, Z be concurrent. From Z and Y draw perps to AB, AC; these perps, since they cannot be parallel, will meet at point O. Ax. 12. Join OX. It is required to prove that OX is perp. to BC. Join OA, OB, OC. In the As OYA, OYC, Because Pr Def. 7. YA = YC, Because \ , • and OY is common to both; also the Z OYA=the Z OYC, being rt. L . -. OA = OC. Similarly, from the A8 OZA, OZB, i t may be proved that OA = OB. Hence OA, OB, OC are a l l equal. Again, in the A8 OXB, OXC BX = CX, and XO i s common to both; also OB = OC: -. the Z OXB = the Z OXC; but these are adjacent Z s ; . -. they are rt. L s ; that is, O X is perp. to B C . Hence the three perps O X , O Y , O Z meet in the point O. Q. E. 1) 2. The bisectors of the angles of a triangle are concurrent. Let ABC be a A. Bisect the Z 8 ABC, A BCA, by straight lines which must meet at some point O. Ax. 12. Join A O . It is required to prove that A O bisects the Z BAC. From O draw OP, OQ, OR perp. to the sides of the a . Then in the A8 OBP, OBR, the Z OBP=the Z Because 1 and the Z OPB Hyp. Hyp. i. 8. OBR, the Z ORB, being rt. L and OB is-common ; . -. OP = OR. I. 26. 1 0 4 EUCLID'S ELEMENTS. Similarly from the A'OCP, OCQ, i t may be shewn that OP = OQ, . -. OP, OQ, OR are all equal. Again in the A' OR A, OQA, .the Z'ORA, OQA are rt. i _ " , „ land the hypotenuse OA i s use< common, ( alsoOR = OQ; Proved. B . • . the L RAO = the z QAO. Ex. 12, p. 91. That is, AO is the bisector of the L BAC. Hence the bisectors of the three Z B meet at the point O. Q. E. D. 3. The bisectors of two exterior angles of u, triangle and the bisector of the third angle are concurrent. Let ABC be a A, of which the sides AB, A C are produced to any points D and E. Bisect the Z' D B C , E C B by straight lines which must meet at some point O. Ax. 12. Join AO. It is required to prove that A O bisects the angle BAC. From O draw OP, O Q , O R perp. to the sides of the A. Then in the As O B P , O B R , /the Z O B P = the Z O B R , Constr. \ also the L O P B = the Z O R B , Because • being rt. L • Because and O B is common; . -. OP=OR. Similarly in the A8 OCP, OCQ, i t may be shewn t h a t OP=OQ: . -. OP, OQ, O R are all equal. Again in the A8 OR A, OQA, the z " ORA, OQA are rt. L", and the hypotenuse OA is common, also OR = O Q ; . -. the L RAO = the z QAO. Proved. Ex. 12, p. 91. That is, AO is the bisector of the z BAC. the biseotors of the two exterior z • DBC, ECB, and of the interior L BAC meet at the point O. Q.K.IJ. THEOREMS AND EXAMPLES ON BOOK I . 1 0 5 4. The medians of a triangle are concurrent. Let ABC be a A. Let BY and CZ be two of its medians, and let them intersect at O. Join AO, and produce it to meet B C in X. It is required to shew that A X is the remaining median of the A. Through C draw C K parallel to BY: produce A X to meet C K at K. Join BK. In the A AKC, because Y is the middle point of AC, and Y O is parallel to CK, . • . O is the middle point of AK. Again in the A ABK, since Z and O are the middle points of AB, AK, . -. Z O is parallel to BK, Ex. 2, p. 96. that is, O C is parallel to BK: . -. the figure B K C O is a par". But the diagonals of a par" bisect one another, Ex. 5, p. 64. . • . X is the middle point of BC. That is, A X is a median of the a . Hence the three medians meet at the point O. q.e.d. Ex. 1, p. Corollary. The three medians of a triangle cut one another at a point of trisection, the greater segment in each being towards the angular point. For in the above figure it has been proved that A O = OK, also that O X is half of O K ; . -. O X is half of O A : that is, O X is one third of AX. Similarly O Y is one third of BY, and O Z is one third of C Z . Q.E.D. By means of this Corollary i t may be shewn that in any triangle the shorter median bisects the greater side. [The point of intersection of the three medians of a triangle is called the centroid. It is shewn in mechanics that a thin triangular plate will balance in any position about this point: therefore the centroid of a triangle is also its centre of gravity.] 1 0 6 EUCLID'S ELEMENTS. " ; . The perpendiculars drawn from t l i e vertices of a triangle to the opposite sides are concurrent. Let A B C be a A, and AD, BE, C F the three perp" drawn from the vertices to the opposite sides: then shall these perp" be concurrent. Through A, B, and C draw straight lines MN, NL, LM parallel to the opposite sides of the a . Then the figure B A M C is a par™ Def. 26. . -. AB=MC. 1.34. Also the f i g u r e BACL i s a par™. . -. AB = LC, . -. LC = C M : that is, C is the middle point of LM. So also A and B are the middle points of M N and N L. Hence AD, BE, C F are the perp8 to the sides of the A L M N from their middle points. Ex. 3, p. 54. But these perp' meet in a point: Ex. 1, p. 103. that is, the perp8 drawn from the vertices of the a A B C to the opposite sides meet in a point. q.e.d. [For another pro if see Theorems and Examples on Book in.] Definitions. (i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its ortho-centre. (ii) The triangle formed by joining the feet of the perpen-diculars is called the pedal triangle. THEOREMS AND EXAMPLES ON BOOK I . 1 0 7 K VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS. No general rules can be laid down for the solution of problems in this section; but in a few typical cases we give constructions, which the student will find little difficulty in adapting to. other questions of the same class. 1. Construct a right-angled triangle, having given the hypotenuse and t l i e sum of the remaining sides. [It is required to construct a rt. angled A,. having its hypotenuse equal to the given straight line K, and the sum of its remaining sides equal to AB. From A draw A E making with BA an Z equal to half a rt. \| _ . From centre B, with radius equal to K, de-scribe a circle cutting A E in the points C, C. From C and C draw perp8 CD, CD' to AB; and join CB, CB. Then either of the A" C D B , C D ' B will satisfy the given conditions. Note. If the given hypotenuse K be greater than the perpendicu-lar drawn from B to AE, there will be two solutions. If the line K be equal to this perpendicular, there will be one solution; hut i f less, the problem is impossible.] 2. Construct a right-angled triangle, having given the hypotenuse and the difference of the remaining sides. 3. Construct an isosceles right-angled triangle, having given the sum of the hypotenuse and one side. 4. Construct a triangle, l i a v i r i g given t l i e perimeter and the angles at the base. P [Let A B be the perimeter of the required A, and X and Y the Z ' the base. From A draw AP, making the L BAP equal to half the Z X. From B draw BP, making the Z A B P equal to half the Z Y. From P draw PQ, making the / A P Q equal to the L BAP.-From P draw PR, making the z BPR equal to the L ABP. Then shall P Q R be the required A .] a t 1 0 8 EUCLID'S ELEMENTS. 5. Construct a right-angled triangle, having given the perimeter and one acute angle. 6. Construct an isosceles triangle of given altitude, so that its bass may bo in a given straight line, and its two equal sides may pass through two fixed points. [See Ex. 7, p. 49.] 7. Construct an equilateral triangle, having given the length of the perpendicular drawn from one of the vertices to the opposite side. 8. Construct an isosceles triangle, having given the base, and the difference of one of the remaining sides and the perpendicular drawn from the vertex to the base. [See Ex. 1, p. 88.] 9. Construct a triangle, having given the base, one of the angles at the base, and the sum of the remaining sides. 10. Construct a triangle, having given the base, one of the angles at the base, and the difference of the remaining sides. 11. Construct a triangle, having given the base, tlie difference of the angles at the base, and the difference of the remaining sides. 1 \ [Let A B be the given base, X the difference of the Z • at tlie base, and K the difference of the remaining sides. Draw BE, making the Z A B E equal to half the z X. From centre A, with radius equal to K, describe a circle cutting B E in D and D'. Let D be the point of intersection nearer to B. Join A D and produce it to C. Draw BC, making the z D B C equal to the z B D C . Then shall C A B be Jhe A required. Ex. 7, p. 101. Note. This problem is possible only when the given difference K is greater than the perpendicular drawn from A to BE.] 12. Construct a triangle, having given the base, the difference of the angles at the base, and the sum of the remaining sides. 13. Construot a triangle, having given tlie perpendioular from the vertex on the base, and the difference between each side and t l i e adjacent segment of tho base. THEOREMS A N D EXAMPLES ON BOOK I. 109 14. Construot a triangle, having given two sides and the median which bisects the remaining side. [See Ex. 18, p. 102.] 15. Construct a triangle, having given one side, and the medians which bisect the two remaining sides. [See Fig. to Ex. 4, p. 105. Let B C be the given side. Take two-thirds of each of the given medians; hence construct the triangle B O C . The rest of the con-struction follows easily.] 16. Construct a triangle, having given its three medians. [See Fig. to Ex. 4, p. 105. Take two-thirds of each of the given medians, and construct the triangle O K C . The rest of the construction follows easily.] vni. ON AREAS. See Propositions 35—48. It must be understood that throughout this section the word equal as applied to rectilineal figures will be used as denoting equality of area unless otherwise stated. 1. Shew that a parallelogram is bisected by any straight line which passes through the middle point of one of its diagonals, [i. 29, 26.] 2. Bisect a parallelogram by a straight hne drawn through a given point. 3. Bisect a parallelogram by a straight line drawn perpendicular to one of its sides. 4. Bisect a parallelogram by a straight line drawn parallel to a given straight line. 5. ABCD is a trapezium in which the side AB is parallel to DC. Shew that its area is equal to the area of a parallelogram formed by drawing through X, the middle point of BC, a straight line parallel to AD. [i. 29, 26.] 6. A trapezium is equal to a parallelogram whose base is half the sum of the parallel sides of the given figure, and whose altitude is equal to the perpendicular distance between them. 7. ABCD is a trapezium in which the side AB is parallel to DC; shew that it is double of the triangle formed by joining the extremities of A D to X, the middle point of BC. 8. Shew that a trapezium is bisected by the straight line which Joins the middle points of its parallel sides. [i. 38.] 110 EUCLID'S ELEMENTS. In the following group of Exercises the proofs depend chiefly on Propositions 37 and 38, and the two converse theorems. 9. If two straight lines AB, CD intersect at X, and if the straight lines A C and BD, which join their extremities are parallel, shew that the triangle A X D is equal to the triangle B X C . 10. If two straight lines AB, CD intersect at X, so that the triangle A X D is equal to the triangle X C B , then A C and B D are parallel. 11. ABCD is a parallelogram, and X any point in the diagonal A C produced; shew that the triangles X B C , X D C are equal. [See Ex. 13, p. 64.] 12. ABC is a triangle, and R, Q the middle points of the sides AB, A C ; shew that if B Q and C R intersect in X, the triangle B X C is equal to the quadrilateral A Q X R . [See Ex. 5, p. 73.] 13. If the middle points of the sides of a quadrilateral be joined in order, the parallelogram so formed [see Ex. 9, p. 97] is equal to half the given figure. 14. Two triangles of equal area stand on the same base but on opposite aides of it: shew that the straight line joining their vertices is bisected by the base, or by the base produced. 15. The straight line which joins the middle points of the dia^ gonals of a trapezium is parallel to each of the two parallel sides. 16. (i) A triangle is equal to the sum or difference of two triangles on the same base (or on equal bases), if the altitude of the former is equal to the sum or difference of the altitudes of t l i e latter. (ii) A triangle is equal to the sum or difference of two triangles of the same altitude if the base of the former is equal to the sum or differ-ence of the bases of the latter. Similar statements hold good of parallelograms. 17. ABCD is a parallelogram, and O is any point outside it; shew that the sum or difference of the triangles O A B . O C D is equal to half the parallelogram. Distinguish between the two eases. On the following proposition depends an important theorem in Mechanics: we give a proof of tho first ease, leaving the second case to be deduced by a similar method. THEOREMS AND EXAMPLES ON BOOK I . Ill 18. (i) ABCD is a parallelogram, and O is any point without the angle BAD and its opposite vertical angle; shew t l i a t the triangle O A C i s equal t o the sum of the triangles OAD, OAB. (ii) If O is within the angle BAD or its.opposite vertical angle the triangle O A C i s equal to the difference of the triangles OAD, OAB. Case I. If O is without the z DAB and i t s opp. vert, z, then O A is with-out the par" A B C D : therefore the perp. drawn from C to OA is equal to the sum of the perp8 drawn from B and D to OA. [See Ex. 20, p. 99.] Now the A8 OAC, OAD, O A B are upon the same base O A; and the altitude of the A O A C with respect to this base has been shewn to be equal to the sum of. the altitudes of the As OAD, OAB. Therefore the A O A C i s equal to the sum of the AB OAD, OAB. [See Ex. 16, p. 110.] q.e.b. 19. ABCD is a parallelogram, and through O, any point within i t , straight lines are drawn parallel to the sides of the parallelogram; shew that the difference of the parallelograms DO, BO is double of the triangle AOC. [See preceding theorem ( i i ) . ] 20. The area of a quadrilateral is equal to the area of a triang having two of i t s sides equal to the diagonals of the given figure, and the included angle equal to either -of the angles between the dia-gonals. 21. A B C i s a triangle, and D i s any point in AB: i t i s required t o draw through D a straight line D E to meet BC produced in E, so that the triangle D B E may be equalto the triangle ABC. [Join DC. Through A draw AE parallel to DC. i . 31. Join DE. The A EBD shall be equal to the A ABC. i . 37.] 112 euclid's elements, 22. On a base of given length describe a triangle equal to a given triangle and having an angle equal to an angle of the given triangle. 23. Construct a triangle equal in area to a given triangle, and having a given altitude. 24. On a base of given length construct a triangle equal to a given triangle, and having its vertex on a given straight line. 25. On a base of given length describe (i) an isosceles triangle; ( i i ) a right-angled triangle, equal to a given triangle. 26. Construct a triangle equal to the sum or difference of two given triangles. [See Ex. 16, p. 110.] 27. ABC is a given triangle, and X a given point: describe a triangle equal to A B C , having its vertex at X, and its base in the same straight line as BC. 28. ABCD is a quadrilateral: on the base AB construct a triangle equal in area to A B C D , and having the angle at A common with the quadrilateral. [Join BD. Through C draw C X parallel to BD, meeting A D pro-duced in X ; join BX.] 29. Construct a rectilineal figure equal to a given rectilineal figure, and having fewer sides by one than t l i e given figure. Hence shew how to construct a triangle equal to a given rectilineal figure. 30. ABCD is a quadrilateral: it is required to construct a triangle equal in area to A B C D , having its vertex at » given point X in DC, and its base in the same straight line as AB. 31. Construct a rhombus equal to a given parallelogram. 32. Construct a parallelogram which shall have the same area and perimeter as a given triangle. 33. Bisect a triangle by a straight line drawn through one of i t s angular points. 34. Trisect a triangle by straight lines drawn through one of its angular points. [See Ex. 19, p. 102, and i. 38.] 35. Divide a triangle into any number of equal parts bv straight lines drawn through one of its angular points. \|See Ex. 19, p. 99, and r . 38.] THEOREMS AND EXAMPLES ON BOOK I . 1 1 3 36. Bisect a triangle by a straight line drawn through a given point in one of its sides. [Let ABC be the given A, and P the given point in the side AB. Bisect A B at Z ; and join C Z , CP. Through Z draw Z Q parallel to CP. Join PQ. Then shall P Q bisect the a . See Ex. 21, p. 111.] 37. Trisect a triangle by straight lines drawn from a given point in one of its sides. [Let ABC be the given a , and X the given point in the side BC. Trisect B C at the points P, Q. Ex. 19, p. 99. Join AX, and through P and Q draw PH and Q K parallel to AX. Join X H , XK. These straight lines shall trisect the A; as may be shewn by joining AP, AQ. See Ex. 21, p. 111.] P X Q C 38. Cut o f f from a given triangle a fourth, fifth, sixth, or any part required by a straight line drawn from a given point in one of i t s sides. [See Ex. 19, p. 99, and Ex. 21, p. 111.] 39. Bisect a quadrilateral by a straight line drawn through an angular point. [Two constructions may be given for this problem: the first will be suggested by Exercises 28 and 33, p. 112. The second method proceeds thus. Let A B C D be the given quadrilateral, and A the given angular point. Join AC, BD, and bisect B D in X. Through X draw P X Q parallel to AC, meeting B C in P; join AP. Then shall A P bisect the quadrilateral. Join AX, CX, and use i. 37, 38.] 40. Cut off from a given quadrilateral a third, a fourth, a fifth, o any part required, by a straight line drawn through a given angular point. [See Exercises 28 and 35, p. 112.] H. E. 114 Euclid's elements. [The following Theorems depend on i. 47.] 41. In the figure of I. 47, shew that ( i ) the sum of the squares on A B and A E is equal to the sum of the squares on A C and AD. ( i i ) the square on E K is equal to the square on A B with four times the square on AC. (hi) the sum of the squares on EK and FD is equal to five times the square on BC. 42. If a straight line be divided into any two parts the square on the straight line is greater than the square on the two parts. 43. If the square on one side of a triangle is less than the squares on the remaining sides, the angle contained by these sides is acute; if greater, obtuse. 44. A B C is a triangle, right-angled at A; the sides AB, A C are intersected by a straight line PQ, and BQ, P C are joined: shew that the sum of the squares on BQ, P C is equal to the sum of the Bquares on BC, PQ. 45. In a right-angled triangle four times the sum of the squares on the medians which bisect the sides containing the right angle is equal to five times the square on the hypotenuse. 46. Describe a square whose area shall be three times that of a given square. 47. Divide a straight line into two parts such that the sum of their squares shall be equal to a given square. IX. on loci. It is frequently required in the course of Plane Geometry to find the position of a point which satisfies given conditions. N o w all problems of this type hitherto considered have been found to be capable of definite determination, though some admit of more than one solution: this however will not be the case if only one condition is given. For example, if we are asked to find a point which shall be at a given distance from a given point, we observe at once that the problem is indeterminate, that is, that it admits of an indefinite number of solutions; for the condition stated is satisfied by any poiut on the circumference of the circle described from the given point as centre, with a radius equal to the given distance: moreover this condition is satisfied by no other point within or without the circle. Again, suppose that it is required to find a point at a given distance from a given straight line. THEOREMS AND EXAMPLES ON BOOK L 115 Here, too, it is obvious that there are an infinite number of such points, and that they lie on the two parallel straight lines which m a y be drawn on either side of the given straight line at the given distance from it: further, no point that is not on one or other of these parallels satisfies the given condition. Hence we see that when one condition is assigned it is not sufficient to determine the position of a point absolutely, but it may have the effect of restricting it to some definite line or lines, straight or curved. This leads us to the following definition. Definition. The Locus of a point satisfying an assigned condition consists of the line, lines, or part of a line, to which the point is thereby restricted; provided that the condition is satisfied by every point on such line or lines, and by no other; A locus is sometimes defined as the path traced out by a point which moves in accordance with an assigned law. Thus the locus of a point, which is always at a given distance from a given point, is a circle of which the given point is the centre: and the locus of a point, which is always at a given distance from a given straight line, is a pair of parallel straight lines. We now see that in order to infer that a certain line, or system of bines, is the locus of a point under a given condition, it is necessary to prove (i) that any point which fulfils the given condition is on the supposed locus; ( i i ) that every point on the supposed locus satisfies the given condition. 1. Find the locus of a point which is always equidistant from two given points. Let A, B be the two given points. (a) Let P be any point equidistant from A and B, so that A P = B P . Bisect A B at X, and join PX. Then in the as AXP, BXP, ( • A X = BX, Constr. Because -j and PX is common to both, [ alsoAP = BP, Hyp. . -. the Z PXA=the z PXB; 1.8. and they are adjacent z B ; . -. PX is perp. to AB. r. Any point which is equidistant from A and B is on the straight line which bisects A B at right angles. 8—2 116 EUCLID'S ELEMENTS. \p) Also every point in this line is equidistant from A and B. For let Q be any point in this line. Join AQ, BQ. Then in the a" A X Q , B X Q , ( • AX=BX, Because \ and X Q i s common to both; ( a l s o the Z AXQ=the Z BXQ, being r t . L; . . AQ=BQ. 1 . 4 . That is, Q is equidistant from A and B. Hence we conclude that the locus of the point equidistant from two given points A, B is the straight line which bisects A B at right angles. 2. To find the locus of the middle point of a straight line drawn from a given point to meet a given straight line of unlimited length. B F X Y C E p / a - - ' / Let A be the given point, and B C the given straight line of un-limited length. (a) Let AX be any straight line drawn through A to meet BC, and'let P be its middle point. Draw AF perp. to BC, and bisect AF at E. Join EP, and produce i t indefinitely. Since A F X is a A, and E, P the middle points of the two sides AF, AX, -• . E P is parallel to the remaining side FX. Ex. 2. p. 96. . -. P is on the straight line which passes through the fixed point E. and is parallel to BC. (/3) Again, every point in EP, or EP produced, fulfils the required condition. For, in this straight line take any point Q. Join AQ, and produce i t to meet B C in Y. Then FAY is a a , and through E, the middle point of the side AF, E Q is drawn parallel to the side FY, . • . Q is the middle point of AY. Ex. 1, p. 96. Hence the required locus is the straight line drawn parallel to BC, and passing through E, the middle point of the perp. from A to BC. THEOREMS AND EXAMPLES ON BOOK I. 117 3. Find the locus of a point equidistant from two given inter-secting straight lines. [ S e e j^ 3 > p # 49_ ] 4. Find the locus of a point at a given radial distance from the circumference of a given circle. 5. Find the locus of a point which moves so that the sum of its distances from two given intersecting straight lines of unlimited length is constant. 6. Find the locus of a point when the differences of its distances from two given intersecting straight lines of unlimited length is constant. 7. A straight rod of given length slides between two straight rulers placed at right angles to one another: find the locus of its middle point. [See Ex. 2, p. 100.] 8. On a given base as hypotenuse right-angled triangles are described: find the locus of their vertices. 9. AB is a given straight line, and AX is the perpendicular drawn from A to any straight fine passing through B: find the locus of the middle point of AX. 10. Find the locus of the vertex of a triangle, when the base and area are given. 11. Find the loeus of the intersection of the diagonals of a paral-lelogram, of which the base and area are given. 12. Find the loeus of the intersection of the medians of a triangle described on a given base and of given area. X. ON THE INTERSECTION OE LOCI. It appeal's from various problems which have already been considered, that w e are often required to find a point, the position of which is subject to two given conditions. The method of loci is very useful in the solution of problems of this kind: for corresponding to each condition there will be a locus on which the required point must lie; hence all points which are common to these two loci, that is, all the points of intersection of the loci, will satisfy both the given conditions. 118 E u c l i d ' s elements. Example 1. To construct a triangle, having given the base, the altitude, and t l i e length of the median which bisects the base. Let AB be the given base, and P and Q the lengths of the altitude and median respectively: then the triangle is known i f its vertex is known. ( i ) Draw a straight line C D parallel to AB, and at a distance from i t equal to P: then the required vertex must l i e o r e CD. ( i i ) Again, from the middle point of A B as centre, with radius equal to Q, describe a circle: t l i e n the required vertex must l i e on this circle. Hence any points which are common to C D and the circle, satisfy both the given conditions: that is to say, i f C D intersect the circle in E, F each of the points of intersection might be the vertex of the required triangle. This supposes the length of the median Q to be greater than the altitude. Example 2. To find a point equidistant from three given points A, B, C, which are not in the same straight line. (i) The locus of points equidistant from A and B is the straight line PQ, which bisects AB at right angles. Ex. 1, p. 115. (ii) Similarly the locus of points equidistant from B and C is the straight line RS which bisects B C at right angles. Hence the point common to P Q and R S must satisfy both con-ditions : that is to say, the point of intersection of P Q and RS will be equidistant from A, B, and C. These principles may also be used to prove the theorems relating to concurrency already given on page 103. Example. To prove that the bisectors of the angles of a triangle are concurrent. Let A B C be a triangle. Bisect the z s ABC, B C A by straight lines BO, C O : these must meet at some point O. Ax. 12. Join OA. Then shall O A bisect the z BAC. Now B O is the locus of points equi-distant from BC, BA; Ex. 3, p. 49. . -. O P = OR. Similarly C O is the loous of points equidistant from BC, CA. . -. O P = O Q ; hence O R - O Q . . -. O is on the loous of points equidistant from AB and A C : that is OA is the bisector of the z BAC. Honoe tho bisectors of the throe z » meet at the point O. THEOREMS AND EXAMPLES ON BOOK I. 119 It may happen that the data of the problem are so related to one another that the resulting loci do not intersect: in this case the problem is impossible. For example, if in Ex. 1, page 118, the length of the given median is less than the given altitude, the straight line C D will not be intersected by the circle, and no triangle can fulfil the conditions of the problem. If the length of the median is equal to the given altitude, one point is common to the two loci; and consequently only one solution of the problem exists: and we have seen that there are two solutions, if the median is greater than the altitude. In examples of this kind the student should make a point of investigating the relations which must exist among the data, in order that the problem m a y be possible; and he must observe that if under certain relations two solutions are possible, and under other relations no solution exists, there will always be some intermediate relation under which one and only one solution is possible. EXAMPLES. 1. Find a point in a given straight line which is equidistant from two given points. 2. Find a point which is at. given distances from each of two given straight lines. H o w many solutions are possible? 3. On a given base construct a triangle, having given one angle at the base and the length of the opposite side. Examine the relations which must exist among the data in order that there may be two solu-tions, one solution, or that the problem may be impossible. 4. On the base of a given triangle construct a second triangle equal in area to the first, and having its vertex in a given straight line. 5. Construct an isosceles triangle equal in area to a given triangle, and standing on the same base. 6. Find a point which is at a given distance from a given point, and is equidistant from two given parallel straight lines, B O O K II. Book II. deals witli the areas of rectangles and squares. Definitions. 1. A Rectangle is a parallelogram which has one of its angles a right angle. It should be remembered that if a parallelogram has one right angle, all i t s angles are right angles. [Ex. 1, p. 64.] 2. A rectangle is said to be contained by any two of its sides which form a right angle : for it is clear that both the form and magnitude of a rectangle are fully determined when the lengths of two such sides are given. Thus the rectangle ACDB is said A B t o be contained by AB, A C ; or by CD, DB : and i f X and Y are two straight lines equal respectively to AB and AC, then the rectangle contained by X and Y i s equal to the rectangle contained by AB, AC. [See Ex. 12, p. 64.] C x -Y -After Proposition 3, we shall use the abbreviation red. AB, A C to denote the rectangle contained by AB and AC. 3. In any parallelogram the figure formed by either of the parallelograms about a diagonal together with the two complements is called a gnomon. Thus the shaded portion of the annexed figure, consisting of the parallelogram EH together with the complements AK, KC i s the gnomon AHF. The other gnomon in the figure i s that which is made up of AK, GF and FH, namely the gnomon AFH. introductory. 121 Introductory. Pure Geometry makes no use of number to estimate the magnitude of the hues, angles, and figures with which it deals: hence it requires no -u n i t s of magnitude such as the student is familiar with in Arithmetic. For example, though Geometry is concerned with the relative lengths of straight lines, it does not seek to express those lengths in terms of yards, feet, or inches: similarly it does not ask how many square yards or square feet a given figure contains, nor how many degrees there are in a given angle. This constitutes an essential difference between the method of Pure Geometry and that of Arithmetic and Algebra; at the same time a close connection exists between the results of these two methods. In the case of Euclid's Book II., this connection rests upon the fact that the number of units of area in a rectangular figure i s found by multiplying together the numbers of units of length in c For example, if the two sides AB, AD of the rectangle A B C D are respectively four and three inches long, and if through the points of division parallels are drawn as in the annexed figure, it is seen that the rectangle is divided into three rows, each containing four square inches, or into four columns, each containing three square inches. Hence the whole rectangle contains 3x4, or 12, square inches. Similarly if AB and AD contain m and n units of length respectively, it follows that the rectangle A B C D will contain m n units of area: further, i f AB and A D are equal, each containing m units of length, the rectangle becomes a square, and contains •m? units of area. [ I t must be understood that this explanation implies that the lengths of the straight fines AB, A D are commensurable, that is, that they can be expressed exactly in terms of some common unit. This however is not always the case: for example, i t may be proved that the side and diagonal of a square are so related, that i t i s impossible to divide either of them into equal parts, of which the other contains an exact number. Such lines are said to be lncommen-122 euclid's elements. surable. Hence if the adjacent sides of a rectangle are incommen-surable, we cannot choose any linear unit in terms of which these sides may be exactly expressed; and thus i t will be impossible to sub-divide the rectangle into squares of unit area, as illustrated in the figure of the preceding page. W e do not here propose to enter further into the subject of incommensurable quantities: i t is s u f f i -cient to point out that further knowledge of them will convince the student that the area of a rectangle may be expressed to any required degree of accuracy by the product of the lengths of two adjacent sides, whether those lengths are commensurable or not. ] From the foregoing explanation we conclude that the rectangle contained by two straight lines in Geometry corresponds to the product of two numbers in Arithmetic or Algebra; and that the square described on a straight line corresponds to the square of a number. Accordingly it will be found in the course of Book II. that several theorems relating to the areas of rectangles and squares are analogous to well-known algebraical formute. In view of these principles the rectangle contained by two straight lines AB, BC is sometimes expressed in the form of a product, as AB.BC, and the square described on AB as AB2. This notation, together with the signs + and —, will be employed in the additional matter appended to this book; but i t is not admitted into EiiclicFs text because it is desirable in the first instance to emphasize the distinction between geometrical mag-nitudes themselves and the numerical equivalents by which they may be expressed arithmetically. Proposition 1. Theorem. If there are two straight lines, one of which is divided into any number of parts, the rectangle contained by t l i e two straight lines is equal to the sum of the rectangles con-tained by the undivided straight line and the several parts of the divided line. Let P and AB be two straight lines, and let AB be divided into any number of parts AC, CD, D B : then shall the rectangle contained by P, A B be equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. BOOK I I . PROP. 1. A C D b K L H 1 2 3 From A draw AF perp. to AB ; i . 11. and make AG equal to P. I . 3. Through G draw G H par1 to AB; i . 31. and through C, D, B draw CK, DL, BH par1 to AG. N o w the fig. AH is made up of the figs. AK, CL, DH : and of these, the fig. AH is the rectangle contained by P, AB; for the fig. AH is contained by AG, AB ; and AG = P : and the f i g . AK is the rectangle contained by P, AC ; for the f i g . AK is contained by AG, AC; and AG = P : also the f i g . CL is the rectangle contained by P, CD ; for the fig. CL is contained by CK, C D ; and CK =the opp. side AG, and AG = P : I . 34. similarly the fig. DH is the rectangle contained by P, DB. . -. the rectangle contained by P, AB is equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. Q.E.D. CORRESPONDING ALGEBRAICAL FORMULA. In accordance with the principles explained on page 122, the result of this proposition may be written thus: P . AB= P . AC+ P. CD + P . DB. Now if the line P contains p units of length, and if AC, C contain a, 5, c units respectively, then A B = a + Z > + e, and we have p(a + b + c)=pa+pb+pc. 1 2 4 e u c l i d s e l e m e n t s . Proposition 2. Theorem. If a straight line is divided into any two parts, t l i e . square on the whole line is equal to the sum of the rectangles contained by the v j l i o l e line and each of the parts. Let tlie straight line AB be divided at C into the two parts AC, CB : then shall the sq. on AB be equal to the sum of the rects. contained by AB, AC, and by AB, BC. On AB describe the square ADEB. i . 4G. Through C draw CF par1 to AD. I . 31. N o w the fig. AE is made up of the figs. AF, CE : and of these, t l i e fig. AE is the sq. on AB : Constr. and the fig. AF is the rectangle contained by AB, AC; for the fig. AF is contained by AD, AC ; and AD = AB ; also the fig. CE is the rectangle contained by AB, BC; for the fig. C E is contained by BE, BC; and BE = AB. . -. the sq. on AB = the sum of the rects. contained by AB, AC, and by AB, BC. Q.E.D. CORRESPONDING ALGEBRAICAL FORMULA. The result of this proposition may be written AB2 = AB. AC + AB.BC. Let AC contain a units of length, and l e t CB contain b units, then AB = a + J , and we have (a + b)^=(a + b)a + {a+b) b. book i i . p r o p . 3 . Proposition 3. Theorem. 1 2 5 If a, straight line is divided into any two parts, the rectangle contained by the whole and one of the parts is equal to t l i e square on that part together with the rectangle contained by the two parts. Let the straight line AB be divided at C into the two parts AC, CB: then shall the rect. contained by AB, AC be equal to the sq. on AC together with the rect. contained by AC, CB. O n AC describe the square AFDC ; I . 46. and through B draw BE par1 to AF, meeting FD produced in E. I . 31. N o w the fig. AE is made up of the figs. AD, C E ; and of these, the fig. AE = the rect. contained by AB, AC ; for AF = A C ; and the fig. AD is the sq. on AC ; Constr. also the fig. C E is the rect. contained by AC, CB ; for CD = AC. But the fig. AE is made up of the figs. AD, CE. . -. the rect. contained by AB, AC is equal to the sq. on AC together with the rect. contained by AC, CB. q.e.d. CORRESPONDING ALGEBRAICAL FORMULA. This result may be written AB . AC= AC2+ AC . CB. Let AC, CB contain a and 6 units of length respectively, then AB = a + 6 , and we have (a + b)a = a?+ ab. Note. It should be observed that Props. 2 and 3 are special cases of Prop. 1. 1 2 6 e u c l i d ' s e l e m e n t s . Proposition 4. Theorem. If a straight line is divided into any tvjo parts, the square on the whole line is equal to t l i e sum of t l i e squares on t l i e two parts together with twice t l i e rectangle contained by the two parts. G Let the straight line AB be divided at C into the two parts AC, CB : then shall the sq. on AB be equal to the sum of the sqq. on AC, CB, together with twice the rect. AC, CB. On AB describe the square ADEB ; i . 4G. and join BD. Through C draw CF par1 to BE, meeting BD in G. i . 31. Through G draw H G K par1 to AB. It is first required to shew that the fig. CK is the sq. on BC. Because the straight line BGD meets the parls CG, AD, . -. tho ext. angle CGB = the int. opp. angle ADB. I . 29. But AB = AD, being sides of a square ; the angle ADB = the angle ABD ; i . f i . the angle CGB = the angle CBG. . -. CB = CG. i . G. And the opp. sides of t l i e par™ CK are equal; i . 34. . ' . the fig. CK is equilateral; and the angle CBK is a right angle ; I . Ui. Cor. CK is a square, and i t is described on BC. Drfi 28. Similarly the fig. HF i s the sq. on HG, that is, the sq. on AC, for HG = the opp. side A C \ . 34. BOOK I I . PROP. 4. 127 Again, the complement AG = the complement GE. I. 43. But the fig. AG =the rect. AC, CB ; for CG = CB. . -. the two figs. AG, GE = twice the rect. AC, CB. Now tlie sq. on AB = the fig. AE = the figs. HF, CK, AG, GE = the sqq. on AC, CB together with twice the rect. AC, CB. the sq. on AB = the sum of the sqq. on AC, CB with twice the rect. AC, CB. q.e.d. For the purpose of oral work, this step of the proof may conveniently be arranged as follows : Now the sq. on AB is equal to the fig. AE, that is, to the figs. HF, CK, AG, G E ; that is, to the sqq. on AC, CB together with twice the rect. AC, CB. Corollary. Parallelograms about the diagonals of a square are themselves squares. CORRESPONDING ALGEBRAICAL FORMULA. The result of this important Proposition may be written thu AB2=AC2 + CB2 + 2AC . CB. Let AG-a, and CB = i; then AB = a + b, and we have ( a + 5)2=ag + 6 2 + 2a&. 1 2 8 e u c l i d ' s e l e m e n t s . Proposition 5. Theorem. If a straight line is divided, equally and also unequally, ( l i e rectangle contained by the unequal parts, and the square on the line between t l i e points of section, are together equal to t l i e square on half the line. Let tlie straight line AB be divided equally at P, and unequally at Q : then the rect. AQ, Q B and the sq. on P Q shall be to-gether equal to the sq. on PB. O n PB describe the square PCDB. Join BC. Through Q draw Q E par1 to BD, cutting BC in F , Through F draw LFHG par1 to AB. Through A draw A G par1 to BD. N o w the complement PF = the complement FD : to each add the fig. QL; then the fig. PL = tlie fig. QD. But the fig. PL = the fig. AH, for they are par equal bases and between the same par1'. the fig. A H = the fig. QD. To each add the fig. PF; then the fig. AF = the gnomon PLE. N o w the fig. AF = the rect. AQ, QB, for Q B = QF : . ' . the rect. AQ, Q B = the gnomon PLE. To each add the sq. on PQ, that is. the fig. H E ; then the rect. AQ, Q B witli the sq. on P Q = the gnomon PLE with the fig. H E I . 46. i . 31. I . 43. " ™ on i . 36. n. 4. : the whole which is t l i ' lig. sq. PD, on PB, book n. prop. 5. 129 That is, the rect. AQ, QB and the sq. on PQ are together equal to the sq. on PB. q.e.d. Corollary. From this Proposition it follows that the difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. For let X and Y be the given a P Q b s t . lines, of which X i s the greater. ' ' Draw AP equal to X, and pro- X duce i t to B, making PB equal to Y AP, that i s to X. From PB cut o f f PQ equal to Y. Then A Q i s equal to the sum of X and Y, and QB i s equal to the d i f f e r e n c e of X and Y. Now because AB i s divided equally at P and unequally at Q, . -. the rect. AQ, QB with sq. on PQ=the sq. on PB; n. 5. that is, the d i f f e r e n c e of the sqq. on PB, PQ=the rect. AQ, QB, or, the d i f f e r e n c e of the sqq. on X and Y=the r e c t . contained by the sum and the d i f f e r e n c e o f A and Y. CORRESPONDING ALGEBRAICAL FORMULA. This result may he written AQ.QB+PQ2=PB2. Let AB=2a; and l e t PQ=6; then AP and PB each=a. Also A Q = a + 6 ; andQB=a-6. Hence we have (a+b)(a-b) + V>=a\ or {a + b){.a-b) = a?-V. EXERCISE. In the above figure shew that AP is half the sum of AQ and and that PQ i s half t h e i r d i f f e r e n c e . 1 3 0 e u c l i d ' s e l e m e n t s . Proposition 6. Theorem. If a straight Ume is bisected and produced to any point, the rectangle contained by the wlwle line thus produced, and the part of i t produced, together with the square on half the line bisected, is equal t o the square on the straight line made up of the half and the part produced. G H F 4 6 . 31. 1.43. Let the straight line AB be bisected at P, and pro-duced to Q: then the rect. AQ, QB and the sq. on PB shall be to-gether equal to the sq. on P Q On PQ describe the square P C D Q Join QC. Through B draw BE par1 to QD, meeting Q C in F . Through F draw LFHG par1 to AQ. Through A draw AG par1 to QD. N o w the complement PF = the complement FD. But the fig. PF = the fig. A H ; for they are par™ on equal bases and between the same par18. i . 36. . -. the fig. AH = the fig. FD. To each add the fig. PL; then the fig. AL = the gnomon PLE. N o w the fig. AL = the rect. AQ, QB, for QB = QL : . ' . the rect. AQ, QB = the gnomon PLE. To each add the sq. on PB, that is, the fig. HE ; then the rect. AQ, QB with the sq. on PB = the gnomon PLE with the f i g . HE = the whole l i g . PD, which is the square on PQ. That is, the rect. AQ, QB and the sq. on PB are together equal to the sq. on PQ. O j . e . d . BOOK I I . PROP. 6 . 131 CORRESPONDING ALGEBRAICAL FORMULA. This result may he written AQ.QB + PB2=PQ2. LetAB = 2a; andIetPQ=6; then AP and PB each=a. AlsoAQ=a+6; andQB=6-a. Hence we have (a+b)(b-a) + a?=W, or (b+a)(b-a) = b2-a2. Definition. If a point X is taken in a straight line AB, or in A B produced, the distances of the point of section from the ex- _ tremities of A B are said to be ~ ~— the segments into which A B is divided at X. ' -In the former case A B is divided internally, in the latter casfc externally. Thus in the annexed figures the segments into which AB is divided at X are the lines X A and X B . This definition enables us to include Props. 5 and 6 in a single Enunciation. If a straight line is bisected, and also divided (internally or ex-ternally) into two unequal segments, the rectangle contained by the un-equal segments is equal to the difference of the squares on half the line, and on the line between the points of section. EXERCISE. Shew that the Enunciations of Props. 5 and 6 may take the following form: The rectangle contained by two straight lines is equal to the differ-ence of the squares on half their sum and on half t l i e i r difference. [See Ex., p. 129.] 9—2 1 3 2 e u c l i d ' s elements. Proposition 7 . Theorem. If a straight line is divided into any two parts, the sum of the squares on the whole line and on one of t l i e parts is equal to twice the rectangle contained by the whole and that part, together with the square on the other part. G Let the straight line AB be divided at C into the two parts AC, C B : then shall the sum of the sqq. on AB, BC be equal to twice the rect. AB, BC together with the sq. on AC. O n AB describe the square ADEB. Join BD. Through C draw CF par1 to BE, meeting BD in G. Through G draw H G K par1 to AB. i . 46. 1.31. i.43. N o w the complement AG = the complement G E : to each add the fig. CK: then the fig. AK = the fig. CE. But the fig. AK = the rect. AB, B C ; for B K = BC. the two figs. AK, C E = twice the rect. AB, BC. But the two figs. AK, C E make up the gnomon AKF and the fig. CK : . -. the gnomon AKF with tlie fig. CK = twice the rect. AB, BC. To each add the fig. H F, which is tho sq. on AC : then the gnomon AKF with the figs. CK, HF = twice tho rect. AB, BC with the sq. on AC. N o w the sqq. on AB, BC = the figs. AE, CK = the gnomon AKF with tho figs. CK, HF = twice t l i e rect. AB, BC with the sq. on AC. BOOK II. PROP. 7. 1 3 3 CORRESPONDING ALGEBRAICAL FORMULA. The result of this proposition may be written AB2 + BC2 = 2AB . BC + AC2. Let AB = a, and BC = 6; thenAC=a-6. Hence we have a? + b2=2ab + (a - 6 ) 2 , or (a-bf=as-2ab + b2. Proposition 8. Theorem. If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line which is made up of the whole and that part. [As this proposition i s of l i t t l e importance we merely give the figure, and the leading points in Euclid's proof.] Let AB be divided at C. A C B D Produce AB to D, making BD equal to BC. O n AD describe the square AEFD; and complete the construction as in-dicated in the figure. Euclid then proves (i) that the figs. CK, BN, GR, KO are all equal. (ii) that the figs. AG, MP, PL, RF are all equal. Hence the eight figures named above are four times the sum of the figs. AG, C K ; that is, four times the fig. AK; that is, four times the rect. AB, BC. But the whole fig. AF is made up of these eight figures, together with the fig. XH, which is the sq. on AC : hence the sq. on AD = four times the rect. AB, BC, together with the sq. on AC. Q.E.D. A CBD G P E t K R i M n The accompanying figure will suggest a l e s s cumbrous proof, which we leave as an Exercise to the student. C -t— 1 3 4 E u c l i d ' s e l e m e n t s . Proposition 9. Theorem. If a straight Une is divided equally and also unequally, the sum of the squares on the tnvo unequal parts is twice the sum of the squares on half t l i e line and on t l i e line between the points of section. Let the straight line AB be divided equally at P, and unequally at Q : then shall the sum of the sqq. on AQ, Q B be twice the sum of the sqq. on AP, P Q A t P draw PC at rt. angles to A B ; and make PC equal to AP or PB. Join AC, BC. Through Q draw Q D par1 to PC; and through D draw DE par1 to AB. Join AD. Then since PA = PC, . • . the angle PAC=the angle PCA. A n d since, in the triangle APC, the angle APC is a rt. angle, Constr. the sum of the angles PAC, PCA is a rt. angle: I . 32. hence each of the angles PAC, PCA is half a rt. angle. So also, each of the angles PBC, PCB is half a rt. angle. . • . the whole angle ACB is a rt. angle. Again, the ext. angle C E D = the int. opp. angle CPB, I. 2ih . • . the angle C E D is a rt. angle : and the angle ECD is half a rt. angle. Proved. . -. also the angle EDC is half a rt. angle ; I . 32. . -. tho angle ECD = the angle E D O ; . ' . EC = ED. i . 6. I . 11. I . 3. i. 31. Constr. i . 5. BOOK I I . PROP. 9 . 135 Again, the ext. angle DQB = the int. opp. angle CPB. I. 29 . ' . the angle DQB is a rt. angle. A n d the angle QBD is half a rt. angle; Proved. . -. also the angle QDB is half a rt. angle : i . 32. . -. the angle QBD = the angle QDB ; . -. QD = QB. I . 6, N o w the sq. on AP = the sq. on PC ; for AP = PC. Constr. But the sq. on AC = the sum of \he sqq. on AP, PC, for the angle APC is a rt. angle. I . 47. . ' . the sq. on AC is twice the sq on AP. So also, the sq. on CD is twice the sq. on ED, that is, twice the sq. on the opp. side PQ. I . 34. N o w the sqq. on AQ, QB = the sqq. on AQ, QD = the sq. on AD, for AQD is a rt. angle; I . 47. = the sum of the sqq. on AC, CD, for ACD is a rt. angle; I . 47. = twice the sq. on AP with twice the sq. on PQ. Proved. That is, the sum of the sqq. on AQ, QB = twice the sum of the sqq. on AP, PQ. Q.E.D. CORRESPONDING ALGEBRAICAL FORMULA. The result of this proposition may be written AQ2+QB2=2(AP2+PQ2). LetAB = 2a; andPQ=6; then AP and PB each=a. AlsoAQ=a + 6; andQB=a-o. Hence we have ( a + 6)2+(a-6)2=2(a3 + 62). 1 3 6 e u c l i d ' s elements, Proposition 10. Theorem. If a straight line is bisected and produced to any point, the sum of the squares on the whole line thus produced, and on the part produced, is twice the sum of the squares on half the line bisected and on the line made up of the half and the part produced. Let the st. line AB be bisected at P, and produced to Q: then shall the sum of the sqq. on AQ, QB be twice the sum of the sqq. on AP, PQ. At P draw PC at right angles to AB; I . 11. and make PC equal to PA or PB. L . 3. Join AC, BC. Through Q draw QD par1 to PC, to meet CB produced in D ; i . 31. and through D draw DE par1 to AB, to meet CP produced in E. Join AD. Then since PA = PC, Constr. . -. the angle PAC = the angle PCA. I . o. And since in the triangle APC, the angle APC is a rt. angle, . . the sum of the angles PAC, PCA is a rt. angle. I . 32. Hence each of the angles PAC, PCA is half a rt. angle. So also, each of the angles PBC, PCB is half a rt. angle. . • . the whole angle ACB is a rt. angle. Again, the ext. angle CPB = the int. opp. angle CED : I . 29. . • . the angle CED is a rt. angle : and the angle ECD is half a rt. angle. Proved. . . the angle EDC is half a rt. angle. I . 32. . -. the angle ECD = the angle EDC : . • . EC = ED. I . 6 . BOOK I I . PROP. 10. 137 Again, the angle DQB = the alt. angle CPB. i. 29. . • . the angle DQB is a rt. angle. Also the angle QBD = the vert. opp. angle CBP ; i . 15. that is, the angle QBD is half a rt. angle. . -. the angle QDB is half a rt. angle ; i . 32. the angle QBD = the angle QDB ; . -. QB = QD. I . 6. N o w the sq. on AP = the sq. on PC ; for AP = PC. Constr. But the sq. on AC = the sum of the sqq. on AP, PC, for the angle APC is a rt. angle. I . 47. . • . the sq. on AC is twice the sq. on AP. So also, the sq. on CD is twice the sq. on ED, that is, twice the sq. on the opp. side PQ. I . 34. N o w the sqq. on AQ, QB =the sqq. on AQ, QD = the sq. on AD, for AQD i s a rt. angle; I . 47. = the sum of the sqq. on AC, CD, for ACD is a rt. angle; I . 47. = twice the sq. on AP with twice the sq. on PQ. Proved. That is, the sum of the sqq. on AQ, QB is twice the sum of the sqq. on AP, P Q q.e.d. corresponding ALGEBRAICAL FORMULA. The r e s u l t of t h i s proposition may be written AQ2+ BQ2=2 (AP2+ PQ2). Let AB=2a; and PQ=S ; then AP and PB each=a. Also AQ=a + 6; and BQ=6-a. Hence we have (a + 6)2 + (b - a)2 = 2 (a2 + 62). EXERCISE. Shew that the enunciations of Props. 9 and 10 may take the following form: The sum of the squares on two straight lines is equal to twice the sum of the squares on half their sum and on half their difference. 1 3 8 e u c l i d ' s e l e m e n t s . Proposition 11. Problem. To divide a given straight Une into two parts, so t l i a t the rectangle contained by the whole and one part may be equal to the square on the other pa/rt. F A E G H Let AB be the given straight line. It is required to divide it into two parts, so that the rectangle contained by the whole and one part may be equal to the square on the other part. On AB describe the square ACDB. I. 46. Bisect AC at E. I . 10. Join EB. Produce CA to F, making EF equal to EB. I . 3. O n AF describe the square AFGH. i . 46. Then shall AB be divided at H, so that the rect. AB, BH i s equal to the sq. on AH. Produce G H to meet CD in K. Then because CA is bisected at E, and produced to F, . -. the rect. CF, FA with the sq. on AE = the sq. on FE II. 6 . = the sq. on EB. Constr. But the sq. on EB = the sum of the sqq. on AB, AE, for the angle EAB is a rt. angle. I . 47. . -. t l i e rect. CF, FA with the sq. on AE = the sum of the sqq. on AB, AE. From theso take the sq. on AE : then t l i e rect. CF, FA = the sq. on AB. BOOK II. prop. 11. 139 But the rect. CF, FA = the fig. FK; for FA = FG; and the sq. on A B = the fig. AD. Constr. .: the fig. FK = the fig. AD. F r o m these take the c o m m o n fig. AK, then the remaining fig. FH = the remaining fig. HD. But the fig. H D = the rect. AB, BH ; for B D = A B ; and the fig. FH is the sq. on AH. . -. the rect. AB, BH =the sq. on AH. q.e.f. Definition. A straight line is said to be divided in Medial Section when the rectangle contained by the given line and one of its segments is equal to the square on the other segment. The student should observe that this division may be internal or external. Thus if the straight line A B is divided internally at H, and ex-ternally at H', so that ( i ) A B . B H =AH2, ' a H R ( i i ) A B . B H ' = A H t! £ 2 we shall in either case consider that A B is divided in medial section. The case of internal section is alone given in Euclid n. 11; but a straight line may be divided externally in medial section by a similar process. See Ex. 21, p. 146. ALGEBRAICAL ILLUSTRATION. It is required to find a point H in AB, or AB produced, such that A B . B H = AH2. Let A B contain a units of length, and let A H contain x units; then H B = a - x: and x must be such that a (a — x)=x2, or x2+ax-a!!=0. Thus the construction for dividing a straight line in medial section corresponds to the algebraical solution of this quadratic equation. EXERCISES. In the figure of n. 11, shew that ( i ) if C H is produced to meet BF at L, C L is at right angles to BF: ( i i ) if B E and C H meet at O, A O is at right angles to C H : (hi) the lines BG, DF, A K are parallel: (iv) C F is divided in medial section at A. 1 4 0 e u c l i d ' s elements. Proposition 12. Theorem. In an obtuse-angled triangle, if a perpendicular is drawn from either of the acute angles to t l i e opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing t l i e obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the line inter-cepted without the triangle, between the perpendicular and the obtuse angle, rfA Let ABC be an obtuse-angled triangle, having the obtuse angle at C ; and let AD be drawn from A perp. to BC produced: then shall the sq. on AB be greater than the sqq. on BC, CA, by twice the rect. BC, CD. Because BD is divided into two parts at C, . -. the sq. on BD =the sum of the sqq. on BC, CD, with twice the rect. BC, CD. H. 4. To each add the sq. on DA. Then the sqq. on BD, DA = the sum of the sqq. on BC, CD, DA, with twice the rect. BC, CD. But the sum of the sqq. on BD, DA = the sq. on AB, for the angle at D is a rt. angle. I . 47. Similarly the sum of the sqq. on CD, DA = the sq. on CA. . • . the sq. on AB = the sum of the sqq. on BC, CA, with twice the rect. BC, CD. That is, the sq. on AB is greater than the sum of the sqq. on BC, CA by twice the rect. BC, CD. Q.E.D. [For alternative Enunciations to Props. 12 and 13 and Exercise boo p. 142.] BOOK I I . PROP. 13. 1 4 1 Proposition 13. Theorem. In every triangle the square on t l i e side subtending an acute angle, is l e s s than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpen-dicular l e t fall on i t from the opposite angle, and the acute angle. Because Let ABC be any triangle having the angle at B an acute angle; and let AD be the perp. drawn from A to the opp. side BC : then shall the sq. on AC be less than the sum of the sqq. on AB, BC, by twice the rect. CB, BD. N o w AD may f a l l within the triangle ABC, as in Fig. 1, or without it, as in Fig. 2. (in Fig. 1. BC is divided into two parts at D, (in Fig. 2. BD is divided into two parts at C, . • . in both cases, the sum of the sqq. on' CB, BD = twice the rect. CB, BD with the sq. on CD. I I . 7. To each add the sq. on DA. Then the sum of the sqq. on CB, BD, DA = twice the rect. CB, BD with the sum of the sqq. on CD, DA. But the sum of the sqq. on BD, DA = the sq. on AB, for the angle ADB is a rt. angle. I . 47. Similarly the sum of the sqq. on CD, DA = the sq. on AC. . -. the sum of the sqq. on AB, BC, = twice the rect. CB, BD, with the sq. on AC. That is, the sq. on AC is less than the sqq. on AB, BC by twice the rect. CB, BD. Q.e.d. 142 EUCLID'S ELEMENTS. 06s. If the perpendicular AD coincides with AC, that is, if ACB is a right angle, i t may be shewn that the proposition merely repeats the result of i . 47. Note. The result of Prop. 12 may be written AB2 = BC2+ CA2 + 2 B C . CD. Bemembering the definition of the Projection of a straight, line given on page 97, the student will see that this proposition may be enunciated as follows: In an obtuse-angled triangle the square on t l i e side opposite t l i e obtuse angle is greater than the sum of the squares on t l i e sides contain-ing t l i e obtuse angle by twice t l i e rectangle contained by either of those sides, and t l i e projection of the other side upon it. Prop. 13 may be written AC2=AB2+ BC2 - 2CB . BD, and i t may also be enunciated as follows: In every triangle the square on the side subtending an acute angle, i s l e s s than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the projection of the other side upon it. EXERCISES. The following theorem should be noticed; it is proved by the help of II. 1. 1. If four points A, B, C, D are taken in order on a straight line, then will A B . C D + B C . A D = A C . BD. ON II. 12 AND 13. 2. If from one of the base angles of an isosceles triangle a per-pendicular is drawn to the opposite side, then twice the rectangle contained by that side and the segment adjacent to the base is equal to the square on the baBe. 3. If one angle of a triangle is one-third of two right angles, shew that the square on the opposite side is less than the sum of the squares on the sides forming that angle, by the rectangle contained by these two sides. [See Ex. 10, p. 101.] 4. If one angle of a triangle is two-thirds of two right angles, shew that the square on the opposite side is greater than tlie squares on the sides forming that angle, by the rectangle contained by these Bides. [See Ex. 10, p. 101.] book i i . p r o p . 1 4 . Proposition 14. Problem. 1 4 3 To describe a square that shall be equal to a given recti lineal figure. Let A be the given rectilineal figure. It is required to describe a square equal to A. Describe the parm BCDE equal to the fig. A, and having the angle CBE a right angle. i . 45. Then i f BC = BE, the fig. BD is a square; and what was required is done. But i f not, produce BE to F, makirig EF equal to ED; i . 3. and bisect BF at G. I . 10. From centre G, with radius GF, describe the semicircle BHF: produce DE to meet the semicircle at H. Then shall the sq. on EH be equal to the given fig. A. Join GH. Then because BF is divided equally at G and unequally at E, . -. the rect. BE, EF with the sq. on G E = the sq. on GF II. 5. = the sq. on GH. But the sq. on GH = the sum of the sqq. on GE, EH ; for the angle HEG is a rt. angle. I . 47. . • . the rect. BE, EF with the sq. on GE = the sum of the sqq. on GE, EH. From these take the sq. on G E : then the rect. BE, EF = the sq. on HE. But the rect. BE, EF = the fig. BD; for EF = ED; Constr. and the fig. BD = the given fig. A. Constr. . ' . the sq. on EH = the given fig. A. q.e.f. 144 EUCLID'S ELEMENTS. THKORKMS AND EXAMPLES ON BOOK II. ON II. 4 AND 7. 1. Sliew by n. 4 tliat the square on a straight line is four times the square on half the line. [This result is constantly used in solving examples on Book n, especially those which follow from n. 12 and 13.] 2. If a straight line is divided into any three parts, the square on the whole line is equal to the sum of the squares on the three parts together with twice the rectangles contained by each pair of these parts. Shew that the algebraical formula corresponding to this theorem is (a + b + c)2=a2+b2 + c2+2bc + 2ca + 2ab. 3. In a right-angled triangle, if a perpendicular is drawn from the right angle to t l i e hypotenuse, t l i e square on this perpendicular is equal to the rectangle contained by t l i e segments of t l i e hypotenuse. 4. In an isosceles triangle, if a perpendicular be drawn from one of the angles at the base to the opposite side, shew that the square on the perpendicular is equal to twice the rectangle contained by the segments of that side together with the square on the segment adjacent to the base. 5. Any rectangle is half the rectangle contained by the diagonals of the squares described upon its two sides. 6. In any triangle if «• perpendicular is drawn from the vertical angle to the base, the sum of the squares on the sides forming that angle, together with twice the rectangle contained by the segments of the base, is equal to the square on the base together with twice the square on the perpendicular. ON II. 5 A N D 6. The student is reminded that these important propositions are both included in the following enunciation. The difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. 7. In a right-angled triangle the square on one of the sides form-ing the right angle is equal to the rectangle oontained by tho sum and difference of the hypotenuse and tho other side. [i. 47 and n. 5.] theorems and examples ON BOOK I I . 1 4 5 8 . _ The difference of the squares on two sides of a triangle is equal to twice the rectangle contained by the base and the intercept between the middle point of the base and the foot of the perpendicular drawn from the vertical angle to the base. Let ABC be a triangle, and l e t P be the middle point of the base BC: l e t A Q be drawn perp. to BC. Then shall AB2-AC2=2BC . PQ. First, l e t A Q f a l l within the triangle. Now AB2 = BQ2 + QA2, i . 47. also AC2=QC2 + QA2, AB2-AC2=BQ2-QC2 Ax. 3. = (BQ + QC) (BQ - QC) n. 5. = BC.2PQ Ex. 1, p. 129. = 2BC.PQ. Q.E.D. The case in which AQ falls outside the triangle presents no difficulty. 9. The square on any straight line drawn from the vertex of an isosceles triangle to the base is less than the square on one of the equal sides by the rectangle contained by the segments of the base. 10. The square on any straight fine drawn from the vertex of an isoBceles triangle to the base produced, is greater than the square on one of the equal sides by the rectangle contained by the segments into which the base is divided externally. 11. If a straight line is drawn through one of the angles of an equilateral triangle to meet the opposite side produced, bo that the rectangle contained by the segments of the base is equal to the square on the side of the triangle; shew that the square on the line so drawn is double of the square on a side of the triangle. 12. If XY be drawn parallel to the base BC of an isosceles triangle A B C , then the difference of the squares on B Y and C Y is equal to the rectangle contained by B C , X Y . [See above, Ex. 8.] 13. In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the square on either side forming the right angle is equal to the rectangle contained by the hypotenuse and the segment of it adjacent to that side. H. E. 10 146 EUCLID'S ELEMENTS. ON II. 9 AND 10. 14. Deduce Prop. 9 from Props. 4 and 5, using, also the theorem that the square on a straight line 18 four times the square on half the line. 15. Deduce Prop. 10 from Props. 7 and 6, using also the theorem mentioned in the preceding Exercise. 16. If a straight line is divided equally and also unequally, the squares on the two unequal segments are together equal to twice the rectangle contained by these Begments together with four times the square on the line between the points of section. ON II. 11. 17. If a straight line is divided internally in medial section, and from the greater segment apart be taken equal to the less; shew that the greater segment is also divided in medial section. 18. If a straight line is divided in medial section, the rectangle contained by the Bum and difference of the segments is equal to the rectangle contained by the segments. 19. If AB is divided at H in medial section, and if X is the middle point of the greater segment AH, shew that a triangle whose sides are equal to A H , X H , B X respectively must be right-angled. 20. If a Btraight line AB is divided internally in medial section at H, prove that the sum of the squares on AB, B H is three times the square on AH. 21. Divide a straight line externally in medial section. [Proceed as in n. 11, but instead of drawing EF, make EF' equal to E B in the direction remote from A; and on AF' describe the square AF'G'H' on the side remote from AB. Then A B will be divided exter-nally at H as required.] ON II. 12 A N D 13. 22. In a triangle ABC the angles at B and C are acute; if E and F are the feet of perpendiculars drawn from the opposite angles to the Bides AC, AB, shew that the square on B C is equal to the sum of the rectangles AB, BF and AC, CE. 23. ABC is a triangle right-angled at C, and DE is drawn from a point D in A C perpendioular to A B : shew that the reotangle AB, A E is equal to the reotangle AC, AD. THEOREMS AND EXAMPLES ON BOOK II. 147 24. In any triangle the sum of the squares on two sides is equal twice the square on half the third side together with twice t l i e square on the median which bisects the third side. Q C Let A B C be a triangle, and AP the median bisecting the side BC. Then shall AB2 + AC2=2BP2 + 2AP2. Draw A Q perp. to BC. Consider the case in which A Q f a l l s within the triangle, but does not coincide with AP. Then of the angles APB, APC, one must be obtuse, and the other acute: l e t APB be obtuse. Then in the A APB, AB2=BP2+AP2 + 2 BP. PQ. ii. 12. Also in the A APC, AC2 = CP3 + AP2-2CP . PQ. n. 13. ButCP=BP, . -. CP2=BP2; and the rect. BP, PQ=the rect. CP, PQ. Hence adding the above results AB2 + AC2=2.BP2+2.AP2. q.e.d. The student will have no d i f f i c u l t y in adapting this proof to the cases in which A Q f a l l s without the triangle, or coincides with A P. 25. The sum of the squares on the sides of a parallelogram is eq t o the sum of the squares on the diagonals. 26. In any quadrilateral the squares on the diagonals are toge-ther equal to twice the sum of the squares on the straight lines join-ing the middle points of opposite sides. [See Ex. 9, p. 97.] 27. If from any point within a rectangle straight lines are draw to the angular points, the sum of the squares on one pair of the lines drawn to opposite angles is equal to the sum of the squares on the other pair. 28. The sum of the squares on the sides of a quadrilateral i s greater than the sum of the squares on i t s diagonals by four times the square on the straight line which joins the middle points of the diagonals. 29. O is the middle point of a given straight line AB, and from 0 as centre, any circle is described: i f P be any point on i t s circum-ference, shew that the sum of the squares on AP, BP i s constant. 148 EUCLID'S ELEMENTS. 30. Given the base of a triangle, and the sum of the squares on the sides forming the vertical angle; find the loeus of the vertex. 31. ABC is an isosceles triangle in which AB and AC are equal. A B is produced beyond the base to D, so that B D is equal to AB. Shew that the square on C D is equal to the square on A B together with twice the square on BC. 32. In a right-angled triangle the sum of the squares on the straight lines drawn from the right angle to the points of bi-section of the hypotenuse is equal to five times the square on the line between the points of trisection. 33. Three timqs the sum of the squares on the sides of a tri-angle is equal to four times the sum of the squares on the medians. 34. A B C is a triangle, and O the point of intersection of its medians: shew that A B2 + BC2 + C A 2 = 3 (OA2 + O B2 + OC2). 35. ABCD is a quadrilateral, and X the middle point of the Btraight line joining the bisections of the diagonals ; with X as centre any circle is described, and P is any point upon this circle: shew that PA2+ PB2+ PC2+ PD2 is constant, being equal to X A2 + X B2 + X C 2 + X D2 + 4X P2. 36. The squares on the diagonals of a trapezium are together equal to the sum of the squares on its two oblique sides, with twice the rectangle contained by its parallel sides. PROBLEMS. 37. Construct a rectangle equal to the difference of two squares. 38. Divide a given straight line into two parts so that the rect-angle contained by them may be equal to the square described on a given straight line which is less than half the straight line to be divided. 39. Given a square and one side of a rectangle which is equal to the square, find the other side. 40. Produce a given straight line so that the rectangle contained by the whole line thus produced, and the part produced, may be equal to the square on half the line. 41. Produce a given straight line so that the rectangle con-tained by the whole line thus produced and the given line shall be equal to a given square. 42. Divide a straight line A B into two parts at C, such that the rectangle contained by B C and another line X may be equal to the square on AC. OAMBUIDQU : PKIHTKlJ BY 0 . 1 . CLAI, M.A. ASB SONS, AT TUH DMVSBSIT? Pa P A R T II. BOOK III. Book III. deals with the properties of Circles. Definitions. 1. A circle is a plane figure bounded by one line, which is called the circum-ference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another: this point is called the centre of the circle. 2. A radius of a circle is a, straight line drawn from the centre to the circumference. 3. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 4. A semicircle is the figure bounded by a diameter of a circle and the part of the circumference cut off" by the diameter. From these definitions we draw the following inferences: ( i ) The distance of a point from the centre of a circle is less than' the radius, i f the point is within the circumference: and the distance of a point from the centre is greater than the radius, if the point is without the circumference. ( i i ) A point is within a circle i f its distance from the centre is less than the radius: and a point is without a circle i f its distance from the centre is greater than the radius. (hi) Circles of equal radius are equal in all respects; that is to say, their areas and circumferences are equal. (iv) A circle is divided by any diameter into two parts which are equal in all respects. H. E. 11 1 5 0 EUCLID'S ELEMENTS. 5. Circles which have the same centre are said to be concentric. 6. An arc of a circle is any part of the circumference. 7. A chord of a circle is the straight line which joins any two points on the circumference. From these definitions i t may be seen that a chord of a circle, which does not pass through the centre, divides the circumference into two unequal arcs; of these, the greater is called the major arc, and the less the minor arc. Thns the major arc is greater, and the minor arc l e s s than the semicireumference. The major and minor arcs, into which a c i r -cumference is divided by a chord, are said to be conjugate to one another. 8. Chords of a circle are said to be ec[uidistant from the centre, when-the perpendiculars drawn to them from the centre are equal: and one chord is said to be further from the centre than another, when the per-pendicular drawn to it from the centre is greater than the perpendicular drawn to the other. 9. A secant of a circle is a straight line of indefinite length, which cuts the circumference in two points. 10. A tangent to a circle is a straight line which meets the circumference, but being produced, does not cut it. Such a line is said to touch the circle at a point; and the point is called the point of contact. DEFINITIONS. If a secant, which outs a circle at the points P and Q, gradually changes its position in such a way that P remains fixed, the point Q will ultimately approach the fixed point P, until at length these points may be made to coincide. When the straight line P Q reaches this limiting position, it beeomes the tangent to the circle at the point P. Hence a tangent may be defined as a straight line which passes through two coinci-dent points on the circumference. 1 5 1 11. Circles are said to touch one another w h e n they meet, but do not cut one another. When each of the circles which meet is outside the other, they are said to touch one another externally, or to have external contact: when one of the circles is within the other, they are said to touch one another internally, or to have Internal contact. 12. A segment of a circle is the figure bounded by a chord and one of the two arcs into which the chord divides the circumference. The chord of a segment is sometimes called its base. 1 1 — 2 152 EUCLID S ELEMENTS. 13. An angle in a segment is one formed by two straight lines drawn from any point in the arc of the segment to the extremities of its chord. [It will be shewn in Proposition 21, that all angles in the sam segment of a c i r c l e are equal.] 14. A n angle at the circumference of a circle is one formed by straight lines drawn from a point on the circumference to the extremities of an arc: such an angle is said to stand upon the arc, which it subtends. 15. Similar segments of circles are those which contain equal angles. 16. A sector of a circle is a figure bounded by two radii and the arc inter-cepted between them. Symbols and Abbreviations. In addition to the symbols and abbreviations given on page 10, wc shall use the following. O for circle, Oco for circumference. book i i i . prop. 1 . ] 5 3 Proposition 1. Problem. To find the centre of a given circle. E^~ ^ B Let ABC be a given circle: i t is required to find its centre. In the given circle draw any chord AB, and bisect AB at D. I . 10. From D draw DC at right angles to AB; i . 11. and produce DC to meet the Oce at E and C. Bisect EC at F. i . 10. Then shall F be the centre of the O ABC. First, the centre of the circle must be in EC : for i f not, let the centre be at a point G without EC. Join AG, DG, BG. Then in the As GDA, GDB, C D A = D B , Constr. Because J . and G D is common; [ and GA = GB, for by supposition they are radii; . " . the z.GDA = the A G D B ; i . 8. . ' . these angles, being adjacent, are rt. angles. But the L CDB is a rt. angle; Constr. . ' . the l GDB = the a CDB, Ax. 11. the part equal to the whole, which is impossible. . ' . G is not the centre. So it may be shewn that no point outside EC is the centre; . ' . the centre lies in EC. . . F, the middle point of the diameter EC, must be the centre of the 0 ABC. q.e.f. Corollary. The straight line which bisects a chord of a circle at right angles passes through the centre. [For Exercises, see page 156.] 154 E U C L I D ' S ELEMENTS. Proposition 2. Theorem. If any two points are taken in the circwmference of a circle, t l i e chord which joins them falls within the circle. Let A B C be a circle, and A and B any two points on its Oce: then shall the chord AB fall within the circle. Find D, the centre of the © A B C ; iil 1. and in AB take any point E. Join DA, DE, DB. In the A DAB, because D A = DB, m . Def. 1. . ' . the l D A B = t h e l DBA. i. 5. But the ext. a DEB is greater than the int. opp. a DAE; i. 16. . ' . also the l DEB is greater than the A D B E ; . ' . in the A D E B , the side DB, which is opposite the greater angle, is greater than D E which is opposite the less: I. 19. that is to say, D E is less than a radius of the circle; . ' . E falls within the circle. So also any other point between A and B m a y be shewn to fall within the circle. . ' . AB falls within the circle. Q. kd. Definition. A part of a curved line is said to be concave to a point when, any chord being taken in it, a l l straight lines drawn from the given point to the intercepted aro are out by t l i e chord: i f , when any ohord i s taken, no straight line drawn from the given point to the intercepted aro is cut by the ohord, the curve is said to be convex to that point. Proposition 2 proves that the whole circumference of a c i r c l e i s oonoave to i t s oentre, book i i i . prop. 3 . 155 Proposition 3. Theorem. If a straight line drawn through the centre of a circle bisects a chord which does not pass through the centre, i t shall cut i t at right angles : and, conversely, if i t cut i t at right angles, i t shall bisect i t . B 'C Let ABC be a circle; and let CD be a st. line drawn through the centre, and AB a chord which does not pass through the centre. First. Let CD bisect AB at F : then shall CD cut AB at rt. angles. Find E, the centre of the circle; in. 1. and join EA, EB. Then in the A8 AFE, BFE, ( AF=BF, Hyp. Because J . and FE i s common; ( and AE = BE, being r a d i i o f the c i r c l e ; . ' . the L AFE = the L BFE; I . 8 . . ' . these angles, being adjacent, are rt. angles, that is, DC cuts AB at rt. angles. q.e.d. Conversely. Let CD cut AB at rt. angles : then shall C D bisect AB at F. As before, find E the centre; and join EA, EB. In the A EAB, because EA = EB, in. Def. 1. .".the L EAB =the L EBA. I . 5. Then in the A3 EFA, EFB, ( the A EAF = the L EBF, Proved. Because -] and the L EFA = the A EFB, being rt. angles; Hyp. ( and EF is common; . -. AF = BF. 1.26. Q.E.D. [For Exercises, see page 156.] 150 EUCLID'S ELEMENTS. EXERCISES. on Proposition 1. 1. If two circles intersect at the points A, B, shew that the line which joins their centres bisects their common chord A B at right angles. 2. AB, AC are two equal chords of a, circle; shew that the straight line which bisects the angle B A C passes through the centre. 3. Two chords of a circle are given in position and magnitude: find the centre of the circle. 4. Describe a circle that shall pass through three given points, which are not in the same straight line. 5. Find the locus of the centres of circles which pass through two given points. 6. Describe a circle that shall pass through two given points, and have a given radius. on Proposition 2. 7. A straight line cannot cut a circle in more than two points. on Pboposition 3. 8. Through a given point within a circle draw a chord which shall be bisected at that point. 9. The parts of a straight line intercepted between the circum-ferences of two coneentrio circles are equal. 10. The line joining the middle points of two parallel chords of a circle passes through the centre. 11. Find the locus of the middle points of a system of parallel chords drawn in a circle. 12. If two ciroles out one another, any two parallel straight lines drawn through the points of intersection to cut tlie circles, are equal 13. PQ and XY are two parallel chords in a circle: shew that the points of intersection of PX, Q Y , and of PY, Q X , lie on the straight lino whioh passes through the middle points of the given chords. BOOK I I I . PROP. 4. 157 Proposition 4. Theorem. If in a circle two chords cut one anotlier, which do not both pass through the centre, they cannot both be bisected at their point of intersection. Let ABCD be a circle, and AC, BD two chords which intersect at E, but do not both pass through the centre: then AC and BD shall not be both bisected at E. Case I. If one chord passes through the centre, it is a diameter, and the centre is its middle point; . " . i t cannot be bisected by the other chord, which by hypo-thesis does not pass through the centre. Case II. If neither chord passes through the centre; then, i f possible, let E be the middle point of both; that is, let A E = EC; and B E = ED. Find F, the centre of the circle: in. 1. Join EF. Then, because FE, which passes through the centre, bisects the chord AC, Hyp. . ' . the A FEC is a rt. angle. in. 3. And because FE, which passes through the centre, bi-sects the chord BD, Hyp. . ' . the A FED is a rt. angle. . . the A FEC = the A FED, the whole equal to i t s part, which is impossible. . ' . AC and BD are not both bisected at E. Q. e.d. [For Exercises, see page 158.] 158 Euclid's elements. Proposition 5. Theorem. If two circles cut one another, they cannot have the same centre. Let the two ©B AGO, BFC cut one another at C: then they shall not have the same centre. For, i f possible, let the two circles have the same centre; and let it be called E. Join EC; and from E draw any st. line to meet the O063 at F and G. Then, because E is the centre of the ©AGC, Hyp. :. EG = EC. And because E is also the centre of the © BFC, Hyp. : . EF=EC. . ' . EG = EF, the whole equal to its part, which is impossible. . ' . t l i e two circles have not the same centre. Q. E. D. EXERCISES. on Proposition 4. 1. If a parallelogram can be inscribed in a oircle, the point of intersection of its diagonals must be at the centre of the circle. 2. Beotangles are the only parallelograms that can be inscribed in a circle. on Phoposition 5. 3. Two oiroles, whioh intersect at one point, must also intersect at another. BOOK I I I . PROP. 6 . 159 Proposition 6 . Theorem. If two circles touch one another internally, they cannot have the same centre. Let the two © s ABC, D E C touch one another internally at C: then they shall not have the same centre. For, if possible, let the two circles have the same centre; and let it be called F. Join FC; and from F draw any st. line to meet the Oces at E and B. Then, because F is the centre of the ©ABC, Hyp. :. FB = FC. And because F is the centre of the 0 DEC, Hyp. :. FE = FC. . ' . FB = FE; the whole equal to its part, which is impossible. . ' . the two circles have not the same centre. Q. e.d. Note. From Propositions 5 and 6 i t i s seen that circles, whose circumferences have any point in common, cannot be concentric, unless they coincide entirely. Conversely, the circumferences of concentric c i r c l e s can have no point in common. 1 6 0 E u c l i d ' s e l e m e n t s . Proposition 7. Theorem. If from any point within a circle which is not the centre, straight lines are drawn to the circumference, the greatest is that which passes through the centre ; and the least is that which, when produced backwards, passes through the centre : and of all other such lines, that which is nearer to the greatest is always greater than one more remote : also two equal straight lines, and only two, can be drawn from t l i e given point to the circumference, one on each side of the diameter. A, Let ABCD be a circle, within which any point F is taken, which is not the centre: let FA, FB, FC, FG be drawn to the bco, of which FA passes through E the centre, and FB is nearer than FC to FA, and FC nearer than FG: and let FD be the line which, when produced backwards, passes through the centre: then of all these st. lines (i) FA shall be the greatest; ( i i ) FD shall be the least; ( i i i ) FB shall be greater than FC, and FC greater than FG; (iv) also two, and only two, equal st. lines can be drawn from F to the Oce. Join EB, EC, EG. (i) Then in the A FEB, the two sides FE, EB are together greater than the third side FB. I . 20. But EB = EA, being radii of tlie circle; . ' . FE, EA are together greater than FB; that is, FA is greater than FB. book m. prop. 7 . 1 6 1 Similarly FA may be shewn to be greater than any ot s t . l i n e drawn from F to the Oce; . ' . FA is the greatest of all such lines. (ii) In the A EFG, the two sides EF, FG are together greater than EG; l . 20. and EG = ED, being radii of the circle; . ' . EF, FG are together greater than ED. Take away the common part EF; then FG is greater than FD. Similarly any other st. line drawn from F to the Oce may be shewn to be greater than FD. . ' . FD is the least of all such lines. (iii) In the A8 BEF, CEF, t BE = CE, I I I . Def. 1. Because < and EF is common; (but the A BEF is greater than the A CEF; . ' . FB is greater than FC. I . 24. Similarly i t may be shewn that FC is greater than FG. (iv) At E in FE make the A FEH equal to the a FEG. I . 23. Join FH. Then in the A8 GEF, HEF, ( GE=HE, in. Def. 1 . Because • ] and EF i s common; ( a l s o the A GEF = the A HEF; Constr. . \ FG = FH. I . 4. And besides FH no other s t r a i g h t l i n e can be drawn from F to the Ooe equal to FG. For, i f p o s s i b l e , l e t FK= FG. Then, because FH = FG, Proved. . \ FK=FH, that is, a line nearer to FA, the greatest, i s equal to a line which is more remote; which i s impossible. Proved. :. two, and only two, equal st. lines can be drawn from FtotheO06. Q-E-D-1 6 2 e u c l i d ' s e l e m e n t s . Proposition 8. Theorem. If from any point without a circle straight Vines are drawn to the circumference, of those which fall on the concave cir-cumference, t l i e greatest is t l i a t which passes through the centre; and of others, t l i a t which is nearer to the greatest is always greater than one more remote: further, of those which fall on the convex circumference, the least is that which, w/ien produced, passes through the centre; and of others that which is nearer to the least is always less than one more remote : lastly, from the given point there can be drawn to the circumference two, and only two, equal straight lines, one on each side of the shortest line. G F Let B G D be a circle of which C is the centre; and let A be any point outside the circle: let ABD, AEH, AFG, be st. lines drawn from A, of which AD passes through C, the centre, and AH is nearer than AG to A D : then of st. lines drawn from A to the concave Oce. (i) AD shall be the greatest, and (ii) AH greater than AG : and of st. lines drawn from A to the convex Offi, ( i i i ) AB shall be the least, and (iv) A E less than AF. (v) Also two, and only two, equal st. lines can be drawn from A to the Ooe. Join CH, CG, CF, CE. (i) Then in the A ACH, the two sides AC, C H are together greater than AH: I . 20. but C H = CD, being radii of the circle; . ' . AC, C D are together greater than AH: that is, A D is greater than AH. Similarly AD may be shewn to be greater than any other st. line drawn from A to the concave O0^.; . " . AD is the greatest of all such lines. BOOK I I I . PROP. 8. 163 (ii) In the A8 HCA, GCA, ( HC = GC, in. Def. 1. Because -1 and CA is common; ( but the a HCA is greater than the a GCA; . ' . AH is greater than AG. I . 24. (iii) In the A AEC, the two sides AE, EC are together greater than AC : I . 20. but E C = B C ; in. Def. 1. . ' . the remainder AE is greater than the remainder AB. Similarly any other st. line drawn from A to the convex Oce may be shewn to be greater than AB; . ' . AB is the least of all such lines. (iv) In the A AFC, because AE, EC are drawn from the extremities of the base to a point E within the triangle, . ' . AF, FC are together greater than AE, EC. I . 21. But FC = EC, in. Def. 1. . ' . the remainder AF is greater than the remainder AE. (v) At C, in AC, make the A ACM equal to the A ACE. Join AM. Then in the two A8 EC A, MCA, ( E C = M C , I I I . Def. 1. Because < and CA is common; ( also the A ECA = the A MCA; Constr. .'.AE = AM; 1.4. and besides AM, no st. line can be drawn from A to the Oce, equal to AE. For, i f possible, let AK = AE : then because AM = AE, Proved. AM=AK; that is, a line nearer to the shortest line i s equal to a line which is more remote; which is impossible. Proved. : . two, and only two, equal st. lines can be drawn from AtotheOce- Q-E.D. Where are the limits of that part of the circumference whi concave to the point A? 1G4 Euclid's elements. Obs. Of the following proposition Euclid gave two distinct proo the f i r s t of which has the advantage of being direct. Proposition 9. Theorem. [First Proof.] If from a point within a circle more than two equal straight lines can be drawn to the circumference, that point is the centre of the circle. Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the O^, namely DA, DB, DC: then D shall be the centre of the circle. Join AB, BC : and bisect AB, BC at E and F respectively. I . 10. Join DE, DF. Then in the As DEA, DEB, [ EA = EB, ( ' o r n l r . Because < and DE is common; I and D A = DB; Hyp. :. the A DEA = the a DEB; I . S. . ' . these angles, being adjacent, are rt. angles. Hence ED, which bisects the chord AB at rt. angles, must pass through the centre. m . 1. Cor. Similarly i t may be shewn t l i a t FD passes through the centre. . " . D, which is the only point common to ED and FD, must be the centre. q.e.d. book i i i . p r o p . 9 . 1 6 5 Proposition 9. Theorem. [Second Proof.] If from a point within a circle more than two equal straight lines can be drawn to the circumference, that point is the centre of the circle. Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the Oce, namely DA, DB, DC : then D shall be the centre of the circle. For, i f not, suppose E to be the centre. Join DE, and produce it to meet the Ooe at F, G. Then because D is a point within the circle, not the centre, and because DF passes through the centre E; . ' . DA, which is nearer to DF, is greater than DB, which is more remote : in. 7. but this-is impossible, since by hypothesis, DA, DB, are equal. . ' . E is not the centre of the circle. And wherever w e suppose the centre E to be, other-wise than at D, two at least'of the st. lines DA, DB, DC may be shewn to be unequal, which is contrary to hypo-thesis. . " . D is the centre of the © ABC. Q.E;D. Note. For example, if the centre E were supposed to be withi the angle BDC, then DB would be greater than DA; i f within the angle ADB, then D B would be greater than DC; ifononeof the three straight lines, as DB, then DB would be greater than both DA and DC. H. E. 12 166 Euclid's elements. Obs. Two proofs of Proposition 10, both indirect, were given Euclid. Proposition 10. Theorem. [First Proof.] One circle cannot cut another at more than two points. If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Join AB, BC. Draw FH, bisecting AB at rt. angles; I . 10, 11. and draw G H bisecting BC at rt. angles. Then because AB is a chord of both circles, and FH bisects it at rt. angles, . ' . the centre of both circles lies in FH. in. 1. Cor. Again, because BC is a chord of both circles, and GH bisects it at right angles, . ' . the centre of both circles lies in GH. in. l.Cor. Hence H, the only point common to FH and GH, is the centre of both circles; which is impossible, for circles which cut one another cannot have a common centre. m . 5. . ' . one circle cannot cut another at more than two points. Q.E.D. Corollaries, (i) Two circles cannot meet {» three points without coinciding entirely. ( i i ) Two circles cannot have a common arc without coinciding entirely. ( i i i ) Only one circle can be described through three points, which are not in the same straight line. BOOK I I I . PROP. 10. 167 Proposition 10. Theorem. [Second Proof.] One circle cannot cut another at more than two points. D If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Find H, the centre of the 0 DABC, m . 1. and join HA, HB, HC. Then since H is the centre of the © DABC, . ' . HA, HB, H C are all equal. in. Def. 1. A n d because H is a point within the © EABC, from which more than two equal st. lines, namely HA, HB, H C are drawn to the Oce, . ' . H is the centre of the 0 EABC : in. 9. that is to say, the two circles have a common centre H ; but this is impossible, since they cut one another, in. 5. Therefore one circle cannot cut another in more than two points. Q.e.d. Note. This proof is imperfect, because i t assumes that the centre of the c i r c l e DABC must f a l l within the c i r c l e EABC; whereas i t may be conceived to f a l l either without the c i r c l e EABC, or on i t s circumference. Hence to make the proof complete, two additional cases are required. 12—2 1 6 8 e u c l i d ' s elements. Proposition 11. Theorem. If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. Let A B C and A D E be two circles which touch one another internally at A; let F be the centre of the 0 ABC, and G the centre of the 0 A D E : then shall FG produced pass through A. If not, let it pass otherwise, as FGEH. Join FA, GA. Then in the A FGA, the two sides FG-, G A are together greater than FA : I - "". but FA = FH, being radii of the Q A B C : Hyp. . ' . FG, G A are together greater than FH. Take away the common part FG ; then G A is greater than GH. But G A = GE, being radii of the © A D E : Hyp. :. G E is greater than GH, the part greater than the whole; which is impossible. . ' . FG, when produced, must pass through A. Q.E.D. EXERCISES. 1. If the distance between the centres of two circles is equal t the difference of their radii, then the oircles must meet in one point, but in no other; that is, they must touch one another. 2. If two c i r c l e s whose centres are A and B touch one another internally, and a straight line be drawn through their point of contact, cutting t l i e circumferences at P and Q; shew that the radii AP and BQ a re parallel. book in. PROP. 12. 169 Proposition 12. Theorem. If two circles touch one another externally, the straight line which joins their centres shall pass through the point of contact. Let A B C and A D E be two circles which touch one another externally at A; let F be the centre of the 0 ABC, and G the centre of the © A D E : then shall FG pass through A. If not, let FG pass otherwise, as FHKG. Join FA, GA. Then in the A FAG, the two sides FA, GA are together greater than FG : I . 20. but FA = FH, being radii of the © A B C ; Hyp. and GA = GK, being radii of the 0 A D E ; Hyp. . ' . FH and G K are together greater than FG; which is impossible. . ' . FG must pass through A. Q.E.D. exercises. 1. Find the locus of the centres of all circles which touch a g c i r c l e at a given point. 2. Find the locus of the centres of all c i r c l e s of given radius, which touch a given c i r c l e . 3. If the distance between the centres of two c i r c l e s i s equal to the sum of their radii, then the c i r c l e s meet in one point, but in no other; that is, they touch one another. 4. If two c i r c l e s whose centres are A and B touch one another externally, and a straight l i n e be drawn through their point of contact cutting the circumferences at P and Q; shew that the radii AP and B Q are parallel. 1 . 7 0 EUCLID'S ELEMENTS. Proposition 13. Theorem. T w o circles cannot touch one another at more than one point, whether internally or externally. Fig. 1 Fig. 2 If possible, let ABC, E D F be two circles which touch one another at more than one point, namely at B and D. Join BD; and draw GF, bisecting B D at rt. angles. I. 10,11. Then, whether the circles touch one another internally, as in Fig. 1, or externally as in Fig. 2, because B and D are on the O " of both circles, . ' . B D is a chord of both circles : . ' . the centres of both circles lie in GF, which bisects BD at rt. angles. in. 1. Cor. Hence G F which joins the centres must pass through a point of contact; in. 11, and 12. which is impossible, since B and D are without GF. . ' . two circles cannot touch one another at more than one point. Q..E.D. Note. It must be observed that the proof here given applies, by virtue of Propositions 11 and 12, to both the above figures: we have therefore omitted the separate discussion of Fig. 2, which finds a place in most editions based on Simson's text. BOOK III. PROP. 13. 171 EXERCISES ON PROPOSITIONS 1 13. 1. Describe a circle to pass through two given points and have its centre on a given straight line. W h e n is this impossible ? 2. All circles which pass through a, fixed point, and have their centres on a given straight line, pass also through a second fixed point. 3. Describe a circle of given radius to touch a given circle at a given point. H o w many solutions will there be? W h e n will there be only one solution? 4. From a given point as centre describe a circle to touch a given circle. H o w many solutions will there be? 5. Describe a circle to pass through a given point, and touch a given circle at a given point. [See Ex. 1, p. 169 and Ex. 5, p. 156.] W h e n is this impossible? 6. Describe a circle of given radius to touch two given circles. [See Ex. 2, p. 169.] H o w many solutions will there be ? 7. Two parallel chords of a circle are six inches and eight inches in length respectively, and the perpendicular distance between them is one inch: find the radius. 8. If two circles touch one another externally, the straight lines, which join the extremities of parallel diameters towards opposite parts, must pass through the point of contact. 9. Find the greatest and least straight lines which have one extremity on each of two given circles, which do not intersect. 10. In any segment of a circle, of all straight lines drawn at right angles to the • chord and intercepted between the chord and the arc, the greatest is that which passes through the middle point of the chord; and of others that which is nearer the greatest is greater than one more remote. 11. If from any point on the circumference of a circle straight fines be drawn to the circumference, the greatest is that which passes through the centre; and of others, that which is nearer to the greatest is greater than one more remote; and from this point there can be drawn to the circumference two, and only two, equal straight lines. 1 7 2 E u c l i d ' s elements. Proposition 14. Theorem. Equal chords in a circle are equidistant from the cent/re: and, conversely, chords which are equidistant from the centre are equal. A vC Let ABC be a circle, and let AB and CD be chords, of which the perp. distances from the centre are EF and EG. First, LetAB = C D : then shall AB and CD be equidistant from the centre E. Join EA, EC. Then, because EF, which passes through the centre, i s perp. to the chord AB; Hyp. . ' . EF bisects AB ; in. 3. that is, AB is double of FA. For a similar reason, CD is double of GC. But AB = CD, . ' . FA = GC. N o w EA = EC, being radii of the circle; . ' . the sq. on EA = the sq. on EC. But the sq. on EA = the sqq. on EF, FA; for the a EFA is a rt. angle. And the sq. on EC = the sqq. on EG, G C ; for the a EGC is a rt. angle. . " . the sqq. on EF, FA = t l i e sqq. on EG, GC. N o w of these, the sq. on FA = the sq. on G C ; for FA = GC. . ' . the sq. on EF = the sq. on EG, . ' . E F = E G ; that is, tho chords AB, CD are equidistant from the centre. Q.E.D. Hyp. 1.47 BOOK I I I . PROP. 14. 173 Conversely. Let AB and CD be equidistant from the centre E; that is, let EF = EG : then shall AB = CD. For, the same construction being made, i t may be shewn as before that AB is double of FA, and CD double of G C ; and that the sqq. on EF, FA = the sqq. on EG, GC. N o w of these, the sq. on EF = the sq. on EG, for EF = EG : Hyp. . ' . the sq. on FA = the square on G C ; /. FA = G C ; and doubles of these equals are equal; that is, AB = CD. Q.E.D. EXERCISES. 1. Find the locus of the middle points of equal chords of a circle. 2. If two chords of a circle cut one another, and make equal angles with the straight line which joins their point of intersection to the centre, they are equal. 3. If two equal chords of a circle intersect, shew that the segments of the one are equal respectively to t l i e segments of the other. 4. In a given circle draw a chord which shall be equal to one given straight line (not greater' than the diameter) and parallel to another. 5. P Q is a fixed chord in a circle, and A B is any diameter: shew that the difference of the perpendiculars let fall from A and B on P Q is constant, that is, the same for all positions of AB. 1 7 4 E u c l i d ' s elements. Proposition 15. Theorem. Tlie diameter is the greatest chord in a circle; and of others, that which is nearer to the centre is greater than one more remote: conversely, the greater chord is nearer to the centre than the l e s s . Let ABCD be a circle, of which AD is a diameter, and E the centre; and let BC and FG be any two chords, whose perp. distances from the centre are EH and EK : then (i) AD shall be greater than BC : ( i i ) i f EH is less than EK, BC shall be greater than FG: ( i i i ) i f BC is greater than FG, EH shall be less than EK. (i) Join EB, EC. Then in the A BEC, the two sides BE, EC are together greater than BC; I . 20. but BE = AE, I I I . Def. 1. and EC = ED ; . . AE and ED together are greater than BC; that is, AD is greater than BC. Similarly AD may be shewn to be greater than any other chord, not a diameter. (ii) Let EH be less than EK; then BC shall be greater than FG. Join EF. Since EH, passing through the centre, is perp. to the chord BC, . ' . EH bisects BC ; in. 3. BOOK I I I . PROP. 15. 175 that is, BC is double of HB. For a similar reason FG is double of KF. N o w EB = EFJ m . Def. 1. . " . the sq. on EB = the sq. on EF. But the-sq. on EB = the sqq. on EH, HB; for the a EH B is a rt. angle; I . 47. also the sq. on EF = the sqq. on EK, KF; for the A EKF is a rt. angle. . ' . the sqq. on EH, HB = the sqq. on EK, KF. But the sq. on EH is less than the sq. on EK, for EH is less than EK; Hyp. . ' . the sq. on HB is greater than the sq. on KF ; . " . HB is greater than KF : hence BC is greater than FG. (iii) Let BC be greater than FG ; then EH shall be less than EK. For since BC is greater than FG, Hyp. . ' . HB is greater than KF : . ' . the sq. on HB is greater than the sq. on KF. But the sqq. on EH, HB = the sqq. on EK, KF : Proved. . ' . the sq. on EH is less than the sq. on EK; . ' . EH is less than EK. Q.E.D. EXERCISES. 1. Through a given point within u, circle draw the least poss chord. 2. AB i s a fixed chord of a c i r c l e , and XY any other chord having i t s middle point Z on AB: what i s the greatest, and what the l e a s t length that XY may have? Shew that XY increases, as Z approaches the middle point of AB. 3. In a given c i r c l e draw a chord of given length, having i t s middle point on a given chord. When i s this problem impossible? 1 7 6 EUCLID'S ELEMENTS. Obs. Of the following proofs of Proposition 16, the second (by reductio ad absurdum) is that given by Euclid; but the first is to be preferred, as i t is direct, and not less simple than the other. P r o p o s i t i o n 16. Theorem. [Alternative Proof.] The straight line drawn at right angles to a diameter of a circle at one of its extremities is a tangent to the circle: and every oilier straight line drawn through this point cuts the circle. Let A KB be a circle, of which E is the centre, and AB a diameter; and through B let the st. line CBD be drawn at rt. angles to AB: then (i) CBD shall be a tangent to the circle; ( i i ) any other st. line through B, as BF, shall cut the circle. (i) In CD take any point G, and join EG. Then, in the A EBG, the A EBG is a rt. angle; . ' . the A EGB is less than a rt. angle; . ' . the A EBG is greater than the a EGB; . ' . EG is greater than EB: that is, EG is greater than a radius of tlie circle; . ' . the point G is without the circle. Similarly any other point in CD, except B, may be shewn to 1 > o outside the circle : hence CD meets the does not cut it; that is Hyp. i . 17. i. 19. at B, but being produced, C D is a tangent to tlie circle. in.i)r/!10. BOOK I I I . PROP. 16. 177 (ii) Draw EH perp. to BF. i. 12. Then in the A EH B, because the A EH B is a rt. angle, . ' . the A EBH is less than a rt. angle; I . 17. . ' . EB is greater than EH; i . 19. that is, EH is less than a radius of the circle: . ' . H, a point in BF, is within the circle; . ' . BF must cut the circle. q.e.d. Proposition 16. Theorem. [Euclid's Proof.] The straight line drawn at right angles to a diameter of a circle at one of its extremities, is a tangent to the circle: and no other straight line can be drawn through this point so as not to cut the circle. Let ABC be a circle, of which D is the centre, and AB a diameter; let AE be drawn at rt. angles to BA, at its extremity A: (i) then shall AE be a tangent to the circle. For, i f not, let AE cut the circle at C. Join DC. Then in the A DAC, because DA = DC, in. Def. 1. . ' . the a DAC = the a DCA. But the A DAC is a rt. angle; Hyp. . ' . the A DCA is a rt. angle; that is, two angles of the A DAC are together equal to two rt. angles; which is impossible. I . 17. Hence AE meets the circle at A, but being produced, does not cut it; that is, AE is a tangent to the circle, in. Def. 10. 1 7 8 EUCLID'S ELEMENTS. (ii) Also through A no other straight line but AE can be drawn so as not to cut the circle. For, i f possible, let AF be another st. line drawn through A so as not to cut the circle. From D draw DG perp. to AF; I . 12. and let DG meet the 0 M at H. Then in the A DAG, because the A DGA is a rt. angle, . . the a DAG is less than a rt. angle; 1.17. . " . DA is greater than DG. I . 19. But DA = DH, I I I . Def. 1. . ' . DH is greater than DG, the part greater than the whole, which is impossible. . ' . no st. line can be drawn from the point A, so as not to cut the circle, except AE. Corollaries, (i) A tangent touches a circle at one point only. ( i i ) There can be but one tangent to a circle at a given point. book m. P R O P . 1 7 . Proposition 17. Problem. 1 7 9 To draw a tangent to a circle from a given point either on, or without the circumference. Fig. 2 Let BCD be the given circle, and A the given point: i t i s required to draw from A a tangent to the © CDB. Case I. If the given point A is on the Oce. Find E, the centre of the circle. in. 1. Join EA. At A draw AK at rt. angles to EA. I . 11. Then AK being perp. to a diameter at one of its extremities, i s a tangent to the circle. in. 16. Case II. If the given point A is without the Oce. Find E, the centre of the circle; in. 1. and join AE, cutting the © BCD at D. From centre E, with radius EA, describe the ©AFG. At D, draw GDF at rt. angles to EA, cutting the © A F G at F and G. I . 11. Join EF, EG, cutting the © BCD at B and C. Join AB, AC. Then both AB and AC shall be tangents to the ©CDB. For in the A8 AEB, FED, r A E = FE, being radii of the 0 G A F ; Because /and EB = ED, being radii of the © BDC; (and the included angle AEF is common; , \ the A ABE = the A FDE, i . 4. 1 8 0 EUCLID'S ELEMENTS. But the a F D E is a rt. angle, Constr. . ' . the a A B E is a rt. angle; hence AB, being drawn at rt. angles to a diameter at one of its extremities, is a tangent to the © BCD. in. 16. Similarly it m a y be shewn that A C is a tangent, q. E. F. Corollary. If two tangents are drawn to a circle from an external point, then (i) they are equal; ( i i ) they subtend equal angles at the centre ; ( i i i ) they make equal angles with the straight line which joins the given point to the centre. For, in the above figure, Since ED is perp. to FG, a chord of the © FAG, /. DF = DG. Then in the A8 DEF, DEG, [DE is common to both, Because -j and EF = EG ; and D F = DG ; . ' . the A DEF = the a DEG. Again in the A8 AEB, AEC, [AE is common to both, Because J . and EB = EC, (and the A AEB = the _ AEC ; . ' . AB = AC: and the A EAB = the _ EAC. in. 3. in. Def. 1. Proved. i . 8. Proved. 1.4. Q.E.D. If the given point A is within the circle, no solution i s Note. possible. Henoe we see that this problem admits of t i r o solutions, one solu-tion, or no solution, aocording as the given point A is without, on, or within the circumference of a eirole. For a simpler method of drawing a tangent to a circle from a given point, see page 202. book i i i . prop. 18. 181 Proposition 18. Theorem. _ The straight line drawn from the centre of a circle to the point o f contact of a tangent is perpendicular to the tangent. Let ABC be a circle, of which F is the centre; and let the st. line DE touch the circle at C : then shall FC be perp. to DE. For, i f not, suppose FG to be perp. to DE, i . 12. and let it meet the Oceat B. Then in the A FCG, because the A FGC is a rt. angle, Hyp. . " . the a FCG is less than a rt. angle : I . 17. . ' . the a FGC is greater than the A FCG ; . ' . FC is greater than FG : I . 19. but FC = FB; . ' . FB is greater than FG, the part greater than the whole, which is impossible. . ' . FC cannot be otherwise than perp. to DE : that is, FC i s perp. to DE. q.e.d. exercises. 1. Draw a tangent to a circle (i) parallel to, (ii) at right angles to a given straight line. 2. Tangents drawn to a circle from the extremities of a diameter are parallel. 3. Circles which touch one another internally or externally have a common tangent at their point of contact. 4. In two concentric circles any chord of the outer circle which touches the inner, is bisected at the point of contact. 5. In two concentric circles, all chords of the outer circle which touch the inner, are equal. h.e. 13 1 8 2 e u c l i d ' s elements. Proposition 19. Theorem. The straight line d r a w n perpendicular to a tangent to a circle f r o m the point of contact passes through the centre. Let ABC be a circle, and DE a tangent to i t at the point C; and let CA be drawn perp. to DE : then shall CA pass through the centre. For i f not, suppose the centre to be outside CA, as at F . Join CF. Then because DE is a tangent to the circle, and FC i s drawn from the centre F to the point of contact, . ' . the a FCE i s a rt. angle. in. 18. But the a ACE is a rt. angle ; Hyp. :. the A FCE = the a ACE; the part equal to the whole, which is impossible. . ' . the centre cannot be otherwise than in CA; that is, CA passes through the centre. Q.E.D. exercises on the tangent. Propositions 16, 17, 18, 19. 1. The centre of any circle which touches two intersecting straight lines must l i e an t l i e bisector of the angle between them. 2. A B and A C are two tangents to a circle whose centre is O; Bhow that A O bisects the ohord of oontaot B C at right angles. BOOK III. PROP. 19. 183 3. If two circles are concentric all tangents drawn from points on the circumference of the outer to the inner circle are equal. 4. The diameter of a circle bisects all chords which are parallel to the tangent at either extremity. 5. Find the locus of the centres of all circles which touch a given straight line at a given point. 6. Find the locus of the centres of all circles which touch each of two parallel straight lines. 7. Find the locus of the centres of all circles which touch each of two intersecting straight lines of unlimited length. 8. Describe a circle of given radius to touch two given straight lines. 9. Through a given point, within or without a circle, draw a chord equal to a given straight line. In order that the problem m a y be possible, between what limits must the given line he, when the given point is (i) without the circle, ( i i ) within it ? 10. T w o parallel tangents to a circle intercept on any third tan-gent a segment which subtends a right angle at the centre. 11. In any quadrilateral circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. 12. Any parallelogram which can be circumscribed about a circle, must be equilateral. 13. If a quadrilateral be described about a circle, the angles sub-tended at the centre by any two opposite sides are together equal to two right angles. 14. A B is any chord of a circle, A C the diameter through A, and A D the perpendicular on the tangent at B: shew that A B bisects the angle D A C . 15. Find the locus of the extremities of tangents of fixed length drawn to a given circle. 16. In the diameter of a circle produced, determine a point such that the tangent drawn from it shall be of given length. 17. In the diameter of a circle produced, determine a point such that the two tangents drawn from it m a y contain a given angle. 18. Describe a circle that shall pass through a given point, and touch a given straight line at a given point. [See page 183. Ex. 5. ] 19. Describe a circle of given radius, having its centre on a given straight line, and touching another given straight line. 20. Describe a circle that shall have a given radius, and touch a given •circle and a given straight line. H o w many such circles can be drawn? 13—2 1 8 4 e u c l i d ' s e l e m e n t s . Proposition 20. Theorem. The angle at the centre of a circle is double of an angle at the circumference, standing on the same arc. Fig. 2 Let ABC be a circle, of which E is the centre; and let BEC be an angle at the centre, and BAC an angle at the O " ' , standing on the same arc BC : then shall the a BEC be double of the a BAC. Join AE, and produce it to F. Case I. When the centre E is within the angle BAC. Then in the A EAB, because EA = EB, . ' . the A EAB = the A EBA; I . 5. . ' . the sum of the a8 EAB, EBA = twice the _ EAB. But the ext. a BEF = the sum of the a ° EAB, EBA; I . 32. . ' . the a BEF = twice the a EAB. Similarly the u FEC = twice the a EAC. . ' . the sum of the a8 BEF, FEC = twice the sum of the A 8 EAB, EAC ; that is, the a BEC = twice the ^ BAC. Case II. When the centre E is without the _ BAC. A s before, it may be shewn that the _ FEB = twice the a FAB; also the l. FEC = twice the _ FAC; . ' . the difference of the a 8 FEC, FEB = twice the difference of the i . 6 FAC, FAB : that is, the a BEC = twice the _ BAC. < j . E . D . BOOK I I I . PROP. 21. 1 8 5 Note. I f the arc BFC, on whioh the angles stand, is greater than a semi-circumference, i t i s clear that the angle BEC at the centre will be reflex: but i t may s t i l l be shewn as, in Case I . , that the r e f l e x i BEC is double of the / BAC at the o™, standing on the same arc BFC, P r o p o s i t i o n 21. Theorem. Angles in the same segment of a circle are equal. Let ABCD be a circle, and let BAD, BED be angles in the same segment BAED: then shall the A BAD = the A BED. Find F, the centre of the circle. i l l . 1. Case I. When the segment BAED is greater than a semicircle. Join BF, DF. Then the a BFD at the centre = twice the a BAD at the Om, standing on the same arc BD: in. 20. and similarly the a BFD = twice the a BED. in. 20. . ' . the A BAD = the A BED. Case II. When the segment BAED is not greater than a semicircle. 1 8 6 Join AF, and produce it to meet the O" at C. Join EC. Then since AEDC is a semicircle; . ' . the segment BAEC is greater than a semicircle: . ' . the a BAC = the a BEC, in this segment. Case 1. Similarly the segment C A E D is greater than a semicircle; . " . the a C A D = the a CED, in this segment. . ' . the sum of the a s BAC, C A D = the sum of the a 3 BEC, C E D : that is, the a BAD = the A BED. Q.E.D. exercises. 1. P is any point on the arc of a segment of which AB is the chord. Shew that the sum of the angles PAB, PBA is constant. 2. P Q and RS are two chords of a circle intersecting at X: prove that the triangles PXS, R X Q are equiangular. 3. Two circles intersect at A and B ; and through A any straight line PAQ is drawn terminated by the circumferences: shew that PQ subtends a constant angle at B. 4. Two circles intersect at A and B: and through A any two straight lines PAQ, X AY are drawn terminated by the circumferences: shew that the arcs PX, QY subtend equal angles at B. 5. P is any point on the arc of a segment whose chord i s AB: and tho angles PAB, PBA are bisected by straight lines which intersect at O. Find the locus of the point O. BOOK III. PROP. 21. 1 8 7 Note. If the extension of Proposition 20, given in the note on page 185, is adopted, a , separate treatment of « the second case of the present proposition is unnecessary. For, as in Case I., the reflex L BFD = twiee the I B A D ; in. 20. also the reflex / BFD = twice the Z BED; . -. the / B A D = the / BED. The converse of Proposition 21 is very important. For the con-struction used in its proof, viz. To describe a circle about a given triangle, the student is referred to Book rv. Proposition 5. [Or see Theorems and Examples on Book i. Page 103, No. 1.] Converse op Proposition 21. Equal angles standing on the same base, and on the same side of it, have their vertices on an arc of a circle, of which the given base is the chord. Let BAC, BDC be two equal angles standing on the same base B C : then shall the vertices A and D lie upon a segment of a circle having B C as its chord. Describe a circle about the A B A C : iv. 5. then this circle shall pass through D. For, i f not, i t must cut B D, or B D produced, at some other point E. Join EC. Then the l BAC=the / BEC, in the same segment: in.21. but the / B A C = the L BDC, by hypothesis; . -. the I BEC=the / B D C ; that is, an ext. angle of a triangle = an int. opp. angle; which is impossible. 1.16. / . the circle which passes through B, A, C, cannot pass otherwise than through D. That is, the vertices A and D are on an arc of a circle of which the chord is BC. Q. e.d. The following corollary is important. All triangles drawn on t l i e same base, and with equal vertical angles, have their vertices on an arc of a circle, of which the given base is the chord. Ok, The locus of the vertices of triangles drawn on the same base with equal vertical angles is an arc of a circle. 1 8 8 E u c l i d ' s elements, Proposition 22. Theorem. Tlie opposite angles of amy quadrilateral inscribed in a circle are together equal to two right angles. Let ABCD be a quadrilateral inscribed in the © A B C ; then shall, (i) the A8 ADC, ABC together = two rt. angles; (ii) the a8 BAD, BCD together = two rt. angles. Join AC, BD. Then the l. ADB = the A ACB, in the segment ADCB; III. 21. also the a C D B - the A CAB, in the segment CDAB. . ' . tlie a A D C = the sum of the l. ° ACB, CAB. To each of these equals add the a ABC: then tlie two A 8 ADC, ABC together = the three a ° ACB, CAB, ABC. But the As ACB, CAB, ABC, being the angles of a triangle, together = two rt. angles. I . 32. . ' . the a8 ADC, ABC together = two rt. angles. Similarly it may be shewn that the a 8 BAD, BCD together = two rt. angles. Q.E.D. EXERCISES. 1. If a circle can be described about a parallelogram, the parallelogram must be rectangular. 2. ABC i s an isosceles triangle, and X Y i s drawn parallel to the base BC: shew that the four points B, C, X, Y l i e on a c i r c l e . 3. If one s i d e of a quadrilateral inscribed i n a c i r c l e i s produced, the exterior angle i s equal t o the opposite i n t e r i o r angle of the quadri-l a t e r a l . BOOK III. PROP. 22. 189 Proposition 22. [Alternative Proof.] Let ABCD be a quadrilateral inscribed in the 0 ABC: then shall the Z »ADC, A B C together=two rt. angles. Join FA, FC. Then the z AFC at the centre = twice the L A D C at the Oce, standing on the same arc ABC. m . 20. Also the reflex angle AFC at the centre =twice the z A B C at the o08, standing on the same arc ADC. in. 20. Hence the z 8 ADC, A B C are together half the sum of the Z AFC and the reflex angle AFC; but these make up four rt. angles: x . 15. Cor. 2. . -. the Z s ADC, ABC together=two rt. angles. q.b.u. Definition. Four or more points through which a circle may be described are said to be concyclic. Converse op Proposition 22. If a pair of opposite angles of a quadrilateral are together equ two right angles, i t s vertices are concyclic. Let ABCD be a quadrilateral, in which the opposite angles at B and D together=two rt. angles; then shall the four points A, B, C, D be concyclic. Through the three points A, B, C describe a circle: iv. 5. then this circle must pass through D. For, i f not, i t will cut AP, or A D produced, at some other point E. Join EC. Then since the quadrilateral A B C E is inscribed in a circle, . -. the Z s ABC, A E C together=two rt. angles. in. 22. But the z 8 ABC, A D C together=two rt. angles; Hyp. hence the Z s ABC, AEC = the z 8 ABC, ADC. Take from these equals the z ABC: then the Z AEC=the Z ADC; that is, an ext. angle of a triangle=an int. opp. angle; which is impossible. 1.16. . . the circle which passes through A, B, C cannot pass otherwise than through D: that is the four vertices A, B, C, D are concyclic. q.e.ju. 190 EUCLID'S ELEMENTS. Definition. Similar segments of circles are those which contain equal angles. Proposition 23. Theorem. On the same chord and on the same side of it, there cannot be two similar segments of circles, not coinciding with one another. D If possible, on the same chord AB, and on the same side of it, let there be two similar segments of circles ACB, ADB, not coinciding with one another. Then since the arcs ADB, A C B intersect at A and B, . ' . they cannot cut one another again; in. 10. . ' . one segment falls within the other. In the outer arc take any point D : join AD, cutting the inner arc at C: join CB, DB. Then because the segments are similar. . ' . the l. A C B = the a A D B ; in. Def. that is, an ext. angle of a . triangle = an int. opp. angle; which is impossible. I. 16. Hence the two similar segments ACB, ADB, on the same chord AB and on the same side of it, must coincide. q.e.d. exercises on proposition 22. 1. The straight linos which bisect any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet on the circumference. 2. A triangle i s inscribed in a circle: shew that the sum of the angles in the three segments exterior to the triangle is equal to four right angles. 3. Divide a oirole into two segments, so that the angle contained by tho one shall be double of the angle contained by the ether. book i i i . p r o p . 24. 1 9 1 Proposition 24. Theorem. Similar segments of circles on equal chords are equal to ; e another. Let AEB and CFD be similar segments on equal chords AB, CD: then shall the segment ABE = the segment CDF. For i f the segment ABE be applied to the segment CDF,. so that A falls on C, and AB falls along CD; then since AB = CD, . ' . B must coincide with D. . ' . the segment AEB must coincide with the segment CFD ; for if not, on the same chord and on the same side of it there would be two similar segments of circles, not co-inciding with one another : which is impossible. in. 23. . ' . the segment AEB = the segment CFD. Q. e.d. exercises. 1. Of two segments standing on the same chord, the greater segment contains the smaller angle. 2. A segment of a c i r c l e stands on a , chord AB, and P is any point on the same side of AB as the segment: shew that the angle APB i s greater or less than the angle in the segment, according as P i s within or without the segment. 3. P, Q, R are the middle points of the sides of a triangle, and X i s t l i e foot of the perpendicular l e t fall from one vertex on the opposite side: shew that the four points P, Q, R, X are concyclic. [See page 96, Ex. 2: also page 100, Ex. 2.] 4. Use the preceding exercise t o shew that the middle points of the sides of a triangle and the feet of the perpendiculars l e t fall from the v e r t i c e s on the opposite sides, are concyclic. 1 9 2 e u c l i d ' s elements. Proposition 25. Problem. A n arc of a circle being given, to describe the whole cir-cumference of which the given arc is a part. Let A B C be an arc of a circle: it is required to describe the whole O m of which the arc ABC is a part. In the given arc take any three points A, B, C. Join AB, BC. D r a w DF bisecting AB at rt. angles, I . 10. 11. and draw EF bisecting BC at rt. angles. Then because DF bisects the chord AB at rt. angles, . ' . the centre of the circle lies in DF. in. 1. Cor. Again, because EF bisects the chord B C at rt. angles, . ' . the centre of the circle lies in EF. in. 1. Cor. . ' . the centre of the circle is F, the only point common to DF, EF. Hence the Oce of a circle described from centre F, with radius FA, is that of which the given arc is a part. Q. e. f. Note. Euclid gave this proposition a somewhat different form, as follows: A segment of a c i r c l e being given, t o describe the circle of which i t i s a segment. Let ABC be the given segment standing on the chord AC. Draw DB, bisecting A C at rt. angles. 1.10. Join AB. At A, in BA, make the z BAE equal to the zABD. 1.23. Let AE meet BD, or BD produced, at E. Then E shall bo the centre of the required oirole. [Join EC; and prove ( i ) EA = EC; i . 4. (ii)EA = EB. i. 6 . ] book in. prop. 26. 193 Proposition 26. Theorem. In equal circles the arcs which subtend equal angles, whether at t l i e centres or at the circumferences, shall be equal. Let ABC, DEF be equal circles and let the A 8 BGC, EHF, at the centres be equal, and consequently the a s BAC, EDF at the Oces equal: in. 20. then shall the arc BKC = the arc ELF. Join BC, EF. Then because the 0s ABC, DEF are equal, . ' . their radii are equal. Hence in the A8 BGC, EHF, t BG = EH, Because -j and GC = HF, (and the a BGC = the A EHF; Hyp. . \ BC = EF. I . 4. Again, because the a BAC = the A EDF, Hyp. . ' . the segment BAC is similar to the segment EDF; in. Def. 15. and they are on equal chords BC, EF; . ' . the segment BAC = the segment EDF. in. 24. But the whole © ABC = the whole 0 DEF; . ' . the remaining segment BKC = the remaining segment ELF. . ' . the arc BKC = the arc ELF. Q. E. D. [For an Alternative Proof and Exercises see pp. 197, 198.] 1 9 4 e u c l i d ' s e l e m e n t s . Proposition 27. Theorem. Tn equal circles the angles, whether at tlie centres or the circumferences, which stand on equal arcs, shall be equal. Let ABC, DEF be equal circles, and let the arc BC = the arc EF: then shall the A BGC = the A EHF, at the centres; and also the A BAC = the A EDF, at the Gcw If the A8 BGC, EHF are not equal, one must be t l i e greater. If possible, let the A BGC be the greater. At G, in BG, make the a BGK equal to the _ EHF. I . 23. Then because in the equal ©s ABC, DEF, the A BGK = the A EHF, at the centres; Constr. . ' . the arc BK = the arc EF. in. 26. But the arc BC = the arc EF, Hyp. . ' . the arc BK = the arc EF, a part equal to the whole, which is impossible. . ' . the u BGC is not unequal to the ^ EHF; that is, the a BGC = the a EHF. And since the a BAC at the Ooc is half the j _ BGC at the centre, i l l . 20. and likewise the A EDF is half the a EHF, . ' . the A BAC = the A EDF. Q.E.D. [For Exorcises see pp. 197, 198.] book i n . prop, 2 8 . 1 9 5 Proposition 28. Theorem. In equal circles the arcs, which are cut off by equal chords, shall be equal, the major arc equal t o the major arc, and the minor to the minor. Let ABC, DEF be two equal circles, and let the chord BC = --t h e chord EF: then shall the major arc BAC = the major arc EDF; and the minor arc BGC = the minor arc EHF. Find K and L the centres of the 0 s ABC, DEF: in. 1. and join BK, KC, EL, LF. Then because the 0s ABC, DEF are equal, . ' . their radii are equal. Hence in the AB BKC, ELF, t BK = EL, Because < KC = LF, (and BC = EF; Hyp. :. the a BKC = the a ELF; I . 8. . ' . the arc BGC =the arc EHF; in. 26. and these are the minor arcs. But the whole Oce ABGC = the whole Oce DEHF; Hyp. . ' . the remaining arc BAC = the remaining arc EDF: and these are the major arcs. q. e. d. [For Exercises see pp. 197, 198.] 1 9 6 e u c l l d ' s e l e m e n t s . Proposition 29. Theorem. In equal circles the chords, which cut off equal arcs, shall be equal. A D Let ABC, DEF be equal circles, and let the arc B G C = the arc EHF: then shall the chord BC = the chord EF. Find K, L the centres of the circles. in. 1. Join BK, KC, EL, LF. Then in the equal © s ABC, DEF, because the arc B G C = the arc EHF, . ' . the A BKC = the a ELF. in. 27. Hence in the As BKC, ELF, \| B K = EL, being radii of equal circles: Because -! KC = LF, for the same reason, (and tlie A BKC = the a ELF; Proved. : . BC = EF. I . 4. Q.E.D. EXERCISES ON PROPOSITIONS 26, 27. 1. If two chords of a circle are parallel, they intercept equa 2. The straight lines, whioh join the extremities of two equal arcs of a c i r c l e towards the same parts, are parallel. 3. In a c i r c l e , or in equal c i r c l e s , sectors arc equal i f their angles at the centres are equal. EXERCISES ON PROPS. 28, 29. 197 4. If two chords of a circle intersect at right angles, the opposite arcs are together equal to a seniicircuniference. 5. If two chords intersect within a circle, they form an angle equal to that subtended at the circumference by t l i e sum of t l i e arcs they cut off. 6. If two chords intersect without a circle, they form an angle equal to that subtended at the circumference by the difference of the arcs they cut off. 7. If AB is a fixed chord of a circle, and P any point on one of the arcs cut off by it, then the bisector of the angle A P B cuts the conjugate arc in the same point, whatever be the position of P. 8. T w o circles intersect at A and B; and through these points straight lines are drawn from any point P on the circumference of one of the circles: shew that when produced they intercept on the other circumference an arc which is constant for all positions of P. 9. A triangle ABC is inscribed in a circle, and the bisectors of the angles meet the circumference at X, Y, Z. Find each angle of the triangle X Y Z in terms of those of the original triangle. ON PROPOSITIONS 28, 29. 10. The straight lines which j oin the extremities of parallel chords in a circle (i) towards the same parts, ( i i ) towards opposite parts, are equal. 11. Through A, a point of intersection of two equal circles two straight lines PAQ, X A Y are drawn: shew that the chord P X is equal to the chord Q Y . 12. Through the points of intersection of two circles two parallel straight lines are drawn terminated by the circumferences: shew that the straight lines which join their extremities towards the same parts are equal. 13. Two equal circles intersect at A and B; and through A any straight line P A Q is drawn terminated by the circumferences: shew that B P = B Q . 14. ABC is an isosceles triangle inscribed in a circle, and the bisectors of the base angles meet the circumference at X and Y. Shew that the figure B X A Y C must have four of its sides equal. What relation must subsist among the angles of the triangle A B C , in order that the figure B X A Y C may be equilateral? n. E. 14 198 EUCLID H ELEMENTS. Note. We have given Euclid's demonstrations of Propositions 26, 27, 28, 29; but it should be noticed that all these propositions also admit of direct proof by the method of superposition. To illustrate this method we will apply it to Proposition 26. Proposition 26. [Alternative Proof.] In equal circles, the arcs which subtend equal angles, whether at t l i e centres or circumferences, shall be equal. D Let ABC, D E F be equal circles, and let the Z s B G C , E H F at the centres be equal, and consequently the L SAC, E D F at the G ^ equal: m . 20. then shall the arc B K C = the arc ELF. For i f the © A B C be applied to the © DEF, so that the centre G may fall on the centre H, then because the circles are equal, Hyp. ; . their OCM must coincide; hence by revolving the upper circle about its centre, the lower circle remaining fixed, B may be made to coincide with E, and consequently G B with H E. And because the z B G C = the z EHF, . . G C must coincide with HF: and since G C = H F, Hyp. . . C must fall on F. Now B coinciding with E, and C with F, and the oc0 of the © A B C with the O™ of t i i e , - DEF, . • . the arc B K C must coincide with the arc ELF. . -. tho aro B K C = the arc ELF. Q.K.D, book in. prop. 30. 1 9 9 Proposition 30. Problem. To bisect a given arc. D Let ADB be the given arc: it is required to bisect it. Join AB; and bisect it at C. i . 10. A t C draw C D at rt. angles to AB, meeting the given arc at D. i . 11. Then shall the arc A D B be bisected at D. Join AD, BD. Then in the As ACD, BCD, ( A C = BC, Constr. Because < and C D is common; (and the a ACD = the A BCD, being rt. angles: . ' . A D = BD. I . 4. A n d since in the © A D B , the chords AD, BD are equal, . ' . the arcs cut off by them are equal, the minor arc equal to the minor, and the major arc to the major: in. 28. and the arcs AD, BD are both minor arcs, for each is less than a semi-circumference, since DC, bisecting the chord AB at rt. angles, must pass through the centre of the circle. ill. 1. Cor. . ' . the arc AD = the arc BD : that is, the arc A D B is bisected at D. q. e. d. EXERCISES. 1. I f a tangent to a c i r c l e i s parallel to » chord, the point of contact will bisect the arc cut o f f by the chord. 2. Trisect a quadrant, or the fourth part of the circumference, of a c i r c l e . 14—2 200 EUCLID'S ELEMENTS. Proposition 31. Theorem. The angle in a semicircle is a right angle .-t l i e angle in a segment greater than a semicircle is l e s s than a right angle : and the angle in a segment l e s s tlian a semicircle is greater than a right angle. Let ABCD be a circle, of which BC is a diameter, and E the centre; and let AC be a chord dividing the circle into the segments ABC, ADC, of which the segment ABC i s greater, and the segment is ADC less than a semicircle: then (i) the angle in the semicircle BAC shall be a rt. angle; ( i i ) the angle in the segment ABC shall be less than a rt. angle; ( i i i ) the angle in the segment ADC shall be greater than a rt. angle. In the arc ADC take any point D; Join BA, AD, DC, AE; and produce BA to F. ( i ) Then because EA = EB, m . Def. 1 . . " . the A EAB = the z _ EBA. I . 5. And because EA = EC, . " . the A EAC = the a ECA. . ' . t l i e whole A BAC = the sum of the as EBA, ECA: but the ext. a FAC = the sum of the two int. a s CBA, BCA; . " . the A BAC -the l . FAC; , ' . those angles, being adjacent, are r t -. angles. . ' . the A BAC, in t l i e semicircle BAC, is a rt. ansrlc. book m. prop. 31. 201 (ii) In the A ABC, because the two A s ABC, BAC are together less than two rt. angles; I . 17. and of these, the z . BAC is a rt. angle; Proved. . " . the a ABC, which is the angle in the segment ABC, i s less than a rt. angle. ( i i i ) Because ABCD is a quadrilateral inscribed in the ©ABC, . " . the a 8 ABC, ADC together = two rt. angles; in. 22. and of these, the A ABC is less than a rt. angle: Proved. . " . the a ADC, which is the angle in the segment ADC, i s greater than a rt. angle. q. e. d. exercises. 1. A circle described on the hypotenuse of a right-angled triangle as diameter, passes through the opposite angular point. 2. A system of right-angled triangles is described upon a given straight line as hypotenuse: find the locus of the opposite angular points. 3. A straight rod of given length slides between two straight rulers placed at right angles to one another: find the locus of its middle point. 4. T w o circles intersect at A and B; and through A two diameters AP, A Q are drawn, one in each circle: shew that the points P, B, Q are collinear. [See Def. p. 102.] 5. A circle is described on one of the equal sides of an isosceles triangle as diameter. Shew that it passes through the middle point of the base. 6. Of two circles which have internal contact, the diameter of the inner is equal to the radius of the outer. Shew that any chord of the outer circle, drawn from the point of contact, is bisected by the circumference of the inner circle. 7. Circles described on any two sides of a triangle as diameters intersect on the third side, or the third side produced. 8. Find the locus of the middle points of chords of d circle drawn through a fixed point. Distinguish between the cases when the given point is within, on, or without the circumference. 9. Describe a square equal to the difference of two given squares. 10. Through one of the points of intersection of two circles draw a chord of one circle which shall be bisected by the other. 11. O n a given straight line as base a system of equilateral four-sided figures is described: find the locus of the intersection of their diagonals. 202 EUCLID'S ELEMENTS. Note 1. The extension of Proposition 20 to straight and reflex angles furnishes a simple alternative proof of the first theorem contained in Proposition 31, viz. The angle in a semicircle is a right angle. For, in the adjoining figure, the angle at the centre, standing on the arc B H C , is double the angle at the oc°, standing on the same arc. Now the angle at the centre is the straight angle BEC; . . the z B A C is half of the straight angle B E C : and a straight angle = two rt. angles; . . the z B A C = one half of two rt. angles, = one rt. angle. q.e.d. Note 2. From Proposition 31 we m a y derive a simple practical solution of Proposition 17, namely, To draw a tangent to a circle from a given external point. Let BCD be the given circle, and A the given exter-nal point: it is required to draw from A a tangent to the © BCD. Find E, the centre of the circle, and join AE. On A E describe the semi-circle A B E , to cut the given circle at B. Join AB. Then A B shall be a tangent to the 0 B C D . For the Z ABE, being in a semicircle, is a rt. angle. in. 31. . -. A B is drawn at rt. angles to tlie radius EB, from its ex-tremity B; . -. A B is a tangent to the circle. in. l ' i . Q.E.F. Since the semicircle might be described on either side of AE. i t is clear that there will be a second solution of tlie problem, as shewn by the dotted lines of the figure. BOOK I I I . PROP. 32. 2 0 3 Proposition 32. Theorem. If a straight line touch a circle, and from, the point of contact a chord be drawn, the angles which this chord makes with the tangent shall be equal to the angles in the alternate segments of the c i r c l e . Let EF touch the given © A B C at B, and let BD be a chord drawn from B, the point of contact: then shall (i) the A DBF = the angle in the alternate segment BAD: ( i i ) the a DBE = the angle in the alternate segment BCD. From B draw BA perp. to EF. Take any point C in the arc BD; and join AD, DC, CB. I . 11. (i) Then because BA is drawn perp. to the tangent EF, at its point of contact B, . ' . BA passes through the centre of the circle: in. 19. . " . the A ADB, being in a semicircle, i s a rt. angle: i l l . 31. . ' . in the A ABD, the other a8 ABD, BAD together = a rt. angle; . I . 32. that is, the A 8 ABD, BAD together = the A ABF. From these equals take the common a ABD; . ' . the A DBF = the A BAD, which is in the alternate seg-ment. 204 EUCLID'S ELEMENTS. A (ii) Because A B C D is a quadrilateral inscribed in a circle, .'.the A8 BCD, B A D together = two rt. angles: in. 22. but the ab DBE, D B F together = two rt. angles; I. 13. . ' . the A8 DBE, DBF together = the a3 BCD, B A D : and of these the A D B F = the a B A D ; Proved. . ' . the a D B E = the A DCB, which is in the alternate seg-ment. Q. E. D. EXERCISES. 1. State and prove the converse of this proposition. 2. Use this Proposition to Bhew that the tangents drawn to a circle from an external point are equal. 3. If two circles touch one anpther, any straight line drawn through the point of contact cuts off similar segments. Prove this for ( i ) internal, ( i i ) external contact. 4. If two circles touch one another, and from A, the point of con-tact, two chords APQ, A X Y are drawn: then PX and Q Y are parallel. Prove this for (i) internal, ( i i ) external contact. 5. Two circles intersect at the points A, B: and one of them passes through O, the centre of the other': prove that O A bisects the angle between the common ohord and the tangent to the first circle at A. 6. Two circles intersect at A and B; and through P, any point on the circumference of one of them, straight lines PAC, PBD are drawn to cut the other circle at C and D: shew that C D is parallel to the tangent at P. 7. If from the point of oontact of a tangent to a circle, a chord be drawn, the perpendiculars dropped on the tangent and chord from the middle point of eithor arc out off by the chord are equal. book in. prop. 33. 205 Proposition 33. Problem. On a given straight line to describe a segment of a c i r c l e which shall contain an angle equal to a given angle. H H \ Let AB be the given st. line, and C the given angle: i t i s required to describe on AB a segment of a circle which shall contain an angle equal to C. At A in BA, make the a BAD equal to the a C. i . 23. From A draw AE at rt. angles to AD. I . 11. Bisect AB at F; I . 10. and from F draw FG at rt. angles to AB, cutting AE at G. Join GB. Then in the A8 AFG, BFG. ( AF = BF, Constr. Because J , and FG is common, (and the A AFG = the a BFG, being rt. angles; . ' . GA = GB : I . 4. . ' . the circle described from centre G, with radius GA, will pass through B. Describe this circle, and call i t ABH: then the segment AHB shall contain an angle equal to C. Because AD i s drawn at rt. angles to the radius GA from i t s extremity A, . ' . AD is a tangent to the circle: in. 16. and from A, its point of contact, a chord AB i s drawn; . ' . the a BAD = the angle in the alt. segment AHB. i l l . 32. But the A BAD = the A C: Constr. . ' . the angle in the segment AHB = the a C. . ' . AHB is the segment required. q.e.f. 206 euclid's elements. Note. In the particular case when the given angle C is a rt. angle, the segment required will be the semicircle described on the given st. line A B ; for the angle in a semicirclo is a rt. angle. in. 31. EXERCISES. [The following exercises depend on the corollary to Proposition 21 given on page 187, namely The locus of the vertices of triangles which stand on the same base and have a given vertical angle, is the arc of the segment standing on this base, and containing an angle equal to the given angle. Exercises 1 and 2 afford good illustrations of the method of find-ing required points by the Intersection of Loci. See page 117.] 1. Describe a triangle on a given base, having a given vertical angle, and having its vertex on a given straight line. 2. Construct a triangle, having given the base, the vertical angle and Ii) one other side. ( i i ) the altitude. ( i i i ) the length of the median which bisects the base. (iv) the point at which the perpendicular from the vertex •meets t l i e base. 3. Construct a triangle having given the base, the vertical angle, and the point at which the base is cut by the bisector of the vertical angle. [Let A B be the base, X the given point in it. and K the given angle. On A B describe a segment of a circle containing an angle equal to K; complete the Oc° by drawing the arc APB. Bisect the arc A P B at P: join PX, and produce i t to meet the O™ at C. Then A B C shall be the required triangle.] 4. Construct a triangle having given the base, the vertical angle, and the sum of the remaining sides. [Let A B be the given base, K the given angle, and H the given line equal to the sum of the sides. On A B describe a segment containing an angle equal to K, also another segment containing an angle equ.il to half the z K. From centre A, with radius H, describe a circle cutting the last drawn segment at X and Y. Join A X (or AY) cutting the first segment at C. Then A B C shall be the required triangle.] 5. Construct a triangle having given the base, the vertical angle, and the difference of the remaining sides. book in. p r o p . 3 4 . 207 Proposition 34. Problem. From a given circle to cut off a segment which shall contain an angle equal to a given angle. Let ABC be the given circle, and D the given angle: it is required to cut off from the © ABC a segment which shall contain an angle equal to D. Take any point B on the Oce, and at B-draw the tangent EBF. in. 17. A t B, in FB, make the a FBC equal to the a D. i . 23. Then the segment BAC shall contain an angle equal to D. Because EF is a tangent to the circle, and from B, its point of contact, a chord BC is drawn, . ' . the a FBC = the angle in the alternate segment BAC. in. 32. But the a FBC = the a D; Constr. . ' . the angle in the segment BAC = the a D. Hence from the given © A B C a segment BAC has been cut off, containing an angle equal to D. q. e. p. exercises. 1. The chord of a given segment of a circle is produced to a. point: on this straight line so produced draw a segment of a c i r c l e similar to the given segment. 2. Through a given point without a c i r c l e draw a straight line that will cut o f f a segment capable of containing an angle equal to a given angle. 208 Euclid's elements. Proposition 35. Theorem. If two chords of a c i r c l e cut one another, the rectangle contained by the segments of one shall be equal to the red angle contained by the segments of the other. Let AB, CD, two chords of the ©ACBD, cut one another at E: then shall the rect. AE, EB = the rect. CE, ED. Find F the centre of the © A C B : in. 1. From F draw FG, FH perp. respectively to AB, CD. 1.12. Join FA, FE, FD. Then because FG i s drawn from the centre F perp. to AB, . ' . AB is bisected at G. IIL 3. For a similar reason CD i s bisected at H. Again, because AB i s divided equally at G, and unequally at E, . ' . the rect. AE, EB with the sq. on EG = the sq. on AG. n. o. To each of these equals add the sq. on GF; then the rect. AE, EB with the sqq. on EG, GF = the sum of the sqq. on AG, GF. But the sqq. on EG, GF = the sq. on FE; l . 47. and the sqq. on AG, GF = the sq. on AF; for t l i e angles at G are rt. angles. . ' . the rect. AE, EB with the sq. on FE = the sq. on AF. Similarly i t may be shewn that the rect. CE, ED with the sq. on FE = the sq. on FD. But the sq. on AF = the sq. on FD; for AF = FD. . ' . the rect. AE, EB with the sq. on FE = the rect. CE, ED with the sq. on FE. From these equals take the sq. on FE: then the rect. AE, EB-the rect. CE, ED. q.kd. book in. prop. 35. 209 Corollary. If through a fixed point within a circle any number of chords are drawn, t l i e rectangles contained by their segments are all equal. Note. The following special cases of this proposition deserve notice. ( i ) when the given chords both pass through the centre: ( i i ) when one chord passes through the centre, and cuts the other at right angles: (hi) when one chord passes through the centre, and cuts the other obliquely. In each of these cases the general proof requires some modifica-tion, which may be l e f t as an exercise to the student. EXERCISES. 1. Two straight lines AB, CD intersect at E, so that the rectangl AE, EB is equal to the rectangle CE, ED: shew t l i a t the four points A, B, C, D are concyclic. 2. The rectangle contained by the segments of any chord drawn through a given point within, a circle is equal to the square on half the shortest chord which may be drawn through that point. 3. A B C is a triangle right-angled at C; and from C « . perpen-dicular C D is drawn to the hypotenuse: shew that the square on C D i s equal to the rectangle AD, DB. 4. A B C is a triangle; and AP, B Q the perpendiculars dropped from A and B on the opposite sides, intersect at O: shew that the rectangle AO, O P is equal to the rectangle BO, OQ. 5. Two circles intersect at A and B, and through any point in AB their common chord two chords are drawn, one in each circle; shew that their four extremities are concyclic. 6. A and B are two points within a circle such that the rectangle contained by the segments of any chord drawn through A is equal to the rectangle contained by the segments of any chord through B: shew that A and B are equidistant from the centre. 7. If through E, a point without a circle, two secants EAB, E C D are drawn; shew that the rectangle EA, EB is equal t o the rectangle EC, ED. [Proceed as in m . 35, using n. 6.] 8. Through A, a point of intersection of two circles, two straight lines CAE, DAF are drawn, each passing through a centre and termi-nated by the circumferences: shew that the rectangle CA, A E is equal to the rectangle DA, AF. 2 1 0 e u o l i d ' s elements. Proposition 36. Theorem. If from any point without a circle a tangent and a secant be drawn, then the rectangle contained by the whole secant and the part of i t without the circle shall be equal to the square on the tangent. Let ABC be a circle; and from D a point without it, l e t there be drawn the secant DCA, and the tangent DB: then the rect. DA, DC shall be equal to the sq. on DB. Find E, the centre of the © ABC: m . 1. and from E, draw EF perp. to AD. 1.12. Join EB, EC, ED. Then because EF, passing through the centre, is perp. to the chord AC, . " . AC is bisected at F. in. 3. A n d since AC is bisected at F and produced to D, .'.the rect. DA, DC with the sq. on FC =the sq. on FD. il 6. To each of these equals add the sq. on EF: then the rect. DA, DC with the sqq. on EF, FC = the sqq. on EF, FD. But the sqq. on EF, FC = the sq. on EC; for EFC i s a rt. angle; = the sq. on EB. And the sqq. on EF, FD = the sq. on ED; for EFD is a rt. angle; = the sqq. on EB, BD; for EBD is a rt. angle. m . IS. . ' . t l i e rect. DA, DC with the sq. on EB = the sqq. on EB, BD. From these equals take the sq. on EB: then the rect. DA, DC = the sq. on DB. q. e.d. Notk. This proof may easily be adapted to the case when tlie secant passes through tho oontre of the c i r c l e . BOOK III. PROP. 36. 211 Corollary. If from a given point without a circle any number of secants are drawn, the rectangles contained by the whole secants and the parts of them without the circle are all equal; for each of these rectangles is equal to the square on the tangent drawn from the given point to the circle. For instance, in the adjoining figure, each of the rectangles PB, PA and PD, PC and PF, PE is equal to the square on the tangent P Q : . " . the rect. PB, PA = the rect. PD, PC = the rect. PF, PE. Note. Eemembering that the segments into which the chord AB i s divided at P, are the lines PA, PB, (see Part I. page 131) we are enabled to include the corollaries of Propositions 35 and 36 in a single enunciation. If any number of chords of a circle are drawn through a given point within or without a circle, the rectangles contained by the segments of the chords are equal. EXERCISES. 1. Use this proposition to shew that tangents drawn to a circle from an external point are equal. 2. If two circles intersect, tangents drawn to them from any point in their common chord produced are equal. 3. If two circles intersect at A and B, and P Q is a tangent to both circles; shew that AB produced bisects PQ. 4. If P is any point on the straight line AB produced, shew that the tangents drawn from P to all circles which pass through A and B are equal. 5. A B C is a triangle right-angled at C, and from any point P in AC, a perpendicular PQis drawn to the hypotenuse: shew that the rectangle AC, A P is equal to the rectangle AB, AQ. 6. A B C is a triangle right-angled at C, and from C a perpen-dicular C D is drawn to the hypotenuse: shew that the rect. AB, A D is equal to the square on AC. 212 euclid's elements. Proposition 37. Theorem. If f r o m a point without a circle there be d r a w n two straight lines, one of which cuts the circle, a n d the oilier meets it, a n d if the rectangle contained by the whole line which cuts the circle a n d the part of it untlicnit tlie circle be equal to the square o n the line which meets the circle, then the line which meets the circle shall be a tangent to it. Let ABC be a circle; and from D, a point without i t , l e t there be drawn two st. lines DCA and DB, of which DCA cuts the circle at C and A, and DB meets it; and l e t the rect. DA, DC = the sq. on DB: then shall DB be a tangent to the c i r c l e . From D draw DE to touch the ©ABC: in. 7. let E be the point of contact. Find the centre F, and join FB, FD, FE. in. 1 . Then since DCA is a secant, and DE a tangent to the c i r c l e , . ' . the rect. DA, DC = the sq. on DE, in. 36. But, by Hypothesis, the rect. DA, DC = the sq. on DB; . ' . the sq. on DE = the sq. on DB, . " . DE = DB. Hence in the A8 DBF, DEF. [ DB = DE, Proved. Because J . and BF = EF; m. D- r \ 1. [ and DF is common: . " . the A DBF = the a DEF. i . 8 . But DEF is a rt. angle ; in. 18. . ' . DBF is also a rt angle; and since BF is a radius, . ' . DB touches the © A B C at the point B. .i. K.tf. THEOREMS AND EXAMPLES ON BOOK III. 219 It follows from hypothesis that the point B is outside the circle used in the construction: . • . two tangents such as B C may always be drawn to it from B ; hence two common tangents may always be drawn to the given circles by the above method. These are called the direct common tangents. W h e n the given circles are external to one another and do not intersect, two more common tangents may be drawn. For, from centre A, with a radius equal to the sum of the radii of the given circles, describe a circle. From B draw a tangent to this circle; and proceed as before, but draw B E in the direction opposite to AD. It follows from hypothesis that B is external to the circle used in the construction; / . two tangents may be drawn to it from B. Hence two more common tangents may be drawn to the given circles: these will be found to pass between the given circles, and are called the transverse common tangents. Thus, in general, four common tangents may be drawn to two given circles. The student should investigate for himself the number of common tangents which may be drawn in the following special cases, noting in each case where the general construction fails, or is modified:— ( i ) W h e n the given circles intersect: ( i i ) W h e n the given circles have external contact: (hi) W h e n the given circles have internal contact: (iv) W h e n one of the given circles is wholly within the other. 18. Draw the direct common tangents to two equal circles, 19. If the two direct, or the two transverse, common tangents are drawn to two circles, the parts of the tangents intercepted be-tween the points of contact are equal. 20. If four common tangents are drawn to two circles external to one another; shew that the two direct, and also the two transverse, tangents intersect on the straight line which joins the centres of the circles. 21. Two given circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that P Q subtends a right angle at the point A. 22. Two circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that a circle described on P Q as diameter is touched at A by the straight Hue which joins the centres of the circles. 220 Euclid's elements. 23. Two circles whose centres are C and C have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that the bisectors of the angles PCA, QC'A meet at right angles in PQ. And i f R is the point of intersection of the bisectors, shew that RA is also a common tangent to the circles. 24. Two circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that the square on P Q is equal to the rectangle contained by the diameters of the circles. 25. Draw a tangent to a given circle, so that the part of i t intercepted by another given circle may be equal to a given straight line. When is this impossible? 26. Draw a secant to two given circles, so that the parts of it intercepted by the circumferences may be equal to two given straight lines. Pboblems on Tangency. The following exercises are solved by the Method of Inter-section of Loci, explained on page 117. The student should begin by making himself familiar with the following loci. (i) Tlie locus of the centres of circles which pass through two given points. ( i i ) Tlie locus of the centres of circles which touch a given straight line at a given point. ( i i i ) The locus of the centres of circles which touch a given circle at a given point. (iv) The locus of the centres of circles which touch a giren straight line, and have a given radius. (v) The locus of t l i e centres of circles which toiuh a given circle, and have a given radius. (vi) The locus of the centres of circles ivhich touch two given straight lines. In each exercise the student should investigate the limits and relations among the data, in order that the problem niav be possible. 27. Describe a cirole to touoh three given straight lines. 28. Describe a cirole to pass through a given point and touoh a given straight line at a given point. 29. Describe a circle to pass through a given point, and touch a given circle at a givon point. THEOREMS AND EXAMPLES ON BOOK III. 2 2 1 30. Describe a circle of given radius to pass through a given point, and touch a given straight line. 31. Describe a circle of given radius to touch two given circles. 32. Describe a circle of given radius to touch two given straight lines. 33. Describe a circle of given radius to touch a given circle and a given straight line. 34. Describe two circles of given radii to touch one another and a given straight line, on the same side of it. 35. If a circle touches a given circle and a given straight line, shew that the points of contact and an extremity of the diameter of the given circle at right angles to the given line are collinear. 36. To describe a circle to touch a given circle, and also to touch a given straight line at a given point. Let DEB he the given circle, PQ the given st. line, and A the given point in it: i t is required to describe a circle to touch the © DEB, and also to touch P Q at A. At A draw A F perp. to P Q : 1.11. then the centre of the required circle must lie in AF. in. 19. Find C, the centre of the © DEB, I . 1. and draw a diameter B D perp. to P Q : join A to one extremity D, cutting the ooe at E. Join C E , and produce i t to cut AF at F. Then F is the centre, and FA the radius of the required circle. [Supply the proof: and shew that a second solution is obtained by joining AB, and producing it to meet the oQ": also distinguish between the nature of the contact of the circles, when P Q cuts, touches, or is without the given circle.] 37. Describe a circle to touch a given straight line, and to touch a given circle at a given point. 38. Describe a circle to touch a given circle, have its centre in a given straight line, and pass through a given point in that straight hue. [For other problems of the same class see page 235.] 222 EUCLID'S ELEMENTS. OllTHOGONAL ClECLES. Definition. Circles which intersect at a point, so that the two tangents at that point are at right angles to one another, are said to be orthogonal, or to cut one another ortho-gonally. 39. In two intersecting circles the angle between the tangents at one point of intersection is equal to the angle between the tangents at the other. 40. If two circles cut one another orthogonally, the tangent to each circle at a point of intersection will pass through the centre of the other circle. 41. If two circles cut one anotlier orthogonally, the square on tlie distance between their centres is equal to t l i e sum of the squares on their radii. 42. Find the locus of the centres of all circles which cut a given circle orthogonally at a given point. 43. Describe a circle to pass through a given point and cut a given circle orthogonally at a given point. III. ON ANGLES IN SEGMENTS, AND ANGLES AT THE CENTRES AND CIRCUMFERENCES OF CIRCLES. See Propositions 20, 21, 22; 26, 27, 28, 29: 31, 32, 33, 34. 1. If two chords intersect within a circle, they form an angle equal to that at the centre, subtended by half the sum of ihe arcs t l i e y cut o f f . Let A B and C D be two chords, intersecting at E within the given © A D B C : then shall the z A E C be equal to the angle at the centre, subtended by half the sum of the arcs AC, BD. Join AD. Then the ext. Z A EC=the sum of the int. opp. Z'EDA, E A D ; that is, the sum of tho Z • CDA, BAD. But the Z'CDA, B A D are the angles at the o™ subtended by the arcs AC, B D ; . -. their sum = half the sum of tho angles at the centre subtended by the same arcs; or, tho Z A E C = the anglo at the centre subtended by half the sum of tho arcs AC, BD. y, u. n. THEOREMS AND EXAMPLES ON BOOK III. 223 2. If two chords when produced intersect outside a circle, they form an angle equal to that at the centre subtended by half the difference oj the arcs they cut off. 3. The sum of the arcs cut off by two chords of a circle at right angles to one another is equal to the semi-circumference. 4. AB, AC are any two chords of a circle; and P, Q are the middle points of the minor arcs cut off by them: i f P Q is joined, cutting A B and A C at X, Y, shew that A X = AY. 5. If one side of a quadrilateral inscribed in u, circle is produced, the exterior angle is equal to the opposite interior angle. 6. If two circles intersect, and any straight lines are drawn, one through each point of section, terminated by the circumferences; shew that the chords which join their extremities towards the same parts are parallel. 7. ABCD is a quadrilateral inscribed in a circle; and the opposite sides AB, D C are produced to meet at P, and CB, D A to meet at Q: i f the circles circumscribed about the triangles PBC, Q A B intersect at R, shew that the points P, R, Q are collinear. 8. If a circle is described on one of the sides of a right-angled triangle, then the tangent drawn to it at the point where it cuts the hypotenuse bisects the other side. 9. Given three points not in the same straight line: shew how to find any number of points on the circle which passes through them, without finding the centre. 10. Through any one of three given points not in the same straight line, draw a tangent to the circle which passes through them, without finding the centre. 11. Of two circles which intersect at A and B, the circumference of one passes through the centre of the other: from A any straight hue A C D is drawn to cut them both; shew that C B = C D . 12. Two tangents AP, AQ are drawn to a circle, and B is the middle point of the arc PQ, convex to A. Shew that PB bisects the angle A P Q . 13. Two circles intersect at A and B; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D: if CB, B D are joined, shew that the triangles A B C , D B A are equi-angular to one .another. 14. Two segments of circles are described on the same chord and on the same side of it; the extremities of the common chord are joined to any point on the arc of the exterior segment: shew that the arc intercepted on the interior segment is constant. 224 EUCLID'S ELEMENTS. 15. If a series of triangles are drawn standing on a fixed base, and having a given vertical angle, shew that the bisectors of the verti-cal angles all pass through a fixed point. 16. ABC is a triangle inscribed in a circle, and E the middle point of the arc subtended by B C on the side remote from A: i f through E a diameter E D is drawn, shew that the angle D E A is half the difference of the angles at B and C. [See Ex. 7, p. 101.] 17. Lf two circles touch each other internally at a point A, any ehord of the exterior circle which touches the interior is divided at i t s point of contact into segments which subtend equal angles at A. 18. If two circles touch one another internally, and a straight line is drawn to cut them, the segments of it intercepted between the circumferences subtend equal angles at the point of contact. The Obthocentee or . - Tbiangle. 19. Tlie perpendiculars drawn from the vertices of " triangle to the opposite sides are concurrent. In the A ABC, let AD, BE be the perp" drawn from A and B to the oppo-site sides; and let them intersect at O. Join C O ; and produce it to meet AB at F. It is required to shew that C F i s perp. t o AB. Join DE. Then, because the z "OEC, O D C are rt. angles, Hyp. . : the points O, E, C, D are concyclic: . • . tho L DEC=the Z D O C , in the same segment; = the vert. opp. z FOA. Again, beoause the Z " AEB, ADB are rt. angles. Hyp. . -. the points A, E, D, B are concyclic: . -. the Z D E B = the L DAB, in the same segment. . . the sum of the z " FOA, F A O = the sum of the L' DEC, DEB = a rt. angle: Hyp-: . the remaining L A F O = art. angle: i. 32. that is, C F is perp. to AB. Honco tho three perp AD, BE, C F meet at the point O. g. K.n. [For an Alternative Proof see page 106.] theorems and examples on book i i i . 225 Definitions. (i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its ortho-centre. ( i i ) _ The triangle formed by joining the feet of the perpendi-culars is called the pedal or orthocentric triangle. 20. In an acute-angled triangle the perpendiculars drawn from the vertices t o the opposite sides b i s e c t the angles of the pedal triangle through which they pass. In the acute-angled A ABC, let AD, BE, CF he the perpB drawn from the vertices to the opposite sides, meeting at the orthocentre O; and l e t DEF. be the pedal triangle: then shall AD, BE, CF bisect respect-ively the Z" FDE, DEF, EFD. For, as in the last theorem, i t may be shewn that the points O, D, C, E are concyclic; . • . the z O D E = the L OCE, in the same segment. Similarly the points O, D, B, F are concyclic; / . the z ODF = the z OBF, in the same segment. But the z OCE = the z OBF, each being the comp' of the z BAC. . -. the L O D E = the L ODF. Similarly it may be shewn that the Z B DEF, EFD are bisected by BEandCF. Q. &. d. Cobollaby. (i) Every two sides of the pedal triangle are equall inclined t o that side of the original triangle in which they meet. For the z EDC = the comp'of the L O D E = the comp'of the z O C E =the z BAC. Similarly it may be shewn that the z FDB = the z BAC, . -. the L EDC = the Z FDB = the z A. In like manner i t may be proved that the L DEC=the Z FEA = the Z B, and the t DFB = the z EFA=the L C. Cobollaby. ( i i ) The triangles DEC, AEF, DBF are equiangular t o one another and t o the triangle ABC. Note. If the angle BAC i s obtuse, then the perpendiculars BE, CF b i s e c t externally the corresponding angles of the pedal triangle. 226 Euclid's elements. 21. In any triangle, if the perpendiculars drawn from the vertices on the opposite sides are produced to meet the circumscribed circle, then each side bisects tliat portion of the line perpendicular to it which lies between t l i e orthocentre and the circumference. Let ABC be a triangle in which the perpen-diculars AD, BE are drawn, intersecting at O the orthocentre; and l e t A D be produced to meet the oco of the circumscribing c i r c l e at G : then shall D O = D G . Join BG. Then in the two A" OEA, ODB, the z OEA=the L ODB, being rt. angles; and the Z EOA = the vert. opp. z DOB; . -. the remaining L EAO=the remaining Z DBO. i . 32. But the z CAG = the z CBG, in the same segment; . -. the Z DBO=the Z DBG. Then in the A" DBO, DBG, (the L DBO=the z DBG, Proved. Because the Z B D O = the z BDG, ( and BD is common; . -. D O = DG. i . 2 1 3 . Q. K. U. 22. In an acute-angled triangle the three sides are the external bisectors of the angles of the pedal triangle: and in an obtuse-angled triangle the sides containing the obtuse angle are the internal bisectors of the corresponding angles of t l i e pedal triangle. 23. If O is the orthocentre of the triangle ABC, shew that the angles B O C , B A C are supplementary. 24. If O is the orthocentre of the triangle ABC, then any one of the four points O, A, B, C is the orthocentre of the triangle whose vertices are the other three. 25. The three circles which pass through two vertices of a triangle and its orthocentre are each equal to the circle circumscribed about the triangle. 26. D, E are taken on the oiroumference of a semicircle described on a given straight line A B : the chords A D , B E and AE, B D intersect (produced if necessary) at F and G : shew that F G is per-pendicular to A B . 27. ABCD is a parallelogram; AE and CE are drawn at right angles to A B , and C E respectively: shew that E D . if produced, will bo perpendicular to A C . THEOREMS AND EXAMPLES ON BOOK III. 227 28. ABC is a triangle, O is its orthocentre, and AK a diameter of the circumscribed circle: shew that B O C K is a parallelogram. 29. The orthocentre of a triangle is joined to the middle point of the base, and the joining line is produced to meet the circumscribed circle: prove that it will meet it at the same point as the diameter which passes through the vertex. 30. The perpendicular from the vertex of a triangle on the base, and the straight line joining the orthocentre to the middle point of the base, are produced to meet the circumscribed circle at P and Q : shew that P Q is parallel to the base. 31. The distance of each vertex of a triangle from the orthocentre is double of the perpendicular drawn from the centre of the circum-scribed circle on the opposite side, 32. Three circles are described each passing through the ortho-centre of a triangle and two of its vertices: shew that the triangle formed by joining their centres is equal in all respects to the original triangle. 33. ABC is a triangle inscribed in a circle, and the bisectors of its angles which intersect at O are produced to meet the circumference in P Q R : shew that O is the orthocentre of the triangle PQR. 34. Construct a triangle, having given a vertex, the orthocentre, and the centre of the circumscribed circle. Loci. 35. Given the base and vertical angle of a triangle, find the locus of its orthocentre. Let B C be the given base, and X the given angle; and let B A C be any triangle on the base BC, having its vertical Z A equal to the Z X. Draw the perpB BE, CF, intersecting at the orthocentre O. It is required to find the locus of O. Since the Z "OFA, O E A are rt. angles, . • . the points O, F, A, E are concyclic; B C .•.the Z F O E is the supplement of the z A: in. 22. . • . the vert. opp. z B O C is the supplement of the L A. But the z A is constant, being always equal to the L X; . • . its supplement is constant; that is, the A B O C has a fixed base, and constant vertical angle; hence the locus of its vertex O is the arc of a segment of which B C is the chord. [See p. 187.] 228 EUCLID H ELEMENTS. 36. Given tlie base and vertical angle of a triangle, find the loc of the intersection of t l i e bisectors of i t s angles. Let BAC be any triangle on the given base BC, having its vertical angle equal to the given z X; and let Al, Bl, CI be the bisectors of its angles: [see Ex. 2, p. 103.] i t is required to find the locus of the point I . Denote the angles of the A A B C by A, B, C; and let the z BIC be denoted by I . Then from the A BIC, B ( i ) I + B + £C = two rt. angles, and from the A ABC, A + B + C=:two rt. angles; i . 32. ( i i ) so that J A + £B + JC = one rt. angle, . ' . , taking the differences of the equals in ( i ) and ( i i ) . I - \A = one rt. angle : or, I = one rt. angle-)-J,A. But A is constant, being always equal to the z X; . . 1 is constant: : . , since the base BC is fixed, the locus of I is the arc of a segment of which BC is the chord. 37. Given tlie base and vertical angle of a triangle, find the loc of the centroid, that is, the intersection of the medians. Let B A C be any triangle on the given base BC, having its vertical angle equal A to the given angle S; let the medians AX, „ ' i BY, C Z intersect at the centroid G [see A \ Ex.4, p. 105]: / \ Z \Y i t is required to find the locus of the point G. ' \ G -~^\ Through G draw GP, G Q par1 to A B y .^ , \ \ and A C respectively. , -" ~ " " \ \ Then Z G is a third part of Z C ; B P X Q C Ex. 4, p. 105. and since G P is par1 to ZB, . -. BP is a third part of BC. Ex. 19. p. 99. Similarly Q C i s a third part of BC; . • . P and Q are fixed points. Now since PG, G Q are par1 respectively to BA, AC, Constr. : . the L PGQ=the z BAC, , . 2 ! . = the z S, that is, the z P G Q is constant; and sinco the base P Q is fixed, . . the loous of G is the aro of a segment of which P Q i s the chord. THEOREMS AND EXAMPLES ON BOOK III. 229 Obs. In this problem the points A and G move on the arcs of similar segments. 38. Given the base and the vertical angle of a triangle; find the locus of the intersection of the bisectors of the exterior base angles. 39. Through the extremities of a given straight line A B any two parallel straight lines AP, B Q are drawn; find the locus of the inter-section of the bisectors of the angles PAB, QBA. 40. Find the locus of the middle points of chords of a circle drawn through a fixed point. Distinguish between the cases when the given point is within, on, or without the circumference. 41. Find the locus of the points of contact of tangents drawn from a fixed point to a system of concentric circles. 42. Find the locus of the intersection of straight lines which pass through two fixed points on a circle and intercept on its circumference an arc of constant length. 43. A and B are two fixed points on the circumference of a circle, and P Q is any diameter: find the locus of the intersection of PA and QB. 44. BAC is any triangle described on the fixed base BC and having a constant vertical angle; and BA is produced to P, so that BP is equal to the sum of the sides containing the vertical angle: find the locus of P. 45. AB is a fixed chord of a circle, and AC is a moveable chord passing through A : if the parallelogram C B is completed, find the locus of the intersection of its diagonals. 46. A straight rod PQ slides between two rulers placed at right angles to one another, and from its extremities PX, Q X are drawn perpendicular to the rulers: find the locus of X. 47. Two circles whose centres are C and D, intersect at A and B : through A, any straight line P A Q is drawn terminated by the circum-ferences ; and PC, Q D intersect at X: find the locus of X, and shew that it passes through B. [Ex. 9, p. 216.] 48. Two circles intersect at A and B, and through P, any point on the circumference of one of them, two straight lines PA, PB are drawn, and produced if necessary, to cut the other circle at X and Y: find the locus of the intersection of "AY and BX. 49. Two circles intersect at A and B; HAK is a fixed straight lino drawn through A and terminated by the circumferences, and P A Q is any other straight line similarly drawn: find the locus of the intersection of H P and QK. [Ex. 3, p. 186.] H. E. 10 230 EUCLID'S ELEMENTS. 50. Two segments of circles are on the same chord AB and on the same side of it; and P and Q are any points one on each arc: find the locus of the intersection of the bisectors -o f the angles PAQ, PBQ. 51. Two circles intersect at A and B; and through A any straight line PAQ is drawn terminated by the circumferences: find the loeus of the middle point of PQ. Miscellaneous Examples on Angles in a Circle. 52. ABC is a triangle, and circles are drawn through B, C, cutting the sides in P, Q, P ' , Q', . . . : shew that PQ, P'Q'... are parallel to one another and to the tangent drawn at A to the circle circumscribed about the triangle. 53. Two circles intersect at B and C, and from any point A, on the circumference of one of them, AB, A C are drawn, and produced i f necessary, to meet the other at D and E: shew that D E is parallel to the tangent at A. 54. A secant PAB and a tangent PT are drawn to a circle from an external point P; and the bisector of the angle A T B meets AB at C: shew that PC i s equal to PT. 55. From a point A on the circumference of a circle two chords AB, A C are drawn, and also the diameter AF: i f AB, A C are produced to meet the tangent at F in D and E, shew that the triangles ABC, AED are equiangular to one another. 56. O is any point within a triangle ABC, and OD, OE, OF are drawn perpendicular to BC, CA, AB respectively : shew that the angle B O C is equal to the sum of the angles BAC, EDF. 57. If two tangents are drawn to a circle from an external point, shew that they contain an angle equal to the difference of the angles in the segments cut o f f by the chord of contact. 58. Two circles intersect, and through a point of section a straig line is drawn bisecting the angle between the diameters through that point: shew that this straight line cuts o f f similar segments from the two circles. 59. Two equal circles intersect at A and B ; and from centre A, with any radius less than AB a third circle is described cutting t l i e given circles on the same side of AB at C and D: shew that the points B, C, D are collinear. GO. ABC and A'B'C are two triangles inscribed in a oircle, so that AB, A C aro respectively p o r t i l l o l to A'B', A'C : shew that B C i s parallel to B'C. THEOREMS A N D EXAMPLES ON BOOK III. 23l 61. Two circles intersect at A and B, and through A two straight lines HAK, P A Q are drawn terminated by the circumferences : if H P and K Q intersect at X, shew that the points H, B, K, X are eoncyclic, 62. Describe a circle touching a given straight line at a given point, so that tangents drawn to it from two fixed points in the given line may be parallel. [See Ex. 10, p. 183.] 63. C is the centre of a circle, and CA, CB two fixed radii: if from any point P on the arc A B perpendiculars PX, PY are drawn to C A and CB, shew that the distance X Y is constant. 64. AB is a chord of a circle, and P any point in its circum-ference ; P M is drawn perpendicular to AB, and A N is drawn perpen-dicular to the tangent at P : shew that M N is parallel to PB. 65. P is any point on the circumference of a circle of which AB is a fixed diameter, and PN is drawn perpendicular to A B ; on A N and BN as diameters circles are described, which are cut by AP, BP at X and Y: shew that X Y is a common tangent to these circles. 66. Upon the same chord and on the same side of it three seg-ments of circles are described containing respectively a given angle, its supplement and a right angle: shew that the intercept made by the two former segments upon any straight line drawn through an ex-tremity of the given chord is bisected by the latter segment. 67. Two straight lines of indefinite length touch a given circle, and any chord is drawn so as to be bisected by the chord of contact: i f the former chord is produced, shew that the intercepts between the circumference and the tangents are equal. 68. Two circles intersect one another: through one of the points of contact draw a straight line of given length terminated by the cir-cumferences. 69. On the three sides of any triangle equilateral triangles are described remote from the given triangle: shew that the circles de-scribed about them intersect at a point. 70. On BC, CA, AB the sides of a triangle ABC, any points P, Q, R are taken; shew that the circles described about the triangles AQR, BRP, C P Q meet in a point. 71. Find a point within a triangle at which the sides subtend equal angles. 72. Describe an equilateral triangle so that its sides may pass through three given points. 73. Describe a triangle equal in all respects to a given triangle, and having its sides passing through three given points. 16—2 232 Euclid's elements. Simmon's Link. 74. If from any point on the circumference of the circle circum-scribed about a triangle, perpendiculars are drawn to t l i e three sides, the feet of these perpendiculars are eollinear. Let P be any point on the o00 of the circle circumscribed about tho A A B C ; and let PD, PE, PF be the perp" drawn from P to the three sides. It is required to prove that the points D, E, F are eollinear. Join FD and D E : then F D and D E shall be in the same st. lino. Join PB, PC. Because the Z " PDB, PFB are rt. angles, Hyp. : . the points P, D, B, F are eoncyiic: . ' . the Z PDF=the Z PBF, in the same segment. m . 21. But since B A C P is a quad1 inscribed in a circle, having one of i t s sides A B produced to F, . -. the ext. Z PBF = the opp. int. L ACP. Ex. 3, p. 188. . -. the z P D F = the z ACP. To each add the z P D E : then the Z'PDF, PDE=the Z8ECP, PDE. But since t l i e Z • PDC, P E C are rt. angles, . ' . the points P, D, E, C are eoncyiic ; . ' . the Z " ECP, P D E together=two rt. angles: . ' . the Z'PDF, P D E together = two rt. angles; / . FD and D E are in the same st. line; 1.14. that is, the points D, E, F are eollinear. q.e.p. [This theorem is attributed to Eobert Simson; and accordingly the straight line F D E is sometimes spoken of as the Simson's Line of tho triangle A B C for the point P: some writers also call i t the Pedal of the triangle A B C for the point P.] 75. ABC is a triangle inscribed in a circle; and from any point P on tho circumference PD, PFare drawn perpendicular to B C and AB: i f FD, or FD produced, cuts A C at E, shew that PE is perpendicular to AC. 76. Find the locus of a, point which moves so that if perpen-diculars are drawn from i t to the sides of a given triangle, their feet are eollinear. 77. ABC and A'B'C are two triangles having a common vertical angle, and tho circles circumscribed about them meet again at P: shew that the feet of perpendiculars drawn from P to the four lines AB, AC, BC, B'C are collinoar. THEOREMS AND EXAMPLES ON BOOK III. 2 3 3 78. A triangle is inscribed in a circle, and any point P on the cir-cumference is joined to the orthocentre of the triangle : shew that this joining line is bisected by the pedal of the point P. IV. ON T H E CIRCLE IN CONNECTION WITH RECTANGLES. See Propositions 35, 36, 37. 1. If from any external point P two tangents are drawn to a given circle whose centre is O, and if O P meets the chord of contact at Q; then the rectangle OP, O Q is equal to the square on the radius. Let PH, PK be tangents, drawn from the external point P to the © HAK, whose , -• --, centre is O; and let O P meet HK the /' \ chord of contact at Q, and the O00 at A: then shall the rect. OP, OQ=the sq. on OA. On H P as diameter describe a circle : this circle must pass through Q, since the Z H Q P is a rt. angle. in. 31. Join OH. Then since PH is a tangent to the 0 HAK, . . the z O H P is a rt. angle. And since HP is a diameter of the © HQP, . • . O H touches the © H Q P at H. . -. the reet. OP, OQ=the sq. on OH, = the sq. on OA. in. 16. m. 36. Q. E. D. 2. A B C is a triangle, and AD, BE, C F the perpendiculars drawn from the vertices to the opposite sides, meeting in the orthocentre O : shew that the reet. AO, O D = t h e rect. BO, OE=the rect. C O , OF. 3. ABC is a triangle, and AD, BE the perpendiculars drawn from A and B on the opposite sides: shew that the rectangle CA, C E is equal to the rectangle CB, CD. 4. ABC is a triangle right-angled at C, and from D, any point in the hypotenuse AB, a straight line D E is drawn perpendicular to A B and meeting B C at E: shew that the square on D E is equal to the difference of the rectangles AD, D B and CE, EB. 5. From an external point P two tangents are drawn to a given circle whose centre is O, and O P meets the ohord of contact at Q: shew that any circle which passes through the points P, Q will cut the given circle orthogonally. [See Def. p. 222.] 234 EUCLID'S ELEMENTS. 6. A series of circles pass through two given points, and from a fixed point in the common chord produced tangents are drawn to all the circles: shew that the points of contact l i e on u . circle which cuts all the given circles orthogonally. 7. All circles which pass through a fixed point, and cut a given circle ortlwgonally, pass also through a second fixed point. 8. Find the locus of the centres of all circles which pass through a given point and cut a given circle orthogonally. 9. Describe a circle to pass through two given points and cut a given circle orthogonally. 10. A, B, C, D are four points taken in order on a given straight line: find a point O between B and C such that the rectangle OA, O B may be equal to the rectangle OC, OD. 11. AB is a fixed diameter of a circle, and CD a fixed straight line of indefinite length cutting A B or AB produced at right angles; any straight line is drawn through A to cut C D at P and the circle at Q: shew that the rectangle AP, A Q is constant. 12. AB is a fixed diameter of a circle, and CD a fixed chord at right angles to AB ; any straight line is drawn through A to cut C D at P and the circle at Q: shew that the rectangle AP, A Q i s equal to the square on AC. 13. A is a fixed point and CD a fixed straight line of indefinite length; AP is any straight line drawn through A to meet C D at P; and in AP a point Q is taken such that the rectangle AP, A Q i s constant: find the locus of Q. 14. Two circles intersect orthogonally, and tangents are drawn from any point on the circumference of one to touch the other: prove that the first circle passes through the middle point of the chord of contact of the tangents. [Ex. 1, p. 233.] 15. A semicircle is described on AB as diameter, and any two chords AC, BD are drawn intersecting at P: shew that AB= = A C . A P + B D . B P . 16. Two circles intersect at B and C, and the two direct common tangents A E and DF are drawn: i f the common chord is produced t o meet the tangents at G and H, shew that GH'J= AE- + BC-. 17. If from a point P, without a cirole, PM is drawn perpendicular to a diameter AB, and also a secant PCD, shew t l i a t PM»=PC.PD.i-.AM . M B , according as P M intersects the circle or not. THEOREMS AND EXAMPLES ON BOOK III. 235 18. Three circles intersect at D, and their other points of intersection are A, B, C; A D cuts the circle B D C at E, and EB, EC cut the circles ADB, A D C respectively at F and G : shew that the points F, A, G are eollinear. 19. A semicircle is described on - given diameter BC, and from B and C any two chords BE, CF are drawn intersecting within the semicircle at O; BF and C E are produced to meet at A: shew that the sum of the squares on AB, A C is equal to twice the square on the tangent from A together with the square on BC. 20. X and Y are two fixed points in the diameter of a circle equidistant from the centre C : through X any chord P X Q is drawn, and its extremities are joined to Y; shew that the sum of the squares on the sides of the triangle P Y Q is constant. [See p. 147, Ex. 24.] Pboblems on Tangency. 21. To describe a circle to pass through two given points and to touch a given straight line. Let A and B be the given points, and C D the given st. line: i t is required to describe a circle to pass through A and B and to touch CD. Join BA, and produce i t to meet C D at P. Describe a square equal to the rect. PA, PB; n. 14. and from PD (or PC) cut o f f P Q equal to a side of this square. Through A, B and Q describe a circle. Ex. 4, p. 156. Then since the rect. PA, PB = the sq. on PQ, . \ the © A B Q touches C D at Q. in. 37. Q. a. Y. Note. (1) Since PQ may be taken on either side of P, it is clear that there are in general two solutions of the problem. (ii) When AB is parallel to the given line CD, the above method i s not applicable. In this ease a simple construction follows from m. 1, Cor. and in. 16; and i t will be found that only one solution exists. 2 3 6 EUCLID H ELEM ENTS. 22. V ' o describe a circle to pass through tin given, points and to touch a given circle. Let A and B be t l i e given points, and C R P the given circle: i t is required to describe a circle to pass through A and B, and to touch the 0 C R P . Through A and B de-scribe any circle to cut the given circle at P and Q. Join AB, PQ, and pro-duce them to meet at D. P . -' ~ --~~ B D From D draw D C to touch the given circle, and let C be the point of contact. Then the circle described through A, B, C will touch the given circle. For, from the 0 A B Q P , the rect. DA, DB=the rect. DP, D Q : and from the 0 P Q C , the rect. DP, D Q = t h e sq. on D C ; in. 36. . -. the rect. DA, D B = the sq. on O C : . -. D C touches the © A B C at C. m . 37. But D C touches the © P Q C at C; Con-tr. . . the © A B C touches the given circle, and it passes through the given points A and B. q.e.f. Note, (i) Since two tangents may be drawn from D to the given circle, it follows that there will be two solutions of the problem. ( i i ) The general construction fails when the straight line bisect-ing A B at right angles passes through the centre of the given circle: the problem then becomes symmetrical, and tlie scAttion is obvious. 23. To describe a circle to pass through a given point and to touch two given straight lines. Let P be the given point, and AB, A C the given straight lines: i t is required to describe a circle to pass through P and to touch AB, AC. Now tho centre of every circle which touches A B and A C must l i e on tho bisector of the Z BAC. Ex. 7, p. 183. Hence draw A E bisecting the L BAC. From P draw PK perp. to AE. and produce i t to P , making KP' equal to PK. THEOREMS AND EXAMPLES ON BOOK III. 237 Then every circle which has its centre in AE, and passes through P, must also pass through P'. Ex. 1, p. 215. Hence the problem is now reduced to drawing a circle through P and P' to touch either A C or AB. Ex. 21, p. 235. Produce P'P to meet A C at S. Describe a square equal to the rect. SP, SP'; n. 4. and cut off S R equal to a side of the square. Describe a circle through the points P', P, R; then since the rect. SP, SP' = the sq. on SR, Constr. . ' . the circle touches A C at R ; in. 37. and since its centre is in AE, the bisector of the Z BAC, i t may be shewn also to touch AB. q. e. y. Note, ( i ) Since S R may be taken on either side of S, i t follows that there will be two solutions of the problem. ( i i ) If the given straight lines are parallel, the centre lies on the parallel straight line mid-way between them, and the construction proceeds as before. 24. To describe u circle to touch two given straight lines and a given circle. Let AB, A C be the two given st. lines, and D the centre of the given circle: i t is required to describe a circle to touch AB, A C and the circle whose centre is D. Draw EF, G H par1 to A B and A C respectively, on the sides remote from D, and at distances from them equal to the radius of the given circle. Describe the © M N D to touch EF and G H at M and N, and to pass through D. Ex. 23, p. 236. Let O be the centre of this circle. Join O M , O N , O D meeting AB, A C and the given circle at P, Q and R. Then a circle described from centre O with radius O P will touch AB, A C and the given circle. For since O is the centre of the © M N D , . -. OM = ON = OD. But PM = QN = RD ; C o n s t r . . . OP=OQ=OR. . -. a circle described from centre O, with radius OP, will pass througli Q and R. And since the z B at M and N are rt. angles, in. 18. . • . the z • at P and Q are rt. angles ; t. 29. . -. the © P Q R touches A B and AC. 238 Euclid's elements. And since R, the point in which the circles meet, is on the line of centres OD, , -. t l i e t -j PQR touches the given circle. Q. e. t: Note. There will be two solutions of this problem, since two circles may be drawn to touch EF, G H and to pass through D. 25. To describe a circle to pass through a given point and touch a given straight line and a given circle. Let P be the given point, AB the given st. line, and D H E the given circle, of which C is the centre: i t is required to describe a circle to pass through P, and to touch AB and the © D H E . Through C draw D C E F perp. to AB, cutting the circle at the points D and E, of which E is between C and AB. Join DP; and by describing a circle through F, E, and P, find a point K in D P (or DP produced) such that the rect. DE, DF = therect. DK, DP. Describe a circle to pass through P, K and touch AB : Ex. 21, p. 2 This circle shall also touch the given © DHE. For let G be the point at which this circle touches AB. Join DG, cutting the given circle D H E at H. Join HE. Then the z D H E is a rt. angle, being in a semicircle. m . 31. also the angle at F is a rt. angle; Constr. . -. the points E, F, G, H are concyclic: . -. the reet. DE, DF = the rect. DH, D G : m . 36. but the rect. DE, DF = the rect. DK, DP : Constr. . : the rect. DH, D G =the rect. DK, D P : . -. the point H is on the © PKG. Let O be the centre of the © PHG. Join OG, OH, CH. Then O G and DF are par1, sinoe they are both perp. to A B : and D G meets them. . -. the Z O G D = the Z G DC. i . 21). But since O G = OH, and C D = CH, . thezOGH = thezOHG ; and the z C D H = the L C H D : . -. thezOHG = the/CHD; . -. O H and C H aro in ono st. line. . . the © P H G touches tho given . - DHE. g. e. , . THEOREMS A N D EXAMPLES ON BOOK III. 239 Note. (I) Since two circles may be drawn to pass through P, K and to touch AB, i t follows that there will be two solutions of the present problem. ( i i ) Two more solutions may be obtained by joining PE, and proceeding as before. The student should examine the nature of the contact between the oircles in each case. 26. Describe a circle to pass through a given point, to touch a given straight line, and to have its centre on another given straight line. 27. Describe a circle to pass through a given point, to touch a given circle, and to have its centre on a given straight line. 28. Describe a circle to pass through two given points, and to intercept an arc of given length on a given circle. 29. Describe a circle to touch a given circle and a given straight line at a given point. 30. Describe a circle to touch two given circles and a given straight line. V. ON MAXIMA AND MINIMA. We gather from the Theory of Loci that the position of an angle, line or figure is capable under suitable conditions of gradual change ; and it is usually found that change .of position involves a corresponding and gradual change of magnitude. Under these circumstances we may be required to note if any situations exist at which the magnitude in question, after increasing, begins to decrease; or after decreasing, to increase: in such situations the Magnitude is said to have reached a M a x i m u m or a M i n i m u m value; for in the former case it is greater, and in the latter case less than in adjacent situations on either side. In the geometry of the circle and straight line we only meet with such cases of continuous change as admit of one transition from an increasing to a decreasing state—or vice versa—so that in all the problems with which we have to deal (where a single circle is involved) there can be only one M a x i m u m and one M i n i m u m — t h e M a x i m u m being the greatest, and the Minimum being the least value that the variable magnitude is callable of taking. 240 EUCLID'S ELEMENTS. Thus a variable geometrical magnitude reaches its maximum or minimum value at a turning point, towards which the magni-tude may mount or descend from either side: it is natural there-fore to expect a m a x i m u m or minimum value to occur when, in the course of its change, the magnitude assumes a symmetrical form or position; and this is usually found to ? j e the case. This general connection between a symmetrical form or posi-tion and a maximum or minimum value is not exact enough to constitute a proof in any particular problem; but by means of i t a situation is suggested, which on further examination may be shewn to give the maximum or minimum value sought for. For example, suppose it is required to determine t l i e greatest straight line that may be drawn perpen-dicular to t l i e chord of a segment of a circle and intercepted between t l i e chord and the arc: we immediately anticipate that the greatest perpendicular is that which occupies a symmetrical position in the figure, namely the perpendicular which passes through the middle point of the chord; and on further examination this m a y be proved to be the case by means of I. 19, and I. 34. Again we are able to find at what point a geometrical magni-tude, varying under certain conditions, assumes its Maximum or Minimum value, if we can discover a construction for drawing the magnitude so that it m a y have an assigned value: for we m a y then examine between what limits the assigned value must lie in order that the construction may be possible; and the higher or lower limit will give the M a x i m u m or Minimum sought for. It was pointed out in the chapter on the Intersection of Loci, [see page 119] that if under certain conditions existing among the data, two solutions of a problem are possible, and under other conditions, no solution exists, there will always be some inter-mediate condition under which one and only one distinct solution is possible. Under these circumstances this single or limiting solution will always be found to correspond to the m a x i m u m or minimum value of the magnitude to be constructed. 1. For example, suppose i t is required t o divide a given straight line so that the rectangle contained by the two segments may be a -ma.rimuin. W e may first attempt to divide t l i e given straight line so that the rectangle contained by i t s segmonts may have a given area—that i s . be equal to the square on a given straight line. THEOREMS AND EXAMPLES ON BOOK III. 2 4 1 Let A B be the given straight line, and K the side of the given square: M B i t is required to divide the st. line A B at a point M, so that the rect. A M , M B may be equal to the sq. on K. Adopting a construction suggested by n. 14, describe a semicircle on A B ; and at any point X in AB, or AB produced, draw X Y perp. to AB, and equal to K. Through Y draw Y Z par1 to AB, to meet the arc of the semicircle at P. Then if the perp. P M is drawn to AB, i t may be shewn after the manner of n. 14, or by in. 35 that the rect. A M , M B = the sq. on PM. = the sq. on K. So that the rectangle A M , M B increases as K increases. Now if K is less than the radius C D , then Y Z will meet the arc of the semicircle in two points P, P'; and i t follows that A B may be divided at two points, so that the rectangle contained by its segments may be equal to the square on K. If K increases, the st. line Y Z will recede from AB, and the points of intersection P, P' will con-tinually approach'one another; until, when K is equal to the radius CD, the st. line Y Z (now in the position Y'Z') will meet the aro in two coincident points, that is, will touch the semicircle at D; and there will be only one solution of the problem. If K is greater than C D , the straight line Y Z will not meet the semicircle, and the problem is impossible. Hence the greatest length that K may have, in order that the con-struction may be possible, is the radius C D . . -. the rect. A M , M B is a maximum, when i t is equal to the square on C D ; that is, when P M coincides with DC, and consequently when M is the middle point of AB. 06s. The special feature to he noticed in this problem iB that the maximum is found at the transitional point between two solutions and no solution; that is, when the two solutions coincide and become identical. 242 EUCLID'S ELEMENTS. The following example illustrates the same point. 2. To find at what paint in a given straight line tlie angle subtended by the line joining two given points, which are on the same side of the given straight line, is a maximum. Let C D be the given st. line, and A, B the given points on the same side of C D : i t is required to find at what point in C D the angle subtended by the Bt. line A B iB a maximum. First determine at what point in C D , the st. line A B subtends a given angle. This is done as follows:— On A B describe a segment of a circle containing an angle equal to the given angle. m . 33. If the arc of this segment intersects C D , two points in C D are found at which A B subtends the given angle: but if the arc does not meet C D , no solution is given. In accordance with the principles explained above, we expect that a maximum angle is determined at the limiting position, that is, when the arc touches C D ; or meets it at two coincident points. [See page 213.] This we may prove to be the case. Describe a circle to pass through A and B, and to touch the st. line CD. [Ex. 21, p. 235.] Let P be the point of contact. Then shall the Z A P B be greater than any other angle subtended by A B at a point / -—-;;< g in C D on the same side of A B as P. / For take Q, any other point in CD, on 1 / , . -the same side of A B as P ; \ . -. ' ^ and join AQ, QB. £ Since Q is a point in the tangent other than the point of contact, i t must be with-out the circle, . ' . either B Q or A Q must meet the arc of the segment APB. Let B Q meet the arc at K : join AK. Then the Z A P B = the z AKB, in the same segment: but tho ext. z A K B is greater than the int. opp. z A Q B . . -. the z A P B is greater than A Q B . Similarly the z A P B may be shewn to be greater than any other anglo subtended by A B at a point in C D on the same side of A B : that is, tho L A P B is tho greatest of all such angles. q.z.d. Notk. Two circles may be described to pass through A and B, and to touch C D , the points of oontact being on opposite sides of AB; THEOREMS AND EXAMPLES ON BOOK III. 2 4 3 hence two points in C D may be found such that the angle subtended by A B at each of them is greater than the angle subtended at any other point in C D on the same side of AB. We add two more examples of considerable importance. 3. In a straight line of indefinite length find a point such that the sum of its distances from two given points, on t l i e same side of the given line, shall be a minimum. Let CD be the given st. line of i n d e f i n i t e length, and A, B the given points on the same side of C D : i t is required to find a point P in CD such that the sum of AP, PB i s a minimum. Draw AF perp. to CD; and produce AF to E, making FE equal to AF. Join EB, cutting C D at P. Join AP, PB. Then of a l l lines drawn from A [ 5 and B to a point in CD, the sum of AP, PB shall be the l e a s t . For, l e t Q be any other point in CD. Join AQ, BQ, EQ. Now in the abAFP, EFP, ( AF = EF, Because -jand FP is common; (and the zAFP=the z EFP, . -. AP=EP. Similarly it may be shewn that A Q = E Q . Now in the A EQB, the two sides EQ, QB are together greater than EB; hence, A Q , Q B are together greater than EB, that is, greater than AP, PB. Similarly the sum of the st. lines drawn from A and B to any other point in C D may be shewn to be greater than AP, PB. : . the sum of AP, PB is a minimum. Constr. being rt. angles. I. 4. Note. It follows from the above proof that the L A P F = the z EPF =the Z BPD. i . 4. I . 15. Thus the sum of AP, PB is a minimum, when these lines are equally inclined to C D . 244 EUCLIj/s ELEMENTS. 4. Given two intersecting straight lines AB, AC, and a point P between them; shew t l u i t of a i l , straight Lines which pass through P ami are terminated hy AB, AC, t l i a t which i s bisected at P cuts off the triangle of minimum area. Let EF be the st. line, terminated by AB, AC, which is bisected at P: then the A FAE shall be of mini-m u m area. For let H K be any other st. line passing through P: through E draw E M par1 to AC. Then in the A'HPF, MPE, ( the Z HPF = the L MPE. x. 15. Because \and the Z H F P = the L M E P , i. 29. ( andFP = EP; Hyp. . . the A HPF = the A M P E . x . 26, Cor. But the A MPE is less than the A KPE; . -. the A HPF is less than the A KPE: to each add the fig. A H P E ; then the A FAE is less than the A HAK. Similarly it may be shewn that the A FAE is less than any other triangle formed by drawing a st. line through P: that is, the A FAE is a minimum. Examples. 1. Two sides of a triangle aie given in length; how must they be placed in order that the area of the triangle may be a maximum? 2 Of all triangles of given base and area, the isosceles is that which has the least perimeter. 3. Given the base and vertical angle of a triangle; construct it so that its area may be a maximum. 4. Find a point in a given straight line such that the tangents drawn from i t to a given oirolo contain the greatest angle possible. 5. A straight rod slips between two straight rulers placed at right angles to one another; in what position i s t l i e triangle intercepted bctwcon tho rulers and rod a maximum ? THEOREMS AND EXAMPLES ON BOOK III. 245 6. Divide a given straight line into two parts, so that the sum of the squares on the segments may (i) be equal to a given square, ( i i ) may be a minimum. 7. Through a point of intersection of two circles draw a straight line terminated by the circumferences, (i) so that it may be of given length, ( i i ) so that it m a y be a maximum. 8. Two tangents to a circle out one another at right angles; find the point on the intercepted arc such that the sum of the perpendiculars drawn from it to the tangents may be a minimum. 9. Straight lines are drawn from two given points to meet one another on the circumference of a given circle: prove that their sum is a minimum when they make equal angles with the tangent at the point of intersection. 10. Of all triangles of given vertical angle and altitude, the isosceles is that which has the least area. 11. Two straight lines CA, C B of indefinite length are drawn from the centre of a circle to meet the circumference at A and B ; then of all tangents that may be drawn to the circle at points on the arc AB, that whose intercept is bisected at the point of contact cuts off the triangle of minimum area. 12. Given two intersecting tangents to a circle, draw a tangent to the convex are so that the triangle formed by it and the given tan-gents may be of maximum area. 13. Of all triangles of given base and area, the isosceles is that which has the greatest vertical angle. 14. Find a point on the circumference of a circle at which the straight line joining two given points (of which both are within, or both without the circle) subtends the greatest angle. 15. A bridge consists of three arches, whose spans are 49 ft., 32 ft. and 49 ft. respectively: shew that the point on either bank of the river at which the middle arch subtends the greatest angle is 63 feet distant from the bridge. 16. From a given point P without a circle whose centre is C, draw a straight line to cut the circumference at A and B, so that the triangle A C B may be of maximum area. 17. Shew that the greatest rectangle which can be inscribed in a circle is a square. 18. A and B are two fixed points without a circle: find a point P on the circumference such that the sum of the squares on AP, PB may be a minimum. [See p. 147, Ex. 24.] H. E. 17 246 EUCLID'S ELEMENTS. 19. A segment of a circle is described on the chord AB: find a point P on its arc so that the sum of AP, B P may be a maximum. 20. Of all triangles that can be inscribed in a circle that which has the greatest perimeter is equilateral. 21. Of all triangles tliat can be inscribed in u, given circle that which has the greatest area is equilateral. 22. Of all triangles tliat can be inscribed in a given triangle that which has the least perimeter is the triangle formed by joining the feet of the perpendiculars drawn from the vertices on opposite sides. 23. Of all rectangles of given area, the square has the least peri-meter. 24. Describe the triangle of maximum area, having its angles equal to those of a given triangle, and its sides passing through three given points. VI. HARDER MISCELLANEOUS EXAMPLES. 1. AB is a diameter of a given circle; and AC, BD, two chords on the same side of AB, intersect at E : shew that the circle which passes through D, E, C cuts the given circle orthogonally. 2. Two circles whose centres are C and D intersect at A and B, and a straight line P A Q is drawn through A and terminated l > y the circumferences: prove that ( i ) the angle PBQ=the angle C A D ( i i ) the angle BPC=the angle B Q D . 3. Two chords AB, CD of a circle whose centre is O intersect at right angles at P : shew that (i) PAa+ PB8+ PC2 + PD2=4 (radius)3. ( i i ) AB2 + CD2+40P2 = 8 (radius)2. 4. Two parallel tangents to a circle intercept on any third tangent a portion which is so divided at its point of contact that the rectangle contained by its two parts is equal to the square on the radius. 5. Two equal circles move between two straight lines placed at right angles, so that each straight line is touched by one circle, and the two circles touch one another: find the locus of the point of contact. 6. A B is a given diameter of a circle, and C D is any parallel ohord: if any point X in A B is joined to tlie extremities of CD, Bhew that XC2 + X D a = X A a + XB-. THEOREMS AND EXAMPLES ON BOOK III. 247 7. PQ is a fixed chord in a circle, and PX, QY any two parallel chords through A and B: shew that X Y touches a fixed concentric circle. 8. Two equal circles intersect at A and B; and from C any point on the circumference of one of them a perpendicular is drawn to AB, meeting the other circle at O and O': shew that either O or O' is the orthocentre of the triangle A B C . Distinguish between the two cases. 9. Three equal circles pass through the same point A, and their other points of intersection are B, C, D : shew that of the four points A, B, C, D, each is the orthocentre of the triangle formed by joining the other three. 10. From a. given point without a circle draw a straight line to the concave circumference so as to be bisected by the convex circumference. W h e n is this problem impossible ? 11. Draw i straight hue cutting two concentric circles so that the chord intercepted by the circumference of the greater circle may be double of the chord intercepted by the less. 12. ABC is a triangle inscribed in a circle, and A', B', C are the middle points of the arcs subtended by the sides (remote from the opposite vertices): find the relation between the angles of the two triangles A B C , A'B'C ; and prove that the pedal triangle of A'B'C is equiangular to the triangle A B C . 13. The opposite sides of a quadrilateral inscribed in a circle are produced to meet: shew that the bisectors of the two angles so formed are perpendicular to one another. 14. If a quadrilateral can have one circle inscribed in it, and another circumscribed about it; shew that the straight lines joining the opposite points of contact of the inscribed circle are perpendicular to one another. 15. Given the base of a triangle and the sum of the remaining sides; find the loeus of the foot of the perpendicular from one extremity of the base on the bisector of the exterior vertical angle. 16. Two circles touch each other at C, and straight lines are drawn through C at right angles to one another, meeting the circles at P, P' and Q, Q' respectively: if the straight line which joins the centres is terminated by the circumferences at A and A', shew that P'P2 + Q'Q2=A'A2. 17. Two circles cut one another orthogonally at A and B; P is any point on the arc of one circle intercepted by the other, and PA, P B are produced to meet the circumference of the second circle at C and D : shew that C D is a diameter. 17—2 248 Euclid's elements. 18. ABC is a triangle, and from any point P perpendiculars PD, PE, PF are drawn to the sides: if Sv S2, S3 are the centres of the circles circumscribed about the triangles DPE, EPF, FPD, shew that the triangle SjS.^Sj is equiangular to the triangle ABC, and that the sides of the one are respectively half of the sides of the other. 19. Two tangents PA, PB are drawn from an external point P to a given circle, and C is the middle point of the chord of contact AB: if X Y is any chord through P, shew that A B bisects the angle XCY. 20. Given the sum of two straight lines and the rectangle con-tained by them (equal to a given square): find the lines. 21. Given the sum of the squares on two straight lines and the rectangle contained by them: find the lines. 22. Given the sum of two straight lines and the sum of the squares on them: find the lines. 23. Given the difference between two straight lines, and the rect-angle contained by them: find the lines. 24. Given the difference between two straight lines and the differ-ence of their squares: find the lines. 25. ABC is a triangle, and the internal and external bisectors of the angle A meet BC, and B C produced, at P and P": if O is the middle point of PP', shew that O A is a tangent to the circle circum-scribed about the triangle A B C . 26. ABC ib a triangle, and from P, any point on the circum-ference of the circle circumscribed about it, perpendiculars are drawn to the sides BC, CA, A B meeting the circle again in A', B'. C ; prove that ( i ) the triangle A'B'C is identically equal to the triangle ABC. ( i i ) AA', B B \ C C are parallel. 27. Two equal circles intersect at fixed points A and B, and from any point in A B a perpendicular4s drawn to meet the circumferences on the same side of A B at P and Q : shew that P Q is of constant length. 28. The straight lines which join the vertices of a triangle to the centre of its circumscribed circle, are perpendicular respectively to the sides of the pedal triangle. 29. P is any point on the oireumferencc of a circle circumscribed about a triangle A B C ; and perpendiculars PD, P E are drawn from P to the sides BC, CA. Find tho loous of the centre of the circle circum-Boribcd about the trianglo PDE, THEOREMS A N D EXAMPLES ON BOOK III. 249 30. P is any point on the circumference of a circle circumscribed about a triangle A B C ; shew that the angle between Simson's Line for the point P and the side BC, is equal to the angle between A P and the diameter of the circumscribed circle. 31. Shew that the orthocentres of the four triangles formed by two pairs of intersecting straight lines are eollinear. 32. Shew that the circles circumscribed about the four triangles formed by two pairs of intersecting straight lines meet in a point. On the Construction of Triangles. 33. Given the vertical angle, one of the sides containing it, and the length of the perpendicular from the vertex on the base: construct the triangle. 34. Given the feet of the perpendiculars drawn from the vertices on the opposite sides : construct the triangle. 35. Given the base, the altitude, and the radius of the circum-scribed circle: construct the triangle. 36. Given the base, the vertical angle, and the sum of the squares on the sides containing the vertical angle: construct the triangle. 37. Given the base, the altitude and the sum of the squares on the sides containing the vertical angle: construct the triangle. 38. Given the base, the vertical angle, and the difference of the squares on the sides containing the vertical angle: construct the tri-angle. 39. Given the vertical angle, and the lengths of the two medians drawn from the extremities of the base: construct the triangle. 40. Given the base, the vertical angle, and the difference of the angles at the base: construct the triangle. 41. Given the base, and the position of the bisector of the vertical angle: construct the triangle. 42. Given the base, the vertical angle, and the length of the bisector of the vertical angle: construct the triangle. 43. Given the perpendicular from the vertex on the base, the bisector of the vertical angle, and the median which bisects the base: construct the triangle. 44. Given the bisector of the vertical angle, the median bisect-ing the base, and the difference of the angles at the base: construct the triangle. BOOK IV. Book IV. consists entirely of problems, dealing with various rectilineal figures in relation to the circles which pass through their angular points, or are touched by their sides. Definitions. 1. A Polygon is a rectilineal figure bounded by more than four sides. A Polygon of five sides is called a Pentagon, six sides „ Hexagon, seven sides , . Heptagon, eight sides . , Octagon, ten sides Decagon, twelve sides Dodecagon, fifteen sides „ Quindecagon. 2. A Polygon is Regular when all its sides are equal, and all its angles are equal. 3. A rectilineal figure is said to be inscribed in a circle, when all its angular points are on the circumference of the circle: and a circle is said to be circumscribed about a rectilineal figure, when the circum-ference of the circle passes through all the angular points of the figure. 4. A rectilineal figure is said to be circumscribed about a circle, when each side of the figure is a tangent to tlie circle : and a circle is said to be inscribed in a reet i -lineal figure, when the circumference of the circle is touched by each side of the figure. 5. A straight line is said to be placed in a circle its extremities are on the circumference of the circle. when BOOK IV. PROP, 1 . 2 5 1 Proposition 1. Problem. In a given circle to place a chord equal to a given straight line, which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight Line not greater than the diameter of the circle : i t is required to place in the 0 A B C a chord equal to D. Draw CB, a diameter of the ©ABC. Then i f CB = D, the thing required is done. But i f not, CB must be greater than D. Prom CB cut off CE equal to D : and from centre C, with radius CE, describe the cutting the given circle at A. Join CA. Then CA shall be the chord required. For CA = CE, being radii of the O AEF : and CE = D : / . CA = D. Hyp. I . 3. 0AEF, Constr. Q.E.F. EXERCISES. 1. In a given circle place a chord of given length so. as to pass through a given point (i) without, (ii) within the circle. W h e n is this problem impossible ? 2. In a given circle place a chord of given length so that it may be parallel to a given straight line. 252 EUCLID B ELEMENTS. Proposition 2. Problem. In a given circle to inscribe a triangle equiangular to a given triangle. A E F Let ABC be the given circle, and DEF the given triangle: i t is required to inscribe in the 0 ABC a triangle equiangular to the A DEF. At any point A, on the Oce of the 0ABC, draw the tangent GAH. m . 17. A t A make the a GAB equal to the _ DFE ; I . 23. and make the a HAC equal to the a DEF. I . 23. Join BC. Then ABC shall be the triangle required. Because G H is a tangent to the 0 A B C , and from A i t s point of contact the chord AB is drawn, . ' . the l GAB = the a ACB in the alt. segment: hi. 32. . ' . the A ACB = the /.DFE. Comtr. Similarly the a HAC = the a ABC, in the alt. segment: . ' . the A ABC = the L DEF. Constr. Hence the third a BAC = the third _ EDF, for the three angles in each triangle are together equal to two rt. angles. i . 32. . ' . the A ABC is equiangular to the A DEF, and i t i s inscribed in the 0 A B C . Q. e. v. BOOK IV. PROP. 3. 253 Proposition 3. Problem. About a given c i r c l e t o circumscribe a triangle equi angular to a given triangle. G E F H M B Let ABC be the given circle, and DEF the given A : i t i s required to circumscribe about the © A B C a triangle equiangular to the A DEF. Produce EF both ways to G and H. Find K the centre of the 0ABC, m . 1 . and draw any radius KB. At K make the a BKA equal to the a DEG ; I . 23. and make the a BKC equal to the A DFH. Through A, B, C draw LM, MN, NL perp. to KA, KB, KC. Then LMN shall be the triangle required. Because LM, MN, NL are drawn perp. to radii at their extremities, . ' . LM, MN, NL are tangents to the circle, hi. 16. And because the four angles of the quadrilateral AKBM together = four rt. angles; I . 32. Cor. and of these, the l s KAM, KBM, are rt. angles; Consbr. . ' . the As AKB, AMB, together = two rt. angles. But the a" DEG, DEF together = two rt. angles; I . 13. . ' . the A s AKB, A M B = the A 8 DEG, DEF; and of these, the A AKB = the A DEG ; Constr. . ' . the A A M B = the A DEF. Similarly it may be shewn that the a LNM = the a DFE. . ' . the third a MLN =the third a EDF. i . 32. . ' . the A LMN is equiangular to the A DEF, and it i s circumscribed about the ©ABC. q.e.v. 2 5 4 E u c l i d ' s e l e m e n t s . P r o p o s i t i o n 4 . Problem. To inscribe a circle in a given triangle. A Let A B C be the given triangle: it is required to inscribe a circle in the A ABC. Bisect the A s ABC, A C B by the st. lines Bl, CI, which intersect at I . I . 9. From I draw IE, IF, IG perp. to AB, BC, CA. I . 12. Then in the A 8 EIB, FIB, I the A EBI = the A FBI; Conslr. Because -<and the a BEI = the A BFI, being rt. angles ; ( and Bl is c o m m o n ; . ' . IE = IF. i . 26. Similarly it m a y be shewn that IF = IG. . ' . IE, IF, IG are all equal. From centre I , with radius IE, describe a circle: this circle must pass through the points E, F, G ; and it will be inscribed in the A ABC. For since IE, IF, IG are radii of the © E F G ; and since the a s at E, F, G are rt. angles; . ' . the O EFG is touched at these points by AB, BC, CA: m. 1 6 . the © EFG is inscribed in the A ABC. Q. E. F. Note. From page 103 it, is seen that if Al bo joined, then Al bisects the angle BAC. BOOK IV. PROP. 4. 2 5 5 Hence i t follows t l i a t the bisectors of the angles of a triangle are concurrent, the point of intersection being the centre of t l i e inscribed circle. The centre of the circle inscribed in a triangle i s sometimes called its in-centre. Definition. A circle which touches one side of a triangle and the other two sides produced is said to be an escribed circle of the triangle. To draw an escribed circle of a given triangle. Let A B C be the given triangle, of which the two sides AB, A C are produced to E and F: i t is required to describe a circle touching BC, and AB, A C produced. Bisect the / » CBE, BCF by the st. lines Blj, Clj, which intersect at lx. i . 9. From l j draw IjG, \1H, lxK perp. to AE, BC, AF. i . 12. Then in the a" lxBG, IjBH, {the z l1BG = the z IjBH, Constr. and the Z IjGB^the L IjHB, being rt. angles; also IjB is common; . -. I1G = I1H. Similarly i t may be shewn that IjH = lxK; . -. IjG, IjH, lxK are all equal. From centre l x with radius lxG, describe a circle: this circle must pass through the points G, H, K : and i t will be an escribed circle of the A ABC. For since IjH, IjG, lxK are radii of the © HGK, and since the angles at H, G, K are rt. angles, . . the © G H K is touched at these points by BC, and by AB, AC produced: . -. the © G H K is an escribed circle of the A ABC. q.e.i\ It is clear that every triangle has three escribed circles. Note. From page 104 it is seen that if Al2 be joined, then Al, bisects the angle BAC: hence i t follows that The bisectors of two exterior angles of a triangle and the bisector of the third angle are concurrent, the point of intersection being the centre of an escribed circle. Because 25fi E u c l i d ' s elements. Proposition 5. Problem. To circumscribe a circle about a given triangle. A^ Let ABC be the given triangle : i t is required to circumscribe a circle about the A ABC. Draw DS bisecting AB at rt. angles : I . 11. and draw ES bisecting AC at rt. angles; then since AB, AC are neither par1, nor in the same st. line, . ' . DS and ES must meet at some point S. Join SA; and i f S be not in BC, join SB, S O Then in the A8 ADS, BDS, ( AD = BD Because <and DS is common to both ; (and the A ADS = the A BDS, being rt. angles; . ' . SA = SB. Similarly it may be shewn that SC = SA. . ' . SA, SB, SC are all equal. From centre S, with radius SA, describe this circle must pass through the points A therefore circumscribed about the A ABC. a circle : B, C, and i s Q.E.F. It follows that (i) when the centre of the circumscribed circle f a l l s within the triangle, each of its angles must be acute, for each angle is then in « segment greater than a semicircle: ( i i ) when t l i e centre falls on one of the sides of the triangle, the angle opposite to this side must be a right angle, for it is t l i e angle in a semicircle : BOOK IV. PROP. 5. 257 (iii) when the centre falls without the triangle, the angle opposite to the side beyond which the centre falls, must be obtuse, for it is the angle in a segment less than a semicircle. Therefore, conversely, if the given triangle be acute-angled, the centre of the circumscribed circle falls within it: if it be a right-angled triangle, the centre falls on the hypotenuse : if it be a n obtuse-angled triangle, the centre falls without the triangle. Note. From page 103 it is seen that if S be joined to the middle point of BC, then the joining line is perpendicular to BC. Hence the perpendiculars drawn to the sides of a triangle from their middle points are concurrent, t l i e point of intersection being the centre of the circle circumscribed about the triangle. The centre of the circle circumscribed about a triangle is some-times called its circum-centre. EXERCISES. On the Inscribed, Circumscribed, and Escribed Circles of a Triangle. 1. An equilateral triangle is inscribed in a circle, and tangents are drawn at its vertices, prove that ( i ) the resulting figure is an equilateral triangle: ( i i ) its area is four times that of the given triangle. 2. Describe a circle to touch two parallel straight lines and a third straight line which meets them. Shew that two such circles can be drawn, and that they are equal. 3. Triangles which have equal bases and equal vertical angles have equal circumscribed circles. 4. I is the centre of the circle inscribed in the triangle ABC, and is the centre of the circle w h shew that A, I , l x are eollinear. 5. If t l i e inscribed and circumscribed circles of a triangle are con-centric, shew that the triangle is equilateral; and that the diameter of the circumscribed circle is double that of the inscribed circle. 6. ABC is a triangle; and I, S are the centres of the inscribed and circumscribed circles; if A, I , S are eollinear, shew that A B = AC. 258 euclid'b elements. 7. The sum of the diameters of the inscribed and circumscribed circles of a right-angled triangle is equal to the sum of the sideB containing the right angle. 8. If the circle inscribed in a triangle A B C touches the sides at D, E, F, shew that the triangle D E F is acute-angled; and express its angles in terms of the angles at A, B, C. 9. If I is the centre of the circle inscribed in the triangle ABC, and l x the centre of the escribed circle which touches B C ; shew that I , B, lj, C are concyclic. 10. In any triangle the difference of two sides is equal to the dif-ference of the segments into which the third side is divided at the point of contact of the inscribed circle. 11. In the triangle ABC the bisector of the angle BAC meets the base at D, and from I the centre of the inscribed circle a perpendicular IE is drawn to B C : shew that the angle Bl D is equal to the angle CIE. 12. In the triangle A B C , I and S are the centres of the inscribed and circumscribed circles: shew that IS subtends at A an angle equal to half the difference of the angles at the base of the triangle. 13. In a triangle A B C , I and S are the centres of the inscribed and circumscribed circles, and A D is drawn perpendicular to BC: shew that AI is the bisector of the angle D A S . 14. Shew that the area of a triangle is equal to the rectangle contained by its semi-perimeter and the radius of the inscribed circle. 15. The diagonals of a quadrilateral A B C D intersect at O : shew that the centres of the circles circumscribed about the four triangles A O B , B O C , C O D , D O A are at the angular points of a parallelogram. IG. In any triangle A B C , if I is the centre of the inscribed circle, and if Al is produced to meet the circumscribed circle at O ; shew that 0 is the centre of the circle circumscribed about the triangle BIC. 17. Given the base, altitude, and the radins of the circumscribed circle; construct the triangle. 18. Describe a circle to intercept equal chords of given length on three given straight lines. 19. In an equilateral triangle the radii of the circumscribed and escribed circles are respectively double and treble of the radius of the inscribed circle. 20. Two circles whoso centres are A, B, C touch one another externally two by two at D, E, F: shew that the inscribed circle ef tho triangle A B O is the circumscribed circle of tho triangle DEF. book i v . p r o p . 6 . 2 5 9 Proposition 6. Problem. To inscribe a square in a given circle. A Let ABCD be the given circle : i t is required to inscribe a square in the 0 A B C D . Find E the centre of the circle : in. 1. and draw two diameters AC, BD perp. to one another, i . 11. Join AB, BC, CD, DA. Then the fig. A B C D shall be the square required. For in the As BE A, DEA, / BE = DE, Because 1 and EA is common; [and the a BEA = the a DEA, being rt. angles; /. BA = DA. I . 4. Similarly it may be shewn that C D = DA, and that BC = CD. . ' . the fig. ABCD is equilateral. A n d since BD is a diameter of the © A B C D , . ' . BAD is a semicircle; . " . the a BAD is a rt. angle. in. 31. Similarly the other angles of the fig. ABCD are rt. angles. . ' . the fig. ABCD is a square, and it is inscribed in the given circle. Q. E. P. [For Exercises see page 263.] 2fiO EUCLID'S ELEMENTS. P r o p o s i t i o n 7 . Problem. To c/trcuinscrihe a square about <(. given circle. G A F H K Let ABCD be t l i e given circle : i t is required to circumscribe a square about i t . Find E the centre of the © A B C D : in. 1 . and draw two diameters AC, BD perp. to one another. I . 11. Through A, B, C, D draw FG, GH, HK, KF perp. to EA, EB, EC, ED. Then the fig. GK shall be the square required. Because FG, GH, HK, KF are drawn perp. to radii at their extremities, . ' . FG, GH, HK, KF are tangents to the circle, in. 16. And because the a- AEB, EBG are both rt. angles. Constr. . ' . G H is par1 to AC. I . 28. Similarly FK is par1 to AC : and in like manner GF, BD, HK are par'. Hence the figs. GK, GC, AK, GD, BK, G E are par™5. . ' . GF and HK each = BD ; also G H and FK each = AC : but AC = BD ; . ' . GF, FK, KH, HG are all equal : that is, the fig. G K is equilateral. And since the fig. G E is a par™, . " . the l . BGA = the _ BEA ; I . 34. but the l . BEA is a rt. angle; Constr. . ' the l . at G i s a rt. angle. Similarly the s at F, K, H are rt. angles. . ' . tho f i g . GK is a square, and i t has been cireumseriVwd about t l i e uABCD. Q. E.F. BOOK IV. PROP. 8 . 2 6 1 Proposition 8. Problem. To inscribe a circle in a given square. AED K r . B H Let ABCD be the given square : i t i s required to inscribe a circle in the sq. ABCD. Bisect the sides AB, AD at F and E. I . 10. Through E draw EH par1 to AB or DC : I . 31. and through F draw FK par1 to AD or BC, meeting EH at G. N o w AB = AD, being the sides of a square; and their halves are equal; Constr. :. AF = AE. Ax. 7. But the fig. AG i s a par"; Constr. . ' . AF = GE, and AE = GF; . ' . GE = GF. Similarly i t may be shewn that GE = GK, and GK = GH : . ' . GF, GE, GK, GH are a l l equal. From centre G, with radius G E, describe a circle; this circle must pass through the points F, E, K, H : and i t will be touched by BA, AD, DC, CB; in. 16. for GF, GE, GK, GH are radii; and the angles at F, E, K, H are rt. angles. I . 29. Hence the © FEKH is inscribed in the sq. ABCD. Q. e. P. [For Exercises see p. 263.] H. E. IS •>(\'l E U C L I D ' S ELEMENTS. Proposition 9 . Problem. To circumscribe a circle about a given square. A, i . Def. 28. i. Def 38. I . 8. Let ABCD be the given square : i t is required to circumscribe a circle about the sq. ABCD. Join AC, BD, intersecting at E. Then in the A8 BAC, DAC, t BA = DA, Because -< and AC is common ; ( and BC = DC ; . ' . the A BAC = the A DAC : tliat is, the diagonal AC bisects the _ BAD. Similarly the remaining angles of t l i e square arc bisected by the diagonals AC or BD. Hence each of the a " EAD, EDA i s half a rt. angle; . ' . the a EAD = the _ EDA: . ' . EA = ED. I . 6 . Similarly it may be shewn that ED = EC, and LC = EB. . ' . EA, EB, EC, ED are all equal. From centre E, with radius EA, describe a circle : this circle must pass through the points A, B, C, D, and i s therefore circumscribed about the sq. ABCD. Q.E.F. BOOK IV. PROP. 9 . 263 Definition. A rectilineal figure about which a circle may be described is said to be Cyclic. exercises on propositions 6—9. 1. If a circle can be inscribed in a quadrilateral, shew that the sum of one pair of opposite sides is equal to the sum of the other pair. 2. If the sum of one pair of opposite sides of a quadrilateral is equal to the sum of the other pair, shew tliat a circle may be inscribed in the figure. [Bisect two adjacent angles of the figure, and so describe a circle to touch three of its sides. Then prove indirectly by means of the last exercise that this cirole must also touch the fourth side.] 3. Prove that a rhombus and a square are tlie only parallelograms in which a circle can be inscribed. 4. All cyclic parallelograms are rectangular. 5. The greatest rectangle which can be inscribed in a given circle is a square. 6. Circumscribe a rhombus about a given circle. 7. All squares circumscribed about a given circle are equal. 8. The area of a square circumscribed about a circle is double of the area of the inscribed square. 9. ABCD is a square inscribed in a circle, and P is any point on the arc A D : shew that the side A D subtends at P an angle three times as great as that subtended at P by any one of the other sides. 10. Inscribe a square in a given square ABCD so that one of its angular points should be at a given point X in AB. 11. In a given square inscribe the square of minimum area. 12. Describe (i) a circle, ( i i ) a square about a given rectangle. 13. Inscribe (i) a circle, (ii) a square in a given quadrant. 14. In a given circle inscribe a rectangle equal to a given recti-lineal figure. 15. ABCD is a square inscribed in a circle, and P is any point on the circumference; shew that the sum of the squares on PA, PB, PC, PD is double the square on the diameter. [See Ex. 24, p. 147.] 18—2 264 EUCLID'S ELEMENTS. Proposition 10. Problem. To describe an isosceles triangle having each of the angles at the base double of the third angle. Take any straight line AB. Divide AB at C, so that the rect. BA, BC -- the sq. on AC. II. 11. From centre A, with radius AB, describe the © B D E : and in it place the chord BD equal to AC. iv. 1. Join DA. Then ABD shall be the triangle required. Join C D ; and about the A ACD circumscribe a circle. iv. • % Then the rect. BA, BC = the sq. on AC Constr. = the sq. on BD. Constr. Hence BD is a tangent to the 0 A C D : in. 37. and from the point of contact D a chord D C is drawn ; . ' . the A BDC = the a CAD in the alt. segment, m . 32. , To each of these equals add the _ CDA : then the whole a BDA -= the sum of tlie _ 3 CAD, CDA. But t l i e ext. a BCD = the sum of the _ s CAD, CDA; I . 32. . " . the A BCD = the _ BDA. And since AB = AD, being radii of the © B D E . . ' . the A DBA - the _ BDA : I . , \ . ' . the A DBC = the _ DCB; BOOK IV. PROP. 10. 265 .". DC = DB ; I. 6. that is, DC = CA : Constr. . ' . the A CAD = the A CDA; I . 5. . " . the sum of the a s CAD, CDA = twice the angle at A. But the a ADB = the sum of the a 8 CAD, CDA; Proved. each of the a s ABD, A D B = twice the angle at A. Q. E. F. EXERCISES ON PROPOSITION 10. 1. In an isosceles triangle in which each of the angles at the base is double of the vertical angle, shew that the vertical angle is one-fifth of two right angles. 2. Divide a right angle into five equal parts. 3. Describe an isosceles triangle whose vertical angle shall be three times either angle at the base. Point out a triangle of this kind in the figure of Proposition 10. 4. In the figure of Proposition 10, if the two circles intersect at F shew that B D = DF. 5. In the figure of Proposition 10, shew that the circle ACD is equal to the circle circumscribed about the triangle ABD. 6. In the figure of Proposition 10, if the two circles intersect at F shew that ( i ) BD, D F are sides of a regular decagon inscribed in the circle EBD. ( i i ) AC, CD, D F are sides of a regular pentagon inscribed in the circle ACD. 7. In the figure of Proposition 10, shew that the centre of the circle circumscribed about the triangle D B C is the middle point of the are CD. 8. In the figure of Proposition 10, if I is the centre of the circle inscribed in the triangle ABD, and I ' , S' the centres of the inscribed and circumscribed circles of the triangle D B C , shew that S ' l = ST. 266 euclid's elements. Proposition 11. Problem. To inscribe a regular pento/jon in a given c i r c l e . A C Let ABC be a given circle : i t i s required to inscribe a regular pentagon in the © ABC. Describe an isosceles AFGH, having each of the angles at G and H double of the angle at F. iv. 10. In the © A B C inscribe the A ACD equiangular to the AFGH, IV. 2. so that each of the A8 ACD, ADC i s double of the _ CAD. Bisect the _ 3 ACD, ADC by CE and DB, which meet the Oco at E and B. i . 9 . Join AB, BC, AE, ED. Then ABODE shall be the required regular pentagon. Because each of the A s ACD, ADC = twice the _ CAD ; and because the A 8 ACD, ADC are bisected by CE, DB, . -. the five a- ADB, BDC, CAD, DCE, ECA are a l l equal. . ' . the five arcs AB, BC, CD, DE, EA are all equal. I I I . 26. . " . the five chords AB, BC, CD, DE, EA are a l l equal, in. 29. . ' . the pentagon ABODE i s equilateral. Again the arc AB = the arc DE ; Proved. to each of these equals add the are BCD; . ' . the whole arc ABCD =the whole arc BCDE : hence the angles at the Oce which stand upon these equal arcs are equal; in. 27. that is, the a AED - the a BAE. In like manner the remaining angles of the pentagon may be shewn to be equal; . ' . the pentagon i s equiangular. Hence the pentagon, being both equilateral and equi-angular, i s regular; and i t i s inscribed in the ©ABC. Q.K.f. book i v . p r o p . 1 2 . 2 6 7 Proposition 12. Problem. To circumscribe a regular pentagon about a given circle. Let ABCD be the given circle : i t i s required to circumscribe a regular pentagon about i t . Inscribe a regular pentagon in the ©ABCD, iv. 11. and let A, B, C, D, E be its angular points. At the points A, B, C, D, E draw GH, HK, KL, LM, MG, tangents to the circle. I l l , 17. Then shall GHKLM be the required regular pentagon. Find F the centre of the © A B C D ; in. 1. and join FB, FK, FC, FL, FD. Then in the two A8 BFK, CFK, ! B F = CF, being radii of the circle, and FK is common : and KB = KC, being tangents to the circle from the same point K. m . 17. Cor. . \ the A BFK = the A CFK, l . 8. also the A BKF = the A CKF. I . 8. Cor. Hence the a BFC = twice the a CFK, and the A BKC = twice the a CKF. Similarly i t may be shewn that the a CFD = twice the a CFL, and that the a CLD = twice the a CLF. But since the arc BC = the arc CD, iv. 11. . ' . the A BFC = the A CFD ; I I I . 27. and the halves of these angles are equal, that is, the a CFK = the a CFL. 2 6 8 EUCLID'S ELEMENTS. G Then in the A8 CFK, CFL, the a CFK = the A CFL, Proved. Because \ and the a FCK = the a FCL,beingrt. angles, in.18. and FC is common; . -. CK = CL, i . 26. and the A FKC =--the A FLC. Hence KL is double of KC; similarly HK is double of KB. And since KC = KB, m . 17. Cor. :. KL = HK. In the same way it may be shewn that every two con-secutive sides are equal; . ' . the pentagon G H K L M is equilateral. Again, it has been proved that the i _ FKC = the a FLC, and that the A 8 HKL, KLM are respectively double of these angles : / . the A HKL = the A KLM. In the same way it may be shewn that ever}- two con-secutive angles of the figure are equal; . ' . the pentagon G H K L M is equiangular. . ' . the pentagon is regular, and it is circumscribed about the © A B C D . Q.E.F. Corollary. Similarly it may be proved thai if tangents are drawn at the vertices of any regular polygon inscribed in a circle, they will form another regular polygon of the same species circumscribed about t l i e circle. [For Exercises see p. 276.] BOOK IV. PROP. 1 . 3 . 2 6 9 Proposition 13. Problem. To inscribe a circle in a given regular pentagon. i . 9 . i. 12. Hyp. Constr. i . 4. Let ABCDE be the given regular pentagon : i t is required to inscribe a circle within i t . Bisect two consecutive a8 BCD, CDE by CF and DF which intersect at F. Join FB; and draw FH, FK perp. to BC, CD. Then in the A8 BCF, DCF, I BC = DC, Because -I and CF i s common to both; [ and the A BCF = the A DCF ; . " . the A CBF -the A CDF. But the A CDF is half an angle of the regular pentagon : . ' . also the a CBF is half an angle of the regular pentagon : that is, FB bisects the a ABC. So i t may be shewn that i f FA, FE were joined, these lines would bisect the A" at A and E. Again, in the A8 FCH, FCK, the A FCH = the A FCK, Constr. and the a FHC = the A FKC being rt. angles : also FC is common; . -. FH = FK. L 26. Similarly i f FG, FM, FL be drawn perp. to BA, AE, ED, i t may be shewn that the five perpendiculars drawn from F to the sides of the pentagon are all equal. Because 2 7 0 EUCLID'S ELEMENTS. From centre F, with radius FH, describe a, circle: this circle must pass through the points H, K, L, M, G ; and it will be touched at these points by the sides of the pentagon, for the A s at H, K, L, M, G are rt. A ' . Constr. . ' . the © H K L M G is inscribed in the given pentagon. Q.E.F. COROLLARY. Tlie bisectors of tlie angles of a regular pentagon meet at a point. In the same way it may be shewn that the bisectors of the angles of any regular polygon meet at a point. [See Ex. 1, p. 274.] [For Exercises on Begular Polygons see p. 270.] MISCELLANEOUS EXERCISES. 1. Two tangentB AB, A C are drawn from an external point A to a given circle: describe a circle to touch AB, A C and the convex arc intercepted by them on the given cirole. 2. ABC is an isosceles triangle, and from the vertex A a straight line ia drawn to meet the base at D and the circumference of the c i r -cumscribed circle at E: shew that AB is a tangent to the circle circumscribed about the triangle BDE. 3. An equilateral triangle is inscribed in u given circle: shew that twice the square on one of its sides is equal to three times the area of the square inscribed in the same circle. 4. ABC is an isosceles triangle in whioh each of the angles at B and C is double of tho angle at A: shew that the square on AB i s equal to t l i e roctanglo AB, B C with t l i e square on BC. book i v . p r o p . 1 4 . 2 7 1 Proposition 14. Problem. To circumscribe a circle about a given regular pentagon. A Let ABODE be the given regular pentagon : i t is required to circumscribe a circle about i t . Bisect the a s BCD, CDE by CF, DF intersecting at F. i. 9. Join FB, FA, FE. Then in the A8 BCF, DCF, I BC = DC, Hyp. Because J and CF is common to both; ( and the a BCF = the a DCF; Constr. . ' . the A CBF = the A CDF. I . 4. But the a CDF is half an angle of the regular pentagon . . ' . also the a CBF is half an angle of the regular pentagon : that is, FB "bisects the a ABC. So it may be shewn that FA, FE bisect the a " at A and E. N o w the ^s FCD, FDC are each half an angle of the given regular pentagon; . " . the a FCD = the a FDC, iv. Def. . ' . F C = FD. I . 6. Similarly it may be shewn that FA, FB, FC, FD, FE are all equal. From centre F, with radius FA describe a circle: this circle must pass through the points A, B, C, D, E, and therefore is circumscribed about the given pentagon. Q.E.F. In the same way a c i r c l e may be circumscribed about any regular polygon. 272 euclid's elements. Proposition 15. Problem. To inscribe a regular hexagon in a given c i r c l e . Let ABDF be the given circle : i t is required to inscribe a regular hexagon in i t . Find G the centre of the © A B D F ; in. 1. and draw a diameter AGD. From centre D, with radius DG, describe the © EGCH. Join CG, EG, and produce them to cut the O08 of the given circle at F and B. Join AB, BC, CD, DE, EF, FA. Then ABCDEF shall be the required regular hexagon. N o w GE = GD, being radii of the © ACE; and DG = DE, being radii of the © EHC : . ' . GE, ED, DG are all equal, and the A E G D is equilateral. Hence the A EGD = one-third of two rt. angles. I . 32. Similarly the a DGC = one-third of two rt. angles. But the A8 EGD, DGC, CGB together = two rt. angles : l . 13. . ' . the remaining a CGB = one-third of two rt. angles. . ' . the three A8 EGD, DGC, CGB are equal to one another. And to these angles the vert. opp. a3 BGA, AGF, FGE are respectively equal: . ' . the A8 EGD, DGC, CGB, BGA, AGF, FGE are all equal; . ' . the arcs ED, DC, CB, BA, AF, FE are all equal; in. 26. . ' . t l i e chords ED, DC, CB, BA, AF, FE are a l l equal: I I I . 29. . ' . the hexagon i s equilateral. Again the arc FA = the arc DE : Proved. to eaeli of these equals add the arc ABCD ; then the whole arc FABCD =the whole arc ABODE : henco the angles at the O"' which stand on these equal ares are equal, BOOK IV. PROP. 15. 273 that is, the _ FED = the a AFE. in. 27. In Like manner the remaining angles of the hexagon may be shewn to be equal. . ' . the hexagon is equiangular : . ' . the hexagon is regular, and it is inscribed in the © ABDF. Q. E. F. Corollary. Tlie side of a regular hexagon inscribed in a circle is equal to the radius of the circle. P r o p o s i t i o n 16. Problem. To inscribe a regular quindecagon in a given circle. A w / ft / EYv C NL \ -\ il y D Let ABCD be the given circle : i t is required to inscribe a regular quindecagon in it. In the © A B C D inscribe an equilateral triangle, iv. 2. and let A C be one of its sides. In the same circle inscribe a regular pentagon, iv. 11. and let AB be one of its sides. Then of such equal parts as the whole Ooe contains fifteen, the arc AC, which is one-third of the Oce, contains five; and the arc AB, which is one-fifth of the Oce, contains three; . ' . their difference, the arc BC, contains two. Bisect the arc BC at E : in. 30. then each of the arcs BE, EC is one-fifteenth of the Oce-. ' . if BE, EC be joined, and st. lines equal to them be placed successively round the circle, a regular quindecagon will be inscribed in it. Q. E. f. 274 EUCLID'S ELEMENTS. NOTE ON REGULAR POLYGONS. The following propositions, proved by Euclid for a regular penta-gon, hold good for all regular polygons. 1. The bisectors of tlw angles of any regular polygon are con-current. Let D, E, A, B, C be consecutive angular points of a regular polygon of any number of sides. Bisect the L EAB, A B C by AO, BO, which intersect at O. Join EO. It is required to prove that EO bisects the z DEA. For in the a" EAO, BAO, ;E A = B A, being sides of a regular polygon; and A O is common; and the Z EAO = the z B A O ; Constr. : . the z O E A = the z O B A. i. 4. But the L OBA is half the Z ABC; Constr. also the L A B C = the Z DEA, since the polygon i regular; . -. the Z O E A is half the / DEA: that is, EO bisects the Z DEA. Similarly if O be joined to the remaining angular points of tlie polygon, i t may be proved that each joining line bisects the angle to whose vertex i t is drawn. That is to say, the bisectors of the angles of the polygon meet at the point O. Q. k.d. Corollames. Since the z EAB = the ZABC; Hyp. and since the Z " OAB, OBA are respectively half of the L ' EAB. ABC; . -. the z O A B = the z OBA. . -. O A = OB. . . G. Similarly O E = OA. Hence The bisectors of the angles of a regular polygon are all equal: and a circle described from the centre O, with radius OA, will be circumscribed about the polygon. Also i t , may be shewn, as in Proposition 13. that perpendiculars drawn from O to the sideB of the polygon are all equal; therefore a circle described from centre O with any one of these perpendiculars as radius will be inscribed in t l i e polygon. BOOK IV. NOTE ON REGULAR POLYGONS. 275 2. If a polygon inscribed in u circle is equilateral, it is also equiangular. Let AB, BC, CD be consecutive sides of an equilateral polygon inscribed in the © A D K ; then shall this polygon be equiangular. Because the chord AB=the chord DC, Hyp. : . the minor are A B = the minor arc DC. m . 28. To each of these equals add the arc A K D : then the arc B A K D = the arc A K D C ; . -. the angles at the 0», which stand on these equal arcs, are equal; that is, the z B C D = the z A B C . in. 27. Similarly the remaining angles of the polygon may be shewn to he equal: . . the polygon is equiangular. q.e.d. 3. If a polygon inscribed in a circle is equiangular, it is also equilateral, provided that the number of its sides is odd. [Observe that Theorems 2 and 3 are only true of polygons inscribed in a circle. The accompanying figures are sufficient to shew that otherwise a polygon may be equilateral without being equiangular, Fig. 1; or equiangular without being equilateral, Fig. 2.] Fig. I Fig. 2 Note. The following extensions of Euclid's constructions for Begular Polygons should be noticed. By continual bisection of arcs, we are enabled to divide the circumference of a circle, by means of Proposition 0, into 4, 8,16,..., 2 . 2™,... equal parts; by meaDS of Proposition 15, into 3, 6, 12,..., 3 . 2",... equal parts; by means of Proposition 11, into 5,10,20,..., 5 . 2™,... equal parts; by means of Proposition 16, into 15, 30, 60,..., 15 . 2™,... equal parts. Hence we can inscribe in a circle a regular polygon the number of whose sides is included in any one of the formulae 2 . 2", 3 . 2™, 5 . 2", 15 . 2™, n being any positive integer. In addition to these, it has been shewn that a regular polygon of 2n + 1 sides, provided 2n + l is a prime number, may be inscribed in a circle. 276 Euclid's elements. EXERCISES ON PROPOSITIONS 11 — 1 6 . 1. Express in terms of a right angle the magnitude of an angle of t l i e following regular polygons: (i) a pentagon, ( i i ) a hexagon, (iii) an octagon, (iv) a decagon, (v) a quindecagon. 2. The angle of a regular pentagon is trisected by the straight lines which join it to the opposite vertices. 3. In a polygon of n sides the straight lines which join any angular point to the vertices not adjacent to it, divide the angle into n - 2 equal parts. 4. Shew how to construct on a given straight line (i) a regular pentagon, ( i i ) a regular hexagon, (hi) a regular octagon. 5. An equilateral triangle and a regular hexagon are inscribed in a given circle; shew that ( i ) the area of the triangle is half that of the hexagon; ( i i ) the square on the side of the triangle is three times the Bquare on the side of the hexagon. 6. ABODE is a regular pentagon, and AC, BE intersect at H: shew that ( i ) A B = C H = EH. (ii) A B is a tangent to the circle circumscribed about the triangle B H C . ( i i i ) A C and B E cut one another in medial section. 7. The straight lines which join alternate vertices of a regular pentagon intersect so as to form another regular pentagon. 8. The straight lines which join alternate vertices of a regular polygon of n sides, intersect so as to form another regular polygon of n sides. If m = 6, shew that tho area of the resulting hexagon is one-third of the given hexagon. 9. By means of rv. 16, inscribe in a circle .v triangle whose angles are as the numbers 2, 5, 8. 10. Shew that the area of a regular hexagon inscribed in a circle is three-fourths of that of tho corresponding circumscribed hexagon. THEOREMS A N D EXAMPLES ON BOOK IV. I . ON THE TRIANGLE AND ITS CIRCLES. 1. D, F, E are the points of contact of the inscribed circle of the triangle A B C , and Dlt Flt Ej the points of contact of the escribed circle, which touches B C and the other sides produced: a, b, c denote the lengths of the sides B C , C A , A B ; s the semi-perimeter of the triangle, and r, r2 the radii of the inscribed and escribed circles. Prove the following equalities: ( i ) (ii) (iii) (iv) ( v ) (vi) AE=AF=s-a, BD = BE=s-6, CD = CF = s-c. AE1=AF1 = s . CD1 = CF1 = s-b, BD1=BEI=s-c. CD = BDX and BD = EE1=. FF1 = a. = CD!. The area of the a ABC B, Ei/" Y as,\ i i \D 1 k =rs=r1 (s-a). H. E. 1 9 278 EUCLID 8 ELEMENTS. 2. Iii the triangle A B C , I is the centre of tlie inscribed circle, and lj, l2, I, the centres of tlie escribed circles touching respectively the sides B C , C A , A B and the other sides produced. I3\:--. >u \ y svl > x - ^\ /X x i - ^ \ \Di m z'\v c Prowc the following properties :— (i) The points A, I, l2 arc eollinear; so are B, I, I.,; and C, I. I... (ii) The points L, A, l3 are eollinear; so are l . t , B, I,; and lj, C, l2. (iii) The triangles BI,C, CI„A, AI3B are equiangular to one another. (iv) The triangle l,U3 is equiangular to the triangle formed by joining the points of contact of the inscribed circle. (v) 0/ the four points I, lj, l„, I . , <vir/i is the orthocentre of the triangle whose vertices are the other three. (vi) The four circles, each of which passes through three of the points I, In l„, I . , , are all equal. THEOREMS AND EXAMPLES ON BOOK IV. 279 3. With the notation of page 277, shew that in a triangle ABC, i f the angle at C is a right angle, r = s - c ; r1=s-b; r2=s-a; rs=s. 4. "With the figure given on page 278, shew that if the circles whose centres are I, \x, l2, l3 touch B C at D, Dj, D2, D3, then (i) DDa=D1D,=6. (ii) DD3=D1D2=c. ( i i i ) D2D3=& + c. (iv) DD1=6~c. 5. Shew that the orthocentre and vertices of a triangle are the centres of the inscribed and escribed circles of the pedal triangle. [See Ex. 20, p. 225.] 6. Given the base and vertical angle of a triangle, find the locus of the centre of the inscribed circle. [See Ex. 36, p. 228.] 7. Given the base and vertical angle of a triangle, find the locus of the centre of the escribed circle which touches the base. 8. Given the base and vertical angle of a triangle, shew that the centre of the circumscribed circle is fixed. 9. Given the base BC, and the vertical angle A of a triangle, find the locus of the centre of the escribed circle which touches A C . 10. Given the base, the vertical angle, and the radius of the inscribed circle; construct the triangle. 11. Given the base, the vertical angle, and the radius of the escribed circle, (i) which touches the base, (ii) which touches one of the sides containing the given angle; construct the triangle. 12. Given the base, the vertical angle, and the point of contact with the base of the inscribed circle; construct the triangle. 13. Given the base, the vertical angle, and the point of contact with the base, or base produced, of an escribed circle; construct the triangle. 14. From an external point A two tangents AB, AC are drawn to a given circle; and the angle B A C is bisected by a straight line which meets the circumference in I and lj: shew that I is the centre of the circle inscribed in the triangle A B C , and l 3 the centre of one of the escribed circles. 15. I is the centre of the circle inscribed in a triangle, and lX) l2, l the centres of the escribed circles; shew that \ \ l t ll2, ll3 are bisected by the circumference of the circumscribed circle. 16. ABC is a triangle, and l2, l3 the centres of the escribed circles which touch A C , and A B respectively: shew that the points B, C, l2, l3 he upon a circle whose centre is on the circumference of the circle circumscribed about A B C . 19—2 280 Euclid's elements. 17. With three given points as centres describe three circles touching one another two by two. H o w many solutions will there be? 18. Two tangents AB, AC are drawn to a given circle from an external point A; and in AB, A C two points D and E are taken so that D E is equal to the sum of D B and E C : shew that D E touches the circle. 19. Given the perimeter of a triangle, and one angle in magnitude and position: shew that the opposite side always touches a fixed circle. 20. Given the centres of the three escribed circles; construct the triangle. 21. Given the centre of the inscribed circle, and the centres of two escribed circles; construct the triangle. 22. Given the vertical angle, perimeter, and the length of the bisector of the vertical angle; construct the triangle. 23. Given the vertical angle, perimeter, and altitude ; construct the triangle. 24. Given the vertical angle, perimeter, and radius of the in-scribed circle; construct the triangle. 25. Given the vertical angle, the radius of the inscribed circle, and the length of the perpendicular from the vertex to the base; construct the triangle. 26. Given the base, the difference of the sides containing the vertical angle, and the radius of the inscribed circle; construct the triangle. [See Ex. 10, p. 258.] 27. Given the base and vertical angle of a triangle, find the locus of the centre of the circle which passes through the three escribed centres. 28. In a triangle ABC, I is the centre of the inscribed circle; shew that the centres of the circles circumscribed about the triangles BIC, CIA, AIB lie on the circumference of the circle circumscribed about the given triangle. 29. In a triangle ABC, the inscribed circle touches the base BC at D; and r, rx are the radii of the inscribed circle and of the escribed circle which touches B C : shew that r. rx= B D . D C . 30. ABC is a triangle, D, E, F the points of contact of its inscribed circle; and D'E'F' is the pedal triangle of the triangle D E F : shew that the sides of the triangle D'E'F' are parallel to those of A B C . 31. In a triangle ABC the inscribed circle touches BC at D. Shew that the circles inscribed in the triangles A B D , A C D touch one another. THEOREMS AND EXAMPLES ON BOOK IV. 2 8 1 On the Nine-Points Chicle. 32. In any triangle the middle points of the sides, the feet of the perpendiculars drawn from the vertices to the opposite sides, and the middle points of the lines joining the orthocentre to the vertices are concyclic. In the A ABC, let X, Y, Z be the middle points of the sides BC, CA, A B ; l e t D, E, F be the feet of the perp8 drawn to these sides from A, B, C; let O be the orthocentre, and a, /3, 7 the middle points of OA, OB, OC: then shall the nine points X, Y, Z, D, E, F, u . , /3, 7 be concyclic. Join XY, X Z , Xa, Yo, Zo. Nowfromthe A ABO, since A Z = ZB, and Aa=aO, Hyp. . : Za is par1 to BO. Ex. 2, p. 96. And from the A ABC, since B Z = ZA, andBX = XC, Hyp. : . Z X is par1 to AC. But B O makes a rt. angle with A C ; . . the Z X Z a is a rt. angle. Similarly, the Z XYa is a rt. angle. x . 29. / . the points X, Z, a, Y are concyclic : that is, a lies on the oco of the circle, which passes through X, Y, Z ; and Xa is a diameter of this circle. Similarly i t may be shewn that / 3 and y l i e on the Oco of the circle which passes through X, Y, Z. Again, since aDX is a rt. angle, Hyp. . " . the circle on Xa as diameter passes through D. Similarly i t may be shewn that E and F he on the circumference of the same circle. . • . the points X, Y, Z, D, E, F, a, p, y are concyclic. q.e.d. Hyp. From this property the circle which passes through the middle points of the sides of a triangle is called the Nine-Points Circle; many of its properties may be derived from the fact of its being the circle circumscribed about the pedal triangle. 2 8 2 EUCLID'S ELEMENTS. 33. To prove that ( i ) the centre of the nine-points circle is the middle point of t l i e straight line which joins the orthocentre to the circumscribed centre: ( i i ) the radius of the nine-points circle is half the radius of the circumscribed circle: ( i i i ) the centroid is eollinear with the circumscribed centre, the nine-points centre, and the orthocentre. In the A ABC, let X, Y, Z be the middle points of the sides; D, E, F the feet of the perp"; O the ortho-centre; S and N the centres of the circumscribed and nine-points circles respectively. ( i ) To prove that N is the middle point of SO. It may be shewn that the perp. to X D from its middle point bisects SO; Ex. 14, p. 98. Similarly the perp. to EY at i t s middle point bisects SO: that is, these perp" intersect at the middle point of SO: And since X D and EY are chords of the nine-points circle, . ' . the intersection of the lines which bisect X D and EY at rt. angles i s i t s centre: in. 1. . . the centre N i s the middle point of SO. ( i i ) To prove that the radius of the nine-points circle i s half the radius of the circumscribed circle. By the last Proposition, Xa is a diameter of the nine-points c i r c l e . . • . the middle point of Xa is i t s centre: but the middle point of S O i s also the centre of the nine-points c i r c l e . (Proved.) Hence Xa and S O bisect one another at N. Then from the A» SNX, ONa ( SN = OIM, Because • } and N X = Na, (and the z S N X = the zONa; . -. SX = Oa = Aa. And SX is also par' to Aa, . -. SA = Xa. But SA is a radius of t l i e circumsoribed circle: and Xa is a diameter of the nine-points circle; . ' . tho radius of the nine-points circle is half t l i e radius of the circum-scribed circle. i.la. 1.4. 3 3 . THEOREMS AND EXAMPLES ON BOOK IV. 283 (iii) To prove that the centroid is eollinear with points S, N, O. Join A X and draw ag par1 to SO. Let A X meet S O at G. Then from the A A G O , since Aa = aO and ag is par1 to OG, . -. Ag=gG. Ex. 13, p. 98. And from the A Xag, since aN = NX, and N G is par1 to ag, : . <7G = G X . Ex. 13, p. 98. . -. A G = 1 of A X ; . ' . G is the centroid of the triangle ABC. That is, the centroid is eollinear with the points S, N, O. q.e.d. 34. G-iven the base and vertical angle of a triangle, find the locus of the centre of the nine-points circle. 35. The nine-points circle of any triangle ABC, whose centre is 0, is also the nine-points circle of each of the triangles A O B , BOC, COA. 36. If I, lj, l2, l3 are the centres of the inscribed and escribed circles of a triangle ABC, then the circle circumscribed about A B C is the nine-points circle of each of the four triangles formed by joining three of the points I , l l t l2, l3. 37. All triangles which have the same orthocentre and the same Eircumseribed circle, have also the same nine-points circle. 38. If S, I are the centres, and R, r the radii of the circumscribed and inscribed circles of a triangle, and if N is the centre of t l i e nine-points circle ; prove t l i a t ( i ) SI2=Rs-2Rr, ( i i ) Nl =JR-r. And establish corresponding properties for the escribed circles. 39. Employ the preceding theorem to shew that the nine-points circle touches the inscribed and escribed circles of a triangle. II. MISCELLANEOUS EXAMPLES. 1. If four circles are described to touch every three sides of a quadrilateral, shew that their centres are concyclic. 2. If the straight lines which bisect the angles of a rectilineal figure are concurrent, a circle may be inscribed in the figure. 3. Within a given circle describe three equal circles touching one another and the given circle. 4. The perpendiculars drawn from the centres of the three escribed circles of a triangle to the sides which they touch, are eon-current. 284 Euclid's elements. 5. Given an angle and the radii of the inscribed and circumscribed circles; construct the triangle. 6. Given the base, an angle at the base, and the distance between the centre of the inscribed circle and the centre of the escribed circle which touches the base; construct the triangle. 7. In a given circle inscribe a triangle such that two of its sides may pass through two given points, and the third side be of given length. 8. In any triangle ABC, I, \L, l2, l3 are the centres of the in-scribed and escribed circles, and Sx, S„, S3 are the centres of the circles circumscribed about the triangles BIC, CIA, AIB: shew that the triangle SjS2S3has its sides parallel to those of the triangle l i l 2 l 3 , and is one-fourth of i t in area: also that the triangles A B C and SjS2S3 have the same circumscribed circle. 9. O is the orthocentre of a triangle ABC: shew that A O S + B C ^ B O S + C A ^ C O ' ^ + AB-^d2, where d is the diameter of the circumscribed circle. 10. If from any point within a regular polygon of n sides perpen-diculars are drawn to the sides the sum of the perpendiculars is equal to n times the radius of the inscribed circle. 11. The sum of the perpendiculars drawn from the vertices of a regular polygon of n sides on any straight line is equal to n times the perpendicular drawn from the centre of the inscribed circle. 12. The area of a cyclic quadrilateral is independent of the order in which the sides are placed in the circle. 13. Of all quadrilaterals which can be formed of four straight lines of given length, that which is cyclic has the maximum area. 14. Of all polygons of a given number of sides, which may be inscribed in a given circle, that which is regular has the maximum area and the maximum perimeter. 15. Of all polygons of a given number of sides circumscribed about a given circle, that which is regular has the minimum area and the minimum perimeter. 16. Given the vertical angle of a triangle in position and magni-tude, and the sum of the sides containing it: find t l i e locus of the centre of the circumscribed circle. 17. P is any point on the circumference of a circle circumscribed about an equilateral triangle A B C : shew that PA- + PB--i- PC2 is constant. BOOK V. Book V. treats of Batio and Proportion. INTRODUCTORY. The first four books of Euclid deal with the absolute equality or inequality of Geometrical magnitudes. In the Fifth Book magnitudes are compared by considering their ratio, or relative greatness. The meaning of the words ratio and proportion in their simplest arithmetical sense, as contained in the following defini-tions, is probably familiar to the student: The ratio of one number to another is the multiple, part, or parts that the first number is of the second; and it may therefore be measured by the fraction of which the first number is the numerator and the second the denominator. Pour numbers are in proportion when the ratio of the first to the second is equal to that of t l i e third to the fourth. But it will be seen that these definitions are inapplicable to Geometrical magnitudes for the following reasons: (1) Pure Geometry deals only with concrete magnitudes, re-presented by diagrams, but not referred to any c o m m o n unit in terms of which they are measured: in other words, it makes no use of number for the purpose of comparison between different magnitudes. (2) It commonly happens that Geometrical magnitudes of the same kind are incommensurable, that is, they are such that it is impossible to express them exactly in terms of some common unit. For example, we can make comparison between the side and diagonal of a square, and we may form an idea of their relative great-ness, but i t can be shewn that it is impossible to divide either of them into equal parts of which the other contains an exact number. And as the magnitudes we meet with in Geometry are more often incom-mensurable than not, i t is clear that i t would not always be possible to exactly represent such magnitudes by numbers, even i f reference to a common unit were not foreign to the principles of Euclid. It is therefore necessary to establish the Geometrical Theory of Proportion on a basis quite independent of Arithmetical principles. This is the aim of Euclid's Fifth Book. 286 euclid's elements. We shall employ the following notation. Capital l e t t e r s , A, B, C,... will be used to denote the magnitudes themselves, not any numerical or algebraical measures of them, and small l e t t e r s , m, n, p,... will be used to denote whole numbers. Also i t will be assumed t l i a t multiplication, in the sense of repeated addition, can be applied to any magnitude, so that m. A or mA w i l l denote the magnitude A taken m times. The symbol > will be used for the words greater than, and < f o r l e s s than. Definitions. 1. A greater magnitude is said to be a multiple of a less, when the greater contains the less an exact number of times. 2. A less magnitude is said to be a submultiple of a greater, when the less is contained an exact number of times in the greater. The following properties of multiples will be assumed as self-(1) mA > = or < m B according as A > = or < B; and conversely. ( 2 ) mA + mB + . . . = m ( A + B + ...). ( 3 ) If A > B , thenmA-?7iB=ro(A-B). ( 4 ) mA + nA+... — (m + n +...) A. (5) If m>n, then mA -nA = (m - n) A. ( 6 ) m. nA = mn.A = nm. A = n. m A. 3. The Ratio of one magnitude to another of the same kind is the relation which the first bears to the second in respect of quanluplicity. The ratio of A to B is denoted thus, A B; and A is called the antecedent, B the consequent of the ratio. The term quantuplicity denotes the capacity of the first magnitude to contain the second with or without remainder. If the magnitudes are commensurable, their quantuplicity may be expressed numerically by observing what multiples of the two magnitudes are equal to one another. Thus i f A = ma, and B = na, i t follows that nA~mB. In this case A = - B, and the quantuplicity of A with respect to B is the B meticul traction —. n definitions. 287 But if the magnitudes are incommensurable, no multiple of the first can be equal to any multiple of the second, and therefore the quantuplicity of one with respect to the other cannot exactly be expressed numerically: in this ease it is determined by examining how the multiples of one magnitude are distributed among the multiples of the other. Thus, let all the multiples of A be formed, the scale extending ad infinitum; also let all the multiples of B be formed and placed in their proper order of magnitude among the multiples of A. This forms the relative scale of the two magnitudes, and the quantuplicity of A with respect to B is estimated by examining how the multiples of A are distributed among those of B in their relative scale. In other words, the ratio of A to B is known, if for all integral values of m we know the multiples nB and (n + 1) B between which m A lies. Li the case of two given magnitudes A and B, the relative scale of multiples is definite, and is different from that of A to C, if C differs from B by any magnitude however small. For let D be the difference between B and C ; then however small D may be, it will be possible to find a number m such that m D > A . In this case, m B and m C would differ by a magnitude greater than A; and therefore could not lie between the same two multiples of A; so that after a certain point the relative scale of A and B would differ from that of A and C. [It is worthy of notice that we can always estimate the arithmetical ratio of two incommensurable magnitudes within any required degree of accuracy. For suppose that A and B are incommensurable; divide B into m equal parts each equal to /S, so that B=mj3, where m is an integer. Also suppose § is contained in A more than n, times and less than (n+1) times; then A nB (ra+l)/S =->-^and<^ -lc\ B m/3 m 8 that is, = lies between — and ; B m m An 1 bo that -= - differs from - by a quantity less than — . And since we B m J ^ J m can choose / 3 (our unit of measurement) as small as we please, m can be made as great as we please. Hence — can be made as small as we e m please, and two integers n and m can be found whose ratio will express that of a and b to any required degree of accuracy.] 288 euclid's elements. 4. The ratio of one magnitude to another is equal to that of a third magnitude to a fourth, when if any equi-multiples whatever of the antecedents of the ratios are taken, and also any equimultiples whatever of the con-sequents, the multiple of one antecedent is greater than, equal to, or less than that of its consequent, according as the multiple of the other antecedent is greater than, equal to, or less than that of its consequent. Thus the ratio A to B is equal to that of C to D when m G > = or < n D according as m A • > - or < nB, whatever whole numbers m and n m a y be. Again, l e t m be any whole number whatever, and n another whole number determined in such a way that either mA is equal to nB, or mA l i e s between »B and (n +1) B; then the definition asserts that the ratio of A to B i s equal to that of C to D i f m C = nD when mA = nB; or i f m C l i e s between nD and (n + 1) D when mA l i e s between nB and (n + l)B. In other words, the ratio of A to B i s equal to that of C to D when the multiples of A are distributed among those of B in the same manner as the multiples of C are distributed among those of D. 5. W h e n the ratio of A to B is equal to that of C to D the four magnitudes are called proportionals. This is ex-pressed by saying " A is to B as C is to D", and the proportion is written A : B : : C : D, or A : B = C : D. A and D are called the extremes, B and C the means: also D is said to be a fourth proportional to A, B, and C. T w o terms in a proportion are said to be homologous when they are both antecedents, or both consequents of the ratios. [ I t will be useful here to compare the algebraical and geometrical definitions of proportion, and to shew that each may be deduced from the other. According to the geometrical definition A, B, C, D are in propor-tion, when m C > = o i D according as m A > = = < h D according as m A > = = < n D according as m A > = — ; then it will be possible to find some fraction -t > U m which lies between them, n and m being positive integers. Hence - > — (1); B m K ' and 5 < « (2). D m v ; From (1), m A > n B ; from (2), mC D the proposition.] 6. The ratio of one magnitude to another is greater than that of a third magnitude to a fourth, w h e n it is possible to find equimultiples of the antecedents and equi-multiples of the consequents such that while the multiple of the antecedent of the first ratio is greater than, or equal to, that of its consequent, the multiple of the antecedent of the second is not greater, or is less, than that of its consequent. 290 kuclid's elements. This definition asserts that if whole numbers m and n can be fo such that while mA i s greater than nB, roC is not greater than nD, or while mA=nB, m C i s less than nD, then the ratio of A to B i s greater than that of C to D. 7. If A is equal to B, the ratio of A to B is called a ratio of equality. If A is greater than B, the ratio of A to B is called a ratio of greater inequality. If A is less than B, the ratio of A to B is called a ratio of less inequality. 8. Two ratios are said to be reciprocal when the ante-cedent and consequent of one are the consequent and ante-cedent of the other respectively; thus B : A is the reciprocal of A : B. 9. Three magnitudes of the same kind are said to be proportionals, when the ratio of the first to the second is equal to that of the second to the third. Thus A, B, C are proportionals i f A : B : : B : C. B is called a mean proportional to A and C, and C is called a third proportional to A and B. 10. Three or more magnitudes are said to be in con-tinued proportion when the ratio of the first to the second is equal to that of tlie second to the third, and the ratio of the second to the third is equal to that of the third to the fourth, and so on. 11. When there are any number of magnitudes of the same kind, the first is said to have to the last the ratio compounded of the ratios of the first to the second, of the second to the third, and so on up to the ratio of tlie last but one to tlie last magnitude. For example, i f A, B, C, D, E be magnitudes of the same kind, A : E is the ratio compounded of the ratios A : B, B : C, C ; D, and D : E. DEFINITIONS. 291 This is sometimes expressed by the following notation: A : E = A : B B : C C : D D : E. 12. If there are any number of ratios, and a set of magnitudes is taken such that the ratio of the first to the second is equal to the first ratio, and the ratio of the second to the third is equal to the second ratio, and so on, then the first of the set of magnitudes is said to have to the last the ratio compounded of the given ratios. Thus, i f A : B, C : D, E : F be given ratios, and i f P, Q, R, S be magnitudes taken so that P : 0. : : A : B, Q : R : : C : D, R : S : : E : F; 'A:B then P : S = -< C : D E : F. 13. When three magnitudes are proportionals, the first i s said to have to the third the duplicate ratio of that which it has to the second. Thus i f A : B : : B : C, then A is said to have to C the duplicate ratio of that which i t has to B. Since A:C=\|b-C i t is clear that the ratio compounded of two equal ratios is the dupli-cate ratio of either of them. 14. When four magnitudes are in continued proportion, the first is said to have to the fourth the triplicate ratio of that which it has to the second. It may be shewn as above that the ratio compounded of three equal ratios is the triplicate ratio of any one of them. 292 euclid's elements. Although an algebraical treatment of ratio and proportion when applied to geometrical magnitudes cannot be considered exact, i t will perhaps be useful here to summarise in algebraical form the principal theorems of proportion contained in Book V. The student will then perceive that i t s leading propositions do not introduce new ideas, but merely supply rigorous proofs, based on the geometrical definition of proportion, of results already familiar in the study of Algebra. W e shall only here give those propositions which are afterwards referred to in Book VI. It will be seen that in their algebraical form many of them are so simple that they hardly require proof. Summary of Principal Theorems of Book V. P r o p o s i t i o n 1 . Ratios which are equal to the same ratio are equal to one another. That is, if A : B = X : Y and C : D = X • Y; then A : B = C : D. Proposition 3. If four magnitudes are proportionals, they are also proportionals when taken inversely. That is, if A : B = C : D. then B : A = D : C . This inference is referred to as invertendo or inversely. Proposition 4. (i) Equal magnitudes have the same ratio to the same magnitude. For i f A = B, then A : C = B : C . (ii) The same magnitude has the same ratio to equal magnitudes. For i f A = B, then C : A = C : B. summaky of principal theorems of book v. 293 Proposition 6. (i) Magnitudes which have the same ratio to the same magnitude are equal to one another. That is, if A:C=B:C, then A=B. (ii) Those magnitudes to which the same magnitude has the same ratio are equal to one another. That is, i f n C : A = C : B. A = B. Proposition 8. Magnitudes have the same ratio to one another which their equi-multiples have. That is, A : B=mA : mB, where m is any whole number. Proposition 11. If four magnitudes of the same kind are proportionals, they are proportionals when taken alternately. I f then shall For since A : B = C : A :C=B : A C B ~D' D D R A R O B . -. multiplying by - , we have - ^ = D ' C ' A B t h a t i s , c = D ' or A:C=B:D. This inference is referred to as alternando or alternately. H. E. 20 294 EUCLID'S ELEMENTS. Proposition 12. If any number of magnitudes of t l i e same kind are proportionals, then as one of the antecedents is to its consequent, so is t l i e sum of the antecedents to the sum of the consequents. Let A: B = C : D = E : F = . . . ; then shall A : B = A + C+E+... : B + D + F+.... A C E For put each of the equal ratios b , = , -=,... equal to / . -; then A = Bk, C ^Dk, E=Fk,... A + C+E+... Bk+Dk+Fk+L..,A_C_E ' ' " BTD+TT77. ~ ~B"+ D + F +..." ~ ~ ~~B ~ D ~ F ; . -. A: B = A + C + E-... . B + D + F+.... This inference i s sometimes referred to as addendo. Proposition 13. ( i ) If four magnitudes are proportionals, the sum of the first and second is to the second as the sum of the third a nd fourth is to the fourth. L e t then shall For since that is, A: B = C : D, A+B:B=C+D: A C B~ D' • • • B + 1=D + l ! A+B C+D B : D~"' D or A + B : B = C + D : D, This inference is referred to as componendo. ( i i ) If four magnitudes are proportionals, the difference of the first and second is to the second as the difference of the third and fourth is to the fourth. That is, i f A : B = C : D, then A ~ B : B = C ~ D : D. The proof i s similar to that of the former case. This inference iB referred to as divldendo. summary of principal theorems of book v. 295 Proposition 14. If there are two sets of magnitudes, such that the first is to the second of the first set as the first to t l i e second of the other set, and the second to the third of the first set as the second to the third of the other, and so on to the last magnitude: then the first is to the last of the first set as the first to the last of the other. First let there be three magnitudes, A, B, C, of one set, and three, P, Q, R, of another set, and let a n d then shall For since that is, o r Similarly if A : B=P : Q, B : C = Q : R; A : C = P : R. A P , B Q B = Q ' a n d C = R A B P Q " B " C-Q ' R' A P C~R' A : C = P : R. A : B=P : Q, B : C = Q: R, = , L: M = Y • Z; it can be proved that A : M = P : Z. This inference is referred to as ex sequaU. Corollary. If A : B=P : Q, and B : C = R : P; then shall A : C = R : Q. u • A P , B R Forsmce B = Q' C=P; A B P B ' C_Q' A R ' • C_Q' R P A : C=R : Q. 20—2 296 piuclid's elements. Proposition 15. / / A:B = C:D, and E : B = F : D; thenshall A + E:B = C + F:D. „ . A C , E F For since B = D'andB = D; A+E_C+F " " B : D ' that is, A + E : B = C + F:D. Proposition 16. If two ratios are equal, their duplicate ratios are equal; and conversely. Let A : B = C : D; then shall the duplicate ratio of A : B be equal to the duplicate ratio of C : D. Let X be a third proportional to A, B; so that A : B = B : X; B _ A " X~B; B A_A A " X ' B ~ B ' B' A A2 that i s , X = B5' But A : X is the duplicate ratio of A : B; : . the duplicate ratio of A : B = A2 : B-. But since A : B = C : D: A C " B~D' A_G> • ' • B3_D2' or A2 : B2 = C- : D-; that is, the duplicate ratio of A : B=the duplicate ratio of C : D. Conversely, let the duplicate ratio of A : B be equal to the dupli-cate ratio of 0 : D; then shall A : B = C : D, for sinco A8 : Bs = C- : P-, . . A : B = C : D. proofs of the propositions of book v. 297 Proofs of the Propositions of Book V. derived from THE GEOMETRICAL DEFINITION OF PROPORTION. Obs. The Propositions of Book V. are all theorems. Proposition 1. Ratios which are equal to the same ratio are equal to one another. Let A : B : : P : Q, and also C : D : : P : Q; then shall A : B : : C : D. For it i s evident that two scales or arrangements of multiples which agree in every respect with a third scale, will agree with one another. Proposition 2. If two ratios are equal, the antecedent of the second is greater than, equal to, or l e s s than i t s consequent according as the antecedent of the first is greater than, equal to, or l e s s than i t s consequent. Let A : B :: C : D, then C > = or < D, according as A > = or < B. This follows at once from Def. 4, by taking m and n each equal to unity. 298 kuclid's elements. Proposition 3. If two ratios are equal, their reciprocal ratios are equal. Let A : B :: C : D, then shall B : A : : D : C. For, by hypothesis, the multiples of A are distributed among those of B in the same manner as the multiples of C are among those of D; therefore also, the multiples of B are distributed among those of A in the same manner as the multiples of D are among those of C. That is, B : A :: D : C. Note. This proposition is sometimes enunciated thus If four magnitudes are proportionals, they are also proportionals when taken inversely, and the inference i s referred to as Invertendo or Inversely. Proposition 4. Equal magnitudes have the same ratio to the same mag-nitude; and the same magnitude has the same ratio to equal magnitudes. Let A, B, C be three magnitudes of the same kind, and let A be equal to B; then shall A : C : : B : C and C : A : : C : B. Since A = B, their multiples are identical and therefore are distributed in the same way among the multiples of C. .'. A : C :: B : C, Def. 4. .'. also, invertendo, C : A :: C : B. v. 3. PROOFS of the propositions of book v. 299 Proposition 5. Of two unequal magnitudes, the greater has a greater ratio to a third magnitude than the l e s s has; and t l i e same magnitude has a greater ratio to the l e s s of two magnitudes than i t has t o the greater. First, let A be > B; then shall A : C be > B : C. Since A > B, i t will be possible to find m such that m A exceeds m B by a magnitude greater than C; hence i f m A l i e s between reC and (n + 1) C, m B < nC: and i f m A = nG, then m B < nC; . " . A : C > B : C. Def. 6. Secondly, let B be < A; then shall C : B be > C : A. For taking m and n as before, nC > mB, while nG is not > mA; . " . C : B > C : A. Def. 6 . Proposition 6. Magnitudes which have the same ratio t o the same mag-nitude are equal t o one another; and those t o which the same magnitude has the same ratio are equal t o one another. First, let A : C : : B : C; -then shall A= B. For i f A > B, then A : C > B : C, and i f B > A, then B : C > A : C, v. 5. which contradict the hypothesis; . ' . A=B. 300 eucud's elements. Secondly, let C : A :: C : B; then shall A = B. Because C : A : : C : B, . ' . , invertendo, A : C : : B : C, v. 3 . A=B, by the first part of t l i e proof. Proposition 7. That magnitude which has a greater ratio than another l i a s t o the same magnitude is the greater of the two; and that magnitude t o which the same l i a s a greater ratio than i t l i a s t o another magnitude is t l i e l e s s of t l i e two. First, let A : C be > B : C; then shall A be = - B. For i f A = B, then A : C : : B : C, v. 4. which i s contrary to the hypothesis. And i f A < B, then A : C < B : C; v. o. which i s contrary to the hypothesis: . " . A > B. Secondly•, let C : A be > C : B; then shall A be < B. For i f A = B, then C : A : : C : B, v. 4. which is contrary to the hypothesis. And i f A > B, then C : A < C : B; v. 5, which i s contrary to the hypothesis; . " . A = ov = or < qB. But m.pA=p . mA, and m. qB = q .mB; . ' . p. mA > = or = or < qB; . ' . A : B : : mA : mB. Def. 4 . Cor. Let A : B : : C : D . Then s i n c e A : B : : mA : mB, and C : D : : nC : nD; ' . ' . mA : mB : : nC : nD. V. 1 . Proposition 9. If two ratios are equal, and any equimultiples of t a n t e c e d e n t s and a l s o of t h e consequents are taken, t h e m u l t i p l e of t h e first antecedent has t o t h a t of i t s consequent t h e same r a t i o as t h e multiple of t h e o t h e r antecedent has t o t h a t of i t s c o n s e q u e n t . Let A : B : : C : D; then s h a l l mA : nB : : m C : nD. Let p, q be any two whole numbers, then because A : B : : C : D, pm . C > = or < qn. D according as pm. A > = or < qn . B, Def. 4 . that i s , p. m C > • = or = or < q. nB ; : . mA : nB : : m C : nD. Def. 4. 302 euclid's elements. Proposition 10. If four magnitudes of the same kind are proportionals, the first is greater than, equal to, or l e s s than the third, according as the second, i s greater than, equal to, or l e s s than the fourth. Let A, B, C, D be four magnitudes of the same kind such that A : B : : C : D; then A > = or < C according as B > = or < D. If B > D, then A : B < A : D; v. 5. but A : B : : C : D; . " . C D< A : D; . " . A : D > C : D; . ' . A > C. v. 7. Similarly i t may be shewn that i f B < D, then A < C, and i f B = D, then A = C. Proposition 11. If four magnitudes of tlie same hind are proportionals, they are also proportionals wlien taken alternately. Let A, B, C, D be four magnitudes of the same kind such that A : B : : C : D; then shall A : C : : B : D. Because A : B : : m A : mB, v. t > . and C : D : : m C : «D; . ' . m A : m B : : nC : ? ; D . v. 1 . . ' . m A > = or <?!C according as »iB> —or <«D: v. 10. and m and n are any whole numbers: . ' . A : C : : B : D. Def. 4. Note. This i n f e r e n c e i s usually r e f e r r e d t o as alternando o r alternately. proofs of the propositions of book v. 303 Proposition 12. If any number of magnitudes of the same kind are pro-portionals, as one of the antecedents is t o its consequent, so is the sum of the antecedents to t l i e sum of the consequents. Let A, B, C, D, E, F,... be magnitudes of the same kind such that A : B : : C : D : : E : F : : ; then shall A : B : : A + C + E+... : B + D + F + . . . . Because A:B::C:D::E:F::..., . ' . according as wiA> = or = or < nD, and m E > = or <: nF, .'. so is mA + mC + mE +...> = or -= nB + nD + nF + o r m ( A + C + E+...)> = or<7i.(B + D + F+...); and m and n are any whole numbers; . ' . A:B::A + C + E+... :B + D+F+.... Def. 4 . Note. This inference is usually referred to as addendo. Proposition 13. If four magnitudes are proportionals, the sum or differ-ence of t l i e first and second is to the second as the sum or difference of the third and fourth is to the fourth. Let A : B : then s h a l l A + B : B : and A ~ B : B : C: D; C + D : D, C ~ D : D. I f m be any whole number, i t i s p o s s i b l e to find another number n such that mA = nB, or l i e s between nB and («+l)B, . ' . mA + mB =mB + nB, or l i e s between mB + nB and mB + (n+ 1) B. 304 E U C L I D ' S ELEMENTS. But mA + mB = m(A + B), and mB + nB = (m + n) B; . ' . m(A + B) = (m + n) B, or l i e s between ( r n + n)B and (m + n+ 1)B. Also because A : B : : C : D, . " . mC = wD, or l i e s between nD and (n+ 1)D; Def. 4 . . ' . m(C + D) = (m + n) D or l i e s between (m + n)D and (m + n + 1)D; that i s , the multiples o f C + D are d i s t r i b u t e d among t h o s e o f D in the same way as the multiples o f A + B among those o f B; : . A+B : B : : C + D : D. In t l i e same way i t may be proved that A - B : B : : C - D : D , or B-A:B::D — C:D, according as A i s > or < B. Note. These inferences are referred to as componendo and d i r t -dendo respectively. Proposition 14. If there are two sets of magnitudes, such that tlie first is to the second of the first set as the first to t l i e second of t l i e other set, and the second to the third of t l i e first s e t as t l i e second to the third of t l i e other, and so on to the last magni-tude: then the first i s to the last of the first set as the first t o the last of the other. First, let there be three magnitudes A, B, C, of one set and three, P, Q, R, of another set, and let A : B : : P : Q, and B : C : : Q : R; then shall A : C : : P : R. Because A : B : : P : Q, . ' . m A : m B : : m P : uiQ.; v. 8, Cor. and because B : C : : Q, : R, . ' . m B : M.C : : m Q : «R, v. 9 . . ' . , invertendo, n C : m B : : nR : mQ.. v. 3. PROOFS OF THE PROPOSITIONS OF BOOK V. 305 Now, if mA > nC, then m A : roB>uC : m B ; v. 5. . ' . roP : « Q > mR : mQ, and . " . m P > n R . v: 7. Similarly i t may be shewn that m P = or < ? R , according as m A = or < nC, . . A : C : : P : R. Def. 4. Secondly, let there be any number of magnitudes, A, B, C,...L, M, of one set, and the same number P, Q, R, ...Y, Z, of another set, such that A : B : : P : Q, B : C : : O. : R, Proved. Hyp. •'•by L then shall A For A and C the first case A and so on, until finally Note. A This inference i s Corollary. If and then : M : : : M : : : C : : : D : : : D : : : M = Y P P R P P • z; -. z. : R, : S; :S, : Z. referred to as ex A : B B : C A : C P : a, R : P: R : Q. squall. Proposition 15. If A : and E then shall A + E For since E , invertendo, B Also A : : B : : B : : B : : B : : E : B : : C : D, : F : D; : C+F : F : D, : D : F. : C : D, Hyp. v . ' 3. ex cequaii, A : E : : C : F, v. 14. 306 Euclid's elements. .'., componendo, A + E : E :: C + F : F. v. 13. Again, E : B : : F : D, Hyp. . ' . , ex atquali, A + E : B : : C + F : D. v. 14. Proposition 16. If two ratios are equal, their duplicate ratios are equal; and conversely, if t l i e duplicate ratios of two ratios are equal, the ratios themselves are equal. Let A : B :: C : D; then shall the duplicate ratio of A to B be equal to that of C to D. Let X be a third proportional to A and B, and Y a third proportional to C and D, so that A : B : : B : X, and C : D : : D : Y: then because A : B : : C : D, . ' . B : X : : D : Y: . ' . , ex aiquali, A : X : : C : Y. But A : X and C : Y are respectively the duplicate ratios of A : B and C : D, Def. 13. . ' . the duplicate ratio of A : B = that of C : D. Conversely, let the duplicate ratio of A : B =that of C : D; then shall A : B : : C : D. Let P be such that A : B : : C : P, . ' . , invertendo, B : A : : P : C. Also, by hypothesis, A : X : : C : Y, . " . , ex wquedi, B : X : : P : Y: but A : B : : B : X, . ' . A : B : : P : Y; v. 1. . ' . C : P : : P : Y; v. 1. that is, P is the mean proportional between C and Y. . -. P=D, . " . A : B : : C : D. B O O K YI. D e f i n i t i o n s . 1. T w o rectilineal figures are said to be equiangular when the angles of the first, taken in order, are equal respectively to those of the second, taken in order. Each angle of the first figure is said to correspond to the angle to which it is equal in the second figure, and sides adjacent to corresponding angles are called corresponding sides. 2. Rectilineal figures are said to be similar when they are equiangular and have the sides about the equal angles proportionals, the corresponding sides being homologous. [See Def. 5, page 288.] Thus the two quadrilaterals ABCD, EFGH are similar i f the angles at A, B, C, D are respee- o t i v e l y equal to those at E, F, G, H, and i f the following proportions hold AB : BC : : EF : FG, A< BC : CD:: FG : GH, CD: DA:: G H : HE, DA : AB : : HE : EF. 3. Two figures are said to have their sides about two of their angles reciprocally proportional when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first. 4. A straight line is said to be divided in extreme and mean ratio when the whole is to the greater segment as the greater segment is to the less. 5. T w o similar rectilineal figures are said to be similarly Situated with respect to two of their sides when these sides are homologous. 3 0 8 EUCLID 8 ELEMENTS. Proposition 1. Theorem. The areas of triangles of the same altitude are to one another as their bases. Let ABC, ACD be two triangles of the same altitude, namely the perpendicular from A to BD: then shall the A ABC : the A ACD : : BC : CD. Produce BD both ways, and from CB produced cut off any number of parts BG, GH, each equal to BC; and from CD produced cut off any number of parts DK, KL, LM each equal to CD. Join AH, AG, AK, AL, AM. Then the A3 ABC, ABG, AGH are equal in are;i, for they are of the same altitude and stand on the equal bases CB, BG, GH, I . 38. . ' . the A AHC i s the same multiple of the A ABC that HC i s of BC; Similarly t l i e A ACM is the same multiple of ACD that CM i s of CD. And i f HC = CM, the A AHC = the A ACM; I . 38. and i f HC i s greater than CM, the A AHC i s greater than the A ACM ; I . 38. Cor. and i f HC i s less than CM, the A AHC is less than the A ACM. j. 38, Cor. N o w since there are four magnitudes, namely, the As ABC, ACD, and tho bases BC, CD; and of the antecedents, any equimultiples have been taken, namely, t l i e A AHC BOOK VI. PROP. 1. 309 and the base HC; and of the consequents, any equi-multiples have been taken, namely the A A C M and the base C M ; and since it has been shewn that the A A H C is greater than, equal to, or less than the A ACM, according as HC is greater than, equal to, or less than C M ; . ' . the four original magnitudes are proportionals, v. Def. 4. that is, the A ABC : the A A G D : : the base BC : the base CD. q.e.d. Corollary. Tlie areas of parallelograms of the same altitude are to one another as their bases. B Let EC, CF be par™ of the same altitude; then shall the par™ EC • the par™ CF : : BC : CD. Join BA, AD. Then the A ABC : the A ACD : : BC : CD; Proved. but the par™ EC is double of the A ABC, and the par™ CF is double of the A ACD; . ' . the par™ EC : the par™ CF : : BC : CD. v. 8. Note. Two straight lines are cut proportionally when the seg-ments of one line are in the same ratio as the corresponding segments of the other. [See definition, page 131.] F i g . 1 Fig.2 A X B A B X Y D C D Y Thus AB and CD are cut proportionally at X and Y, i f AX : XB : : CY : YD. And the same definition applies equally whether X and Y divide AB, CD internally as in Fig. 1 or externally as in Fig. 2. n. e . 21 3 1 0 EUCLID'S ELEMENTS. Proposition 2. Theorem. If a straight line be drawn parallel to one side of a triangle, i t shall cut the other sides, or tlwse sides produced, proportionally: Conversely, if the sides or the sides produced be cut pro-portionally, the straight line which joins the points of section, shall be parallel to the remaining side of the triangle. A A Y X_ B O Let XY be drawn par1 to BC, one of the sides of the A ABC: then shall BX : XA : : CY : YA. Join BY, CX. Then the A BXY = the A CXY, being on the same base XY and between the same parallels XY, BC; I . 37. and AXY is another triangle; . -. the A BXY : the A A X Y : : the A C X Y : the AAXY. v. 4. But the A BXY : the A A X Y : : BX : XA, and the A C X Y : the A A X Y : : CY : YA, . . BX : XA : : CY : YA. n. 1. v. 1. Conversely, let BX : XA : : CY : YA, and let XY be joined: then shall XY be par1 to BC. As before, join BY, CX. By hypothesis BX : XA : : CY : YA; but BX : XA : : the A BXY : the AAXY, vi. 1. and CY : YA : : the A C X Y : the A A X Y ; . -. the A BXY : the A AXY : : the A CXY : the A AXY. v. 1. . • . the A BXY = the A C X Y ; v. 6. and they are triangles on the same base and on the same side of it; . ' . XY is par1 to BC. I. 39. Q.K.D. BOOK VI. PROP. 2. 311 EXERCISES. 1. Shew that every quadrilateral is divided by its diagonals into torn' triangles proportional to each other. 2. If any two straight lines are cut by three parallel straight line they are cut proportionally. 3. From a point E in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, B D at F, G : shew that FG is parallel to CD. 4. In a triangle ABC the straight line DEF meets the sides BC, CA, A B at the points D, E, F respectively, and i t makes equal angles with A B and7\C: prove that B D : C D : : BF : CE. 5. If the bisector of the angle B of a triangle ABC meets AD at right angles, shew that a line through D parallel to B C will bisect AC. 6. From B and C, the extremities of the base of a triangle ABC, lines BE, C F are drawn to the opposite sides so as to intersect on the median from A: shew that EF is parallel to BC. 7. From P, a given point in the side AB of « triangle ABC, draw a straight line to A C produced, so that i t - will be bisected by BC. 8. Find a point within a triangle such that, if straight lines be drawn from it to the three angular points, the triangle will be divided into three equal triangles. 21-312 EUCLID'S ELEMENTS. Proposition 3. Theorem. If the vertical angle of a triangle be bisected by a strai line which cuts the base, the segments of the base shall have t o one another the same ratio as the remaining sides of t l i e triangle: Conversely, if the base be divided so that its segments have to one another the same ratio as the remaining sides of the triangle have, the straiglU line drawn from the vertex t o the point of section shall b i s e c t the vertical angle. In the A ABC let the a BAC be bisected by AX, which meets the base at X; then shall BX : XC : : BA : AC. Through C draw CE par1 to XA, to meet BA produced at E. i . 31. Then because XA and CE are par1, . " . the a BAX = the int. opp. a AEC, i . 29. and the A XAC = the alt. a ACE. 1.29. But the A BAX = the a XAC; Hyp. : . the a AEC = the a ACE; AC = AE. I . 6 . Again, because XA i s par1 to CE, a side of the A BCE, . ' . BX : XC : : BA : AE; vi. 2. that is, BX : XC : : BA : AC. BOOK VI. PROP. 3. 3i3 Conversely, let BX : XC :: BA : AC; and let AX be joined: then shall the A BAX = a XAC. For, with the same construction as before, because XA is par1 to CE, a side of the A BCE, . " . BX : XC : : BA : AE. VI. 2. But by hypothesis BX : X C : : BA : AC; . ' . BA : AE : : BA : AC; v. 1. . ' . AE = AC; . " . the a ACE = the a AEC. i . 5. But because XA is par1 to CE, . " . the A XAC = the alt. a ACE. I . 29. and the ext. a BAX = the int. opp. a AEC; I . 29. . ' . the a BAX = the A XAC. Q.E.D. EXERCISES. 1. The side BC of a triangle ABC is bisected at D, and the ADB, ADC are bisected by the s t r a i g h t l i n e s DE, DF, meeting AB, AC at E, F respectively: shew that EF i s p a r a l l e l to BC. 2. Apply Proposition 3 to trisect a given finite straight line. 3. If the line bisecting the vertical angle of a triangle be divided into parts which are to one another as the base to the sum of the sides, the point of division is the centre of the inscribed circle. 4. A B C D is a quadrilateral: shew that if the bisectors of the angles A and C meet in the diagonal B D, the bisectors of the angles B and D will meet on A C . 5. Construct a triangle having given the base, the vertical angle, and the ratio of the remaining sides. 6. Employ this proposition to shew that the bisectors of the angles of a triangle are concurrent. 7. AB is a diameter of a circle, CD is a chord at right angles to it, and E any point in C D : A E and B E are drawn and produced to out the circle in F and G : Bhew that the quadrilateral C F D G has any two of its adjacent sides in the same ratio as the remaining two. 314 EUCLID'S ELEMENTS. Proposition A. Theorem. If one side of a triangle be produced, and the exterior angle so formed be bisected by a straight l i r i e which cuts the base produced, the segments betvjeen the bisector and the extremities of the base shall have to one another the same ratio as the remaining sides of the triangle have: Conversely, if the segments of the base prod/weed have t o one another the same ratio as the remaining sides of the t r i -angle have, the straight line drawn from the vertex to the point of section shall b i s e c t the exterior vertical angle. In the A ABC let BA be produced to F, and let the exterior A CAF be bisected by AX which meets the base produced at X: then shall BX : XC : : BA : AC. Through C draw CE par1 to XA, I. 31. and let CE meet BA at E. Then because AX and CE are par1, . ' . the ext. a.FAX = the int. opp. a AEC, and the a XAC = the alt a ACE. i . 20. But the a FAX = the L.XAC; Hyp. : . the AAEC = the a ACE; . ' . AC = AE. I . 6. Again, because XA i s par1 to CE, a side of the A BCE, Constr. . ' . BX : XC : : BA : AE; vi. 2. t l i a t is, BX : XC : : BA : AC. BOOK VI. PROP. A. 315 Conversely, let BX : XC :: BA : AC, and let AX be joined: then shall the A FAX = the A XAC. For, with the same construction as before, because AX is par1 to CE, a side of the A BCE, . ' . BX : XC : : BA : AE. vi. 2. But by hypothesis BX : XC : : BA : AC; . -. BA : AE : : BA : AC; v. 1. . ' . AE = AC, .•.the A ACE = the A AEC. 1.5. But because AX is par1 to CE, . ' . the / . XAC = the alt. A ACE, and the ext. a FAX = the int. opp. A AEC; I . 29. . " . the AFAX=the a XAC. Q.e.d. Propositions 3 and A may be both included in one enunciation as follows: If the interior or exterior vertical angle of a triangle be bisected by a straight line which also cuts t l i e base, t l i e base shall be divided internally or externally into segments which have the same ratio as the sides of the triangle : Conversely, if the base be divided internally or externally into seg-ments which have the same ratio as the sides of the triangle, the straight line drawn from the point of division to the vertex will bisect the interior or exterior vertical angle. EXERCISES. 1. In the circumference of a circle of which AB is a diameter, a point P is taken; straight lines PC, P D are drawn equally inclined to A P and on opposite sides of it, meeting A B in C and D ; shew that A C : C B : : A D : DB. 2. From a point A straight lines are drawn making the angles BAC, C A D , D A E , each equal to half a right angle, and they are cut by a straight line B C D E , which makes B A E an isosceles triangle: shew that B C or D E is a mean proportional between B E and C D . 3. By means of Propositions 3 and A, prove that the straight lines' bisecting one angle of a triangle internally, and the other two externally, are concurrent. 3 1 6 EUCLID 8 ELEMENTS. Proposition 4. Theorem. If two triangles be equiangular to one anotlier, the sides about the equal angles shall be proportionals, those which are opposite t o equal angles being Iwmologous. B C E Let the A ABC be equiangular to the A DCE, having the A ABC equal to the a DCE, the A BCA equal to the a CED, and consequently the a CAB equal to the a EDC: I . 32. then shall the sides about these equal angles be propor-tionals, namely AB BC and AB BC : CA : AC : : DC : CE : DC CE ED, DE Let the A DCE be placed so that i t s side CE may be contiguous to BC, and in the same straight line with it Then because the l . s ABC, ACB are together less than two rt. angles, I . 17. and the a ACB = the l . DEC; Hyp. . ' . the A- ABC, DEC are together less than two rt. angles; . ' . BA and ED will meet i f produced. Ax. 12. Let them be produced and meet at F. Then because the a ABC = t l i e a DCE, . " . BF i s par' to CD; and because t l i e a ACB = the . ' . AC i s par1 to FE, . ' . FACD is a par"; . ' . AF = CD, and AC ^ ~ FD. DEC, Hyp. i . 28. Hyp. i . 28. i . 34. BOOK VI. PROP. 4. 317 Again, because CD is par1 to BF, a side of the A EBF, . " . BC : CE : : FD : DE; vi. 2. but FD= AC; . " . BC : CE : : AC DE; and, alternately, BC : CA : : CE . ED. v. 11. Again, because AC is par1 to FE, a side of the A FBE, vi. 2. v. 11. Proved. v. 14. Q. E. D. . ' . BA : AF : but AF = . ' . BA : CD : and, alternately, AB : BC : Also BC : CA : . ' . , ex cequali, AB : AC : : BC : = CD; : BC : : DC : : CE : : DC : CE; CE; : CE. : ED; : DE. [For Alternative Proof see Page 320.] EXERCISES. 1. If one of the parallel sides of a trapezium is double the other, shew that the diagonals intersect one another at a point of t r i s e c t i o n . 2. In the side AC of a triangle ABC any point D i s taken : shew that i f AD, DC, AB, BC are bisected in E, F, G, H respectively, then EG i s equal to HF. 3. AB and C D are two parallel straight lines; E i s the middle point of C D ; AC and BE- meet at F, and AE and BD meet at G : shew that FG i s parallel to AB. 4. ABODE is a regular pentagon, and AD and BE intersect in F : shew that AF : AE : : AE : AD. 5. In the figure of i. 43 shew that EH and GF are parallel, an that FH and G E will meet on CA produced. 6 . Chords AB and C D of a c i r c l e are produced towards B and D respectively to meet in the point E, and through E, the line EF i s drawn parallel to AD to meet CB produced in F. Prove that EF i s a mean proportional between FB and FC. 318 e u c l i d ' s elements. Proposition 5. Theorem. If t h e s i d e s of two t r i a n g l e s , t a k e n in order about each of t h e i r a n g l e s , b e proportionals, t l i e t r i a n g l e s s h a l l be e q u i -angular t o one another, having t h o s e angles equal v j l i i c h a r e o p p o s i t e t o t h e homologous s i d e s , Let the A" ABC, DEF have t h e i r s i d e s proportionals, s o that AB : BC : : DE : EF, BC : CA : : EF : FD, and consequently, ex c e q u a l i , AB : CA : : DE : FD. Then s h a l l the t r i a n g l e s be equiangular. At E in FE make the A FEG equal t o the a ABC; and at F in EF make the A EFG equal to the a BCA; i . 23. then the remaining a EG F = the remaining ABAC. 1.32. . ' . the A GEF i s equiangular to the A ABC; . ' . GE : EF : : AB : BC. v i . 4 . But AB : BC : : DE : EF; Hyp. : . GE : EF : : DE : EF; v . 1 . . ' . GE= DE. S i m i l a r l y GF= DF. Then in the t r i a n g l e s GEF, DEF I GE=DE, Because < GF = DF, (and EF i s common; . " . the a GEF = the _ DEF, i . 8 . and the a GFE = the a DFE, and the A EGF = the A EDF. But the a GEF = the _ABC; Constr. . ' . the a DEF = the a ABC. S i m i l a r l y , tho a EFD = the a BCA, BOOK VI. PROP. 6 . 319 .'. the remaining a FDE = the remaining a CAB; I. 32. that is, the A DEF is equiangular to the A ABC. Q. E.D. Proposition 6. Theorem. If two triangles have one angle of tlie one equal to one angle of t l i e other, and the sides about the equal angles pro-portionals, the triangles shall be similar. In the As ABC, DEF let the A BAC = the a EDF, and let BA : AC,:: ED : DF. Then shall the As ABC, DEF be similar. At D in FD make the A FDG equal to one of the Aa EDF, BAC: at F in DF make the A DFG equal to the a ACB; l . 23. . ' . the remaining a FGD = the remaining a ABC. i . 32. Then the A ABC i s equiangular to the A DGF; / . BA But BA . • - GD - . AC : : GD AC : : ED : DF : : ED GD = ED. DF. DF; DF, vi. 4 Hyp Then in the As GDF, EDF, G D = ED, Because -] and DF is common; [and the A GDF = the A EDF; Constr. the A3 GDF, EDF are equal in all respects, I . 4. so that the A EDF is equiangular to the A GDF; but the A GDF is equiangular to the ABAC; Constr. . ' . the A EDF is equiangular to the A BAC; . ' . their sides about the equal angles are proportionals, vi. 4. that is, the As ABC, DEF are similar. Q . e . d. 320 EUCLID S ELEMENTS. Note 1. From Definition 2 i t i s seen that two conditions are necessary for similarity of rectilineal figures, namely (1) the figures must be equiangular, and (2) the sides about the equal angles must be proportionals. In the ease of triangles we learn from Props. 4 and 5 that each of these conditions follows from the other: this how-ever i s not necessarily the case with rectilineal figures of mere than three sides. Notb 2. We have given Euclid's demonstrations of Propositions 4, 6, 6; but these propositions also admit of easy proof by the method of superposition. Aa an illustration, we will apply this method to Proposition 4. Proposition 4. [Alternative Proof.] If two triangles be equiangular to one another, the sides about equal angles shall be proportionals, those sides which are opposite t o equal angles being homologous. Let the A ABC be equiangular to the a DEF, having the L ABC equal to the L DEF, the / BCA equal to the L EFD, and conse-quently the Z CAB equal to the L FDE: x . 32. then shall the sides about these equal angles be proportionals. Apply the A ABC to the A DEF, so that B f a l l s on E and BA along ED: then BC will f a l l along EF, since the i ABC=the L DEF. Hyp. Let G and H be the points in ED and EF, on which A and C f a l l . Join GH. Then because the / EGH = the L EDF, Hyp. : . G H i s par1 to DF: . -. DG : GE:: FH : HE; . ' . , componendo, DE : G E : : FE : HE, v. 13. . -. , alternately, DE : FE : : G E : HE. , . 11. that is, DE : EF : : AB : BC. Similarly by applying the A ABC to the A DEF, so that the point C may f a l l on F, i t may be proved that EF : FD : : BC : CA. . -. , ex cequali, DE : DF : : AB : AC. BOOK VI. PROP, 7 . 3 2 1 Proposition 7. Theorem. If two triangles have one angle of t l i e one equal to one angle of the other and the sides about one other angle in each proportional, so that the sides opposite to the equal angles are homologous, then the third angles are either equal or sup-plementary ; and in the former case the triangles are similar. Let ABC, DEF be two triangles having the a ABC equal to the a DEF, and the sides about the angles at A and D pro-portional, so that BA : AC : : ED : DF; then shall the as ACB, DFE be either equal or supple-mentary, and in the former case the triangles shall be similar. If the ^ BAC = the a EDF, then the a BCA = the A EFD; I . 32. and the As are equiangular and therefore similar. VI. 4, But i f the a BAC is not equal to the A EDF, one of them must be the greater. Let the A EDF be greater than the A BAC. At D in ED make the a EDF' equal to the a BAC. I . 23. Then the As BAC, EDF' are equiangular, . -. BA : AC : : ED : DF'; but BA : AC : : ED : DF; . -. ED : DF : : ED : DF', . -. DF = DF', . ' . the ADFF' = the A DF'F. But the a a DF'F, DF'E are supplementary, . ' . the As DFF', DF'E are supplementary: at is, the ls DFE, A C B are supplementary. Constr. vi. 4. Hyp. v. 1. I. 5. i . 13. Q.E.D. 322 euclid's elements. Corollaries to Proposition 7. A Three cases of this theorem deserve special attention. I t has been proved that i f the angles ACB, DFE are not supple-mentary, they are equal: and we know that of angles which are supplementary and unequal, one must be acute and the other obtuse. Hence, in addition to the hypothesis of this theorem, ( i ) If the angles ACB, DFE, opposite to the two homologous sides AB, DE are both acute, both obtuse, or i f one of them i s a right angle, i t follows that these angles are equal; and therefore the triangles are similar. ( i i ) If the two given angles are right angles or obtuse angles, i t follows that the angles ACB, DFE must be both acute, and therefore equal, by (i): so that the triangles are similar. ( i i i ) If in each triangle the side opposite the given angle i s not less than the other given side; that is, i f A C and DF are not l e s s than AB and DE respectively, then the angles ACB, DFE cannot be greater than the angles ABC, DEF, respectively; therefore the angles ACB, DFE, are both acute; hence, as above, they are equal; and the triangles ABC, DEF similar. BOOK VI. PROP. 7. 323 EXERCISES. on Propositions 1 to 7. 1. Shew that the diagonals of a trapezium cut one another in the same ratio. 2. If three straight lines drawn from a point cut two parallel straight lines in A, B, C and P, Q, R respectively, prove that A B : B C : : P Q : QR. 3. From a point O, a tangent OP is drawn to a given circle, and O Q R is drawn cutting i t in Q and R; shew that OQ: OP : : OP ; OR. 4. If two triangles are on equal bases and between the same paralle any straight line parallel to their bases will cut off equal areas from the two triangles. 5. If two straight lines PQ, XY intersect in a point O, so that PO : O X : : Y O : O Q , prove that P, X, Q, Y are concyclic. 6. On the same base and on the same side of it two equal triangles ACB, A D B are described; A C and BD intersect in O, and through O lines parallel to DA and C B are drawn meeting the base in E and F. Shew that A E = BF. 7. BD, CD are perpendicular to the sides AB, AC of a triangle ABC, and C E is drawn perpendicular to AD, meeting AB in E : shew that the triangles ABC, A C E are similar. 8. AC and BD are drawn perpendicular to a given straight line CD from two given points A and B; A D and B C intersect in E, and EF is perpendicular to C D ; shew that AF and BF make equal angles with CD. 9. ABCD is a parallelogram; P and Q are points in a straight line parallel to A B ; PA and Q B meet at R, and P D and Q C meet at S: shew that R S is parallel to AD. 10. In the sides AB, AC of a triangle ABC two points D, E are taken such that B D is equal to C E ; i f DE, B C produced meet at F, shew that A B : A C : : EF : DF. 11. Find a point the perpendiculars from which on the sides of a given triangle shall be in a given ratio. 324 E U C L I D ' S ELEMENTS. Proposition 8 . Theorem. In a right-angled triangle if a perpendicular be drawn from the right angle to the hypotenuse, t l i e triangles on each side of it are similar to the whole triangle and to one another. A B D Let ABC be a triangle right-angled at A, and let AD be perp. to BC: then shall the As DBA, DAC be similar to the A ABC and to one another. In the A8 DBA, ABC, the a BDA = the a BAC, being rt. angles, and the a ABC is common to both; . ' . the remaining a BAD = the remaining _ BCA, I . 32. that is, the A 8 DBA, ABC are equiangular; . ' . they are similar. vi. 4. In the same way it may be proved that the A DAC, ABC are similar. Hence the A3 DBA, DAC, being equiangular to the same A ABC, are equiangular to one another; . ' . they are similar. vi. 4. Q. E. D. Corollary. Because the A8 BDA, A D C are similar, . ' . BD : DA : : DA : DC; and because the A8 CBA, ABD are similar. . " . CB : BA : : BA : BD; and because the A8 BCA, ACD are similar, . ' . BC : CA : : CA : CD. EXERCISES. 1. Prove that the hypotenuse i s to one s i d e as the second side i s t o the perpendioular. 2. Sliew that the radius of a c i r c l e i s a mean proportional between the segments of any tangent between i t s point of contact and a pair of parallel tangents. BOOK VI. PROP. 9 . 325 Definition. A less magnitude is said to be a sub-multiple of a greater, when the less i s contained an exact number of times in the greater. [Book v. Def. 2.] Proposition 9. Problem. From a given straight line to cut off any required sub-multiple. Q, E^ A F Let AB be the given straight line. It is required to cut off a certain submultiple from AB. From A draw a straight line AG of indefinite length making any angle with AB. In AG take any point D; and, by cutting off successive parts each equal to AD, make AE to contain AD as many times as AB contains the required submultiple. Join EB. Through D draw DF par1 to EB, meeting AB in F. Then shall AF be the required submultiple. Because DF is par1 to EB, a side of the A AEB, . " . BF : FA : : ED : DA; vi. 2. . ' . , componendo, BA : AF : : EA : AD. v. 13. But AE contains A D the required number of times; Constr. . ' . AB contains AF the required number of times; that is, AF is the required submultiple. q.e.f. exercises. 1. Divide a straight line into five equal parts. 2. Give a geometrical construction for cutting o f f two-sevenths of a given straight line. H. E. 22 326 kuclid's elements. Proposition 10. Problem. To divide a straight line similarly to a given divided straight line. A f/N\| , E Let AB be the given straight line to be divided, and AC the given straight line divided at the points D and E. It is required to divide AB similarly to AC. Let AB, AC be placed so as to form any angle. Join CB. Through D draw DF par1 to CB, I . 31. and through E draw EG par1 to CB, and through D draw DHK par' to AB. Then AB shall be-divided at F and G similarly to AC. For by construction each of the figs. FH, HB is a par"; . ' . DH = FG, and HK = GB. 1.34. N o w since HE is par1 to KC, a side of the A DKC, . ' . KH : HD : : CE : ED. vi. 2. But KH = BG, and HD = GF; . ' . BG : GF : : CE : ED. v. 1. Again, because FD is par1 to GE, a side of t l i e A AGE, . ' . GF : FA : : ED : DA, vi. 2. and i t has been shewn that BG : GF : : CE : ED, . ' . , ex cequali, BG : FA : : CE : DA : v. 14. . ' . AB i s divided similarly to AC. q . E.F. exercise. Divide a straight line internally and externally in a given ratio. Is this always possible > book vi. prop. 11. 327 Proposition 1.1. Problem. To find a third proportional to two given straight lines. K BA D Let A, B be two given straight lines. It is required to find a third proportional to A and B. Take two st. lines DL, DK of indefinite length, containing any angle: from DL cut off D G equal to A, and G E equal to B; and from DK cut off DH equal to B. I . 3. Join GH. Through E draw EF par1 to GH, meeting DK in F. I . 31. Then shall HF be a third proportional to A and B. Because GH is par1 to EF, a side of the A DEF; . " . D G : G E : : D H : HF. VI. 2. But DG = A ; and GE, D H each= B; Constr. ;. A : B : : B : H F ; that is, HF is a third proportional to A and B. Q. E. F. exercises. 1. AB i s a diameter of a c i r c l e , and through A any straight line i s drawn to cut the circumference in C and the tangent at B in D : shew that A C i s a third proportional to A D and AB. 2. A B C i s an isosceles triangle having each of the angles at the base double of the vertical angle BAC ; the bisector of the angle BCA meets AB at D. Shew that AB, BC, BD are three proportionals. 3. Two c i r c l e s intersect at A and B; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D: shew that i f CB, BD are joined, BD i s a third proportional to CB, BA. 22—2 3 2 8 e u c l i d ' s elements. Proposition 12. Problem. To find a fourth proportional to three given straight line r> A B C Let A, B, C be the three given straight lines. It is required to find a fourth proportional to A, B, C. Take two straight lines DL, DK containing any angle: from DL cut off DG equal to A, G E equal to B; and from DK cut off DH equal to C. I . 3. Join GH. Through E draw EF par1 to GH. L 31. Then shall HF be a fourth proportional to A, B, C. Because GH is par1 to EF, a side of the A DEF; . " . DG : GE : : DH : HF. VI. 2. But DG = A, G E = B, and DH = C; Constr. :. A : B : : C : HF; that is, HF i s a fourth proportional to A, B, C. Q. E. F. EXERCISES. 1. If from D, one of the angular points of a parallelogram A B C D , a straight line is drawn meeting A B at E and C B at F ; shew that C F is a fourth proportional to EA, AD, and AB. 2. In a triangle ABC the bisector of the vertical angle BAC meets the base at D and the oiroumference of the circumscribed circle at E: shew that BA, AD, EA, A C are four proportionals. 3. From a point P tangents PQ, PR are drawn to a circle whose centre is C, and Q T is drawn perpendicular to R C produced: shew that Q T is a fourth proportional to PR, RC, and RT. book vi. prop. 13. 329 Proposition 13. Problem. To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines. It i s required to find a mean proportional between them. Place AB, BC in a straight line, and on AC describe the semicircle ADC. From B draw BD at rt. angles to AC. I . 11. Then shall BD be a mean proportional between AB and BC. Join AD, DC. Now the A ADC being in a semicircle is a rt. angle; in. 3 and because in the right-angled A ADC, DB i s drawn from the rt. angle perp. to the hypotenuse, . ' . the As ABD, DBC are similar; vi. 8. /. AB : BD : : BD : BC; that is, BD i s a mean proportional between AB and BC. Q. E. F. EXERCISES. 1. If from one angle A of a parallelogram a straight line be drawn cutting the diagonal in E and the sides in P, Q, shew that A E is a mean proportional between PE and EQ. 2. A, B, C are three points in order in a straight line: find a point P in the straight line so that PB may be a mean proportional between PA and PG. 3. The diameter AB of a semicircle is divided at any point C, and C D is drawn at right angles to A B meeting the circumference in D ; D O is drawn to the centre, and C E is perpendicular to O D : shew that D E is a third proportional to A O and DC. 330 EUCLID'S ELEMENTS. 4. AC is the diameter of a semicircle on which a point B is taken bo that B C is equal to the radius: shew that A B is a mean propor-tional between B C and the sum of BC, CA. 5. A is any point in a semicircle on BC as diameter; from D any point in B C a perpendicular is drawn meeting AB, AC, and the cir-cumference in E, G, F respectively; shew that D G is a third propor-tional to D E and DF. 6. Two circles touch externally, and a common tangent touches them at A and B: prove that A B is a mean proportional between the diameters of the circles. [See Ex. 21, p. 219.] 7. If a straight line be divided in two given points, determine u . third point such that its distances from the extremities may be proportional to its distances from the given points. 8. AB is a straight line divided at C and D so that AB, AC, AD are in continued proportion; from A a line A E is drawn in any direc-tion and equal to A C ; shew that B C and C D subtend equal angles at E. 9. In a given triangle draw a straight line parallel to one of the sides, so that i t may be a mean proportional between the segments of the base. 10. On the radius OA of a quadrant OAB, a semicircle ODA is described, and at A a tangent A E is drawn; from O any line O D F E i s drawn meeting the circumferences in D and F and the tangent in E: i f D G is drawn perpendicular to OA, shew that OE, OF, O D , and O G are in continued proportion. 11. From any point A, in the circumference of the circle ABE, as centre, and with any radius, a circle B D C is described cutting the former circle in B and C ; from A any line A F E i s drawn meeting the chord B C in F, and the circumferences BDC, A B E in D, E respec-tively: shew that A D is a mean proportional between AF and AE. Definition. T w o figures are said to have their sides about two of their angles reciprocally proportional, when a side of the first is to a side of the second as tlie remaining side of the second is to the remaining side of the first. [Hook vi. Def. 3.] book v i . prop. 1 4 . 331 Proposition 14. Theorem. Parallelograms which are equal in area, and which have one angle of the one equal to one angle of t l i e oilier, have their sides about the equal angles reciprocally proportional : Conversely, parallelograms which have one angle of the one equal to one angle of the other, and the sides about t l i e s e angles reciprocally proportional, are equal in area. A F / Let the par™3 AB, BC be of equal area, and have the A DBF equal to the A GBE: then shall the sides about these equal angles be reciprocally proportional, that is, DB : BE : : G B : BF. Place the par™ so that DB, BE may be in the same straight line; . ' . FB, BG are also in one straight line. I . 14. Complete the par™ FE. Then because the par™ AB = the par™ BC, Hyp. and FE is another par™, . " . the par™ AB : the par™ FE : : the par™ BC : the par"' FE; but the par™ AB : the par™ FE : : DB : BE, vi. 1. and the par™ BC : the par™ FE : : G B : BF, . " . DB : BE : : G B : BF. V. 1. Conversely, let the a DBF be equal to the a GBE, and let DB : BE : : G B : BF. Then shall the par™ AB be equal in area to the par™ BC. For, with the same construction as before, by hypothesis DB : BE : : G B : BF; but DB : BE : : the par"' AB : the par™ FE, VI. 1, and G B BF : : the par™ BC : the par™ FE, . ' . the par™ AB : the par™ FE : : the par™ BC : the par™ FE; v. 1. . ' . the par™ AB =the par™ BC. (J.E. D. 332 Euclid's elements. Proposition 15. Theorem. Tria/ngles which are equal in area, and which liave one angle of the one equal t o one angle of t l i e o i l i e r , have their sides about the equal angles reciprocally proportional: Conversely, triangles which have one angle of t l i e one equal t o one angle of the other, and t l i e sides about t h e s e angles reciprocally proportional, are equal in area. Let the A8 ABC, ADE be of equal area, and have the A CAB equal to the A EAD: then shall the sides of the triangles about these angles be reciprocally proportional, that is, CA : AD : : EA : AB. Place the As so that CA and AD may be in the same st. line; . ' . BA, AE are also in one st. line. I . 14. Join BD. Then because the A CAB = the A EAD, Hyp. and ABD i s another triangle; . ' . the A C A B : the A A B D : : the A E A D : the AABD; but the A C A B : the A A B D : : CA : AD, vi. 1. and the A E A D : the A A B D : : EA : AB, . " . CA : AD : : EA : AB. v. 1. Conversely, let the a CAB be equal to the . _ EAD, and let CA : AD : : EA : AB. Then shall the A CAB = A EAD. For, with the same construction as before, by hypothesis CA : AD : : EA : AB; but CA : AD : : the A C A B : the AABD, VI. 1. and EA : AB : : the A E A D • the AABD, . ' . the A C A B : the A A B D : : the A E A D : the A A B D ; v. 1. ' . " . the A CAB-tlie AEAD. q . e.d. BOOK VI. PROP. 15. 333 EXERCISES. on Pkopositions 14 and 15. 1. Parallelograms which are equal in area and which have their sides reciprocally proportional, have their angles respectively equal. 2. Triangles which are equal in area, and which have the sides about a pair of angles reciprocally proportional, have those angles equal or supplementary. 3. AC, BD are the diagonals of a trapezium which intersect in 0 ; if the side A B is parallel to C D , use Prop. 15 to prove that the triangle A O D is equal to the triangle B O C . 4. From the extremities A, B of the hypotenuse of a right-angled triangle A B C lines AE, B D are drawn perpendicular to AB, and meeting B C and A C produced in E and D respectively: employ Prop. 15 to shew that the triangles A B C , E C D are equal in area. 5. On AB, AC, two sides of any triangle, squares are described externally to the triangle. If the squares are A B D E , ACFG, shew that the triangles DAG, F A E are equal in area. 6. ABCD is a parallelogram; from A and C any two parallel straight lines are drawn meeting D C and A B in E and F respectively; EG, which is parallel to the diagonal AC, meets A D in G : shew that the triangles DAF, G A B are equal in area. 7. Describe an isosceles triangle equal in area to a given triangle and having its vertical angle equal to one of the angles of the given triangle. 8. Prove that the equilateral triangle described on the hypotenuse of a right-angled triangle is equal to the sum of the equilateral triangles described on the sides containing the right angle. [Let ABC be the triangle right-angled at C ; and let BXC, CYA, A Z B be the equilateral triangles. Draw C D perpendicular to A B ; and join D Z . Then shew by Prop. 15 that the A AYC=the A D A Z ; and similarly that the A B X C = the A BDZ.] 3 3 4 E u c l i d ' s e l e m e n t s . Proposition 16. Theorem. If four straight lines a/re proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means: Conversely, if the rectangle contained, by the extremes i s equal to the rectangle contained by the mea/ns, the four straight lines are proportional. B D E G Let the st. lines AB, CD, EF, G H be proportional, so that AB : CD : : EF : GH. Then shall the rect. AB, G H =the rect. CD, EF. From A draw AK perp. to AB, and equal to GH. I . 11, 3. From C draw CL perp. to CD, and equal to EF. Complete the par™' KB, LD. Then because AB : CD : : EF : G H ; Hyp. and EF = CL, and GH = AK; Constr. : . AB : CD : : CL : AK; that is, the sides about equal angles of par"" KB, LD are reciprocally proportional; . ' . KB = LD. VI. 14. But KB is the rect. AB, GH, for AK = GH, Constr. and LD i s the rect. CD, EF, for C L = EF; . ' . the rect. AB, GH = the rect. CD, EF. Conversely, let the rect. AB, GH = the reet. CD, EF: then shall AB : CD : : EF : GH. For, with the same construction as before, because the rect. AB, GH=the reet. CD, EF; Hyp. and the rect. AB, GH • - KB, for G H = AK, Constr. and t l i e reet, CD, EF = LD, for EF = CL ; . ' . KB = LD: BOOK VI. PROP. 17. 3 3 5 that is, the par™3 KB, LD, which have the angle at A equal to the angle at C, are equal in area; . ' . the sides about the equal angles are reciprocally proportional: that is, AB : CD : : CL : AK; . " . AB : CD : ; EF : GH. Q. E. D. Proposition 17. Theorem. If three straight lines are proportional the rectangle con-tained by the extremes is equal to the square on the mean: Conversely, if the rectangle contained by the. extremes is equal to the square on the mean, t l i e three straight lines are proportional. -C -D -B -A B VI. D = B; Let the three st. lines A, B, C be proportional, so that A : B : : B : C. Then shall the rect. A, C be equal to the sq. on B. Take D equal to B. Then because A : B : : B : C, and D = B; . -. A : B : : D : C; . ' . the rect. A, C = the rect. B, D; but the rect. B, D = the sq. on B, for . ' . the rect. A, C =the sq. on B. Conversely, let the rect. A, C = the sq. on B: then shall A : B : : B : C. For, with the same construction as before, because the rect. A, C = the sq. on B, and the sq. on B = the rect. B, D . ' . the rect. A, C = the rect. B, D . " . A : B : : D : C, that is, A : B : : B : C. 16. Hyp-for D = B; vi. 16. Q. E. D. 336 EUCLID'S ELEMENTS. EXERCISES. on Propositions 16 and 17. 1. Apply Proposition 16 to prove that if two chords of a circle intersect, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. 2. Prove that the rectangle contained by the sides of a right-angled triangle is equal to the rectangle contained by the hypotenuse and the perpendicular on it from the right angle. 3. On a given straight line construct a rectangle equal to a given rectangle. 4. ABCD is a parallelogram; from B any straight line is drawn cutting the diagonal A C at F, the side D C at G, and the side A D pro-duced at E: shew that the rectangle EF, F G is equal to the square on BF. 5. On a given straight line as base describe an isosceles triangle equal to a given triangle. 6. AB is a diameter of a circle, and any line ACD cuts the circle in C and the tangent at B in D ; shew by Prop. 17 that the rectangle AC, A D is constant. 7. The exterior angle A of a triangle ABC is bisected by a straight line which meets the base in D and the circumscribed circle in E: shew that the rectangle BA, A C is equal to the rectangle EA, AD. 8. If two chords AB, AC drawn from any point A in the cir-cumference of the circle A B C be produced to meet the tangent at the other extremity of the diameter through A in D and E, shew that the triangle A E D is similar to the triangle A B C . 9. At the extremities of a diameter of a circle tangents are drawn; these meet the tangent at a point P in Q and R : shew that the rect-angle QP, PR is constant for all positions of P. 10. A is the vertex of an isosceles triangle ABC inscribed in a circle, and A D E is a straight line which cuts the base in D and the circle in E; shew that the rectangle EA, A D is equal to the square on AB. 11. Two circles touch ono another externally in A; a straight line touches the circles at B and C, and is produced to meet the straight line joining their centres at S: shew that the rectangle SB, S C is equal to the square on SA. 12. Divide a triangle into two equal parts by a straight line at right angles to one uf the sides. BOOK VI. PROP. 18. 337 Definition. Two similar rectilineal figures are said to be similarly situated with respect to two of their sides when these sides are homologous. [Book vi. Def. 5 . 1 Proposition 18. Problem. On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. Let AB be the given st. line, and CDEF the given rectil. figure: first suppose CDEF to be a quadrilateral. It is required to describe on the st. line AB, a rectil. figure similar and similarly situated to CDEF. Join DF. At A in BA make the a BAG equal to the a DCF, I . 23. and at B in AB make the a ABG equal to the a CDF; . . the remaining a AGB = the remaining a CFD; I . 32. and the A A G B is equiangular to the A CFD. Again at B in GB make the a GBH equal to the a FDE, and at G in BG make the a BGH equal to the A DFE; I . 23. . ' . the remaining a BHG =the remaining a DEF; I . 32. and the A BHG is equiangular to the A DEF. Then shall ABHG be the required figure. (i) To prove that the quadrilaterals are equiangular. Because the A AG B = the a CFD, and the a BGH = the A DFE; Constr. . ' . the whole a AGH = the whole A CFE. Ax. 2. Similarly the A ABH = the a C D E ; and the angles at A and H are respectively equal to the angles at C and E ; Constr. . ' . the fig. ABHG is equiangular to the fig. CDEF. 338 Euclid's elements. ( i i ) To prove that the quadrilaterals have the sides about their equal angles proportional. Because the A8 BAG, DCF are equiangular; . ' . AG : GB : : CF : FD. VI. 4. And because the A8 BGH, DFE are equiangular; . " . BG : GH : : DF : FE, . ' . , ex cequali, AG GH : : CF : FE. v. 14. Similarly i t may be shewn that AB : BH : : CD : DE. Also BA : AG : : DC : CF, vi. 4. and GH : HB : : FE : ED; . ' . the figs. ABHG, CDEF have their sides about the equal angles proportional; . " . ABHG i s similar to CDEF. Def. 2. In like manner the process of construction may be extended to a figure of five or more sides. Q.E.F. Definition. W h e n three magnitudes are proportionals the first i s said to have to the third the duplicate ratio of that which i t has to the second. [Book v. Def. 13.] book vi. prop. 19. 339 Proposition 19. Theorem. Similar triangles are to one another in the duplicate rat of their homologous sides. D Let ABC, DEF be similar triangles, having the ^ABC equal to the a DEF, and let BC and EF be homologous sides: then shall the A A B C be to the A DEF in the duplicate ratio of BC to EF. To BC and EF take a third proportional BG, so that BC : EF : : EF : BG. vi. 11. Join AG. Then because the A° ABC, DEF are similar, Hyp. . ' . AB : BC : : DE : EF; . ' . , alternately, AB : DE : : BC : EF; v. 11. but BC : EF : : EF : BG; Constr. . ' . AB : DE : : EF : BG; v. 1. that is, the sides of the As ABG, DEF about the equal angles at B and E are reciprocally proportional; . ' . the A ABG = the A DEF. vi. 15. Again, because BC : EF : : EF : BG, Constr. . " . BC : BG in the duplicate ratio of BC to EF. Def. But the A A B C : the A ABG : : BC : BG, vi. 1. . ' . the A A B C : the A ABG in the duplicate ratio of BC to EF: V. 1. and the A ABG = the A DEF; Proved. . ' . the A A B C : the A DEF in the duplicate ratio of BC : EF. Q-E-D-340 euclid's elements. Proposition 20. Theorem. Similar polygons may be divided into the same number of similar triangles, having the same ratio each to each that the -polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides. H Let ABODE, FGHKL be similar polygons, and let AB be the side homologous to FG; then (i) the polygons may be divided into the same number of similar triangles; ( i i ) these triangles shall have each to each the same ratio that the polygons have; ( i i i ) the polygon ABODE shall be to the polygon FGHKL in the duplicate ratio of AB to FG. Join EB, EC, LG, LH. ( i ) Then because the polygon ABCDE is similar to the polygon FGHKL, Hyp. :. the A EAB = the A LFG, and EA : AB : : LF FG ; VI. D>f. 2. . ' . the A EAB is similar to the A LFG ; vi. 6 . . -. the A ABE = the A FGL. But, because the polygons are similar, Hyp. : . the £.ABC^the a FGH, vi. Def 2. . ' . tho remaining a EBC--the remaining a LGH. And because the As ABE, FGL are similar. Pn-vcd. :. EB : BA : : LG : GF; and because the polygons are similar. Hyp. : . AB : BC = FG : GH; vi. Def 2. / . , ex t r q i t a l i , EB : BC : : LG GH, v. 14. that is, the sides about the equal a 3 EBC, LGH are proportionals; . ' . the A EBC i s similar to the A LGH. vi. 6. book vi. prop. 20. 341 In the same way it may be proved that the A ECD is similar to the ALHK. . " . the polygons have been divided into the same number of similar triangles. (ii) Again, because the A ABE is similar to the A FGL, . ' . the A ABE is to the A FGL in the duplicate ratio of EB : LG; VI. 19. and, in like manner, the A EBC is to the A LGH in the duplicate ratio of EB to LG; .'.the A A B E : the A FGL : : the A EBC : the A LGH. v. 1. In like manner it can be shewn that the A EBC : the A LGH : : the A EDC : the ALKH, . ' . the A A B E : the A F G L : : the A E B C : the A L G H : : the A EDC : the ALKH. But when any number of ratios are equal, as each ante-cedent is to its consequent so i s the sum of all the ante-cedents to the sum of all the consequents; v. 12. . ' . the A A B E : the A L F G : : the fig. ABODE : thefig. FGHKL. ( i i i ) N o w the A EAB : the A LFG in the duplicate ratio of AB : FG, and the A EAB : the A L F G : : thefig. ABODE : thefig. FGHKL; . ' . the fig. ABODE : the fig. FGHKL in the duplicate ratio of AB : FG. Q.E.D. Corollary 1. Let a third proportional X be taken to AB and FG, then AB i s to X in the duplicate ratio of AB : FG; but the f i g . ABODE : the fig. FGHKL in the duplicate ratio of AB : FG. Hence, if three straight lines are proportionals, as the first i s t o the third, so is any rectilineal figure described on the first t o a similar and similarly described rectilineal figure on the second. Corollary 2. It follows that similar rectilineal figures are t o one another as the squares on their homologous sides. For squares are similar figures and therefore are to one another in the duplicate ratio of their sides. H. E. 23 342 EUCLID'S ELEMENTS. Proposition 21. Theorem. Rectilineal figures which are similar to the same recti-lineal figure, are also similar to each other. Let each of the rectilineal figures A and B be similar to then shall A be similar to B. For because A is similar to C, Hyp. . ' . A is equiangular to C, and the sides about their equal angles are proportionals. vi. Def. 2. Again, because B is similar to C, Hyp. . ' . B i s equiangular to C, and the sides about their equal angles are proportionals. vi. Def. 2. . . A and B are each of them equiangular to C, and have the sides about the equal angles proportional to the cor-responding sides of C; . ' . A is equiangular to B, and the sides about their equal angles are proportionals; v. 1 . . ' . A is similar to B. Q.E.D. book vi. prop. 22. 343 Proposition 22. Theorem. If four straight lines be proportional and a pair of similar rectilineal figures be similarly described on the first and second, and also a pair on the third and fourth, these figures shall be proportional: Conversely, if a rectilineal figure on the first of four straight lines be to the similar and similarly described figure on the second as a rectilineal figure on the third is t o the similar and similarly described figure on the fourth, the four straight lines shall be proportional. Let AB, CD, EF, GH be proportionals, so that AB : CD : : EF : GH ; and let similar figures KAB, LCD be similarly described on AB, CD, and also let similar figs. MF, NH be similarly described on EF, GH: then shall the fig. KAB : the fig. LCD : : the fig. MF : the fig. NH. To AB and CD take a third proportional X, vi. 11. and to EF and GH take a third proportional O; then AB : CD : and EF : GH : But AB : CD : . ' . CD : X : . ' . , ex cequali, AB : X : But AB : X : : thefig. KAE and EF : O : : thefig. MF 3 % . KAB : the fig. LCD : : : CD : X, : GH : O. : EF : GH; : GH : 0, : EF : O. Constr. Hyp. v. 1. v, 14. ; : the fig. LCD, vi. 20, Cor. : thefig. NH: the fig. MF : > the fig. NH. v. 1. 23—2 344 E u c l i d ' s elements. F G H O P R Conversely, let the fig. KAB : the fig. LCD : : the fig. M F : the fig. N H ; then shall AB : C D : : EF : GH. To AB, CD, and EF take a fourth proportional PR: vi. 12. and on PR describe the fig. SR similar and similarly situated to either of the figs. MF, NH. vi. 18. Then because AB : C D : : EF : PR, Constr. . ' . , by the former part of the proposition, the fig. KAB : the fig. LCD : : the fig. M F : the fig. SR. But the fig. KAB : the fig. LCD : : the fig. M F : the fig. NH. Hyp. :. thefig. M F : the fig. SR : : the fig. M F • the fig. NH, v. 1. . ' . the fig. SR = the fig. NH. A n d since the figs. SR and N H are similar and similarly situated, /. PR=GH. Now AB : CD : : EF : PR; Constr. . \ AB : CD : : EF : GH. Q. E. D. Euclid here assumes that if two similar and similarly situate figures are equal, their homologous sides are equal. The proof i s easy and may be l e f t as an exercise for the student. Definition. When there are any number of magnitudes of the same kind, the first is said to have to the last the ratio compounded of the ratios of the first to the second, of the second to the third, and so on up to the ratio of the last but one to the last magnitude. [Book v. Def. 12.] book v i . prop. 2 3 . 3 4 5 Proposition 23. Theorem. Parallelograms which are equiangular to one another have to one another t l i e ratio which is compounded of the ratios of their sides. A V B K L M E F Let the par™ AC be equiangular to the par™ CF, having the A BCD equal to the A ECG: then shall the par™ AC have to the par™ CF the ratio com-pounded of the ratios BC : CG and DC : CE. Let the par™3 be placed so that BC and CG are in a st.line; then DC and CE are also in a st. line. I . 14. Complete the par™ DG. Take any st. line K, and to BC, CG, and K find a fourth proportional L; vi. 12. and to DC, CE, and L take a fourth proportional M; then BC : CG : : K : L, and DC ; CE : : L : M. But K : M is the ratio compounded of the ratios K : L and L : M, v. Def. 12. that is, K : M is the ratio compounded of the ratios BC : CG and DC : CE. N o w the par™ AC : the par™ CH : : BC : CG vi. 1. : : K : L, Constr. and the par™ CH : the par™ CF : : DC : CE. VI. 1. : : L : M, Constr. . ' . , ex cequali, the par™ AC : the par™ CF : : K : M. v. 14. But K : M is the ratio compounded of the ratios of the sides; . " . the par™ AC has to the par™ CF the ratio compounded of the ratios of the sides. Q.E.D. EXERCISE. The areas of two triangles or parallelograms are to one another in the r a t i o compounded of the ratios of their bases and of their a l t i t u d e s . 34g euclid's elements. Proposition 24. Theorem. Parallelograms about a diagonal of any parallelogram are similar to the whole parallelogram and to one another. A E B c / \ / / / \ D K J C Let ABCD be a par™ of which AC is a diagonal; and let EG, HK be par™ about AC: then shall the par™8 EG, HK be similar to the par™ ABCD, and to one another. For, because DC is par1 to GF, . ' . the a ADC = the a AGF; i . 29. and because BC is par1 to EF, /. the AABC = the AAEF; i . 29. and each of the a s BCD, EFG i s equal to the opp. a BAD, . ' . the A BCD = the A EFG; [ i . 34. . ' . the par™ ABCD i s equiangular to the par™ AEFG. Again in the A" BAC, EAF, because the a ABC = the a AEF, i . 29. and the a BAC i s common; . ' . As BAC, EAF are equiangular to one another; I . 32. . ' . AB : BC : : AE : EF. vi. 4. But BC = AD, and EF = AG; I . 34. . " . AB : AD : : AE : AG; and DC : CB : : GF : FE, and CD : DA : : FG : GA, . ' . the sides of the par"" ABCD, AEFG about their equal angles are proportional; . ' . the par™ ABCD is similar to the par™ AEFG. vi. Dei. 2. In the same way i t may be proved that the par™ ABCD i s similar to the par™ FHCK, . " . each of the parIM EG, HK i s similar to the whole par™; . ' . the par™ EG i s similar to the par™ HK. vi. 21. Q.E.D. book v i . p r o p . 25. 3 4 7 Proposition 25. Problem. To describe a rectilineal figure which shall be equal to one and similar to another rectilineal figure. ! Z 2 _ i Let E and S be two rectilineal figures: it is required to describe a figure equal to the fig. E and similar to the fig. S. On AB a side of the fig. S describe a par™ ABCD equal to S, and on B C describe a par™ C B G F equal to the fig. E, and having the A C B G equal to the a DAB: I. 45. then A B and BG are in one st. line, and also D C and CF in one st. line. Between AB and BG find a mean proportional HK; vi. 13. and on H K describe the fig. P, similar and similarly situated to the fig. S: vi. 18. then P shall be the figure required. Because AB : HK :: HK : BG, Constr. : . A B : BG : : the fig. S : the fig. P. vi. 20, Cor. But A B : BG : : the par™ A C : the par™ BF; . ' . the fig. S : the fig. P : : the par™ A C : the par™ BF; v. I . and the fig. S = the par™ AC; Constr. : . the fig. P =the par™ BF = the fig. E. Constr. A n d since, by construction, the fig. P is similar to the fig. S, . ' . P is the rectil. figure required. Q. E. F. 348 EUCLID'S ELEMENTS. Proposition 26. Theorem, If two similar parallelograms have a common angle, and be similarly situated, they are about t l i e same diagonal. Let the par™ ABCD, AEFG be similar and similarly situated, and have the common angle BAD: then shall these par™" be about the same diagonal. Join AC. Then i f AC does not pass through F, let i t cut FG, or FG produced, at H. Join AF; and through H draw HK par1 to AD or BC. I . 31. Then the par™8 BD and KG are similar, since they are about the same diagonal AHC; VL 24. . " . DA : AB : : GA : AK. But because the par™5 BD and EG are similar; Hyp. : . DA : AB : : GA : AE; VI. Def. 2. . ' . GA : AK : : GA : AE; . . AK = AE, which is impossible; . ' . AC must pass through F; that is, the par™ BD, EG are about the same diagonal. Q.E.D. book vi. prop. 30. 349 Obs. Propositions 27, 28, 29 being cumbrous in form and of li value as geometrical results are now very generally omitted. Definition. A straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. [Book vi. Def. 4.] Proposition 30. Problem. To divide a given straight line in extreme and mean ratio. A C B Let AB be the given st. line: it is required to divide it in extreme and mean ratio. Divide AB in C so that the rect. AB, BC may be equal to the sq. on AC. II. 11. Then because the rect. AB, BC = the sq. on AC, . " . AB : AC : : AC : BC. vi. 17. Q.E.F. EXERCISES. 1. ABODE is a regular pentagon; if the lines BE and AD inter-s e c t in O, shew that each of them i s divided in extreme and mean ratio. 2. If the radius of a circle is cut in extreme and mean ratio greater segment i s equal to the side of a regular decagon inscribed in the c i r c l e . 3 5 0 EUCLID'S ELEMENTS. Proposition 31. Theorem. In a right-angled triangle, any rectilineal figure described on the hypotenuse is equal to t l i e s u m of t l i e two similar and similarly described figures on t l i e sides containing the right angle. Let ABC be a right-angled triangle of which BC i s the hypotenuse; and let P, Q, R be similar and similarly described figures on BC, CA, AB respectively: then shall the fig. P be equal to the sum of the figs. Q and R. Draw AD perp. to BC. Then the . ' . CB : . ' . CB : . ' . , inversely, BD : In like manner D C : . ' . the sum of BD, D C vi. 8. A 8 CBA, A B D are similar; BA : : BA : B D ; BD : : the fig. P : thefig. R,vi.20.Cor. BC : : the fig. R : the fig. P. v. 2. BC : : thefig. Q : the fig. P; : BC : : the sum of fia^. R, Q but BC = the sum of BD, D C ; the fig. P = the sum of the figs. R and Q. hs. P; v. 15. Q.E.F. Note. This proposition i s a generalization of the 47th Prop, of Book i . It will be a useful exercise for the student to deduce the general theorem from tho particular case with the aid of Pron -' 0 Cor. 2. ~ ' exercises on prop. 31. 351 EXERCISES. 1. In a right-angled triangle if a perpendicular be drawn from the right angle to the opposite side, the segments of the hypotenuse are in the duplicate ratio of the sides containing the right angle. 2. If, in Proposition 31, the figure on the hypotenuse is equal to the given triangle, the figures on the other two sides are each equal to one of the parts into which the triangle is divided by the perpen-dicular from the right angle to the hypotenuse. 3. AX and BY are medians of the triangle ABC which meet in G : if X Y be joined, compare the areas of the triangles A G B , X G Y . 4. Shew that similar triangles are to one another in the duplicate ratio of (i) corresponding medians, ( i i ) the radii of their inscribed circles, ( i i i ) the radii of their circumscribed circles. 5. DEF is the pedal triangle of the triangle ABC; prove that the triangle A B C is to the triangle D B F in the duplicate ratio of A B to BD. Hence shew that the fig. A F D C : the A B F D : : AD2 : BD2. 6. The base BC of a triangle ABC is produced to a point D such that B D : D C in the duplicate ratio of BA : AC. Shew that A D is a mean proportional between B D and DC. 7: Bisect a triangle by a line drawn parallel to one of its sides. 8. Shew how to draw a line parallel to the base of a triangle so as to form with the other two sides produced a triangle double of the given triangle. 9. If through any point within a triangle lines be drawn from the angles to cut the opposite sides, the segments of any one side will have to each other the ratio compounded of the ratios of the segments of the other sides. 10. Draw a straight line parallel to the base of an isosceles tri-angle so as to cut off a triangle which has to the whole triangle the ratio of the base to a side. 11. Through a given point, between two straight lines containing a given angle, draw a line which shall cut off a triangle equal to a given rectilineal figure. Obs. The 32nd Proposition as given by Euclid is de-fective, and as it is never applied, w e have omitted it. 3 5 2 Euclid's elements. Proposition 33. Theorem. In equal circles, angles, whether at the centres or the c i r -cumferences, have the same ratio as the arcs on which they stand: so also have the sectors. Let ABC and DEF be equal circles, and let BGC, EHF be angles at the centres, and BAC and EDF angles at the O18; then shall A EHF : : the arc BC : the arc EF, EDF : : the arc BC : the arc EF, the sector EHF : : the arc BC : the ( i ) the a BGC : the ( i i ) the a BAC : the ( i i i ) the sector BGC arc EF, Along the Oce of the © A B C take any number of arcs CK, KL each equal to BC; and along the O m of the © D E F take any number of arcs FM, MN, NR each equal to EF. Join GK, GL, HM, HN, HR. (i) Then the AB BGC, CGK, KGL are all equal, for they stand on the equal arcs BC, CK, KL: i l l . 27. . ' . the a BGL i s the same multiple of the a BGC that the arc BL is of the arc BC. Similarly the a EHR i s the same multiple of the _ EHF that the arc ER is of the arc EF. And i f the arc BL = the arc ER, the a BGL = the a EHR; in. 27. and i f the arc BL i s greater than the arc ER, the a BGL is greater than the _ EHR; and i f the arc BL is less than the are ER, the a BGL is less than the _ EHR. BOOK VI. PROP. 33. 3 5 3 Now since there are four magnitudes, namely the As BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the a BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the a EHR and the arc ER: and i t has been proved that the a BGL i s greater than, equal to, or less than the a EHR according as BL i s greater than, equal to, or less than ER; . . the four magnitudes are proportionals; v. Def 4. that is, the a BGC : the A EHF : : the arc BC : the arc EF. ( i i ) And since the a BGC = twice the A BAC, in. 20. and the a EHF = twice the A EDF; . " . the A BAC : the A EDF : : the arc BC : the arc EF. v. 8. ( i i i ) Join BC, CK; and in the arcs BC, CK take any points X, O. Join BX, XC, CO, OK. Then in the AB BGC, CGK, ( BG = CG, Because] GC = GK, [and the A BGC = the A CGK; . " . BC = CK; and the A BG C = the A CG K. And because the arc BC = the arc CK, . " . the remaining arc BAC = the remaining arc CAK: . ' . the ABXC=the A COK; in. 27. . " . the segment BXC i s similar to the segment COK; in. Def. and they stand on equal chords BC, CK; . -. the segment BXC = the segment COK. I I I . 24. And the A BGC = the A C G K ; . ' . the sector BGC = the sector CGK. in. 27. i . 4. Constr. 3 5 4 EUCLID'S ELEMENTS. Similarly it may be shewn that the sectors BGC, CGK, KGL are all equal; and likewise the sectors EHF, FHM, MHN, NHR are all equaL . ' . the sector BGL is the same multiple of the sector BGC that the arc BL is of the arc BC; and the sector EHR is the same multiple of the sector EHF that the arc ER is of the arc EF: And i f the arc BL = the arc ER, the sector BGL = the sector EHR: Proved. and i f the arc BL is greater than the arc ER, the sector BGL is greater than the sector EHR: and i f the arc BL is less than the arc ER, the sector BGL is less than the sector EHR. N o w since there are four magnitudes, namely, the sec-tors BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the sector BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the sector EHR and the arc ER: and it has been shewn that the sector BGL is greater than, equal to, or less than the sector EHR according as the arc BL is greater than, equal to, or less than the arc ER; . ' . the four magnitudes are proportionals; v. Def. 4. that is, the sector BGC : the sector EHF : : the arc BC : the ate EF. Q. E.D. BOOK VI. PROP. B. 355 Proposition B. Theorem. If the vertical angle of a triangle be bisected by a straight line which cuts the base, the rectangle contained by the sides of tlie triangle shall be equal to the rectangle contained by the segments of the base, together with t l i e square on the straight line which bisects the angle. Let ABC be a triangle having the A BAC bisected by AD: then shall the rect. BA, AC = the rect. BD, DC, with the sq. on AD. Describe a circle about the AABC, iv. 5. and produce AD to meet the Oce in E. Join EC. Then in the As BAD, EAC, because the a BAD = the a EAC, Hyp. and the A ABD = the a AEC in the same segment; in. 21. . ' . the A BAD is equiangular to the A EAC. i . 32. . • . BA : AD : : EA : AC; vi. 4. A the rect. BA, AC = the rect. EA, AD, vi. 16. = the rect. ED, DA, with the sq. on AD. n. 3. But the rect. ED, DA = the rect. BD, DC; in. 35. . ' . the rect. BA, AC = the rect. BD, DC, with the sq. on AD. Q.E.D. EXERCISE. If the vertical angle B A C be externally bisected by a straight line which meets the base in D, shew that the rectangle contained by BA, A C together with the square on A D is equal to the rectangle contained by the segments of the base. 356 EUCLID'S ELEMENTS. Proposition C. Theorem. If from the vertical angle of a triangle a straight Une be drawn perpendicular to the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle con-tained by the perpendicular and the diameter of the c i r c l e described about the triangle. Let ABC be a triangle, and let AD be the perp. from A to BC: then the rect. BA, AC shall be equal to the rect. contained by AD and the diameter of the circle circumscribed about the AABC. Describe a circle about the AABC; iy. 5. draw the diameter AE, and join EC. Then in the As BAD, EAC, the rt. angle BDA = the rt. angle ACE, in the semicircle ACE, and the a ABD = the a AEC, in the same segment; in. 21. . " . the A BAD is equiangular to the A EAC; I . 32. . ' . BA : AD : : EA : AC; VI. 4. . ' . the rect. BA, AC = the rect. EA, AD. vi. 16. Q.E.D. book vi. prop. d. 357 Proposition D. Theorem. The rectangle contained by the diagonals of a quadri-l a t e r a l inscribed in a c i r c l e is equal t o t l i e sum of the two rectangles contained by i t s opposite sides. Let ABCD be a quadrilateral inscribed in a c i r c l e , and l e t AC, BD be i t s diagonals: then the reet. AC, BD shall be equal to the sum of the rect-angles AB, CD and BC, AD. Make the a DAE equal to the a BAC; I . 23; to each add the a EAC, then the A DAC = the A BAE. Then in the A8 EAB, DAC, the a EAB = the a DAC, and the A ABE = the A ACD in the same segment; in. 21. . . the triangles are equiangular to one another; I . 32. . ' . AB : BE : : AC : CD; vi. 4. . ' . the rect. AB, CD = the rect. AC, EB. vi 16, Again in the A8 DAE, CAB, the a DAE = the ACAB, Constr. and the a ADE = the A ACB, in the same segment, in. 21. . ' . the triangles are equiangular to one another; I . 32. /. AD : DE : : AC : CB; VI. 4. . ' . the rect. BC, AD = the rect. AC, DE. vi. 16. But the rect. AB, CD = the rect. AC, EB. Proved. . ' . the sum of the rects. BC, AD and AB, CD = the sum of the rects. AC, DE and AC, EB; that is, the sum of the rects. BC, AD and AB, CD = the rect. AC, BD. n. 1 . Q.E.D. h. e. 24 358 EUCLID'S ELEMENTS. Note. Propositions B, C, and D do not occur in Euclid, but were added by Eobert Simson. Prop. D is usually known as Ptolemy's theorem, and it is the par-ticular case of the following more general theorem: The rectangle contained by the diagonals of a quadrilateral is less than the sum of t l i e rectangles contained by its opposite sides, unless a circle can be circumscribed about the quadrilateral, in which case it is equal to that sum. EXERCISES. 1. ABC is an isosceles triangle, and on the base, or base pro-duced, any point X is taken: shew that the circumscribed circles of the triangles A B X , A C X are equal. 2. From the extremities B, C of the base of an isosceles triangle A B C , straight lines are drawn perpendicular to AB, A C respectively, and intersecting at D : shew that the rectangle BC, A D is double of the rectangle A B , DB. 3. If the diagonals of a quadrilateral inscribed in a circle are at right angles, the sum of the rectangles of the opposite sides is double the area of the figure. 4. A B C D is a quadrilateral inscribed in a circle, and the diagonal B D bisects A C : shew that the rectangle A D , A B is equal to the rect-angle DC, CB. 5. If the vertex A of a triangle ABC be joined to any point in the base, it will divide the triangle into two triangles such that their circumscribed circles have radii in the ratio of A B to A C . 6. Construct a triangle, having given the base, the vertical angle, and the rectangle contained by the sides. 7. Two triangles of equal area are inscribed in the same circle: Bhew that the rectangle contained by any two sides of the one is to the rectangle contained by any two sides of the other as the base of the second is to the base of the first. 8. A circle is desoribed round an equilateral triangle, and from any point in the circumference straight lines are drawn to tlie angular points of the triangle: shew that one of these straight lines is equal to the sum of the other two. 9. A B C D is a quadrilateral insoribed in a circle, and B D bisects the angle A B C : i f the points A and C are fixed on the circumference of the circle and B is variable in position, Bhew that the sum of A B and B C has a constant ratio to BD. THEOREMS A N D EXAMPLES ON BOOK VI. I . on haemonic s e c t i o n . 1. To divide a given straight line internally and externally so that its segments may be in a given ratio. H, L M A p\/B Q Let A B be the given st. line, and L, M two other st. lines which determine the given ratio: it is required to divide A B internally and externally in the ratio L : M . Through A and B draw any two par1 st. lines AH, BK. Prom A H cut off Aa equal to L, and from B K cut off Bb and Bb' each equal to M, B6' being taken in the same direction as Aa, and Bb in the opposite direction. Join ab, cutting A B in P; join ab', and produce it to cut A B externally at Q. Then A B is divided.internally at P and externally atQ, bo that A P : P B = L : M , and A Q : Q B = L : M . The proof follows at once from Euclid vr.,4. Obs. The solution is singular; that is, only one internal and one external point can be found that will divide the given straight line into segments which have the given ratio. 24—2 360 EUCLID'S ELEMENTS. DEFINITION. A finite straight line is said to be cut harmonically when it is divided internally and externally into segments which have the same ratio. A P B O Thus AB is divided harmonically at P and Q, if AP : PB = A Q : QB. P and Q are said to be harmonic conjugates of A and B. If P and Q divide AB internally and externally in the same ratio, i t i s easy to shew that A and B divide P Q internally and externally in the same ratio: hence A and B are harmonic conjugates of P and Q. Example. The base of a triangle is divided harmonically by the internal and external bisectors of the vertical angle: for in each case the segments of the base are in the ratio of the other sides of the triangle. [Euclid vi. 3 and A.] Obs. We shall use the terms Arithmetic, Geometric, and Harmonic Means in their ordinary Algebraical sense. 1. If AB is divided internally at P and externally at Q in the same ratio, then AB is the harmonic mean between A Q and AP. For by hypothesis A Q : Q B = AP : PB; . -. , alternately, A Q : A P = Q B : PB, that is, A Q : A P = A Q - A B : A B - A P , which proves the proposition. 2. If AB is divided harmonically at P and Q, and O is the middle point of AB; then shall OP . OQ=OA2. A O P B Q For since AB is divided harmonically at P and Q . -. AP : PB = AQ: QB; . -. AP-PB : AP + PB = A Q - Q B : AQ + QB, or, 20P : 20A = 20A : 20Q; . -. OP.OQ=OAa. Conversely, if OP . OQ=OAs, i t may be shewn that AP : PB = AQ:QB; that i s , that AB i s divided harmonioally at P and Q. THEOREMS AND EXAMPLES ON BOOK VI. 3 6 1 3. The Arithmetic, Geometric and Harmonic means of t i o o straight lines may be thus represented graphically. In the adjoining figure, two tan- y \ gents AH, AK are drawn from any external point A to the circle P H Q K ; HK is the chord of contact, and the st. line joining A to the centre O cuts the oM at P and Q. Then ( i ) A O is the Arithmetic mean between AP and A Q : for clearly AO=i(AP + AQ). ( i i ) AH is the Geometric mean between AP and A Q : for AH3=AP.AQ. m . 36. ( i i i ) AB is the Harmonic mean between AP and AQ: for O A . O B = OPs. Ex. 1, p. 233. / . AB is cut harmonically at P and Q. Ex. 1, p . - 360. That is, AB is the Harmonic mean between AP and AQ. ' And from the similar triangles OAH, HAB, O A : A H = AH : AB, : . A O . A B = AH2; vi. 17. . " . the Geometric mean between two straight lines i s the mean propor-tional between their Arithmetic and Harmonic means. 4. Given the base of a triangle and the ratio of the other sides, t o find the locus of the vertex. Let BC be the given base, and l e t BAC be any triangle standing upon it, such that BA : AC=the given ratio: i t is required to find the locus of A. Bisect the / BAC internally and externally by AP, AQ. Then BC is divided internally at P, and externally at Q, bo that BP : PC = B Q : Q C = the given ratio; / . P and Q are fixed points. And since AP, A Q are the internal and external bisectors of the L BAC, : . the L P A Q i s a rt. angle; . • . the locus of A is a circle described on P Q as diameter. Exercise. Given three points B, P, C in a straight line: find the locus of points at which BP and PC subtend equal angles. 362 EUCLID'S ELEMENTS. DEFINITIONS. 1. A series of points in a straight line is called a range. If the range consists of four points, of which one pair are har-monic conjugates with respect to the other pair, it is said to be a harmonic range. 2. A series of straight lines drawn through a point is called a pencil. The point of concurrence is called the vertex of the pencil, and each of the straight lines is called a ray. A pencil of four rays drawn from any point to a harmoDic range is said to be a harmonic pencil. 3. A straight line drawn to cut a system of lines is called a transversal. 4. A system of four straight lines, no three of which are concurrent, is called a complete qnadrilateraL These straight lines will intersect two and two in six points, called the vertices of the quadrilateral; the three straight lines which join opposite vertices are diagonals. Theobems on Habmonio Section. 1. If a transversal is drawn parallel to one ray of a harmonic pencil, the o t l i e r three rays intercept equal parts upon it: and con-versely. 2. Any transversal is cut harmonically by the rays of a harmonic pencil. 3. In a harmonic pencil, if one ray bisect the angle between the o t l i e r pair of rays, i t is perpendicular to i t s conjugate ray. Conversely if one pair of rays form a right angle, then they bisect internally and externally the angle between the other pair. i. If A, B, C, D and a, b, e, d are harmonic ranges, one on each of two given straight lines, and if Aa, Bb, Cc, the straight lines which join-three pairs of corresponding points, meet at S; then will Dd also pass through S. 6. If two straight lines intersect at O, and if O, C, B, D and O, d are two harmonic ranges one on each straight line (the points corre-sponding as indicated by the letters), then Co, Bb, Dd i r i 7 Z be con-current: also Cd, Bb, Do will be concurrent. 6. Use Theorem 5 to prove that in a complete quadrilateral in which the three diagonals are drawn, the straight line joining any pair of opposite vertices is cut harmonically by the other two diagonals. THEOREMS AND EXAMPLES ON BOOK VI. 363 1 1 . On centres of similarity and similitude. 1. If any two unequal similar figures are placed so thai their homologous sides are parallel, the lines joining corresponding points in the two figures meet in a point, whose distances from any two corre-sponding points are in the ratio of any pair of homologous sides. Let A B C D , A'B'C'D' be two similar figures, and let them be placed bo that their homologous sides are parallel; namely, AB, BC, CD, DA parallel to A'B', B'C, CD', D'A'respectively: then shall AA', BB', C C , DD' meet in a point, whose distances from any two corresponding points shall be in the ratio of any pair of homologous sides. Let AA' meet BB', produced if necessary, in S. Then because A B is par1 to A'B'; Hyp. . : the Aa SAB, SA'B' are equiangular; . ' . SA : SA' = A B : A'B'; vi. 4. . • . AA' divides BB', externally or internally, in the ratio of A B to A'B'. Similarly i t may be shewn that C C divides BB' in the ratio of BC to B'C. But since the figures are similar, BC : B ' C = A B : A'B'; . ' . AA' and C C divide BB' in the same ratio; that is, AA', BB', C C meet in the same point S. In like manner i t may be proved that DD' meets C C in the point S. . -. AA', BB', C C , DD' are concurrent, and each of these lines is divided at S in the ratio of a pair of homologous sides -o f the two figures. Q.ji.i). Coe. If any line is drawn through S meeting any pair of homolo-gous sides in K and K', the ratio SK : SK' is constant, and equal to the ratio of any pair of homologous sides. Note. It will be seen that the lines joining corresponding points are divided externally or internally at S according as the correspond-ing sides are drawn in the same or in opposite directions. In either ease the point of concurrence S is called a centre of similarity of the two figures. 364 Euclid's elements. 2. A common tangent STT' to two circles whose centres are C, C, meets the line of centres in S. If through S any straight line is drawn meeting these two circles in P, Q, and P', Q', respectively, then the radii CP, C Q shall be respectively parallel to CP', C'Q'. Also the rectangles S Q . SP', S P . SQ' shall each be equal to the rectangle S T . ST'. Join CT, CP, C Q and C'T', CP', C'Q'. Then since each of the L CTS, C'T'S i s a right angle, in. 18. . -. C T i s par1 to C'T'; . -. the A" SCT, SCT' are equiangular; . -. SC : S C = C T : C'T' = CP : C'P'; . -. the A" SCP, SCP' are similar; vi. 7 . . -. the l SCP=the z SCP^; . -. CP i s par1 to CP'. Similarly C Q is par1 to C'Q'. Again, it easily follows that TP, TQ are par1 to T'P', T'Q' respectively; . -. the A" STP, ST'F" are similar. Now the rect. SP . SQ=the sq. on ST; m. 37. A SP : ST = ST : SQ, vi. 16. and SP : ST = SP' : ST'; : . ST : SQ=SF: ST'; : . the rect. S T . ST'= S Q . SP'. In the same way it may be proved that the rect. S P . SQ'=the rect. S T . ST' Q. E. I). Con. 1. It has been proved that SC : SC'=CP : CP'; thus the external common tangents to the two circles meet at a point S which divides the line of centres externally in the ratio of the radii. Similarly it may be shewn that the transverse common tangents meet at a point S' which divides the line of centres internally in the ratio of the radii. Con. 2. CC is divided harmonioally at S and S'. Definition. The points S and S' wliioh divide externally and internally the line of centres of two circles in the ratio of their radii are called the external and Internal centres of similitude respectively. THEOREMS AND EXAMPLES ON BOOK VI. 365 EXAMPLES. 1. Inscribe a square in a given triangle. 2. In a given triangle inscribe a triangle similar and similarly situated to a given triangle. 3. Inscribe a square in a given sector of circle, so that two angular points shall be on the are of the sector and the other two on the bounding radii. 4. In the figure on page 278, if Dl meets the inscribed circle in X, shew that A, X, Dx are eollinear. Also if A\x meets the base in Y shew that I \ x is divided harmonically at Y and A. 5. With the notation on page 282 shew that'O and G are respec-tively the external and internal centres of similitude of the circum-scribed and nine-points circle. 6. If a variable circle touches two fixed circles, the line joining their points of contact passes through a centre of similitude. Distinguish between the different cases. 7. Describe a circle which shall touch two given circles and pass through a given point. 8. Describe a circle which shall touch three given circles. 9. Cj, C2, C3 are the centres of three given circles; \v E1 are the internal and external centres of similitude of the pair of circles whose centres are C2, C3, and l2, Eg, l3, E3, have similar meanings with regard to the other two pairs of circles: shew that (i) liCj, l2C2, l3C3 are concurrent; ( i i ) the six points l l f l2, l3, Elt E2, E3, lie three and three on four straight lines. III. ON POLE A N D POLAR. DEFINITIONS. (i) If iii any straight line drawn from the centre of a circle two points are taken such, that the rectangle contained by their distances from the centre is equal to the square on the radius, each point is said to be the inverse of the other. Thus in the figure given below, if O is the centre of the circle, and if O P . O Q = (radius)2, then each of the points P and Q is the inverse of the other. It is clear that if one of these points is within the circle the other must be without it. 360 EUCLID'S ELEMENTS. (ii) The polar of a given point with respect to a given circle is the straight line drawn through the inverse of the given point at right angles to the line which joins the given point to the centre: and with reference to the polar the given point is called the pole. Thus in the adjoining figure, if O P . OQ=(radius)2, and if through P and Q, L M and H K are drawn perp. to O P ; then H K is the polar of the point P, and P is the pole of the st. line H K : also L M is the polar of the point Q, and Q the pole of LM. It is clear that the polar of an external point must intersect the circle, and that the polar of an internal point must fall without it: also that the polar of a point o r e the circumference is the tangent at that point. 1. N o w i t has been proved [see Ex. 1, page 233] that i f from an external point P two tangents PH, PK are drawn to a circle, of which O is the centre, then O P cuts the chord of contact H K at right angles at Q, so that O P . O Q = (radius)3, . • . H K is the polar of P with respect to the circle. Def. 2. Hence we conclude that The Polar of an external point with reference to a circle is the chord of contact of tangents drawn from the given point to the circle. The following Theorem is known as tlie Reciprocal Property of Pole and Polar. THEOREMS AND EXAMPLES ON BOOK VI. 3 6 7 2. If A and P are any two points, and if the polar of A with respect to any circle passes through P, then the polar of P must pass through A. Let B C be the polar of the point A with respeot to a circle whose oentre is O, and let B C pass through P; then shall the polar of P pass through A. Join O P ; and from A draw A Q perp. to OP. W e shall shew that A Q is the polar of P. Now since B C is the polar of A, . -. the z A B P is a rt. angle; Def. 2, page 360. and the I A Q P is a rt. angle: Constr. . -. the four points A, B, P, Q are concyclic; . -. OQ.OP=OA.OB hi. 3 6 . = (radius)2, for C B is the polar of A: . • . P and Q are inverse points with respect to the given circle. And since A Q is perp. to OP, , \ A Q is the polar of P. That is, the polar of P passes through A. Q.,E. D. A similar proof applies to the ease when the. given point A i s without the circle, and the polar B C cuts i t . 3. To prove that the locus of the intersection of tangents drawn t o a circle at the extremities of all chords which pass through a given point i s the polar of that point. Let A be the given point within the q circle, of which O is the centre. Let H K be any chord passing through A; and let the tangents at H and K intersect at P: i t is required to prove that the locus of P is the polar of the point A. I. To shew that P lies on the polar of A. Join O P cutting H K in Q. Join OA: and in O A produced take the point B, so that O A . O B = (radius)2, n. 14. Then since A is fixed, B is also fixed. Join PB. H \ \ Q; 0 _ _ , -/ A\ . --S M /'/ • / / i B 368 EUCLID'S ELEMENTS. Then since H K is the chord of contact of tangents from P, . -. O P . OQ=(radius)2. Ex. I. p. 233. But O A . O B = (radius)2; Constr. : . OP.OQ=OA.OB: . -. the four points A, B, P, Q are concyclic. . -. the z • at Q and B together=two rt. angles. in. 22. But the Z at Q is a rt. angle; Constr. : . the z at B is a rt. angle. And since the point B is the inverse of A; Constr. : . PB is the polar of A; that is, the point P lies on the polar of A. II. To shew that any point on the polar of A satisfies the- given conditions. Let B C be the polar of A, and let P be any point on it. Draw tangents PH, PK, and let H K be the chord of contact. N o w from Ex. 1, p. 366, we know that the chord of Contact H K is the polar of P, and we also know that the polar of P must pass through A; for P is on BC, the polar of A: Ex. 2, p. 367. that is, H K passes through A. : . P is the point of intersection of tangents drawn at the ex-tremities of a chord passing through A. From I. and H. we conclude that the required locus is the polar of A. Note. If A is without the circle, the theorem demonstrated in Part I. of the above proof s t i l l holds good; but the converse theorem in Part II. is not true for all points in BC. For if A is without the circle, the polar B C will intersect it; and no point on that part of the polar which is within the circle can be the point of intersection of tangents. W e now see that (i) The Polar of an external point with respect to a circle is the chord of contact of tangents drawn from i t . ( i i ) The Polar of an internal point i s the locus of the intersections of tangents drawn at the extremities of all chords which p<iss through it. ( i i i ) The Polar of a point on tho circumference ts the tangent at that point. THEOREMS AND EXAMPLES ON BOOK VI. 3 6 9 Ex. 1, p. 366. The following theorem is known as the Harmonic Property of Pole and Polar. 4. Any straight line drawn through a point is cut harmonically by the point, i t s polar, and the circumference of the circle. Let A H B be a circle, P the given point and H K its polar; let Paqb be any straight line drawn through P meeting the polar at q and the om of the circle at a and b: then shall P, a, q, 6 be a harmonic range. In the case here considered, P is an external point. Join P to the centre O, and let PO cut the o00 at A and B: let the polar of P cut the O00 at H and K, and PO at Q. Then PH is a tangent to the e AHB. From the similar triangles OPH, HPQ, O P : PH = PH : PQ, . -. P Q . P O = PH2 = Pa. Pb. : . the points O, Q, a, b are concyclic: . -. the ZoQA=the z abO = the z Oab i . 5. =the Z OQ&, in the same segment. And since Q H is perp. to AB, . -. the ZaQH=the z&QH. . • . Qg and Q P are the internal and external bisectors of the Z aQb: . ' . P, a, q, b is a harmonic range. Ex. 1, p. 360. The student should investigate for himself the case when P is an internal point. Conversely, it may be shewn that if through a fixed point P any secant is drawn cutting the circumference of a given circle at a and b, and if q is the harmonic conjugate at P with respect to a, b; then the locus ofqis the polar of P with respect to the given circle. [For Examples on Pole and Polar, see p. 370.] DEFINITION. A triangle so related to a circle that each side is the polar of the opposite vertex is said to be self-conjugate with respect to the circle. 370 EUCLID'S ELEMENTS. EXAMPLES ON POLE AND POLAR. 1. The straight line which joins any two points is the polar with respect to a given circle of the point of intersection of their polars. 2. The point of intersection of any two straight lines is the pole of the straight line which joins their poles. 3. Find the locus of the poles of all straight lines which past through a given point. 4. Find the locus of the poles, with respect to a given circle, of tan-gents drawn to a concentric circle. 5. If two circles cut one another orthogonally and PQ be any diameter of one of them; shew that the polar of P with regard to the other circle passes through Q 6. If two circles cut one another orthogonally, the centre of each circle is the pole of their common chord with respect to the other circle. 7. Any two points subtend at the centre of a circle an angle equal to one of the angles formed by the polars of the given points. 8. O is the centre of a given circle, and AB a fixed straight line. P is any point i r e A B ; find the locus of the point inverse to P with respect to the circle. 9. Given a circle, and a fixed point O on its circumference: P is any point on the circle: find the locus of the point inverse to P with respect to any circle whose centre is O. 10. Given two points A and B, and a circle whose centre u O; shew that the rectangle contained by O A and t l i e perpendicular from B on the polar of A is equal to the rectangle contained by O B and the perpendicular from A on the polar of B. 11. Four points A, B, C, D are taken in order on the circumference of a circle; DA, C B intersect at P, A C , B D at Q and BA, C D in R: shew that the triangle P Q R is self-conjugate with respect to the circle. 12. Give a linear construction for finding the polar of a given point with respect to a given circle. Hence find a linear construction for drawing a tangent to a circle from an external point. 13. If a triangle is self-conjugate with respect to a circle, the centre of the circle is at the orthocentre of the triangle. 14. The polars, with respect to a given circle, of the four points of u harmonic range form a harmonic pencil.- and conversely. THEOREMS AND EXAMPLES ON BOOK VI. 3 7 1 IV. ON T H E RADICAL AXIS. 1. To find the locus of points from which the tangents drawn to two given circles are equal. Fig. 1. Fig. 2. Let A and B be the centres of the given circles, whose radii are a and 6; and let P be any point such that the tangent P Q drawn to the circle (A) is equal to the tangent PR drawn to the circle (B): i t is required to find the locus of P. Join PA, PB, A Q , BR, A B ; and from P draw PS perp. to AB. Then because P Q = P R , . • . PQ2=PR2. But PQ2=PA2-AQ2; and PR2 = PB2-BR2: i . 47. . -. PA2-AQ2=PB2-BR2; that is, PS2+AS2-a2=PS2+SB2-62; x . 47. or, AS2-a2=SB2-62. Hence AB i s divided'at S, so that AS2- SB2=a2-62: . ' . S is a , fixed point. Hence all points from which equal tangents can be drawn to the two circles he on the straight line which cuts A B at rt. angles, so that the difference of the squares on the segments of A B is equal to the difference of the squares on the radii. Again, by simply retracing these steps, i t may be shewn that in Fig. 1 every point in SP, and.in Fig. 2,every point in S P exterior to' the circles, is such that tangents drawn from it to the two circles are equal. Hence we conclude that in Fig. 1 the whole line SP is the required locus, and in Fig. 2 that part of SP which i s . without the circles. In either case S P is said to be the Radical Axis of the two circles. 372 EUCLID'S ELEMENTS. Corollary. If tlie circles cut one another as in Fig. 2, it is clear t l i a t the Radical Axis is identical with the straight line which passes through the points of intersection of the circles; for it follows readily from in. 36 that tangents drawn to two intersecting circles from any point in the common chord produced are equal. 2. Tlie Radical Axes of three circles taken in pairs are concurrent. Z, Let there be three circles whose centres are A, B, C Let O Z be the radical axis of the ©" (A) and (B); and O Y the Badical Axis of the 0s (A) and (C), O being the point of their intersection: then shall the radical axis of the & (B) and (C) pass through O. It will be found that the point O is either without or within all the circles, I. When O is without the circles. From O draw OP, O Q , O R tangents to the & (A), (B), (C). Then because O is a point on the radical axis of (A) and (B); Hyp. : . OP = OQ. And because O is a point on the radical axis of (A) and (C), Hyp. . : OP=OR, . -. OQ=OR; / . O is a point on the radical axis of (B) and yC). i.e. the radical axis of (B) and (C) passes through O. II. If the circles intersect in such a way that O is within them all; the radical axes are then the oommon chords of the three circle? taken two and two; and it is required to prove that these common chorda are concurrent. This may be shewn indireotly by rn. 35. Definition. The point of intersection of the radical axes of three oiroles taken in pairs is called tlie radical centre. THEOREMS A N D EXAMPLES ON BOOK VI. 3 7 3 8. To draw the- radical axis of two given circles. Let A and B be the centres of the given circles: it is required to draw their radical axis. If the given circles intersect, then the st. line drawn through their points of intersection will be the radical axis. [Ex. 1, Cor. p. 372.] But if the given circles do not intersect, describe any circle so as to cut them in E, F and G, H: Join E F and H G , and produce them to meet in P. Join A B ; and from P draw P S perp. to AB. Then P S shall be the radical axis of the ©• (A), (B). Definition. If each pair of circles in a given system have the same radical axis, the circles are said to be co-axal. EXAMPi.:::: 1. Shew that the radical axis of two circles bisects any one of their common tangents. 2. If tangents are drawn to two circles from any point on their radical axis; shew that a circle described with this point as centre and any one of the tangents as radius, cuts both the given circles ortho-gonally. 3. O is the radical centre of three circles, and from O a tangent O T is drawn to any one of them: shew that a circle whose centre is O and radius O T cuts all the given circles orthogonally. 4. If three circles touch one another, taken two and two, shew that their common tangents at the points of contact are concurrent. H. E. 25 374 E u c l i d ' s elements. 5. If circles are described on the three sides of u. triangle as diameter, their radical centre is the orthocentre of the triangle. 6. All circles which pass through a fixed point and cut a given circle orthogonally, pass through a second fixed point. 7. Find the locus of the centres of all circles which pass through a given point and cut a given circle orthogonally. 8. Describe a circle to pass through two given points and cut a given circle orthogonally. 9. Find the locus of the centres of all circles which cut two given circles orthogonally. 10. Describe a circle to pass through a given point and cut two given circles orthogonally. 11. The difference of the squares on the tangents drawn from any point to two circles is equal to twice the rectangle contained by the straight line joining their centres and the perpendicular from the given point on their radical axis. 12. Ire a system of co-axal circles which do not intersect, any point is taken on the radical axis; shew that a circle described from this point as centre with radius equal to the tangent drawn from it to any one of the circles, will meet the line of centres in two fixed points. [These fixed points are called tlie Limiting Points of the system.] 13. In a system of co-axal circles the two limiting points and the points in which any one circle of t l i e system cuts the line of centres form a harmonic range. 14. In u, system of co-axal circles a limiting point has the same polar with regard to all the circles of the system. 15. If two circles are orthogonal any diameter of one is cut harmonically by the otlier. Obs. In the two following theorems w e are to suppose that the segments of straight lines are expressed numerically in terms of some c o m m o n unit; and the ratio of one such sesiient to another will be denoted by the fraction of which the first is the numerator and the second the denominator. THEOREMS AND EXAMPLES ON BOOK VI. 375 V. ON TRANSVERSALS. Definition. A straight line drawn to cut a given system of lines is called a transversal. 1. If three concurrent straight lines are drawn from the angular points of a triangle to meet the opposite sides, then, the product of three alternate segments taken in order is equal to the product of the other three segments. F . B D C B Let AD, BE, CF be drawn from the vertices of the A ABC to intersect at O, and out the opposite sides at D, E, F: then shall BD . C E . A F = D C . EA . FB. By similar triangles i t may be shewn that BD : DC=the alt. of a A O B : the alt. of A AOC; BD aAOB similarly, and DC A AOC CE A BOC EA- ABOA' AF aCOA FB aCOB Multiplying these ratios, we have BD C E AF D C ' E A " F B - ' or, BD .CE . A F = D C . EA. FB. q. e. d. The converse of this theorem, which may be proved indirectly, is very important: i t may be enunciated thus: If three straight lines drawn from the vertices of a triangle cut the opposite sides so that the product of three alternate segments taken in order is equal to the product of the other three, then the three straight lines are concurrent. That is, if BD . C E . A F = D C . EA . FB, then AD, BE, CF are concurrent. 3 7 6 EUCLID'S ELEMENTS. 2. If a transversal is drawn to cut the sides, or the sides produced, of a triangle, t l i e product of three alternate segments taken in order is equal to the product of the other three segments. Let A B C be a triangle, and l e t a transversal meet the sides BC, CA, AB, or these aides produced, at D, E, F: then Bhall BD . C E . A F = D C . EA . FB. Draw A H par1 to BC, meeting the transversal at H. Then from the similar A' DBF, HAF, BD _ HA. FB_AF: and from the similar a" DCE, HAE, CE _EA DC~HA: ... .. .. BD CE EA . -. , by multiplication, pg- . ^ = -^; that is, BD.CE.AF = 1 , DC.EA.FB" or, BD.CE. AF=DC.EA.FB. Q.E. D. Note. In this theorem the transversal must either meet two sides and the third side produced, as in Fig. 1; or all three sides pro-duced, as in Fig. 2. The converse of this Theorem m a y be proved indirectly: If three points are taken in two sides of a triangle and the third side produced, or in all three sides produced, so that the product of three alternate segments taken in order is equal to the product of the other three segments, the three points are eollinear. The propositions given on pagea 103—106 relating to the concur-rence of straight lines in a triangle, m a y be proved by the method of transversals, and in addition to these the following important theorems may be established. MISCELLANEOUS EXAMPLES ON BOOK VI. 377 DEFINITIONS. (i) If two triangles are such that three straight lines joining corresponding vertices are concurrent, they are said to be co-polar. (ii) If two triangles are such that the points of intersection of corresponding sides are eollinear, they are said to be co-axial. Theobems to be peoved bt Teansveebals. 1. The straight lines which join the vertices of a triangle to the points of contact of the inscribed circle (or any of the three inscribed circles) are concurrent. 2. The middle points of the diagonals of a complete quadrilateral are eollinear. 3. Go-polar triangles are also co-axial; and conversely co-axial triangles are also co-polar. 4. The six centres of similitude of three circles lie three by thr on four straight lines. MISCELLANEOUS EXAMPLES ON BOOK VI. 1. Through D, any point in the base of a triangle ABC, straight lines DE, DF are drawn parallel to the sides AB, AC, and meeting the sides at E, F: shew that the triangle AEF is a mean proportional between the triangles FBD, EDC. 2. If two triangles have one angle of the one equal to one angle of the other, and a second angle of the one supplementary to a second angle of the other, then the sides about the third angles are proportional. 3. A E bisects the vertical angle of the triangle A B C and meets the base in E; shew that i f circles are described about the triangles ABE, ACE, the diameters of these circles are to each other in the Bame ratio as the segments of the base. 4. Through a fixed point O draw a straight line so that the parts intercepted between O and the perpendiculars drawn to the straight line from two other fixed points may have a given ratio. 25—3 378 EUCLID'S ELEMENTS. 5. The angle A of a triangle ABC is bisected by AD meeting BC in D, and A X is the median bisecting BC: shew that X D has the same ratio to X B as the difference of the sides has to their sum. 6. AD and AE bisect the vertical angle of a triangle internally and externally, meeting the base in D and E; shew that if O is the middle point of BC, then O B is a mean proportional between O D and OE. 7. P and Q are fixed points; AB and CD are fixed parallel straight lineB; any straight line is drawn from P to meet A B at M, and a straight line is drawn from Q parallel to P M meeting C D at N: shew that the ratio of P M to Q N is constant, and thence shew that the straight line through M and N passes through a fixed point 8. C is the middle point of an are of a circle whose chord is AB; D is any point in the conjugate arc: shew that AD + DB : DC :: AB : AC. 9. In the triangle ABC the side AC is double of BC. If CD, C E bisect the angle A C B internally and externally meeting A B in D and E, shew that the areas of the triangles C B D , A C D , ABC, C D E are as 1, 2, 3, 4. 10. AB, AC are two chords of a circle; a line parallel to the tangent at A cuts AB, A C in D and E respectively: shew that the rectangle AB, A D is equal to the rectangle AC, AE. 11. If from any point on the hypotenuse of a right-angled triangle perpendiculars are drawn to the two sides, the rectangle contained by the segments of the hypotenuse will be equal to the sum of the rectangles contained by the segments of the sides. 12. D is a point in the side AC of the triangle ABC, and E is a point in AB. If BD, C E divide each other into parts in the ratio 4 : 1, then D, E divide CA, BA in the ratio 3 : 1. 13. If the perpendiculars from two fixed points on a straight line passing between them be in a given ratio, the straight line must pass through a third fixed point. 14. PA, PB are two tangents to a circle; PCD any chord through P: shew that the rectangle contained by one pair of opposite sides of the quadrilateral A C B D is equal to the rectangle contained by the other pair. IS. A, B, C are any three points on a oirole, and the tangent at A mootB B C produced in D: shew tliat the diameters of the circles oiroumsoribed about ABD, A C D are as A D to CD. MISCELLANEOUS EXAMPLES ON BOOK VI. 379 16. AB, CD are two diameters of the circle ADBC at right angles to each other, and EF is any chord; CE, C F are drawn meeting A B produced in G and H : prove that the rect. CE, H G = the rect. EF, CH. 17. From the vertex A of any triangle ABC draw a line meeting B C produced in D so that A D may be a mean proportional between the segments of the base. 18. Two circles touch internally at O; AB a chord of the larger circle touches the smaller in C whioh is cut by the lines OA, O B in the points P, Q : shew that O P : O Q : : A C : CB. 19. AB is any chord of a circle; AC, BC are drawn to any point C in the circumference and meet the diameter perpendicular to A B at D, E: if O be the centre, shew that the rect. O D , O E is equal to the square on the radius. 20. YD is a tangent to a circle drawn from a point Y in the diameter A B produced; from D a perpendicular D X is drawn to the diameter: shew that the points X, Y divide A B internally and ex-ternally in the same ratio. 21. Determine a point in the circumference of a circle, from which lines drawn to two other given points shall have a given ratio. 22. O is the centre and OA a radius of a given circle, and V is the middle point of O A ; P and Q are two points on the circum-ference on opposite sides of A and equidistant from it; Q V is pro-duced to meet the circle in L: shew that, whatever be the length of the are PQ, the chord LP will always meet O A produced in a fixed point. 23. EA, EA' are diameters of two circles touching each other externally at E ; a chord A B of the former circle, when produced, touches the latter at C , while a chord A'B of the latter touches the former at C : prove that the rectangle, contained by A B and A'B', is four times as great as that contained by B C and B'C. 24 If a circle be described touching externally two given circles, the straight line passing through the points of contact will intersect the line of centres of the given circles at a fixed point. 25. Two circles touch externally in C; if any point D be taken without them so that the radii AC, B C subtend equal angles at D, and DE, D F be tangents to the circles, shew that D C is a mean proportional between D E and DF. 380 euclid's elements. 26. If through the middle point of the base of a triangle any line be drawn intersecting one side of the triangle, the other produced, and the line drawn parallel to the base from the vertex, it will be divided harmonically. 27. If from either base angle of a triangle a line be drawn intersecting the median from the vertex, the opposite side, and the line drawn parallel to the base from the vertex, it will be divided harmonically. 28. Any straight line drawn to cut the arms of an angle and its internal and external bisectors is cut harmonically. 29. P, Q are harmonic conjugates of A and B, and C is an external point: if the angle P C Q is a right angle, shew that CP, C Q are the internal and external bisectors of the angle A C B . 30. From C, one of the base angles of a triangle, draw a straight line meeting A B in G, and a straight line through A parallel to the base in E, so that C E may be to E G in a given ratio. 31. P is a given point outside the angle formed by two given lines AB, A C : shew how to draw a straight line from P such that the parts of i t intercepted between P and the lines AB, A C may have a given ratio. 32. Through a given point within a given circle, draw a straight line such that the parts of i t intercepted between that point and the circumference may have a given ratio. H o w many solutions does the problem admit of? 33. If a common tangent be drawn to any number of circles which touch each other internally, and from any point of this tangent as a centre a circle be described, cutting the other circles; and if from this centre lines be drawn through the intersections of the circles, the segments of the lines within each circle shall be equal. 34. APB is a quadrant of a circle, SPT a line touching it at P: C is the centre, and P M is perpendicular to CA: prove that the A S C T : the A A C B : : the a A C B : the A C M P . 35. ABC is a triangle inscribed in a circle. AD, AE are lines drawn to the base B C parallel to t l i e tangents at B, C respectively; shew that A D = A E , and B D : C E : : ABS : AC'-. 36. AB is the diameter of a cirole, E the middle point of the radius O B ; on AE, E B as diameters circles are described; P Q L is a common tangent meeting the circles at P and Q, and A B produced at L: shew that BL is equal to the radius of the smaller oircle. MISCELLANEOUS EXAMPLES ON BOOK VI. 381 37. _ The vertical angle C of a triangle is bisected by a straight line which meets the base at D, and is produced to a point E, suoh that the reotangle contained by C D and C E is equal to the rectangle contained by A C and C B : shew that if the base and vertical angle be given, the position of E is invariable. 38. ABC is an isosceles triangle having the base angles at B and C each double of the vertical angle: if B E and C D bisect the base angles and meet the opposite sides in E and D, shew that D E divides the triangle into figures whose ratio is equal to that of A B to BC. 39. If AB, the diameter of a semicircle, be.bisec'ted in C and on A C and C B circles be described, and in the space between the three circumferences a circle be inscribed, shew that its diameter will be to that of the equal circles in the ratio of two to three. 40. O is the centre of a circle inscribed in a quadrilateral ABCD; a line E O F is drawn and making equal angles with A D and BC, and meeting them in E and F respectively: shew that the triangles- A E O , B O F are similar, and that A E : E D = C F : FB. 41. From the last exercise deduce the following: The inscribed circle of a triangle A B C touches A B in F; X O Y is drawn through the centre making equal angles with A B and A C , and meeting them in X and Y respectively: shew that B X : X F = A Y • YC. 42. Inscribe a square in a given semicircle. 43. Inscribe a square in a given segment of a circle. 44. Describe an equilateral triangle equal to a given isosceles triangle. 45. Describe a square having given the difference between a diagonal and a side. 46. Given the vertical angle, the ratio of the sides containing it, and the diameter of the circumscribing circle, construct the triangle. 47. Given the vertical angle, the line bisecting the base, and the angle the bisector makes with the base, construct the triangle. 48. In a given circle inscribe a triangle so that two sides may pass through two given points and the third side be parallel to a given straight line. 49. In a given circle inscribe a triangle so that the sides may pass through three given points. 382 EUCLID'S ELEMENTS. 50. A, B, X, Y are four points in a straight line, and O is such a point in it that the rectangle OA, O B is equal to the rectangle O X , O Y : if a circle be described with centre O and radius equal to a mean proportional between O A and O B , shew that at every point on this circle A B and X Y will subtend equal angles. 51. O is a fixed point, and OP is any line drawn to meet a fixed straight line in P; if on O P a point Q is taken so that O Q to O P is a constant ratio, find the locus of Q. 52. O is a fixed point, and OP is any line drawn to meet the circumference of a fixed circle in P; if on O P a point Q is taken so that O Q to O P is a constant ratio, find the locus of Q. 53. If from a given point two straight lines are drawn including a given angle, and having a fixed ratio, find the locus of the extremity of one of them when the extremity of the other lies on a fixed straight line. 54. On a straight line PAB, two points A and B are marked and the line P A B is made to revolve round the fixed extremity P. C is a fixed point in the plane in which P A B revolves; prove that if C A and C B be joined and the parallelogram C A D B be completed, the loous of D will be a circle. 55. Find the Iocub of a point whose distances from two fixed points are in a given ratio. 56. Find the loous of a point from which two given circles sub-tend the same angle. 57. Find the locus of a point such that its distances from two intersecting straight lines are in a given ratio. 58. In the figure on page 364, shew that QT, P'T' meet on the radioal axis of the two circles. 59. Through two given points draw a circle cutting another circle so that their common chord may be equal to a given straight line. 60. ABC is any triangle, and on its sides equilateral triangles are described externally: if X, Y, Z are the oentres of their inscribed circles, shew that the triangle X Y Z is equilateral. •AMBRIDOB: PRINTED BY C. J . CLAY, M.A. A SONS. AT THB I ' . N I V K R S I T Y PRSSS. M A C M I L L A N A N D CO.'S PUBLICATION'S. Globe 8vo. Part I., Second Udition, Books I. and II p r i c e 2s. A TEXT-BOOK OF EUCLID'S ELEMENTS, in-cluding Alternative Proofs, together with additional Theorems and E zeroises, classified and arranged. By H. S. Hall, M.A., and F. H. Stevens, M. A., Masters of the Military and Engineering Side, Clifton College. The Athenceum says:—"In this first instalment of a complete Euclid for school use the deviations from ordinary text-books in the definitions, axioms, enunciations, and proofs are decided improve-ments. The definitions besides being stated with greater distinctness and equal precision, are in many instances accompanied by explana-tory remarks which bring useful light to bear upon them. The few changes in the enunciations and proofs at once shorten and simplify what was before cumbrous and obscure.... But the chief value of the work consists in the extensive and carefully arranged collection of exercises inserted at suitable points in. the course.. . . Under the heading ' Theorems and Examples on Book I.' there is a large amount of valuable supplementary matter which ought not to escape mention." The School Guardian says:—" The editors are not only themselves mathematicians and practical teachers, but, what is much more to the point, they are also fully alive to the true province and aims of geometry in ordinary school work.... A large amount of explanatory matter is furnished throughout the present volume, while simple questions are attached to many of the propositions, and exercises of a more d i f f i c u l t type are given at the end of each book." The Tutorial Matriculation Directory says:—" Judging from the Books (I. and H.) before us, this edition of Euclid gives promise of being perhaps the very best we have seen. The exercises are admirable." The Schoolmaster says:—"This new edition of Euclid is the first instalment of the full work, and opens very favourably.... Alternative proofs are given to several propositions, and numerous deductions accompany the various numbers. The notes are clear, and add much interest to the study of the subject. W e cordially recommend the book to our readers." ARITHMETICAL EXERCISES AND EXAMINA-TION PAPEBS. With an Appendix containing Questions in Logarithms and Mensuration. By H. S. Hall, M.A., and 8. B. Knight, B.A., Authors of "Elementary Algebra," "Al-gebraical Exercises and Examination Papers," and "Higher Algebra," &c. Globe 8vo. 2s. 6d. TheScotsmansays:—" The exercises and questions are well selected, and form an excellent book of discipline in arithmetic. An appendix of questions in Logarithms and Mensuration adds to the value of the work." MACMILLAN AND CO., LONDON. MACMILLAN AND CO.'S PUBLICATIONS. WORKS BY H. S. HALL, M.A., FOEMBBI.Y SCHOLAR OP CHEIBT'S COLLEGE, CAMBRIDGB, MASTBE OP THB MILITARY AND K.NGISEEIIISO 81DB, CLIPT05 COLLBGB; AND S. B. KNIGHT, B.A., FORMERLY SCHOLAR OP TRIBITY COLLEGE, CAMBBIDGB, LATE ASSISTANT MASTBE AT MARLBOROUGH COLLEGE. ELEMENTARY ALGEBRA FOR SCHOOLS. Fourth Edition. Eevised and enlarged. G-lobe 8vo. (bound in maroon coloured cloth), 3s. 6d. With Answers (bound in green coloured cloth), 4s. 6<2. The Schoolmaster says:—"Has so many points of excellence, as compared with predecessors, that no apology i s needed for i t s issue ...The plan always adopted by every good teacher of frequently recapitulating and making additions at every recapitulation, l a well carried out. The Educational Times says:—"...A very good book. The explanations are concise aud clear, and the examples both numerous and well chosen." Nature says:—" This is, in our opinion, the best Elementary Algebra for school use... W e confidently recommend i t to mathematical teachers, who, we f e e l sure, will find i t the best book of i t s kind for teaching purposes." The Academy says:—"Buy or borrow the book for yourselves and judge, or write a better...A higher text-book i s on i t s way. This occupies s u f f i c i e n t ground for the generality of boys." ALGEBRAICAL EXERCISES AND EXAMINA-T I O N P A P E B S . To accompany Elementary Algebra. Second Edition. Globe 8vo. 2s. 6d. The Saturday Review says:—" To the exercises, one hundred and twenty in number, are added a large s e l e c t i o n of examination papers set at the principal examinations, which require a knowledge of algebra. These papers are intended c h i e f l y as an aid to teachers, who no doubt w i l l find them useful as a criterion of t l i e amount of proficiency to which they must work up their pupils before they can send them in to the several examinations with any certainty of success." The Schoolmaster says:—" As useful a collection of examination papers in algebra as we ever met with... .Each 'exercise' i s calculated to occupy about an hour m i t s solution. Besides these, there arc 35 examination papers set at various competitive examinations during the l a s t three years. The answers are at the end of the book. W e can strongly recommend the volume to teachers seeking a well-arranged series of t e s t s in algebra," The Educational Times says:—"It i s only a few months since we spoke in high praise of an Elementary Algebra by tho same authors of t l i e above papers. W e can Bpeak also in high praise about t h i s l i t t l e book. It consists of over a hundred progressive miscellaneous exercises, followed by a collection of papers set at recent examinations. The exercises are timed, as a rule, to take an hour. . . . Messrs Hall and Knight nave had plenty of experience, and have put that experience to good use." HIGHER ALGEBRA. A Sequel to "Elementary Algebra for Schools." Second Edition, revised. Crown Svo. 7s. (ki. The Athenaeum says:—"The Elementary Algebra by tho same authors, which has already reached a third edition, I s a work of such exceptional merit t l i a t those acquainted with i t w i l l form high expectations of the sequol to i t now issued. Nor will they be dis-appointed. Of tho authors' Higher Algebra, as of their Elementary Algebra, wo un-hosltatlngly assort that i t i s by far the best work of i t s kind with which wo are acquainted. I t supplies a want much f e l t by toachers," Trq Academy says:—"Is as admirably adapted for College students as i t s predecessor was for schools. It I s a well arranged and well-reasoned-out treatise, and contains much that wo have not mot with bofore in similar works. For instance, wo noto as specially good the a r t i c l e s on Convergency and Divergency of Tories, on the treatment of Series generally, and tho treatment of Contlnuod Fractions,..The book I s almost indispensable and wllfbo found to improve upon acquaintance." MACMILLAN AND CO., LONDON. THE EISENHOWER LIBRARY
6291	https://jillianstarrteaching.com/10-tens-and-ones-activities/	Skip to main content Teaching with Jillian Starr Making Sense of Math for Elementary Teachers 10 Tens and Ones Activities to Teach Place Value Understanding place value by practicing tens and ones activities is a great way for our students to get comfortable with this vital math concept. After all, place value is a building block for so many math concepts! Today, I’m sharing some of my favorite tens and ones activities. Great news, my students love these, too! Ten More This is a simple and fun tens and ones activity. All you need is a game board, a die, and some colored counters. Students take turns rolling the die, and adding ten to that number. Students cover the numbers on the game board until one player gets 4 in a row. Ten More is perfect for students who are beginning to make the connection between single-digit numbers and teen numbers. Race to 50 Race to 50uses a place value mat and reinforces tens and ones with Unifix cubes. Students love rolling the die and adding cubes to their mat. This version provides the visual guide for students to build the numbers on a ten frame and see when it’s time to turn ten ones into a ten stick. Students continue playing until one player reaches 50. Not only is this tens and ones activity is a class favorite, but it’s perfect for students that benefit from the repetitive practice of counting by tens and then ones. Race to 100 This tens and ones activity gets students racing to get 100 on the hundred chart. The more students play, the more they start to notice the patterns of plus and minus one and plus and minus ten on the hundred chart. And don’t worry, there’s a Race to 99 version if you use a 0-99 board with your students. In this game, you’ll need one counter for each player, and you’ll need to create a die with +10, -10, +1, and -1. I make these dice by using dot stickers on top of a regular die. Students start at zero and roll the die. They move up, down, left, or right on the hundred chart according to their roll. The first one to 100 (or 99) wins. This game has the suspense of Chutes and Ladders- you can never tell who’s going to win. One minute you’re getting close to one hundred, and the next you roll that dreaded -10! Roll Build & Write This is one of my favorite tens and ones activities for giving my students independent practice. Roll, Build & Write allows students the fun of rolling dice, along with the opportunity to practice building a number with tens and ones. This activity also reinforces writing out the numbers in expanded form. It’s the whole package! And the beauty of using dice? This is a never-ending tens center: students can keep going creating new numbers for their entire time at this center! Make It Big My students absolutely LOVE this tens and ones activity. Make It Big gives kids a chance to build numbers using tens and ones and practice comparing numbers using <, >, or =. For this game, you’ll need a deck of 0-9 cards, a Make It Big mat for each student, and a recording page for each student. Students take turns drawing cards. Upon drawing a card, they decide whether to place that card in the tens or ones space. On the next draw, students place the card in the other space (tens or ones) to create a two-digit number. When both players have made their numbers, the students compare the numbers and record their numbers along with a comparison number sentence on the recording page. The player that made the bigger number wins that round. I love watching this game go from a game of chance to a game of strategy with my students as they start to think more strategically about where they’ll place their cards. I Have/Who Has In this classic class game with a tens and ones twist, students practice seeing the visual of the tens and ones and translating it to say the number out loud. If you’re new to I Have/ Who Has, this game is played with an I Have/ Who Has deck. Each student gets one card. The person with “start” begins by reading their card. All cards follow the pattern, “I have_____. Who has___?” So a student with 3 tens and 4 ones on their card, would say, “I have 34, who has 15?” Then the student with one ten and five ones would read their card. Play continues until all students have had a turn or you reach the card that says end. I love this tens and ones activity because it encourages students to focus on listening to their peers and mentally translating visuals of tens and ones into numbers. And my students love this game because it’s so fun! Tens and Ones Pack Need games and activities to support your students understanding of place value? This Tens & Ones Pack has you covered! GIVE IT TO ME Ten More/Ten Less Ten More/ Ten Less is a great independent activity. This is another never-ending center because students use die to create their starting teen numbers. Once the number is created, students practice adding ten and twenty, and then subtracting ten and zero from that number. This activity really helps students see the relationships between adding and subtracting tens to teen numbers. It’s also a great opportunity to use manipulatives to support solving problems. 4 Part Number Puzzles Four-Part Number puzzles are a great way for students to see numbers in multiple ways, including: number form expanded form placement on a hundred chart base ten blocks This tens and ones activity can be done independently, with a partner, or with a small group. The beauty of 4 Part Puzzles is that they are self-correcting, so students can independently figure out if they’ve solved the problem correctly or should try again. I love how these puzzles pull it all together and help students get comfortable with seeing and understanding two-digit numbers in multiple forms. You can grab your free set right here! FREE Tens & Ones Puzzles FREE Tens and Ones Activities to help reinforce a beginning understanding of place value AND have fun doing it! Give it to me! FREE Tens & Ones Puzzles FREE Tens and Ones Activities to help reinforce a beginning understanding of place value AND have fun doing it! Get to that Number I love this activity for those in-between times when I have 5 minutes before dismissal or after a special. All you need for this game is a set of number cards (or some two-digit numbers written on index cards) and enough space for your students to stand in a circle. To start, flip a card with a 2 digit number. Students go around the circle, first count by tens and then switch to ones to get to that number. When it gets to the student who says the number, they have to sit down. (e.g. 43: 10, 20, 30, 40, 41, 42, 43- sit down). And then flip a new card with a new number. Continue playing until everyone is sitting down. It’s so great for all of our students to hear this way of counting up and on by tens and ones. I love watching my students pick up on the counting patterns and follow the counting around the circle, making predictions about who will say the number. Tens and Ones Pictures This tens and ones activity incorporates fine motor skills and creativity into our math block, as students use their imagination to create a picture using tens and ones. I let students decide to make an animal, a shape, a design… the possibilities are endless! For this activity, you’ll need images of ten sticks and ones, a piece of construction paper, and scissors and glue. The only rules are that they can only use ten sticks and ones to create their picture. Students cut out the pieces they’ll need to create their picture and glue the pieces onto the construction paper. I leave space or give them a template to share how many tens and ones and the total units they used to make their picture. Some students really go all-in on this activity and challenge themselves to make an elaborate picture or make multiple pictures to turn into a book. This offers the opportunity to differentiate so that students can work with their just-right amount of tens and ones. An added bonus? Students love seeing each other’s designs and figuring out how many units their peers used in their designs. You can grab free place vale block visuals for free in my printable manipulatives pack. Free Printable Math Manipulatives Need to keep students supplies separate, but still want them to have access to ALL the math manipulatives? Grab these free printables and problem solved! Give it to me! Free Printable Math Manipulatives Need to keep students supplies separate, but still want them to have access to ALL the math manipulatives? Grab these free printables and problem solved! Bonus: Incorporate Tens and Ones into Classroom Organization and Routines There are so many ways that tens and ones fit into our daily routines. I love providing my students with opportunities to work with tens and ones outside of our math block. This got me thinking, where else could students work with tens and ones as part of our classroom routines? A few ideas that my students love: attendance board putting away Unifix Cubes our calendar routine …the possibilities are endless! What are your favorite ways to practice tens and ones with your students? Place Value Resources Tens & Ones Math Centers Place Value Math Centers Free Number Talk Slides Free Hundred Chart Puzzles You May Also Enjoy These Posts: Number Magnitude: What Is It and Why Should I Teach It?5 Easy Ways to Add Ten Frames to Your Classroom Routines3 Meaningful Ways to Introduce Place Value Ready to go deeper? JOIN MEANINGFUL MATH Become the math teacher you want to be! Were you given just one course on teaching math and then expected to know how to teach it? You are not alone. Meaningful Math promises you ah-ha moments, deeper understanding, short-term wins, and long-term, meaningful change. I can’t wait to support you on your math journey! hello I'm Jillian I’m so happy you’re here. I want every child to feel confident in their math abilities, and that happens when every teacher feels confident in their ability to teach math. In my fifteen years of teaching, I sought every opportunity to learn more about teaching math. I wanted to know HOW students develop math concepts, just like I had been taught how students learn to read. I want every teacher to experience the same math transformation I did, and have the confidence to teach any student that steps foot in their classroom. I’m excited to be alongside you in your math journey!
6292	https://math.stackexchange.com/questions/638539/did-i-go-about-determining-the-coplanarity-of-these-three-vectors-wrong	Did I go about determining the coplanarity of these three vectors wrong? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Did I go about determining the coplanarity of these three vectors wrong? Ask Question Asked 11 years, 8 months ago Modified7 years, 8 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I asked this question a few days ago, where the question was this: I have a task stating this: Determine if the following vectors are coplanar. Assume that v 1 v 1, v 2 v 2 and v 3 v 3 are not coplanar. w 1=4 v⃗1+3 v⃗2 w 1=4 v→1+3 v→2 w 2=v⃗2+4 v⃗3 w 2=v→2+4 v→3 w 3=−v⃗1−3 v⃗3 w 3=−v→1−3 v→3 I don't quite understand how I'll do this when I do not know the values of any of the vectors. Also, what significance does the information "v 1 v 1, v 2 v 2 and v 3 v 3 are not coplanar" have in terms of the solution? I'm guessing knowing that helps decide whether they're coplanar or not, but I can't see how. In that question we arrived at the conclusion that they are coplanar. However, since then, I've attempted to find a way to mathmatically prove that they are coplanar, but instead I seem to arrive at the conclusion that they are in fact NOT coplanar. What I've done looks like this: If the sum of the three vectors equals the zero vector in any way other than by multiplying every vector by zero, then they are coplanar. So if I can prove that one of the vectors can be represented as a multiple of the other, then they are coplanar. So to try and prove this, I’ll try to add two vectors. w⃗1+w⃗2=(4 v⃗1+3 v⃗2)+(v⃗2+4 v⃗3)w→1+w→2=(4 v→1+3 v→2)+(v→2+4 v→3) From there, I realized I could do this instead: w⃗1−3 w⃗2=(4 v⃗1+3 v⃗2)−3(v⃗2+4 v⃗3)w→1−3 w→2=(4 v→1+3 v→2)−3(v→2+4 v→3) w⃗1−3 w⃗2=4 v⃗1+3 v⃗2−3 v⃗2−12 v⃗3 w→1−3 w→2=4 v→1+3 v→2−3 v→2−12 v→3 w⃗1−3 w⃗2=4 v⃗1−12 v⃗3 w→1−3 w→2=4 v→1−12 v→3 Now the two vectors are written in the same terms as the third equation. However, this poses the problem that I need to something to multiply (4−12)(4−12) with to make it (−1−3)(−1−3). Which simply isn't possible, meaning there is no way I can represent w⃗1 w→1 and w⃗2 w→2 as a linear combination of w⃗3 w→3. This again ultimately means the vectors are linearly independent, and thus also not coplanar. So, did I do something wrong here? Am I trying to prove this the wrong way, or was the answer for the other question simply wrong? As far as I can tell now, they are not coplanar, but in the other question we concluded that they were. vectors Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:20 CommunityBot 1 asked Jan 14, 2014 at 20:47 ThreethumbThreethumb 969 6 6 gold badges 25 25 silver badges 37 37 bronze badges 4 I'm sorry, but your statement and example are wrong. Of course are 0,e 1,e 2 0,e 1,e 2 coplanar since you can write 100⋅0+0⋅e 1+0⋅e 2=0.I n s t e a d o f 100⋅0+0⋅e 1+0⋅e 2=0.I n s t e a d o f 100$ you can use any non-zero constant. We also have the result that any three non-zero vectors are linear dependent if and only if they are coplanar. (See portal.tpu.ru/SHARED/k/KONVAL/Sites/English_sites/V/…)Marc –Marc 2014-01-14 21:20:28 +00:00 Commented Jan 14, 2014 at 21:20 @Marc: My apologies, I was out to lunch! I read collinear instead of coplanar. I will delete my comment.copper.hat –copper.hat 2014-01-15 05:24:29 +00:00 Commented Jan 15, 2014 at 5:24 @Marc: The question seems a little bogus to me. Any three vectors must be coplanar. Clearly S=sp{v 1−v 3,v 2−v 3}S=sp⁡{v 1−v 3,v 2−v 3} is at most a two dimensional space, hence a plane (or contained on one), and S+{v 3}S+{v 3} is just a translated plane.copper.hat –copper.hat 2014-01-15 05:54:40 +00:00 Commented Jan 15, 2014 at 5:54 No, three vectors must not be coplanar. Coplanarity of 3 vectors in 3D space is identical to linear dependency of 3 vectors in 3D space. If 3 vectors can be linearly independent in 3D space, then they can be non-coplanar too.Threethumb –Threethumb 2014-01-15 10:05:48 +00:00 Commented Jan 15, 2014 at 10:05 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. You did the right thing in general, but trying out specific cases is a bad way to go -- you might miss the one combination of the w w s that actually comes out to zero. Instead, what you want to do is say this: Let's suppose that the w w are coplanar. Then there are numbers a a, b b, and c c such that a w 1+b w 2+c w 3=0 a w 1+b w 2+c w 3=0 Now I can substitute in the expressions for the w w s to get that a(4 v 1+3 v 2)+b(v 2+4 v 3)+c(−v 1−3 v 3)=0 a(4 v 1+3 v 2)+b(v 2+4 v 3)+c(−v 1−3 v 3)=0 Expanding, you get 4 a v 1+3 a v 2+b v 2+4 b v 3−c v 1−3 c v 3=0(4 a−c)v 1+(3 a+b)v 2+(4 b−3 c)v 3=0 4 a v 1+3 a v 2+b v 2+4 b v 3−c v 1−3 c v 3=0(4 a−c)v 1+(3 a+b)v 2+(4 b−3 c)v 3=0 So IF you could find a (not all zeros) combination of the w w s that's zero, you'd also have a combination of the v v s that's zero. You might think that'd mean that the v v s are coplanar, but it only means that if the coefficients aren't all zero. What would make all three of those coefficients be zero? Let's do some algebra. We'll assume that 4 a−c=0 3 a+b=0 4 b−3 c=0 4 a−c=0 3 a+b=0 4 b−3 c=0 Multiply the first equation by 3 to get 12 a−3 c=0 3 a+b=0 4 b−3 c=0 12 a−3 c=0 3 a+b=0 4 b−3 c=0 Subtract the thrid equation from the first to get 12 a−4 b=0 3 a+b=0 4 b−3 c=0 12 a−4 b=0 3 a+b=0 4 b−3 c=0 Multiply the second equation by 4 4 to get 12 a−4 b=0 12 a+4 b=0 4 b−3 c=0 12 a−4 b=0 12 a+4 b=0 4 b−3 c=0 Subtract the first from the second to get 12 a−4 b=0 8 b=0 4 b−3 c=0 12 a−4 b=0 8 b=0 4 b−3 c=0 So b=0 b=0, which (from the first equation) says that a=0 a=0, which from the third equation tells us c=0 c=0. In summary: if the three w w s were coplanar, we'd have found a,b,c a,b,c, not all zero, with a w 1+b w 2+c w 3=0 a w 1+b w 2+c w 3=0. That would in turn make a combination of the v v s be zero. That latter combination has all-zero coeffs only if the former does, so you'd have found a not-all-zero combination of the v v s that's zero, which would imply the v v s are coplanar. Since they're not, we've got a contradiction. The assumption that the w w s are coplanar must be wrong. Hence they're non-coplanar. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 14, 2014 at 20:58 John HughesJohn Hughes 102k 4 4 gold badges 88 88 silver badges 160 160 bronze badges 6 Ah yes, I actually did think of this problem too. So I tried every combination of the w w s to make sure I didn't miss anything. Your method is a lot more foolproof though, and leaves no doubt that they are not coplanar. Thanks!Threethumb –Threethumb 2014-01-14 21:04:35 +00:00 Commented Jan 14, 2014 at 21:04 Could you expand a bit on why you make every of the parenthesized expressions equal zero? I don't quite get how doing that proves that there are no non-zero scalars to multiply the terms with to make their sum equal zero.Threethumb –Threethumb 2014-01-14 21:29:09 +00:00 Commented Jan 14, 2014 at 21:29 The big structure of this argument is this: If the w w s are coplanar, then the v v s are coplanar. To show that, I assume the w w s are coplanar, and I get that a certain combination of the v v s is zero. That's of no use to me unless I know that it's a nontrivial combination (i.e., at least one of the three coeffs is nonzero). So I ask "Is it possible that all three are zero? If they were, what would that imply?" To answer that rhetorical question, I set all three to zero, and discover that it would mean that a,b,c a,b,c were all zero. So since a,b,c a,b,c are not all zero, nor are all three coeffs.John Hughes –John Hughes 2014-01-14 21:33:20 +00:00 Commented Jan 14, 2014 at 21:33 Ah, I get the argument you're posing now. I'm still not 100% how you determine that the a,b,c a,b,c are not all zero though.Threethumb –Threethumb 2014-01-14 21:46:32 +00:00 Commented Jan 14, 2014 at 21:46 We ASSUME that a,b,c a,b,c not all zero (i.e., that the w w s are coplanar). From this, we conclude (by the argument in my last comment) that the coeffs in the v v combination are not all zero. That makes the v v s coplanar, a contradiction. So our assumption (that there exist a,b,c a,b,c not all zero, such that a w 1+b w 2+c w 3=0 a w 1+b w 2+c w 3=0) was false, i.e., our assumption that the w w s are coplanar was false.John Hughes –John Hughes 2014-01-14 22:03:10 +00:00 Commented Jan 14, 2014 at 22:03 \|Show 1 more comment This answer is useful 0 Save this answer. Show activity on this post. I think you are correct: if one is a linear combination of the other two, it can be any one. You picked w⃗3 w→3 and showed that it is not. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 14, 2014 at 20:54 user66081user66081 4,057 18 18 silver badges 29 29 bronze badges 1 That statement (that one vector being a combination of two others is unaffected by the vector you pick) is not generally true: if v 1 v 1 is a combination of v 2 v 2 and v 3 v 3, it's possible that v 2 v 2 is not a combination of v 1 v 1 and v 3 v 3. Example: v 1=[2,0,0];v 2=[0,1,0],v 3=[1,0,0]v 1=[2,0,0];v 2=[0,1,0],v 3=[1,0,0]. You then have v 1=0 v 2+2 v 3 v 1=0 v 2+2 v 3 but v 2 v 2 cannot be expressed as a combination of v 1 v 1 and v 3 v 3.John Hughes –John Hughes 2014-01-15 12:26:48 +00:00 Commented Jan 15, 2014 at 12:26 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions vectors See similar questions with these tags. 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6293	https://www.ck12.org/flexi/cbse-math/venn-diagrams-and-operations-of-sets/define-venn-diagram./	Define Venn Diagram. - Examples \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Venn Diagrams and Operations of Sets Question Define Venn Diagram. Flexi Says: A Venn Diagram is a diagrammatic representation of sets. It is used to illustrate the relationships between different sets, showing all possible logical relations between these sets. The sets are represented by circles or ovals, and the position and overlap of these circles indicate the relationships between the sets. For example, if we have a set A, a Venn Diagram can show all the elements that are a part of A, given a universal set U. i Analogy / Example Try Asking: Define the Property of Ⲫ/ Identity Law in set theory.What is the range of a set?What are the properties of the complement of a set? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
6294	https://www.geeksforgeeks.org/engineering-mathematics/hyperbolic-function/	Hyperbolic Function Last Updated : 12 Jul, 2025 Suggest changes Like Article Hyperbolic Functions are similar to trigonometric functions but their graphs represent the rectangular hyperbola. These functions are defined using hyperbola instead of unit circles. Hyperbolic functions are expressed in terms of exponential functions ex. In this article, we will learn about the hyperbolic function in detail, including its definition, formula, and graphs. Table of Content What are Hyperbolic Functions? Hyperbolic Functions Formulas Domain and Range of Hyperbolic Functions Properties of Hyperbolic Functions Hyperbolic Trig Identities Inverse Hyperbolic Functions Hyperbolic Definition The six basic hyperbolic functions are, Hyperbolic sine or sinh x Hyperbolic cosine or cosh x Hyperbolic tangent or tanh x Hyperbolic cosecant or cosech x Hyperbolic secant or sech x Hyperbolic cotangent or coth x Hyperbolic functions are defined using exponential functions. They are represented as, sinh x which is read as hyperbolic sinh x. Then the sinh x is defined as, sinh x = (ex + e-x)/2 Similarly, other hyperbolic functions are defined. Hyperbolic Functions Formulas Various hyperbolic function formulas are, sinh(x) = (ex - e-x)/2 \| Function \| Definition \| --- \| \| Hyperbolic Cosine (cosh x) \| cosh(x) = (ex + e-x)/2 \| \| Hyperbolic Sine (sinh x) \| \| Hyperbolic Tangent (tanh x) \| tanh(x) = sinhx/coshx = (ex - e-x)/(ex + e-x) \| \| Hyperbolic Cotangent (coth x) \| coth(x) = cosh x/sin hx = (ex + e-x)/(ex - e-x) \| \| Hyperbolic Secant (sech x) \| ech(x) = 1/cosh x = 2/(ex + e-x) \| \| Hyperbolic Cosecant (csch x) \| csch(x) = 1/sinh x = 2/(ex - e-x) \| Domain and Range of Hyperbolic Functions Domain and Range are the input and output of a function, respectively. The domain and range of various hyperbolic functions are added in the table below: \| Hyperbolic Function \| Domain \| Range \| --- \| sinh x \| (-∞, +∞) \| (-∞, +∞) \| \| cosh x \| (-∞, +∞) \| [1, ∞) \| \| tanh x \| (-∞, +∞) \| (-1, 1) \| \| coth x \| (-∞, 0) U (0, + ∞) \| (-∞, -1) U (1, + ∞) \| \| sech x \| (-∞, + ∞) \| (0, 1] \| \| csch x \| (-∞, 0) U (0, + ∞) \| (-∞, 0) U (0, + ∞) \| Learn, Domain and Range of a Function Properties of Hyperbolic Functions Various properties of the hyperbolic functions are added below, sinh (-x) = – sinh(x) cosh (-x) = cosh (x) tanh (-x) = - tanh x coth (-x) = - coth x sech (-x) = sech x csc (-x) = - csch x cosh 2x = 1 + 2 sinh2(x) = 2 cosh2x - 1 cosh 2x = cosh2x + sinh2x sinh 2x = 2 sinh x cosh x Hyperbolic functions are also derived from trigonometric functions using complex arguments. Such that, sinh x = - i sin(ix) cosh x = cos(ix) tanh x = - i tan(ix) coth x = i cot(ix) sech x = sec(ix) Hyperbolic Trigonometric Identities There are various identities that are related to hyperbolic functions. Some of the important hyperbolic trigonometric identities are, sinh(x ± y) = sinh x cosh y ± coshx sinh y cosh(x ± y) = cosh x cosh y ± sinh x sinh y tanh(x ± y) = (tanh x ± tanh y)/ (1± tanh x tanh y) coth(x ± y) = (coth x coth y ± 1)/(coth y ± coth x) sinh x – sinh y = 2 cosh [(x+y)/2] sinh [(x-y)/2] sinh x + sinh y = 2 sinh [(x+y)/2] cosh[(x-y)/2] cosh x + cosh y = 2 cosh [(x+y)/2] cosh[(x-y)/2] cosh x – cosh y = 2 sinh [(x+y)/2] sinh [(x-y)/2]) cosh2x - sinh2x = 1 tanh2x + sech2x = 1 coth2x - csch2x = 1 2 sinh x cosh y = sinh (x + y) + sinh (x - y) 2 cosh x sinh y = sinh (x + y) – sinh (x – y) 2 sinh x sinh y = cosh (x + y) – cosh (x – y) 2 cosh x cosh y = cosh (x + y) + cosh (x – y) Also, Check Trigonometric Identities Hyperbolic Functions Derivative Derivatives of Hyperbolic functions are used to solve various mathematical problems. The derivative of hyperbolic cos x is hyperbolic sin x, i.e. d/dx (cosh x) = sinh x = d(cosh(x))/dx = d((ex + e-x)/2)/dx = 1/2(d(ex + e-x)/dx) = 1/2(ex - e-x) = sinh(x) Similarly, derivatives of other hyperbolic functions are found. The table added below shows the hyperbolic functions. \| Derivatives of Hyperbolic Functions \| \| --- \| \| Hyperbolic Function \| Derivative \| \| sinh x \| cosh x \| \| cosh x \| sinh x \| \| tanh x \| sech2 x \| \| coth x \| -csch2 x \| \| sech x \| -sech x.tanh x \| \| csch x \| -csch x.coth x \| Learn, Derivative in Maths Integration of Hyperbolic Functions Integral Hyperbolic functions are used to solve various mathematical problems. The integral of hyperbolic cos x is hyperbolic sin x, i.e. ∫ (cosh x).dx = sinh x + C The table added below shows the integration of various hyperbolic functions. \| Integral of Hyperbolic Functions \| \| --- \| \| Hyperbolic Function \| Integral \| \| sinh x \| cosh x + C \| \| cosh x \| sinh x + C \| \| tanh x \| ln (cosh x) + C \| \| coth x \| ln (sinh x) + C \| \| sech x \| arctan (sinh x) + C \| \| csch x \| ln (tanh(x/2)) + C \| Learn, Integration Inverse Hyperbolic Functions Inverse hyperbolic functions are found by taking the inverse of the hyperbolic function, i.e. if y = sinh x then, x = sinh-1 (y) this represents the inverse hyperbolic sin function. Now the inverse of various hyperbolic function are, sinh-1x = ln (x + √(x2 + 1)) cosh-1x = ln (x + √(x2 - 1)) tanh-1x = ln [(1 + x)/(1 - x)] coth-1x = ln [(x + 1)/(x - 1)] sech-1x = ln [{1 + √(1 - x2)}/x] csch-1x = ln [{1 + √(1 + x2)}/x] Learn More: Inverse Trigonometric Identities Hyperbolic Functions Examples Example 1: Find the value of x solving, 4sinh x - 6cosh x - 2 = 0. Solution: We know that, sinh x = (ex - e-x)/2 cosh x = (ex + e-x)/2 Given, 4sinh x - 6cosh x + 2 = 0 ⇒ 4[(ex - e-x)/2] - 6[(ex + e-x)/2] + 6 = 0 ⇒ 2(ex - e-x) - 3(ex + e-x) + 6 = 0 ⇒ 2ex - 2e-x - 3ex - 3e-x + 6 = 0 ⇒ -ex - 5e-x + 6 = 0 ⇒ -e2x - 5 + 6ex = 0 ⇒ e2x - 6ex + 5 = 0 ⇒ e2x - 5ex - ex + 5 = 0 ⇒ ex(ex - 5) - 1(ex - 5) = 0 ⇒ (ex - 1)(ex - 5) = 0 (ex - 1) = 0 ex = 1 x = 0 (ex - 5) = 0 ex = 5 x = ln 5 Example 2: Prove, cosh x + sinh x = ex Solution: LHS = cosh x + sinh x = (ex - e-x)/2 + (ex + e-x)/2 = (ex - e-x + ex + e-x)/2 = 2ex / 2 = ex = RHS Hyperbolic Functions Practice Questions Q1: Find the value of x solving, sinh x + 5cosh x - 4 = 0 Q2: Find the value of x solving, 2sinh x - 6cosh x - 5 = 0 Q3: Find the value of x solving, 9sinh x + 6cosh x + 11 = 0 Q4: Find the value of x solving, sinh x - cosh x - 3 = 0 Related : Hyperbola Trigonometric Functions Unit Circle Differentiation Formulas Integration Formulas Trigonometric Formulas Next Article Matrices M MandarBapat Improve Article Tags : Engineering Mathematics Maths-Class-12 Similar Reads Engineering Mathematics Tutorials Engineering mathematics is a vital component of the engineering discipline, offering the analytical tools and techniques necessary for solving complex problems across various fields. Whether you're designing a bridge, optimizing a manufacturing process, or developing algorithms for computer systems, 3 min read Linear Algebra Matrices Matrices are key concepts in mathematics, widely used in solving equations and problems in fields like physics and computer science. A matrix is simply a grid of numbers, and a determinant is a value calculated from a square matrix.Example: \begin{bmatrix} 6 & 9 \ 5 & -4 \ \end{bmatrix}_{2 3 min readRow Echelon Form Row Echelon Form (REF) of a matrix simplifies solving systems of linear equations, understanding linear transformations, and working with matrix equations. A matrix is in Row Echelon form if it has the following properties:Zero Rows at the Bottom: If there are any rows that are completely filled wit 4 min readEigenvalues and Eigenvectors Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis and data analysis (e.g., PCA). They are associated with a square matrix and provide insights into its properties.Eigen value and Eigen vectorTable 10 min readSystem of Linear Equations A system of linear equations is a set of two or more linear equations involving the same variables. Each equation represents a straight line or a plane and the solution to the system is the set of values for the variables that satisfy all equations simultaneously.Here is simple example of system of 5 min readMatrix Diagonalization Matrix diagonalization is the process of reducing a square matrix into its diagonal form using a similarity transformation. This process is useful because diagonal matrices are easier to work with, especially when raising them to integer powers.Not all matrices are diagonalizable. A matrix is diagon 8 min readLU Decomposition LU decomposition or factorization of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. It is a fundamental technique in linear algebr 6 min readFinding Inverse of a Square Matrix using Cayley Hamilton Theorem in MATLAB Matrix is the set of numbers arranged in rows & columns in order to form a Rectangular array. Here, those numbers are called the entries or elements of that matrix. A Rectangular array of (mn) numbers in the form of 'm' horizontal lines (rows) & 'n' vertical lines (called columns), is calle 4 min read Sequence & Series Mathematics \| Sequence, Series and Summations Sequences, series, and summations are fundamental concepts of mathematical analysis and it has practical applications in science, engineering, and finance.Table of ContentWhat is Sequence?Theorems on SequencesProperties of SequencesWhat is Series?Properties of SeriesTheorems on SeriesSummation Defin 8 min readBinomial Theorem Binomial theorem is a fundamental principle in algebra that describes the algebraic expansion of powers of a binomial. According to this theorem, the expression (a + b)n where a and b are any numbers and n is a non-negative integer. It can be expanded into the sum of terms involving powers of a and 15+ min readFinding nth term of any Polynomial Sequence Given a few terms of a sequence, we are often asked to find the expression for the nth term of this sequence. While there is a multitude of ways to do this, In this article, we discuss an algorithmic approach which will give the correct answer for any polynomial expression. Note that this method fai 4 min read Calculus Limits, Continuity and Differentiability Limits, Continuity, and Differentiation are fundamental concepts in calculus. They are essential for analyzing and understanding function behavior and are crucial for solving real-world problems in physics, engineering, and economics.Table of ContentLimitsKey Characteristics of LimitsExample of Limi 10 min readCauchy's Mean Value Theorem Cauchy's Mean Value theorem provides a relation between the change of two functions over a fixed interval with their derivative. It is a special case of Lagrange Mean Value Theorem. Cauchy's Mean Value theorem is also called the Extended Mean Value Theorem or the Second Mean Value Theorem.According 7 min readTaylor Series A Taylor series represents a function as an infinite sum of terms, calculated from the values of its derivatives at a single point.Taylor series is a powerful mathematical tool used to approximate complex functions with an infinite sum of terms derived from the function's derivatives at a single poi 8 min readInverse functions and composition of functions Inverse Functions - In mathematics a function, a, is said to be an inverse of another, b, if given the output of b a returns the input value given to b. Additionally, this must hold true for every element in the domain co-domain(range) of b. In other words, assuming x and y are constants, if b(x) = 3 min readDefinite Integral \| Definition, Formula & How to Calculate A definite integral is an integral that calculates a fixed value for the area under a curve between two specified limits. The resulting value represents the sum of all infinitesimal quantities within these boundaries. i.e. if we integrate any function within a fixed interval it is called a Definite 8 min readApplication of Derivative - Maxima and Minima Derivatives have many applications, like finding rate of change, approximation, maxima/minima and tangent. In this section, we focus on their use in finding maxima and minima.Note: If f(x) is a continuous function, then for every continuous function on a closed interval has a maximum and a minimum v 6 min read Probability & Statistics Mean, Variance and Standard Deviation Mean, Variance and Standard Deviation are fundamental concepts in statistics and engineering mathematics, essential for analyzing and interpreting data. These measures provide insights into data's central tendency, dispersion, and spread, which are crucial for making informed decisions in various en 10 min readConditional Probability Conditional probability defines the probability of an event occurring based on a given condition or prior knowledge of another event. It is the likelihood of an event occurring, given that another event has already occurred. In probability, this is denoted as A given B, expressed as P(A \| B), indica 12 min readBayes' Theorem Bayes' Theorem is a mathematical formula used to determine the conditional probability of an event based on prior knowledge and new evidence. It adjusts probabilities when new information comes in and helps make better decisions in uncertain situations.Bayes' Theorem helps us update probabilities ba 13 min readProbability Distribution - Function, Formula, Table A probability distribution is a mathematical function or rule that describes how the probabilities of different outcomes are assigned to the possible values of a random variable. It provides a way of modeling the likelihood of each outcome in a random experiment.While a Frequency Distribution shows 13 min readCovariance and Correlation Covariance and correlation are the two key concepts in Statistics that help us analyze the relationship between two variables. Covariance measures how two variables change together, indicating whether they move in the same or opposite directions. Relationship between Independent and dependent variab 6 min read Practice Questions Last Minute Notes - Engineering Mathematics GATE CSE is a national-level engineering entrance exam in India specifically for Computer Science and Engineering. It's conducted by top Indian institutions like IISc Bangalore and various IITs. In GATE CSE, engineering mathematics is a significant portion of the exam, typically constituting 15% of 15+ min readEngineering Mathematics - GATE CSE Previous Year Questions Solving GATE Previous Year's Questions (PYQs) not only clears the concepts but also helps to gain flexibility, speed, accuracy, and understanding of the level of questions generally asked in the GATE exam, and that eventually helps you to gain good marks in the examination. Previous Year Questions h 4 min read We use cookies to ensure you have the best browsing experience on our website. 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6295	https://www.quora.com/How-do-you-find-90-of-a-number	How to find 90% of a number - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Percentage Calculator Decimal Fractions (mathematics) Basic Math 3 Calculations Number and Arithmetic Percentage (%) Simple Mathematics 5 How do you find 90% of a number? All related (43) Sort Recommended Assistant Bot · 1y To find 90% of a number, you can follow these steps: Convert the percentage to a decimal: 90% as a decimal is 0.90. Multiply the number by the decimal: 90% of a number=number×0.90 90% of a number=number×0.90 Example: If you want to find 90% of 50: 90%of 50=50×0.90=45 90%of 50=50×0.90=45 So, 90% of 50 is 45. Upvote · James Angelini NADA from High School Only (Graduated 1968) · Author has 4.3K answers and 3.5M answer views ·4y Here is a fast and accurate method that takes all of the mystery out of the process . Whatever your number might be divide it by ten . With that answer simply multiply that number by 9 . Your result is 90% of your starting number. Now for a few examples : your number is 80 .. divide by ten equals 8 ; multiply by 9 ; equals 72 . your number is 50 .. divide by ten equals 5 ; multiply by 9 ; equals 45 . your number is 25 .. divide by ten ; equals 2.5 ; multiply by 9 = 22.5 Upvote · 9 4 Promoted by Webflow Metis Chan Works at Webflow ·Aug 12 What are the best AI website builders now? When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real Continue Reading When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real time, streamline feedback, and manage all your content in one place. Publish instantly on enterprise-grade hosting with built-in SEO, security, and the flexibility to scale as you grow. Many other tools offer AI-powered templates or quick site launches, but Webflow stands out by letting you take control—so your site never feels generic, and you can easily update, expand, or redesign as your needs change. Want to see what AI-powered site building can really do? Try Webflow AI Site Builder for free today. Upvote · 99 15 Hunter Gagnon Mathematician · Author has 162 answers and 375.3K answer views ·4y Multiply a given number by 0.9 to find 90% of a number. For instance, 90% of 100 is 0.9 x 100, or 90. Another easier way is to find 10% of a number, and multiply it by 9. For instance, 10% of 1000 is 100, and 9 times 100 is 900. 900 is 90% of 1000. Upvote · 9 1 Related questions More answers below How do you find 100 percent of a number? If 60% of a number is 90, what is the number? What is 90% of 90%? How do you get 15 percent of a number? 90 is 30% of what number? Johannes Devries Former Retired, former position: property manager (2014–2017) · Author has 6.3K answers and 1.3M answer views ·4y Multiply the number by .9 Or, if you wish: divide the number by 10 and then multiply the result by 9. Upvote · 9 6 John Rourke Engineer (2013–present) · Author has 77 answers and 75.1K answer views ·4y All the other answers involving multiplying by 0.9 may be correct, but some may find it easier to use subtraction rather than multiplication - so here’s how I do it: Make a second number that’s 1/10 of the first (if the number is 56, the second number would be 5.6) Subtract the second number from the first number (in the example, that would be 56–5.6; which is 50.4) Upvote · 9 1 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 206 Lance Berg Author has 28K answers and 54.7M answer views ·4y Of is an English language word meaning “multiply” 90% is a mathematical phrase meaning 0.90 “A number” is a variable, you can use x, or z, or q, or n, or whatever you like to represent it, in this case, I prefer n since the word number happens to start with it. 90% of a number = 0.9n Upvote · 9 1 Related questions More answers below How do I find 12.5% of any number easily? How do you find percentage of number, find 17% of 10000? What is the number whose 15% is 90? How can I find the percentage of any number easily? 10 is 50 percent of what number? Mike Kidwell Author has 138 answers and 79.8K answer views ·4y Simply deduct 10% or multiply by 0.9 e.g. 100 -10% (i.e. 10) 100–10 =90 100 x 0.9 = 90 25 - 10 % (I.e. 2.5) 25–2.5 = 22.5 25 x 0.9 = 22.5 5 - 10% 5–0.5 = 4.5 5 x 0.9 = 4.5 Upvote · Sponsored by USA Corporate Incorporate in the US correctly and avoid costly mistakes later. Download our free e-book to learn about visas, LLCs, and more from a leading US incorporation service. Download 999 116 Chris Robida Former 40 Years Teaching Elem., Middle Sch. & College (1973–2016) · Author has 4.8K answers and 2.6M answer views ·4y Probably the easiest way to find the percent of any number is to multiply the number by .90 (which is the same as 90%), or by .9…and ta da…you’ll have your answer. Upvote · 9 1 Joshua Ruhl Has a PhD in One Foot in Front of the Other ·4y Divide by 100 and multiply by 90, or if you simplify the equation, you can simply divide by 10 and multiply by 9. Upvote · 9 1 Sponsored by Felix Health Ozempic: Am I eligible? Curious about Ozempic? Talk to a Canadian health professional online - find out if you're eligible today! Learn More 999 725 Philip Wu Studied Mathematics&Sociology at University of California, Berkeley · Author has 1.1K answers and 980.5K answer views ·5y Related How do you find 100 percent of a number? If you are given a percentage (other than 100%) then just divide the amount by the percentage (so that you find “1% of it”) and then multiply by 100 (so that you find 100%). For example, if it is said 15 equals 20%, then divide 15 by 20 to get 1% (which equals 3/4 or 0.75), then multiply it by 100 which equals 75. Upvote · 99 34 9 5 Aftab Naseer B.Sc Electrical Engineering from National University of Sciences and Technology, Pakistan ·4y Note that 1% is 1/100, therefor 90% will be 90/100 or 9/10. So if you want to 90% of number just multiply it with 90/100 or 9/10, the result will be the 90% of that number. Upvote · Murali Krishna Former Retired Senior Lecturer in DIET at Government (General) (1990–2004) · Author has 6.5K answers and 6.7M answer views ·4y Let thd number be say 150 90% 0f 150 is 150 ×90/100 = =15×9= 135 90 % 0f X is 90/100 × X = 9X/10 Upvote · Malavika Sajeev 5y Originally Answered: How do you solve 90% of 90? · 90% 90 =90/10090 =8100/100 =81 Upvote · 9 1 9 1 Related questions How do you find 100 percent of a number? If 60% of a number is 90, what is the number? What is 90% of 90%? How do you get 15 percent of a number? 90 is 30% of what number? How do I find 12.5% of any number easily? How do you find percentage of number, find 17% of 10000? What is the number whose 15% is 90? How can I find the percentage of any number easily? 10 is 50 percent of what number? If 24 of a number equals 90%, what is 100% of the number? What's the number of digits in the number 100? 48 is 32 percentage of what number? Which number is 10% less than 100? How do you solve 90% of 90? Related questions How do you find 100 percent of a number? If 60% of a number is 90, what is the number? What is 90% of 90%? How do you get 15 percent of a number? 90 is 30% of what number? How do I find 12.5% of any number easily? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
6296	https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas?srsltid=AfmBOoq-12dxUC34jYVPToQskuJP6RM5tXzN6Ivml6ZQecavjBZmYXmx	Art of Problem Solving Vieta's Formulas - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Vieta's Formulas Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Vieta's Formulas In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients. It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments. Contents 1 Statement 2 Proof 3 Problems 3.1 Introductory 3.2 Intermediate 4 Advanced 5 See also Statement Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots. Vieta’s formulas then state that This can be compactly summarized as for some such that . Proof Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant . Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof. Problems Here are some problems with solutions that utilize Vieta's quadratic formulas: Introductory 2005 AMC 12B Problem 12 2007 AMC 12A Problem 21 2010 AMC 10A Problem 21 2003 AMC 10A Problem 18 2021 AMC 12A Problem 12 Intermediate 2017 AMC 12A Problem 23 2003 AIME II Problem 9 2008 AIME II Problem 7 2021 Fall AMC 12A Problem 23 2019 AIME I Problem 10 Advanced 2020 AIME I Problem 14 See also Polynomial Retrieved from " Categories: Algebra Polynomials Theorems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6297	https://byjus.com/chemistry/tin-oxide/	What is Tin Oxide? The inorganic compound tin(IV) oxide, also known as stannic oxide, has the formula SnO2. Cassiterite is a tin oxide mineral, SnO2, and it is the most common tin ore. Tin metal is burned in the air to create synthetic tin(IV) oxide. The annual production is in the kilotons scale. In a reverberatory furnace at 1200–1300 °C, SnO2 is reduced to metal with carbon. Tin(iv) oxide is a crystalline solid or powder that is white or off-white. Sublime: 1800-1900°C, mp: 1127°C, density: 6.95 g/cm3 Water doesn’t remove it. It can dissolve in hydrochloric acid and concentrated sulphuric acid. Other names – Stannic oxide \| \| \| --- \| \| SnO2 \| Tin Oxide \| \| Density \| 6.95 g/cm³ \| \| Molecular Weight/ Molar Mass \| 150.71 g/mol \| \| pH \| (6–12) \| \| Melting Point \| 1,630 °C \| \| Chemical Formula \| SnO2 \| Table of Contents Tin Oxide Structure Physical Properties of Tin Oxide Chemical Properties of Tin Oxide Uses of Tin Oxide Frequently Asked questions Tin Oxide Structure – SnO2 Physical Properties of Tin Oxide – SnO2 \| \| \| --- \| \| Odour \| Odourless \| \| Appearance \| Yellowish or light grey powder \| \| Covalently-Bonded Unit \| 2 \| \| Heavy Atom Count \| 3 \| \| Hydrogen Bond Acceptor \| 2 \| \| Solubility \| insoluble in water \| Chemical Properties of Tin Oxide – SnO2 Sulfate is formed when SnO2 dissolves in sulphuric acid. The chemical equation is given below. SnO2 + 2 H2SO4 → Sn(SO4)2 + 2 H2O When tin(IV) oxide reacts with sodium hydroxide, stannate (IV) sodium and water are formed. At a temperature of 350-400°C, this reaction takes place. The chemical equation is given below. SnO2 + 2NaOH → Na2SnO3 + H2O Uses of Tin Oxide – SnO2 Tin oxide, which has a +4 oxidation state, can be used to make ceramic bodies opaque, as a mild abrasive, and as a fabric weighting agent. The cosmetics ingredient review has determined that tin oxide is a healthy ingredient, based on the fact that it is not easily absorbed through the skin. Tin oxide is used in dentistry and to render a precious metal polishing paste for high polishing of amalgam and precious metals when combined with water. Frequently Asked Questions – FAQs Is SnO2 tin oxide? Tin(IV) oxide, also known as stannic oxide, is the inorganic compound with the formula SnO2. Q2 How is SnO2 formed? Crystalline SnO2 starts to form when annealed at 200 °C and becomes dominant only after annealing at 300 °C. It is thus suggested that the SnO2 shell formation, from decomposition/oxidation of hydrates, is a result of a strong interaction between the metal oxide and SiO2 surface. Q3 Is SnO2 acidic or basic? SnO2 reacts with acid as well as the base. So SnO2 is amphoteric. Q4 How do you dissolve SnO2? Tin dioxide, SnO2, can be reactively dissolved by soaking in hot aqueous solutions of HBr or HCl (approx. 6 N) to which metallic chromium and/or zinc are also added (as needed). Test Your Knowledge On Tin Oxide! Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs
6298	https://thirdspacelearning.com/us/math-resources/topic-guides/geometry/translation-math/	High Impact Tutoring Built By Math Experts Personalized standards-aligned one-on-one math tutoring for schools and districts Request a demo In order to access this I need to be confident with: 2D shapes Polygons Quadrilateral Scale factor Coordinate plane Decimals What is translation math? Common Core State Standards How to use translation Translation math examples Example 1: translate a 2D shape Example 2: translate a 2D shape Example 3: translate a 2D shape How to describe translations Describing translations examples ↓ Example 4: describe a translation Example 5: describe a translation Example 6: describe a translation Teaching tips for translation math Easy mistakes to make Related transformations lessons Practice translation math questions Translation math FAQs Next lessons Still stuck? Math resources Geometry Transformations Translation math Translation math Here you will learn about translation math, including how to translate 2D shapes and how to describe translations. Students will first learn about translation math as part of geometry in 8 th grade. What is translation math? A translation in math is a type of transformation that refers to moving a shape or object from one location to another without changing its size, shape, or orientation. A translation moves a shape in a horizontal direction (left and right) and in a vertical direction (up and down). A translation is typically described using a column vector to help record the movement. For example, Shape A has been translated to shape B by the column vector \begin{pmatrix} \; 3 \;\ \; 2 \; \end{pmatrix} . The column vector gives instructions on how to move each point of the original shape (also known as the pre-image). \begin{pmatrix} \; 3 \;\ \; 2 \; \end{pmatrix} \text{ is } \; \begin{matrix} 3 \ \text{right}\ 2 \ \text{up}\ \end{matrix} The inverse transformation would translate shape B back to shape A using the column vector: \begin{pmatrix} \; -3 \;\ \; -2 \; \end{pmatrix} \text{ is } \; \begin{matrix} 3 \ \text{left}\ 2 \ \text{down}\ \end{matrix} The object or pre-image is the name of the original shape (or original figure). The image is the name of the shape after it had been translated. When an object is translated, the object and the image have the same orientation. There is no rotation or reflection. The object and the image are congruent because they are the same shape and the same size. Using graph paper or tracing paper can be useful when translating shapes. What is translation math? Common Core State Standards How does this relate to 8 th grade math and high school math? Grade 8 – Geometry (8.G.A.3)Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates. High School – Geometry – Congruence (HS.G.CO.A.5) Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, example, graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another. High School – Geometry – Congruence (HS.G.CO.B.6) Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent. [FREE] Common Core Practice Tests (Grades 3 to 8) Prepare for math tests in your state with these Grade 3 to Grade 8 practice assessments for Common Core and state equivalents. 40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics! DOWNLOAD FREE x [FREE] Common Core Practice Tests (Grades 3 to 8) Prepare for math tests in your state with these Grade 3 to Grade 8 practice assessments for Common Core and state equivalents. 40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics! DOWNLOAD FREE How to use translation In order to translate a 2D shape: Choose a vertex of the shape. Move it left or right using the top value in the column vector. Move it up or down using the bottom value in the column vector. Repeat for each vertex. Translation math examples Example 1: translate a 2D shape Translate shape A by the column vector \begin{pmatrix} \; 4 \;\ \; 1 \; \end{pmatrix} and label the image B. Choose a vertex of the shape. Choose one of the vertices of the shape (a coordinate) as the point you are going to move. Let’s use the bottom right hand point of the shape. 2Move it left or right using the top value in the column vector. The top number in the column vector is 4 , so move the point \bf{4} squares to the right. 3Move it up or down using the bottom value in the column vector. The bottom number in the column vector is 1 , so move the point \bf{1} square up. 4Repeat for each vertex. You can move each of the other points in the same way and draw in the rest of the 2D shape. Example 2: translate a 2D shape Translate shape A by the column vector \begin{pmatrix} \; -1 \;\ \; 3 \; \end{pmatrix} and label the image B. Choose a vertex of the shape. Choose one of the vertices of the shape as the point you are going to move. Let’s use the bottom left hand corner of the shape. Move it left or right using the top value in the column vector. The top number in the column vector is - \, 1 , so move the point \bf{1} square to the left. Move it up or down using the bottom value in the column vector. The bottom number in the column vector is 3 , so move the point \bf{3} squares up. Repeat for each vertex. You can move each of the other points in the same way and draw in the rest of the 2D shape. Example 3: translate a 2D shape Translate shape A by the column vector \begin{pmatrix} \; 0 \;\ \; -4 \; \end{pmatrix} and label the image B. Choose a vertex of the shape. Choose one of the vertices of the shape as the point you are going to move. Let’s use the bottom left hand corner of the shape. Move it left or right using the top value in the column vector. The top number in the column vector is 0 , so there is no horizontal movement of the point. Move it up or down using the bottom value in the column vector. The bottom number in the column vector is - \, 4 , so move the point \bf{4} squares down. Repeat for each vertex. You can move each of the other points in the same way and draw in the rest of the 2D shape. How to describe translations In order to describe a translation of a shape: Pair up two identical vertices. Work out the horizontal movement. Work out the vertical movement. State the column vector. Describing translations examples Example 4: describe a translation Describe the following translation of shape P to shape Q. Pair up two identical vertices. Pair up a point from the object and the corresponding point on the image. Here one pair of points is chosen. It should not matter which pair of points are chosen. From the object P the point is (1, \, 1). For the image Q the point is (3, \, 2). Work out the horizontal movement. Using the grid or the x -coordinates, work out the horizontal movement. There is a shift of \bf{2} to the right. Work out the vertical movement. Using the grid or the y -coordinates, work out the vertical movement. There is a shift of \bf{1} upwards. State the column vector. Now you need to write the horizontal shift and the vertical shift in a column vector. \begin{matrix} 2 \ \text{right}\ 1 \ \text{up}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; 2 \;\ \; 1 \; \end{pmatrix} Shape P has been translated to shape Q by the column vector \begin{pmatrix} \; 2 \;\ \; 1 \; \end{pmatrix} . Example 5: describe a translation Describe the translation of shape A to shape B. Pair up two identical vertices. Pair up a point from the object and the corresponding point on the image. Here one pair of points is chosen. It should not matter which pair of points are chosen. From the object A the point is (6, \, 4). For the image B the point is (1, \, 2). Work out the horizontal movement. Using the grid or the x -coordinates, work out the horizontal movement. There is a shift of \bf{3} to the left. Work out the vertical movement. Using the grid or the y -coordinates, work out the vertical movement. There is a shift of \bf{4} downwards. State the column vector. Now you need to write the horizontal shift and the vertical shift in a column vector. \begin{matrix} 3 \ \text{left}\ 4 \ \text{down}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; -3 \;\ \; -4 \; \end{pmatrix} Shape A has been translated to shape B by the column vector \begin{pmatrix} \; -3 \;\ \; -4 \; \end{pmatrix} . Example 6: describe a translation Describe the translation of shape A to shape B. Pair up two identical vertices. Pair up a point from the object and the corresponding point on the image. Here one pair of points is chosen. It should not matter which pair of points are chosen. From the object A the point is (1, \, 1). For the image B the point is (1, \, 6). Work out the horizontal movement Using the grid or the x -coordinates, work out the horizontal movement. There is no horizontal shift. Work out the vertical movement. Using the grid or the y -coordinates, work out the vertical movement. There is a shift of \bf{5} upwards. State the column vector. Now you need to write the horizontal shift and the vertical shift in a column vector. \begin{matrix} 0 \ \text{left or right}\ 5 \ \text{up}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; 0 \;\ \; 5 \; \end{pmatrix} Shape A has been translated to shape B by the column vector \begin{pmatrix} \; 0 \;\ \; 5 \; \end{pmatrix} . Teaching tips for translation math Start by introducing translations in the coordinate plane. Explain that translating a shape means moving it horizontally (along the x -axis) or vertically (along the y -axis), while the shape remains the same in size and orientation. Connect translation examples to everyday actions, like sliding a book across a table or moving a piece on a board game, to make it relatable. Connect translation to their previous knowledge of coordinate systems from 6 th and 7 th grade. Explain that just like they used coordinates to locate points, they will now use translation vectors to move these points to new coordinates. Once students are comfortable with single translations, give worksheets that include multiple translations, first a horizontal translation then a vertical translation or vice versa. This reinforces the idea of movement along both the x and y axes. Introduce the translation rule, which is typically written as (x, \, y) \rightarrow(x+a, \, y+b), where a is the horizontal shift (left or right), and b is the vertical shift (up or down). Break down translations into steps, calculating each of the new coordinates based on the translation rule. This helps students understand how each axis is affected independently. Easy mistakes to make Not checking the scale of the coordinate grid Be careful to consider the scaling on the axes. For example, What is the column vector for the translation of shape P to shape Q? If you count the squares, it may appear that shape P has been translated by the column vector \begin{pmatrix} \; 6 \;\ \; 2 \; \end{pmatrix}. But if by considering the scale on the axes, the correct column vector is \begin{pmatrix} \; 3 \;\ \; 1 \; \end{pmatrix} . Misinterpreting the column vector Remember, the top number is for horizontal movement. A positive number moves the shape to the right and a negative number moves the shape to the left. The bottom number is for vertical movement. A positive moves the shape upwards and a negative number moves the shape downwards. Confusing object and image The original shape is the object or pre-image and the translated shape is the image. Make sure you know which shape is the original shape and start there when describing transformations such as translations. Related transformations lessons Transformations Rotations Reflection in math Dilations Center of dilation Practice translation math questions Translate the shaded shape by the column vector \begin{pmatrix} \; 3 \;\ \; 2 \; \end{pmatrix} . Choose a point on the vertex of the shape and move it 3 to the right and 2 up. Do the same with all the points, or carefully draw in the rest of the image. \begin{pmatrix} \; 3 \;\ \; 2 \; \end{pmatrix} \text{ is } \; \begin{matrix} 3 \ \text{right}\ 2 \ \text{up}\ \end{matrix} Translate the shaded shape by the column vector \begin{pmatrix} \; -4 \;\ \; 1 \; \end{pmatrix} . Choose a point on the vertex of the shape and move it 4 to the left and 1 up. Do the same with all the points, or carefully draw in the rest of the image. \begin{pmatrix} \; -4 \;\ \; 1 \; \end{pmatrix} \text{ is } \; \begin{matrix} 4 \ \text{left}\ 1 \ \text{up}\ \end{matrix} Translate the shaded shape by the column vector \begin{pmatrix} \; 3 \;\ \; 0 \; \end{pmatrix} . Choose a point on the vertex of the shape and move it 3 to the right and 0 up or down. Do the same with all the points, or carefully draw in the rest of the image. \begin{pmatrix} \; 3 \;\ \; 0 \; \end{pmatrix} \text{ is } \; \begin{matrix} 3 \ \text{right}\ 0 \ \text{up or down}\ \end{matrix} Describe the transformation of shape A to shape B. Translation by \begin{pmatrix} \; 2 \;\ \; 3 \; \end{pmatrix} Translation by \begin{pmatrix} \; -3 \;\ \; -2 \; \end{pmatrix} Translation by [ \begin{pmatrix} \; 3 \;\ \; 2 \; \end{pmatrix} Translation by \begin{pmatrix} \; -2 \;\ \; -3 \; \end{pmatrix} Choose a point on the object shape A and its corresponding point on the image shape B. Count the shift in the horizontal direction. Count the shift in the vertical direction. \begin{matrix} 2 \ \text{right}\ 3 \ \text{up}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; 2 \;\ \; 3 \; \end{pmatrix} Describe the transformation of shape A to shape B. Translation by \begin{pmatrix} \; 4 \;\ \; 3 \; \end{pmatrix} Translation by \begin{pmatrix} \; -3 \;\ \; 4 \; \end{pmatrix} Translation by \begin{pmatrix} \; 4 \;\ \; -3 \; \end{pmatrix} Translation by \begin{pmatrix} \; -3 \;\ \; -4 \; \end{pmatrix} Choose a point on the object shape A and its corresponding point on the image shape B. Count the shift in the horizontal direction. Count the shift in the vertical direction. \begin{matrix} 4 \ \text{right}\ 3 \ \text{down}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; 4 \;\ \; -3 \; \end{pmatrix} Describe the transformation of shape A to shape B. Translation by \begin{pmatrix} \; 4 \;\ \; -1 \; \end{pmatrix} Translation by \begin{pmatrix} \; 1 \;\ \; -4 \; \end{pmatrix} Translation by \begin{pmatrix} \; 4 \;\ \; 1 \; \end{pmatrix} Translation by \begin{pmatrix} \; 1 \;\ \; 4 \; \end{pmatrix} Choose a point on the object shape P and its corresponding point on the image shape Q. Count the shift in the horizontal direction. Count the shift in the vertical direction. \begin{matrix} 1 \ \text{right}\ 4 \ \text{down}\ \end{matrix} \; \text{ is } \begin{pmatrix} \; 1 \;\ \; -4 \; \end{pmatrix} Translation math FAQs What is a translation in math? A translation in math is a type of transformation where a shape or object is moved from one position to another on the Cartesian plane without changing its size, shape, or orientation. Each point of the shape moves the same distance in the same direction. How do you describe a translation? A translation is described by a vector (a, \, b), where a represents the horizontal shift (positive for right, negative for left) and b represents the vertical shift (positive for up, negative for down). What are the coordinates of a point after translation? If point A \, (x, \, y) is translated by a vector (a, \, b), the new coordinates of point A’ are given by x'=x+a and y'=y+b. What are common mistakes when performing translations? Some common mistakes include mixing up the x -axis and y -axis when applying the translation, forgetting to apply the translation to every point of the shape to find the new points, misinterpreting the direction of negative or positive values in the translation vector. Are there any real-world examples of translations? Yes, real-world examples of translations in math include sliding a book across a table, shifting a piece on a chessboard, or moving an image across a computer screen. The next lessons are Geometric proofs Area 3D shapes Still stuck? At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts. Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. Find out how we can help your students achieve success with our math tutoring programs. Introduction What is translation math? Common Core State Standards How to use translation Translation math examples ↓ Example 1: translate a 2D shape Example 2: translate a 2D shape Example 3: translate a 2D shape How to describe translations Describing translations examples ↓ Example 4: describe a translation Example 5: describe a translation Example 6: describe a translation Teaching tips for translation math Easy mistakes to make Related transformations lessons Practice translation math questions Translation math FAQs Next lessons Still stuck? x [FREE] Common Core Practice Tests (3rd to 8th Grade) Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents. Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers! Download free
6299	https://www.mathdoubts.com/relation-between-logarithms-and-exponents/	Algebra Trigonometry Geometry Calculus Relation between Logarithms and Exponents Math Doubts Logarithms Basics The logarithmic and exponential systems both have mutual direct relationship mathematically. So, the knowledge on the exponentiation is required to start studying the logarithms because the logarithm is an inverse operation of exponentiation. Example The number $9$ is a quantity and it can be expressed in exponential form by the exponentiation. $9 \,=\, 3 \times 3$ $\implies 9 \,=\, 3^2$ In this case, $3$ is a quantity and $2$ is the number of its multiplying factors. The inverse operation of $9 \,=\, 3^2$ is written in logarithmic form. $\log_{3}{(9)} \,=\, 2$ The logarithmic system represents that the number of multiplying factors is $2$ when the quantity $9$ is written as multiplying factors on the basis of number $3$. The mutual inverse mathematical relationship between exponential and logarithmic systems is written in mathematics as follows. $9 \,=\, 3^2$ $\,\, \Leftrightarrow \,\,$ $\log_{3}{(9)} \,=\, 2$ Algebraic form $y$ is a quantity. It is written in terms of $b$ and the total number of multiplying factors is $x$. The relation between three of them is written in mathematical form as per exponentiation. $y \,=\, b^{\, \displaystyle x}$ It is written in logarithmic form in the following way to find the total number of multiplying factors by expressing y as multiplying factors of b.. $\log_{b}{(y)} \,=\, x$ The relationship between logarithmic and exponential systems is written in algebraic form as follows. $y \,=\, b^{\, \displaystyle x}$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{(y)} \,=\, x$ Latest Math Topics Sep 18, 2025 What is a Point in geometry? Sep 13, 2025 What is a Factor in Higher mathematics? Sep 09, 2025 X-Axis of Two dimensional cartesian coordinate system Sep 04, 2025 Distributive property over Addition of Complex conjugates Aug 28, 2025 Origin of Two dimensional cartesian coordinate system May 31, 2025 What is Zero angle? Latest Math Problems Jul 19, 2025 Evaluate $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize (x^3-5x+6)}$ by Direct substitution Feb 14, 2025 Factorize $8x^3-\dfrac{1}{27y^3}$ Jan 08, 2025 Evaluate $\displaystyle \int{\dfrac{\sin{x}}{\sqrt{3+2\cos{x}}}}\,dx$ Dec 26, 2024 Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cos{3x}-\cos{4x}}{x\sin{2x}}}$ by using formulas Dec 20, 2024 Factorize $4x^2y^3$ $-$ $6x^3y^2$ $-$ $12xy^2$ Dec 15, 2024 Evaluate $\displaystyle \int{\dfrac{\sqrt{x}}{x+1}}\,dx$ Arithmetic Pre Algebra Algebra Trigonometry Geometry Calculus Factors Fractions Exponents Logarithms Inequalities Matrices Quadratic Equations Hyperbolic functions Angles Straight Lines Circles Coordinate Geometry Limits Differentiation integration Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions Math Worksheets The math worksheets with answers for your practice with examples. Practice now Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now Subscribe us Get the latest math updates from the Math Doubts by subscribing us. Learn more A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects. About us Terms of Use Privacy policy Contact us Copyright © 2012 - 2025 Math Doubts, All Rights Reserved

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