output_mds/page_34.md

This exceeds the yield strain for Grade-60 steel ($\epsilon_y = 0.00207$, previously calculated), so the assumption that the tension steel is yielding is confirmed.

2. Compute the nominal moment strength, $M_n$. As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones for

$$M_n=A_s{f}_y(d-\frac{a}{2})=4.0{\text{ in.}}^2\times 60\text{ ksi}(17.5\text{ in.}-\frac{5.88\text{ in.}}{2})M\_n\; =\; A\_s\; f\_y\; \backslash left(d\; -\; \backslash frac\{a\}\{2\}\backslash right)\; =\; 4.0\; \backslash text\{\; in.\}^2\; \backslash times\; 60\; \backslash text\{\; ksi\}\; \backslash left(17.5\; \backslash text\{\; in.\}\; -\; \backslash frac\{5.88\; \backslash text\{\; in.\}\}\{2\}\backslash right)$$ $$M_n=3490\text{ k-in.}=291\text{ k-ft}M\_n\; =\; 3490\; \backslash text\{\; k-in.\}\; =\; 291\; \backslash text\{\; k-ft\}$$

3. Confirm that tension steel area exceeds $A_{s,\min}$. Although this is seldom a problem with most beam sections, it is good practice to make this check. The expression for $A_{s,\min}$ is given in Eq. (4-11) and includes a numerator that is to be taken equal to $3\sqrt{f_c'}$, but not less than 200 psi. As was shown in Example 4-1, the value of 200 psi governs for beams constructed with 4000 psi concrete. Thus,

$$A_{s,\min }=\frac{200\text{ psi}}{f_y}{b}_{w}d=\frac{200}{60,000}\times 12\text{ in.}\times 17.5\text{ in.}=0.70{\text{ in.}}^{2}A\_\{s,\backslash min\}\; =\; \backslash frac\{200\; \backslash text\{\; psi\}\}\{f\_y\}\; b\_w\; d\; =\; \backslash frac\{200\}\{60,000\}\; \backslash times\; 12\; \backslash text\{\; in.\}\; \backslash times\; 17.5\; \backslash text\{\; in.\}\; =\; 0.70\; \backslash text\{\; in.\}^2$$

Clearly, $A_s$ for this section satisfies the ACI Code requirement for minimum tension reinforcement.

4. Compute the strength reduction factor, $\phi$, and the resulting value of $\phi M_n$. As stated previously, for a single layer of tension reinforcement, $\epsilon_t$ is equal to $\epsilon_s$, which was calculated in step 1. Because $\epsilon_t$ is between 0.002 and 0.005, this is a transition-zone section. Thus, Eq. (4-28) is used to calculate $\phi$:

$$\varphi =0.65=0.65+(0.00459-0.002)\frac{250}{3}=0.87\backslash phi\; =\; 0.65\; =\; 0.65\; +\; (0.00459\; -\; 0.002)\backslash frac\{250\}\{3\}\; =\; 0.87$$

Then,

$$\varphi M_n=0.87\times 291\text{ k-ft}=253\text{ k-ft}\backslash phi\; M\_n\; =\; 0.87\; \backslash times\; 291\; \backslash text\{\; k-ft\}\; =\; 253\; \backslash text\{\; k-ft\}$$

Beam 2: Same as Beam 1, except that $f_c' = 6000$ psi. As shown in Fig. 4-12, changing the concrete compressive strength will not produce a large change in the nominal moment strength, but it does increase the ductility of the section. Thus, increasing the concrete compressive strength might change the beam section in Fig. 4-26 from a transition-zone section to a tension-controlled section.

1. Compute $a$, $c$, and $\epsilon_s$. Again, assume that the tension steel is yielding, so $f_s = f_y$. For this compressive strength, Eq. (4-14b) is used to determine that $\beta_1 = 0.75$. Then, using Eq. (4-16),

$$a={\beta }_{1}c=\frac{A_s{f}_y}{0.85f_c^{\mathrm{\prime }}b}a\; =\; \backslash beta\_1\; c\; =\; \backslash frac\{A\_s\; f\_y\}\{0.85\; f\_c\text{'}\; b\}$$ $$=\frac{4.00{\text{ in.}}^2\times 60\text{ ksi}}{0.85\times 6\text{ ksi}\times 12\text{ in.}}=3.92\text{ in.}=\; \backslash frac\{4.00\; \backslash text\{\; in.\}^2\; \backslash times\; 60\; \backslash text\{\; ksi\}\}\{0.85\; \backslash times\; 6\; \backslash text\{\; ksi\}\; \backslash times\; 12\; \backslash text\{\; in.\}\}\; =\; 3.92\; \backslash text\{\; in.\}$$

Thus, $c = a/\beta_1 = 5.23$ in., and using strain compatibility as expressed in Eq. (4-18), find

$${ϵ}_s=\left(\frac{d-c}{c}\right){ϵ}_{cu}=\left(\frac{17.5-5.23}{5.23}\right)0.003=0.00704\backslash epsilon\_s\; =\; \backslash left(\backslash frac\{d\; -\; c\}\{c\}\backslash right)\; \backslash epsilon\_\{cu\}\; =\; \backslash left(\backslash frac\{17.5\; -\; 5.23\}\{5.23\}\backslash right)\; 0.003\; =\; 0.00704$$

This exceeds the yield strain for Grade-60 steel ($\epsilon_y = 0.00207$), confirming the assumption that the tension steel is yielding.

2. Compute the nominal moment strength, $M_n$. As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones:

$$M_n=A_s{f}_y(d-\frac{a}{2})=4.0{\text{ in.}}^2\times 60\text{ ksi}(17.5\text{ in.}-\frac{3.92\text{ in.}}{2})M\_n\; =\; A\_s\; f\_y\; \backslash left(d\; -\; \backslash frac\{a\}\{2\}\backslash right)\; =\; 4.0\; \backslash text\{\; in.\}^2\; \backslash times\; 60\; \backslash text\{\; ksi\}\; \backslash left(17.5\; \backslash text\{\; in.\}\; -\; \backslash frac\{3.92\; \backslash text\{\; in.\}\}\{2\}\backslash right)$$ $$M_n=3730\text{ k-in.}=311\text{ k-ft (7\% increase from Beam 1)}M\_n\; =\; 3730\; \backslash text\{\; k-in.\}\; =\; 311\; \backslash text\{\; k-ft\}\; \backslash text\{\; (7\backslash \%\; increase\; from\; Beam\; 1)\}$$