text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
) need not be
proportional.
An alternative form of this solution can be given in terms of absolute position and the distance
propagated in a given time:
p(t, x) = g+ x − ct
(
)+ g− x + ct
(
), and
v x (t, x) =
1
z0
(
g+(x − ct) − g−(x + ct)
) ,
(2.10)
(2.11)
Equations 2.8&9 define vx(x,t) and p(x,t) in terms of two functions (f+ and f–) that depend on the
sound source and the boundary conditions at the two ends of our one-dimensional system. In the case of
a completely open space the sound produced by a source propagates along its one dimensional axis as a
forward traveling wave, and there is no backward traveling wave:
)
(
f +(t − x /c)
.
p(t,x) = f + t − x /c
v x (t, x) =
), and
(2.12)
1
z0
(
Why are these Forward Traveling waves?
14-Sept-2004
page 5
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
Why are these Forward Traveling waves?
Think about how sound travels. Assume (1) the sound pressure in this room is zero at all negative
times and (2) at t=0 we generate a one dimensional plane-wave pulse of pressure of 1 Pa peak.
To be precise:
p(t,x) = 0 when t < 0;
p(0,0) = 1;
p(t,0) = 0 at t > 0
.
(2.13)
These boundary conditions when applied to Equation 2.12 suggest that the we can define the one
dimensional forward traveling wave: as:
p(t,x) = f + ζ( ); where ζ = t − x /c
)
with f+(ζ) = 0 for ζ< 0, and ζ> 0
and f+(ζ) = 1 for ζ= 0 .
(
(2.14)
How does pressure vary within the room at | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
f+(ζ) = 1 for ζ= 0 .
(
(2.14)
How does pressure vary within the room at t=0, t=1 ms, t=3 ms, t=5 ms ?
1.0
1.0
0.5
0.5
0
0
x=0
x=0
1 meter
1 meter
x=Distance from the Doorway
2 meters
2 meters
The pulse propagates as a “wave front” of the traveling wave.
14-Sept-2004
page 6
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
5. Sinusoidal Traveling Waves:
A sinusoidal steady state solution for the wave equation also depends on the summation of a
forward and backward going traveling wave
p(t, x) = Real P+e jω(t− x /c) + P−e jω(t+ x /c)
{
},
v x (t, x) = Real
⎧
⎨
⎩
1
z0
(
P+e jω(t− x /c) − P−e jω(t+ x /c)
⎫
)
,
⎬
⎭
(2.15)
(2.16)
For those of you still not comfortable with the exponential notation, remember that Eqn. 2.15 is
equivalent to:
(
p(t, x) = P+ cos ω(t − x /c) + ∠P+
)+ P− cos ω(t + x /c) + ∠P−
(
).
How does this equivalence come about?
Equations 2.15 and 2.16 can also be written in terms of the variable k= ω/c = 2π/λ, where k has units of
radians per meter and is sometimes called the wave number, length constant or spatial frequency :
p(t, x) = Real P+e j (ωt−kx) + P−e j (ωt+kx)
v x (t, x) = Real
(
P+e j (ωt−kx) − P−e j (ωt+kx)
{
⎧
⎨
� | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
P+e j (ωt−kx) − P−e j (ωt+kx)
{
⎧
⎨
⎩
1
z0
}, and
⎫
)
⎬
⎭ .
(2.17)
(2.18)
Suppose the sound pressure source in one of the walls produces a steady-state sinusoidal
variation in pressure with radial frequency ω = 2π 170 Hz :
p(t,0) = cos(ωt) = Real{ejωt}.
The sinusoid also "travels" across the room at a velocity of c, i.e.
p(t,x)=cos(ωζ); where ζ=(t - x/c) .
How does sound pressure vary across the room at time 0 and at fractions of a period later? (Hint: What
is the wavelength of a sound of 170 Hz?)
14-Sept-2004
page 7
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
p(t,x)=cos(ωζ); where ζ=(t - x/c) .
t=0
t=T/8
t=T/4
t=T/2
t=3T/4
)
a
P
(
E
R
U
S
S
E
R
P
D
N
U
O
S
1.0
0.5
0.0
-0.5
-1.0
0.00
0.25
0.50
1.00
X = DISTANCE FROM REFERENCE POINT
1.50
1.25
0.75
1.75
2.00
As time progresses, the “wave-front” (here defined as the location of maximum pressure) travels across
the room with a velocity c.
Now suppose we place a microphone at various locations in the room. How does the sound pressure
vary with time at x = 0, x=0.5 meters, x=1 meters, and x=2 meters?
x=0 m
x=0.5 m
x=1 m
)
a
P
(
E
R
U
S
S
E
R
P | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
0.5 m
x=1 m
)
a
P
(
E
R
U
S
S
E
R
P
D
N
U
O
S
1.0
0.5
0.0
-0.5
-1.0
0.000
0.001
0.004
0.003
0.002
time in sec (1/170 = 0.006)
0.005
0.006
0ne-dimensional wave propagation depends on both time and space. The events that occur in the
present at location x=0, predict the events that will occur further away from the source at a later time.
14-Sept-2004
page 8
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
6. The separation of time and space dependence.
The time and space dependence of traveling waves can be separated from each other. In cases
where the temporal dependence of the wave is well defined, such a separation allows us to concentrate
on the spatial dependence. In the case of the sinusoidal steady state:
we can factor out ejωt,
p(t, x) = Real P+e j (ωt−kx) + P−e j (ωt+kx)
{
}
{
p(t, x) = Real e jωt P(x)
P(x) = P+e− jkx + P−e jkx
}, where
)
(
(2.19)
(2.20)
In a wide open environment with no reflection, we can define the spatial dependence of a forward
traveling plane wave, as
) .
If P+=1, how do the magnitude and angle of P(x) vary in space?
(
P x( ) = P+e− jkx
(2.21)
14-Sept-2004
page 9
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
1.0
0.8
0.6
0.4
0.2
0.0
)
a
P
(
e
d
u
t
i
n
g
a
M
e
r
u | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
0
)
a
P
(
e
d
u
t
i
n
g
a
M
e
r
u
s
s
e
r
P
d
n
u
o
S
x=0
λ/4
3λ/4
λ/2
x = Distance from the Reference Point
)
s
n
a
d
a
R
i
(
l
e
g
n
A
e
r
u
s
s
e
r
P
d
n
u
o
S
π
−π
−2π
x=0
λ/2
λ/4
3λ/4
x = Distance from the Reference Point
λ
λ
14-Sept-2004
page 10
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
7. Reflections at Rigid Boundaries
Suppose our propagating plane wave hits a rigid wall placed orthogonally to the direction of
propagation, where the wall dimensions are much larger than the wavelength. The interaction will
produce a reflected wave that appears as a backward traveling wave in a one-dimensional system.
At the rigid boundary, the reflected wave acts as a continuation of the original wave, but its direction is
altered. In the steady state, the sound pressure at each location is the sum of the two waves:
}
p(t, x) = Re P+eωt−kx) + P−eωt+kx) +
{
In the case of rigid boundary reflection in a one dimensional system:
(1) The amplitude and angle of the incident and reflected waves are equal P+|=|P–.
(2) The value of the incident and reflected pressure at the boundary is equal at all times
p+(t,0)=p-(t,0).
(3) The two waves always cancel at nλ/4, (n=1, 3, 5, …) distance from the wall.
(4) The sum of the two waves has a magnitude of 2|P+| at distances mλ/2 (m=0, 1, 2, …) from the wall.
(5) At times when the incident wave is | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
λ/2 (m=0, 1, 2, …) from the wall.
(5) At times when the incident wave is in ±sine phase at the wall, the summed pressure is
0 everywhere.
14-Sept-2004
page 11
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
8. The Spatial Dependence of the Total Sound Pressure in Rigid Wall Reflection
Earlier, we described the total pressure at any location and time in terms of the sum of the
forward and backward going waves:
We also separated out the temporal and location dependence, i.e.
p(t, x) = Real P+e j (ωt−kx) + P−e j (ωt+kx)
{
}
where:
}
{
p(t, x) = Real e jωt P(x)
P(x) = P+e− jkx + P−e jkx
(2.22)
(2.23)
In the case of a forward traveling wave with a rigid boundary at x=0 where P+ = P− , as in Figure 2.6,
Equation 2.23 simplifies via Euler’s equations to
P(x) = 2P+ cos kx(
),
(2.24)
Note that Equation 2.24:
(a) Is dependent on x, ω (k=ω/c) but independent of t, this is a standing wave.
(b) The sound pressure at the rigid boundary (x=0) is twice the amplitude of the traveling
waves P(0) = 2P+ .
(c) When x=–λ/4; kx=π/2 and P(−λ/ 4) = 0 ; this zero is repeated at x=–3λ/4, –5λ/4, –7λ/4 …
(d) ∠P(x) = ∠P+ and is invariant in space.
9. The Spatial Dependence of the Specific Acoustic Impedance in Rigid Wall Reflection
Rigid-walled reflection, where the angle of incidence is 90° relative to the boundary, also
produces standing waves in particle velocity.
We can define Vx(x) starting from Equation 2.16 | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
90° relative to the boundary, also
produces standing waves in particle velocity.
We can define Vx(x) starting from Equation 2.16:
(
P+e jω(t− x /c) − P−e jω(t+ x /c)
v x (t, x) = Real
⎧
⎨
⎩
1
z0
where:
⎧
⎪
1
vx (t,x) = Real
⎨
⎩ ⎪
z0
e jωt P(x)
⎫
⎪
, with
⎬
⎭ ⎪
⎫
)
⎬
⎭
(2.16)
V x(x) =
1
z0
(
P+e− jkx − P−e jkx
)= −2 j
sin(kx)
P+
z0
(
Note that for x < 0; ∠Vx(x) =( π 2 + ∠P+
of 2 |P+|/z0 at x=–λ/4, –3λ/4, –5λ/4, –7λ/4 …
The ratio of P(x) and Vx(x) defines the spatially varying specific acoustic impedance ZS(x).
In the case of rigid boundary reflection:
(2.25)
), has a magnitude of 0 at x=0, and has a magnitude maximum
Z S(x) =
P(x)
V x (x)
=
2P+ cos(kx)
2P+
z0
sin(kx)
− j
= jz0 cot(kx)
(2.26)
At a position λ/4 away from the reflector, i.e. x = –λ/4, –3λ/4, –5λ/4, –7λ/4 … ; ZS=0.
14-Sept-2004
page 12
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
When x =0, –λ/2, –λ, –3λ/2 … ; ZS=∞.
14-Sept-2004
page 13 | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
Harvard-MIT Division of Health Sciences and Technology
HST.951J: Medical Decision Support, Fall 2005
Instructors: Professor Lucila Ohno-Machado and Professor Staal Vinterbo
6.873/HST.951 Medical Decision Support
Spring 2005
Variable Compression:
Principal Components Analysis
Linear Discriminant Analysis
Lucila Ohno-Machado
Variable Selection
• Use few variables
• Interpretation is easier
Ranking Variables Univariately
• Remove one variable
from the model at a
time
• Compare
performance of [n-1]
model with full [n]
model
• Rank variables
according to
performance
difference
Screenshots removed due to copyright reasons.
Figures removed due to copyright reasons.
Please see:
Khan, J., et al. "Classification and diagnostic prediction of cancers using gene expression
profiling and artificial neural networks." Nat Med 7, no. 6 (Jun 2001): 673-9.
Variable Selection
•
Ideal: consider all variable combinations
– Not feasible in most data sets with large number of n variables:
n
2
• Greedy Forward:
– Select most important variable as the “first component”, Select
other variables conditioned on the previous ones
– Stepwise: consider backtracking
• Greedy Backward:
– Start with all variables and remove one at a time.
– Stepwise: consider backtracking
• Other search methods: genetic algorithms that optimize
classification performance and # variables
Variable compression
• Direction of maximum variability
– PCA
• PCA regression
– LDA
– (Partial Least Squares)
correlatio
n
t
_ coefficien
r =
σ XY = ρ
σ σ Y
X
VARIANCE
COVARIANCE
n
∑ ( X −
X
i
)( i −
Y
Y
)
σ XY = i= 1
n − 1
n
∑ ( X −
X
i
)(
X
i − X )
σ =
XX
i= 1
n − 1
st _ deviation
n
∑
X
(
i
1
=
− | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
i= 1
n − 1
st _ deviation
n
∑
X
(
i
1
=
−
X
)(
X
i
i
−
X
)
n
−
1
Xσ
=
Y
Covariance and
Correlation Matrices
⎡σXX σXY ⎤
cov = ⎢
⎥
σYX σYY ⎦
⎣
⎡ 1 ρ⎤
⎣ρ 1 ⎥
⎦
corr = ⎢
n
n
σ =
XY
i=1
σ = i=1
XX
∑( X −
i
X
Y
)( −
i
Y
)
∑( X −
i
X
)(
X
i − X )
n − 1
n − 1
0
X
Slope from linear regression is asymmetric,
covariance and ρ are symmetric
β 0 = y − β x
1
y
Σ ( x − x)(
−
Σ ( x − x)2
β 1 =
y
y
)
x
y = β + β 1 x
0
y = 2 + 4 x
x = y / 4 − 2
⎥ Σ =
cov = ⎢
corr = ⎢
.0 35 ⎤
⎡ .0 86
⎣ .0 35 15.69⎦
.0 96 ⎤
⎡ 1
⎥
1 ⎦
⎣ .0 96
Principal Components Analysis
• Our motivation: Reduce the number of variables
so that we can run interesting algorithms
• The goal is to build linear combinations of the
variables (transformation vectors)
• First component should represent the direction
with largest variance
• Second component is orthogonal to
(independent of) the first, and is the next one
with largest variance
• and so on…
Y
X and Y are not
in | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
dependent of) the first, and is the next one
with largest variance
• and so on…
Y
X and Y are not
independent
(covariance is not 0)
Y =
( X * 4)
+
e
cov
=
≠
0
σ
XY
σ
YY
⎤
⎥
⎦
σ
XX
≠ 0
⎡
σ
⎢
⎣
.0 86
XY
.0 35
.0 35 15.69
⎤
⎥
⎦
=
⎡
⎢
⎣
.0 96
cov
=
ρ
0
X1
Eigenvalues
I is the identity matrix.
A is a square matrix (such as the covariance matrix).
|A - λ I| = 0
λ is called the eigenvalue (or characteristic root) of A.
⎡σ XX σ XY ⎤
⎥
⎢
⎣σ XY σ YY
⎦
⎡ 1 0⎤
−λ
⎥
⎢
⎣ 0 1⎦
= 0
Eigenvectors
σ
⎡
⎢
σ
⎣
σ
⎡
⎢
σ
⎣
XX
XY
XX
XY
XY
σ
σ
YY
XY
σ
σ
YY
a
⎤
⎡
⎥
⎢
b
⎣
⎦
c
⎤
⎡
⎥
⎢
d
⎣
⎦
q=
⎤
⎥
⎦
m=
⎤
⎥
⎦
a
⎡
⎢
b
⎣
⎤
⎥
⎦
c
⎡
⎢
d
⎣
⎤
⎥
⎦
q and m are eigenvalues
[a b]T and [c d]T are eigenvectors, | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
⎥
⎦
q and m are eigenvalues
[a b]T and [c d]T are eigenvectors, they are orthogonal
(independent of each other, do not contain redundant information)
The eigenvector associated with the largest eigenvalue will point in
the direction of largest variance in the data.
If q > m, then [a b]T is PC1
Principal Components
27
.1
⎡
⎢
.
.5 12 21
⎣
.1
⎡
⎢
.
.5 12 21
⎣
27
.5 12
⎡ ⎤
⎢
⎥
65
⎣
⎦
.0 23⎤
⎥
.0 97
⎦
.0 97
.5 12
⎡
⎤
⎥
⎢
65 − .0
⎦
⎣
⎤
⎥
23
⎦
=
22.87
⎡
⎢
⎣
=
. 0 05
.0 23⎤
⎥
.0 97
⎦
. 0 97
⎡
⎢
⎣
−
⎤
⎥
⎦
23
.0
Total variance is 21.65 + 1.27 = 22.92
X2
0
X1
Transformed data
x
=
x
21
x22
O1 O2 O3
x11
⎡
⎢
x12
⎣
b
d
x31
⎤
⎥
x32
⎦
bx12
+
dx12
+
x
⎤
⎥
⎦
=
ax11
⎡
⎢
cx11
⎣
97.0
⎤
⎥
23.0
−
⎦
x
a
c
⎡
⎢
⎣
⎡
� | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
.0
−
⎦
x
a
c
⎡
⎢
⎣
⎡
⎢
⎣
23.0
97.0
, PC
=
a
b
⎡
⎢
⎣
ax
cx
+
+
21
21
c
⎤
⎥
⎦
d
bx
dx
22
22
=
23.0
⎡
⎢
97.0
⎣
x11
x11
+
−
97.0
23.0
x
12
x
12
P
C
2
ax31
cx
31
+
+
23.0
97.0
bx32
⎤
⎥
dx32
⎦
97.0
x
+
21
23.0
x
−
21
x
x
22
22
23.0
97.0
x31
x31
P
C
1
+
−
97.0
.0
23
x32
⎤
⎥
32 x
⎦
X2
0
X1
Total variance is 22.92
Variance of PC1 is 22.87, so it captures 99% of the variance.
PC2 can be discarded with little loss of information.
PC1 is not at the regression line
• y=4x
• [a b]T = [0.23 0.97]
• Transformation is
0.23x+0.97y
• PC1 goes thru
•
(0,0) and (0.23,0.97)
Its slope is
– 0.97/0.23 = 4.217
PC
Reg
Y
0
X
PCA regression
• Reduce original
dimensionality n (number of
variables) finding n PCs, such
that n<d
• Perform regression on PCs
• Problem: Direction of greater
overall variance is not
necessarily best for
classification
• Solution: Consider also
direction of greater separation
between two classes
(not so good) idea:
Class mean separation
• Find means of each
category | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
Consider also
direction of greater separation
between two classes
(not so good) idea:
Class mean separation
• Find means of each
category
• Draw the line that passes
through the 2 means
• Project points on the line
(a.k.a. orthogonalize points with
respect to the line)
• Find point that best
separates data
Fisher’s Linear Discriminant
• Use classes to define discrimination line,
but criterion to maximize is:
– ratio of (between classes variation) and
(within classes variation)
• Project all objects into the line
• Find point in the line that best separates
classes
Sw is the sum of scatters
within the classes
)(
x
)
−
μ
μ
1
1
i
=
n
1
−
x
(
i
−
T
scatter
x
(
i
−
=
x
)(
−
μ
i
2
n
1
−
T
μ
2
)
4
⎤
1
⎡
⎥
⎢
4 16
⎦
⎣
9.3
1.1
⎡
= ⎢
⎣
9.3
⎤
⎥
14
⎦
cov1
=
cov2
S
w
=
k
∑
j 1 =
p j cov j n
( −
1)
(99)
1
4
⎤
⎥
4 16
⎦
9.3
⎤
⎥
14
⎦
1.1
(99)
S1
=
5.
⎡
⎢
⎣
2
S
=
5.
⎡
⎢
⎣
Sw =
S1
+
S
2
9.3
Sb is scatter between the classes
Sb = (μ − μ )(μ − μ )
2
2
1
1
T
• Maximize Sb/Sw (a square d x d matrix,
where d is the number of dimensions)
• Find maximum eigenvalue and respective
eigenv | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
b/Sw (a square d x d matrix,
where d is the number of dimensions)
• Find maximum eigenvalue and respective
eigenvector
• This is the direction that maximizes Fisher’s
criterion
• Points are projected over this line
• Calculate distance from every projected point
to projected class means and decide class
corresponding to smaller distance
• Assumption: class distributions are normal
Eigenvector with
max eigenvalue
Classification Models
• Quadratic Discriminant Analysis
• Partial Least Squares
– PCA uses X to calculate directions of greater
variation
– PLS uses X and Y to calculate these directions
• It is a variation of multiple linear regression
Var(Xα),
Corr2(y,Xα)Var(Xα)
PCA maximizes
PLS maximizes
• Logistic Regression
PCA, PLS, Selection
• 3 data sets
– Singh et al. (Cancer Cell, 2002: 52 cases of benign and
malignant prostate tumors)
– Bhattachajee et al. (PNAS, 2001: 186 cases of different types of
lung cancer)
– Golub et al. (Science, 1999: 72 cases of acute myeloblastic and
lymphoblastic leukemia)
• PCA logistic regression
• PLS
• Forward selection logistic regression
• 5-fold cross-validation
Missing Values: 0.001% -
0.02%
Screenshots removed due to copyright reasons.
Classification of Leukemia with Gene Expression
PCA
Variable
Selection
Variable selection from ~2,300 genes | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
2.094— Finite Element Analysis of Solids and Fluids
— Fall ‘08 —
MIT OpenCourseWare
Contents
1 Large displacement analysis of solids/structures
1.1 Project Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Large Displacement analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Mathematical model/problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 Requirements to be fulfilled by solution at time t . . . . . . . . . . . . . . . . . . .
1.2.3 Finite Element Method
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.4 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Finite element formulation of solids and structures
2.1 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Finite element formulation for solids and structures
4 Finite element formulation for solids and structures
5 F.E. displacement formulation, cont’d
6 Finite element formulation, example, convergence
3
3
4
4
5
5
6
7
8
8
10
14
19
23
6 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
convergence
3
3
4
4
5
5
6
7
8
8
10
14
19
23
6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.1 F.E. model
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.2 Higher-order elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
7 Isoparametric elements
8 Convergence of displacement-based FEM
9 u/p formulation
10 F.E. large deformation/general nonlinear analysis
11 Deformation, strain and stress tensors
12 Total Lagrangian formulation
1
28
33
37
41
45
49
MIT 2.094
13 Total Lagrangian formulation, cont’d
14 Total Lagrangian formulation, cont’d
15 Field problems
Contents
53
57
61
15.1 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
15.1.1 Differential formulation
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
15.1.2 Principle of virtual temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
15.2 Inviscid, incompressible, irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
ational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
16 F.E. analysis of Navier-Stokes fluids
17 Incompressible fluid flow and heat transfer, cont’d
65
71
17.1 Abstract body
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.2 Actual 2D problem (channel flow)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.3 Basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
17.4 Model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
17.5 FSI briefly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
18 Solution of F.E. equations
76
18.1 Slender structures
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
19 Slender structures
20 Beams, plates, and shells
21 Plates and shells
81
85
90
2
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 1 - Large displacement analysis of solids/structures
Prof. K.J. Bathe
MIT OpenCourseWare
1.1 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
08
Lecture 1 - Large displacement analysis of solids/structures
Prof. K.J. Bathe
MIT OpenCourseWare
1.1 Project Example
Physical problem
Reading:
Ch. 1 in
the text
“Simple” mathematical model
More complex mathematical model
•
analytical solution
•
F.E. solution(s)
•
holes included
•
large disp./large strains
•
F.E. solution(s)
⇒
•
How many finite elements?
We need a good error measure (especially for FSI)
“Even more complex” mathematical model
The “complex mathematical model” includes Fluid
Structure Interaction (FSI).
You will use ADINA in your projects (and homework) for structures and fluid flow.
3
MIT 2.094
1. Large displacement analysis of solids/structures
1.2 Large Displacement analysis
Lagrangian formulations:
• Total Lagrangian formulation
• Updated Lagrangian formulation
Reading:
Ch. 6
1.2.1 Mathematical model/problem
Given
Calculate
the original configuration of the body,
the support conditions,
the applied external loads,
the assumed stress-strain law
the deformations, strains, stresses of the body.
Question
for large displacement/strain.
Is there a unique solution? Yes, for infinitesimal small displacement/strain. Not necessarily
For example:
Snap-through problem
The same load. Two different deformed configurations.
4
MIT 2.094
1. Large displacement analysis of solids/structures
Column problem, statics
Not physical
tR is in “direction” of bending moment Not in
equilibrium.
⇒
1.2.2 Requirements to be fulfilled by solution at time t
I. Equilibrium of stresses (Cauchy stresses, forces per unit area in tV and on tSf ) with the applied
body forces tf B and surface tractions tf Sf | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
V and on tSf ) with the applied
body forces tf B and surface tractions tf Sf
II. Compatibility
III. Stress-strain law
1.2.3 Finite Element Method
I. Equilibrium condition means now
• equilibrium at the nodes of the mesh
• equilibrium of each finite element
II. Compatibility satisfied exactly
III. Stress-strain law satisfied exactly
5
MIT 2.094
1. Large displacement analysis of solids/structures
1.2.4 Notation
Cauchy stresses (force per unit area at time t):
tτij
i, j = 1, 2, 3
tτij = tτji
(1.1)
6
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 2 - Finite element formulation of solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
Assume that on tSu the displacements are zero (and tSu is constant). Need to satisfy at time t:
Reading:
Ch. 1, Sec.
6.1-6.2
•
Equilibrium of Cauchy stresses tτij with applied loads
�
tτ T =
tτ11
tτ22
tτ33
tτ12
tτ23
tτ31
�
(For i = 1, 2, 3)
tτij,j + tfi
tτij
(e.g. tfi
t nj =
Sf =
B = 0 in tV (sum over j)
tfi
tτi1
Sf on tSf (sum over j)
t n2 + tτi3
t n1 + tτi2
t n3 )
And: tτ11
tn1 + tτ12
Sf
tn2 = tf1
• Compatibility The displacements tui need to be continuous and zero on tSu.
•
Stress-Strain law | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
• Compatibility The displacements tui need to be continuous and zero on tSu.
•
Stress-Strain law
tτij = function t uj
�
�
7
(2.1)
(2.2)
(2.3)
(2.4)
(2.5)
MIT 2.094
2. Finite element formulation of solids and structures
2.1 Principle of Virtual Work∗
�
tV
tτij teij d tV =
�
tV
where
tf B
i ui d tV +
�
tSf
tf Sf
i u Sf
i d tSf
=
�
∂ui
∂ txj
1
2
+
�
∂uj
∂ txi
=
0
teij
�
�
with ui
�
tSu
2.2 Example
Assume “plane sections remain plane”
�
tf B
1 u1 d tV +
tPr u Sf
1 d tSf
Principle of Virtual Work
�
�
tτ11 te11 d tV =
tV
t V
Derivation of (2.9)
tτ11,1 + tf B
1 = 0 by (2.2)
� tτ11,1 + tf B
�
u1 = 0
1
∗or Principle of Virtual Displacements
tSf
8
(2.6)
(2.7)
(2.8)
(2.9)
(2.10)
(2.11)
MIT 2.094
2. Finite element formulation of solids and structures
Hence,
�
�
tV
tτ11,1 + tf1
�
B u1 d tV = 0
11u
�
1 �
tτ
�
��
Sf tτ
u1
11
t
t
S
S
f
u
−
�
tS
f
�
u | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
tτ
u1
11
t
t
S
S
f
u
−
�
tS
f
�
u1,1
tV ����
te11
t
tτ11 d V +
u1
B
tf1
d tV = 0
�
tV
where tτ11|tSf
= tPr.
Therefore we have
�
tV
te11
tτ11 d tV =
�
tV
u1
tf1
B d tV + u S
f tPr
1
tSf
(2.12)
(2.13)
(2.14)
From (2.12) to (2.14) we simply used mathematics. Hence, if (2.2) and (2.3) are satisfied, then (2.14)
must hold. If (2.14) holds, then also (2.2) and (2.3) hold!
Namely, from (2.14)
or
�
tV
�
tV
u1,1
tτ11d V = u1
t
tSf
�
�
tτ11 tS u
�
−
tV
u1
tτ11,1d V =
t
�
tV
u1
B t
Sf
tf1 d V + u1
�
�
Sf
tτ11,1 + tf1 d V + u1
B
t
�
�
tSf = 0
tPr − tτ11
u1
x
Now let u1 = x 1 − t
L
�
�
�
tτ11,1 + tf1
�
B , where tL = length of bar.
Hence we must have from (2.16)
tτ11,1 + tf1
B = 0
and then also
tP = tτ
r
11
9
tPr
tSf
(2.15)
(2.16)
(2.17)
(2.18)
2 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Pr
tSf
(2.15)
(2.16)
(2.17)
(2.18)
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 3 - Finite element formulation for solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
We need to satisfy at time t:
•
Equilibrium
t∂ τij + tf B = 0 (i = 1, 2, 3) in tV
t
∂ xj
i
tτij n t
t Sf
j = f
i
(i = 1, 2, 3) on tSf
• Compatibility
• Stress-strain law(s)
Principle of virtual displacements
�
tV
tτij teij d Vt =
�
tV
ui
tf B
i d Vt +
�
tSf
ui|tSf
ft Sf
i d St
f
teij =
�
∂
u
i
t
x
∂
j
1
2
+
�
∂u
j
tx
∂
i
Reading:
Sec. 6.1-6.2
(3.1)
(3.2)
(3.3)
(3.4)
•
If (3.3) holds for any continuous virtual displacement (zero on tS ), then (3.1) and (3.2) hold and
vice versa.
u
•
Refer to Ex. 4.2 in the textbook.
10
MIT 2.094
Major steps
I. Take (3.1) and weigh with ui:
�
tτij,j + tfi
�
B ui = 0.
II. Integrate (3.5a) over volume tV :
�
�
tV
tτij,j + tfi
�
B ui d tV = 0
3. Finite element formulation for solids and structures
( | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
,j + tfi
�
B ui d tV = 0
3. Finite element formulation for solids and structures
(3.5a)
(3.5b)
III. Use divergence theorem. Obtain a boundary term of stresses times virtual displacements on tS =
tSu ∪ tSf .
IV. But, on tSu the ui = 0 and on tSf we have (3.2) to satisfy.
Result: (3.3).
Example
�
tV
tτ
t
e11 d V =
11 t
�
St
f
One element solution:
t Sf
ui f1 d Sf
t
(3.6)
11
MIT 2.094
3. Finite element formulation for solids and structures
u(r) =
t u(r) =
u(r) =
1
2
1
2
1
2
(1 + r) u1 +
1
2
(1 + r) t u1 +
1
2
(1 + r) u1 +
1
2
(1 − r) u2
(1 − r) t u2
(1 − r) u2
Suppose we know tτ11, tV , tSf , tu ... use (3.6).
For element 1,
te11 =
∂u
∂ tx
= B(1)
�
�
u1
u2
T
t
te11t τ11d V −→
for el. (1)
�
[u1 u2]
� tV
B(1)T
��
= tF (1)
tτ11 d tV
�
for el. (1)
−→
⎡
= ⎣ U 1
����
u2
[u1 u2]
U 2
����
u1
tF (1)
⎤
�
U 3⎦
�
tFˆ (1)
0
�
tV
where
tFˆ
1
tFˆ
2 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
tFˆ (1)
0
�
tV
where
tFˆ
1
tFˆ
2
(1) = tF2
(1) = tF1
(1)
(1)
⎡
For element 2, similarly,
⎤
U 3 ⎦
����
u1
= ⎣U 1 U 2
����
u2
�
�
0
t ˆ
F (2)
R.H.S.
�
�
⎡
�
U 1 U 2 U 3 ⎣
�
��
U
T
(unknown reaction at left)
0
t Sf
f ·
f
1
tS
⎤
⎦
Now apply,
= �
U
T
1 0 0 �
then,
then,
T
U
�
= 0 1 0
T
U
�
= 0 0 1
�
�
12
(3.7)
(3.8)
(3.9)
(3.10)
(3.11)
(3.12)
(3.13)
(3.14)
(3.15)
(3.16)
(3.17)
(3.18)
(3.19)
(3.20)
MIT 2.094
This gives,
3. Finite element formulation for solids and structures
�
t ˆF (1)
0
�
�
+
0
t ˆF (2)
�
⎡
= ⎢
⎣
unknown reaction
0
t tSf
· tSf
f
1
⎤
⎥
⎦
We write that as
tF = tR
�
tF = fn
�
tU1, tU2, tU3
(3.21)
(3.22)
(3.23)
13
2.094 — Finite Element Analysis | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(3.21)
(3.22)
(3.23)
13
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 4 - Finite element formulation for solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
We considered a general 3D body,
Reading:
Ch. 4
The exact solution of the mathematical model must satisfy the conditions:
• Equilibrium within tV and on tSf ,
•
Compatibility
•
Stress-strain law(s)
I. Differential formulation
II. Variational formulation (Principle of virtual displacements) (or weak formulation)
We developed the governing F.E. equations for a sheet or bar
We obtained
tF
tR=
(4.1)
where tF is a function of displacements/stresses/material law; and tR is a function of time.
Assume for now linear analysis: Equilibrium within 0V and on 0Sf , linear stress-strain law and small
displacements yields
tF = K · tU
We want to establish,
KU (t) = R(t)
14
(4.2)
(4.3)
MIT 2.094
4. Finite element formulation for solids and structures
Consider
Uˆ T = �
U1 V1 W1 U2
· · ·
WN
�
(N nodes)
where Uˆ T is a distinct nodal point displacement vector.
Note: for the moment “remove Su ”
We also say
Uˆ T = �
U1 U2 U3
· · ·
Un
�
(n = 3N )
We now assume
(m)
u
= H (m)Uˆ , u
(m)
⎤(m)
⎡
u
= ⎣ v ⎦
w
where H (m) is 3 x n and Uˆ is n x 1.
�(m) = B(m)Uˆ
where B(m) is 6 x n, and
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
x 1.
�(m) = B(m)Uˆ
where B(m) is 6 x n, and
�(m)T
�xx
= �
∂v
∂x
�zz γxy γyz γzx
�
�yy
∂u
∂y
e.g. γxy =
+
We also assume
u(m) = H (m)Uˆ
�(m) = B(m)Uˆ
15
(4.4)
(4.5)
(4.6a)
(4.6b)
(4.6c)
(4.6d)
MIT 2.094
4. Finite element formulation for solids and structures
Principle of Virtual Work:
�
V
�
�T τ dV = Uˆ f B dV
V
T
(4.7) can be rewritten as
� �
m
V (m)
�(m)T
τ (m)dV (m) =
� �
m
V (m)
Uˆ
(m)T
f B(m)
dV (m)
Substitute (4.6a) to (4.6d).
�
T �
�
Uˆ
B(m)T
�
τ (m) dV (m) =
m
�
T �
V (m)
�
Uˆ
m
V (m)
H (m)T
f B (m)
�
dV (m)
τ (m) = C(m)�(m) = C(m)B(m)Uˆ
Finally,
�
T �
�
�
Uˆ
�
B(m)T
C(m)B(m) dV (m) Uˆ =
�
V (m)
m
�
�
T �
�
Uˆ
�
m
V (m)
H (m)T
f B(m)
�
dV (m)
with
�(m)T
ˆ
= U
T
B(m)T
KUˆ | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
dV (m)
with
�(m)T
ˆ
= U
T
B(m)T
KUˆ = RB
where K is n x n, and RB is n x 1.
Direct stiffness method:
�
K = K(m)
m
�
(m)
RB = RB
m
�
K(m) =
B(m)T
C(m)B(m) dV (m)
(m) =
RB
V (m)
�
V (m)
H (m)T
f B(m)
dV (m)
16
(4.7)
(4.8)
(4.9)
(4.10)
(4.11)
(4.12)
(4.13)
(4.14)
(4.15)
(4.16)
(4.17)
MIT 2.094
4. Finite element formulation for solids and structures
Example 4.5 textbook
E = Young’s Modulus
Mathematical model Plane sections remain plane:
F.E. model
⎤
⎡
U = ⎣ U2 ⎦
U1
U3
Element 1
�
u(1)(x) = 1 − 100
�
x
��
H (1)
⎤
⎡
� U1
100 0 ⎣ U2 ⎦
�
U3
x
(4.18)
(4.19)
17
MIT 2.094
4. Finite element formulation for solids and structures
(4.20)
(4.21)
(4.22)
(4.23)
(4.24)
(4.25)
(4.26)
xx (x) = �
�(1)
�
− 1
100
1
100
��
B(1)
⎡
0 �
⎣
�
U1
U2
U3
⎤
⎦
Element 2
u(2)(x) = �
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
U1
U2
U3
⎤
⎦
Element 2
u(2)(x) = �
�
xx (x) = �
�(2)
�
0
0
x
80
�
U
�
1
80
�
U
�
1 − x
80
��
H (2)
− 1
80
��
B(2)
Then,
K =
⎡
⎣
E
100
1
−1
0
−1
1
0
0
0
0
⎤
⎦ +
13E
240
⎡
⎣
0
0
0
⎤
⎦
0
1
−1
0
−1
1
where,
�
�
AE
L
�
≡
�
E(1)
100
E · 13
3 · 80
E
80
=
13
3
� �� �
A∗
�
�
A∗ < A
�
η=80
4.333 < 9
�
�
A
�
η=0
1
<
<
18
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 5 - F.E. displacement formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
For the continuum
•
Differential formulation
•
Variational formulation (Principle of Virtual Displacements)
Reading:
Ch. 4
Next, we assumed infinitesimal small displacement, Hooke’s Law, linear analysis
· · ·
Un
�
, (n = all d.o.f. of element assemblage)
KU = R
u
(m) = H (m)U
�
K = K(m)
m
�
R = R(m)
m
B
�(m) = B(m)U
�
U T = U1 U2
�
K(m) =
B(m)T
C(m) B(m | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
U T = U1 U2
�
K(m) =
B(m)T
C(m) B(m) dV (m)
(m) =
RB
V (m)
�
V (m)
H (m)T
f B(m)
dV (m)
Surface loads
Recall that in the principle of virtual displacements,
“surface” loads =
�
T
Sf
U
f Sf dSf
u S (m)
H S (m)
Sf
= H S (m)
U
�
�
�
evaluated at the surface
= H (m)
19
(5.1a)
(5.1b)
(5.1c)
(5.1d)
(5.1e)
(5.1f)
(5.1g)
(5.1h)
(5.2)
(5.3)
(5.4)
MIT 2.094
5. F.E. displacement formulation, cont’d
Substitute into (5.2)
�
T
U
S(m)
H S(m) T
f S(m)
dS(m)
for element (m) and one surface of that element.
m) =
R(
s
�
S(m)
H S(m) T
f S(m)
dS(m)
Need to add contributions from all surfaces of all loaded external elements.
KU = RB + RS + Rc
where Rc are concentrated nodal loads.
Assume
• (5.7) has been established without any displacement boundary conditions.
• We, however, know nodal displacements Ub (rewriting (5.7)).
� �
�
�
�
�
Kaa
Kba
Kab
Kbb
Ua
Ub
=
Ra
Rb
KU = R ⇒
Solve for Ua:
KaaUa = Ra − KabUb
where Ub is known!
Then use
KbaUa + KbbUb = Rb + Rr
where Rr are unknown reactions.
Example 4.6 textbook | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
baUa + KbbUb = Rb + Rr
where Rr are unknown reactions.
Example 4.6 textbook
⎞
⎛
τxx
⎝ τyy ⎠ =
τxy
E
1 − ν2
⎡
1 ν
⎣ ν 1
0
0
�xx
⎤ ⎛
⎞
0
0 ⎦ ⎝ �yy ⎠
ν
1−
2
γxy
20
(5.5)
(5.6)
(5.7)
(5.8)
(5.9)
(5.10)
(5.11)
MIT 2.094
5. F.E. displacement formulation, cont’d
�
�
u(x, y)
v(x, y)
⎛
= H
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
u1
u2
u3
u4
v1
v2
v3
v4
If we can set this relation up, then clearly we can get H (1), H (2), H (3), H (4).
u(m) = H (m)U
Also want �(m) = B(m)U . We want H. We could proceed this way
u(x, y) = a1 + a2x + a3y + a4xy
v(x, y) = b1 + b2x + b3y + b4xy
Express a1. . . a4, b1. . . b4 in terms of the nodal displacements u1. . . u4, v1. . . v4.
(e.g.) u(1, 1) = a1 + a2 + a3 + a4 = u1.
(5. | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
.
(e.g.) u(1, 1) = a1 + a2 + a3 + a4 = u1.
(5.12)
(5.13)
(5.14)
(5.15)
h1(x, y) = 1
for node 1.
4 (1 + x)(1 + y) interpolation function
h2(x, y) = 1 (1 − x)(1 + y)
4
21
MIT 2.094
5. F.E. displacement formulation, cont’d
h3(x, y) = 1 (1 − x)(1 − y)
4
h4(x, y) = 1 (1 + x)(1 − y)
4
u(x, y) = h1u1 + h2u2 + h3u3 + h4u4
v(x, y) = h1v1 + h2v2 + h3v3 + h4v4
h1 h2 h3 h4
0
0
0
0 h1 h2 h3 h4
0
0
0
0
��
H (2x8)
�
u(x, y)
v(x, y)
�
�
=
�
We also want,
⎛
�
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
�
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
u1
u2
u3
u4
v1
v2
v3
v4
⎛
⎛
⎝
�xx
�yy
γxy
⎡⎞
⎠ = ⎣ 0
h1,x h2,x h3,x h4,x
0
0
0
⎤
0
h1,y h2,y h3,y h4,y ⎦
0
0
0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
⎤
0
h1,y h2,y h3,y h4,y ⎦
0
0
0
h1,y h2,y h3,y h4,y h1,x h2,x h3,x h4,x
��
B (3x8)
�
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
�
⎜
⎝
�xx =
�yy =
γxy =
∂u
∂x
∂v
∂y
∂u
∂y
+
∂v
∂x
22
(5.16)
(5.17)
(5.18)
(5.19)
(5.20)
(5.21)
(5.22)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
u1
u2
u3
u4
v1
v2
v3
v4
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 6 - Finite element formulation, example, convergence
Prof. K.J. Bathe
MIT OpenCourseWare
6.1 Example
t = 0.1, E, ν plane stress
KU = R; R = RB + Rs + Rc + Rr
�
�
B(m)T
C(m) B(m) d V (m)
K
= K(m); K (m) =
�
m
�
RB = R(m)
m
B
(m)
; R =
B
V (m)
V (m)
H (m)T
f B(m) d V (m)
Reading:
Ex. 4.6 in
the text
(6.1)
(6.2)
(6.3)
6.1.1 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
.6 in
the text
(6.1)
(6.2)
(6.3)
6.1.1 F.E. model
u1 u2 u3 u4 v1 v2 v3 v4
↓
↓
↓
↓
↓
↓
⎤
� � �
× × × × ×
. . .
↓
↓
�
�
K
�
el. (2)
⎡
= ⎢
⎢
⎢
⎣
⎥
⎥
⎥
⎦
23
←
u1
. . .
(6.4)
6. Finite element formulation, example, convergence
MIT 2.094
In practice,
�
�
K
�
el
=
�
V
⎛
BT CB dV ; � = B
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
u1
. . .
u4
v1
. . .
v4
where K is 8x8 and B is 3x8.
Assume we have K (8x8) for el. (2)
⎡
↓
↓
· · ·
↓
· · ·
↓
U11
↓
U1 U2 U3 U4 U5
↓
↓
↓
U18
↓
× × × × × × × × ×
.
.
.
.
.
.
�
.
.
.
.
.
.
.
.
.
.
.
.
�
.
.
.
.
.
.
.
.
.
.
.
.
�
.
.
.
.
.
.
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
��
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
× × × × × × × × ×
K =
����
assemblage
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
Consider,
�
RS = H S T
S
�
f S d S; H S = H
�
�
on surface
�
H =
h1 h2 h3 h4
0
0
0
0 h1 h2 h3 h4
0
0
0
0
u =
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
u1
u2
. . .
u4
v1
. . .
v4
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
�
←
←
u(x, y)
v(x, y)
24
U1←
. . .
.
. .
U11←
. . .
. . .
U18←
(6.5)
(6.6)
(6.7)
(6.8)
(6.9)
MIT 2.094
6. Finite element formulation, example, convergence
⎤
⎦
(6.10)
(6.11)
(6.12)
(6.13)
�
H S = H
�
�
y=+1
1
2 (1 + x)
⎡
= ⎣
1
2 (1 − x) 0 0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
1
2 (1 + x)
⎡
= ⎣
1
2 (1 − x) 0 0
0
0
0 0
0
0
0 0
1 (1 + x)
2
1 (1 − x)
2
0
0
�
⎤
⎥
⎥
⎦
0
−p(x)
�
(0.1) dx
� �� �
thickness
From (6.7);
RS
=
⎡
⎢
⎢
⎣
� +1
−1
1 (1 + x)
2
1 (1 − x)
2
0
0
0
0
1 (1 + x)
2
1
(1 − x)
2
RS
=
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
0
0
−p0(0.1)
−p0(0.1)
6.1.2 Higher-order elements
Want h1, h2, h3, h4, h5
u(x, y) = �
5
=1 hiui.
i
hi = 1 at node i and 0 at all other nodes.
h5 = 1 (1 − x2)(1 + y)
2
25
MIT 2.094
6. Finite element formulation, example, convergence
(6.14)
(6.15)
(6.16)
(6.17)
(6.18)
(6.19)
(6.20)
(6.21)
(6.22)
(6.23)
h1 =
h2 =
h3 =
h1 =
1
4
1
4
1
4
1
4
(1 + x)(1 + y) −
h5
(1 − x)(1 + y) −
h5
1
2
1
2
(1 − x)(1 − y)
(1 + x)(1 − y)
Note:
�
hi = 1
We | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(1 − x)(1 − y)
(1 + x)(1 − y)
Note:
�
hi = 1
We must have
�
i hi = 1 to satisfy the rigid body mode condition.
u(x, y) =
�
hiui
i
Assume all nodal point displacements = u∗. Then,
u(x, y) =
�
hiu∗ = u∗
�
hi = u∗
i
i
From (6.1),
� ��
m
V (m)
�
�
m
�
K(m) U = R
�
B(m)T
C(m)B(m)dV (m) U = R
where C(m)B(m)U = τ (m). (Assume we calculated U .)
� �
m
V (m)
B(m)T
τ (m) d V (m) = R
�
F (m) = R; F (m) =
m
�
V (m)
B(m)T
τ (m) d V (m)
Two properties
I. The sum of the F (m)’s at any node is equal to the applied external forces.
26
MIT 2.094
6. Finite element formulation, example, convergence
II. Every element is in equilibrium under its F (m)
�
T
T
ˆU
F (m) = ˆU
�
�
=
V (m)
��
=�(m)T
�(m)T
B(m)T
τ (m) d V (m)
�
τ (m) d V (m)
V (m)
= 0
where Uˆ T
= virtual nodal point displacement.
(6.24)
(6.25)
(6.26)
Apply rigid body displacement.
If we move the element virtually in the rigid body modes, �(m) is zero. Therefore the virtual work
obtained due to virtual motion of the element is zero. Then the | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
m) is zero. Therefore the virtual work
obtained due to virtual motion of the element is zero. Then the element is in equilibrium under its F (m).
27
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 7 - Isoparametric elements
Prof. K.J. Bathe
MIT OpenCourseWare
We want K = V BT CB dV , RB = V H T f B dV . Unique correspondence (x, y) ⇔ (r, s)
�
�
(r, s) are natural coordinate system or isoparametric coordinate system.
4
�
hixi
x =
i=1
4
�
hiyi
i=1
y =
where
h1 = (1 + r)(1 + s)
(1 − r)(1 + s)
1
4
1
4
h2 =
. . .
u(r, s) =
4
�
hiui
v(r, s) =
i=1
4
�
hivi
i=1
28
Reading:
Sec. 5.1-5.3
(7.1)
(7.2)
(7.3)
(7.4)
(7.5)
(7.6)
MIT 2.094
7. Isoparametric elements
� = Buˆ uˆT = u1 u2
�
�
· · ·
v4
⎡
⎢
⎢
⎢
⎢
⎣
⎛
⎝
⎛
⎝
⎤
⎥
⎥
⎥ = Buˆ
⎥
⎦
∂u
∂x
∂v
∂y
∂u + ∂v
∂x
∂y
⎞ ⎡
⎠ = ⎣
�
⎞
⎠ = J −1
∂x
∂r
∂y
∂r | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
��
�
⎞
⎠ = J −1
∂x
∂r
∂y
∂r
∂x
∂s
∂y
∂s
��
J
⎛
∂
∂r
⎝
∂
∂s
∂
∂r
∂
∂s
∂
∂x
∂
∂y
⎞
⎠
∂
∂x
∂
∂y
⎤ ⎛
⎦ ⎝
�
⎞
⎠
(7.7)
(7.8)
(7.9)
(7.10)
J must be non-singular which ensures that there is unique correspondence between (x, y) and (r, s).
Hence,
K =
Also, RB =
� 1 � 1
−1 −1
� 1 � 1
−1 −1
BT CB t det(J ) dr ds
�
�
��
d V
H T f B t det(J ) dr ds
Numerical integration (Gauss formulae) (Ch. 5.5)
� �
K ∼
= t
BT
ij CBij
det(Jij ) × (weight i, j)
i
j
2x2 Gauss integration,
(i = 1, 2)
(j = 1, 2) (weight i, j = 1 in this case)
29
(7.11)
(7.12)
(7.13)
(7.14)
(7.15)
MIT 2.094
9-node element
7. Isoparametric elements
9
�
hixi
x =
i=1
9
�
hiyi
i=1
9
�
hiui
i=1
9
�
hivi
i=1
y =
u =
v =
Use 3x3 Gauss integration
For rectangular elements, J = const
Consider the following element,
Note, here we could use hi(x, y) directly. | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
3 Gauss integration
For rectangular elements, J = const
Consider the following element,
Note, here we could use hi(x, y) directly.
J =
� �
�
3 = 6
2
0
�
0
3
2
(7.16)
(7.17)
(7.18)
(7.19)
(7.20)
Then, we can determine the number of appropriate integration points by investigating the maximum
order of BT CB.
For a rectangular element, 3x3 Gauss integration gives exact K matrix. If the element is distorted,
a K matrix which is still accurate enough will be obtained, (if high enough integration is used).
30
MIT 2.094
7. Isoparametric elements
Convergence Principle of virtual work:
�
V
�T C� dV = R(u)
Reading:
Sec. 5.5.5,
4.3
(7.21)
Find u, solution, in V, vector space (any continuous function that satisfies boundary conditions),
satisfying
�
V
�T C� dV = a(u, v) = (f , v)
� �� �
R(v)
� �� �
bilinear form
Example:
for all v, an element of V.
(7.22)
Finite Element problem Find uh ∈ Vh, where Vh is F.E. vector space such that
a(uh, vh) = (f , vh) ∀vh ∈ Vh
(7.23)
Size of Vh ⇒ # of independent DOFs (here it’s 12).
Note:
a(w, w)
� �� �
2x (strain energy when imposing w)
> 0 for w ∈ V (w = 0)
Also,
a(wh, wh) > 0 for wh ∈ Vh
(Vh ⊂ V, wh = 0)
31
�
�
MIT 2.094
7. Isoparametric elements
Property I Define | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
0)
31
�
�
MIT 2.094
7. Isoparametric elements
Property I Define: eh = u − uh.
From (7.22), a(u, vh) = (f , vh)
From (7.23), a(uh, vh) = (f , vh)
Hence,
a(u − uh, vh) = 0
a(eh, vh) = 0
(error is orthogonal in that sense to all vh in F.E. space).
Property II
a(uh, uh) ≤ a(u, u)
Proof:
a(u, u) = a(uh + eh, uh + eh)
= a(uh, uh) + ������� 0 by Prop. I
2a(uh, eh)
+ a(eh, eh)
� �� �
≥0
∴ a(u, u) ≥ a(uh, uh)
(7.24)
(7.25)
(7.26)
(7.27)
(7.28)
(7.29)
(7.30)
(7.31)
32
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 8 - Convergence of displacement-based FEM
Prof. K.J. Bathe
(A) Find
u ∈ V such that a(u, v) = (f , v) ∀v ∈ V (Mathematical model)
a(v, v) > 0 ∀v ∈ V, v =� 0.
where (8.2) implies that structures are supported properly. E.g.
(B) F.E. Problem Find
uh ∈ Vh such that a(uh, vh) = (f , vh) ∀vh ∈ Vh
a(vh, vh) > 0 ∀vh ∈ Vh,
vh �= 0
Properties eh = u − uh
(I) a(eh, vh) = 0 ∀vh ∈ Vh
(II) a(uh, uh) ≤ a(u, u)
MIT OpenCourseWare
(8.1)
(8.2) | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
II) a(uh, uh) ≤ a(u, u)
MIT OpenCourseWare
(8.1)
(8.2)
(8.3)
(8.4)
(8.5)
(8.6)
33
MIT 2.094
8. Convergence of displacement-based FEM
�
�
(C) Assume Mesh
�
h1
�
�
“is contained in” Mesh
�
h2
�
�
e.g. Mesh
�
�
�
not contained in Mesh
�
h2
h1
We assume (C), but need another property (independent of (C))
(III) a(eh, eh) ≤ a(u − vh, u − vh) ∀vh ∈ Vh
uh minimizes! (Recall eh = u − uh)
Proof: Pick wh ∈ Vh.
a(eh + wh, eh + wh) = a(eh, eh) + ������� 0
2a(eh, wh)
(
)
+ a w wh, h
�
�
��
≥0
Equality holds for (wh = 0)
a(eh, eh) ≤ a(eh + wh, eh + wh)
= a(u − uh + wh, u − uh + wh)
Take wh = uh − vh.
a(eh, eh) ≤ a(u − vh, u − vh)
(8.7)
(8.8)
(8.9)
(8.10)
(8.11)
Using property (III) and (C), we can say that we will converge monotonically, from below, to a(u, u):
34
MIT 2.094
8. Convergence of displacement-based FEM
Pascal triangle (2D)
⇒
4-node element is complete to k = 1.
⇒
9-node element is complete to k = 2.
(Ch. 4.3)
error in displacement ∼ C ·
hk+1
(C is a constant determined by the exact solution, material property. . . )
error in stresses ∼ C hk
·
error in strain energy ∼ C h2k
· | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
exact solution, material property. . . )
error in stresses ∼ C hk
·
error in strain energy ∼ C h2k
·
←
(
these C are different)
Hence,
E − Eh = C h2k
·
(roughly equal to)
By theory,
log (E − Eh) = log C + 2k log h
(8.12)
(8.13)
(8.14)
(8.15)
(8.16)
35
MIT 2.094
8. Convergence of displacement-based FEM
By experiment, we can evaluate log(E − Eh) for different meshes and plot log(E − Eh) vs. log h
We need to use graded meshes if we have high stress gradients.
Example Consider an almost incompressible material:
�V = vol. strain
or
� · v → very small or zero
We can “see” difficulties:
p = −κ�V κ = bulk modulus
As the material becomes incompressible (ν = 0.3
→
0.4999)
�
κ → ∞
→
0
�V
p
→
finite number
(8.17)
(8.18)
(8.19)
(8.20)
(Small error in �V results in huge error on pressure as κ → ∞, the constant C in (8.15) can be very large
⇒
locking)
36
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 9 - u/p formulation
Prof. K.J. Bathe
We want to solve
I. Equilibrium
�
B = 0
τij,j + fi
Sf
τij nj = fi
in Volume
on Sf
II. Compatibility
III. Stress-strain law
Use the principle of virtual displacements
�
V
�T C� dV = R
We recognize that if ν
0.5→
�V
→
0 (�V = �xx + �yy + �zz) | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
We recognize that if ν
0.5→
�V
→
0 (�V = �xx + �yy + �zz)
κ =
E
3(1 − 2ν)
→ ∞
p = −κ�V must be accurately computed
Solution
τij = κ�V δij + 2G��
ij
where
δij = Kronecker delta =
�
1
0
i = j
j
i =�
Deviatoric strains:
��
ij = �ij −
�V
3
δij
τij = −pδij + 2G��
ij
�
p = −
τkk
3
�
(9.2) becomes
�
V
�
��T C��� dV +
�
�V κ�V dV = R
V
�
��T C��� dV −
�T
V p dV = R
V
V
MIT OpenCourseWare
Reading:
Sec. 4.4.3
(9.1)
(9.2)
(9.3)
(9.4)
(9.5)
(9.6)
(9.7)
(9.8)
(9.9)
(9.10)
(9.11)
37
MIT 2.094
9. u/p formulation
We need another equation because we now have another unknown p.
p + κ�V = 0
p (p + κ�V ) dV = 0
�
p
κ
dV = 0
p �V +
�
�
� V
−
V
For an element,
u = Huˆ
�� = BDuˆ
�V = BV uˆ
p = Hppˆ
Example
�V = �xx + �yy
⎤
⎡
1
�xx −
3
⎢ �yy − (�xx + �yy) ⎥
1
⎥
⎢
3
⎣
γxy
⎦
(�xx + | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
⎥
1
⎥
⎢
3
⎣
γxy
⎦
(�xx + �yy)
�� =
− 1
3
(�xx + �yy)
Note: �zz = 0 but �� = 0!
zz
p = Hppˆ = [1]{p0}
p(x, y) = p0
We obtain from (9.11) and (9.14)
�
� �
�
�
�
Kuu Kup
Kpu Kpp
uˆ
pˆ
=
R
0
Kuu =
DC�BD dV
Kup = − BV
T Hp dV
Kpu = − Hp
T BV dV
�
BT
V
�
V
�
V
�
V
T
Kpp = − Hp
1
κ
Hp dV
38
(9.12)
(9.13)
(9.14)
(9.15)
(9.16)
(9.17)
(9.18)
Plane strain (�zz
4/1 element
= 0)
Reading:
Ex. 4.32 in
the text
(9.19)
(9.20)
(9.21)
(9.22)
(9.23)
(9.24a)
(9.24b)
(9.24c)
(9.24d)
�
MIT 2.094
9. u/p formulation
In practice, we use elements that use pressure interpolations per element, not continuous between
elements. For example:
Then, unless ν = 0.5 (where Kpp = 0), we can use static condensation on the pressure dof’s.
Use pˆ equations to eliminate pˆ from the uˆ equations.
�
Kuu − KupK−1
�
pp Kpu uˆ = R
(In practice, ν can be 0.499999. . . )
The “best element” is the 9/3 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
� = R
(In practice, ν can be 0.499999. . . )
The “best element” is the 9/3 element. (9 nodes for displacement and 3 pressure dof’s).
p(x, y) = p0 + p1x + p2y
The inf-sup condition
(9.25)
(9.26)
Reading:
Sec. 4.5
⎤
⎡
=�V
� �� �
�
Vol qh � · vh dVol
⎥
⎥
⎥
v
h
⎦
q
� � �
h
��
�
for normalization
⎢
⎢
⎢
⎣
�
�
sup
inf
����
����
qh∈Qh vh∈Vh
≥ β > 0
(9.27)
Qh: pressure space.
If “this” holds, the element is optimal for the displacement assumption used (ellipticity must also be
satisfied).
Note:
infimum = largest lower bound
supremum = least upper bound
For example,
inf {1, 2, 4} = 1
sup {1, 2, 4} = 4
inf {x ∈ R; 0 < x < 2} = 0
sup {x ∈ R; 0 < x < 2} = 2
(9.23) rewritten (κ = ∞, full incompressibility). Diagonalize using eigenvalues/eigenvectors.
For a mesh of element size h we want
βh > 0
as we refine the mesh, h
→
0
39
MIT 2.094
9. u/p formulation
For
(entry [3,1] in matrix) assume the circled entry is the minimum (inf) of
.
Also, all entries in the matrix not shown are zero.
Case 1 βh = 0
⎧
⎨
⎩ ���� ·
α
=0
⇒
·
·
uh + 0 ph = Rh
0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
⎩ ���� ·
α
=0
⇒
·
·
uh + 0 ph = Rh
0 uh|i = 0 (from the bottom equation)
(from the top equation)
|j
|i
|i
⇒ no equation for ph|j
⇒
spurious pressure! (any pressure satisfies equation)
Case 2 βh = small = �
� · uh|i = 0 ⇒ uh|i = 0
∴ � ·
ph|j + uh|i · α = Rh|i
Rh|
i
�
ph|j =
⇒
�
⇒
displ. = 0
pressure
→
large
�
as � is small
The behavior of given mesh when bulk modulus increases: locking, large pressures. See Example
4.39 textbook.
40
�
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 10 - F.E. large deformation/general nonlinear analysis
Prof. K.J. Bathe
MIT OpenCourseWare
We developed
�
tτij teij d Vt = Rt
tV
et ij =
1
2
�
∂ui
∂ xt
j
+
�
∂uj
∂ xt
i
�
tV
tτij δteij d Vt = Rt
δteij =
�
∂(δui)
∂ txj
1
2
+
�
∂(δuj )
∂ txi
(≡ teij )
In FEA:
tF = tR
In linear analysis
tF = K tU KU = R
⇒
In general nonlinear analysis, we need to iterate. Assume the solution is known “at time t”
t
x = x + u
0
t
Hence tF is known. Then we consider
t+Δt
F =
t+Δt
R | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Hence tF is known. Then we consider
t+Δt
F =
t+Δt
R
Consider the loads (applied external loads) to be deformation-independent, e.g.
41
Reading:
Ch. 6
(10.1)
(10.2)
(10.3)
(10.4)
(10.5)
(10.6)
(10.7)
(10.8)
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
Then we can write
t+ΔtF = tF + F
t+ΔtU = tU + U
where only tF and tU are known.
F ∼
= tK ΔU ,
tK = tangent stiffness matrix at time t
From (10.8),
tK ΔU = t+ΔtR − tF
(10.9)
(10.10)
(10.11)
(10.12)
We use this to obtain an approximation to U . We obtain a more accurate solution for U (i.e. t+ΔtF )
(10.13)
(10.14)
(10.15)
(10.16)
(10.17)
(10.18)
(10.19)
using
t+ΔtK (i−1)ΔU (i) = t+ΔtR − t+ΔtF (i−1)
t+ΔtU (i) = t+ΔtU (i−1) + ΔU (i)
Also,
t+ΔtF (0) = tF
t+ΔtK (0) = tK
t+ΔtU (0) = tU
Iterate for i = 1, 2, 3 . . . until convergence. Convergence is reached when
�
�
ΔU (i) < �D
�
�
�
�
2
�
t+ΔtF (i−1 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�U (i) < �D
�
�
�
�
2
�
t+ΔtF (i−1)
�
�
2
< �F
�
�
�
t+ΔtR
−
Note:
�a�2 =
��
(ai)2
i
ΔU (i) = U
�
i=1,2,3...
ΔU (1) in (10.13) is ΔU in (10.12).
(10.13) is the full Newton-Raphson iteration.
How we could (in principle) calculate tK
Process
•
Increase the displacement tUi by �, with no increment for all tUj , j = i
calculate t+�F
•
• the i-th column in tK = ( t+�F − tF ) /� = ∂ tF .
t∂ Ui
42
�
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
So, perform this process for i = 1, 2, 3, . . . , n, where n is the total number of degrees of freedom.
Pictorially,
tK =
⎡
⎢
⎢
⎢
⎣
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
⎤
⎥
⎥
⎥
⎦
· · ·
A general difficulty: we cannot “simply” increment Cauchy stresses.
•
t+Δtτ
ij
referred to area at time t + Δt
• tτij referred to area at time t.
tSij , where 0 in the leading subscript
We define a new stress measure, 2nd Piola - Kirchhoff stress, t+Δ
0
refers to original configuration. Then,
t+Δt
0Sij = 0
tSij + 0S | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
configuration. Then,
t+Δt
0Sij = 0
tSij + 0Sij
(10.20)
The strain measure energy-conjugate to the 2nd P-K stress 0
Then,
�
tSij δ 0
0
t�ij d V = R
0
t
tSij is the Green-Lagrange strain 0
t�ij
(10.21)
(10.22)
0V
Also,
�
0V
Example
t+Δt
S δ
0 ij
t+Δt
0
� d V =
0 ij
t+Δt
R
43
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
(10.23)
(10.24)
(10.25)
(10.26)
(10.27)
(10.28)
tF = tR
t+Δt
F =
R
t+Δt
(every time it is in equilibrium)
(10.13) and (10.14) give:
i = 1,
t+ΔtK (0) ΔU (1) = t+ΔtR
−
t+ΔtF (0) ≡ fn(tU )
t+ΔtU (1) = t+ΔtU (0) + ΔU (1)
i = 2,
t+ΔtK (1) ΔU (2) = t+ΔtR − t+ΔtF (1)
t+ΔtU (2) = t+ΔtU (1) + ΔU (2)
44
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 11 - Deformation, strain and stress tensors
Prof. K.J. Bathe
MIT OpenCourseWare
We stated that we use
�
tV
tτij δ eij d Vt =
t
�
0V
tS | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
that we use
�
tV
tτij δ eij d Vt =
t
�
0V
tS
t�
0 ij δ 0 ij d
0V = tR
The deformation gradient We use txi = 0xi + tui
tX0
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∂ t x1
∂ 0x1
∂ t x2
0
x
∂
1
∂ x t
3
∂ x0
1
∂ t x1
∂ 0x2
∂ t x2
0
x
∂
2
∂ x t
3
∂ x0
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
∂ t x1
∂ 0x3
∂ t x2
x
∂
3
0
∂ x t
3
∂ x0
3
dt x
=
⎡
⎣
⎡
d0 x
=
⎣
⎤
⎦
⎤
⎦
td x1
td x
2
td x3
0d x1
0
d x2
0d x3
Implies that
dt x = 0
tX d0 x
Reading:
Ch. 6
(11.1)
(11.2)
(11.3)
(11.4)
(11.5)
tX is frequently denoted by 0
(0
force vector)
tF or simply F , but we use F for
We will also use the right Cauchy-Green deformation tensor
tC
0
=
tX T tX
0
0
(11.6)
Some applications
45
MIT 2.094
11. Deformation, strain and stress tensors
The stretch of a | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
11.6)
Some applications
45
MIT 2.094
11. Deformation, strain and stress tensors
The stretch of a fiber (tλ):
�
�2
tλ =
t
d xT d x
t
d0xT d0x
=
�
�2
t
d s
d0s
The length of a fiber is
�
�
d s = d x T d x
0
0
0
1
2
�
�2
tλ =
� �
�
d0xT tX T
tX d0x
0
0
0
·
d s d s
0
�
,
from (11.5)
Express
�
d x = d s n
�
0
0
0
0
0
n = unit vector into direction of d x
�
tλ �2 = 0 n T
tC 0 n
0
⇒
�
∴ tλ = 0 n T
tC 0 n
0
�
1
2
(11.7)
(11.8)
(11.9)
(11.10)
(11.11)
(11.12)
(11.13)
Also,
�
t �T · �
d ˆx
t �
d x
= �
d ˆs t � �
d st �
cos tθ,
(a · b = �a��b� cos θ)
(11.14)
From (11.5),
�
d ˆx0 T ˆtX T
0
�
�
tX0
cos tθ =
d ˆts d st
d x0 �
�
ˆtX0 ≡ tX0
�
=
d ˆs0 ˆn0 T tC0 n0 d s0
d ˆts · d st
∴
cos tθ =
tC 0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
n0 d s0
d ˆts · d st
∴
cos tθ =
tC 0n
0nˆ T
0
tλˆ tλ
Also,
tρ =
0ρ
det 0
tX
(see Ex. 6.5)
Example
46
(11.15)
(11.16)
(11.17)
(11.18)
Reading:
Ex. 6.6 in
the text
MIT 2.094
11. Deformation, strain and stress tensors
1
4
(1 + x1)(1 + x2)
0
0
h1 =
. . .
t
t
0
xi = xi + ui
4
�
=
hk
t xi
k ,
(i = 1, 2)
k=1
where txi
k are the nodal point coordinates at time t (tx1
1 = 2, tx2
1 = 1.5)
Then we obtain
�
tX =
0
1
4
5 + 0x2
2 (1 + 0x2)
1
1 + 0x1
2 (9 + 0x1)
1
�
At 0x1 = 0, 0x2 = 0,
�
�
tX �
�
0x =0x2=0
i
0
1
4
=
�
5
1
2
1
9
2
The Green-Lagrange Strain
t� =
0
1 �
2 0
tX T
�
tX − I =
0
�
tC − I
1 �
2 0
t
∂ xi
∂ 0x
j
=
�
0
�
∂ xi + ui
∂ 0x
t
j
= δij +
t
∂ ui
∂ 0x
j
We find that
1 �
2
t� =
0 ij
t u + t u + | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
j
We find that
1 �
2
t� =
0 ij
t u + t u + t u
0 j,i
0 i,j
�
t u
0 k,i 0 k,j
where
t∂ u
i
t ui,j = 0
∂ xj
0
,
sum over k = 1, 2, 3
47
(11.19)
(11.20)
(11.21)
(11.22)
(11.23)
(11.24)
(11.25)
(11.26)
(11.27)
MIT 2.094
11. Deformation, strain and stress tensors
Polar decomposition of
tX
0
tX
0
=
tR tU
0
0
where tR
0
is a rotation matrix, such that
tRT
0
tR
0 = I
and
tU
0
is a symmetric matrix (stretch)
Ex. 6.9 textbook
tX =
0
� √
3
2
1
2
� �
�
4
3
0
0
3
2
−
√
1
2
3
2
Then,
tC
0
=
t�
0
=
�2
tX T tX
0
0
��
1
tU
0
2
�
�
tU
= 0
�
2
I−
(11.28)
(11.29)
(11.30)
(11.31)
(11.32)
This shows, by an example, that the components of the Green-Lagrange strain are independent of a
rigid-body rotation.
48
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 12 - Total Lagrangian formulation
Prof. K.J. Bathe
MIT OpenCourseWare
We discussed:
�
tX =
0
�
i
j
⇒ dt x = tX d0 x,
0
d0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
tX =
0
�
i
j
⇒ dt x = tX d0 x,
0
d0 x = � tX �−1 dt x
0
t
∂ x
0
x
∂
tC = tX T tX
0
0
0
d0 x = 0
tX dt x
where 0
tX = �
tX �−1 =
0
�
�
0xi
∂
t
∂ xj
The Green-Lagrange strain:
t� =
0
1 �
tX − I =
tX T
0
0
2
�
�
1 �
tC − I
0
2
Polar decomposition:
tX = 0
0
tR0
tU ⇒ 0
t� =
�
1 ��
�2
tU − I
0
2
(12.1)
(12.2)
(12.3)
(12.4)
(12.5)
We see, physically that:
where dt+Δtx and dtx are the same lengths
⇒
change.
the components of the G-L strain do not
Note in FEA
0
xi =
0 k
� ⎫
hk xi ⎪⎪⎬
t k ⎪⎪
hk ui ⎭
k
�
k
t
ui =
t
xi = xi + ui →
0
t
for an element
for any particle
Hence for the element
�
�
t
xi =
0 k
hk xi +
t k
hk ui
=
=
k
�
hk
k
�
k
k + t ui
�
0 xi
k
�
hk
k
t xk
k
49
(12.6)
(12.7)
(12.8)
(12.9)
(12.10)
MIT 2.094
E.g., k = 4
12. Total | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
tF (k−1)
t+ΔtR − t+Δ
0
In the full N-R iteration, we use
�
(k−1) ΔU (k)
�
t+Δt
(k−1)
t+Δt
+
0KN L
0KL
=
t+Δt
R −
52
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 13 - Total Lagrangian formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
Example truss element. Recall:
Principle of virtual displacements applied at some time t + Δt:
�
t+ΔtV
�
0V
t+Δt
τij δ t+Δteij d
t+Δt
V =
t+Δt
R
t+Δt
S δ
0 ij
t+Δt
0
� δ V =
0 ij
t+Δt
R
t+Δ
tSij = 0
tSij + 0Sij
0
t�ij = 0
t+Δ
t�ij + 0�ij
0
0�ij = 0eij + 0ηij
where 0
tSij and 0
t�ij are known, but 0Sij and 0�ij are not.
0eij =
0ηij =
t uk,i 0uk,j + 0
�
t uk,j 0uk,i
1 �
2 0ui,j + 0uj,i + 0
�
1 �
0uk,i 0uk,j
2
Substitute into (13.2) and linearize to obtain
�
0V
δ 0e 0C
ij
ijrs
0ers
0
d V
�
+
0V
F.E. discretization gives
tS
0
ij δ ηij d V =
0
0
�
t+Δt
R − | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
gives
tS
0
ij δ ηij d V =
0
0
�
t+Δt
R −
0V
�
tKL + 0
0
�
tKN L ΔU = t+ΔtR − 0
tF
53
(13.1)
(13.2)
(13.3)
(13.4)
(13.5)
(13.6)
(13.7)
tS
δ 0eij
0
0
ij d V
(13.8)
(13.9)
MIT 2.094
13. Total Lagrangian formulation, cont’d
tK
0 L =
tK =
0 N L
tF =
0
�
0V
�
0V
�
0V
tBT
0
0 L 0C 0 Ld V
tB
T
tB
0 N L 0
tBN Ld V
0
0
tS
����
matrix
tB tSˆ d V
T
0
0 L 0
����
vector
The iteration (full Newton-Raphson) is
�
t+ΔtK (i−1) + t+ΔtK (i−1) ΔU (i) = t+ΔtR
0 N L
0 L
�
−
t+ΔtF (i−1)
0
t+ΔtU (i) = t+ΔtU (i−1) + ΔU (i)
Truss element example (p. 545)
Here we have to only deal with 0
tS11, 0e11, 0η11
0e11 =
0η11 =
∂u1 +
∂ 0x1
�
∂uk
1
2 ∂ 0x1
t
∂ uk
∂ 0x1
·
·
∂uk
∂ 0x1
∂uk
∂ 0x1
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
��
tK
0 L
⎤
⎥
⎥
⎦
�
(13.22)
(13.23)
(13.24)
⎡
+
tP
0L
⎢
⎢
⎣
�
0 −1
1
0
1
0
1
0
−1
0
0 −1
��
tK0 N L
0
−1
0
1
⎤
⎥
⎥
⎦
�
When θ = 0, 0
tKL doesn’t give stiffness corresponding to u2
2, but 0
tKN L does.
56
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 14 - Total Lagrangian formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
Truss element. 2D and 3D solids.
�
t+ΔtV
�
0V
t+Δt
τij δ t+Δteij d
t+Δt
V =
t+Δt
R
t+Δt
0Sij δ
t+Δt
R
t+Δt
0
0�ij δ V =
⏐
⏐
� linearization
�
t+ΔtR −
0
ij δ V +
tSij δ ηij δ V =
0
0
0
tS
0 ij δ 0e
0
ij δ V
0V
�
0V
�
0V
0Cijrs 0ers 0e
δ
Note:
δ
0eij
t
= δ �
0 ij
varying with respect to the configuration at time t.
F.E. discretization
0
xi =
�
0 k
hk xi
k
�
t k
hk ui
t
ui =
k
(14.4) into (14.3) gives
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
t k
hk ui
t
ui =
k
(14.4) into (14.3) gives
�
tKL + 0
0
�
tKN L U = t+ΔtR − 0
tF
t
xi =
�
t k
hk xi
t+Δt
xi =
�
hk
t+Δt k
xi
t+Δt
ui =
k
�
hk
t+Δt k
ui
k
�
k
hkui
ui =
k
k
57
(14.1)
(14.2)
(14.3)
(14.4a)
(14.4b)
(14.5)
MIT 2.094
Truss
14. Total Lagrangian formulation, cont’d
ΔL � 1 small strain assumption:
0L
tK =
0
0E A
0L
⎡
cos θ sin θ
sin2 θ − sin θ cos θ
cos2 θ
sin θ cos θ
− cos2 θ − cos θ sin θ
− sin2 θ
sin θ cos θ
sin2 θ
=
⎢
⎢
⎣
cos2 θ
cos θ sin θ
− cos2 θ
− cos θ sin θ
⎡
+
tP
0L
⎢
⎢
⎣
− cos θ sin θ
− sin2 θ
0
−1
0
1
0 −1
0
1
1
0
0
0 −1
1
0
−1
⎤
⎥
⎥
⎦
(notice that the both matrices are symmetric)
�
u1
v1
�
�
=
cos θ
sin θ
− sin θ cos θ
� �
�
u1
v1
Corresponding to the u and v displacements we have:
tK =
0
0E A
0L | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
v1
Corresponding to the u and v displacements we have:
tK =
0
0E A
0L
⎡
=
⎢
⎢
⎣
1 0 −1 0
0
0 0
0
1 0
−1 0
0
0
0
0
⎤
⎥
⎥
⎦
⎡
+
tP
0L
⎢
⎢
⎣
1
0
−1
0
0 −1
0
1
1
0
0
−1
⎤
⎥
⎥
⎦
0
−1
0
1
58
⎤
⎥
⎥
⎦
(14.6)
(14.7)
(14.8)
(14.9)
MIT 2.094
14. Total Lagrangian formulation, cont’d
0
Q L =
tP
·
Δ
⇒
Q =
tP
0L
·
Δ
(14.10)
where the boxed term is the stiffness. In axial direction, 0L is not very important because usually
E 0A
0L � 0L . But, in vertical direction, 0L is important.
tP
tP
tP
tF
0 =
⎡
− cos θ
⎢ − sin θ
tP ⎢
⎣
⎤
⎥
⎥
cos θ ⎦
sin θ
2D/3D (e.g. Table 6.5) 2D:
0�11 =
0u1,1 + 0
�
t u1,1 0u
1,1 + 0
��
0e11
t u2,1 0u2,1
+
�
1 ��
2
�
�2
�
0u1,1 + | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
21)
How do we assess the accuracy of an analysis?
Reading:
Sec. 4.3.6
• Mathematical model ∼ u
• F.E. solution ∼ uh
Find �u − uh� and �τ − τh�.
References
[1] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - on Mesh Selection.” Computers
& Structures, 21:257–264, 1985.
[2] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - Stress Band Plots and the
Evaluation of Finite Element Meshes.” Journal of Engineering Computations, 3:178–191, 1986.
60
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 15 - Field problems
Prof. K.J. Bathe
MIT OpenCourseWare
Heat transfer, incompressible/inviscid/irrotational flow, seepage flow, etc.
Reading:
Sec. 7.2-7.3
•
Differential formulation
•
Variational formulation
•
Incremental formulation
•
F.E. discretization
15.1 Heat transfer
Assume V constant for now:
S = Sθ ∪ Sq
θ(x, y, z, t) is unkown except θ|Sθ = θpr. In addition, q |Sq
s
is also prescribed.
15.1.1 Differential formulation
I. Heat flow equilibrium in V and on Sq.
II. Constitutive laws qx = −k ∂θ .∂x
qy = −k
qz = −k
∂θ
∂y
∂θ
∂z
(15.1)
(15.2)
III. Compatibility: temperatures need to be continuous and satisfy the boundary conditions.
61
MIT 2.094
15. Field problems
Heat flow equilibrium | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
temperatures need to be continuous and satisfy the boundary conditions.
61
MIT 2.094
15. Field problems
Heat flow equilibrium gives
�
�
�
k
∂θ
∂x
∂
∂x
+
∂
∂y
�
k
∂θ
∂y
+
∂
∂z
�
�
k
∂θ
∂z
B
= −q
(15.3)
where qB is the heat generated per unit volume. Recall 1D case:
unit cross-section
dV = dx (1)
·
�
q|x − q|x +
q x − q x+dx + q B dx = 0
|
�
dx + q B dx = 0
|
∂qx
∂x
�
−
∂
∂x
−k
∂θ
∂x
�
��dx + q dx = 0
B ��
�
�
k
∂θ
∂x
∂
∂x
B
= −q
We also need to satisfy
k
∂θ
∂n
= q S
on Sq.
15.1.2 Principle of virtual temperatures
�
� �
∂
k
θ
∂x
∂θ
∂x
( θ� = 0 and θ to be continuous.)
�
+ q B = 0
· · ·
+
�
Sθ
�
V
�
θ
�
k
∂
∂x
∂θ
∂x
�
+
�
+ q B dV = 0
· · ·
62
(15.4)
(15.5)
(15.6)
(15.7)
(15.8)
(15.9)
(15.10)
(15.11)
MIT 2.094
15. Field problems
Transform using divergence theorem (see Ex 4.2, 7.1)
�
�
�
dV =
θqB dV +
Sq
θ
q S dSq
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
1)
�
�
�
dV =
θqB dV +
Sq
θ
q S dSq
�T
θ
V
kθ�
����
heat flow
V
Sq
⎛
⎞∂θ
∂x
⎟
⎜
⎟
⎜
∂θ ⎟
θ� = ⎜
⎜ ∂y ⎟
⎠
⎝
∂θ
∂z
⎡
⎤
k 0
k = ⎣ 0 k 0 ⎦
0
0
0 k
Convection boundary condition
�
�
q S = h θe − θS
where θe is the given environmental temperature.
Radiation
q S = κ∗ �
(θr)4 − �
= κ∗ �
(θr)2 + �
�
�
= κ θr − θS
θS �4 �
θS �2 � �
θr + θS � �
θr − θS �
where κ = κ(θS ) and θr is given temperature of source. At time t + Δt
�
�T
θ
t+Δtk t+Δtθ�dV =
θ t+Δt q B dV +
S t+Δt q S dSq
θ
�
�
V
V
Sq
Let
t+Δtθ = tθ + θ
or
with
t+Δtθ(i) = t+Δtθ(i−1) + Δθ(i)
t+Δtθ(0) = tθ
From (15.19)
�
�T
θ
�
V
=
k
t+Δt (i−1)Δθ�(i)
�
θ t+Δt q B dV −
�
V
S t+Δth(i−1)
θ
dV
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
+Δt q B dV −
�
V
S t+Δth(i−1)
θ
dV
�T
θ
�
V
+
t+Δtk(i−1) t+Δtθ�(i−1)dV
t+Δt e
θ
−
�
t+Δt
S(i−1)
θ
+
ΔθS(i)
��
dSq
Sq
where the ΔθS(i)
term would be moved to the left-hand side.
We considered the convection conditions
S t+Δth � t+Δtθe − t+ΔtθS �
θ
�
dSq
Sq
The radiation conditions would be included similarly.
63
(15.12)
(15.13)
(15.14)
(15.15)
(15.16)
(15.17)
(15.18)
(15.19)
(15.20)
(15.21)
(15.22)
(15.23)
(15.24)
MIT 2.094
F.E. discretization
15. Field problems
for 4-node 2D planar element
t+Δtθ�
t+Δt
t+Δt
θˆ
θ = H1x4 ·
4x1
t+Δtθˆ
2x1 = B2x4 ·
t+ΔtθS = H S t+Δtθˆ
·
4x1
For (15.23)
�
V
�
V
�
V
�
�T
θ
t+Δt (i−1)Δθ�(i)
k
�
⎛
dV = ⎝ BT
t+Δt (i−1)
k
����
��
�
4x2
2x2
gives
⇒
V
⎞
B dV ⎠ Δ θˆ(i)
����
4x1
� ����
2x4
θ t+Δt q B dV
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�ˆ(i)
����
4x1
� ����
2x4
θ t+Δt q B dV
⇒
�
V
H T t+Δt q B dV
�T
θ
t+Δtk(i−1) t+Δtθ�(i−1)dV
⇒
��
V
BT t+Δtk(i−1)BdV
�
t+Δtθˆ(i−1)
��
�
�
known
S T
θ
t+Δth(i−1)
�
t+Δtθe −
��
+ ΔθS(i)
�
t+ΔtθS(i−1)
⎛
⎛
Sq
�
H S T t+Δth(i−1) H S
����
� �� �
Sq 4x1
1x4
⎝
−
⎝
t+Δtθˆe
� �� �
4x1
t+Δtθˆ(i−1) +Δ θˆ(i)
����
�
�
4x1
��
4x1
dSq = ⇒
⎞⎞
⎠⎠ dSq
15.2
Inviscid, incompressible, irrotational flow
2D case: vx, vy are velocities in x and y directions.
or
� · v = 0
∂vy = 0
∂y
∂vy
∂x
= 0
∂vx +
∂x
∂vx
∂y
−
(incompressible)
(irrotational)
Use the potential φ(x, y),
vx =
∂φ
∂x
vy =
∂φ
∂y
⇒
∂2φ
∂x2
+
∂2φ
∂y2
= 0
in V
(Same as the heat transfer equation with k = 1, qB = 0)
64
(15.25)
(15 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(Same as the heat transfer equation with k = 1, qB = 0)
64
(15.25)
(15.26)
(15.27)
(15.28)
(15.29)
(15.30)
(15.31)
(15.32)
(15.33)
(15.34)
(15.35)
(15.36)
Reading:
Sec. 7.3.2
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 16 - F.E. analysis of Navier-Stokes fluids
Prof. K.J. Bathe
MIT OpenCourseWare
Reading:
Sec.
7.1-7.4,
Table 7.3
(16.1)
(16.2)
(16.3)
(16.4)
Incompressible flow with heat transfer
We recall heat transfer for a solid:
Governing differential equations
(kθ,i),i + q B = 0
in V
�
�
θ
�
Sθ
is prescribed, k
�
∂θ
�
�
∂n
Sq
= q S
�
�
�
Sq
Sθ ∪ Sq = S
Sθ ∩ Sq = ∅
Principle of virtual temperatures
�
�
�
θ,ikθ,idV =
θqB dV +
S
q S dSq
θ
V
V
Sq
for arbitrary continuous θ(x1, x2, x3) zero on Sθ
For a fluid, we use the Eulerian formulation.
65
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
�
ρcpv θ|x −
ρcpv θ|x +
∂
∂x
�
(ρcpvθ)dx
+ conduction + etc
In general 3D, we have an additional term for the left hand side of (16.1):
−� · (ρcp | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
etc
In general 3D, we have an additional term for the left hand side of (16.1):
−� · (ρcpvθ) = −ρcp� · (vθ) = −ρcp(���
� · v )θ − ρcp (v · �) θ
�
�
��
term (A)
where � · v = 0 in the incompressible case.
� · v = vi,i = div(v) = 0
So (16.1) becomes
(kθ,i),i + q B = ρcpθ,ivi ⇒ (kθ,i),i + �
�
q B − ρcpθ,ivi
= 0
Principle of virtual temperatures is now (use (16.4))
�
�
�
�
θ,ikθ,idV +
θ (ρcpθ,ivi) dV =
θqB dV +
S
q S dSq
θ
V
V
V
Sq
Navier-Stokes equations
• Differential form
τij,j + f B
i = ρvi,j vj
with ρvi,j vj like term (A) in (16.6) = ρ(v · �)v in V .
τij = −pδij + 2µeij
eij =
�
∂vi
∂xj
1
2
�
∂vj
∂xi
+
• Boundary conditions (need be modified for various flow conditions)
(16.5)
(16.6)
(16.7)
(16.8)
(16.9)
(16.10)
(16.11)
(16.12)
τij nj = f Sf
i on Sf
Mostly used as fn = τnn
inflow conditions).
= prescribed, ft = unknown with possibly ∂vn
∂n = ∂vt
∂n = 0 (outflow or
And vi prescribed on Sv, and Sv | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
∂vt
∂n = 0 (outflow or
And vi prescribed on Sv, and Sv ∪ Sf = S and Sv ∩ Sf = ∅.
66
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
•
Variational form
�
viρvi,j vj dV +
V
eij τij dV =
�
V
p� · vdV = 0
�
V
�
V
vifi
B dV +
�
Sf
Sf fi
vi
Sf dSf
(16.13)
(16.14)
•
F.E. solution
We interpolate (x1, x2, x3), vi, vi, θ, θ, p, p. Good elements are
×: linear pressure
: biquadratic velocities
◦
(Q2, P1), 9/3 element
9/4c element
Both satisfy the inf-sup condition.
So in general,
Example:
For Sf e.g.
τnn = 0,
∂vt = 0;
∂n
67
(16.15)
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
and ∂vn is solved for. Actually, we frequently just set p = 0.
∂t
Frequently used is the 4-node element with constant pressure
It does not strictly satisfy the inf-sup condition. Or use
3-node element with a bubble node.
Satisfies inf-sup condition
1D case of heat transfer with fluid flow, v = constant
Re =
vL
ν
Pe =
vL
α
α =
k
ρcp
•
Differential equations
kθ�� = ρcpθ�v
θ|x=0 = θL
θ|x=L = θR
In non-dimensional form
1
Pe
θ�� = θ�
(now θ�� and θ� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
= θR
In non-dimensional form
1
Pe
θ�� = θ�
(now θ�� and θ� are non-dimensional)
⇒
θ − θL
θR − θL
=
�
�
exp Pe x
L − 1
exp (Pe) − 1
68
Reading:
Sec. 7.4
Reading:
Sec. 7.4.3
(16.16)
Reading:
p. 683
(16.17)
(16.18)
(16.19)
(16.20)
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
• F.E. discretization
θ�� = Peθ�
� 1
�
θ
0
� 1
θ�dx + Pe
0
θθ�dx = 0 + { effect of boundary conditions = 0 here}
(16.22)
(16.21)
Using 2-node elements gives
1
(h∗)2 (θi+1 − 2θi + θi−1) =
Pe
2h∗
(θi+1 − θi−1)
Pe =
vL
α
Define
Pee = Pe
·
h
L
=
vh
α
�
�
�
�
−1 −
Pee
2
what is happening when Pee is large? Assume two 2-node elements only.
− 1 θi+1 = 0
θi−1 + 2θi +
Pee
2
θi−1 = 0
θi+1 = 1
1
2
θi =
�
1 −
�
Pee
2
69
(16.23)
(16.24)
(16.25)
(16.26)
(16.27)
(16.28)
(16.29)
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
�
1 −
�
Pee
2 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
2.094
16. F.E. analysis of Navier-Stokes fluids
�
1 −
�
Pee
2
1
2
θi =
For Pee > 2, we have negative θi (unreasonable).
(16.30)
70
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 17 - Incompressible fluid flow and heat transfer, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
17.1 Abstract body
Reading:
Sec. 7.4
Fluid Flow
Sv, Sf
Sv ∪ Sf
= S
Sv ∩ Sf = 0
Heat transfer
Sθ, Sq
Sθ ∪ Sq
= S
Sθ ∩ Sq = 0
17.2 Actual 2D problem (channel flow)
71
MIT 2.094
17. Incompressible fluid flow and heat transfer, cont’d
17.3 Basic equations
P.V. velocities
�
V
�
viρvi,j vj dV +
V
τij eij dV =
�
V
Continuity
�
pvi,idV = 0
V
P.V. temperature
vifi
B dV +
�
Sf
Sf fi
vi
Sf dSf
�
V
�
θρcpθ,ividV +
V
θ,ikθ,idV =
�
V
�
θqB dV +
S
q S dS
θ
Sq
F.E. solution
k
hkvi
hkx k
i
�
�
�
hkθk
�
h˜
kpk
xi =
vi =
θ =
p =
⎛ ⎞
v
⇒
F (u) = R
u = ⎝ p ⎠ nodal variables
θ
17.4 Model problem
1D equation,
dθ | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
= ⎝ p ⎠ nodal variables
θ
17.4 Model problem
1D equation,
dθ
ρcpv = k
dx
d2θ
dx2
(v is given, unit cross section)
Non-dimensional form (Section 7.4)
dθ
Pe =
dx
d2θ
dx2
72
(17.1)
(17.2)
(17.3)
(17.4)
(17.5)
(17.6)
(17.7)
(17.8)
(17.9)
(17.10)
MIT 2.094
17. Incompressible fluid flow and heat transfer, cont’d
θ∗ is non-dimensional
�
�
exp Pe
x − 1
L
exp (Pe) − 1
θ − θL
θR − θL
=
Pe =
vL
,
α
α =
k
ρcp
(17.10) in F.E. analysis becomes
�
V
θPe
dV +
dθ
dx
�
V
dθ dθ
dx dx
dV = 0
Discretized by linear elements:
h∗ =
h
L
ξ
h∗
θi
�
θ(ξ) = 1 −
�
ξ
h∗
θi−1 +
For node i:
−θi−1 −
Pee
2
θi−1 + 2θi − θi+1 +
Pee
2
θi+1 = 0
where
Pee =
vh
α
�
�
h
= Pe
L
This result is the same as obtained by finite differences
�
� =
�
�
i
�
θ� =
�
�
i
(θi+1 − 2θi + θi−1)
1
(h∗)2
θi+1 − θi−1
2h∗
73
(17.11)
(17.12)
( | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
2
θi+1 − θi−1
2h∗
73
(17.11)
(17.12)
(17.13)
(17.14)
(17.15)
(17.16)
(17.17)
(17.18)
MIT 2.094
17. Incompressible fluid flow and heat transfer, cont’d
Considered θi+1 = 1, θi−1 = 0. Then
θi =
1 − (Pee/2)
2
(17.19)
Physically unrealistic solution when Pee > 2. For this not to happen, we should refine the mesh—a very
fine mesh would be required. We use “upwinding”
�
dθ
�
�
dx i
=
θi − θi−1
h∗
The result is
(−1 − Pee) θi−1 + (2 + Pee) θi − θi+1 = 0
Very stable, e.g.
�
θi−1 = 0
θi+1 = 1
⇒ θi
=
1
2 + Pee
(17.20)
(17.21)
(17.22)
Unfortunately it is not that accurate. To obtain better accuracy in the interpolation for θ, use the
function
�
�
exp Pe x
L − 1
exp (Pe) − 1
The result is Pee dependent:
(17.23)
This implies flow-condition based interpolation. We use such interpolation functions—see references.
References
[1] K.J. Bathe and H. Zhang. “A Flow-Condition-Based Interpolation Finite Element Procedure for
Incompressible Fluid Flows.” Computers & Structures, 80:1267–1277, 2002.
[2] H. Kohno and K.J. Bathe. “A Flow-Condition-Based Interpolation Finite Element Procedure for | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Kohno and K.J. Bathe. “A Flow-Condition-Based Interpolation Finite Element Procedure for
Triangular Grids.” International Journal for Numerical Methods in Fluids, 51:673–699, 2006.
74
MIT 2.094
17. Incompressible fluid flow and heat transfer, cont’d
17.5 FSI briefly
Lagrangian formulation for the structure/solid
Arbitrary Lagrangian-Eulerian (ALE) formulation Let f be a variable of a particle (e.g. f = θ).
Consider 1D
�
f˙
�
�
particle
=
∂f
∂t
+
∂f
∂x
v
where v is the particle velocity. For a mesh point,
�
f ∗
�
�
mesh point
=
∂f
∂t
+
∂f
∂x
vm
where vm
is the mesh point velocity. Hence,
�
f˙
�
�
particle
�
= f ∗
�
�
mesh point
+
∂f
∂x
(v − vm)
(17.24)
(17.25)
(17.26)
Use (17.26) in the momentum and energy equations and use force equilibrium and compatibility at
the FSI boundary to set up the governing F.E. equations.
References
[1] K.J. Bathe, H. Zhang and M.H. Wang. “Finite Element Analysis of Incompressible and Compressible
Fluid Flows with Free Surfaces and Structural Interactions.” Computers & Structures, 56:193–213,
1995.
[2] K.J. Bathe, H. Zhang and S. Ji. “Finite Element Analysis of Fluid Flows Fully Coupled with
Structural Interactions.” Computers & Structures, 72:1–16, 1999.
[3] K.J. Bathe and H. Zhang. “ | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
& Structures, 72:1–16, 1999.
[3] K.J. Bathe and H. Zhang. “Finite Element Developments for General Fluid Flows with Structural
Interactions.” International Journal for Numerical Methods in Engineering, 60:213–232, 2004.
75
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 18 - Solution of F.E. equations
Prof. K.J. Bathe
In structures,
F (u, p) = R.
In heat transfer,
F (θ) = Q
In fluid flow,
F (v, p, θ) = R
In structures/solids
�
F (m) � �
F =
=
m
m
0V (m)
Elastic materials
Example p. 590 textbook
MIT OpenCourseWare
Reading:
Sec. 8.4
(18.1)
(18.2)
(18.3)
(18.4)
t
0BL
(m)T t ˆ(m) 0
d V
0S
(m)
76
18. Solution of F.E. equations
MIT 2.094
Material law
t
tS
0 11 = E �0
˜
11
In isotropic elasticity:
E˜ =
E (1 − ν)
,
(1 + ν) (1 − 2ν)
(ν = 0.3)
t� =
0
1 ��
tU − I ⇒ 0
0
2
�2
�
t�11 =
�� 0L + tu �2
0L
1
2
�
− 1 =
1
2
��
tu �2
1 + 0L
�
− 1
where 0
tU is the stretch tensor.
tS =
0 11
0ρ
tρ
0X T
tτ
0X
t 11 11 t 11
with
0
tX11 =
0L
,
0 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
0X
t 11 11 t 11
with
0
tX11 =
0L
,
0L + tu
0ρ 0L = tρ tL
⇒ 0
tL
tS11 = 0L
� �2
0L
tL
0L
tτ11 = tL
tτ11
∴
0
L tτ11 = E˜ ·
tL
1
2
��
t
u
1 + 0
L
�2
�
− 1
⇒ tτ11A =
tP =
��
1 +
E˜A
2
tu �2
0L
� �
− 1
1 +
tu �
0L
This is because of the material-law assumption (18.5) (okay for small strains . . . )
Hyperelasticity
tW = f (Green-Lagrange strains, material constants)
0
�
�
tSij =
0
0Cijrs =
t
0 ij
t
1 ∂ W ∂ W
0
2 ∂ �
�
t
1 ∂ S
0
t
2 ∂ �
0 rs
t + 0
t
∂ �
0 ji
t �
∂ S
ij
ij + 0
t
∂ �
0 sr
77
(18.5)
(18.6)
(18.7)
(18.8)
(18.9)
(18.10)
(18.11)
(18.12)
(18.13)
(18.14)
(18.15)
MIT 2.094
Plasticity
18. Solution of F.E. equations
• yield criterion
•
flow rule
• hardening rule
�
tτ = t−Δtτ +
t
dτ
t−Δt
Solution of (18.1) (similarly (18.2) and (18.3))
Newton-Raphson Find U � | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Solution of (18.1) (similarly (18.2) and (18.3))
Newton-Raphson Find U ∗ as the zero of f (U ∗)
f (U ∗) =
−
R
t+Δt
�
t+Δt
t+Δt
F
�
(i−1) +
U
= f
∂f
∂U
�
�
�
�
t+Δt
(i−1)
U
�
U ∗ − t+ΔtU (i−1) + H.O.T.
·
�
where t+ΔtU (i−1) is the value we just calculated and an approximation to U ∗.
Assume t+ΔtR is independent of the displacements.
�
t+ΔtR
0 =
−
t+ΔtF (i−1)
�
−
∂ t+ΔtF
∂U
�
�
�
�
t+Δt
(i−1)
U
· ΔU (i)
We obtain
t+ΔtK (i−1)ΔU (i) = t+ΔtR − t+ΔtF (i−1)
t+ΔtK (i−1) =
∂ t+ΔtF
∂U
Physically
�
�
�
�
t+Δt
(i−1)
U
=
�
∂F
∂U
��
�
�
�
t+Δt
(i−1)
U
t+Δt
(i−1)K11 =
�
(i−1)
�
Δ t+ΔtF1
Δu
78
(18.16)
(18.17)
(18.18)
(18.19)
(18.20)
(18.21)
(18.22)
MIT 2.094
18. Solution of F.E. equations
Pictorially for a single degree of freedom system
i = 1; | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Solution of F.E. equations
Pictorially for a single degree of freedom system
i = 1;
i = 2;
tK Δu(1) = t+ΔtR − tF
t+ΔtK (1)Δu(2) = t+ΔtR
−
t+ΔtF (1)
Convergence Use
�ΔU (i)�2 < �
�a�2 =
��
(ai)2
i
But, if incremental displacements are small in every iteration, need to also use
�t+ΔtR − t+ΔtF (i−1)�2 < �R
18.1 Slender structures
(beams, plates, shells)
t
Li
� 1
79
(18.23)
(18.24)
(18.25)
(18.26)
(18.27)
(18.28)
18. Solution of F.E. equations
MIT 2.094
Beam
t
1
e.g. L = 100
(4-node el.)
The element does not have curvature
a spurious shear strain
→
we have
(9-node el.)
→
→
like ill-conditioning.
We do not have a shear (better)
But, still for thin structures, it has problems
We need to use beam elements. For curved structures also spurious membrane strain can be
⇒
present.
80
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 19 - Slender structures
Prof. K.J. Bathe
MIT OpenCourseWare
Beam analysis, t
L � 1 (e.g.
t
L =
1
1
100 , 1000 , · · · )
Reading:
Sec. 5.4,
6.5
(plane stress)
�
L
2
0
�
0
t
2
(1 − r) (1 + s)
(1 − | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
)
�
L
2
0
�
0
t
2
(1 − r) (1 + s)
(1 − r) (1 − s)
J =
h2 =
h3 =
1
4
1
4
Beam theory assumptions (Timoshenko beam theory):
∗
∗=v
v3
2
= v2
t
= u2 + θ22
t
θ2
= u2 −
2
u∗
3
u∗
2
B∗ =
⎡
⎢
⎢
⎢
⎢
⎣
u
∗
2
1
− 4
(1 + s)
2
L
∗
v
2
0
u
∗
3
− 1
4 (1 − s) 2
L
∗
v3
0
0
1
4
(1
2
)− r t
0
−
1
4
(1
2
)− r t
1
4
(1
2
)− r t
1
− 4
(1 + s)
2
L − 1
4
(1
2
)− r t
− 1
4
(1 − s)
2
L
81
(19.1)
(19.2)
(19.3)
(19.4)
(19.5)
(19.6)
etc
(19.7)
⎤
⎥
⎥
⎥
⎥
⎦
MIT 2.094
19. Slender structures
⎡
Bbeam = ⎢
⎢
⎢
⎢
⎣
u2
v2
−
1
L
· · ·
�
0
0
�
0
θ2
t
2L
s
�
0
0 − 1 − 1 (1 − r)
L
2
⎤
⎥
⎥
⎥
⎥
⎦
∼
⎛
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
⎤
⎥
⎥
⎥
⎥
⎦
∼
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
∂u
∂x
0
∂v��
�∂y
∂u + ∂v
∂y
∂x
(1 − r)v2
v(r) =
u(r) =
1
2
1
2
(1 − r)u2 −
st
4
(1 − r)θ2
at r = −1,
v(−1) = v2
u(−1) = −
θ2 + u2
st
2
Kinematics is
u(r) =
1
2
(1 − r)u2
results into �xx
→ �xx =
∂u 2
·
L
∂r
= −
1
L
u(r, s) = −
(1 − r)θ2
st
4
results into �xx, γxy
→
�xx =
st
2L
γxy =
∂u 2
·
t
∂s
1
= −
2
(1 − r)
v(r) =
1
2
(1 − r)v2
results into γxy
→ γxy = −
1
L
82
(19.8)
(19.9)
(19.10)
(19.11)
(19.12)
(19.13)
(19.14)
(19.15)
(19.16)
(19.17)
(19.18)
(19.19)
MIT 2.094
19. Slender structures
For a pure bending moment, we want
1
−
L
v2 −
1
2
(1 − r)θ2 = 0
(19.20)
for all r!
⇒
Impossible (except for v2 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
r)θ2 = 0
(19.20)
for all r!
⇒
Impossible (except for v2 = θ2 = 0)
⇒
So, the element has a spurious shear strain!
Beam kinematics (Timoshenko, Reissner-Mindlin)
γ =
dw
dx
− β
�
I =
�
1
12
bt3
Principle of virtual work
� L
� L
EI
0
dβ
dx
dβ
dx
dx + AS G
�
dw
dx
� �
�
− β
dx =
� L
0
pwdx
dw
dx
− β
0
As = kA = kbt
To calculate k
�
A
1
2G
(τa)2dA =
�
AS
1
2G
�
�2
V
As
where τa is the actual shear stress:
�
�
τa =
3
2
·
V
A
� t
�2 − y2
2
� t �2
2
and V is the shear force.
⇒
k =
5
6
dAs
83
(19.21)
(19.22)
(19.23)
(19.24)
(19.25)
(19.26)
Reading:
p. 400
Reading:
Ex. 5.23
(19.27)
MIT 2.094
Now interpolate
w(r) = h1w1 + h2w2
β(r) = h1θ1 + h2θ2
Revisit the simple case:
w =
β =
1 + r
2
1 + r
2
w1
θ1
Shearing strain
γ =
w1
L
−
1 + r
2
θ1
19. Slender structures
(19.28)
(19.29)
(19.30)
(19.31)
(19.32)
Shear strain is not zero all along the beam. But, at r = 0, we can have the shear strain | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(19.32)
Shear strain is not zero all along the beam. But, at r = 0, we can have the shear strain = 0.
w1
L
−
θ1
2
can be zero
Namely,
w1
L
−
θ1
2
= 0
for θ1 =
2
L
w1
(19.33)
(19.34)
84
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 20 - Beams, plates, and shells
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory
The fiber moves up and rotates and its length does not change.
Principle of virtual displacement (Linear Analysis)
�T �
− β
dw
dx
� L
EI
� �T
�
β
0
� L �
β�dx + (Ak)G
0
dw
dx
Two-node element:
Three-node element:
For a q-node element,
· · · wq θ1
· · · θq
�T
ˆu = �
w1
w = Hw ˆu
β = Hβ ˆu
Hw = �
h1
Hβ = �
0
dx
dr
J =
· · ·
· · ·
hq 0
0 h1
· · ·
0
· · ·
hq
�
�
85
�
− β dx =
� L
0
w T pdx
(20.1)
(20.2)
(20.3)
(20.4)
(20.5)
(20.6)
(20.7)
MIT 2.094
20. Beams, plates, and shells
dw
dx
dβ
dx
= J −1Hw,r uˆ
� �� �
Bw
= J −1Hβ,r ˆ
u
�
� ��
Bβ
Hence we obtain
� 1
�
EI | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
β,r ˆ
u
�
� ��
Bβ
Hence we obtain
� 1
�
EI
Bβ
T Bβ det(J )dr + (Ak)G
−1
� 1
�
(Bw − Hβ )T (Bw − Hβ ) det(J )dr uˆ
−1
� 1
= Hw
−1
T p det(J )dr
Kuˆ = R
(20.8)
(20.9)
(20.10)
(20.11)
K is a result of the term inside the bracket in (20.10) and R is a result of the right hand side.
For the 2-node element,
w1 = θ1 = 0
w2, θ2 = ?
γ =
w2
L
−
1 + r
2
θ2
(20.12)
(20.13)
(20.14)
We cannot make γ equal to zero for every r (page 404, textbook). Because of this, we need to use
about 200 elements to get an error of 10%. (Not good!)
Recall almost or fully incompressible analysis: Principle of virtual displacements:
�
�
��T C���dV +
�v (κ�v )dV = R
V
V
u/p formulation
�
V
�
�vpdV = R
��T C���dV −
�
�
�
p
p + �v dV = 0
κ
V
V
(20.15)
(20.16)
(20.17)
But now we needed to select wisely the interpolations of u and p. We needed to satisfy the inf-sup
condition
sup
inf
����
����
qh∈Qh vh∈Vh
�
Vol qh� · vhdVol
�qh��vh�
≥ β > 0
86
(20.18) | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
Vol qh� · vhdVol
�qh��vh�
≥ β > 0
86
(20.18)
MIT 2.094
4/1 element:
20. Beams, plates, and shells
We can show mathematically that this element does not satisfy inf-sup condition. But, we can also
show it by giving an example of this element which violates the inf-sup condition.
v1 = �,
above
�
Vol
v2 = 0 ⇒ � · vh for both elements is positive and the same. Now, if I choose pressures as
qh�vhdVol = 0,
hence (20.18) is not satisfied!
(20.19)
9/3 element
9/4-c
satisfies inf-sup
satisfies inf-sup
Getting back to beams
� L
EI
�
β
0
� L �
βdx + (AkG)
0
�
− β γAS dx = R
w
d
dx
�
γAS γ − γAS dx = 0
�
� L
0
where
γ =
dw
dx
− β,
from displacement interpolation
87
(20.20)
(20.21)
(20.22)
MIT 2.094
20. Beams, plates, and shells
Reading:
Sec. 4.5.7
(20.23)
(20.24)
(20.25)
(20.26)
(20.27)
(20.28)
(20.29)
(20.30)
γAS = Assumed shear strain interpolation
2-node element, constant shear assumption. From (20.21),
� L �
0
dw
dx
�
− β �
γAS�
dx =
� L
0
γAS �
γAS�
dx
� +1 �
1 + r �
·
θ2
2
⇒ −
−1
L
2
dr + w2 = γAS · L
� | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
·
θ2
2
⇒ −
−1
L
2
dr + w2 = γAS · L
⇒
γAS =
L
w2 − 2 θ2
L
γAS (shear strain) is equal to the displacement-based shear strain at the middle of the beam.
Use γAS in (20.20) to obtain a powerful element. For “our problem”,
γAS = 0 hence w2 =
L
2
θ2
� L
EI
⇒
0
�
�
β�dx = M β�
x=L
β
�� �2 �
1
L
⇒
EI
·
L θ2 = M
⇒
θ2 =
M L
,
EI
w2 =
M L2
2EI
(exact solutions)
88
MIT 2.094
Plates
20. Beams, plates, and shells
Reading:
Fig. 5.25,
p. 421
⎧
⎪⎨
⎪⎩
w = w(x, y)
v = −zβy(x, y)
u = −zβx(x, y)
is the transverse displacement of the mid-surface
For any particle in the plate with coordinates (x, y, z), the expressions in (20.31) hold!
We use
q
�
hiwi
w =
i=1
q
�
βx = −
i
hiθy
i=1
q
�
βy = +
i
hiθx
(20.31)
(20.32)
(20.33)
(20.34)
i=1
where q equals the number of nodes. Then the element locks in the same way as the displacement-based
beam element.
89
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 21 - Plates and shells
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory, and Reissner-Mindlin | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory, and Reissner-Mindlin plate theory
For plates, and shells, w, βx, and βy as independent variables.
w = displacement of mid-surface, w(x, y)
A = area of mid-surface
p = load per unit area on mid-surface
w = w(x, y)
w(x, y, z) = w(x, y)
The material particles at “any z” move in the z-direction as the mid-surface.
u(x, y, z) = −βxz = −βx(x, y)z
v(x, y, z) = −βyz = −βy(x, y)z
�
∂βx +
∂y
�
∂βy
∂x
∂βx
∂x
∂βy
∂y
�xx =
�yy =
γxy =
∂u
∂x
∂v
∂y
∂u
∂y
= −z
= −z
+
∂v
∂x
= −z
⎛
⎛
⎝
�xx
�yy
γxy
⎞
⎠ = −z
⎜
⎜
⎜
⎜
⎜
⎝
�
∂βx
∂x
∂βy
∂y
∂βx + ∂βy
∂x
∂y
��
κ
⎞
⎟
⎟
⎟
⎟
⎟
⎠
�
90
(21.1)
(21.2)
(21.3)
(21.4)
(21.5)
(21.6)
(21.7)
(21.8)
MIT 2.094
21. Plates and shells
γxz =
γyz =
∂w
∂x
∂w
∂y
+
+
∂u
∂z
∂v
∂z | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
∂x
∂w
∂y
+
+
∂u
∂z
∂v
∂z
=
=
∂w
∂x
∂w
∂y
− βx
− βy
⎞
⎛
τxx
⎝ τyy ⎠ =
τxy
E
1 − ν2
⎡
1 ν
⎣ ν 1
0
0
(plane stress)
⎞
⎤ ⎛
0
�xx
0 ⎦ ⎝ �yy ⎠ = C �
ν
1−
γxy
2
·
�
�
τxz =
τyz
E
2(1 + ν)
�
�
γxz = Gγ
γyz
Principle of virtual work
for the plate:
� �
t
2+
�
A − t
2
�xx
�yy
γ
xy
⎡
⎣
� E
1 − ν2
� �
k
t
2+
�
A
− t
2
γ
xz
γ
Consider a flat element:
1 ν
ν 1
0
0
�
�
G
yz
⎞
⎛
w1
⎜ θ1 ⎟
⎜ x ⎟
y ⎠
⎝
.
. .
⇒
Kb ⎜ θ1 ⎟ , also Kpl. str. ⎝
⎞
⎛
u1
⎜ v1 ⎟
⎠
.
..
where Kb is 12x12 and Kpl. str. is 8x8.
⎞
⎤ ⎛
0
�xx
0 ⎦ ⎝ �yy ⎠dzdA+
1−ν
γxy
2
� �
γxz
γyz
�
�
dzdA = wpdA | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
1−ν
γxy
2
� �
γxz
γyz
�
�
dzdA = wpdA
A
1
0
0
1
91
(21.9)
(21.10)
(21.11)
(21.12)
(21.13)
(21.14)
21. Plates and shells
=
· · ·
(21.15)
MIT 2.094
For a flat element:
�
⇒
0
Kb
0 Kpl. str.
⎛
�
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
w1
θ1
x
θ1
y
. . .
θ4
y
u1
v1
. . .
v4
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
�
�
�
u =
v =
w =
hiui
hivi
hiwi
�
i
hiθy
βx = −
�
βy =
i
hiθx
From (21.13)
�
A
κT
Et3
12 (1 − ν2)
⎡
1 ν
⎣ ν 1
0
0
�
⎤
0
0 ⎦ κdA +
1−ν
2
A
(21.16)
(21.17)
(21.18)
(21.19)
(21.20)
(21.21)
�
γdA
1
0
0
1
�
γT Gt
· k
where k is the shear correction factor.
Next, evaluate ∂w , ∂w , ∂βx , . . | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
where k is the shear correction factor.
Next, evaluate ∂w , ∂w , ∂βx , . . . etc.
∂x ∂y
∂x
⇒
This element, as it is, locks!
This displacement-based element “locks in shear”. We need to change the transverse shear interpola
tions.
γyz = (1 − r)γA + (1 + r)γC
yz
yz
where
γA
yz
=
�
∂w
∂y
− βy
��
�
�
�
evaluated at A
from the w, βy displacement interpolations.
γxz
= (1 − s)γB + (1 + s)γD
xz
xz
1
2
1
2
1
2
1
2
(21.22)
(21.23)
(21.24)
with this mixed interpolation, the element works. Called MITC interpolation (for mixed interpolated-
tensional components)
92
MIT 2.094
21. Plates and shells
Aside: Why not just neglect transverse shears, as in Kirchhoff plate theory?
∂w
• If we do, γxz = ∂x − βx = 0 ⇒ βx = ∂x
�
�
∂w
•
Therefore we have ∂2 w
�
�
∂w
∂x
· · ·
∂x2 , · · ·
in strains, so we need continuity also for
•
w =
1
2
r(1 + r)w1 −
r(1 − r)w2 + �
1
2
1 − r 2�
w3
w2 and w3 never affect w1 (∴ w|r=1 = w1).
But,
∂w
∂r
=
1
2
(1 + 2r)w1 −
1
2
(1 − 2r)w2 | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
r
=
1
2
(1 + 2r)w1 −
1
2
(1 − 2r)w2 − 2rw3
�
∂w �
w2 and w3 affect ∂r
1 ⇒
.
element based on Kirchhoff theory.
This results in difficulties to develop a good
With Reissner-Mindlin theory, we independently interpolate rotations such that this problem does
not arise.
For flat structures, we can superimpose the plate bending and plane stress element stiffness. For
shells, curved structures, we need to develop/use curved elements, see references.
References
[1] E. Dvorkin and K.J. Bathe. “A Continuum Mechanics Based Four-Node Shell Element for General
Nonlinear Analysis.” Engineering Computations, 1:77–88, 1984.
[2] K.J. Bathe and E. Dvorkin. “A Four-Node Plate Bending Element Based on Mindlin/Reissner Plate
Theory and a Mixed Interpolation.” International Journal for Numerical Methods in Engineering,
21:367–383, 1985.
[3] K.J. Bathe, A. Iosilevich and D. Chapelle. “An Evaluation of the MITC Shell Elements.” Computers
& Structures, 75:1–30, 2000.
[4] D. Chapelle and K.J. Bathe. The Finite Element Analysis of Shells – Fundamentals. Springer,
2003.
93
MIT OpenCourseWare
http://ocw.mit.edu
2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
2.092/2.093 — Finite Element Analysis of Solids & Fluids I
Fall ‘09
Lecture 6 - Finite Element Solution Process
Prof. K. J. Bathe
MIT OpenCourseWare
In the last lecture, we used the principle of virtual displacements to obtain the following equations:
KU = R
(1)
K = Σ K(m)
m
; K(m) =
�
V (m)
B(m)T C(m)B(m)dV (m)
R = RB + RS
RB = Σ RB
m
(m)
; RB
(m) =
�
H (m)T f B(m)dV (m)
RS = Σ R(m)
m S
; R(m) = Σ
S
i
H Si(m)T f Si(m)
f
f
dSi(m)
f
V (m)
�
S
i(m)
f
u(m) = H (m)U
↓
ε(m) = B(m)U
(2)
(3)
Note that the dimension of u(m) is in general not the same as the dimension of ε(m).
Example: Static Analysis
Reading assignment: Example 4.5
1
Lecture 6
Finite Element Solution Process
2.092/2.093, Fall ‘09
Assume:
i. Plane sections remain plane
ii. Static analysis
→
no vibrations/no transient response
iii. One-dimensional problem; hence, only one degree of freedom per node
Elements 1 and 2 are compatible because they use the same U2. Next, use a linear interpolation function.
u
(1)(x) =
� �
�
�
x
1 − 100
��
H (1)
x
100 0
ε(1)(x) =
1
− 100
1
100 0
�
�
��
B(1)
⎤
⎡
⎥ �
⎢
⎢
1 ⎥
⎣ 100 ⎦ − 100 | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
�� �
⎢
⎢
1 ⎥
⎣ 100 ⎦ − 100
0
1
− 100
1
� 100
K = E 1
·
·
0
⎡
�
⎣
�
⎡
�
⎣
�
U1
U2
U3
U1
U2
U3
⎤
⎦
⎤
⎦
; u(2)(x) = 0
�
�
�
�
x
1 − 80
��
H (2)
;
ε(2)(x) =
0
�
�
−
1
80
��
B(2)
⎤
⎦
⎡
� U1
x ⎣
U2
80
�
U3
⎤
⎡
� U1
1 ⎣
U2
80
�
U3
⎦
�
0 dx + E
� 80 �
1 +
0
1
100
⎤
⎡
0
⎥ �
x �2 ⎢
⎢
1 ⎥
40 ⎣ − 80 ⎦ 0 − 80
1
80
1
�
dx
1
80
=
E
100
⎡
⎢
⎢
⎣
1
−
1
−
1
0
1
0
⎤
⎥
⎥
⎦
+
0
0
0
13E
3 80·
⎡
⎢
⎢
⎣
0
0
0
⎤
⎥
⎥
⎦
0
0
1 −1
−1
1
The “equivalent cross-sectional area” of element 2 is A = 3 cm . This equivalent area must lie between the
areas of the end faces A = 1 and A = 9.
13
2 | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
3 cm . This equivalent area must lie between the
areas of the end faces A = 1 and A = 9.
13
2
2
Lecture 6
Finite Element Solution Process
2.092/2.093, Fall ‘09
⎡
K =
E
240
⎢
⎢
⎣
2.4 −2.4
0
−2.4 15.4 −13
0
−13
13
⎤
⎥
⎥
⎦
We note:
• Diagonal terms must be positive. If the diagonal terms are zero or negative, then the system is unstable
physically. A positive diagonal implies that the degree of freedom has stiffness at that node.
• K is symmetric.
• K is singular if rigid body motions are possible. To be able to solve the problem, all rigid body modes
must be removed by adequately constraining the structure. i.e. K is reduced by applying boundary
conditions to the nodes.
The K used to solve for U is, then, positive definite (det K > 0). This ensures that the elastic strain energy
is positive and nonzero for any displacement field U . In the analysis, each element is in equilibrium under
its nodal forces, and each node is in equilibrium when summing element forces and external loads.
Homework Problem 2
⎡
⎣
εxx
εzz
⎡⎤
⎦ = ⎣
⎤
∂u
∂x ⎦
u
x
3
Lecture 6
Finite Element Solution Process
2.092/2.093, Fall ‘09
εzz is frequently called the “hoop strain”, εθθ.
εzz =
2π(u + x) − 2πx
2πx
C =
⎡
⎣
E
1 − ν2
�
1 ν
ν 1
�
u
x
=
⎤ | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
�
E
1 − ν2
�
1 ν
ν 1
�
u
x
=
⎤
⎦
f B = ρω2R N/cm3
; R = x
4
MIT OpenCourseWare
http://ocw.mit.edu
2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
6.825 Techniques in Artificial Intelligence
Planning
• Planning vs problem solving
• Situation calculus
• Plan-space planning
Lecture 10 • 1
We are going to switch gears a little bit now. In the first section of the class, we
talked about problem solving, and search in general, then we did logical
representations. The motivation that I gave for doing the problem solving stuff
was that you might have an agent that is trying to figure out what to do in the
world, and problem solving would be a way to do that.
And then we motivated logic by trying to do problem solving in a little bit more
general way, but then we kind of really digressed off into the logic stuff, and so
now what I want to do is go back to a section on planning, which will be the
next four lectures.
Now that we know something about search and something about logic, we can talk
about how an agent really could figure out how to behave in the world. This will
all be in the deterministic case. We're still going to assume that when you take
an action in the world, there's only a single possible outcome. After this, we'll
back off on that assumption and spend most of the rest of the term thinking
about what happens when the deterministic assumption goes away.
1
Planning as Problem Solving
• Planning:
• Start state (S)
• Goal state (G)
• Set of actions
Lecture 10 • 2
In planning, the idea is that you're given some description of a starting state or
states; a goal state or states; and some set of possible actions that the agent can
take. And you want to find the sequence of actions that get you from the start
state to the goal state.
2
Planning as Problem Solving
• Planning:
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
Lecture 10 • 3
It’s pretty clear that you can cast this as a problem solving problem. Remember
when we talked about problem solving, we were given a start state, and we
searched through a tree that was the sequences of actions that you could take,
and we tried to find a nice short plan. So, planning problems can certainly be | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
through a tree that was the sequences of actions that you could take,
and we tried to find a nice short plan. So, planning problems can certainly be
viewed as problem-solving problems, but it may not be the best view to take.
3
Planning as Problem Solving
B
A
• Planning:
DC
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
• But, what if initial state is
not known exactly?
start in bottom row in 4x4
world, with goal being C.
E.g.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
Lecture 10 • 4
Just for fun, let's think about moving around on the squares of this grid. The goal is
to get to state C, but all we know initially is that the agent is in the bottom row,
one of states I, J, K, and L.
4
Planning as Problem Solving
B
A
• Planning:
DC
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
• But, what if initial state is
not known exactly?
start in bottom row in 4x4
world, with goal being C.
• Do search over “sets” of
E.g.
underlying states to find a
set of actions that will reach
C for any starting state.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
Lecture 10 • 5
We could try to solve this using our usual problem solving search, but where our
nodes would now represent sets of states, rather than single states. So, how
would that go?
5
Planning as Problem Solving
B
A
• Planning:
DC
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
• But, what if initial state is
not known exactly?
start in bottom row in 4x4
world, with goal being C.
• Do search | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
what if initial state is
not known exactly?
start in bottom row in 4x4
world, with goal being C.
• Do search over “sets” of
E.g.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
{I, J, K, L}
N
E
{G, J, K, H}
{J, K, L}
underlying states to find a
set of actions that will reach
C for any starting state.
Lecture 10 • 6
The agent starts in the node corresponding to states I,J,K,L, because it doesn’t know
exactly where it is.
Now, let’s say that it can move North, South, East, or West, but if it runs into an
obstacle or the edge of the world, it just stays where it was.
6
Planning as Problem Solving
B
A
DC
• Planning:
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
• But, what if initial state is
E.g.
not known exactly?
start in bottom row in 4x4
world, with goal being C.
• Do search over “sets” of
underlying states.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
{I, J, K, L}
N
E
{G, J, K, H}
{J, K, L}
Lecture 10 • 7
So, if it moves north, it could end up in states G, J, K, or H, so that gives us this
result node.
7
Planning as Problem Solving
B
A
DC
• Planning:
• Start state (S)
• Goal state (G)
• Set of actions
• Can be cast as “problem-
solving” problem
• But, what if initial state is
E.g.
not known exactly?
start in bottom row in 4x4
world, with goal being C.
• Do search over “sets” of
underlying (atomic) states.
E
G
I
J
K | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
with goal being C.
• Do search over “sets” of
underlying (atomic) states.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
{I, J, K, L}
N
E
{G, J, K, H}
{J, K, L}
Lecture 10 • 8
And if it moves East, it could end up in J, K, or L.
You can see how this process would continue. And probably the shortest plan is
going to be to go east three times (so we could only possibly be in state L) and
then continue north until we get to the top, and then go west to the goal.
8
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the
exponential blowup in the number of sets of atomic
states.
Lecture 10 • 9
So, we do have a way to think about planning in the case where you're not
absolutely sure what your starting state is. But, it's really inefficient in any kind
of a big domain because we're planning with sets of states at the nodes. Just to
figure out how the state transition function works on sets of primitive states,
we're having to go through each little atomic state in our set of states, and think
about how it transitions to some other atomic state. And that could be really
desperately inefficient.
9
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the
exponential blowup in the number of sets of atomic
states.
• Logic provides us with a way of describing sets of
states.
Lecture 10 • 10
One of our arguments for moving to logical representations was to be able to have a
compact way of describing a set of states. You should be able to describe that
set of states IJKL by saying, the proposition "bottom row" is true and you might
be able to say, well, if "bottom row", and no obstacles above me, then when I
move, then I'll be in "next to bottom row", or something like that.
10
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
or something like that.
10
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the
exponential blowup in the number of sets of atomic
states.
• Logic provides us with a way of describing sets of
states.
• Can we formulate the planning problem using
logical descriptions of the relevant sets of states?
Lecture 10 • 11
That might mean that if you can set up the logical description of your world in the
right way, you can talk about these sets of states, not by enumerating them, but
by giving logical descriptions of those sets of states. And the idea is that a
logical formula stands for all the interpretations, or world situations, in which it
is true. It stands for all the locations that the agent could be in that would make
the formula true.
11
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the
exponential blowup in the number of sets of atomic
states.
• Logic provides us with a way of describing sets of
states.
• Can we formulate the planning problem using
logical descriptions of the relevant sets of states?
• This is a classic approach to planning: situation
calculus, use the mechanism of FOL to do planning.
Lecture 10 • 12
So, this leads us to the first approach to planning. It's going to turn out not to be
particularly practical, but it's worth mentioning because it's a kind of a classical
idea, and you can get somewhere with it. It’s called situation calculus, and the
idea is to use first-order logic, exactly as you know about it, to do planning.
12
Planning as Logic
• The problem solving formulation in terms of sets of
atomic states is incredibly inefficient because of the
exponential blowup in the number of sets of atomic
states.
• Logic provides us with a way of describing sets of
states.
• Can we formulate the planning problem using
logical descriptions of the relevant sets of states?
• This is a classic approach to planning: situation
calculus, use the mechanism of FOL to do planning.
• Describe states and actions in FOL and use theorem
proving to find a plan.
Lecture 10 • 13
We have to figure out a way to talk | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
in FOL and use theorem
proving to find a plan.
Lecture 10 • 13
We have to figure out a way to talk about states of the world, and start states and
goal states and actions in the language of first order logic, and then somehow
use the mechanism of theorem proving to find a plan. That's the idea. It's
appealing, right? It's always appealing, given a very general solution
mechanism, if you can take a particular problem, and see how it's an instance of
that general solution; then you can apply your general solution to the particular
problem and get an answer out without inventing any more solution methods.
13
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
Lecture 10 • 14
In situation calculus, the main idea is that we're going to "reify" situations. To reify
something is to treat it as if it’s an object. So, we’re going to have variables and
constants in our logical language that will range over the possible situations, or
states of the world. This will allow us to say that world states have certain
properties, but to avoid enumerating them all.
14
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
• At(Robot, Room6, S9) where S9 refers to a particular situation
• Result function: a function that describes the new situation
resulting from taking an action in another situation.
• Result(MoveNorth, S1) = S6
Lecture 10 • 15
So we can say things like "at robot Room Six" in "situations 9". More ordinarily,
we would have said at(robot,room6), implicitly saying something about the
current state of the world. But in planning, we need to be thinking about a
bunch of different situations at the same time, and so we have to be able to name
them.
For any proposition that can change its truth value in different situations, we’ll add
an extra argument, which is the situation in which we’re asserting that the
proposition holds. Then we can talk about how the world changes over time, by
having the | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
which is the situation in which we’re asserting that the
proposition holds. Then we can talk about how the world changes over time, by
having the predicate or function "result". We can say the result of doing action
"move north" in situation one is situation six.
15
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
• At(Robot, Room6, S9) where S9 refers to a particular situation
• Result function: a function that describes the new situation
resulting from taking an action in another situation.
• Result(MoveNorth, S1) = S6
• Effect Axioms: what is the effect of taking an action in the
world
Lecture 10 • 16
Now we need to write down what we know about S6 because of the fact that we got
it by moving north from S1. So, what do we know about S6? In the textbook,
Russell and Norvig use the wumpus world to illustrate situation calculus. We
can describe the dynamics of the world using effect axioms. They say what the
effects in the world are of taking particular actions.
16
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
• At(Robot, Room6, S9) where S9 refers to a particular situation
• Result function: a function that describes the new situation
resulting from taking an action in another situation.
• Result(MoveNorth, S1) = S6
• Effect Axioms: what is the effect of taking an action in the
world
•
∀
x.s. Present(x,s) Æ Portable(x)
Holding(x, Result(Grab, s))
→
Lecture 10 • 17
So, we might say that, for all objects x and situations s, if object x is present in
situation s, and if object s is portable, then in the situation that results from
executing the grab action in situation s, the agent is holding the object x.
Just to be clear, Grab is an action. So Result(Grab, s) is a term that names the
situation that results if | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
.
Just to be clear, Grab is an action. So Result(Grab, s) is a term that names the
situation that results if the agent does a grab action in situation s. And what
we’re saying in the consequent here is that the agent will be holding the object x
in that resulting situation.
17
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
• At(Robot, Room6, S9) where S9 refers to a particular situation
• Result function: a function that describes the new situation
resulting from taking an action in another situation.
• Result(MoveNorth, S1) = S6
• Effect Axioms: what is the effect of taking an action in the
world
•
•
∀
∀
¬
x.s. Present(x,s) Æ Portable(x)
x.s.
→
Holding(x, Result(Drop, s))
Holding(x, Result(Grab, s))
Lecture 10 • 18
Or, we might say that, for all objects x and situations s, that the agent is not holding
object x in a situation that results from doing the drop action in situation s.
You can see the power of the logical representation in these axioms. We’re able to
say things about a very large, or possibly infinite set of situations, without
enumerating them.
18
Situation Calculus
• Reify situations: [reify = name, treat them as objects] and
use them as predicate arguments.
• At(Robot, Room6, S9) where S9 refers to a particular situation
• Result function: a function that describes the new situation
resulting from taking an action in another situation.
• Result(MoveNorth, S1) = S6
• Effect Axioms: what is the effect of taking an action in the
world
•
•
∀
∀
¬
x.s. Present(x,s) Æ Portable(x)
x.s.
→
Holding(x, Result(Drop, s))
Holding(x, Result(Grab, s))
• Frame Axioms: what doesn’t change
Lecture 10 • 19
It's not enough | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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