text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
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 .
(... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
:
(
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(... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
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 th... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
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
... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
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 syst... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
+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 ... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
)
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 a... | 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-... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
n − 1
n
∑ ( X −
X
i
)(
X
i − X )
σ =
XX
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
)... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
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
independent
(covariance is not 0)
Y =
( X * 4)
+
e
cov
=
≠
0
σ
... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
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 varian... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
11
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 li... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
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... | https://ocw.mit.edu/courses/hst-951j-medical-decision-support-fall-2005/11668a5f867641748200d0bfd6a889a3_hst951_7.pdf |
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 maxi... | 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 . . . . . . . . . . . . . . . . . . .... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.1 F.E. model
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
. . . . . . . . . . . 61
15.1.2 Principle of virtual temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
15.2 Inviscid, incompressible, irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
16 F.E. analysis of Navier-Stokes fluids
17 Incompressible fluid flow and heat transfer, c... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 a... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
=
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.
•
Stre... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 �
t... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 +... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 text... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
�
�... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 wit... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(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 (... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 a... | 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)
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. displaceme... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
yy ⎠ =
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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
xy =
∂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 OpenCourseWar... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 ... | 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
. . .
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) ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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) − ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
)
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 (... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
) (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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 ⇒ ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
4)
(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 t... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
! (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 pr... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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)
� ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�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 ano... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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) ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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: ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 Soli... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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)... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
< �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 colum... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
)
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) ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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, ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
δij +
t
∂ ui
∂ 0x
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 decomposi... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 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 = � ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 Lagrangian formulation
2nd Piola-Kirchhoff stress
0ρ
tX tτ 0
0
tS = tρ
0
tX T →
components also independent of a rigid body rotation
(12.1... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
2.25)
(12.26)
(12.27)
(12.28)
(12.29)
MIT 2.094
12. Total Lagrangian formulation
Left hand side as before but using (k − 1) and right hand side is
(12.30)
(12.31)
(12.32)
�
t+Δt
R −
0V
=
gives
t+Δt
S δ
0 ij
t+Δt (k−1) 0
�
0 ij
d V
t+Δt
(k−1)
0 F
tF (k−1)
t+ΔtR − t+Δ
0
In the full N-R iteration,... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
· ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
=u1
t
2
=u2
t
∂u1 +
∂ 0x1
�
0L
0L
∂ u1 ∂u1 +
∂ 0x1 ∂ 0x1
�
cos θ − 0L
�
+ ΔL sin θ
+ ΔL
�
t
∂ u2 ∂u2
∂ 0x1 ∂ 0x1
�
−1 0 1 0 uˆ
�
1
0e11 = 0L
⎛
⎞
⎜
⎜
⎜
⎜
⎜
⎜
⎝
�
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
�
+
0L + ΔL
0L
�
1
0L
−1
0
1
0
�
uˆ
⎟
⎟
⎟
·
cos θ − 1
⎟
⎟
⎟
�
⎠
��
∂ tu1
0∂ x1
⎞
⎟
⎟
⎟
·
sin θ
⎟
⎟
⎟
�
⎠
... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
1
�
uˆ
�
55
�
(13.20g)
(13.21a)
(13.21b)
(13.21c)
(13.21d)
(13.21e)
MIT 2.094
13. Total Lagrangian formulation, cont’d
0C = E
tSˆ = tS
0
0 11
Assume small strains
tK =
0
EA
0L
⎡
cos2 θ cos θ sin θ
sin2 θ
=
⎢
⎢
⎣
�
sym
− cos2 θ
− sin θ cos θ
cos2 θ
− cos θ sin θ
− sin2 θ
sin θ cos θ
sin2... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
teij 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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�
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
� ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(14.12)
(14.13)
(14.14)
(14.15)
(14.16)
(14.17)
(14.18)
(14.19)
MIT 2.094
14. Total Lagrangian formulation, cont’d
t+Δt
� =
0 33
tu + u1
1
0
x1
+
�
1
2
tu + u1
1
0
x1
�2
t�33 − 0
0�33 = t+Δ
0
u1
t�33 = 0x1
tu
+ 0x
1
1
u1
· 0x1
+
�
1 u1
0x1
2
�2
(14.20)
(14.21)
How do we assess the accu... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
∂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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
Fr... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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)
(irro... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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,i... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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;... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
2... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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-dimensio... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
1)
1
(h∗)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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
“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
Triangular Grids.” International Journal for Numerical Methods in Fluids, 51:673–... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
compressible 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. ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
− 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
,
0L + tu
0ρ 0L = tρ tL
⇒ 0
tL
tS11 = 0L
� �2
0L
tL
0L
tτ11 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
∂ �
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.... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
+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 ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
⇒
pr... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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
... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
=
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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
inciple 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
·... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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%. ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
�,
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 − γ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
(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)
... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
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,... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
⎞
⎛
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... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
.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 (... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
+ (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-
tensio... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/116df281a310a9ccdbbb8dd7ec7a2d6e_MIT2_094S11_2094_lectures.pdf |
fficulties 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.
R... | 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)d... | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
) =
1
− 100
1
100 0
�
�
��
B(1)
⎤
⎡
⎥ �
⎢
⎢
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
⎦
�... | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
2 is A = 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 z... | https://ocw.mit.edu/courses/2-092-finite-element-analysis-of-solids-and-fluids-i-fall-2009/117f4cc28ac685fc73a87c6f9dc44378_MIT2_092F09_lec06.pdf |
πx
C =
⎡
⎣
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
representatio... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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
viewed as problem-solving problems, but it ma... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
ving
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.
E
G
I
J
K
F
H
L
Actions: N,S,E,W
{I, J, K, L}
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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
F
H
L
Actions: N,... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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 ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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 d... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/1184a975225bdbab3e3d215bf173bde1_Lecture10FinalPart1.pdf |
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