2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 5 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
03/06/2008
03/13/2008
Problem 1 (10 points):
τ ij = κε vδ ij + 2Gε ij' (a)
τ ij = Cijrsε rs
(b)
τ = Cε
(c)
Let’s start from equation (b). Using γ ij = ε ij + ε ji (i ≠ j ) and Cijrs = λδ ijδ rs + μ (δ irδ js + δ isδ jr ) ,
τ 11 = C1111ε11 + C1122ε 22 + C1133ε 33 + C1112γ 12 + C1123γ 23 + C1131γ 31
= (λ + 2μ )ε11 + λε 22 + λε 33
τ 22 = λε11 + (λ + 2 μ )ε 22 + λε 33
τ 33 = λε11 + λε 22 + (λ + 2μ )ε 33
τ 12 = C1211ε11 + C1222ε 22 + C1233ε 33 + C1212γ 12 + C1223γ 23 + C1231γ 31 = μγ 12
τ 23 = μγ 23
τ 31 = μγ 31
Therefore,
λ
λ
⎡ λ + 2μ
⎢ λ
λ + 2μ
λ
⎢
⎢ λ
λ
λ + 2μ
C=⎢
0
0
⎢ 0
⎢ 0
0
0
⎢
0
0
⎣ 0
Page 1 of 6
0
0
0
0
0
0
μ
0
0
μ
0
0
0⎤
0 ⎥⎥
0⎥
⎥
0⎥
0⎥
⎥
μ⎦
Substituting λ =
Eν
E
and μ =
, we obtain C in Table 4.3.
(1 + ν )(1 − 2ν )
2(1 + ν )
Hence equation (c) is equivalent to equation (b).
Now derive equation (a) from equation (b).
ε
⎞
⎛
⎞
δ rs ⎟ = λδ ijδ rs + μ ( δ irδ js + δ isδ jr ) ⎜ ε rs′ + v δ rs ⎟
3
3
⎠
⎝
⎠
ε
ε
ε
= λδ ijδ rsε rs′ + λδ ijδ rs v δ rs + μδ irδ jsε rs′ + μδ isδ jr ε rs′ + μδ irδ js v δ rs + μδ isδ jr v δ rs
⎛
⎝
τ ij = Cijrsε rs = Cijrs ⎜ ε rs′ +
εv
{
}
3
= λδ ij ε rr′ + λδ ij
εv
3
= λε vδ ij + 2με ij′ +
3
δ rr + με ij′ + με ′ji + μ
εv
3
δ ij + μ
εv
3
3
δ ji
2μ
ε vδ ij
3
2μ ⎞
⎛
= ⎜λ +
⎟ ε vδ ij + 2με ij′
3 ⎠
⎝
= κε vδ ij + 2Gε ij′
Here we used
′ + ε 33
′ = 0 and δ rr = δ11 + δ 22 + δ 33 = 3
ε rr′ = ε11′ + ε 22
Problem 2 (10 points):
h1 =
1
2
1
2
1
2
(1 + x ) ⎛⎜ 1 + y ⎞⎟ , h2 = (1 − x ) ⎛⎜ 1 + y ⎞⎟ , h3 = (1 − x ) ⎛⎜ 1 −
4
4
4
⎝ 3 ⎠
⎝ 3
⎝ 3 ⎠
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1
⎞
⎛ 2
y ⎟ , h4 = (1 + x ) ⎜ 1 −
4
⎠
⎝ 3
⎞
y⎟
⎠
T
Define u = ⎡⎣u1
u2
u3
u4
v1 v2
v3
⎡ h1
v4 ⎤⎦ , then, H = ⎢
⎣0
h2
h3
h4
0
0
0
0
0
0
h1 h2
h3
0⎤
h4 ⎥⎦
The deviatoric strains are
1 ⎤ ⎡ 2
1
1
⎡
⎤
⎢ ε xx − 3 ε v ⎥ ⎢ 3 ε xx − 3 ε yy − 3 ε zz ⎥
⎢
⎥ ⎢
⎥
⎢ε − 1 ε ⎥ ⎢ − 1 ε + 2 ε − 1 ε ⎥
ε ′ = ⎢ yy 3 v ⎥ = ⎢ 3 xx 3 yy 3 zz ⎥ = B D u
⎢
⎥ ⎢
⎥
γ xy
⎢ γ xy ⎥ ⎢
⎥
⎢
1 ⎥ ⎢ 1
1
2 ⎥
⎢ ε zz − ε v ⎥ ⎢ − ε xx − ε yy + ε zz ⎥
3 ⎦ ⎣ 3
3
3 ⎦
⎣
where
ε xx =
∂u
∂v
∂u ∂v
u
+ , and ε zz =
, ε yy =
, γ xy =
∂x
x+5
∂y
∂y ∂ x
Therefore,
1 h1
⎡ 2
⎢ 3 h1, x − 3 x + 5
⎢
⎢ − 1 h − 1 h1
B D = ⎢ 3 1, x 3 x + 5
⎢
h1, y
⎢
⎢ 1
2 h1
⎢ − h1, x +
3 x+5
⎣ 3
2
1 h2
h2, x −
3
3 x+5
1
1 h2
− h2, x −
3
3 x+5
h2, y
1
2 h2
− h2, x +
3
3 x+5
2
1 h3
h3, x −
3
3 x+5
1
1 h3
− h3, x −
3
3 x+5
h3, y
1
2 h3
− h3, x +
3
3 x+5
2
1 h4
h4, x −
3
3 x+5
1
1 h4
− h4, x −
3
3 x+5
h4, y
1
− h1, y
3
2
h1, y
3
h1, x
1
− h2, y
3
2
h2, y
3
h2, x
1
− h3, y
3
2
h3, y
3
h3, x
1
2 h4
− h4, x +
3
3 x+5
1
− h1, y
3
1
− h2, y
3
1
− h3, y
3
The volumetric strain is
ε v = ε xx + ε yy + ε zz = B v u
h
⎡
BV = ⎢ h1, x + 1
x+5
⎣
h2, x +
h2
x+5
h3, x +
h3
x+5
h4, x +
The pressure is
p = H p lp
where
Page 3 of 6
h4
x+5
h1, y
h2, y
h3, y
⎤
h4, y ⎥
⎦
1
⎤
− h4, y ⎥
3
⎥
2
h4, y ⎥
⎥
3
⎥
h4, x ⎥
⎥
1
− h4, y ⎥
3
⎦
lp = ⎡ p ⎤
⎣ 0⎦
Therefore
H p = [1]
And,
⎡ 2G 0 0 0 ⎤
⎢ 0 2G 0 0 ⎥
⎥
C′ = ⎢
⎢ 0
0 G 0 ⎥
⎢
⎥
0 0 2G ⎦
⎣ 0
Problem 3 (20 points):
We can see in convergence curves that the displacement-based elements are bad when a material is
incompressible. However, when the mixed (u/p) elements are used, we can obtain almost optimal convergence
rates. Also, the curves of the mixed elements are shifted much down from those of the displacement-based
elements, which means that the solutions obtained using the mixed elements are much more accurate. (Note that
even the 4/1 u/p element is better than the 9-node displacement-based element in this problem.)
Page 4 of 6
In addition, the mixed elements predict better pressure distributions with reasonable magnitudes.
(4-node displacement-based element)
(9-node displacement-based element)
Page 5 of 6
(4-node u/p element)
(9-node u/p element)
Page 6 of 6
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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