2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Quiz #2 - Solution
Instructor:
Prof. K. J. Bathe
Date:
05/13/2008
Problem 1 (10 points)
a)
⎡7 3
⎡ cos 30° − sin 30° ⎤ ⎡ 7 / 6 0 ⎤ ⎢ 12
t
0X = ⎢
⎥ ⎢ 0 8 / 9⎥ = ⎢
sin
30
cos
30
°
°
⎣
⎦⎣
⎦ ⎢ 7
⎢
⎣ 12
b)
⎡ 13
⎢ 72
1 t Tt
1 t 2
t
=
−
=
−
=
ε
X
X
I
U
I
⎢
0
0
0
0
2
2
⎢0
⎣⎢
(
)
(
)
4⎤
⎥
9⎥
4 3⎥
⎥
9 ⎦
−
⎤
0 ⎥
⎥
17 ⎥
−
162 ⎦⎥
⎛ 13 ⎞
⎛ 13 ⎞
⎜ 72 ⎟
⎡ 1 0 0 ⎤ ⎜ 72 ⎟
⎜
⎟
⎜
⎟
t
⎢
⎥
=
=
S
E
E
0
1
0
17
17
⎜
⎟
⎜
⎟
0
⎢
⎥ −
−
⎜
⎟
⎜
162 ⎟
⎢⎣ 0 0 1/ 2 ⎥⎦ 162
⎜
⎟
⎜
⎟
⎝ 0 ⎠
⎝ 0 ⎠
c)
⎡7 3
⎢
1
1
ρ
T
t
t
t
t
⎢ 12
τ = 0 0t X 0t S 0t X T =
=
0 X 0S 0 X
t
ρ
det 0 X
7 3 4 3 4 7 ⎢ 7
⋅
+ ⋅ ⎢
12
9
9 12 ⎣ 12
t
Page 1 of 3
4 ⎤ ⎡ 13
⎤ ⎡7 3
0 ⎥⎢
⎥⎢ E
9 ⎥ 72
12
⎢
⎥⎢
17 ⎥ ⎢ 4
4 3⎥⎢ 0
−
E −
⎥ ⎣⎢
162 ⎦⎥ ⎢⎣ 9
9 ⎦
−
7 ⎤
⎥
12 ⎥
4 3⎥
⎥
9 ⎦
Problem 2 (10 points)
The governing equation we need is
∫
V
eijτ ij dV = ℜ
(
)
By substituting τ ij = − pδ ij + μ vi , j + v j ,i = − pδ ij + 2μ eij ,
∫
V
eijτ ij dV = ∫ eij ( 2 μ ) eij dV − ∫ eii pdV
V
V
In the following calculations, we need to know the interpolation function and its derivatives only at the node 1.
h1 =
1
x1 (1 + x1 ) x2 (1 + 2 x2 )
2
h1,1 =
1
(1 + 2 x1 ) x2 (1 + 2 x2 )
2
h1,2 =
1
x1 (1 + x1 )(1 + 4 x2 )
2
The volume integration of the element assuming the unit depth into the x3 direction is,
∫
V
( )dV = ∫
+1
−1
∫
+1/ 2
−1/ 2
( )dx1dx2
a) Evaluate K (1,1)
Let
⎛ e11 ⎞ ⎛ v1,1 ⎞
⎟
⎜
⎟ ⎜
e = ⎜ e22 ⎟ = ⎜ v2,2 ⎟ = Bvˆ
⎜ 2e ⎟ ⎜ v + v ⎟
⎝ 12 ⎠ ⎝ 1,2 2,1 ⎠
where vˆ = ⎡⎣ v11
T
v12
v12
v22 " v19
Then,
∫
V
T
eij ( 2μ ) eij dV = vˆ
( ∫ B C BdV ) vˆ
T
V
We only need the first column of B to compute K (1,1) ,
T
B (1) = ⎡⎣ h1,1
0 h1,2 ⎤⎦
Page 2 of 3
⎡ 2μ
⎢
where C = 0
⎢
⎢⎣ 0
0
2μ
0
0⎤
0 ⎥⎥
μ ⎥⎦
v29 ⎤⎦
Therefore,
{
}
K (1,1) = ∫ B (1) C B (1) dV = ∫ 2μ ( h1,1 ) + μ ( h1,2 ) dV
T
V
V
2
2
*** Or, simply from Equation 7.79 in the textbook,
(
)
K μ v1v1 = ∫ 2 μ H ,Tx1 H , x1 + μ H ,Tx2 H , x2 dV
V
{
}
K (1,1) = K μv1v1 (1,1) = ∫ 2 μ ( h1,1 ) + μ ( h1,2 ) dV
V
2
2
b) Evaluate K (1, 21)
Let
eii = e11 + e22 = v1,1 + v2,2 = B v vˆ
⎡ p0 ⎤
x2 ⎤⎦ ⎢⎢ p1 ⎥⎥ = H pˆ
⎢⎣ p2 ⎥⎦
p = p0 + p1 x1 + p2 x2 = ⎡⎣1 x1
Then,
(
)
T
T − ∫ eii pdV = vˆ − ∫ B v HdV
pˆ
V
V
We need the first column of B v and the third column of H to compute K (1, 21) ,
T
B v (1) = ⎡⎣ h1,1 ⎤⎦ and H (3) = ⎣⎡ x2 ⎦⎤
Therefore,
T
K (1, 21) = − ∫ B v (1) H (3) dV = − ∫ h1,1 x2 dV
V
V
*** Or, simply from Equation 7.81 in the textbook,
K v1 p = − ∫ H ,Tx1 H dV
V
K (1, 21) = K v1 p (1,3) = − ∫ h1,1 x2 dV
V
c) The 9/3 element is a suitable element for the incompressible analysis because it does not lock and passes the
inf-sup condition and therefore it shows an optimal convergence.
Page 3 of 3
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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