2.092/2.093
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS I
FALL 2009
Quiz #2-solution
Instructor:
TA:
Prof. K. J. Bathe
Seounghyun Ham
Problem 1 (10 points):
1 ⎛ x ⎞⎛
y⎞
1⎛ x⎞
f
his = ⎜1 − ⎟ (1 − y ) ; hi = ⎜ 1− ⎟ ⎜ 1+ ⎟ .
4 ⎝ 2 ⎠⎝ 2 ⎠
4⎝ 2⎠
his, x = −
1
x
1
(1 − y ) ; his, y = − ⎛⎜1− ⎞⎟ .
4 ⎝ 2 ⎠
8
1⎛
1 ⎛ x ⎞
y⎞
hi,xf = − ⎜ 1+ ⎟ ; hi ,f y = ⎜ 1− ⎟
8⎝ 2 ⎠
8⎝ 2⎠
H
(1)
⎡ hi f
0⎤
(2)
⎥; H = ⎢
his ⎦
⎣0
⎡ his
=⎢
⎣0
T
U = [ui
B
(1)
(1)
vi ] .
⎡ his, x
⎢
=⎢ 0
⎢ his, y
⎣
M =
∫ρH
s
V (1)
0⎤
⎥.
hi f ⎦
0 ⎤
⎥
his, y ⎥ B(2) = ⎡ h f
;
⎣ i,x
s ⎥
hi , x ⎦
hi ,f y ⎤⎦ .
1 2
(1)T
H dV =ρs ∫ ∫ H
(1)
(1)
(1)T
(1)
H dxdy
-1 -2
1 of 3
(2)
M =
∫ρH
2 2
(2)T
f
H dV =ρ f ∫ ∫ H
(2)
(2)
V (2)
(1)
K =
K =
∫B
∫
(2)
H dxdy
-2 -2
1 2
CB dV = ∫ ∫ B
(1)T
(1)
(1)
V (1)
(2)
(2)T
(1)T
(1)
CB dxdy
-1 -2
2 2
B
(2)T
βB dV = ∫ ∫ B
(2)
V (2)
(1)
M=M +M
(2)
(2)T
(2)
βB dxdy
-2 -2
(2)
(1)
;
K=K +K
(2)
.
Problem 2 (10 points):
(a) “Remove Clamps” to eliminate u1 and u3 prior to the use of the central difference method.
�� ⎤ ⎡ 2 −1 0 ⎤ ⎡ U ⎤ ⎡ 0 ⎤
⎡0 0 0⎤ ⎡ U
1
1
⎢0 2 0⎥ ⎢ U
�� ⎥ + ⎢ −1 4 −1⎥ ⎢ U ⎥ = ⎢ R (t) ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥⎢ 2⎥ ⎢ 2 ⎥
��
⎢0
⎣ 0 0⎥⎦ ⎢⎣ U3 ⎥⎦ ⎢⎣ 0 −1 2 ⎦⎥ ⎣⎢ U 3 ⎥⎦ ⎢⎣ 0 ⎦⎥
�� ⎤ ⎡ 2 −1 0 ⎤ ⎡ U ⎤ ⎡ 0 ⎤
⎡0 0 0⎤ ⎡ U
1
1
⎢
⎢0 2 0⎥ U
�� ⎥ + ⎢ 0 3.5 −1⎥ ⎢ U ⎥ = ⎢ R (t) ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥⎢ 2⎥ ⎢ 2 ⎥
��
⎢
⎥
⎣⎢0 0 0⎥⎦ ⎣ U3 ⎦ ⎣⎢0 −1 2 ⎦⎥ ⎣⎢ U 3 ⎦⎥ ⎣⎢ 0 ⎦⎥
�� ⎤ ⎡ 2 −1 0 ⎤ ⎡ U ⎤ ⎡ 0 ⎤
⎡0 0 0 ⎤ ⎡ U
1
1
⎢0 2 0 ⎥ ⎢ U
�� ⎥ + ⎢ 0 3 0 ⎥ ⎢ U ⎥ = ⎢ R (t) ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥⎢ 2⎥ ⎢ 2 ⎥
��
⎢0
⎣ 0 0⎥⎦ ⎢⎣ U3 ⎥⎦ ⎢0
⎣ −1 2⎥⎦ ⎣⎢ U 3 ⎥⎦ ⎢⎣ 0 ⎦⎥
�� +3U =R (t)
2U
2
2
2
Using the central difference method,
�� +3 t U = t R
2t U
2
2
2
(1)
2 of 3
t
t
�� = 1 ( t + Δt U -2 t U + t - Δt U )
U
2
2
2
2
Δt 2
(2)
� = 1 ( t + Δt U − t - Δt U )
U
2
2
2
2Δt
(3)
Substitute (2) into (1)
2
Δt 2
t + Δt
U 2 = t R 2 -(3-
4 t
2
) U 2 - 2
t - Δt U 2
2
Δt
Δt
�� +3U =R (t) at time t=0.
To start the solution, we use 2U
2
2
2
�� =0 since U and R are zero at t=0.
Hence, 0 U
2
2
2
We can obtain
- Δt
U 2 =0 using Eq. (2) and Eq. (3).
�� = 1 ( Δt U -2 0 U + - Δt U )
0 = 0U
2
2
2
2
Δt 2
� 1 ( Δt U − - Δt U )
0 = 0 U=
2
2
2Δt
⎡1⎤
⎡
U ⎤ ⎢2⎥
Determine t + Δt U1 and t + Δt U 3 by ⎢ t + Δt 1 ⎥ = ⎢ ⎥ t + Δt U 2 .
U3 ⎦ ⎢ 1 ⎥
⎣
⎢⎣ 2 ⎥⎦
t + Δt
(b) Δt cr =
2
2
=2 =1.633
ω
3
For stability, Δt should be smaller than Δtcr. The frequency content of the load can be roughly
ω̂
2π
ˆ
where T=80
approximated by ω̂=
so that = 0.0641 . Because this value is quite close to
ω
80
zero, we can assume the response to be almost static. Therefore, Δt=1.6 is a well selected time
step for the stability and the accuracy.
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2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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