2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 6 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
03/13/2008
03/20/2008
Problem 1 (20 points):
t
Let’s define Rˆ =
t
t
R tˆ
F
Δ
t
, F=
and U =
.
2kL
2kL
L
t
Since t Fˆ = t R̂ at equilibrium,
t
⎧
⎫
1
⎪
⎪
t ˆ
t
ˆ
F = R = ⎨ −1 +
⎬ (sin15° − U )
t
t 2 1/2
⎡⎣1 − 2 U sin15° + U ⎤⎦ ⎪
⎪⎩
⎭
The tangent stiffness can be calculated by
t + Δt
t + Δt
K
( i −1)
=
Fˆ ( t + ΔtU (i −1) + ε ) − t + Δt Fˆ ( t + ΔtU (i −1) )
ε
with a sufficiently small ε .
Then the Newton-Raphson iteration is performed
t + Δt
K ( i −1) ΔU ( i ) =
t + Δt
U (i ) =
t + Δt
Rˆ − t + Δt Fˆ (i −1)
t + Δt
U ( i −1) + ΔU (i )
with the initial conditions
t + Δt
U (0) = tU ,
t + Δt
K (0) = t K ,
t + Δt
Fˆ (0) = t Fˆ
The calculated values in each step are listed in Table 1. The number of iterations and the calculated values can be
different from those in Table 1 depending on which value of ε you used for the calculation of the tangent stiffness.
Page 1 of 3
Note that this problem has multiple solutions as shown in the graph in the textbook. Therefore if your stiffness is
not correct, then you may get another solution which is far from the initial equilibrium state. Make sure that the
displacement you calculated is the nearest one from the initial equilibrium state.
Table 1. Calculated values in each iteration ( ε = 0.0001 )
Step
0
1
2
3
t + Δt
t + Δt
U
3.702230E-002
6.112280E-002
6.801735E-002
6.863081E-002
Fˆ
t + Δt
2.000000E-003
2.817306E-003
2.986220E-003
2.999909E-003
K
4.149292E-002
2.649827E-002
2.246174E-002
(Not necessary)
t + Δt
Rˆ − t + Δt Fˆ
1.000000E-003
1.826936E-004
1.377952E-005
9.081258E-008
Problem 2 (10 points):
(a)
h1 =
1
1
1
1
1 + 0 x1 )(1 + 0 x2 ) , h2 = (1 − 0 x1 )(1 + 0 x2 ) , h1 = (1 − 0 x1 )(1 − 0 x2 ) , h1 = (1 + 0 x1 )(1 − 0 x2 )
(
4
4
4
4
Define
0
xˆ T = ⎡⎣ 0 x1
t
xˆ T = ⎡⎣ t x1
0
x2
0 3
x
0
x4
0
y1
0
t
t
x3
t
x4
t
y1
t
x2
y2
y2
0
t
y3
y3
0
t
y 4 ⎤⎦
y 4 ⎤⎦
Then, using t u = t x − 0 x ,
t
⎡ t u ⎤ ⎡ h1 h2 h3
u = ⎢t ⎥ = ⎢
⎣ v⎦ ⎣ 0 0 0
⎡ 3
⎤
1 + 0 x2 ) ⎥
(
⎢
⎥
=⎢ 2
⎢ 1
⎥
0
⎢⎣ − 2 (1 + x2 ) ⎥⎦
h4
0
0
h1
Page 2 of 3
0
h2
0
h3
0⎤ t
( xˆ − 0 xˆ )
h4 ⎥⎦
(b)
⎡0
⎤
3
1 + 0 x2 ) ⎥
(
⎢ x1 +
t
2
⎥
x = 0 x + tu = ⎢
1
⎢ 0
⎥
0
⎢⎣ x2 − 2 (1 + x2 ) ⎥⎦
⎡ ∂ t x1
⎢ 0
∂ x1
t
⎢
0X =
⎢∂tx
⎢ 0 2
⎣⎢ ∂ x1
∂ t x1 ⎤ ⎡
⎥
∂ 0 x2 ⎥ ⎢ 1
=⎢
∂ t x2 ⎥ ⎢
0
⎥
∂ 0 x2 ⎦⎥ ⎢⎣
⎡
⎢ 1
t
t
T t
⎢
0C = 0 X 0 X =
⎢ 3
⎢
⎣ 2
t
ρ=
0
ρ
t
0
det X
=
3⎤
⎥
2 ⎥
1 ⎥
2 ⎥⎦
3⎤
⎥
2 ⎥
⎥
1 ⎥
⎦
0.05
= 0.1
1/ 2
Page 3 of 3
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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