2.092/2.093
FINITE ELEMENT OF SOLIDS AND FLUIDS I
FALL 2009
Homework 5- solution
Instructor:
TA:
Prof. K. J. Bathe
Seounghyun Ham
Assigned:
Due:
Problem 1 (20 points):
y
2
1
b
x
4
3
a
1 ⎛ 2x ⎞ ⎛ 2 y ⎞
h1 = ⎜ 1+ ⎟ ⎜1+
⎟
4⎝
a ⎠⎝
b ⎠
1 ⎛ 2x ⎞ ⎛ 2 y ⎞
h2 = ⎜1− ⎟ ⎜1+
⎟
4⎝
a ⎠⎝
b ⎠
1 ⎛ 2x ⎞⎛ 2 y ⎞
h3 = ⎜ 1− ⎟⎜ 1−
⎟
4⎝
a ⎠⎝
b ⎠
1 ⎛ 2x ⎞⎛ 2 y ⎞
h4 = ⎜1+ ⎟⎜ 1−
⎟
4⎝
a ⎠⎝
b ⎠
u = [ h1
ε xx =
h2
h3
∂u
= ⎡ h1, x
∂x ⎣
⎡ u1 ⎤
⎢u ⎥
h4 ] ⎢ 2 ⎥ , v = [ h1
⎢ u3 ⎥
⎢ ⎥
⎣u4 ⎦
h2
0 h2, x
0 h4, x
0 h3, x
h3
1
⎡ v1 ⎤
⎢v ⎥
h4 ] ⎢ 2 ⎥
⎢ v3 ⎥
⎢ ⎥
⎣ v4 ⎦
0 ⎤⎦ U
Session10
Session12
ε yy =
∂v
= ⎡ 0 h1, y
∂y ⎣
where U T = [u1
0 h2, y
v1 u2
0 h3, y
v2
u3
v3
0 h4, y ⎤⎦ U
v4 ] ; ε zz = 0
u4
As ε V = ε xx + ε yy
∴ ε V = BεV U = ⎡⎣ h1, x
h1, y
h2, x
h2, y
h3, x
h3, y
h4, y ⎦⎤ U
h4, x
Hence
b
2
K =t∫
a
2
∫
β BTεV BεV dxdy
b a
− −
2 2
Problem 2 (20 points):
u2
t
1
R2
L2
2
ө
u1
L1
Assuming tension in the bar as positive, the equilibrium of the joints gives:
Joint 2
Joint 1
t
t
R2
P2
ө
t
t
P1
t
P2
2
ө
t
P1 − t P2 cos tθ = 0
t
P1 =
EA t
u1 ,
L1
t
P2 =
(1) R2 − t P2 sin tθ = 0
EA
δ L2
L2
(2)
(3)
where δ L2 = (5 − t u1 ) 2 + (0.5 + t u2 ) 2 − L2
From the geometry, tan tθ =
0.5 + t u2
,
5 − t u1
t
(4)
u2 = −Δ , R2 = − P (5)
Eq. (1) and (2) are the force equilibrium equations. We use them by assuming a Δ,
solving from equation (1) for tu1, then substituting tu1 and Δ into the equation (2) to
obtain the corresponding P. We can also solve them in different way. We first assume
t
ө and then calculate tu1 and Δ.
P
×104 Vs. Δ
EA
3
Displacement u1 Vs. Δ
P
×104 Vs. Δ Using ADINA
EA
4
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2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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