2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 7 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
04/03/2008
04/10/2008
Problem 1 (20 points):
(a)
⎡1
°
−
°
cos
45
sin
45
⎡
⎤ ⎢2
0
0
0
t X = t R tU = ⎢
⎥⎢
⎣ sin 45° cos 45° ⎦ ⎢ 0
⎣⎢
t
0
X=
( X)
0
t
−1
⎡ 2
⎢
=⎢ 2 2
⎢⎣ − 3
⎤
⎡1
0⎥
1 ⎢2
⎥=
⎢
3⎥
2 ⎢1
⎢⎣ 2
4 ⎦⎥
2 ⎤
⎥
2 2⎥
3 ⎥⎦
(b)
⎡ 17 5 ⎤
⎢ 18 9 ⎥
1 t Tt
t
ε
X
X
I
=
−
=
⎢
⎥
0
0
0
2
⎢ 5 17 ⎥
⎣⎢ 9 18 ⎥⎦
(
)
⎡17 ⎤
⎢ ⎥
⎡11 7 0 ⎤ ⎢ 18 ⎥ ⎡17 ⎤
17
t
⎢
11 0 ⎥⎥ ⎢ ⎥ = ⎢⎢17 ⎥⎥
0S = ⎢ 7
⎢ 18 ⎥
⎢⎣ 0 0 9 ⎥⎦ ⎢ ⎥ ⎢⎣ 5 ⎥⎦
⎢5⎥
⎣⎢ 9 ⎦⎥
ρ 0V (1.5)(1)(thickness) 3
=
=
=
0
ρ tV (2)(2)(thickness) 8
t
Page 1 of 3
3⎤
− ⎥
4
⎥
3 ⎥
4 ⎦⎥
Therefore,
t
τ=
t
0
ρ t t t T ⎡33 0⎤
X S X =⎢
⎥
ρ0 0 0
⎣ 0 8⎦
τ11 = 33,
t
τ 22 = 8,
t
τ12 = 0
t
Hence the Cauchy stress τ given by the program is not correct.
t
We can identify the program error by noting that
33 + 8
= 20.5 , 20.5 + 12.5 = 33 and 20.5 − 12.5 = 8 . Hence a
2
rotation of 45° was wrongly applied. Therefore,
t
τ
=R τ
t
program
⎡ cos 45° − sin 45° ⎤
T
R where R = ⎢
⎥
above
⎣ sin 45° cos 45° ⎦
Problem 2 (10 points):
Since H , h b , we only consider the displacement ur in the x1 -direction with the plane stress assumption.
Total Lagrangian formulation
t
0
f B = t ρ t rω 2
⎛ r − tb ⎞
⎛ r − ta ⎞
⎛ r−b ⎞
⎛ r−a⎞ t
,
(Thickness)
t=H⎜
h
+
=
+
t
H
h
⎜t
⎜t
⎟
⎜
⎟
t ⎟
t ⎟
⎝ a−b⎠
⎝ b−a⎠
⎝ a− b⎠
⎝ b− a⎠
∂u r ∂ t u r ∂ u r
1 ⎛ ∂u ⎞
e
=
+ 0
, 0ηrr = ⎜ 0 r ⎟
0 rr
0
0
2⎝ ∂ r ⎠
∂ r ∂ r ∂ r
ur t u r ur
1 ⎛ ur ⎞
0 eθθ = 0 + 0 2 , 0ηθθ =
⎜ ⎟
2 ⎝ 0r ⎠
r
r
Page 2 of 3
2
2
Therefore,
∫
b
a 0
Cijrs 0 ersδ 0 eij 0 r 0tdr + ∫
t + Δt
= ∫ t + Δt
b
a
t + Δt
b
t
a 0
Sijδ 0ηij 0 r 0tdr
ρ t + Δt rω 2 t + Δt r t + Δt tδ ur dr − ∫
Page 3 of 3
b
t
a 0
Sijδ 0 eij 0 r 0tdr
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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