2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 10 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
05/06/2008
05/13/2008
Problem 1 (10 points):
The governing differential equation is
ρcp
dθ
d 2θ
v=k 2
dx
dx
The non-dimensional form is
dΘ
1 d 2Θ
= e
dX Pe dX 2
where
Θ=
x
vh
vh
θ − θL
e
=
, X =
and Pe =
h
θR − θL
α k / (ρcp )
The principle of virtual temperature for a unit area is
∫Θ
dΘ
1 d 2Θ
1 dΘ dΘ
dX = ∫ Θ e
dX = − ∫ e
dX + (boundary terms)
2
dX
Pe dX
Pe dX dX
Therefore,
⎡ dΘ
1 dΘ dΘ ⎤
dX = (boundary terms )
e
dX dX ⎥⎦
∫ ⎢⎣Θ dX + Pe
If we use two-node elements, then for i-th element
Θ( i ) =
1
1
(1 − r )Θi + (1 + r )Θi +1
2
2
Page 1 of 4
d Θ(i ) d Θ(i ) dr d Θ(i ) 2
=
=
⋅ = − Θi + Θi +1
dX
dr dX
dr 1
Hence
⎡ ( i ) d Θ( i )
1 d Θ( i ) d Θ ( i ) ⎤
∫0 ⎢⎣ Θ dX + Pee dX dX ⎥⎦ dX
1
+1 ⎡
d Θ( i )
1 d Θ( i ) d Θ ( i ) ⎤ 1
= ∫ ⎢ Θ( i )
+ e
⎥ dr
−1
dX
Pe dX dX ⎦ 2
⎣
ˆ ( i )T
=Θ
⎧ ⎛ ⎡1
⎫
⎞
⎤
⎪⎪ +1 ⎜ ⎢ (1 − r ) ⎥
⎟
1 ⎡ −1⎤
1 ⎪⎪ ˆ ( i )
2
⎥ [ −1 1] + e ⎢ ⎥ [ −1 1] ⎟ dr ⎬ Θ
⎨ ∫−1 ⎜ ⎢
Pe ⎣ 1 ⎦
⎟2 ⎪
⎪ ⎜ ⎢ 1 (1 + r ) ⎥
⎜
⎟
⎪⎩ ⎝ ⎣⎢ 2
⎪⎭
⎦⎥
⎠
ˆ ( i )T
=Θ
⎡ 1
⎢− 2 +
⎢
⎢− 1 −
⎢⎣ 2
ˆ ( i ) = ⎡Θ
where Θ
⎣ i
1
Pee
1
Pee
1
−
2
1
+
2
1 ⎤
Pee ⎥ ˆ (i )
⎥Θ
1 ⎥
Pee ⎥⎦
T
Θi +1 ⎤⎦
Therefore, the governing equation for the finite element node i by assembling the element (i-1) and (i) is
⎧⎛ 1
⎫
1 ⎞
1 ⎞ ⎫ ⎧⎛ 1
1 ⎞
1 ⎞
⎛1
⎛1
⎨ ⎜ − − e ⎟ Θi −1 + ⎜ + e ⎟ Θi ⎬ + ⎨⎜ − + e ⎟ Θi + ⎜ − e ⎟ Θi +1 ⎬
⎝ 2 Pe ⎠ ⎭ ⎩⎝ 2 Pe ⎠
⎝ 2 Pe ⎠
⎩ ⎝ 2 Pe ⎠
⎭
1 ⎞
2
1 ⎞
⎛ 1
⎛1
= ⎜ − − e ⎟ Θi −1 + e Θi + ⎜ − e ⎟ Θi +1 = 0
Pe
⎝ 2 Pe ⎠
⎝ 2 Pe ⎠
Finally,
⎛
⎛ Pee
⎞
Pee ⎞
−
−
+
+
− 1⎟ Θi +1 = 0
Θ
Θ
1
2
⎜
⎟ i −1
⎜
i
2 ⎠
⎝
⎝ 2
⎠
Page 2 of 4
Problem 2 (20 points):
v 2
u2
y
x
v 1
u1
⎡ uk ⎤ ⎡ cos 30° − sin 30° ⎤ ⎡uk ⎤ ⎡ 3 / 2
⎢ v ⎥ = ⎢ sin 30° cos 30° ⎥ ⎢ v ⎥ = ⎢
⎦ ⎣ k ⎦ ⎣⎢ 1/ 2
⎣ k⎦ ⎣
−1/ 2 ⎤ ⎡ uk ⎤
⎥⎢ ⎥
3 / 2 ⎦⎥ ⎣ vk ⎦
We define strains and stresses in local coordinate system ( x , y ) .
u = hi ui −
st
hiθi and v = hi vi
2
where ( r , s ) is an iso-parametric coordinate system and t denotes the thickness.
2 ∂hi
st ∂hi
∂u 2 ∂u 2 ∂hi
θi =
ui −
=
=
L ∂r
L ∂r
∂x L ∂r L ∂r
⎛ 3
1 ⎞ st ∂hi
θi
ui − vi ⎟⎟ −
⎜⎜
2 ⎠ L ∂r
⎝ 2
∂u 2 ∂u
2t
=
=−
hiθi = − hiθi
t 2
∂y t ∂s
2 ∂hi
∂v 2 ∂v 2 ∂hi
vi =
=
=
L ∂r
∂x L ∂r L ∂r
⎛1
3 ⎞
vi ⎟
⎜⎜ ui +
2 ⎟⎠
⎝2
∂v 2 ∂v
=
=0
∂y t ∂s
Then,
∂u / ∂x
⎡ε xx ⎤ ⎡
⎤
⎢
⎥
⎢
ε = ⎢γ xy ⎥ = ⎢ ( ∂u / ∂y + ∂v / ∂x ) r = 0 ⎥⎥ = Buˆ
⎢⎣ ε zz ⎥⎦ ⎢⎣
⎥⎦
u/x
Page 3 of 4
where uˆ = ⎣⎡u1
T
v1 θ1 u2
v2 θ 2 ⎦⎤
Hence,
⎡ 3
⎢
⎢ 2L
⎢ 1
B=⎢
⎢ 2L
⎢ h1
⎢ x
⎢⎣
⎡ 3
⎢
⎢ 20
⎢ 1
=⎢
⎢ 20
⎢ h1
⎢ x
⎢⎣
where x = h1 x1 + h2 x2 =
1
2L
3
2L
1
− (1 + r ) r = 0
2
0
0
−
1
20
3
20
−
−
st
2L
st
20
1
−
2
−
0
0
3
20
1
−
20
h2
x
−
3
2L
1
−
2L
h2
x
1
2L
3
−
2L
1
20
3
−
20
st ⎤
⎥
20 ⎥
1⎥
− ⎥
2⎥
⎥
0 ⎥
⎥⎦
−
0
0
⎤
⎥
⎥
⎥
1
− (1 − r ) r = 0 ⎥
2
⎥
⎥
0
⎥
⎥⎦
st
2L
1
1
40 − 5 3 5 3
(1 + r )(20) + (1 − r )(20 − 10 cos 30° ) =
+
r
2
2
2
2
The corresponding stress-strain matrix is (for a plane stress with a hoop strain)
0
⎡1
⎢
1 −ν
E ⎢
0
C=
2
1 −ν ⎢
2
⎢ν
0
⎣
ν⎤
⎥
0⎥
⎥
1 ⎥⎦
Page 4 of 4
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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