2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 9 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
04/17/2008
04/24/2008
Problem 1 (10 points):
H=
S
1
[ (1 + r )(1 + s) (1 − r )(1 + s) (1 − r )(1 − s) (1 + r )(1 − s)]
4
H =H
⎡
= ⎢0 0
s =−1
⎣
(1 − r )
2
(1 + r ) ⎤
2 ⎥⎦
s ⎤
⎡
⎢2
1
2 ⎥
J=⎢
⎥ from y = 2(6 + r ) and z = (6 + 5s + rs)
2
⎢0 5 + r ⎥
⎢⎣
2 ⎥⎦
det J = 5 + r
⎡5+ r
1 ⎢
B=
2
5+ r ⎢
⎣ 0
⎡1 + s
s⎤⎢
− ⎥ 4
2 ⎢
⎥ 1+ r
2 ⎦⎢
⎣⎢ 4
1+ s
4
1− r
4
−
1− s
4
1− r
−
4
−
Page 1 of 6
1− s ⎤
4 ⎥
⎥
1+ r ⎥
−
4 ⎦⎥
⎡k 0⎤
k=⎢
⎥
⎣0 k ⎦
The conductivity matrix is
K = ∫ B k B ydydz = ∫
k
T
V
+1
−1
∫
+1
−1
B k B (5 + r ) {2(6 + r )} drds
T
The convection matrix is
+1
c
sT
s
sT
s ⎛ 17 ⎞
K = ∫ hH H dS = ∫ hH H ⎜⎜
⎟⎟ {2(6 + r )} dr
Sc
−1
2
⎝
⎠
s
because dS = ydl = y det J dr
⎡ ⎛ ∂y ⎞ 2 ⎛ ∂z ⎞ 2 ⎤
where det J = ⎢⎜
⎟ +⎜ ⎟ ⎥
⎢⎣⎝ ∂r ⎠ ⎝ ∂r ⎠ ⎥⎦
1/2
s
=
s =−1
∂y
∂z s
17
= 2 and
=
with
∂r
∂r 2
2
The heat flow load vector is
Q = QB + Q
e
where
Q B = ∫ H q B dV = ∫
T
V
+1
−1
∫
+1
−1
H q B {2(6 + r )} (5 + r )drds
T
⎛ 17 ⎞
+1
e
sT
s
sT
s
Q = ∫ hH H θˆe dS = ∫ hH H θˆe {2(6 + r )} ⎜⎜
⎟⎟ dr
Se
−1
⎝ 2 ⎠
Page 2 of 6
Problem 2 (10 points):
Note that φ is zero on the boundary of the shaft. Therefore we have only one degree of freedom at node 1, φ1 .
Because of the symmetry, we need to consider only one element.
We have for element 1,
1
4
φ = (1 + r )(1 + s)φ1 = H (1) φˆ
⎡ ∂φ ⎤ ⎡ ∂φ ⎤ ⎡ 1
⎤
⎢ ∂x ⎥ ⎢ ∂r ⎥ ⎢ 4 (1 + s ) ⎥
(1)
φ'= ⎢ ⎥ = ⎢
⎥=⎢
⎥ ⎡⎣φ1 ⎤⎦ = B φˆ
⎢ ∂φ ⎥ ⎢ 2 ∂φ ⎥ ⎢ 1 (1 + r ) ⎥
⎦⎥
⎣⎢ ∂y ⎥⎦ ⎣⎢ 3 ∂s ⎦⎥ ⎣⎢ 6
Then FE governing equation for a unit length of the shaft is
⎧
⎡ 1
⎪⎪ +1 +1 ⎡ 1
1
⎤ ⎢ 2G
(1 + r ) ⎥ ⎢
⎨ 4∫−1 ∫−1 ⎢ (1 + s )
6
⎣4
⎦⎢
⎪
0
⎪⎩
⎣⎢
+ 1 +1 1
(1 + r )(1 + s ) det Jdrds
= 4θ ∫ ∫
− 1 −1 4
⎫
⎤ ⎡1
⎤
0 ⎥ ⎢ (1 + s ) ⎥
⎪⎪
4
⎥⎢
⎥ det Jdrds ⎬ φ1
1 ⎥ ⎢1
⎪
(1 + r ) ⎥
⎪⎭
2G ⎦⎥ ⎣⎢ 6
⎦⎥
Page 3 of 6
where det J =
3
2
Hence,
φ1 =
54
Gθ
13
Thus we obtain
T = 4∫
+1
−1
∫
+1
−1
1
648
⎛ 54
⎞
(1 + r )(1 + s ) ⎜ Gθ ⎟ det Jdrds =
Gθ
2
13
⎝ 13
⎠
so that
T
θ
=
648
G
13
Problem 3 (10 points):
(1) coarse mesh (2x12)
(2) fine mesh (10x72)
(3) finest mesh (20x144)
The analytical velocity and pressure distributions are,
vθ (r ) = Ar +
( ω1 − ω2 ) r12 r22
B
ω r 2 − ω1r12
B
=
where A = 2 22
and
r22 − r12
r2 − r12
r
Page 4 of 6
⎛1
⎞
1 B2
p ( r ) = ρ ⎜ A2 r 2 −
+ 2 AB ln(r ) ⎟ + C
2
2 r
⎝2
⎠
where
⎛1
⎞
1 B2
+ 2 AB ln(r1 ) ⎟
C = p r = r − ρ ⎜ A2 r12 −
2
1
2 r1
⎝2
⎠
In this problem,
1 − 2 ) (12 )(22 )
(2)(22 ) − (1)(12 ) 7
(
4
A=
= , B=
=−
2
2
2
2
2 −1
3
2 −1
3
⎛1
⎞
11
1 (−4 / 3) 2
7⎛ 4⎞
2
2
C = 0 − (1) ⎜ ( 7 / 3) (1 ) −
+ 2 ⎜ − ⎟ ln(1) ⎟ = −
2
6
2 1
3⎝ 3⎠
⎝2
⎠
The results are compared with these analytical solutions in the following figures. Note that we obtained the
velocities very close to the analytical ones even with the coarse mesh. However the pressure deviates from the
analytical solution on the boundaries because the pressure is constant in the element we used and we averaged the
pressure at each node for the plot.
Page 5 of 6
4
3.5
Analytical
Coarse Mesh
Fine Mesh
Finest Mesh
Velocity
3
2.5
2
1.5
1
1
1.2
1.4
1.6
1.8
2
1.6
1.8
2
r
5
4.5
4
Analytical
Coarse Mesh
Fine Mesh
Finest Mesh
3.5
Pressure
3
2.5
2
1.5
1
0.5
0
1
1.2
1.4
r
Page 6 of 6
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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