2.094
FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS
SPRING 2008
Homework 3 - Solution
Instructor:
Assigned:
Due:
Prof. K. J. Bathe
02/21/2008
02/28/2008
Problem 1 (10 points):
From the plane strain condition,
1
 zz   ( xx   yy )  0
E

 zz 

 zz   ( xx   yy )
The strains are
 xx 
1
1
 xx   ( yy   zz ) 
(1  2 ) xx  (1   ) yy  4.7667  107
E
E
 yy 
1
1
 yy  ( xx   zz ) 
(1  2 ) yy  (1   ) xx  4.3333  108
E
E




 xy 




1
2(1   )
 xy 
 xy  8.6667  107
G
E
The displacements in the four-node element can be written as
u ( x, y )  a0  a1 x  a2 y  a3 xy
v( x, y )  b0  b1 x  b2 y  b3 xy
(*Note that you can also use the following form of displacements as in the class.
u ( x, y )  h1 ( x, y )u1  h2 ( x, y )u2  h3 ( x, y)u3  h4 ( x, y)u4
v( x, y )  h1 ( x, y )v1  h2 ( x, y )v2  h3 ( x, y )v3  h4 ( x, y )v4
But, then equations are slightly complicate to solve. Please see attached sample solution which uses this method)
Page 1 of 4
Then,
 xy 
 xx 
u
 a1  a3 y  4.7667  107
x
 yy 
v
 b2  b3 x  4.3333  108
y
u v

 a2  a3 x  b1  b3 y  8.6667  107
y x
Since these equations hold for all x and y,
a3  b3  0
a1  4.7667  107
b2  4.3333  108
a2  b1  8.6667  107
Therefore,
u ( x, y )  a0  4.7667  10 7 x  a2 y
v( x, y )  b0  b1 x  4.3333  10 8 y
The boundary conditions with setting the node 3 to the origin of the coordinate system are
u (0, 0)  a0  0
v(0, 0)  b0  0
v(5, 0)  b0  5b1  0
Finally,
u ( x, y )  4.7667  107 x  8.6667 107 y
v( x, y )  4.3333  108 y
The displacements at each node are
Node 1 (x=5, y=8)
Node 2 (x=0, y=8)
Node 3 (x=0, y=0)
Node 4 (x=5, y=0)
u (in.)
9.3167 X 10-6
6.9334 X 10-6
0
2.3833 X 10-6
Page 2 of 4
v (in.)
3.4666 X 10-7
3.4666 X 10-7
0
0
Problem 2 (20 points):
(a) The stresses can be obtained from a bilinear interpolation.
1
1
1
1
 xx(4)  (1  x)(1  y )(298.5)  (1  x)(1  y )(624.0)  (1  x)(1  y )(1.146)  (1  x)(1  y )(324.3)
4
4
4
4
 149.8365  162.7365 x  311.4135 y  0.0135xy
1
1
1
4
4
4
 465.9900  542.4600 x  93.4900 y  0.0600 xy
1
4
 yy(4)  (1  x)(1  y )(914.9)  (1  x)(1  y)(169.9)  (1  x)(1  y)( 16.96)  (1  x)(1  y )(1102)
1
1
4
4
 289.9587  108.9912 x  189.8412 y  0.0087 xy
1
4
1
4
 xy(4)  (1  x)(1  y )(370.8)  (1  x)(1  y )(588.8)  (1  x)(1  y)(209.1)  (1  x)(1  y )(8.865)
Note that the coefficients of ‘ xy ’ in each stress must be zero because of our displacement assumption. But here
we have them due to rounding. (The strains do not have ‘ xy ’ terms because they are derivatives of the
displacements.)
B
(4)
0
0
0
0 
 (1  y )  (1  y )  (1  y ) (1  y )
1
  0
0
0
0
(1  x) (1  x)  (1  x)  (1  x) 
4
 (1  x) (1  x)  (1  x)  (1  x) (1  y)  (1  y)  (1  y) (1  y) 
F (4)
 xx(4) 
1 1
 
   B (4)T  yy(4 )  (0.1)dxdy
1 1
 x(y4) 
 
 0.00127 
 57.99301


 28.02571 


29.96603 


 100.0049 


 6.806910 
 51.18483 


 42.01317 
Page 3 of 4
(b) Consider element 4.
a. Horizontal equilibrium:
0  57.99  28.03  29.97  0
b. Vertical equilibrium:
100  6.81  51.18  42.01  0
c. Moment equilibrium about its local node 3:
100  2  57.99  2  42.01 2  0
Problem 3 (10 points):
For the element A,
𝒖𝟏 , 𝒗𝟏 , 𝒖𝟒 , 𝒗𝟒
𝑻
= 𝑼𝟏 , 𝑼 𝟐 , 𝑼𝟑 , 𝑼𝟒
𝑻
The corresponding components of the stiffness matrix are
𝒂𝟏𝟏
𝒂
𝑲𝑨 = 𝒂𝟐𝟏
𝟕𝟏
𝒂𝟖𝟏
𝒂𝟏𝟐
𝒂𝟐𝟐
𝒂𝟕𝟐
𝒂𝟖𝟐
𝒂𝟏𝟕
𝒂𝟐𝟕
𝒂𝟕𝟕
𝒂𝟖𝟕
𝑼𝟏
𝑼𝟐
𝑼𝟑
𝑼𝟒
𝒂𝟏𝟖
𝒂𝟐𝟖
𝒂𝟕𝟖
𝒂𝟖𝟖
For the element B,
𝒖𝟏 , 𝒗𝟏 , 𝜽𝟏
𝑻
= 𝑼𝟑 , 𝑼 𝟒 , 𝑼𝟓
𝑻
The corresponding components of the stiffness matrix are
𝒃𝟏𝟏
𝑲𝑩 = 𝒃𝟐𝟏
𝒃𝟑𝟏
𝒃𝟏𝟐
𝒃𝟐𝟐
𝒃𝟑𝟐
𝒃𝟏𝟑 𝑼𝟑
𝒃𝟐𝟑 𝑼𝟒
𝒃𝟑𝟑 𝑼𝟓
Then the global stiffness matrix is
𝒂𝟏𝟏
𝒂𝟐𝟏
𝑲 = 𝒂𝟕𝟏
𝒂𝟖𝟏
𝟎
𝒂𝟏𝟐
𝒂𝟐𝟐
𝒂𝟕𝟐
𝒂𝟖𝟐
𝟎
𝒂𝟏𝟕
𝒂𝟐𝟕
𝒂𝟕𝟕 + 𝒃𝟏𝟏
𝒂𝟖𝟕 + 𝒃𝟐𝟏
𝒃𝟑𝟏
𝒂𝟏𝟖
𝒂𝟐𝟖
𝒂𝟕𝟖 + 𝒃𝟏𝟐
𝒂𝟖𝟖 + 𝒃𝟐𝟐
𝒃𝟑𝟐
Page 4 of 4
𝟎
𝟎
𝒃𝟏𝟑
𝒃𝟐𝟑
𝒃𝟑𝟑
𝑼𝟏
𝑼𝟐
𝑼𝟑
𝑼𝟒
𝑼𝟓
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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