Page 56 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers

IntroductIon to FInIte element AnAlysIs • 43
Element (3)

j = 3

m = 5 (3)

i = 2

A= 1 bh
2
1 2
A= (250 )(250 )=31250 mm
2
For element (3), we have coordinates x = 500, y = 0, x = 250, y = 125,
m
i
m
i
x = 500 and y = 250 because the global axes are set up at node 1, and
j
j
b = y − y =125
j
i
m
b = y − y =125
i
m
j
b = y − y =− 250
m
i
j
g = x − x =− 250
m
i
j
g = x − x = 250
m
j
i
g = x − x =0
i
j
m
Therefore, substituting in [B] matrix,
 125 0 125 0 −250 0 
B []= 1   0 −250 0 250 0 0  
62500
 −  250 125 250 125 0 −250  
 1 0 3 . 0 
D []= 210000   03 . 1 0  
091
.
  0 0 035  
.
 125 0 −250
 
 0 −250 125   1 0 3 . 0 
T
B [] []= 210000   125 0 250     03 . 1 0  
D
(.
62500 091)  0 250 125   035
 −250 0 0   0 0 . 
0
 
  0 0 − 250 