Page 58 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers

IntroductIon to FInIte element AnAlysIs • 45

b = y − y =−125
m
j
i
g = x − x =− 500
m
i
j
g = x − x = 250
m
j
i
g = x − x = 250
j
i
m
Therefore, substituting in [B] matrix,
 0 0 125 0 −125 0 
B []= 1   0 −500 0 250 0 250   
62500
 −  500 0 250 125 250 125 
 1 0 3 . 0 
D []= 210000   03 . 1 0  
.
091
  0 0 035  
.
 0 0 −500
 
 0 −500 0   1 0 3 . 0 
T
D
Then, B [] []= 210000   125 0 250     03 . 1 0  
(.
62500 091)  0 250 125   035
 −125 0 250   0 0 . 
0
 
  0 250 − 125 
 0 0 −175 
 
 −150 −500 0 
T
B
D
∴[] []=3 6923   125 37 5 . 875 .  
.
 75 250 43 75 
.
 −125 −37..5 875 
.
 
  75 250 − 43 .75 
The stiffness matrix for element (4) is:
 0 0 −175 
 
 −150 −500 0   0 0 125 0 − −125 0 
)
.
5
k []=()(31250 )(3 6923   125 37 5 . 875 .   × 1   0 −500 0 250 0 250  
.
 75 250 43 75  62500 −  500 0 250 125 250 125 
 − −125 −37 5 . 875 .   
 
  75 250 −43 75  
.