Page 48 - Using ANSYS for Finite Element Analysis A Tutorial for Engineers

IntroductIon to FInIte element AnAlysIs • 35

 21 −21 0 0 0 
 
 −21 84 −21 −21 −21 
3
K = 10 *  0 −21 21 0 0 
 
  0 −21 0 21 0 
  0 −21 0 0 21  
The nodal displacement equations:
{F}=[K ]{d}

 F   d 
x 1
x 1
 F   d 
 2 x   2 x 




K
 F  = [] d 
x
x
3
3
 F    d 
 4 x   4 x 
  F    d 
x 
x 
5
5
 F x 1   21 − 21 0 0 0  0 
   − − 21 − 21 −  d 
 15000   21 84 1 21  2x 




 F 3 x  = 10 * 0 − 21 21 0 0   0 
3
 F   −    
 4 x   0 21 0 21 0   0 


  F 5 x    0 − 21 0 0 21  0 


F =− 21 10 * d 2 x (1)
3
*
x 1
15000 = 84 10 *d 2x (2)
3
*
F =− 21 10 * d
3
*
3 x 2 x (3)
F 4 x =− 21 10 * d (4)
3
*
x
2
Fx =− 21 10 * dx (5)
3
*
5 2
From equation (2), d = 0.17857 mm
2x
From equations (1–3–4–5), the reactions, F = F = F = F = –3.75kN
4x
3x
5x
1x
The force in each element, {f} = [k]{d}
The first element between nodal 1, 2:
 f x 1   21 − 21 d = 0

1 ()

 1 ()  = 10 *    d x 1 
3
 f  2 x    − 21 21  2 x 