Page 92 - Finite Element Modeling and Simulations with ANSYS Workbench

Beams and Frames 77

Reduced FE equation

EI 24 0  v 2   − P


2 θ 
L 3   0 8 L 2   =  M 
 



Solving this, we obtain
 v 2  L   − PL  
2
  = 24  
2 θ 

  EI  3 M  
From the global FE equation, we obtain the reaction forces and moments
3 /
 F Y1   − 12 6 L    P + M L 
2
   2   
 M 1 EI − 6 L 2 L   1  PL + M 


v 2
  = 3   =  
3 /
 
 F Y  L   − 12 − 6 L θ  4 2 P − M L 

3
2
2 
 M 3   6 L 2 L   −PL + M  
     
Stresses in the beam at the two ends can be calculated using the formula
My
σ= σ= −
x
I
Note that the FE solution is exact for this problem according to the simple beam
theory, since no distributed load is present between the nodes. Recall that
4
dv
EI 4 = qx()
dx
If q(x) = 0, then exact solution for the deflection v is a cubic function of x, which is
exactly what described by the shape functions given in Equation 3.6.

EXAMPLE 3.2
y p

1 E, I 2 x
L

Given:
A cantilever beam with distributed lateral load p as shown above.

Find:
The deflection and rotation at the right end, the reaction force and moment at the left
end.