Page 94 - Finite Element Modeling and Simulations with ANSYS Workbench

Beams and Frames 79

From the FE equation, we can calculate the reaction force and moment as


 F Y1  EI − 12 6 L    pL 2 / 
v 2

  = 3  − 6 2 2   =  5 2 / 
2 θ 

 M 1   L  L L    pL 12 
where the result in (A) has been used. This force vector gives the total effective nodal
forces, which include the equivalent nodal forces for the distributed lateral load p
given by
 −pL/2 
 2 
 −pL /12  
The correct reaction forces can be obtained as follows:

/
/
 F Y1   pL 2   − pL 2   pL 
  =  5 2 /  −  − 2 /  =  2 2 


 M 1    pL 12   pL 12   pL /2  
Check the results: Draw the FBD for the FE model (with the equivalent nodal force
vector) and check the equilibrium condition.
pL/2 pL/2

2
2
5pL /12 pL /12

EXAMPLE 3.3
y
P
1 E, I 2
1 2 3 k x
L L 4

Given:
P = 50 kN, k = 200 kN/m, L = 3 m, E = 210 GPa, I = 2 × 10 m .
−4
4
Find:
Deflections, rotations, and reaction forces.

Solution
The beam has a roller (or hinge) support at node 2 and a spring support at node 3. We
use two beam elements and one spring element to solve this problem.
The spring stiffness matrix is given by

v 3 v 4
 k − k
s k =  − 
 k k 