Finite Element analysis of Spring Assembly

1. FINITE ELEMENT ANALYSIS OF
SPRING ASSEMBLY
BY-
Ms. JAPE ANUJA S.
ASSISTANT PROFESSOR,
CIVIL ENGINEERING DEPARTMENT,
SRES, SANJIVANI COLLEGE OF ENGINEERING,
KOPARGAON-423603.
MAID ID: anujajape@gmail.com
japeanujacivil@sanjivani.org.in

2. FINITE ELEMENT ANALYSIS OF SPRING ASSEMBLY
• Springs are 1D structures subjected to axial force only.
• The degree of freedom at each node is one i.e. axial displacement.
• Stiffness matrix for spring element having stiffness constant k is given below which can be
obtained by giving unit displacement one by one at each node.
• Let consider a two noded spring element with ui and uj displacements at each nodes.
• Let unit displacement at node i
k11=k
k21=-k
• Let unit displacement at node j
k12=k
k22=-k
11 12
21 22
1 1
1 1
k k
K
k k
K k
 
  
 
 
   

3. FINITE ELEMENT ANALYSIS OF SPRING ASSEMBLY
    K f 
 f

4. EXAMPLE
 Determine elongations at each node and hence the forces in
springs.
Solution-
1. Discretization
5N
k1=500N/m k2=100N/m
1 2 3
Element k (N/m) Nodes
Displacements
(m)
Boundary
conditions
1 500 1,2 u1, u2 u1=0
2 100 2,3 u2, u3 ----

5. 2. Element stiffness matrices
u1 u2 u2 u3
3. Global stiffness matrix
u1 u2 u3
4. Reduced stiffness matrix
Imposing the boundary condition u1=0, i.e. eliminate first row and first column
Therefore reduced stiffness matrix is
 1 1
1 1 500 500 1
1 1 500 500 2
u
K k
u
    
        
 2 2
1 1 100 100 2
1 1 100 100 3
u
K k
u
    
        
 
500 500 0 1
500 600 100 2
0 100 100 3
u
K u
u
 
    
  
 
600 100
100 100
K
 
   

6. 5. Determine unknown joint displacements
Applying Equation of Equilibrium
6. Calculation of spring force
For Spring 1
(u1=0 and u2=0.01)
For Spring 2, and
    
600 100 2 0
100 100 3 5
K f
u
u
 
     
         
     
 
 
1 11
1
2
1
2
500 500 1
500 500 2
5
5
K f
fu
fu
f N T
f N T
 
     
         
 

 2 5f N T  3 5f N T
 
 
2 0.01
3 0.06
u m
u m
  
 
5N5N
5N5N

10. ASSIGNMENT

11. ASSIGNMENT