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