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ME 475 Computer Aided Design of Structures Finite Element Analysis of Trusses – Part 3

ME 475 Computer Aided Design of Structures Finite Element Analysis of Trusses – Part 3. Ron Averill Michigan State University. Learning Objectives. Describe displacement continuity and force equilibrium at the nodes of a planar truss system

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ME 475 Computer Aided Design of Structures Finite Element Analysis of Trusses – Part 3

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  1. ME 475Computer Aided Design of StructuresFinite Element Analysis of Trusses – Part 3 Ron Averill Michigan State University

  2. Learning Objectives • Describe displacement continuity and force equilibrium at the nodes of a planar truss system • Assemble the system level finite element equations for a planar truss

  3. Displacement Continuity at Nodes Continuity (compatibility) of displacement components must be enforced at each node in a truss. Consider the example below: 2 3 Y 1 2 1 3 4 X Continuity conditions:

  4. Force Equilibrium at Nodes Equilibrium of internal and external force components must be enforced at each node in a truss. Consider the example below: 2 3 Y 1 2 1 3 4 X Equilibrium conditions:

  5. Assembly of Global FE Equations The assembly process will again make use of the Boolean array. Recall that the Boolean array maps each local node to its corresponding global node. It does not directly map local DOFs to global DOFs. So when we have more than one DOF per node, a modified assembly process is needed compared to what we did for one DOF systems. If our nodal DOFs are clustered together in the element and global FE arrays, then we can use the Boolean to assemble in a manner very similar to the case in which we have only DOF per node.

  6. Assembly of Global FE Equations The element level stiffness matrix for a 2-noded truss element is 4x4. If there are 4 nodes in the truss system, then the global stiffness matrix is 8x8.

  7. Assembly of Global FE Equations Because the DOFs at each node are clustered together in and , we can partition the stiffness matrices into 2x2 “chunks” that correspond to the DOFs at each node.

  8. Assembly of Global FE Equations Now, we can assemble these 2x2 chunks in the same way that we did for 1x1 chunks when there was only one DOF per node. 1 1 1 2 2 1 2 2 1 1 1 3 1 2 1 4 2 1 2 3 2 2 2 4 3 1 3 3 3 2 3 4 4 1 4 3 4 2 4 4

  9. Assembly of Global FE Equations Consider an element defined by global nodes 3 and 1. We have: 11 -> 33, 12 -> 31, 21 -> 13, 22 -> 11

  10. Exercise For the truss shown below, assemble the stiffness matrix for element 2 into the global system. 2 3 Y 1 2 1 3 4 X Elements 1 and 3 have length 10ft. For all elements, E=30E6 psi and A = 2 in2. The boolean array is:

  11. Exercise

  12. Solution Element 2 is defined by global nodes 1 and 3. We have: 11 -> 11, 12 -> 13, 21 -> 31, 22 -> 33

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