ME 475 Computer Aided Design of Structures Finite Element Analysis of Trusses – Part 2

1. ME 475Computer Aided Design of StructuresFinite Element Analysis of Trusses – Part 2 Ron Averill Michigan State University

2. Learning Objectives • Describe why a truss element has multiple DOFs per node • Develop the 2D vector transformation equations • Use the 2D vector transformation equations to develop the finite element equations of a plane truss element

3. Plane Truss Element DOFs The nodes of a plane truss element can displace in both the x and y directions, even though each element elongates (strains) in only the local x-direction. The element rotation is a rigid body motion. blue – undeformed red – deformed F

4. Plane Truss Element DOFs In order to describe the motion in two dimensions, we need two components of displacement, or two degrees of freedom (DOFs), per node. We define two local DOFs at each node: is the x-component of displacement at the ithnode (i = 1,2) is the y-component of displacement at the ithnode (i = 1,2) Similarly, there will be two components of force at each node.

5. 2D Plane Truss Elements – DOFs Plane truss elements have two DOFs per node: Y x y e 2 1 X θ

6. 2D Plane Truss Elements – Forces Plane truss elements have two internal force components per node: Y x y e 2 1 X θ

7. Expanded Bar Finite Element Equations To accommodate the extra DOFs and forces, we expand the 1D bar finite element equations as follows: Note the order of the local DOFs and forces. Displacements and forces at the same node are clustered together. These equations are equivalent to the 1D bar equations, and they are still written with respect to local coordinates. So we can express these equations as:

8. Need for a Common Coordinate System To simplify the enforcement of displacement continuity at nodes, we need to express all local DOFs in a common coordinate system. 1 1 Y 2 2 3 3 X 45° The boolean array is: 45°

9. 2D Vector Transformations Our common coordinate system will be the global XY system. Y y x X We can transform the nodal displacements as follows: θ

10. 2D Vector Transformations Let’s define and Then we can write the transformation equations in matrix form as: Inverting: Note that transformations occur at a point – a node in this case. So we will apply these transformations at each node.

11. Transformation of Planar Truss DOFs In terms of the nodal DOFs, we have: Or: Where is called the transformation matrix. Because is orthogonal, it has the special property:

12. Transformation of Planar Truss Forces Similarly for forces, we have: Or: Where is the same transformation matrix as before.

13. 2D Planar Truss FE Equations Recall the expanded 4x4 system of bar equations from page 5: Introducing the transformations just developed: Premultiply both sides by Simplifying:

14. 2D Planar Truss FE Equations Define: So we have the final form: These are the local (element level) finite element equations in terms of the global coordinate system. Just as before: is the stiffness matrix of an element is the element displacement vector is the vector of internal element forces

15. 2D Planar Truss FE Equations We can expand the equations as: These equations describe the behavior of an arbitrary 2D planar linear truss element in terms of the global coordinate system.

16. Exercise Determine the element stiffness matrix for each of the truss elements shown below. 2 3 Y 1 2 1 3 4 P = 10,000 lbsX Elements 1 and 3 have length 10ft. For all elements, E=30E6 psi and A = 2 in2. The boolean array is: 45° 45°

17. Solution First we determine the orientation of each element, based on the direction of local x as defined in the boolean array. 2 3 Y 1 2 1 3 4 P = 10,000 lbsX So the orientations are: 45° 45°

18. Solution In general, recall that: Element 1: E = 30E6 psi, A = 2 in2, h = 120 in, θ = 90°

19. Solution Element 2: E = 30E6 psi, A = 2 in2, h = 120 in, θ = 45° Element 3: E = 30E6 psi, A = 2 in2, h = 120 in, θ = 0°