AERSP 301 Finite Element Method

1. AERSP 301Finite Element Method Jose Palacios July 2008

2. Today • HW 5 has been uploaded • Due Thursday • HW 6 due after midterm…and not included on exam • No Class on Friday • Continue Finite Element Method Bars

3. Axial deformations in barsApplication of Stationary Principle • So far, for the given cantilevered bar: • Obtained 2x2 Element Stiffness Matrix • Obtained 2x1 Element Load Vector • Assembled to get NxN Global Stiffness Matrix (N = number of global DOFs) • Assembled to get Nx1 Global Load Vector • Strain Energy for entire structure • External Work Potential Note – Global Stiffness Matrix, K, is symmetric F is the Global Load Vector

4. Axial deformations in barsApplication of Stationary Principle • Now:  = U – W , Apply Stationary Principle: • Since q is a vector (of global degrees of freedom), this amounts to:

5. Axial deformations in barsApplication of Stationary Principle • Consider

6. Axial deformations in barsApplication of Stationary Principle

7. Axial deformations in barsApplication of Stationary Principle • Repeat process for other qi’s

8. Axial deformations in barsApplication of Stationary Principle • Compact form: • Then for any structure: • We are looking to solve for the q vector, but before inverting the K matrix, we need to apply the geometric boundary conditions.

9. Axial deformations in barsApplication of Stationary Principle

10. Sample Problem: Vertical Column under its own weight • Exact Solution • FEM Solution

11. Exact Solution Problem 1 Equilibrium Equation

12. Exact Solution Problem 1 Plug constant c to obtain the compressive stress (linear with x): Use the strain – displacement, stress - strain relationship: Integrate to get u(x): Bc’s: x = 0  u = 0  C2 = 0

13. FEM Solution Problem 1 q4 We use equal lengths for all the elements q3 3 l q2 For element e: 2 l q1 1 l Because Notice, because l is the same for all the elements, is the same for all the elements, and because f(x) = = constant, Qe are also the same for all the elements

14. FEM Solution Problem 1

15. FEM Solution Problem 1 q1 = 0 Reduced (because q1 = 0)

16. FEM Solution Problem 1 Solution of the 3x3 gives (check at home for yourself): Compare to Exact Solution:

17. FEM Solution Problem 1 • What do you notice when you compared the exact solution and the one obtained from FEM?

18. FEM Solution Problem 1 What does FE model says about the stresses? Recall: Within an element: Stress will be constant in each element

19. FEM Solution Problem 1 Element 1: Element 2: Element 3:

20. FEM Solution Problem 1

21. FEM Solution Problem 1 • In this sample, nodal displacements are exact. Do not expect this to happen in general. Results from FEM are usually not exact • FEM stress solution is exact at the element centroid IN THIS CASE!. Once again, do not expect this to happen in more complicated problems. • Overall, the stress solution is not as accurate as the displacement solution. Displacements are assumed, whereas the stresses are calculated from the derivatives of the displacements. The stresses are therefore necessarily a lower order function, and therefore less likely to be an accurate representation of the actual problem. • FEM stress solutions are more accurate at the element centroid because the slope of the calculated displacements is closest to the slope of the actual displacement at the center.

22. Sample Problem 2: Axial Deformation of Uniform Bar • Determine tip deflection • Exact Solution • FEM Solution • 1 element • 2 elements • Account for a discrete spring

23. Sample Problem 3: FEM of a more complex system (loaded axially)

24. The Rotated Bar Element v1 v2 u1 u2 It will be needed in Hw #5, Problem 3

25. v1 v2 u1 u2 The Rotated Bar Element v’2 u’2 v1 v’1 v2 θ u1 u’1 u2 [To]

26. The Rotated Bar Element