The F inite Element Method

1. The Finite Element Method A Practical Course CHAPTER 3: FEM FOR BEAMS

2. CONTENTS • INTRODUCTION • FEM EQUATIONS • Shape functions construction • Strain matrix • Element matrices • Remarks • EXAMPLE AND CASE STUDY • Remarks

3. INTRODUCTION • The element developed is often known as a beam element. • A beam element is a straight bar of an arbitrary cross-section. • Beams are subjected to transverse forces and moments. • Deform only in the directions perpendicular to its axis of the beam.

4. INTRODUCTION • In beam structures, the beams are joined together by welding (not by pins or hinges). • Uniform cross-section is assumed. • FE matrices for beams with varying cross-sectional area can also be developed without difficulty.

5. FEM EQUATIONS • Shape functions construction • Strain matrix • Element matrices

6. Shape functions construction • Consider a beam element Natural coordinate system:

7. Shape functions construction Assume that In matrix form: or

8. Shape functions construction To obtain constant coefficients – four conditions At x= -a or x = -1 At x= a or x = 1

9. Shape functions construction or  or

10. Shape functions construction Therefore, where in which

11. Strain matrix Eq. (2-47) Therefore, where (Second derivative of shape functions)

12. Element matrices Evaluate integrals

13. Element matrices Evaluate integrals

14. Element matrices

15. Remarks • Theoretically, coordinate transformation can also be used to transform the beam element matrices from the local coordinate system to the global coordinate system. • The transformation is necessary only if there is more than one beam element in the beam structure, and of which there are at least two beam elements of different orientations. • A beam structure with at least two beam elements of different orientations is termed a frame or framework.

16. P=1000 N 0.1 m 0.06 m E=69 GPa =0.33 0.5 m EXAMPLE Consider the cantilever beam as shown in the figure. The beam is fixed at one end and it has a uniform cross-sectional area as shown. The beam undergoes static deflection by a downward load of P=1000N applied at the free end. The dimensions and properties of the beam are shown in the figure.

17. P=1000 N E=69 GPa =0.33 0.5 m EXAMPLE Exact solution: • Step 1: Element matrices Eq. (2.59) = -3.355E10-4 m

18. P=1000 N E=69 GPa =0.33 0.5 m EXAMPLE • Step 1 (Cont’d): • Step 2: Boundary conditions

19. EXAMPLE • Step 2 (Cont’d): • Step 3: Solving FE equation (Two simultaneous equations) Therefore, K d = F, where dT = [ v22] , v2 = -3.355 x 10-4 m 2 = -1.007 x 10-3 rad

20. EXAMPLE • Step 4: Stress recovering v2 = -3.355 x 10-4 m 2 = -1.007 x 10-3 rad Substitute back into first two equations

21. Remarks • FE solution is the same as analytical solution • Analytical solution to beam is third order polynomial (same as shape functions used) • Reproduction property

22. CASE STUDY • Resonant frequencies of micro resonant transducer

23. Number of 2-node beam elements Natural Frequency (Hz) Mode 1 Mode 2 Mode 3 10 4.4058 x 105 1.2148 x 106 2.3832 x 106 20 4.4057 x 105 1.2145 x 106 2.3809 x 106 40 4.4056 x 105 1.2144 x 106 2.3808 x 106 60 4.4056 x 105 1.2144 x 106 2.3808 x 106 Analytical Calculations 4.4051 x 105 1.2143 x 106 2.3805 x 106 CASE STUDY [ K - lM ]f = 0 Section 3.6, pg. 58

24. CASE STUDY

25. CASE STUDY

26. CASE STUDY