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

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 constructionor  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. 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.
P=1000 N
0.5 m
0.06 m
0.1 m
E=69 GPa
=0.33

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

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

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

20. EXAMPLE Step 4: Stress recovering v2 = -3.355 x 10-4 m
2 = 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. CASE STUDY [ K - lM ]f = 0 Section 3.6, pg. 58
Number of 2-node beam elements
Natural Frequency (Hz)
Mode 1
Mode 2
Mode 3 10
x 105
x 106
x 106 20
x 105
x 106
x 106 40
x 105
x 106
x 106 60
Analytical Calculations
x 105
x 106
x 106
[ K - lM ]f = 0 Section 3.6, pg. 58

24. CASE STUDY

25. CASE STUDY

26. CASE STUDY