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

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

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.

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.

FEM EQUATIONS Shape functions construction Strain matrix Element matrices

Shape functions construction Consider a beam element Natural coordinate system:

Shape functions construction Assume that In matrix form: or

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

Shape functions construction or  or

Shape functions construction Therefore, where in which

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

Element matrices Evaluate integrals

Element matrices Evaluate integrals

Element matrices

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.

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

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

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

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 = -3.355 x 10-4 m 2 = -1.007 x 10-3 rad

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

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

CASE STUDY Resonant frequencies of micro resonant transducer

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 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 Analytical Calculations 4.4051 x 105 1.2143 x 106 2.3805 x 106 [ K - lM ]f = 0 Section 3.6, pg. 58

CASE STUDY

CASE STUDY

CASE STUDY