Fundamental Concept of Finite Element Method By

Fundamental Concept of Finite Element Method By Gurunath V Shinde Assistant Professor (DACOE,Karad,India)

“Finite Element Method is a numerical method to solve differential and integral equations since behavior of almost all physical system can be represented by these equations” A number of popular brand of finite element analysis packages are now available commercially. Some of the popular packages are STAAD-PRO, GT-STRUDEL, ABAQUS,NASTRAN, NISA and ANSYS. Using these packages one can analyse several complex structures.

Need of FEA packages: 1. Which elements are to be used for solving the problem in hand. 2. How to discritise to get good results. 3. How to introduce boundary conditions properly. 4. How the element properties are developed and what are their limitations. 5. How the displays are developed in pre and post processor to understand their limitations. 6. To understand the difficulties involved in the development of FEA programs and hence the need for checking the commercially available packages with the results of standard cases.

Finite Element Methods, Finite Difference Method and Boundary Element method are the Numerical Methods of solving engineering problems Boeing has started to use FEM

Applications of FEM:

I ) Steady state problems(Equilibrium Problems): Body under equilibrium can be analyzed and it’s distortion(displacements) predicted it is then possible to derive stress and strain developed in body. Ex. Stress analysis of shafts,gears,Thermal analysis of heat exchangers, stress analysis of aircraft wings and frames etc. II) Eigen Value Problems: Calculation of fundamental characteristic of body such as natural frequency and modes shapes, buckling loads of structures, Ex. Modal analysis of engine components, Natural frequency and bucking loads of building structures etc. III)Transient Problems(Propagation Problems): Loads can be function of time and FEM is used to analyze such systems under forced conditions Ex. Crash test of car, Dynamic Crack propagation, Seismic analysis of building structure etc.

1.Simple Spring Mass Damper System: General Form of Equation of Spring System : K.U =F K-Stiffness U –Nodal Displacements F-Force

2.Heat Conduction through Composite Wall Q= (KA/L) T Q-Heat Flow Rate K-Thermal Conductivity A-Cross Section Area L-Length of Element T-Nodal Temp.

T

T K,A,L,

FEA Model Fj=AE/L Uj Fi=-Fj…………..Tension Fi=AE/L Ui Fj= -Fi………Compression Stress and Strain , σ = E.ε P/A=E.u/L P=AE/L .u

We know, P=K.u Hence, K=AE/L……(Stiffness Constant)

…..Tension ……Compression

FEA Procedure : I ) Pre Processing : 1) Discretization of Problem 2) Selection of Approximation Function(Linear or Non Linear) 3) Formulation of Basic Element Equation( behavior of system) II ) Analysis

4) Formation of Global System Equation 5) Incorporation of boundry conditions 6) Solution of System equations III ) Post Processing 7) List of Results 8) Report formation

Stress Analysis of Stepped Bar:

A=10 mm2 E=2 Gpa L=100mm

A=20 mm2 L=100 mm E=2 Gpa

1

2 FEA Model

10 N

3

I ) Pre Processing : 1) Discretization of Problem Element i j 1 1 2 2 2 3 2) Selection of Approximation Function F=K.u 3) Formulation of Basic Element Equation 1 2 1 -1 1 4 -4 X 104 N/mm K1= (A1 E1/L1 ) -1 1 2 -4 4 K2= (A2 E2/L2 ) 1 -1 2 2 -2 X 104 N/mm -1 1 3 -2 2 2 3

II ) Analysis 4) Formation of Global System Equation [K]=

4 -4 0 -4 4+2 -2 0 -2 2

U1 U2

X 104 N/mm

U3

U1 U2 U3 5) Incorporation of boundary conditions U1=0 F3 =10N

4 -4 0 -4 6 -2 0 -2 2

X 104

0 -U2 U3

=

F1 F2 F3

= 104

F

=

4 -4 0 -4 4+2 -2 0 -2 2 K

U1 U2 U3 u

0 0 10

6) Solution of System equations

4U2 X 104=0 …………..1 (-6 U2-2 U3 ) X 104 =0 …………...2 (2 U2+2 U3) X 104 =10 ……………3

U1 =0, U2 =0.25 X 10-3, U3 = 0.75 X 10-3

ɛ1 =(-u1+u2)/L=(0.25X10-3)/100=2.5X10-6 ɛ2=(-u2+u3)/L=(0.75X10-3)/100=5X10-6 σ1=E ɛ1=0.5 N/mm2 σ2=E ɛ2=1 N/mm2

Theoretical Stresses σ1=P/A1=10/20=0.5N/mm2

σ2=P/A3=10/10=1N/mm2

III ) Post Processing 7) List of Results i. Deformation U1 =0, U2 =0.25 X 10-3, U3 = 0.75 X 10-3 ii. Stress and

Strain

Ɛ1 =2.5X10-6 ɛ2=5X10-6

σ1=0.5 N/mm2 σ2=1 N/mm2