INTRODUCTION TO FINITE ELEMENT METHOD

VIJAYAVITHAL BONGALEDEPARTMENT OF MECHANICAL ENGINEERING

MALNAD COLLEGE OF ENGINEERING

HASSAN - 573 202.

Mobile : 9448821954

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Experimental Investigation:

• Expensive and impossible

• Performance on small scale model

• Small scale model do not simulate all the features of the full scale difficulties of measurement

Theoretical Calculation:

• Low cost and fast

• Complete information

• Ability to simulate realistic condition

• Ability to simulate ideal model

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Methods of Analysis:

• Exact methods( e.g. Separation of variables and Laplace transformation methods)

• Approximate methods ( e.g. Rayleigh –Ritz and Galerkin methods)

Analytical Methods

• Finite Difference method

• Finite Element Method

• Boundary Element Method

• Finite Volume Method

• Spectral Method

• Mesh Free Method

Numerical Methods

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Why FEM?

• In nature every phenomenon is governed by laws of

physics, in terms algebraic, differential or integral

equations relating various quantities of interest.

• Examples

1. Dynamic behavior of the body- Newton’s law of motion

2. Heat conduction for analysis of temperature distribution in

solids -Fourier’s law

3. Study of motion of viscous fluids- Navier-Stokes equations

4. Structural Analysis – Force-stress-strain relations

• Derivation of these governing equations from basics

is easier but their solution by exact methods is a

formidable task

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• Used in problems where analytical

solution not easily obtained.

• Mathematical expressions required for

solution not simple because of complex:

– geometries

– loadings

– material properties

Why FEA?

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What is Finite Element Method?

• FEM is a numerical analysis techniquefor obtaining approximate solutions toa wide variety of engineeringproblems.

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FEA: Basic concept

• Replace continuous geometry with a set of objects with

a finite number of DOF

• Divide body into finite number of simpler units

(elements).

• Elements connected at nodal points

– points common to two or more adjacent

elements

– set of elements referred to as “mesh”

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Example of FEA Mesh

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Applications of FEM:

1. Equilibrium problems or time independent problems.e.g. i) To find displacement distribution and stressdistribution for a mechanical or thermal loading insolid mechanics. ii) To find pressure, velocity,temperature, and density distributions of equilibriumproblems in fluid mechanics.

2. Eigenvalue problems of solid and fluid mechanics.e.g. i) Determination of natural frequencies andmodes of vibration of solids and fluids. ii) Stabilityof structures and the stability of laminar flows.

3.Time-dependent or propagation problems of continuummechanics.e.g. This category is composed of the problems thatresults when the time dimension is added to theproblems of the first two categories.

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Engineering applications of the FEM:

• Civil Engineering structures

• Air-craft structures

• Heat transfer

• Geomechanics

• Hydraulic and water resource engineering and

hydrodynamics

• Nuclear engineering

• Biomedical Engineering

• Mechanical Design- stress concentration problems,

stress analysis of pistons, composite materials,

linkages, gears, stability of linkages, gears and

machine tools. Cracks and fracture problems under

dynamic loads etc8/14/2015 12

Advantages of Finite Element Method

• Model irregular shaped bodies quite easily

• Can handle general loading/ boundary conditions

• Model bodies composed of composite and

multiphase materials because the element equations

are evaluated individually

• Model is easily refined for improved accuracy by

varying element size and type

• Time dependent and dynamic effects can be

included

• Can handle a variety nonlinear effects including

material behavior, large deformation, boundary

conditions etc.

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Disadvantages:

• Needs computer programmes and computer

facilities

• The computations involved are too numerous for

hand calculations even when solving very small

problems

• Computers with large memories are needed to

solve large complicated problems

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Similarities that exists between various

types of engineering problems:

1. Solid Bar under Axial Load

areasectionalcrossisAand

nt,displacemeaxialisu

modulus,sYoung'theisE,Where

0,x

uAE

x

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2. One – dimensional Heat Transfer

areasectionalcrossisAande,temperaturisT

ty,conductivithermaltheisK,Where

equationLaplace0,x

TKA

x

3. One dimensional fluid flow

x

ΦuandareasectionalcrossisAand

,functionpotentialisφ,densitytheisρ

,Where0,x

ΦρA

x

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General Steps to be followed while

solving a structural problem by using

FEM:

1. Discretize and select the element type2. Choose a displacement function3. Define the strain/displacement and stress/ strain

relationships4. Derive the element stiffness matrix and equations

by using direct or variational or Galerkin’s approach5. Assemble the element equations to obtain the

global equations and introduce boundary conditions6. Solve for the unknown degrees of freedom or

generalized displacements7. Solve for the element strains and stresses8. Interpret the results8/14/2015 18

Step 1. Discretize and Select element type

1. Dividing the body into an equivalent

system of finite elements with associated

nodes

2. Choose the most appropriate element

type

3. Decide what number, size and the

arrangement of the elements

4. The elements must be made small enough

to give usable results and yet large

enough to reduce computation effort8/14/2015 19

Step 2. Selection of the displacementfunction

1. Choose displacement function within

the element using nodal values of the

element

2. Linear, quadratic, cubic polynomials can

be used

3. The same displacement function can be

used repeatedly for each element

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Step 3. Define the strain/displacement andstress/strain relationships

1. Strain/displacement and stress/strain

relationships are necessary for deriving the

equations for each element

2. In case of 1-D, deformation, say in x-direction

is given by,

3. Stress / strain law is , Hooke’s law given by,

dx

dux

xxE

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Step 4. Derive element stiffness matrix andequations

1. Following methods can be used

– Direct equilibrium method

– Work or energy methods

– Method of weighted residuals such as

Galerkin’s method

2. Any one of the above methods will produce

the equations to describe the behavior of an

element8/14/2015 22

3. The equations are written conveniently in matrix

form as,

nnnn

n

n

n

nd

d

d

d

kk

kkkk

kkkk

kkkk

f

f

f

f

3

2

1

1

3333231

2232221

1131211

3

2

1

or in compact matrix form as dkf 8/14/2015 23

Step 5. Assemble the element equations to obtain

the global or total equations and introduce

boundary conditions

1. The element equations generated in the step 4

can be added together using the method of

superposition

2. The final assembled or global equations will

be of the matrix form

3. Now introduce the boundary conditions or

supports or constraints

4. Invoking boundary conditions results in a

modification of the global equation

dKF

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Step 6. Solve for the unknown degrees offreedom or generalized displacements

• After introducing boundary conditions, we get a set

of simultaneous algebraic equations and these

equations can be written in the expanded form as

nnnn

n

n

n

nd

d

d

d

KK

KKKK

KKKK

KKKK

F

F

F

F

3

2

1

1

3333231

2232221

1131211

3

2

1

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Step7. Solve for the element strain and stress

The above equations can be solved for unknown

degrees of freedom by using an elimination

method such as Gauss ‘s method or an iteration

method such as the Gauss-Seidel method

Secondary quantities such as strain and stress ,

moment or shear force can now be obtained

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Step 8. Interpret the results

1. The final goal is to interpret and analyze

the results for use in design / analysis

process.

2. Determine the locations where large

deformations and large stresses occur in

the structure

3. Now make design and analysis decisions

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Introduction to FEM –ME 703 Course Contents

Computer Programmes for the FEM

1. Algor2. ANSYS – Engineering Analysis System3. COSMOS/M4. STARDYNE5. IMAGES-3D6. MSC/NASTRAN- NASA Structural Analysis7. SAP90- Structural Analysis Programme8. GT- STRUDL – Structural Design Language9. SAFE- Structural Analysis by Finite Elements10.NISA- Non linear Incremental Structural Analysis etc.

Matrix Algebra:

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Matrix is an mxn array of numbers arranged in m rows

and n columns

mn3m2m1m

n3333231

n2232221

n1131211

aaaa

aaaa

aaaa

aaaa

a

A matrix is represented by

ijaoraora

Matrix types:• Rectangular matrix if m ≠ n

• Row matrix if m = 1 and n > 1

• Column matrix if m > 1 and n = 1

• Square matrix if m=n

Row matrices and rectangular matrices are denoted

by using brackets and column matrices are

denoted by using braces

In FEM

• Force matrices and Displacements matrices are

Column matrices

• Stiffness matrix is a square matrix

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Matrix operations:

• Multification of a matrix by a scalar

• Addition of matrices- Matrices of same order

• Multification of matrices- If two matrices to be

multiplied then the number of columns in one

matrix must be equal to the number of rows in the

other.

• Transpose of a matrix- by interchanging rows and

columns

• Another important relationship that involves

transpose is8/14/2015 31

abbaC

axbbxa

Tjiij

aa

T-

TT

abba

• Symmetric matrix

• Unit matrix or identity matrix – a square matrix with

each element of the main diagonal to 1 and all other

elements equal to zero and is denoted by matrix I

• Diagonal Matrix- A square matrix with non-zero

elements only along the principle diagonal

• Upper triangular matrix- matrix with the elements

below the principle diagonal are all zero

• Differentiating a matrix-differentiating every element

in the conventional manner

• Integrating a matrix- integrating every element in the

conventional manner

• Determinant of a matrix

• Inverse of a matrix-

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T

aa

a

aadj

a

ca

T

1

• Inverse of a matrix is also obtained by row reduction

or Gauss-Jordan method

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Methods for solution of

simultaneous linear equations

• The general form of a set of equations will

be,

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nnnn2n21n1

3n3n232131

2n2n222121

1n1n212111

cxaxaxa

cxaxaxa

cxaxaxa

cxaxaxa

termssiderightknownthearec

andsxunkownstheoftcoefficienthearea

Where,

i

'

jij

• In Structural problems-

aij’ s are the stiffness coefficients kij

’ s ,

xj’ s are the unknown nodal displacements di

’ s and

ci’ s are the known nodal Fi

’ s (forces)

• If c’s are not all zeros, the set of equations is non-

homogeneous and all equations must be

independent to yield a unique solution (Stress

analysis).

• If c’s are all zeros, the set of equations is

homogeneous and no trivial solutions exists only if

all equations are not independent. Buckling and

vibration problems involve homogeneous sets of

equations

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Gaussian Elimination:• It is a method which is easily adapted to the computer

and is based on triangularization of the coefficient

matrix and evaluation of the unknowns by back-

substitution starting from the last equation

• Procedure: Consider the general system of n

equations with ‘n’ unknowns

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)1(

c

c

c

x

x

x

aaa

aaa

aaa

n

2

1

n

2

1

nn2n1n

n22221

n11211

• To solve

1. Eliminate the coefficient of x1 in every equation except

the first one

To do this, select a11 as the pivot and

a) Add the multiple of the first row to the second

b) Add the multiple of the first row to the third row

c) Continue this through the nth row

The system of equations

will then be reduced to

the form,

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11

21

a

a

11

31

a

a

)2(

c

c

c

x

x

x

aa0

aa0

aaa

'

n

'

2

1

n

2

1

'

nn

'

2n

'

n2

'

22

n11211

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2. Eliminate the coefficient of x2 in every equation below

except the second equation

To do this, select a’22 as the pivot and

a) Add the multiple of the second row to the third

b) Add the multiple of the second row to the fourth

row

c) Continue this through the nth row

The system of equations

will then be reduced to

the form,

'

22

'

32

a

a

'

22

'

42

a

a

)3(

c

c

c

c

x

x

x

x

aa00

aa00

aaa0

aaaa

''

n

''

3

'

2

1

n

3

2

1

''

nn

''

3n

''

n3

''

33

'

n2

'

23

'

22

n1131211

• Repeat this process for the remaining rows until we

have the system of equations (called triangular zed) as

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)4(

c

c

c

c

x

x

x

x

a0000

a0000

aa000

aaa00

aa0

aaa

1n

n

''

3

'

2

1

n

3

2

1

1n

nn

''''

n5

'''

n4

'''

44

''

n3

''

34

''

33

'

n2

'

22

n11211

3. Determine xn from the last equation as

And determine the other unknowns by back substitution.

These steps are in general form

where ,

k = 1,2,- - - - , n-1

i = k+1, - - - - , n

j = k, - - - , n+1

where,

ai, n+1 represents the

latest right side c’s given by

equation (4)8/14/2015 40

1n

nn

1n

n

n

a

cx

kk

ik

kjijij

a

aaaa

n

1ir rir1n,i

ii

ixaa

a

1x

Eigen Values and Eigen Vectors

• Given a square matrix [a] if there exists a scalar

λ (real or complex) and a non zero column

vector {X} such that [a] {X} = λ {X} , then λ is

called the Eigen value of [a] and {X} is called

Eigen vector of [a] corresponding to an Eigen

value λ.

Determination of Eigen value and Eigen vector

Consider an identity matrix ‘[I]’ of the same order

of [a], we can write,

{X} = [I] {X}

Therefore, [a] {X} = λ {X} = λ ( [I] {X} )8/14/2015 41

i.e. ([a]- λ [I]) {X} = 0

For example, Let [a] be a 3X3 matrix given

by

We have,

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333231

232221

131211

aaa

aaa

aaa

a

00

00

00

100

010

001

I

8/14/2015 43

333231

232221

131211

aaa

aaa

aaa

Iatherefore

0

x

x

x

aaa

aaa

aaa

XIaand

3

2

1

333231

232221

131211

0xaxaxa

0xaxaxa

0xaxaxa

,getWe

333232131

323222121

313212111

• The solution for the above system of

equations exists the determinant

This equation is called the characteristic

equation of matrix [a]

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0Ia.e.i

0

aaa

aaa

aaa

333231

232221

131211

• Example: Find all the Eigen values and the corresponding Eigen

vectors of the matrix

The characteristic equation for the above matrix will be in the form of

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342

476

268

a

0

342

476

268

.e.i

0Ia

07x2)4x6(2

)4)(2(3x6)6(

)4x4(37)8(.e.i

• We get, after simplification of above equation

• These are called Eigen Values

To determine the Eigen vectors, form the equation based on,

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15and3,0

,Therefore

0153

0

x

x

x

342

476

268

.e.i

0XIa

3

2

1

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0x3x4x2

0x4x7x6

0x2x6x8.e.i

321

321

321

Case 1 If λ= 0

Using the rule of cross multification

2x,2x,1x,getwe

20

x

20

x

10

xe.i

76

68

x

46

28

x

47

26

x

321

321321

2

2

1

X

is0atvectorEigenTherefore

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Case 2 If λ= 3

Using the rule of cross multification

2x,1x,2x,getwe

16

x

8

x

16

xe.i

46

65

x

46

25

x

44

26

x

321

321321

2

1

2

X

is3atvectorEigenTherefore

Case 3 If λ= 15

Using the rule of cross multification

1x,2x,2x,getwe

20

x

40

x

40

xe.i

86

67

x

46

27

x

48

26

x

321

321321

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1

2

2

X

is15atvectorEigenTherefore

Example: Gaussian Elimination Method

Solve the following system of equations using Gaussian Elimination method

6x1x1x1

4x1x2

9x1x2x2.1Ex

321

21

321

)1(

6

4

9

x

x

x

111

012

122

formmatrixIn

3

2

1

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Step 1:

Eliminate the coefficient of x1 in every equation except the first

one. Select a11 = 2 as the pivot

a) Add the multiple of the first row to the second

b) Add the multiple of the first row to the third

We obtain

2

2

a

a

11

21

2

1

a

a

11

31

5)ccxa

a(

and1)aaxa

a(,1)aax

a

a(,0)aax

a

a(

:operationsrowSecond

)2(

2

3

5

9

x

x

x

2

100

110

122

21

11

21

2313

11

21

2212

11

21

2111

11

21

3

2

1

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Step 2:

Eliminate the coefficient of x2 in every equation below the

second equation. In this case, we accomplished this in step 1.

Step 3:

Solve for x3 in the third of equation (2) as

3x2

3x

2

133

Solve for x2 in the second of equation (2) as

2x5x1x1232

Solve for x1 in the first of equation (2) as

12

)3(1)2(29x

1

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6x1x2x2

8x3x2x4

11x3x1x2.3Ex

3x2x1x2

1x4x3

4x5x4x.2Ex

321

321

321

321

32

321

Thank You

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