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Finite Element Method

FUNDAMENTAL FOR

FINITE ELEMENT

METHOD

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CONTENTS STRONG AND WEAK FORMS OF GOVERNING EQUATIONS

HAMILTON’S PRINCIPLE

FEM PROCEDURE

– Domain discretization

– Displacement interpolation

– Formation of FE equation in local coordinate system

– Coordinate transformation

– Assembly of FE equations

– Imposition of displacement constraints

– Solving the FE equations

STATIC ANALYSIS

EIGENVALUE ANALYSIS

TRANSIENT ANALYSIS

REMARKS

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STRONG AND WEAK

FORMS OF GOVERNING

EQUATIONS System equations: strong form, difficult to solve.

Weak form: requires weaker continuity on the dependent variables (u, v, w in this case).

Weak form is often preferred for obtaining an approximated solution.

Formulation based on a weak form leads to a set of algebraic system equations – FEM.

FEM can be applied for practical problems with complex geometry and boundary conditions.

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HAMILTON’S PRINCIPLE

“Of all the admissible time histories of

displacement the most accurate solution makes the

Lagrangian functional a minimum.”

An admissible displacement must satisfy:

– The compatibility equations

– The essential or the kinematic boundary conditions

– The conditions at initial (t1) and final time (t2)

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HAMILTON’S PRINCIPLE

Mathematically

02

1

dtLt

t

where L=T-P+Wf

VUUT T

V

d2

1

VcVΠ T

V

T

V

dd εε2

1σε

2

1

fs

T

S

b

T

V

f SfUVfUWf

dd

(Kinetic energy)

(Potential energy)

(Work done by

external forces)

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FEM PROCEDURE

Step 1: Domain discretization

Step 2: Displacement interpolation

Step 3: Formation of FE equation in local coordinate

system

Step 4: Coordinate transformation

Step 5: Assembly of FE equations

Step 6: Imposition of displacement constraints

Step 7: Solving the FE equations

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Step 1: Domain discretization

The solid body is divided into Ne elements with proper connectivity – compatibility.

All the elements form the entire domain of the problem without any overlapping – compatibility.

There can be different types of element with different number of nodes.

The density of the mesh depends upon the accuracy requirement of the analysis.

The mesh is usually not uniform, and a finer mesh is often used in the area where the displacement gradient is larger.

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Step 2: Displacement interpolation

Bases on local coordinate system, the displacement within

element is interpolated using nodal displacements.

eii

n

i

zyxzyxzyxd

dNdNU ),,( ),,(),,(1

1

2

displacement compenent 1

displacement compenent 2

displacement compenent f

i

n f

d

d

d n

d

1

2

displacements at node 1

displacements at node 2

displacements at node d

e

n dn

d

dd

d

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Step 2: Displacement interpolation

N is a matrix of shape functions

1 2( , , ) ( , , ) ( , , ) ( , , )

for node 1 for node 2 for node

dn

d

x y z x y z x y z x y z

n

N N N N

fin

i

i

i

N

N

N

000

000

000

000

2

1

Nwhere

Shape function

for each

displacement

component at a

node

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Displacement interpolation

Constructing shape functions

– Consider constructing shape function for

a single displacement component

– Approximate in the form

1

( ) ( ) ( )dn

h

i i

i

Tu p

x x p x α

1 2 3 ={ , , , ......, }d

T

n α

pT(x)={1, x, x2, x3, x4,..., xp} (1D)

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Pascal triangle of monomials: 2D

xy x

2

x3

x4

x5

y2

y3

y4

y5

x2y

x3y

x4y x

3y

2

xy2

xy3

xy4

x2y

3

x2y

2

Constant terms: 1

x y

1

Quadratic terms: 3

Cubic terms: 4

Quartic terms: 5

Quintic terms: 6

Linear terms: 2

3 terms

6 terms

10 terms

15 terms

21 terms

2 2( ) ( , ) 1, , , , , ,..., ,T T p px y x y xy x y x y p x p

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Pascal pyramid of monomials : 3D

x

x2

x3

x4

y

y2

y3

y4

xy

z

xz yz

x2y xy2

x2z zy2

z2

xz2 yz2

xyz

z3

x3y

x3z

x2y2

x2z2

x2yz

xy3

zy3

z2y2

xy2z xyz2

xz3

z4 z3y

1 Constant term: 1

Linear terms: 3

Quadratic terms: 6

Cubic terms: 10

Quartic terms: 15

4 terms

10 terms

20 terms

35 terms

2 2 2( ) ( , , ) 1, , , , , , , , , ,..., , ,T T p p px y z x y z xy yz zx x y z x y z p x p

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Displacement interpolation

– Enforce approximation to be equal to the nodal

displacements at the nodes

di = pT(xi) i = 1, 2, 3, …,nd

or

de=P

where

1

2=

d

e

n

d

d

d

d

T

1

T

2

T

( )

( )

( )dn

p x

p xP

p x

,

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Displacement interpolation

– The coefficients in can be found by

e

- 1α P d

– Therefore, uh(x) = N( x) de

1 2

1 1 1 1

1 2

( ) ( ) ( )

1 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

n

T T T T

n

N N N

nN N N

- - - -

x x x

N x p x P p x P p x P p x P

x x x

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Displacement interpolation

Sufficient requirements for FEM shape

functions

1 , 1,2, ,

0 , , 1,2, ,

d

i j ij

d

i j j nN

i j i j n

x1. (Delta function

property)

1

( ) 1n

i

i

N

x2. (Partition of unity property –

rigid body movement)

1

( )dn

i i

i

N x x x

3. (Linear field reproduction property)

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Step 3: Formation of FE equations in local

coordinates

Since U= Nde

Therefore, e = LU e = L N de= B de

Strain matrix

e

T

eΠ kdd2

1or where

(Stiffness matrix)

e

T

Ve

T

e e

T T

e

Ve

T

Ve

V c V c V c Π d d B B d d Bd B d d ) ( 2

1

2

1 ε ε

2

1

V c T

Ve

e d B B k

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Step 3: Formation of FE equations in local

coordinates

Since U= Nde eU Nd

or ee

T

eT dmd 2

1 where

(Mass matrix)

1 1 1d d ( d )

2 2 2e e e

T T T T T

e e e e

V V V

T V V V U U d N Nd d N N d

d

e

T

e

V

V m N N

d

e

T

e

V

V m N N

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Step 3: Formation of FE equations in local

coordinates

e

T

es

T

eb

T

efW FdFdFd

sbe FFf (Force vector)

d d ( d ) ( d )

e e e e

T T T T T T T T

f e b e s e b e s

V S V S

W V S V S d N f d N f d N f d N f

d

e

T

b b

V

V F N f d

e

T

s s

S

S F N f

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Step 3: Formation of FE equations in local

coordinates

0d)(2

1

- te

T

eee

T

eee

T

e

t

tFddkddmd

)(d

d)

d

d( T

e

T

eT

ett

dd

d

ttt ee

t

t

T

eee

t

t

T

e

t

tee

T

eee

t

t

T

e ddd2

1

2

1

2

1

2

1

dmddmddmddmd --

0d)(2

1

-- teeee

T

e

t

tFkddmd

0d)2

1

2

1(

2

1

- te

T

eee

T

eee

T

e

t

tFddkddmd

eeeee fdmdk

FE Equation

(Hamilton’s principle)

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Step 4: Coordinate transformation

eeee fdmkd

x

y

x' y'

y'

x'

Local coordinate

systems

Global

coordinate

systems

ee TDd

eeeee FDMDK

TkTK e

T

e TmTM e

T

e e

T

e fTF , ,

where

(Local)

(Global)

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Step 5: Assembly of FE equations

Direct assembly method

– Adding up contributions made by elements

sharing the node

FDMKD

FKD (Static)

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Step 6: Impose displacement constraints

No constraints rigid body movement

(meaningless for static analysis)

Remove rows and columns corresponding

to the degrees of freedom being constrained

K is semi-positive definite

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Step 7: Solve the FE equations

Solve the FE equation,

for the displacement at the nodes, D

The strain and stress can be retrieved by

using e = LU and s = c e with the

interpolation, U=Nd

FDMKD

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STATIC ANALYSIS

Solve KD=F for D

– Gauss elmination

– LU decomposition

– Etc.

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EIGENVALUE ANALYSIS

0 DMKD (Homogeneous equation, F = 0)

Assume )exp( tiD

0][ 2 - MK

Let 2 0][ - MK

0]det[ -- MKMK

[ K - i M ] i = 0 (Eigenvector)

(Roots of equation are the

eigenvalues)

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EIGENVALUE ANALYSIS

Methods of solving eigenvalue equation

– Jacobi’s method

– Given’s method and Householder’s method

– The bisection method (Sturm sequences)

– Inverse iteration

– QR method

– Subspace iteration

– Lanczos’ method

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TRANSIENT ANALYSIS

Structure systems are very often subjected to transient excitation.

A transient excitation is a highly dynamic time dependent force exerted on the structure, such as earthquake, impact, and shocks.

The discrete governing equation system usually requires a different solver from that of eigenvalue analysis.

The widely used method is the so-called direct integration method.

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TRANSIENT ANALYSIS

The direct integration method is basically using the finite difference method for time stepping.

There are mainly two types of direct integration method; one is implicit and the other is explicit.

Implicit method (e.g. Newmark’s method) is more efficient for relatively slow phenomena

Explicit method (e.g. central differencing method) is more efficient for very fast phenomena, such as impact and explosion.

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Newmark’s method (Implicit)

Assume that

2 1

2t t t t t t tt t

-

D D D D D

1t t t t t tt - D D D D

KD CD MD FSubstitute into

2 1

2

1

t t t t t

t t t t t t t t

t t

t

-

-

K D D D D

C D D D MD F

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Newmark’s method (Implicit)

residual

cm t t t t K D F

where

2

cm t t

K K C M

2residual 1

12

t t t t t t t t tt t t

- - - -

F F K D D D C D D

Therefore, 1

cm

residual

t t t t

-

D K F

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Newmark’s method (Implicit)

Start with D0 and 0D

Obtain 0D KD CD MD Fusing

1

cm

residual

t t t t

-

D K FObtain tD using

Obtain Dt and tD using

2 1

2t t t t t t tt t

-

D D D D D

1t t t t t tt - D D D D

March

forward

in time

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Central difference method (explicit)

int residual - - MD F CD KD F F F

residual- 1D M F (Lumped mass – no need to solve matrix equation)

2t t t t tt - D D D

2t t t t tt - D D D

2

12t t t t t t

t - -

D D D D

2

2t t t t t

tt-

- D D D D

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Central

difference

method

(explicit)

D,

t

x

x

x x

x

t0 t

-t

-t/2

t/2

Find average velocity at time t =

-t/2 using

Find using the average acceleration at

time t = 0.

Find Dt using the average velocity at time t =t/2

Obtain D-t using

D0 and are

prescribed and

can be obtained from

Use to

obtain assuming .

Obtain using

Time marching in half the time step

0D

0D

residual- 1D M F

2

2t t t t t

tt-

- D D D D

/ 2t-D

/ 2 / 2t t t t tt - D D D

/ 2tD

/ 2 / 2t t t t tt - D D D

/ 2 / 2t t t t tt - D D D

/ 2 / 2t t t t tt - D D D

tD / 2 0t D D

tDresidual- 1

D M F

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REMARKS In FEM, the displacement field U is expressed by

displacements at nodes using shape functions N defined over elements.

The strain matrix B is the key in developing the stiffness matrix.

To develop FE equations for different types of structure components, all that is needed to do is define the shape function and then establish the strain matrix B.

The rest of the procedure is very much the same for all types of elements.

THANK YOU

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