Finite Element Method for Professionals

8/13/2019 Finite Element Method for Professionals

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1

Finite Element Method

THE FINITE ELEMENT

METHOD

for readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 3:


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Finite Element Method by G. R. Liu and S. S. Quek2

CONTENTS STRONG AND WEAK FORMS OF GOVERNING EQUATIONS

HAMILTONS 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.

Weakform: requires weaker continuity on thedependent variables (u, v, win this case).

Weak form is often preferred for obtaining anapproximated solution.

Formulation based on a weak form leads to a setof algebraic system equationsFEM.

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


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HAMILTONS PRINCIPLE

Of all the admissibletime 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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FEM PROCEDURE

Step 1: Domain discretization

Step 2: Displacement interpolation

Step 3: Formation of FE equation in local coordinatesystem

Step 4: Coordinate transformation

Step 5: Assembly of FE equations

Step 6: Imposition of displacement constraints Step 7: Solving the FE equations


7/34Finite Element Method by G. R. Liu and S. S. Quek7

Step 1: Domain discretization

The solid body is divided into Ne elements with properconnectivitycompatibility.

All the elements form the entire domain of the problemwithout any overlappingcompatibility.

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

The density of the mesh depends upon the accuracyrequirement of the analysis.

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




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 compenentf

i

n f

d

d

d n

d

1

2

displacements at node 1

displacements at node 2

displacements at noded

e

n dn

d

dd

d




Nis a matrix ofshape 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

NN

000

000

000000

2

1

Nwhere

Shape functionfor each

displacement

component at a

node




Constructing shape functions

Consider constructing shape function for

a single displacement component

Approximate in the form

1

( ) ( ) ( )

dn

hi i

i

Tu p

x x p x

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

d

T

n

p

T

(x)={1,x,x

2

,x

3

,x

4

,...,x

p

} (1D)



Pascal triangle of monomials: 2D

xy

x

2

x3

x4

x5

y

2

y3

y4

y5

x2y

x3y

x4y x y

xy2

xy3

xy

x2y

3

x y

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



Pascal pyramid of monomials : 3D

x

x2

x3

x

4

2

3

4

xy

xz z

x2y xy

2

x2z y

2

2

xz2 z

2

xyz

3

x3y

x3z

x

2

y

2

x2z2x2yz

xy3

y3

2y

2

xy2zxyz

2

xz3

43y

1Constant 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 p

x y z x y z xy yz zx x y z x y z p x p




Enforce approximation to be equal to the nodal

displacements at the nodes

di = pT(xi) i = 1, 2, 3, ,ndor

de=P

where

1

2

=

d

e

n

d

d

d

d

T

1

T

2

T

( )

( )

( )dn

p x

p x

P

p x

,




The coefficients in can be found by

e

- 1 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




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

ii

N

x2. (Partition of unity property

rigid body movement)

1

( )dn

i i

i

N x x x

3. (Linear field reproduction property)



Step 3: Formation of FE equations in local

coordinates

Since U= Nde

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

eTe kdd

2

1or where

(Stiffness matrix)

e

T

Ve

T

ee

TT

e

Ve

T

Ve

VcVcVc ddBBddBdBdd )(2

1

2

1

2

1

VcT

Ve

edBBk

dT V m N N


17/34Finite Element Method by G. R. Liu and S. S. Quek

17


coordinates

Since U= Nde eU Nd

or

ee

T

e

T 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

e

V

V m N N

d

e

T

e

V

V

m N N



18


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 de

Ts s

S

S F N f



19


coordinates

0d)(

2

1

-

te

T

eee

T

eee

T

e

t

t Fddkddmd

)(d

d)

d

d( Te

T

eT

ett

dd

d

ttt eet

t

Teee

t

t

Te

t

tee

Teee

t

t

Te ddd

2

1

2

1

2

1

2

1

dmddmddmddmd --

0d)(2

1

-- teeeeT

e

t

tFkddmd

0d)2

1

2

1(

2

1

- teT

eee

T

eee

T

e

t

tFddkddmd

eeeee fdmdk

FE Equation

(Hamiltons principle)



20

Step 4: Coordinate transformation

eeee fdmkd

x

y

x'y'

y'

x'

Local coordinate

systems

Global

coordinate

systems

ee TDd

eeeee FDMDK

TkTK eT

e TmTM eT

e eT

e fTF , ,

where

(Local)

(Global)



21

Step 5: Assembly of FE equations

Direct assembly method

Adding up contributions made by elements

sharing the node

FDMKD

FKD (Static)



22

Step 6: Impose displacement constraints

No constraints rigid body movement(meaningless for static analysis)

Remove rows and columns correspondingto 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= LUand s = c e with theinterpolation, U=Nd

FDMKD


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STATIC ANALYSIS

Solve KD=Ffor 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 - iM] i= 0 (Eigenvector)(Roots of equation are the

eigenvalues)


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EIGENVALUE ANALYSIS

Methods of solving eigenvalue equation

Jacobis method

Givens method and Householders 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 totransientexcitation.

A transient excitation is a highly dynamic timedependent force exerted on the structure, such asearthquake, impact, and shocks.

The discrete governing equation system usually

requires a different solver from that of eigenvalueanalysis.

The widely used method is the so-called directintegration method.


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TRANSIENT ANALYSIS

The direct integration method is basically usingthefinite difference methodfor time stepping.

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

Implicit method (e.g.Newmarksmethod) is moreefficient for relatively slow phenomena

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


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Newmarks method (Implicit)

Assume that

2 1

2t t t t t t t t t

-

D D D D D

1t t t t t t t - 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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residual

cm t t t t K D F

where

2

cm t t K K C M

2residual1

12

t t t t t t t t t t 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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Start with D0and 0D

Obtain0

D KD CD MD Fusing

1

cm

residual

t t t t

- D K FObtain tD using

Obtain Dt and tD using

2 12t t t t t t t t t -

D D D D D

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

Marchforward

in time


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Central difference method (explicit)

int residual - - MD F CD KD F F F

residual- 1D M F (Lumped massno need to solve matrix equation)

2t t t t t t - D D D

2t t t t t t - 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

D0and are0D


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Central

differencemethod

(explicit)

D,

t

x

x

x x

x

t0 t-t -t/2 t/2

Find average velocity at time t=

-t/2using

Find using the average acceleration at

time t= 0.

FindDtusing the average velocity at time t=t/2

Obtain D-tusing

prescribed and

can be obtained from

Use toobtain assuming .

Obtain using

Time marching in half the time step

0D

residual- 1D M F

2

2t t t t t

tt-

- D D D D

/ 2t-D

/2 /2t t t t t t

-

D D D

/ 2tD

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

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

/2 /2t t t t t t - D D DtD / 2 0t D D

tDresidual- 1D M F


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REMARKS In FEM, the displacement field Uis expressed by

displacements at nodes using shape functions Ndefined over elements.

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

To develop FE equations for different types ofstructure components, all that is needed to do is

define the shape function and then establish thestrain matrix B.

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

Finite Element Method for Professionals

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