foundmentals for finite element method chapter1: the finite element method a practical course the...

FOUNDMENTALS FOR FINITE ELEMENTMETHOD

CHAPTER1:

The FThe Finite Element Methodinite Element MethodA Practical CourseA Practical Course

CONTENTSCONTENTS 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 (reading materials) REMARKS

STRONGSTRONG AND AND WEAKWEAK FORMS OF FORMS OF GOVERNING EQUATIONSGOVERNING EQUATIONS

System equations: strong form (PDE), difficult to solve.

Weak (integral) form: requires weaker continuity on the dependent variables (e.g., u, v, w).

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.

HAMILTON’S PRINCIPLEHAMILTON’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 conditions– The essential or the kinematic boundary conditions

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

HAMILTON’S PRINCIPLEHAMILTON’S PRINCIPLE

Mathematically

02

1 dtL

t

t

where L=T+Wf

VUUT T

V

d2

1

VcVΠ T

V

T

V

dd εε2

1σε

2

1

fsT

Sb

T

Vf SfUVfUW

f

dd

(Kinetic energy)

(Potential energy)

(Work done by external forces)

Lagrangian functional

FEM PROCEDUREFEM PROCEDURE

Step 1: Domain discretization Step 2: Displacement interpolation Step 3: Formation of FE equation in local

coordinates Step 4: Coordinate transformation Step 5: Assembly of FE equations Step 6: Imposition of displacement constraints Step 7: Solving the FE equations

Step 1: Domain Step 1: Domain discretizationdiscretization

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

All the elements form the entire domain of the problem without any gap or 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.

Triangular elements Nodes

Step 2: Displacement Step 2: Displacement interpolationinterpolation

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

nf: Degree of freedoms at a node

x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

A fsx

fsy

nd: number of nodes in an element

Step 2: Displacement interpolationStep 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

Displacement interpolationDisplacement interpolation

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 α

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

Tn α

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

Basis function

Pascal triangle of monomialsPascal triangle of monomials: : 2D2D

xy x2

x3

x4

x5

y2

y3

y4

y5

x2y

x3y

x4y x3y2

xy2

xy3

xy4 x2y3

x2y2

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 monomialsPascal pyramid of monomials : : 3D3D

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


– 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

T1

T2

T

( )

( )

( )dn

p x

p xP

p x

,

Moment matrix


– The coefficients in can be found by

e 1α P d

– Therefore, uh(x) = N( x) de

1 2

1 1 1 11 2

( ) ( ) ( )

1 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

n

T T T Tn

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 ijd

i j j nN

i j i j n

x

1. (Delta function

property)

1

( ) 1n

ii

N

x2. (Partition of unity property – rigid body movement)

1

( )dn

i ii

N x x x

3. (Linear field reproduction property)

Step 3: Formation of FE equations in local Step 3: Formation of FE equations in local coordinatescoordinates

Since U= Nde

Therefore, = LU = L N de= B de

Strain matrix

or where

(Stiffness matrix)

eT

Ve

Tee

TTe

Ve

T

Ve

VcVcVcΠ ddBBddBdBdd )(2

1

2

1εε

2

1

VcT

Ve

e dBBk 1

2Te e e d k d


Since U= Nde eU Nd

or eeTeT dmd

2

1 where

(Mass matrix)

1 1 1d d ( d )

2 2 2e e e

T T T T Te e e e

V V V

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

de

Te

V

Vm N N


eTes

Teb

TefW FdFdFd

sbe FFf (Force vector)

d d ( d ) ( d )e e e e

T T T T T T T Tf 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

de

Tb b

V

VF N f de

Ts s

S

SF N f


)(d

d)

d

d( T

e

TeT

e ttd

dd

ttt ee

t

t

Teee

t

t

Te

t

teeTeee

t

t

Te ddd

2

1

2

1

2

1

2

1

dmddmddmddmd

0d)(2

1

teeee

Te

t

tFkddmd

0d)2

1

2

1(

2

1

te

Teee

Teee

Te

t

tFddkddmd

eeeee fdmdk

FE Equation

(Hamilton’s principle)

2

1

( )d 0

t T T Te e e e e e e et

t d m d d k d d f

Step 4: Coordinate Step 4: Coordinate transformationtransformation

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)

Step 5: Assembly of FE equationsStep 5: Assembly of FE equations

Direct assembly method– Adding up contributions made by elements

sharing the node

FDMKD

FKD (Static)

eeeee FDMDK

Step 6: Impose displacement constraintsStep 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

Step 7: Solve the FE equationsStep 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 = LU and = c with the interpolation, U=Nd

FDMKD

STATIC ANALYSISSTATIC ANALYSIS

Solve KD=F for D

– Gauss elimination– LU decomposition– Etc.

EIGENVALUE ANALYSISEIGENVALUE 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)

EIGENVALUE ANALYSISEIGENVALUE 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

TRANSIENT ANALYSISTRANSIENT 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.

TRANSIENT ANALYSISTRANSIENT ANALYSIS(reading material)(reading material)

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.

REMARKSREMARKS 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.

Newmark’s method (Implicit)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

Typically

= 0.5

= 0.25

Newmark’s method (Implicit)Newmark’s method (Implicit)residual

cm t t t t K D F

where

2

cm t t K K C M

2residual 11

2t t t t t t t t tt t t

F F K D D D C D D

Therefore, 1cm

residualt t t t

D K F

Newmark’s method (Implicit)Newmark’s method (Implicit)

Start with D0 and 0D

Obtain 0D KD CD MD Fusing

1cm

residualt 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

Central difference method (explicit)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

Central Central difference difference

method method (explicit)(explicit)

D,

t

x

x

x x

x

t0t-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

tD residual 1D M F

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