Fem Lanjut

7/28/2019 Fem Lanjut

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F inite Element Methods (FEM )

Suzanne Vogel

COMP 259

Spring, 2002


2/73


3/73

1. Set up a global modelin terms of the world

coordinates of mass points of the object.

These equations will be continuous.

2. Discretize the object into a nodal mesh.3. Discretizethe equations using finite

differences and summations (rather than

derivatives and integrals).4. Use (2) and (3) to write the global equations

as a sti ffness matr ixtimes a vector of

(unknown) nodal values.

Top-Down: Steps in FEM


4/73

Top-Down: Steps in FEM

6. Solve for the nodal values.

Staticnodal values at equilibrium

Dynamicnodal values at next time step

7. Interpolate values between nodal coordinates.

5

2 31 4

678

udiscretize interpolate

+

global model

object

nodal mesh interpolate values between nodes+

local model


5/73

Bottom-Up: Steps in FEM

Nodesare point masses connected with springs.A continuum equation is solved for the nodes,

and intermediate points are interpolated.

A collection of nodes forms an element.

A collection of elements forms the object.

5

2

3

1

4

678

u


6/73

Elements and I nterpolations

Interpolating equations for an element are

determined by the number and distribution of

nodes within the element.

More nodes mean higher degree, for smoother

simulation.


7/73

Example: Hermite as 1D Cubic

I nterpolation Equation1. Assume

u

r

dcubuauur 23)(

44332211 )()()()()( ruNruNruNruNur cubic equation

equation using shape (blending) functions

and


8/73


I nterpolation Equation2. Normalize the element to [0,1] and rewrite

dcubuauur 23)( as a matrix equation

d

c

b

a

d

c

b

a

uuu

uuu

uuu

uuu

r

r

r

r

1111

1272

274

278

13

1

9

1

27

11000

1

1

1

1

4

2

4

3

4

323

33

2

2

2

3

2

1

2

1

3

1

4

3

2

1

or QUR 00


9/73


I nterpolation Equation3. Solve for the coefficients Q

00

1

000 RMRUQQUR H

4. Plug the coefficients into the cubic equation

dcubuauur 23)(

5. Rewrite the cubic equation in the form

44332211 )()()()()( ruNruNruNruNur


10/73


I nterpolation Equation4 + 5. are equivalent to the steps

4

3

2

1

4321

000

)()()()()(

)()()(

r

r

r

r

uNuNuNuNur

RMURMUQUur HH

values at the 4 nodes of the element

shape (blending) functions


11/73


I nterpolation Equation

1

0

U

shape (blending) functions within one elementLet

u

rtuHN ii ,


12/73

1D Elements

(x) (x)

(x)

Example: bungee


13/73

2D Elements

(x,y)

(x,y)

(x,y)

Example: cloth


14/73

3D Elements

(x,y,z)

(x,y,z)

Example: skin


15/73

Staticanalysis is good for engineering, to findjust the end result.

Dynamicanalysis is good for simulation, to

find all intermediate steps.

Static vs. Dynamic FEM


16/73

Types of Global Models[6]

Variational- Find the position function, w(t)that minimizes the some variational integral.

This method is valid only if the position

computed satisfies the governing differentialequations.

Rayleigh-Ritz- Use the variational methodassuming some specific form ofw(t) and

boundary conditions. Find the coefficients and

exponents of this assumed form ofw(t).


17/73

Example of Variational Method[6]

0)(

tf

w

ww

dfwwwcwcwJ 2

2

1)( 1

2

2

cwbwaw

0)(

)(

)(

3

2

1

fcc

bc

ac

Minimizing the variation w.r.t. w of the

variational function

under the conditions

satisfies the governing

equation, Lagranges

Equation


18/73

Galerkin(weighted residual) - Minimize the

residual of the governing differential equation,

F(w,w,w,,t) = 0. The residual is the form ofFthat results by plugging a specific form of the

position function w(t) intoF. Find the

coefficients and exponents of this assumed formofw(t).

Types of Global Models[6]


19/73

We can approximate w(t) using Hookes Law

0)(

tf

w

ww

Example of Galerkin Method[6]

If we use that equation to compute the 1st and

2nd time derivatives ofw, then we can computethe residual as

)(

)(

)(

)(

11

11

1

2

1

2

1

1

00

00

tw

tw

Etf

Etf

LL

LL

E


20/73

Example of Static, Elastic FEM

Problem: If you apply the pressure shown, whatis the resulting change in length?

Object


21/73

First step. Set up a continuum model:

F = force

P = pressure

A = area

L = initial length

E = Youngs modulus

L

LE

A

PF

AE

PLL

duuhwE

PLdu

uwhE

PLdu

AE

PLL

un

u

un

u

un

u 000 )(1

)(

1

Entire length:

Infinitessimal length:



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Since the shape is regular, we can integrate to

find the solution analytically. But suppose we

want to find the solution numerically.

Next step. Discretize the object.



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Discretization of object into

linear elements bounded by nodes1 2 3 4

n1 n2 n3 n4 n5


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Next step. Set up a local model.

Stress-Strain Relationship(like Hookes Law)

0,,0,,

0

,)()( LkrrkLrrkLkL

L

Ejijijijijiji

ji

i

0,, )( Lkrrk jijijiij

Youngs modulus distance between adjacent nodes

stress (elastic force)


25/73


0,, )( Lkrrk jijijii

0,, )( Lkrrk jijijij

j

i

jiji

jiji

jij

jii

ji

ji

j

i

jiji

jiji

j

i

rr

kkkk

LkLk

Lk

Lk

r

r

kk

kk

,,

,,

0,

0,

0,

0,

,,

,,

11,1,

1,1,

01,1

01,

j

j

jjjj

jjjj

jjj

jjj

r

r

kk

kk

Lk

Lk

Next step. Set up a local (element) stiffness matrix.

Rewrite the

above as a

matrix equation.

Same for the

adjacent element.

element stiffness matrix

nodal stressesnodal coordinates


26/73

2

1

2,12,1

2,12,1

02,12

02,11

r

r

kk

kk

Lk

Lk

Example of Static, Elastic FEMNow, all of the element stiffness matrices are as

follows.

3

2

3,23,2

3,23,2

03,23

03,22

r

r

kk

kk

Lk

Lk

4

3

4,34,3

4,34,3

04,34

04,33

r

r

kk

kk

Lk

Lk

5

4

5,45,4

5,45,4

05,45

05,44

r

r

kk

kk

Lk

Lk

1 2 3 4n1 n2 n3 n4 n5

riis the x-coordinate of node ui


27/73


Next step. Set up a global stiffness matrix.

Pad the element stiffness matrices with zeros

and sum them up. Example:

5

4

3

2

1

3,23,2

3,23,2

03,23

03,22

00000

00000000

000

00000

0

0

0

r

rr

r

r

kk

kk

Lk

Lk


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5

4

3

2

1

5,45,4

5,44,35,44,3

4,34,33,23,2

3,23,22,12,1

2,12,1

05,45

05,44,34

04,33,23

03,22,12

02,11

000

00

00

00000

)(

)(

)(

r

r

r

rr

kk

kkkk

kkkk

kkkkkk

Lk

Lkk

Lkk

LkkLk


Final step. Solve the matrix equation for thenodal coordinates.

Global stiffness matrix.

Captures material properties.

Nodal coordinates.

Solve for these!Applied forces


29/73

Elastic FEM

A material is elasticif its behavior depends only

on its state during the previous time step.

Think: Finite state machine

The conditions under which an elastic

material behaves elastically are:

Force is small.

Force is applied slowly and steadily.


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I nelastic FEM

A material is inelasticif its behavior depends onallof its previous states.

A material may behave inelastically if:Force is large - fracture, plasticity.

Force is applied suddenly and released, i.e., is

transient- viscoelasticity.

Conditions for elastic vs. inelastic depend on

the material.


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Examples of Elastici ty

Elasticity

Springs, rubber, elastic, with small, slowly-

applied forces


32/73

Examples of I nelastici ty

InelasticityViscoelasticity

Silly putty bounces under transient force (but

flows like fluid under steady force)

Plasticity

Taffy pulls apart much more easily under

more force (material prop.)

Fracture

Lever fractures under heavy load


33/73

L inear and Nonl inear FEMSimilarly to elasticity vs. inelasticity, there are

conditions for linear vs. nonlinear deformation.

Often these coincide, as in elastoplastic.

0

:L

LeEe

= e


34/73

Hookes Law

Describes spring without damping

Linear range of preceding stress vs. strain graph

eaf 0

Elastic Deformation

Elastic vs. I nelastic FEM

e

e

t

loading unloading

0

:L

LeEe orstress strain

Youngs modulus


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Damped Elastic Deformation

e

e

t

loading unloading

eaeaf 01

viscous linear stress

Rate of deformat ion is c ons tant .

a1e.

a1e.


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Viscoelastic Deformation


e

e

t

loading unloading

.

eaeaeafbfbfb 012012

viscousnew term!

This graph is actual ly viscou s,

but visc oelast ic is prob ably simi lar

Rate of deform at ion is greatest

immed iately after start ing

loading or unloading.

depends on time t

linear stress


37/73

Elastoplastic Deformation


e

This graph is actual ly p last ic ,

but visc oelast ic is prob ably simi lar

f

e

x

x

compare

loading

unloading

loadingx

depends on force f

e

eaf 0


38/73


Fracture

Force response is locally discontinuous

Fracture will propogate if energy release rateis greater than a threshold

e

x

x

loading

unloading

depends on force f


39/73

1. Worldcoordinatesw

in inertial frame

(a frame with

constant velocity)

2. Object

(material)

coordinatesrin non-inertial

frame

r (w,t) = rref(w,t) + e(w,t)

Elastic vs. I nelastic

FEM4,5

world, or

inertial frame

re f

robject, or

non-inertial

frame

origin of

= center of mass in


40/73

Transform

reference

component rrefelastic component e

object frame

w.r.t. world frame



FEM4,5

re f

r


41/73


All these equations are specific for:Elasticity

Viscosity

Viscoelasticity

Plasticity

Elastoplasticity

Fracture

(not mentioned) Elastoviscoplasticity

Ideally: We want a general equation that will

fit all these cases.


42/73


FEM4,5A More General ApproachTo simulate dynamicswe can use Lagranges

equation ofstrainforce. At each timestep, the

force is calculated and used to update the

objects state (including deformation).

stress component

of force

wwwtwf

),(

mass density damping density

elastic potential energyLagranges Equation

w

e

L

E

w

L

w 0

0 )/(


43/73

wwwtwf

),(


FEM4,5

Given:

Mass density and damping density are known.Elastic potential energy derivative w.r.t. rcan be

approximated using one of various equations.

The current position wtof all nodes of the object

are known.

Unknown:

The new position wt+dtof nodes is solved for at

each timestep.vector

vector

matrices

next slide

Lagranges

Equation


44/73

wwwtwf

),(

drr

e

r

e

r

er

,...,,,

3

3

2

2


FEM4,5

For both elastic and inelastic deformation,express elastic potential energy as an integral

in terms ofelastic potential energy density.

elastic potential energy density

elastic potential energy


45/73


FEM4,5Elastic potential energy density can beapproximated using one of various equations

which incorporate material properties.

Elastic deformation: Use tensors called metric

(1D, 2D, 3D stretch), curvature(1D, 2D

bend), and twist (1D twist).

Inelastic deformation: Use controlled-

continuity splines.


46/73

Elastic FEM4

For elastic potential energy density in 2D, use metrictensors G (for stretch)

curvaturetensorsB (for bend)

2020 ||||||||)( BBGGr

||M|| = weighted norm of matrixM


47/73

Elastic FEM4

Overview of derivation of metric tensor

Since the metric tensorG represents stretch, it

incorporates distances between adjacent points.

T

ji

ji

ji

ji

ji ji

dr

dr

dr

dr

GG

GG

drdr

drdrdrdr

drdr

GGGGdrdrG

drdrr

w

r

wdwdwdL

2

1

2

1

2,21,2

2,11,1

22

12

21

11

2,21,22,11,1

2,1,

,

2,1,

2

11

world coordinatesobject coordinates


48/73

Elastic FEM4

Overview of metric and curvature tensors.From the previous slides, we found:

Similarly:

represents stretch

represents bend

Theorem. G andB together determine shape.

jiji r

w

r

w

rwG ))((,

jiji

rrwrwB

2

, ))((


49/73

Elastic FEM4

For elastic FEM, elastic potential energydensity in 2D incorporates changes in the

metric tensorG and the curvature tensorB.

2020 ||||||||)( BBGGr

||M|| = weighted norm of matrixM

weights = material properties


50/73

I nelastic FEM5

For inelastic FEM, elastic potential energydensity is represented as a controlled-

continuity spl ine.

p

m mjjdjj

m

j

j

d

errr

wjjj

m

0

2

||21

21 ...!!...!

!

2

1

For some degreep, dimensionality d, computethe sum of sums of all combinations of

weighted 1st, 2nd,, mth derivatives of strain e

w.r.t. node location r, where m


51/73

p

m mjjdjj

m

j

jd

jdjj

m

jm

errrwjjj

m

rrre 0 || 212121 ...!!...!

!

...1

I nelastic FEM5

Then the elastic potential energy densityderivative w.r.t. strain e is:weighting function = material property

e

r

w

r

e

r

w

r

e

rr

w

rr

er

wr

er

wr

ewe

2

2

2

022

2

2

2

1

2

202

1

2

21

2

11

21

2

2

01

21

10

1

00

!2!0

!2

!0!2

!2

!1!1

!2

!1!0

!1

!0!1

!1

!0!0

!0

Example:p = 2, d= 3


52/73

wwwtwf

),(

p

m mjjdjj

m

j

j

d

jdjj

m

jme

rrrw

jjj

m

rrre 0 ||21

21

21 ...!!...!

!

...1

p

m mj

jdjj

m

j

j

d

e

rrr

w

jjj

m

0

2

||

21

21 ...!!...!

!

2

1

udr

r

e

r

e

r

er ,...,,,

2

2

2

2


FEM4,5

2020 ||||||||)( BBGGr Inelastic

Elastic

RecapLagranges Eqn

total force(includes stress)

elastic

potential energy

elastic potential

energy density

45

5

material properties

How it has beenexpanded and is continuing

to be expanded...


53/73

wwwtwf

),(

drBBGGrji

jiji

2,1,

20

,

20

,)(

Elastic FEM4Continuing

2020 ||||||||)( BBGGr

0

,,,, )(),( jijijiji GGwwr 0

,,,, )(),( jijijiji GGwwr

udr

r

e

r

e

r

er ,...,,,

2

2

2

2

elastic

potential

energy

>0: surface wants to shrink

0: surface wants to flatten


54/73

wwwtwf

),(

I nelastic FEM5Continuing

udr

r

e

r

e

r

er ,...,,,

2

2

2

2

p

m mjjdjj

m

j

j

d

jdjj

m

jm errr

wjjj

m

rrre 0 ||21

21

21 ...!!...!

!

...1

Deformation has been modeled by

approximating elastic potential energy.

elastic potential energy

elastic potential

energy density

strain


55/73

I nelastic FEM5Continuing

Now rigid-body motion and other aspects ofdeformation must be computed using physics

equations of motion.

In this way, both (in)elastic deformation and

rigid-body motion can be modeled, providing a

very general framework.



56/73

),()()()(),( tretcttctrw

I nelastic FEM5

Motion of object (non-inertial) frame w.r.t.world (inertial) frame

drtrwrtc ).,()()(

drtrwdrtredt

dcm

dt

dfv .),(.),()(

drwrdrerdt

dI

dt

df ..)(

ewrerce

dt

dtfe

2)()()(

wwwtwf

),( Combines

dynamics of

deformable

andrigidbodies

elastic

rot

trans


57/73

),()()()(),( tretcttctrw

I nelastic FEM5

Velocity of node of object (non-inertial) framew.r.t. world (inertial) frame (radians / sec) x(radius)

Identically, in another

coordinate system,

r(w,t) = rref(w,t) + e(w,t)

w.r.t. object

velocity of reference

component

velocity of elastic

component

w.r.t. world

wwwtwf

),(

),( trw

)(t)(tc

)(te


58/73

drwrdrerdt

dI

dt

df ..)(

I nelastic FEM5wwwtwf

),(

rot

)(

)(

)(

))(

)(2

3

2

22313

32

2

3

2

112

312123

22

tdw

wwwwww

wwwwww

wwwwww

tI

angular momentum

inertia tensor

Angular momentum is conserved in the absense

of force. So a time-varying angular momentum

indicates the presence of foce.


59/73

drwrdrerdt

dI

dt

df ..)(


),(

rot

indicates changing angle between position and direction of stretch

)(tr

)(te


60/73


),(

ewrerce

dt

dtfe

2)()()(elastic

inertial centripetal Coriolis transverse damping

elastic potential energy strain

restoring

If the reference component has no translation or

rotation, then

ee

dt

dtfe

)()(

e

tfe

)(

Furthermore, if the elastic component has no

acceleration, then


61/73


),(

Recall that non-elastic behavior is characterized

by acceleration of the elastic component

(strain)...

eedt

d

tf

e

)()(

etfe

)(

And elastic behavior is characterized by

constant velocity of strain.

loading x

e

eaf 0


62/73

Now Lagranges equation has been expanded.

Final StepsDiscretize using f inite differences(rather than

derivatives).

Write as a matrixtimes a vector of nodal

coordinates (rather than a single mass point).

Solve for the objects new set of positions of

all nodes.


FEM4,5

wwwtwf

),(


63/73

Discretization of FEM4,5

wwwtwf

),(

wwCwMtf

)(

Discretize Lagranges equation over all nodes

Procedure described in [4] but not [5]


64/73

tttttttt

ttttttt

ttt

ttt

tt

ttttttt

ttt

ttttttttt

tttttttt

tttttt

vCMtwCtMtwwgwKtMtA

wherewwgwA

vCMt

wCt

Mt

f

t

wwCM

twC

tM

tf

wCt

Mt

wMt

fwwKt

Mt

wwKt

wwC

t

wwwM

wwKwCDwDMD

fwwKtwC

twM

2

11

2

13

),(,)(2

11

_),(

2

11

2

13

2

11

2

13

2

112)(

2

11

)(2

2

)()())((

)(

22

2

2

222

2

2

2

Discretization of Elastic FEM4


65/73

Results of Elastic FEM4


66/73



67/73



68/73

3D plasticine bust of Victor Hugo.

180 x 127 mesh; 68,580 equations.

Results of I nelastic FEM5


69/73

Results of I nelastic FEM5

Sphere pushing through 2D mesh.

23 x 23 mesh; 1,587 equations.

Yield limit is uniform, causing linear tears.


70/73


71/73

References

0. David Baraff. Rigid Body Simulation.Physically Based Modeling, SIGGRAPH

Course Notes, August 2001.

1. George Buchanan. Schaums Outlines:

Finite Element Analysis. McGraw-Hill, 1995.

2. Peter Hunter and Andrew Pullan. FEM/BEM

Notes. The University of Auckland, New

Zealand, February 21 2001.
http://www.esc.auckland.ac.nz/Academic/Texts/FEM-BEM-notes.htmlhttp://www.esc.auckland.ac.nz/Academic/Texts/FEM-BEM-notes.htmlhttp://www.esc.auckland.ac.nz/Academic/Texts/FEM-BEM-notes.htmlhttp://www.esc.auckland.ac.nz/Academic/Texts/FEM-BEM-notes.html


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References3. Tom Lassanske. [Slides from class lecture]

4. Demetri Terzopoulost, John Platt, Alan Barr,

and Kurt Fleischert. Elastically Deformable

Models. Computer Graphics, Volume 21,Number 4, July 1987.

5. Demetri Terzopoulos and Kurrt Fleiseher.Modeling Inelastic Deformation:

Viscoelasticity, Plasticity, Fracture. Computer

Graphics, Volume 22, Number 4, August 1988

N i


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Notation

densityenergypotentialelastic

energypotentialelastic

ulussYoungE

f orcestress

stretchstraine

scoordinateworldw

scoordinateobjectr

___

__

mod_'

)_(

)_(

_

_

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