FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_06.pdf · Instituto Tecnológico de Aeronáutica AE-245 4 The exact problem may be stated as: find u∈S such

Instituto Tecnológico de Aeronáutica

AE-245 1

FINITE ELEMENTS I

Class notes


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6. Mixed and Penalty Methods


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• The FE technique is optimal in some sense

• “Best approximation” and error estimative

• Establish convergence of the technique

MIXED AND PENALTY METHODSPreliminaries


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The exact problem may be stated as: find u∈Ssuch that for all w∈V

MIXED AND PENALTY METHODSBest approximation

Γ+= ),(),(),( hwfwuwa

u satisfies the essential (geometric) boundary conditions

w satisfies the corresponding homogeneous essential boundary conditions


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The approximate problem may be stated as: find uh∈Sh such that for all wh∈Vh


Γ+= ),(),(),( hwfwuw hhhha

uh satisfies the essential (geometric) boundary conditions

wh satisfies the corresponding homogeneous essential boundary conditions


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Assumptions


1. Sh ⊂ Sand Vh ⊂ V

2. a(•,•), (•,•) and (•,•)Γ are symmetric and bilinear forms

3. a(•,•) and ||•||m define equivalent norms where

( )2/1

...,...,,, ...

Ω+++= ∫

Ω

dwwwwww ljkiljkijijiiimw


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Theorem: let e = uh – u denote the error in the finite element approximation

hhhh

h

Saa

a

∈−−≤

=

UuUuUee

ew

allfor ),(),( b.

0),( a.


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Pythagorean theorem: assume Sh = Vh.

),(),(),( eeuuuu aaa hh +=

),(),(

),(),(),(

uuuu

uuuuee

aa

aaahh

hh

≤

−=Consequences:

The approximate solution underestimates the strain energy


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The principle of minimum potential energy

wuU εε +=

where ε ∈ ℜ. Every member of S can be represented in the form above for some w ∈ V and ε ∈ ℜ.


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Define potential energy:

Γ−−= ),(),(),(2

1)( hUfUUUU εεεεε aI

The potential energy is stationary if and only if the variational equation is satisfied

The potential energy is minimized at u.

The approximate solution overestimates I


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MIXED AND PENALTY METHODSEffect of numerical quadrature

Theory developed assumes that all integrals are evaluated exactly

This is a serious limitation for isoparametricelements because of numerical integration

Number of Gaussian points should be carefully determined


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MIXED AND PENALTY METHODSSummary

• In practice results obtained via Galerkinformulation are very good.

• The best approximation property is a guarantee of a reasonable convergence.

• There are problems: what if the best approximation is not so good? That is, what if there is no reasonable uh that approximates u?


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MIXED AND PENALTY METHODSIncompressible elasticity and Stokes flow

• Motions that preserve volume locally

• After deformation each elementary portion of the medium has the same volume as before deformation

• In elasticity, materials with ν approaching 1/2 are often modeled as incompressible.


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MIXED AND PENALTY METHODSIncompressible elasticity

)( ,,, ijjiijkkij uuu ++= µδλσ

ννµλ21

2

−= ∞=

→λ

ν 2/1lim

The constitutive equation must be reformulated

)( ,, ijjiijij uup ++−= µδσ

)21)(1( νννλ

−+= E

GE =+

=)1(2 ν

µ


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p is the hydrostatic pressure that is an additional variable. The additional equation is the incompressibility condition:

0)(div , == kkuu


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Given fi, gi and hi find ui and pressure p such that

)(with

on

on

in 0

in 0

,,

,

,

ijjiijij

hiijij

giii

ii

ijij

uup

hn

gu

u

f

++−=

Γ=Γ=Ω=Ω=+

µδσσ

σ


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MIXED AND PENALTY METHODSStokes flow

Equation identical to the incompressible elasticity situation

u is the velocity of the fluid and µ is the fluid viscosity

Stokes flow governs highly viscous flows


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MIXED AND PENALTY METHODSStokes flow

x

ylid

vx = 1, vy = 0

vx = 0vy = 0

vx = 0vy = 0

vx = 0, vy = 0

-300

-200

-100

0

100

p

0.000.20

0.400.60

0.801.00

x

0.000.20

0.400.60

0.801.00

y

x

y

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.10.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0vx

0.9370.8750.8120.7500.6870.6250.5620.5000.4370.3750.3120.2500.1870.1250.062

-0.000-0.063-0.125-0.188-0.250

x

y

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.10.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0vy

0.6010.5510.5020.4520.4020.3520.3020.2530.2030.1530.1030.0540.004

-0.046-0.096-0.145-0.195-0.245-0.295-0.345

0

2

=⋅∇=∇+∇−

v

bv pν


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MIXED AND PENALTY METHODSMixed and penalty methods

Matrix FE problem:

fKd =Alternative formulation: minimize

fdKdd

d TT

F −=2

)(

Total potential energy function


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One parameter family of displacements:

0) (0

=

+=ε

εε

cdFd

d

cd ε+ whereε ∈ ℜ and c is arbitrary

If F is minimized by d


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)() (0

fKdccd −=

+=

TFd

d

εε

ε

Minimization:

c is arbitrary fKd =


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MIXED AND PENALTY METHODSLagrange multiplier method

Consider the function

where

)()(),(* ddd mGFmF +=

0)( =−= gG TQd1d


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0)()(), (0

* =−+−+=

++=

glmlmFd

d TQQ

T d1f1Kdccdε

εεε

Minimization:

c is arbitrary

l is arbitrary

=

gmTQ

Q fd01

1K

Interpretation: m is a force that must exist in order to maintain the constraint valid.


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Mixed equation:

=

gmTQ

Q fd01

1K

The equation above is mixed since it contains displacement and force unknowns.

The “stiffness” matrix is not positive-definite anymore


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MIXED AND PENALTY METHODSPenalty method

Approximation to the Lagrange multiplier method

)(2

)()( 2** ddd Gk

FF +=

wherek is a large known number. Minimization yields

QTQQ kgk 1fd11K +=+ )( gdQ

k=

∞→lim


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MIXED AND PENALTY METHODSPenalty method

gdQk

=∞→

lim

• Approximation of the incompressible case by a slightly compressible formulation

• The larger k the better

• Too large k may lead to numerical instabilities


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Formulation that solves both compressible and incompressible elasticity problems.

λ

µδσp

u

uup

ii

ijjiijij

+=

++−=

,

,,

0

)(


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Strong form

Given fi, gi and hi find ui and p such that

)( where

on

on

in 0/

in 0

,,

,

,

ijjiijij

hiijij

giii

ii

ijij

uup

hn

gu

pu

f

++−=

Γ=Γ=Ω=+Ω=+

µδσσ

λσ


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Weak form

Given fi, gi and hi find ui∈Si and p∈P such that for all wi∈Vi and q∈P

∑ ∫∫

∫∫

= ΓΩ

ΩΩ

Γ+Ω

=Ω+−Ω+

3or 2

1

,,, )/()(2

1

iiiii

iiijijji

hi

dhwdfw

dpuqdww λσ


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Weak form: variational equation

Γ+=

+−− ),(),()(div,)),(div(),( hwfwuwuwλp

qpa

∫Ω

Ω++= duucwwa kllkijklijji )(2

1)(

2

1),( ,,,,uw

)( jkiljlikijklc δδδδµ +=


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Galerkin formulation

Given f, g and h find uh= vh + gh ∈ Sh and ph∈Ph

such that for all wh∈Vh and qh∈Ph

( )))(div,(),(),(),(

/)(div,)),(div(),(hhhhhh

hhhhhhh

qa

pqpa

ggwhwfw

vwvw

−−+

=+−−

Γ

λ


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• Interpolation expressions for ph

• ph must be square integrable; it may be discontinuous across element boundaries

• Wider space of functions allowed for p

• Arbitrary combination of interpolation may lead to poor numerical performance


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Continuous bilinear displ.

Discontinuous constant pressure


Continuous bilinear pressure


Discontinuous bilinear pressure


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Matrix equation

=

b

fp

d

MG

GKT

K is symmetric and positive-definiteM is symmetric and negative-definiteWhen ν = 1/2, M = 0


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• Problems with mesh locking

• Babu ška-Brezzi stability condition

• Approximate method: constraint count



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• Mixed formulation use two field variables: displacement and pressure

• Difficulty associated with the selection of the k factor in penalty methods

• Can displacement based formulations satisfactorily model the nearly incompressible problem?

• Use of reduced and selective integration

MIXED AND PENALTY METHODSReduced and selective integration


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Element stiffness matrix


∫Ω

Ω= djTiij DBBk DDD +=

D

D

Part ofD related to µ

Part ofD related to λijijij kkk +=


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Element stiffness matrix


k ij is the element stiffness matrix that appears in the mixed formulation in the term a.

Nearly incompressible problem: λ/µ >> 1

Terms in k ij tend to be very large when compared to those in k ij.


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Special treatment must be given to k ij

Reduced integration


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Selective reduced integration:

Reduced integration used only on the λ-term

Uniform reduced integration:

Reduced integration used on both λ- and µ-terms


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Equivalence theorem

• Several mixed elements are equivalent to reduced integration elements, that is, both formulations lead to identical stiffness matrices;

• At Gauss points of the reduced integration the pressure agrees with that of the mixed method;

• Reduced integration is a simple way of achieving performance without the complications of mixed type elements.


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MIXED AND PENALTY METHODSPatch test

The reduced integration method does not fully comply with the Galerkin recipe since integration

is not exact.

How can we assess the performance of elements that violate the Galerkin recipe?

Irons proposed the “patch test”.


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MIXED AND PENALTY METHODSPatch test

Example: prescribe displacements on boundaries to represent the states 1, x and y.

1 2 3

4

5

6

7 8 9

y

y

x

x

vu

06

05

04

03

102

011

testTests 1 and 2: 1

Tests 3 and 4: x

Tests 5 and 6: y

Test: node 5 behaves

accordingly?


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MIXED AND PENALTY METHODSRank deficiency

Two-dimensional bilinear elasticity element

∫Ω

×××× Ω= dT83333888 BDBk

Exact 2×2 Gaussian integration k is of rank 5

Reduced 1×1 Gaussian integration k is of rank 3

3 rigid body modes


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MIXED AND PENALTY METHODSRank deficiency

Spurious zero energy modes

Whenk is of rank 3 there are 5 zero eigenvalues: 3 correspond to the rigid body modes and 2 are spurious zero energy modes.

ξ ξ

η η


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MIXED AND PENALTY METHODSNonconforming elements

Approximation in the Galerkin recipe requires:

VV

SSh

h

⊂

⊂

The convergence of the FEM cannot be guaranteed if the conditions above are not met. Elements that do not satisfy

Sh⊂Sand Vh⊂V are referred to as incompatible or nonconforming.


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MIXED AND PENALTY METHODSIncompatible modes element

Motivation: difficulty to represent pure bending

Pure bending

Exact pure bending response

Element in pure bending

Element in pure bending response


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Exact pure bending:

u = −σ0cxy

v = −σ0c[(h/2)2 − x2]/2

Element in pure bending:

u ≈ xy

v = 0


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Remedy proposed: add quadratic terms

∑∑==

+=6

5

4

1

),(),(),(i

iii

iih NN αdu ηξηξηξ

26

25

1),(

1),(

ηηξ

ξηξ

−=

−=

N

NIncompatible modes


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Terms N5 and N6 are used only to obtain the stiffness matrix but not to obtain the loading.

=

0

f

α

d

kk

kk

ααα

α

d

ddd

dkkα dααα1−−= dd αααα kkkkk 1~ −−=

Effective

matrix


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• The assembly process is done with k

• Flaw: for arbitrary quadrilaterals the behavior is very poor

• Improved version exists that pass the patch test

∼

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FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_06.pdf · Instituto Tecnológico de Aeronáutica AE-245 4 The exact problem may be stated as: find u∈S such