FE Review Web

7/27/2019 FE Review Web

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Strong and weak

forms

Finite Elementapproximation

Finite Elementformulation

Implementationissues

SMA01: Advanced numerical modeling of

mechanical structuresFinite Element - Basics

D.Brancherie, A. Rassineux

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Strong and weak

forms




Strong and weak forms

Definitions

A scalar problem : the thermal diffusion problemA vectorial problem : the elasticity problem

linear elasticityquasi-static

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Strong and weak

forms




Strong and weak form

Definitions

Strong form :

set of differential or partial differential equations + boundary

conditions written on each point of a domain

D(u(x)) = 0 x

with boundary conditions

BC(u(x)) = 0 x

weak form (or variational formulation) :

an integral version of the strong form

w(x) D(u(x)) = 0 w(x)

with boundary conditions

BC(u(x)) = 0 x

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Strong and weak

formsFinite Elementapproximation



A scalar problem : the thermal diffusion

The strong form :

Equilibrium

cTt

= divqth + Pth

for stationary problem :

divqth + Pth = 0 x

Behavior : Fouriers law(homogeneous material)

qth = kT

Boundary conditions

qth = qd ontT = Td onu

qd

t

Tdu

Thermal problem

kT = Pth x

with qth = qd on t and T = Td on u4 / 2 7
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Strong and weak

formsFinite Elementapproximation



A sclar problem : the thermal diffusion

Weak form of equilibrium

div qth Pthwd = 0 wuse of the integral theorems :

wkT d

wPth

d

(kT

qthn)w dS = 0 w

remark : weak form and strong form are equivalent

practically, for FEM issues, w can be chosen in KA0

KA0 = {w regular | w = 0 on u}

Weak form for FE issues

find T KA such that

wkT d

wPth

d

(kT

qth

n)w dS = 0 w KA0

where KA = {T regular | T = Td on u}5 / 2 7
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Strong and weakforms




A vectorial problem : the continuum mechanics problem

The strong form :

Equilibrium (quasi-static)div + b = 0 x

n = t = td on t

Kinematics(small perturbation hypothesis)

(u) = 12

u +uT

u = ud on u

Behavior (linear elasticity, isotropic,homogeneous)

= C : (u)ou = tr((u))1 + 2(u)

td

t

udu

Navier equation for elasticity

( + )(divu) + u + b = 0 x

with u = ud on u and n = t = td on t6 / 2 7
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Continuum mechanics problem

Weak form of equilibrium

(div

+ b) w d = 0 w

use of the integral theorems (Green formula) :

: (w) d

b w d

n

tw dS = 0 w

remark : weak form and strong form are equivalent

practically, for FEM issues, w can be chosen in KA0

KA0 = {w regular | w = 0 on u}

Weak form for FE issues

find u KA such that

: (w) d

b w d

t

td w dS = 0 w KA0

where KA = {u regular | u = ud on u} and = ((u))7 / 2 7

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Continuum mechanics problem

Weak form for FE issuesIn the case of linear elastic isotropic homogeneous material, the

solution u solves :

(u) : C : (w) d =

b w d +

t

td w dS w KA0

Remarks :for elastic problems, the solution u can also be defined as theminimum of the potential energy

potential energy :

(v) = 12

(v) : C : (v) d

b v d t

td v dS

u = arg minvKA

(v)

strong /weak form :

derivatives order lower in the weak form than in the strong form8 / 2 7

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Summary (for elasticity problem)

Objectivefind u KA solution of :

(u) : C : (w) d =

bw d+

t

tdw dS w KA0

where u = u(x) is a field defined on

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( )
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Tensor to vector notations (Voigt notations)Second order tensors :

=

xxyyzz

2 xy2 xz2 yz

and =

xxyyzzxyxzyz

Fourth order tensor : elasticity tensor

C =

+ 2 0 0 0 + 2 0 0 0 + 2 0 0 00 0 0 0 00 0 0 0 0

0 0 0 0 0

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S (f )
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(u) : C : (w) d =

bw d+

t

tdw dS w KA0


Weak form in Voigt notations

(w)TC(u) d =

wTb d+

t

wTtd dS w KA0

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S (f l i i bl )
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(u) : C : (w) d =

bw d+

t

tdw dS w KA0


Weak form in Voigt notations

(w)TC(u) d =

wTb d+

t

wTtd dS w KA0

FE purposefind an approximation uh Sh KA of the solution u solvingthe previous equation by choosing a priori the form of thesolution

the weak form is not solved w KA0 but only forw Vh0 KA0

FE mesh, FE approximation9 / 2 7

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O li
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Outline

Discretisation of the domain : mesh

Approximation, shape functions

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M h
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Mesh

Discretisation of the domain into h

Geometry approximation : h approximation of

Mesh :partition of h into elements connected through their nodes

h =

Nele=1

e with e

f(e = f) =

an edgea vertex

node element

1

2 3

Forbidden situation

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M h
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Mesh

Practical definition of the meshThe mesh is fully defined by giving

the node coordinate tabledefinition of the node position

node number x y z

1 x1 y1 z1...

......

...i xi yi zi...

......

...

the connectivity tabledefinition of the elements through their nodes

element number node 1 node 2 node 3

1 node 1(1) node 2(1) node 3(1)...

......

... node 1() node 2() node 3()...

.

.....

.

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FE approximation and shape functions
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FE approximation and shape functions

Main idearepresentation of a field through a minimum number ofparameters (the degrees of freedom)

parameters : the value of the field on the nodes of the mesh

Shape (interpolation) functionsFor each node i of the mesh, a spatial function Ni(x) is definedsuch that :

Ni(xj) = 0 if j = i

1 if j = i

Ni(x)

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FE approximation
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FE approximation

1 2 3 4 5 6 7node

For elasticity, approximated displacement field :

uh(x) =Nnodei=1

Ni(x)di = N(x)d

with di = uh(xi), d the nodal solution vector (displacement,

temperature, ...) andNi(x

) the FE shape functions.

uh(x) =

N1 0 N2 0 ... N Nnode 0

0 N1 0 N2 ... 0 NNnode

N(x)

d1xd1y

.

.

.

dNnodex

dNnodey

d14/27

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FE approximation
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FE approximation

1 2 3 4 5 6 7node

Remarks

the field uh(x) is defined for all x through the definition ofd

d is a vector of parameters, of limited length Nnode defining thesolution

for standard FE, shape functions are polynomial functions (moreelaborated for XFEM or isogeometric elements)

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Outline
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Outline

FE formulation

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From continuum mechanics to FEM
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Construction of the FE problem

Weak form (recall)(u) : C : (w) d =

b w d +

t

td w dS w KA0

FE formulation (Voigt notations)

h

(uh)TC(wh) d =h

wTb d +

t

hwTtd dS wh KA0

FE approximation

solution : uh

(x) =

Nnode

i=1

Ni(x)di = N(x)d

test function : wh(x) =

Nnodei=1

Ni(x)qi = N(x)q

(uh) = Bd where B = LN with L the matrix form ofs

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FE problemfind uh such that

h

(uh)TC(wh) d =

hwTb d +

th

wTtd dS wh KA0

FE formulation

find d such that q s.t. q = 0 on uh

qT

hBTCB d

d = qT

hNTb d +

th

NTtd dS

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FE problem
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FE problem

find d such that q s.t. q = 0 on uh

qT

h BTCB dd h N

Tb d th N

Ttd

dS = 0

find d solution of

hBTCB d

K

d = hNTb d +

thNTtd dS

fext

FE problem

find d solution ofKd = fext

Remarks :K : (Ndof Nnode, Ndof Nnode)fext : (Ndof Nnode, 1)

K is a symmetric, def., positive matrix (with a lot of zeros !)19/27

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Outline
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Outline

Construction ofK and fext

Reference element

Numerical integration

Numerical resolution

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FE problem - construction ofK and fext

K = hBTCB d =

Neleme=1

eBTCB de

Term ij of matrix K

Kij =Neleme=1

eBTi CBj d

e

case 1 : i and j have disconnected

supports : Kij = 0

case 2 : Supp(i) Supp(j) =

Kij =

eSupp(i)Supp(j)

eBTi CBj d

e

only a limited number of elements have a contribution to KijElement stiffness :

computation on each element ofKe =

eBTeCBe d

Stiffness K obtained by assembly procedure of all Ke

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FE problem - construction ofK and fext

K = hBTCB d =

Neleme=1

eBTCB de

Term ij of matrix K

Kij =Neleme=1

eBTi CBj d

e

case 1 : i and j have disconnected

supports : Kij = 0

case 2 : Supp(i) Supp(j) =

Kij =

eSupp(i)Supp(j)eBTi CBj d

e

only a limited number of elements have a contribution to KijElement stiffness :

computation on each element ofKe =

eBTeCBe d

Stiffness K obtained by assembly procedure of all Ke

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Isoparametric finite elements
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p

Reference element ( to limit the number of geometry to consider)introduction of the natural coordinates : ,

geometry described by : x =

Nnodei=1

Ni(, )xi

Example for a quadrangular element :

1 2

4 3

1

3

2

4

Ni(, ) =14

(1 i)(1 i)Example for a linear triangle :N1(, ) = 1 , N2(, ) = , N3(, ) =

Isoparametric elementsthe geometry and the solution are interpolated with the sameshape functions

uh =

Nnode

i=1

Ni(, )di and x =Nnode

i=1

Ni(, )xi

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g

Computation ofKe, fe,ext

Numerical integration : only an approximation of the integrals iscomputed :

ref

g(, ) d Nint=1

g(, )w

(, ) : natural coordinates of integration pointsw : integration weight

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g

Computation ofKe (2D case)

Ke = eBTCB g(x,y)

dxdy

1 change of variable : (x, y) (, )

Ke =

eg(x, y) dx dy =

ref

g (x(, ), y(, )) | detJ(, )| d d

2 numerical integration (quadrature formula)

Ke =eg(x, y) dxdy

=

ref

g (x(, ), y(, )) | detJ(, )| d d

Nint

=1

g (x(, ), y(, )) | detJ(, )|w

Ke =

Nint=1

BT(, )CB(, )| det J(, )|w

only the shape functions and their derivatives with respect to

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Architecture of a FE code
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Global level of FE code

Equilibrium :

Nelem

Ae=1

fe,int fe,ext

= 0 Ku = fext

Element level

Computation :Ke =

eBTCB dx

fe,int =eBTAdx, fe,ext

Integration points level

Computation : = C

= Bu

uK

e

f

e,int

, fe,ex

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Global computation - static
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General form to be solved :

Ku = fDirect method : (Gauss elimination)

triangular decomposition : K = LUwith L lower triangular part, U upper triangular part(diag(L)=1, diag(U)=diag(K))

solution : Uu = yLy = f

2 lower and upper triangular systems to be solved

Iterative method : conjugate gradient or preconditionnedconjugate gradient

use the terms in K directly, no need to decompose K

efficient for large problems

ifK symmetric definite matrix

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Global computation - static
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Gauss elimination : computation cost

system ofn linear algebraic equations

Triangular decomposition : K = LU

additions n3n3

n3multiplications n3n3

divisions n(n1)2Lower triangular system : Ly = f

additions n(n1)2 n2

multiplications n(n1)2

Upper triangular system : Uu = yadditions n(n1)2

n2multiplications n(n1)2divisions n

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