SOLVING LINEAR ELASTICITY BY FEM

SOLID MECHANICS

CONTINUUM MECHANICS

3D (SOLID)

2D (TRUSS)

1D (BAR)

ELEMENT FORMULATION FOR

FEM

ASSUMPTIONS

• Deformations are small• Behavior of material is linear• Dynamic effects are neglected• No gaps or overlaps occur during the

deformation of the solid

1 DIMENSIONAL

Ex: Elastic bar- (linear stress analysis/ linear elasticity)

? To find stress distribution(σ(x)) in bar deformation ε(x)

displacement of points (u(x))

STRONG FORM• Equation and boundary for physical system• Partial differential equations (ordinary

differential equation-1D)

WEAK FORM• Integral equation of strong form• Formulate FEM

STRONG FORM WEAK FORM

DISCRETE EQUATIONS

APPROXIMATION OF FUNCTIONS

• Bar condition1. Equilibrium2. Elastic stress-strain law3. Displacement compatible4. Satisfy strain displacement equation

• Equilibrium eq. of int. force(p(x)) and ext force(b(x)) in x axis direction

lbdx

duAE

x

xxExx

u

x

xuxxux

xxAxpxA

xpx

xbx

xp

xxb

x

xpxxp

xxpxx

xbxp

x0 0)(

)()()(

)()()(

)()()(- )(

)()(

0)()(

0)2

()()(

0)()2

()(

Divide with △x

Limit △x=0

stress

strain

Hook’s Law

2nd order ordinary diff. eq

STRONG FORM--SOLUTIONS• CLUE - either LOAD or DISPLACEMENT at boundary (2 ends)• Essential boundary condition/ Displacement boundary

condition x=l or u(x=l)=u

• Natural boundary condition/traction boundary conditionx=0

unknown is u(x)

(data)given are and ,

)(

)0(

0)(

0

but

ulxu

tdx

duEx

bdx

duAE

dx

d

x

STRONG FORM

tA

p

dx

duE

x

)0(

)0()0(

0

Traction (stress) bound. Cond. (natural)

Disp. bound. Cond. (essential)

Governing strong form eq.

WEAK FORM• What?

- Restated a partial diff eq. into integral form (a trial solution)• How to create ??1. For governing eq. Multiply w(x) and integrate with domain(0,l)2. For trace boundary multiply w(x)A at x=0 (no integral bcoz only

holds at point) w(x) is weight /test/arbitrary function where w(l) =0

wtdx

duEwA

wdxbdx

duAE

dx

dw

x

l

0

0

0

0 (1)

Some restriction on weak form

• denotes that w(x) is an arbitrary fn• weight fn- enforcer to be zero by its arbitrariness• we did not enforce boundary cond. It is easy to construct trial or candidate

solutions u(x) that satisfy displacement condition• Now, transform into the form containing first derivatives. That will lead to a

symmetric stiffness matrix and simplify the treatment of traction boundary• Second derivation of u(x) need smooth trial solution(easy to construct by 1

D) • The trial solution that satisfy WEAK FORM is the solution of the STRONG

FORM• A trial solutions that is smooth and satisfies the essential boundary cond.

admissible• A weight function that is smooth and vanishes on essential bound.

admissible•

w

MATH

l

xlx

ll

l

l ll

dxdx

dwfwfwfdx

dx

dwfwfdx

dx

dfw

dxdx

dwfdxwf

dx

ddx

dx

dfw

dx

dwfwf

dx

d

dx

dfw

dx

dwf

dx

dfwwf

dx

d

0

0

00

0

0 00

)()()(

)(

)( -- )(

Rearrange the Weak Form EQ where f=AE(du/dx)

lll

dxdx

duAE

dx

dw

dx

duwAdx

dx

duAE

dx

dw

000

Rewrite and subtitute stress, σ=AEdu/dx

ll

xlx lwwbdxdxdx

duAE

dx

dwwAwA

00

0 0)( with w 0)()(

For w(l)=0, and Second term of eq (1), (wAt)x=0

l

x

l

lwwbdxtwAdxdx

duAE

dx

dw

0

0

0

0)( with w )(

σ(x=0)=-t

WEAK FORM

Arbitrary Boundary conditionBoundary of 1D which consists of two end points is denoted by ΓBoundary for displacement Γu

Boundary for traction Γt

The traction and displacement both cannot be prescribed at the same boundaryΓu U Γt=0 Γu ∩ Γt=0

Unit normal to the body as, n where n=-1 at x=0 and n=+1 at x=l

Notation of Ω indicates any limits of integration Admissible weight Fn.And H is infinite dimensional set

uon

on

0 0)(

uu

tdx

duEnn

lxbdx

duAE

dx

d

t

l

oUwwbdxtwAdxdx

duAE

dx

dwt

0

)(

uwHxwxwU on 0,)()( 10

STRONG FORM

WEAK FORM

DISCRETIZATION

lT

Tl

lwwbdxwdxdx

duAE

dx

dw

0

0xT

0

0)( with A)t(w-

The procedure1. Evaluate the weak form for FE trial solutions and weight fn.2. By invoking the arbitrariness of the weight fn. Deduce a set of linear algebraic eq.3. Use transpose of weight fn as w(x) is scalar and for consistency

Global The FE weight fn. w(x)≈wh(x)=N(x)w Where N(x)= matrix shape fn w(x)=w1N1(x)+w2N2(x)+w3N3(x)

The trial solution u(x) ≈ uh(x)=N(x)d Where u(x)=u1N1+u2N2+u3N3

Replace the weak form over the entire domain by sum of integrals over the element domains.

02

1

2

1

1

e

e

e

e

el

x

x

eeTeTe

ee

Tx

x

en

eAtwbdxwdx

dx

duEA

dx

dw0x )(-

E to indicate element

element ,e, weight fn. and trial fn

eTeT

TeeTeTeT

eee

eee

dx

dww

dx

duxu

Bw Nw

dB dN

,

,)(

Substitute element weight fn. and trial fn

0)( 01

2

1

2

1

xeeT

x

x

eTeeeex

x

eTn

e

eT

e

e

e

e

e

e

e

el tAbdxdxEA f

fK

NNdBBw

e

e

e

dxEAdxEA eeeeTeeex

x

eTe BBBBK2

1 e

et

e

e

e

tAbdxtAbdx eeTeTx

eeTx

x

eTe

f

f

NNNNfe

)()( 0

2

1

ELEMENT STIFNESS MATRIX & ELEMENT EXTERNAL FORCE MATRIX

Element stiffness matrix Element external force matrix

Element & Global matrix relationship

Element and global matrix relation we=Lew de=Led

Substitutes into weak form of element domains;

01 1

el eln

e

n

e

eeTeeeTT fLdLKLw

Global/assembled stiffness matrix Global/assembled external force matrix

eln

e

eeeT

1

LKLK

eln

e

eeT

1

fLf

Re-subtitute into eq

00

0)(0)(

1

1

ww

lwwwT

T

except rw ,f Kdr

except fKdw

Where r is called the residual

2 Dimensional KINEMATICS (STRAIN –DISPLACEMENT RELATION)

Displacement in matrix and vector form,

juiuuu

uyx

y

x

u ,

Strain and shear strain

2

1

,

y

u

x

ux

u

x

u

xyxy

yyy

xxx

strain vs displacement eq.

y

xss

xy

yy

xx

x u

uu

x

y

y

x

isWhere

S

s

/

/

0

/

0

/

operator matrix gradient symmetric a

STRESS AND TRACTION

Stress component matrix for 2D

yy

xy

xy

xx

Unit normal to the surface, njninn yx

The force equilibrium of triangular body shown in Fig. 9.14

0

dxdydt yx

Dividing by dΓ and noting that dy=nxdΓ and dx=nydΓ

0

yyxx nnt

Multiply by unit vectors i and j

nnnt

nnnt

yyyyxxyy

xyxyxxxx

.

.

Stress and Traction in matrix form

τnt

EQUILIBRIUM

Figure 9.5 shows body force and the surface traction acting in the xy-planeBody force and traction vectors are written as

0),(2

,2

,,2

,2

yxyxbxx

yxxx

yxyyx

xyyx

x yyxx

The equilibrium equation is given by

jtittjbibb yxyx

,

Dividing the above eq. by x y, taking limit as △ △ △x 0, △y0,

0,0

0

yyyx

yx

bb

j ibyx

form; vector the in

and vector unit by multiple and

or 0bσ TS

Equilibrium Eq.

CONSTITUTIVE EQUATION (STRESS & STRAIN EQ)

Hooke’s LawDεσ

Matrix D will depends on whether one assumes a plane stress or a plane strain.

Plane strainthe body is thick relative to xy plane strain normal to plane εz, is zero shear strain that involves angles normal to plane, γxz & γyz are assumed to vanish

Plane stressthe body is thin relative to dimensions in xy plane. no loads are applied on z faces, stress normal to xy plane, σzz is vanish

For isotropic 2D

2/)21(00

01

01

)21)(1(

2/)1(00

01

01

1 2

E

E

D

DPlane stress

Plane strain

STRONG & WEAK FORMSEquilibrium Eq., Kinematics Eq., Constitutive Eq. with a boundary condition as

ttuΓ

uu and on uu ,0

u

tyx

T

on d)

on and c)uD b)

on a)

uu

tntn

bb

yx

yyyx

0,0

STRONG FORM

To obtain weak form, define weight fn. and trial fn. Using Green’s theorem and finally

,,

,,1

0

1

0

u

u

wHU

HUwhere

Uwwww

on 0uu

on uuuu

bddtudD)( TT

s

T

Ωs

t

WEAK FORM

DISCRETIZATIONGlobal nodal displacement matrix

npnp ynxnyxyx uuuuuu ...2211d

Element trial solution

Element weight fn.

Element shape fn.

Weak form over element domains Ωe

Express the strain in terms of element shape fn. and nodal displacement

eeee (x,y)yxyxyx dNuu ),(),().(

eeeTeTT (x,y)yxyxyx Nwww ),(),().(

enen

ee

enen

ee

NNN

NNN

0...00

0...00

21

21N

01

nel

e

eTeTes

eeTs

e et

e

ddd bwtwuDw

eeees

es

e

xy

yy

xx

dBdNu

Strain- displacement matrix, Be

Element stiffness matrix

Element external force matrix

Weak form

e

dEA eeeeTe BBK

eΓ

et

e

f

Γ

f

tNNf

dΓbdx eTeTe

e

F

n

e

eeTn

e

eeeTTelel

-LKL w fLdw

fK

011

x

N

y

N

x

N

y

N

x

N

y

N

x

N

y

N

y

Nx

N

x

N

x

N

en

en

eeee

en

ee

en

ee

ese

enen

en

en

...

0...00

0...00

2211

21

21

NB

Where, weight fn areeTeTTeeTe

s BwwBw )()(

FT

FT

w

w

rw ,f Kdr

fKdw

0

0)(

General Equation

3D2D1DEntity DomainΩ Boundary ΓOne Dimension(1D)Two Dimensions(2D)Three Dimension(3D)

Line SegmentTwo dimensional areaVolume

Two end pointsCurveSurface

3D

0bσ

Ts

yz

xz

xy

zz

yy

xx

Ts

yxz

zxy

zyx

000

000

000

z

y

x

z

y

x

zzzyzx

yzyyyx

xzxyxx

n

n

n

t

t

t

τ

n t

nt T

Equilibrium

Stress-traction relation

Strain-displacement relation

Isotropic Hooke’s Law

zyxT

yxxzxyzzyyxx

s

uuu

u

u

D

Dεσ

2

2100000

02

210000

002

21000

0001

0001

0001

)21)(1(

E