MMJ 1113 FINITE ELEMENT METHOD FEM in Elasticityfkm.utm.my/~taminmn/pdf-files/FEM...

MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity

Task:To derive the element stiffness matrix for a 3-node triangular element under plane stress / plane strain condition in elasticity.

Requirements:Constitutive equations (Generalized stress-strain relations)

For variational approach- Examine the functional (based on potential energy)

Minimize the functional - yield finite element equations: [k]δ = f

Finite Element in Elasticity


General state of stress, strain and displacement at a point in a continuous body

x

τxy

τxz

τyzσx

σyy

z

=

yz

xz

xy

z

y

x

γγγεεε

ε componentsstrain

=

yz

xz

xy

z

y

x

τττσσσ

σ components stress

yw

zv

xw

zu

xu

xz ∂∂

+∂∂

=∂∂

+∂∂

=∂∂

+∂∂

=

∂∂

=∂∂

=∂∂

=

yzxy

zyx

, , xv

yu

zw ,

yv , :ndeformatio smallFor

γγγ

εεε

( )( )( )

=

zyxwzyxvzyxu

,,,,,,

componentsnt Displaceme δ


Elasticity Stress tensor

Hook's law for an isotropic material undergoing infinitesimal deformation.

Strain tensor

zxxz

zyyz

yxxy

zzzyzy

yzyyyx

xzxyxx

ij

ττττττ

στττστττσ

σ===

= where

zxxz

zyyz

yxxy

zzzyzx

yzyy

yx

xzxyxx

ij

γγγγγγ

εγγ

γε

γ

γγε

ε===

= where

22

22

22

==

=

Ε−

Ε+

=

ji if 0ji if 1

:deltakronecker

1

ijδ

σδνσνε kkijijij

⁄


Under Plane stress condition

Under Plane strain condition.

Non-zero strain components

Inverted for stress (Plane stress)

( ) ( ) 0 , 0 , 0 zxxzyz ===== ττττσ zyzz

( ) ( ) 0 , 0 , 0 zxxzyz ===== γγγγε zyzz

( )( )

( )( )

( )( )

xyxy

yyxxzzzz

xxzzyyyy

zzyyxxxx

τνγ

σσνσε

σσνσε

σσνσε

Ε+

=

+−Ε

=

+−Ε

=

+−Ε

=

12

1

1

1

( )

( )

( ) xyxy

zz

yyxxyy

yyxxxx

E γν

τ

σ

ενεν

σ

νεεν

σ

+=

=

+−Ε

=

+−Ε

=

12

01

1

2

2


Constitutive equations - Generalized Hook's Law

For a homogenous, isotropic material:

( )( )

−

−

−−

−−

−+Ε

=

yz

xz

xy

z

y

x

yz

xz

xy

z

y

x

γγγεεε

ν

ν

νννν

νννννν

νν

τττσσσ

22100000

02210000

00221000

000100010001

211

[ ] [ ] matrix elasticityor material a is C whereεσ C=


Finite Element in Elasticity

Scope of lecture:Derivation of element equations based on the minimum potential energy principle.Displacement method.

Combination of boundary disp. and tractions

Equilibrium boundary tractions

Displacement compatibility

Along inter element boundary

Continuous stress and displacement

Continuous and equilibrating stress

Continuous displacement

Inside each element

Reissner’s principleMin. Complementary energy

Min. Potential energyVariable Principle

MixedForce or equilibriumDisplacement or compatibility

Model Method


LET US REVIEW THE DERIVATION OF ELEMENT EQUATIONS FOR CONSTANT STRAIN TRIANGULAR (CST) ELEMENT

Keywords:

Variational approach

Minimum potential energy principle


FE formulation for two-dimensional problems

=

=

3

2

1

3

2

1

6

5

4

3

2

1

vvvuuu

δδδδδδ

δ

12

3

u1u2

u3

v2

v3

v1

y

x

Displacement modelTriangular 3-node element

( ) ( )( )

=yxvyxue

,,

δ


yxvyxu

654

321

αααααα

++=++=

assume a linear variation of displacement field within the element

363543

262542

161541

333213

232212

131211

:3 :2 :1 node

yxvyxvyxvyxuyxuyxu

αααααααααααααααααα

++=++=++=++=++=++=


( ) ( ) ( )yucucucxubububuauauau 332211332211332211 21

21

21

++∆

+++∆

+++∆

=

where

123312231

213132321

122133113223321

c , c , b , b ,

, ,

xxxxxxcyyyyyyb

yxyxayxyxayxyxa

−=−=−=−=−=−=−=−=−=

( ) ( ) ( )yvcvcvcxvbvbvbvavavav 332211332211332211 21

21

21

++∆

+++∆

+++∆

=

area of ∆ element33

22

11

111

21

yxyxyx

=

( ) 1,2,3i , 21

332211

332211

=++∆

=

++=++=

ycxbaN

vNvNvNvuNuNuNu

iiiiNi’s are interpolation functions.


[ ] δδ Ν=

ΝΝΝ

ΝΝΝ=

Ν+Ν+Ν=Ν+Ν+Ν=

~

000000

3

3

2

2

1

1

321

321

332211

332211

vuvuvu

vu

vvvvuuuu

Displacement field

The linear interpolation model satisfies both compatibility andcompleteness requirements


Strain–displacement relationships

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

∂Ν∂

=

∂∂

+∂∂∂∂∂∂

=

=

3

3

2

2

1

1

332211

321

321

vuvuvu

xyxyxy

yyy

xxx

yu

xvyvxu

xy

y

x

γεε

ε

[ ] δε Β=

0 0

0

0

00


Derive the element equations using the minimum potential energy principle.General element equations2-dimensional (plane) elasticity problemsApply these equations for case of plane stress, plane strain and

axisymmetric solids.

The functional in the variational principle is the potential energy of a 2-D elastic body acted by surface and body forces.

The area, A, is divided into M discrete elements.

( ) ( ) ( )∑=

=M

e

e vuvu1

,, ππ

T*

F*

C1

x, u

y, v

t(x,y)= thickness


To ensure convergence as element size decreases, the chosen interpolation functions must satisfy compatibility and completeness requirements.

Examine the potential energy equation:

CompatibilityOnly first–order derivatives of the displacement appear in the integrand of the functional.Thus the interpolation functions must ensure continuous displacement at element interfaces.

i.e. C0 continuity in displacement at interfaces.CompletenessThe interpolation functions must ensure that rigid body displacement (uniform displacement) and constant-strain states (uniform first derivatives of displacement) are represented in the limit as element size is reduced.

i.e. C1 continuity of displacement within elements.

( ) [ ] [ ][ ] [ ] [ ] [ ]

[ ] dSTtdA

tdACBtdABCBvu

cA

oT

AA

T

δδ

εδδδπ

~ ~F-

~221 ~~

21,

1

**

*

∫∫∫

∫∫∫∫−

−=


Polynomial containing at least one constant and the linear terms can serve as interpolation functions satisfying both compatibility and completeness requirements.

For an element (e) with r number of nodes:

( ) ( )( )

( )

( )

( )

( )

( ) [ ] ( )

ntsdisplaceme nodal

~

N

,

,N

,,~

1

1i

↓

=

=

=

=

∑

∑

=

=

ee

e

r

iii

r

ii

ee

N

vNu

vyxN

uyx

yxvyxu

δδ

δ


Minimum Potential Energy Principle

A variational principle that hold throughout the system.Potential Energy = strain energy – work by external force

Strain energy of linear elastic body

Initial strain terms

( ) ( ) ( )wvuWwvuUwvu ,,,,,, −=Π

[ ]

[ ] [ ][ ] dv~ BC~21

dv 21

dv 21

T δδ

εε

σε

B∫∫∫

∫∫∫

∫∫∫

Ω

Ω

Ω

=

=

=

C

U

[ ]

[ ] [ ] dv C~221

dv 221U

0T

0o

εδ

εε

B∫∫∫

∫∫∫

Ω

Ω

=

= C


Work done by external force

∫∫+∫∫∫= Ω 1

~~21

S dSTdVFW δδ

Body Force Traction

ZYXF = zyx TTTT =


The potential energy functional

The equilibrium displacement field (u, v, w) minimizes П and satisfies all the boundary conditions.

[ ] [ ][ ] [ ] [ ] 1s

dS ~T-dV ~F-dV ~2~~21

1

δδεδδδ ∫∫∫∫∫∫∫∫ Β−ΒΒ=ΠΩΩ

oTT CC


Theory of Minimum Potential Energy

The displacement field (u , v) which satisfies the equilibrium, and the conditions at the boundary surface is the one that minimizes the potential energy.

( ) ( )( )

( )( )( ) ( )

0,

0,,

11

1

=∑∂Π∂

+∑∂Π∂

=Π

=∑ Π=Π

==

Μ

=

i

r

i i

e

i

r

i i

ee

e

e

vv

uu

vu

vuvu

δδδ

δδvariation

( ) ( )r1,2,.....,i ; 0 ==

∂∂

=∂∂

i

e

i

e

vuππ

Since δui , δvi are independent variations and not necessarily zero, thus


Look at typical node q of the element, we have

The form of this force-displacement relation for node q is

Additional load vector, if preset, can be added to the resultant external load vector Fq.Example: initial force vector due to initial strains from non-uniform temperature.

( )

( )

[ ] ( )

( )

[ ]( )[ ] ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

( )

0**

0

1

=Ν−Ν−ΒΒ

=

∂∂∂∂

∫∫∫∫∫eee

e

s

eq

eqq

eeq

Aq

eeqep

e

A

Tq

q

eq

e

dSTdApdAtC

v

u

δ

π

π

[ ] elementboundary for -loading surface todue vector force nodal F

vectorforcebody nodal

T

q

=

=

=+=

qB

qqT

qB

qqp

F

FFFk δ

[ ] ( )

( )

[ ]( ) [ ] ( ) ( ) ( )eeepo

e

A

Tq dAtC

e

e *ε∫∫ Β


Plane stress versus plane strain


Plane stress condition - small displacement

( )

( )

( )332211332211

332211

332211

21

21

21

ucucucvbvbvbyu

xv

vcvcvcyv

ubububxu

xy

y

x

+++++∆

=∂∂

+∂∂

=

++∆

=∂∂

=

++∆

=∂∂

=

λ

ε

ε

∆=

=

3

3

2

2

1

1

332211

321

321

21

vuvuvu

bcbcbcccc

bbb

xy

y

x

γεε

ε

[ ] δε Β=

000

000

12

3

u1u2

u3

v2

v3

v1

y

x


[ ] εσ

γεε

νν

ν

ντσσ

C

xy

y

x

xy

y

x

=

−−Ε

=

2100

0101

1 2

Plane stress conditionHomogeneous and isotropic solids

Stress-strain relations

( ) xyxy

yxy

yxx

γν

τ

εν

εν

νσ

εν

νεν

σ

+Ε

=

−Ε

+−Ε

=

−Ε

+−Ε

=

12

11

11

22

22

[ ] ( )( )

lyrespective ratio, sPoisson' and modulus sYoung' are and E where22100

0101

211

conditionstrain PlaneFor

ν

γεε

ννν

νν

νν

−−

−

−+Ε

=

xy

y

x

C


Example:For a plane stress case, and a typical node, i

If the thickness of the element is uniform (t(e) constant)

If the thickness of the element varies, estimate

Using table 5.2 (text book)

[ ] [ ]( )( )[ ]( )[ ]( ) ( ) ( ) 1,2,3ji,

eA== ∫∫ eeeeeT

i

ij dAtBCBk

( ) ( ) ( ) ( )332211 NtNtNtt eeee ++≈

( )( )

( ) ( ) ∆=∫∫ .ee

A

e tdAte

( )( )

( ) ( ) ( ) ( )( )eeee

A

e tttdAte 3213

++∆

=∫∫

( )( )

( )eA

e

ii

i

i

ii

ii dAtbcc

b

bccbE

e∫∫

−

−∆

= 00

2100

0101

014

122 ν

νν

ν 0


( )

( ) ( )

( )

∆

=

= ∫∫

*

*e

ee*

*

A

3t

dA te

i

i

i

ii

iB

YX

YXNF

dS *

*

1

= ∫yi

xi

ci

iT T

TNF

Force vector terms, F(e)

Body forces

Surface traction

k(e)

j

i

Lij

s

y

x

( )

=

=

−=⇒

−=Ν

∫

*

*ij

*

*

0ij

2

*

*

0

2L

2Ls-s

dS 1

1

yi

xi

yi

xi

L

yi

xiL

ij

it

iji

TT

TT

TT

LsF

Lss

ij

ij


Arrangement of termsdof at each node : 2No. of nodes for each element: rElement matrix : Force-displacement equations: [ ]( ) ( ) ( )eee Fk =δ

[ ]( )

[ ] [ ] [ ][ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

=

rrrr

qrqq

r

r

e

kkk

kkk

kkkkkk

k

....................

....................

....

....

21

21

22221

11211

( )

( )

=

=

=

r

qe

r

r

r

e

F

F

FF

vu

vuvu

.....

....F , ........

2

1

2

2

1

1

2

1

δ

δδ

δdiscrete nodal disp.

[k](e)


Equivalent Nodal Force Vector

STEPS:

Calculate the total load acting on each element surface due to applied distributed loading.Split the total load proportionally to each of the node along the element edge.Sum the load vector at each nodeResolve the resultant nodal force into the desired (x-,y-) component.


Example 1

Task: To determine the displacement (deflection) and stress field in the plate subjected to the given loadingsince t is only 0.1% of the next smallest dimension of the plate, assume plane stress condition exist.Due to symmetry (in both geometry and loading) along a vertical plane in the center of the plate, consider only one half of the region for FE discretizationChoose 3-node triangular elements (this will yield a constant strain field within each element).

3020 20 20

w=10kN/cm2

E=20x106N/cm2

υ=0.25t=3cm


Boundary Conditions


Continuum discretization

3-node triangular elementsTotal of 18 elements and

16 nodes.


Result

Both tensile and compressive stresses occurs because plate in bending

Stresses are compressive in y-direction


Example 2

Simply-supported beam


Finite Element Mesh


Results

MMJ 1113 FINITE ELEMENT METHOD FEM in Elasticityfkm.utm.my/~taminmn/pdf-files/FEM...

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