Finite Element Modeling & Analysis

Human-centered CAD Lab.

1 2007-10-24

Finite Element Modeling and Analysis

Structural problems Stress, strain, distribution

Heat transfer problems Temperature distribution

Electrostatic potential problems

Fluid mechanics problems

Vibration analysis problems

2

Finite Element Modeling and Analysis – cont’

When geometry of problem domain is complicated, analytic solution can not be derived Numerical approximate solution

Even for simple geometry, analytic solution cannot be obtained if the object is composed of composite material

3

Problems which cannot be solved analytically

4

Approximation of each object by an assemblage of finite element

5

Analysis Procedure

Input material properties and boundary condition before elements are generated B.C

Known displacement

Known external force

Known temperature

Pre-processor Allow the interactive input of B.C’s

Generate meshes automatically

6

Choice of Element

Select element type among those supported by the FEM package in use

Select element type mostly suitable for the given problem ← “know-how”

Determining element size is also a “know-how”

Use small elements where rapid change in stress or strain is expected

7

Analysis Procedure – cont’

System equations are composed by FEM package

System equation Relation between B.C. and unknown (displacement,

temperature, …) at every node

→ By Solving system equation, unknowns at each node

is obtained

Unknown values (displacement, temperature, stress,strain, …) at arbitrary location is obtained byinterpolating the values at nodes

Post-processor displays color coding or contour plot

8

Formulation of System Equation

9

Formulation of System Equation – cont’

: External surface tractions

: Body forces

: Concentrated external forces

10

SfBfiF

Formulation of System Equation – cont’

Only displacement at any X, Y, Z

needs to be solved

→ Strain and stress are derived from

11

Z)]Y, W(X, Z),Y, V(X, Z),Y, U(X,[UT

ZXYZXYZZYYXXT ε

ZXYZXYZZYYXXT τ

U

Formulation of System Equation – cont’

Principle of virtual displacement total external virtual work

= total internal virtual work

‘ – ’ indicates the quantities caused by virtual displacement

12

i

i

iS

S

SB

V

T

V

T TT

dSdVdv FUfUfU ε

Formulation of System Equation – cont’

Let displacement of all the nodes

where N : Number of nodes

n = 3N : Degree of freedom

13

U

NNN222111T w v u w v u w v u U

U U U U n321

Formulation of System Equation – cont’

Displacement at arbitrary location in an element

→ Is approximated by proper shape function that

interpolates the displacements at nodes of the

element

→ Thus is the unknown to be solved

14

TU

Formulation of System Equation – cont’

Displacement at arbitrary x, y, z in m-element

Strain is obtained by differentiating displacement

Stress – Strain relation

↑

Element initial stress

15

UHu

)z,y,x()z,y,x( )m()m(

UB

)z,y,x()z,y,x(ε )m()m(

I(m)(m))m()m( τ ετ C

Formulation of System Equation – cont’

→ Virtual nodal displacement

16

i

ii

mS

mmSmS

mV

mmBm

mV

mmm

T

m

T

m

T

m

T

dS

dVdV

FUfu

fu

)(

)()(

)()()(

)()()()()()(

τε

FB

fH

fHUUBCBU

)()()(

)()()(

)()()()()()()(

)(

)(

)()(

m

mV

mIm

m

mS

mSmS

m

mV

mBmTmm

mV

mmT

dV

dS

dVdV

m

T

m

T

m

T

m

T

τ

Formulation of System Equation – cont’

17

FB

fH

fHUBCB

)()()(

)()()(

)()()()()()()(

)(

)(

)()(

m

mV

mIm

m

mS

mSmS

m

mV

mBmmm

mV

mm

dV

dS

dVdV

m

T

m

T

m

T

m

T

τ

RUK

Formulation of System Equation – cont’

Role of FEM package is to evaluate the integrals for K, R

→ Solve for displacements at all the nodes

→ Evaluate displacement, stress, strain at any

location in each element using shape function

18

Example of Formulation

Derive the system equations of the plate using a two-element model

Thickness is 1 cm

Load is applied very slowly so that inertial effect is ignored

19

yP

Example of Formulation – cont’

Deriving the displacement transformation matrix of any element

20

yxu 321

yxv 321

131211 131211 yxvyxu

232212 232212 yxvyxu

333213 333213 yxvyxu

The displacements of each node

Example of Formulation – cont’

The constants can be determined in terms of

and

21

ii ,

ii vu , ii yx ,

a

uauaua

2332211

1

a

ububub

2332211

2

a

ucucuc

2332211

3

a

vavava

2332211

1

a

vbvbvb

2332211

2

a

vcvcvc

2332211

3

23132123321 xxcyybyxyxa

31213231132 xxcyybyxyxa

12321312213 xxcyybyxyxa

3212 aaaa

Example of Formulation – cont’

Substituting the values of of element 1 into the previous equations

22

321321 ,,,,, yyyxxx

yvvxvvvv

yuuxuuuu

)4

1

4

1()

10

1

10

1(

)4

1

4

1()

10

1

10

1(

31211

31211

Example of Formulation – cont’

23

8

7

6

5

4

3

2

1

)1(

004

10

10

10)

4

1

10

11(0

0004

10

10

10)

4

1

10

11(

U

U

U

U

U

U

U

U

yxyx

yxyx

v

u

8

7

6

5

4

3

2

1

)2(

)410

1(0)10

1(0)4

1(000

0)410

1(0)10

1(0)4

1(00

U

U

U

U

U

U

U

U

yxxy

yxxy

v

u

Example of Formulation – cont’

Determining the strains or the matrix using the following two-dimensional strain-displacement relations

24

)(mB

y

u

x

v

y

vx

u

xy

y

x

Example of Formulation – cont’

25

UB )1(

8

7

6

5

4

3

2

1

)1(

)1(

0004

1

10

10

10

1

4

1

004

1000

4

10

0000010

10

10

1

U

U

U

U

U

U

U

U

xy

y

x

ε

UB )2(

8

7

6

5

4

3

2

1

)2(

)2(

10

1

4

1

10

100

4

100

4

1000

4

1000

010

10

10

10000

U

U

U

U

U

U

U

U

xy

y

x

ε

Example of Formulation – cont’

The stress-strain relations for a homogeneous, isotropic plane-stress element are given by

26

xy

y

x

xy

y

xm E

)1(2

100

01

01

1 2)(τ

Example of Formulation – cont’

We assumed that there is no initial stress on the structure

The stiffness matrix of each element is derived as follows by substituting the previous results

Since two elements are made of the same material, the same expression for in Equation can be used for both the elements

27

)(mτ)( mC

Example of Formulation – cont’

28

)1()1()1()1()1()1(

dVT

VBCBK )1()1()1()1(

)1(dA

T

ABCB dxx

T

)10

44()1()1(

)1(10

0 BCB

dxx

E

T

)10

44(

0004

1

10

10

10

1

4

1

004

1000

4

10

0000010

10

10

1

2

100

01

01

0004

1

10

10

10

1

4

1

004

1000

4

10

0000010

10

10

1

1

10

02

Example of Formulation – cont’

29

00000000

00000000

0016

100

4016

1

40

00032

1

80

10

80

1

32

1

00080

1

200

10

200

1

80

1

0040

00100

1

40100

1

0016

1

80

1

200

1

40200

1

16

1

80

1

004032

1

80

1

100

1

80

1

32

1

100

1

1

202

)1(

E

K

Example of Formulation – cont’

30

)2()2()2()2()2()2(

dVT

VBCBK )2()2()2()2(

)2(dA

T

ABCB dxx

T

10

4)2()2()2(10

0BCB

200

1

16

1

80

1

200

1

4016

1

80

100

80

1

32

1

100

1

80

1

100

1

4032

100

200

1

80

1

200

100

80

100

40100

10

100

1

40000

16

1

400

4016

1000

80

1

32

1

80

100

32

100

00000000

00000000

1

202

E

Example of Formulation – cont’

31

2001

161

801

2001

40161

80100

801

321

1001

801

1001

4032100

2001

801

2001

16100

801

161

40

4010010

321

1001

8010

801

321

161

400

801

2001

1610

2001

801

801

321

80100

321

1001

401001

00161

801

2001

402001

161

801

004032

180

1100

180

132

1100

1

120

2)2(

E

K

Example of Formulation – cont’

The load vector is equal to because there are only concentrated forces on the nodes

32

CRR

y

y

x

y

x

P

F

F

F

F

0

0

0

3

3

1

1

R

Example of Formulation – cont’

Solving for the unknown nodal point displacements from the following equation

33

y

y

x

y

x

P

F

F

F

F

U

U

U

U

U

U

U

U

kkkkk

kkkkk

kkkkk

kkkkk

0

0

0

3

3

1

1

8

7

6

5

4

3

2

1

8887838281

7877737271

2827232221

1817131211

Example of Formulation – cont’

Applying the boundary conditions as below

34

y

y

x

y

x

P

F

F

F

F

U

U

U

U

kkkkk

kkkkk

kkkkk

kkkkk

0

0

0

0

0

0

0

3

3

1

1

8

7

4

3

8887838281

7877737271

2827232221

1817131211

Example of Formulation – cont’

Partitioned into two parts as follows

35

yPU

U

U

U

kkkk

kkkk

kkkk

kkkk

0

0

0

8

7

4

3

88878483

78777473

48474443

38373433

y

x

y

x

F

F

F

F

U

U

U

U

kkkk

kkkk

kkkk

kkkk

3

3

1

1

8

7

4

3

68676463

58575453

28272423

18171413