MECH4450 Introduction to Finite Element Methods

Chapter 2

Finite Element Analysis (F.E.A.) of 1-D Problems

Historical Background

• Hrenikoff, 1941 – “frame work method”

• Courant, 1943 – “piecewise polynomial interpolation”

• Turner, 1956 – derived stiffness matrices for truss, beam, etc

• Clough, 1960 – coined the term “finite element”

Key Ideas: - frame work method piecewise polynomial approximation

Axially Loaded Bar Review:

Stress:

Strain:

Deformation:

Stress:

Strain:

Deformation:

Axially Loaded Bar Review:

Stress:

Strain:

Deformation:

Axially Loaded Bar – Governing Equations and Boundary

Conditions• Differential Equation

• Boundary Condition Types

• prescribed displacement (essential BC)

• prescribed force/derivative of displacement (natural BC)

Lxxfdx

duxEA

dx

d

0 0)()(

Axially Loaded Bar –Boundary Conditions

• Examples

• fixed end

• simple support

• free end

Potential Energy

• Elastic Potential Energy (PE)- Spring case

- Axially loaded bar

- Elastic body

x

Unstretched spring

Stretched bar

0PE

2

2

1PE kx

undeformed:

deformed:

0PE

L

Adx02

1PE

dvV

Tεσ2

1PE

Potential Energy

• Work Potential (WP)

B

L

uPfdxu 0

WP

P

f

f: distributed force over a lineP: point forceu: displacement

A B

• Total Potential Energy

B

LL

uPfdxuAdx 002

1

• Principle of Minimum Potential Energy For conservative systems, of all the kinematically admissible displacement fields,

those corresponding to equilibrium extremize the total potential energy. If the extremum condition is a minimum, the equilibrium state is stable.

Potential Energy + Rayleigh-Ritz Approach

P

f

A B

Example:

Step 1: assume an admissible displacement field nixau ii

i to1

f is shape function / basis functionn-1 is the order of approximation

Step 2: calculate total potential energy

Find the displacement field x

u x

Assume 1 2u x a a x so 1 21 x

The admissible displacement field must satisfy the essential boundary condition 0 at 0u x

1 0a So and 2u x a x

First write down the elastic potential energy

22 2 2

0 0

1 1 1

2 2 2

L L

Adx Ea a Adx Ea AL

2u x a x 2

dua

dx 2E Ea

Elastic potential:

Assume f is a constant

Potential Energy + Rayleigh-Ritz Approach

P

f

A B

Example:

Step 3:select ai so that the total potential energy is at its minimum or maximum.

Second write down the work energy

2 2

0 0

22 2

WP

1

2

L L

Bu fdx P u a x fdx P a L

a fL Pa L

2 22 2 2

1 1

2 2Ea AL a fL Pa L

22

2

10 0

21 12 2

dEa AL fL PL

da

P fL P fLa u x

EA EA

Galerkin’s Method

P

f

A B

Example:

Pdx

duxEA

xu

xfdx

duxEA

dx

d

Lx

)(

00

0)()(Seek an approximation so

Pdx

udxEA

xu

dVxfdx

udxEA

dx

dw

Lx

V

i

~)(

00~

0)(~

)(

u~

In the Galerkin’s method, the weight function (wi) is chosen to be the same as the shape Function, i.e.,

1 to i ii

u a x i n Let find ai

i iw

Galerkin’s Method

P

f

A B

Example:

0)(~

)(

dVxf

dx

udxEA

dx

dw

V

i0

~)(

~)(

00 0

L

i

L L

ii

dx

udxEAwfdxwdx

dx

dw

dx

udxEA

1 2 3

1

2

3

Let 1 , 0u a x x L 1 and x w x

100 0

1 0 0L L du

EAa dx xfdx w L P w EAdx

12 2

fL fLP P

a u xEA EA

Find the displacement field

FEM Formulation of Axially Loaded Bar – Governing Equations

• Differential Equation

• Weighted-Integral Formulation

• Weak Form

Lxxfdx

duxEA

dx

d

0 0)()(

0)()(0

dxxf

dx

duxEA

dx

dw

L

LL

dx

duxEAwdxxwf

dx

duxEA

dx

dw

00

)()()(0

Finite Element Method – Piecewise Approximation

x

u

x

u

Approximation Methods – Finite Element Method

Example:

Step 1: Discretization

Step 2: Weak form of one element P2P1x1 x2

0)()()()()(2

1

2

1

x

x

x

x dx

duxEAxwdxxfxw

dx

duxEA

dx

dw

0)()()( 1122

2

1

PxwPxwdxxfxw

dx

duxEA

dx

dwx

x

Find the displacement field

Approximation Methods – Finite Element Method

Example (cont):

Step 3: Choosing shape functions - linear shape functions

2211 uuu

lx1 x2

x x=-1x =0x =1x

l

xx

l

xx 12

21 ;

2

1 ;

2

121

11 2

1 ;1

2x

lxxx

l

Approximation Methods – Finite Element Method

Example (cont):

Step 4: Forming element equation

2

1

2

1

1

1 1 1 1

2 2 2 2

2

1 1

1 1

x

x

x

x

fdxu P f PEAu P f Pl

fdx

0)()()( 1122

2

1

PxwPxwdxxfxw

dx

duxEA

dx

dwx

x

Let , 1w 2 2

1 1

2 11 1 2 2 1 1 1

10

x x

x x

u uEA dx f dx x P x P

l l

1121

2

1

Pdxful

EAu

l

EAx

x

Let , 2w 2 2

1 1

2 12 2 2 2 2 1 1

10

x x

x x

u uEA dx f dx x P x P

l l

2221

2

1

Pdxful

EAu

l

EAx

x

E,A are constant

Approximation Methods – Finite Element Method

Example (cont):

Step 5: Assembling to form system equation

Approach 1:

Element 1:

1 1 1

2 2 2

1 1 0 0

1 1 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

I I I

I I II I

I

u f P

u f PE A

l

Element 2:1 1 1

2 2 2

0 0 0 0 0 0 0

0 1 1 0

0 1 1 0

0 0 0 0 0 0 0

II II IIII II

II II IIII

u f PE A

u f Pl

Element 3:

1 1 1

2 2 2

0 0 00 0 0 0

0 0 00 0 0 0

0 0 1 1

0 0 1 1

III III

III III IIIIII

III III III

E Au f Pl

u f P

Approximation Methods – Finite Element Method

Example (cont):

Step 5: Assembling to form system equation

Assembled System:

1 1 1

2 2 2

3 3 3

4 4 4

0 0

0

0

0 0

I I I I

I I

I I I I II II II II

I I II II

II II II II III III III III

II II III III

III III III III

III III

E A E A

l lu f PE A E A E A E Au f Pl l l lu f PE A E A E A E Au f Pl l l l

E A E A

l l

1 1

2 1 2 1

2 1 2 1

2 2

I I

I II I II

II III II III

III III

f P

f f P P

f f P P

f P

Approximation Methods – Finite Element Method

Example (cont):

Step 5: Assembling to form system equation

Approach 2: Element connectivity table

1 2 3

2 3 4

global node index (I,J)

local node (i,j)

eij IJk K

Element 1 Element 2 Element 3

Approximation Methods – Finite Element Method

Example (cont):

Step 6: Imposing boundary conditions and forming condense system

Condensed system:

2 2

3 3

4 4

00

0

0

I I II II II II

I II II

II II II II III III III III

II II III III

III III III III

III III

E A E A E A

l l l u fE A E A E A E A

u fl l l l

u f PE A E A

l l

1 1

2 2 2

3 3 3

4 4 4

0 0

00

0

0 0

I I I I

I I

I I I I II II II II

I I II II

II II II II III III III III

II II III III

III III III III

III III

E A E A

l lF fE A E A E A E A

u F fl l l lu F fE A E A E A E Au F fl l l l

E A E A

l l

1

0

0

P

P

Approximation Methods – Finite Element Method

What if

Step 6: Imposing boundary conditions and forming condense system

Condensed system:

11 12 13 14 1

21 22 23 24 2 2

31 32 33 34 3 3

41 42 43 44 4 4

K K K K a F

K K K K u F

K K K K u F

K K K K u F

1u a

22 23 24 2 2 21

32 33 34 3 3 31

42 43 44 4 4 41

K K K u F K

K K K u F a K

K K K u F K

12 13 14 1 112

22 23 24 2 213

32 33 34 3 314

42 43 44 4 41

K K K F Ku

K K K F Ku a

K K K F Ku

K K K F K

Approximation Methods – Finite Element Method

Example (cont):

Step 7: solution

Step 8: post calculation

dx

du

dx

du

dx

du 22

11

2211 uuu dx

dEu

dx

dEuE 2

21

1

Summary - Major Steps in FEM

• Discretization

• Derivation of element equation

• weak form

• construct form of approximation solution over one element

• derive finite element model

• Assembling – putting elements together

• Imposing boundary conditions

• Solving equations

• Postcomputation

Exercises – Linear Element

Example 1:

E = 100 GPa, A = 1 cm2

1 2 31 2 3 4

Element equation for linear bar element:

2

1

2

1

1

1 1 1 1

2 2 2 2

2

1 1

1 1

x

x

x

x

fdxu P f PEAu P f Pl

fdx

1 1 1

2 2 2

3 3

4 4

0 0

0

0

0 0

I I I I

I I

II I I I II II II II

II I II II

II II II II III III III III

II II III III

III III III III

III III

E A E A

l lu F fE A E A E A E Au F f fl l l lu FE A E A E A E Au Fl l l l

E A E A

l l

1

1 2 1

2 1 2 1

2 2

I

II I II

II III II III

III III

P

P P

f f P P

f P

Global system:

Find the displacement, stressand strain field

Exercises – Linear Element

Example 1:E = 100 GPa, A = 1 cm2

1 1

27

3

4

1 1 0 0 1

1 2 1 0 2 010

0 1 2 1 2 0

0 0 1 1 1 10

Iu P

u

u

u

Force vector:

1 11 1

2 22 1

3 32 1

4 42

1

2 0

2 0

1 10

I I

I II

II III

III

f Pf P

f Pf f

f Pf f

f Pf

Global system: Imposing boundary conditions:

1

27

3

4

01 1 0 0 1

1 2 1 0 2 010

0 1 2 1 2 0

0 0 1 1 1 10

IP

u

u

u

Exercises – Linear Element

Example 1:E = 100 GPa, A = 1 cm2

Condensed system:

Displacement field:

27

3

4

2 1 0 2

10 1 2 1 2

0 1 1 11

u

u

u

27

3

4

15

28 10 m

39

u

u

u

72 1 110 1 16N I Iu P P

7 71 1 2 2 2 2

7 72 1 3 2

7 73 1 4 2

00,1 15 10 15 10 m

12 1

1,2 15 28 10 13 2 10 m1 1

3 22,3 28 39 10 11 6 10 m

1 1

xx u u u u x

x xx u u u x

x xx u u u x

Linear Formulation for Bar Element

2

1

2212

1211

2

1

2

1

u

u

KK

KK

f

f

P

P

2

1

2

1

, x

x

ii

x

x

jiji

ij dxffKdxdx

d

dx

dEAKwhere

x=x1 x=x2

1 f2 f1 1

x

x=x1 x= x2

u1 u2

1P 2Pf(x)

L = x2-x1

u

x

Higher Order Formulation for Bar Element

(x)u(x)u(x)u(x)u 332211

)x(u)x(u)x(u)x(u(x)u 44332211

1 3

u1 u3u

x

u2

2

1 4

u1 u4

2

u

x

u2 u3

3

)x(u)x(u)x(u)x(u)x(u(x)u nn44332211

1 n

u1 un

2

u

x

u2u3

3

u4 ……………

4 ……………

Natural Coordinates and Interpolation Functions

2

1 ,

2

121

Natural (or Normal) Coordinate:

x=x1 x= x2

x=-1 x=1

xx

0x x l

1xxx 1 2

2/ 2

x xx

l

1 32

xx=-1 x=1

1 2

xx=-1 x=1

1 42

xx=-1 x=1

3

2

1 ,11 ,

2

1321

13

11

16

27 ,1

3

1

3

1

16

921

3

1

3

11

16

9 ,1

3

11

16

2743

Quadratic Formulation for Bar Element

2

1

1

1

nd , , 1, 2, 32

x

i i i

x

la f f dx f d i j

2

1

1

1

2

xj ji i

ij ji

x

d dd dwhere K EA dx EA d K

dx dx d d l

3

2

1

332313

232212

131211

3

2

1

3

2

1

u

u

u

KKK

KKK

KKK

f

f

f

P

P

P

x=-1 x=0 x=1

f3f1 f2

Quadratic Formulation for Bar Element

u1 u3u2f(x)P3

P1

P2

x=-1 x=0 x=11x 2x 3x

2

1u11u

2

1u)(u)(u)(u)(u 321332211

2

1 ,11 ,

2

1321

1 1 2 2 3 32 2 1 2 4 2 2 1, ,

d d d d d d

dx l d l dx l d l dx l d l

1 2

2/ 2

x xx

l

2

ld dx

2d

dx l

Exercises – Quadratic Element

Example 2:

E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2

12

Element equations:

Element 11 11 11 112 21 13 3

7 8 1

8 16 83

1 8 7

u PEA

u PL

u P

Element 22 21 12 222 22 23 3

7 8 1

8 16 83

1 8 7

u PEA

u PL

u P

Global system 11 1

1 1 11 1

2 22 3

1 21 1 3 3

4 42 22

5 5

2 23

7 8 0 033 388 16 0 033 3 87 78 3 33 33 3 16 880 0 3 33

8 70 0 3 33

EAEA EALL LEA u PEA EAL u PL L EA EAEA EAEA EA u PL LL LL L u PEA EAEA

u PL LLEA EAEAL LL

Find the displacement, stressand strain field

Exercises – Quadratic Element

Example 2:

Imposing boundary conditions

11 1

1 11 1

22 3

1 21 1 3

42 22

5

2 23

7 8 0 033 38 08 16 0 03 03 3 87 78 53 33 33 3 016 880 0 103 33

8 70 0 3 33

EAEA EALL LEA PEA EAL uL L EA EAEA EAEA EA uL LL LL L uEA EAEA

uL LLEA EAEAL LL

Solutions:2

3

4

5

0.0008

0.0015m

0.0023

0.0030

u

u

u

u

Some Issues

Non-constant cross section:

Interior load point:

Mixed boundary condition:k