power point presentation on finite element methods...finite element methods iii b tech ii semester...

Prepared by

Mr. G S D Madhav, Assistant Professor, Dept of Aeronautical Engg

Ms Y Shwetha, Assistant Professor, Dept of Aeronautical Engg

Power Point Presentation

on

Finite element methods III B Tech II Semester

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous)

Dundigal, Hyderabad - 500 043

1

UNIT - 1

• Introduction to FEM:

– Stiffness equations for a axial bar element in local co-ordinates using Potential Energy approach and Virtual energy principle

– Finite element analysis of uniform, stepped and tapered bars subjected to mechanical and thermal loads

– Assembly of Global stiffness matrix and load vector

– Quadratic shape functions

– properties of stiffness matrix

2

Axially Loaded Bar Review:

Stress:

Strain:

Deformation:

Stress:

Strain:

Deformation:

3

Axially Loaded Bar Review:

Stress:

Strain:

Deformation:

4

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)()(

5

Axially Loaded Bar –Boundary Conditions

• Examples

• fixed end

• simple support

• free end

6

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

Adx0

2

1PE

dvV

Tεσ2

1PE

7

Potential Energy

• Work Potential (WE)

B

L

uPfdxu 0

WP

P

f

f: distributed force over a line P: point force u: displacement

A B

• Total Potential Energy

B

LL

uPfdxuAdx 00

2

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.

8

Potential Energy + Rayleigh-Ritz Approach

P

f

A B

Example:

Step 1: assume a displacement field nixau i

i

i to1

is shape function / basis function n is the order of approximation

Step 2: calculate total potential energy

9

Potential Energy + Rayleigh-Ritz Approach

P

f

A B

Example:

Step 3:select ai so that the total potential energy is minimum

10

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 is chosen to be the same as the shape function.

11

Galerkin’s Method

P

f

A B

Example:

0)(~

)(

dVxf

dx

udxEA

dx

dw

V

i 0~

)(~

)(00 0

L

i

L L

ii

dx

udxEAwfdxwdx

dx

dw

dx

udxEA

1 2 3

1

2

3

12

Finite Element Method – Piecewise Approximation

x

u

x

u

13

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

14

Approximation Methods – Finite Element Method

Example:

Step 1: Discretization

Step 2: Weak form of one element P2 P1 x1 x2

0)()()()()(

2

1

2

1

x

x

x

xdx

duxEAxwdxxfxw

dx

duxEA

dx

dw

0)()()( 1122

2

1

PxwPxwdxxfxw

dx

duxEA

dx

dwx

x 15


Example (cont):

Step 3: Choosing shape functions - linear shape functions

2211 uuu

l x1 x2

x x x1 x0 x1

l

xx

l

xx 12

21 ;

2

1 ;

2

121

x

x

112

1 ;1

2x

lxxx

l

xx

16


Example (cont):

Step 4: Forming element equation

Let , weak form becomes 1w 01

1121112

2

1

2

1

PPdxfdx

l

uuEA

l

x

x

x

x

1121

2

1

Pdxful

EAu

l

EAx

x

Let , weak form becomes 2w 01

1222212

2

1

2

1

PPdxfdx

l

uuEA

l

x

x

x

x

2221

2

1

Pdxful

EAu

l

EAx

x

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 PEA

u P f Plfdx

E,A are constant

17


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 A

u f Pl

u f P

18


Example (cont):


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 A

u f Pl l l l

u f PE A E A E A E A

u 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

19


Example (cont):


Approach 2: Element connectivity table Element 1 Element 2 Element 3

1 1 2 3

2 2 3 4

global node index

(I,J)

local node

(i,j)

e

ij IJk K

20


Example (cont):

Step 6: Imposing boundary conditions and forming condense system

Condensed system:

2 2

3 3

4 4

0

0

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 lu f

E A E A E A E Au f

l l l lu f P

E A E A

l l

21


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

22

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 23

Exercises – Linear Element

Example 1:

E = 100 GPa, A = 1 cm2

24

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

ji

jiij dxffKdx

dx

d

dx

dEAKwhere

x=x1 x=x2

1 2 1 1

x

x=x1 x= x2

u1 u2

1P 2Pf(x)

L = x2-x1

u

x

25

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 u3 u

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

u2 u3

3

u4 ……………

4 ……………

26

Natural Coordinates and Interpolation Functions

2

1 ,

2

121

x

x

Natural (or Normal) Coordinate:

x=x1 x= x2

x=-1 x=1

x

x

0x x l

1xxx

1 2

2/ 2

x xx

lx

1 3 2

x x=-1 x=1

1 2

x x=-1 x=1

1 4 2

x x=-1 x=1

3

2

1 ,11 ,

2

1321

xxxx

xx

13

11

16

27 ,1

3

1

3

1

16

921

xxxxxx

3

1

3

11

16

9 ,1

3

11

16

2743 xxxxxx

27

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 x

2

1

1

1

2

x

j ji iij ji

x

d dd dwhere K EA dx EA d K

dx dx d d l

x

x x

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

3 1 2

28

Quadratic Formulation for Bar Element

u1 u3 u2 f(x) P3 P1

P2

x=-1 x=0 x=1

1x 2x 3x

2

1u11u

2

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

xxxx

xxxxxx

2

1 ,11 ,

2

1321

xxxx

xx

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

x x x

x x x

1 2

2/ 2

x xx

lx

2

ld dxx

2d

dx l

x

29

Exercises – Quadratic Element

Example 2:

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

30

Some Issues

Non-constant cross section:

Interior load point:

Mixed boundary condition:

k

31

UNIT – 2

Finite Element Analysis of Trusses

32

Plane Truss Problems

Example 1: Find forces inside each member. All members have the same length.

F

33

Arbitrarily Oriented 1-D Bar Element on 2-D Plane

Q2 , v2

11 u ,P

q

22 u ,P

P2 , u2

Q1 , v1

P1 , u1

34

Relationship Between Local Coordinates and Global Coordinates

2

2

1

1

2

2

1

1

cossin00

sincos00

00cossin

00sincos

0

0

v

u

v

u

v

u

v

u

qq

qq

qq

qq

35

Relationship Between Local Coordinates and Global Coordinates

2

2

1

1

2

1

cossin00

sincos00

00cossin

00sincos

0

0

Q

P

Q

P

P

P

qq

qq

qq

qq

36

Stiffness Matrix of 1-D Bar Element on 2-D Plane

2

2

1

1

2

2

1

1

cossin00

sincos00

00cossin

00sincos

cossin00

sincos00

00cossin

00sincos

v

u

v

u

K

Q

P

Q

P

ij

qq

qq

qq

qq

qq

qq

qq

qq

0000

0101

0000

0101

L

AEKij

Q2 , v2

11 , uP

q

22 u ,P

P2 , u2

Q1 , v1

P1 , u1

2

2

1

1

22

22

22

22

2

2

1

1

sincossinsincossin

cossincoscossincos

sincossinsincossin

cossincoscossincos

v

u

v

u

L

AE

Q

P

Q

P

qqqqqq

qqqqqq

qqqqqq

qqqqqq

37

Arbitrarily Oriented 1-D Bar Element in 3-D Space

2

2

2

1

1

1

zzz

yyy

xxx

zzz

yyy

xxx

2

2

2

1

1

1

w

v

u

w

v

u

000

000

000

000

000

000

0w

0v

u

0w

0v

u

x, x, x are the Direction Cosines of the bar in the x-

y-z coordinate system

- - -

11 u ,Px

22 u ,P

x

x x

y

z 2

1

- -

-

2

2

2

1

1

1

2

2

2

1

1

1

000

000

000

000

000

000

0

0

0

0

R

Q

P

R

Q

P

R

Q

P

R

Q

P

zzz

yyy

xxx

zzz

yyy

xxx

38

Stiffness Matrix of 1-D Bar Element in 3-D Space

2

2

2

1

1

1

22

22

22

22

22

22

2

2

2

1

1

1

w

v

u

w

v

u

L

AE

R

Q

P

R

Q

P

xxxxxxxxxx

xxxxxxxxxx

xxxxxxxxxx

xxxxxxxxxx

xxxxxxxxxx

xxxxxxxxxx

11 u ,Px

22 u ,P

x

x x

y

z 2

1

- -

-

0w

0v

u

0w

0v

u

000000

000000

001001

000000

000000

001001

L

AE

0R

0Q

P

0R

0Q

P

2

2

2

1

1

1

2

2

2

1

1

1

39

Matrix Assembly of Multiple Bar Elements

2

2

1

1

2

2

1

1

0000

0101

0000

0101

v

u

v

u

L

AE

Q

P

Q

P

Element I

Element I I

Element I I I

3

3

2

2

3

3

2

2

3333

3131

3333

3131

4

v

u

v

u

L

AE

Q

P

Q

P

3

3

1

1

3

3

1

1

3333

3131

3333

3131

4

v

u

v

u

L

AE

Q

P

Q

P

40


3

3

2

2

1

1

3

3

2

2

1

1

000000

000000

000000

000404

000000

000404

4

v

u

v

u

v

u

L

AE

Q

P

Q

P

Q

P

Element I

3

3

2

2

1

1

3

3

2

2

1

1

333300

313100

333300

313100

000000

000000

4

v

u

v

u

v

u

L

AE

Q

P

Q

P

Q

P

3

3

2

2

1

1

3

3

2

2

1

1

330033

310031

000000

000000

330033

310031

4

v

u

v

u

v

u

L

AE

Q

P

Q

P

Q

P

Element I I

Element I I I

41


3

3

2

2

1

1

3

3

2

2

1

1

v

u

v

u

v

u

33333333

33113131

33303000

31301404

33003030

31043014

L4

AE

S

R

S

R

S

R

0v

0u

0v

?u

?v

0u

603333

023131

333300

313504

330033

310435

L4

AE

?S

?R

?S

FR

0S

?R

3

3

2

2

1

1

3

3

2

2

1

1

Apply known boundary conditions

42

Solution Procedures

0v

0u

0v

?u

?v

0u

603333

023131

333300

310435

330033

313504

L4

AE

?S

?R

?S

?R

0S

FR

3

3

2

2

1

1

3

3

2

1

1

2

u2= 4FL/5AE, v1= 0

0v

0u

0vAE5

FL4u

0v

0u

603333

023131

333300

310435

330033

313504

L4

AE

?S

?R

?S

?R

0S

FR

3

3

2

2

1

1

3

3

2

1

1

2

43

Recovery of Axial Forces

05

4

05

4

05

4

0

0

0000

0101

0000

0101

2

2

1

1

2

2

1

1

F

vAE

FLu

v

u

L

AE

Q

P

Q

P

Element I

Element I I

Element I I I

53

51

53

51

0

0

05

4

3333

3131

3333

3131

4

3

3

2

2

3

3

2

2

F

v

u

vAE

FLu

L

AE

Q

P

Q

P

0

0

0

0

0

0

0

0

3333

3131

3333

3131

4

3

3

1

1

3

3

1

1

v

u

v

u

L

AE

Q

P

Q

P

44

Stresses inside members

Element I

Element I I

Element I I I

A

F

5

4

5

41

FP

5

42

FP

FP5

12

FQ5

32

FQ5

33

FP5

13

45

Finite Element Analysis of

Beams

46

Bending Beam

Review

Normal strain:

Pure bending problems:

Normal stress:

Normal stress with bending moment:

Moment-curvature relationship:

Flexure formula:

x

y

yx

Eyx

M M

MydAx

EI

M

1

I

Myx dAyI 2

2

21

dx

ydEIEIM

47

Bending Beam

Review

Deflection:

Sign convention:

Relationship between shear force, bending moment and transverse load:

q(x)

x

y

qdx

dV V

dx

dM

qdx

ydEI

4

4

+ - M M M

+ - V V V

48

Governing Equation and Boundary Condition

• Governing Equation

• Boundary Conditions -----

0at , ? &? &? & ?2

2

2

2

x

dx

vdEI

dx

d

dx

vdEI

dx

dvv

Essential BCs – if v or is specified at the boundary.

Natural BCs – if or is specified at the boundary.

dx

dv

{ Lx

dx

vdEI

dx

d

dx

vdEI

dx

dvv

at , ? &? &? & ?

2

2

2

2

2

2

dx

vdEI

2

2

dx

vdEI

dx

d

,0)()(

2

2

2

2

xq

dx

xvdEI

dx

d 0<x<L

49

Weak Formulation for Beam Element

• Governing Equation

,0)()(

2

2

2

2

xq

dx

xvdEI

dx

d

• Weighted-Integral Formulation for one element

2

1

)()(

)(02

2

2

2x

x

dxxqdx

xvdEI

dx

dxw

• Weak Form from Integration-by-Parts ----- (1st time)

2

1

2

1

2

2

2

2

0

x

x

x

xdx

vdEI

dx

dwdxwq

dx

vdEI

dx

d

dx

dw

21 xxx

50

Weak Formulation

• Weak Form from Integration-by-Parts ----- (2nd time)

2

1

2

1

2

1

2

2

2

2

2

2

2

2

0

x

x

x

x

x

xdx

vdEI

dx

dw

dx

vdEI

dx

dwdxwq

dx

vdEI

dx

wd

V(x2)

x = x1

M(x2)

q(x) y

x x = x2

V(x1)

M(x1)

L = x2-x1

2

1

2

1

2

2

2

2

0

x

x

x

x

Mdx

dwwVdxwq

dx

vdEI

dx

wd

51

Weak Formulation

• Weak Form

24231211 , , , xMQxVQxMQxVQ

4

2

2

1

32112

2

2

2

)()(2

1

Qdx

dwQ

dx

dwQxwQxwdxwq

dx

vdEI

dx

wdx

x

2

1

2

1

2

2

2

2

0

x

x

x

x

Mdx

dwwVdxwq

dx

vdEI

dx

wd

Q3

x = x1

Q4

q(x) y(v)

x x = x2

Q1

Q2

L = x2-x1

52

Ritz Method for Approximation

n

j

jj nxuxvLet1

4 and )()(

4

2

2

1

32112

24

12

2

)()(2

1

Qdx

dwQ

dx

dwQxwQxwdxwq

dx

duEI

dx

wdx

x

j

j

j

Let w(x)= i (x), i = 1, 2, 3, 4

4

2

2

1

32112

24

12

2

)()(2

1

Qdx

dQ

dx

dQxQxdxq

dx

duEI

dx

d iiii

x

x

i

j

j

ji

Q3

x = x1

Q4

q(x) y(v)

x x = x2

Q1

Q2

L = x2-x1

where ; ; ; ;

21

423211

xxxx dx

dvuxvu

dx

dvuxvu

53


i

j

jij

x

i

xi

x

i

xi quKQdx

dQQ

dx

dQ

4

1

4321

2

2

1

1

2

1

2

1

and where2

2

2

2 x

x

ii

x

x

jiij qdxqdx

dx

d

dx

dEIK

Q3

x = x1

y(v)

x x = x2

Q1

Q2

L = x2-x1

Q4

54


4

3

2

1

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

44

44

33

33

22

22

11

11

2

2

1

1

2

2

1

1

2

2

1

1

2

2

1

1

q

q

q

q

u

u

u

u

KKKK

KKKK

KKKK

KKKK

Q

Q

Q

Q

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

xx

xx

xx

xx

xx

xx

xx

xx

jiij KK where

Q3

x = x1

y(v)

x x = x2

Q1

Q2

L = x2-x1

Q4

55

Selection of Shape Function

1000

0100

0010

0001

2

2

1

1

2

2

1

1

2

2

1

1

2

2

1

1

44

44

33

33

22

22

11

11

xx

xx

xx

xx

xx

xx

xx

xx

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

dx

d

4

3

2

1

4

3

2

1

44342414

34332313

24232212

14131211

4

3

2

1

q

q

q

q

u

u

u

u

KKKK

KKKK

KKKK

KKKK

Q

Q

Q

Q

The best situation is -----

Interpolation Properties

56

Derivation of Shape Function for Beam Element – Local Coordinates

1 1 2 2 3 3 4 4( )v u u u ux

How to select i???

1 2 3 41 2 3 4

( )dv d d d du u u u

d d d d d

x

x x x x x

and

where 1 21 1 2 3 2 4

dv dvu v u u v u

d dx x

Let 32 xxx iiiii dcba

Find coefficients to satisfy the interpolation properties.

57

Derivation of Shape Function for Beam Element

How to select i???

e.g. Let 3

1

2

1111 xxx dcba

xx 214

1 2

1

Similarly

114

1

214

1

114

1

2

2

2

3

2

2

xx

xx

xx

58

Derivation of Shape Function for Beam Element

In the global coordinates:

)(2

)()(2

)()( 42

3221

11 xdx

dvlxvx

dx

dvlxvxv

12

1

2

12

11

3

12

1

2

12

1

2

12

11

3

12

1

2

12

1

4

3

2

1

2

23

12

231

xx

xx

xx

xxxx

l

xx

xx

xx

xx

xx

xxxx

l

xx

xx

xx

xx

59

Element Equations of 4th Order 1-D Model

u3

x = x1

u4

q(x) y(v)

x x = x2

u1

u2

L = x2-x1

4

3

2

1

44342414

34332313

24232212

14131211

4

3

2

1

4

3

2

1

u

u

u

u

KKKK

KKKK

KKKK

KKKK

q

q

q

q

Q

Q

Q

Q

2

1

2

1

2

2

2

2 x

x

ii

x

x

ji

jiij qdxqandKdx

dx

d

dx

dEIKwhere

x=x2 x=x1

1 1 1

3 2

4

60

Element Equations of 4th Order 1-D Model

u3

x = x1

u4

q(x) y(v)

x x = x2

u1

u2

L = x2-x1

2

1

x

x

ii qdxqwhere

24

23

12

11

22

22

3

4

3

2

1

4

3

2

1

233

3636

323

3636

2

q

q

u

vu

u

vu

LLLL

LL

LLLL

LL

L

EI

q

q

q

q

Q

Q

Q

Q

61

Finite Element Analysis of 1-D Problems - Applications

F

L L L

Example 1.

Governing equation:

Lxxqdx

vdEI

dx

d

0 0)(

2

2

2

2

Weak form for one element

04322112

2

2

2

21

2

1

Q

dx

dwQxwQ

dx

dwQxwdxwq

dx

vd

dx

wdEI

xx

x

x

where )( 11 xVQ )( 12 xMQ )( 23 xVQ )( 24 xMQ

62

Finite Element Analysis of 1-D Problems Example 1.

Approximation function:

3 2

1 4

x=x1

x=x2

12

1

2

12

11

3

12

1

2

12

1

2

12

11

3

12

1

2

12

1

4

3

2

1

2

23

12

231

xx

xx

xx

xxxx

l

xx

xx

xx

xx

xx

xxxx

l

xx

xx

xx

xx

)(2

)()(2

)()( 42

3221

11 xdx

dvlxvx

dx

dvlxvxv

63

Finite Element Analysis of 1-D Problems Example 1.

Finite element model:

2

2

1

1

22

22

3

4

3

2

1

233

3636

323

3636

2

q

q

v

v

LLLL

LL

LLLL

LL

L

EI

Q

Q

Q

Q

P1 , v1 P2 , v2 P3 , v3 P4 , v4

M1 , q1 M2 , q2 M3 , q3 M4 , q4

I II III

Discretization:

64

Matrix Assembly of Multiple Beam Elements

Element I

Element I I

11

2 2

12

23

2 2

24

3

3

3

4

4

6 3 6 3 0 0 0 0

3 2 3 0 0 0 0

6 3 6 3 0 0 0 0

3 3 2 0 0 0 02

0 0 0 0 0 0 0 00

0 0 0 0 0 0 0 00

0 0 0 0 0 0 0 00

0 0 0 0 0 0 0 00

I

I

I

I

vL LQ

L L L LQ

vL LQ

L L L LQ EI

vL

v

q

q

q

q

1

1

21

2 2

22

3

33

2 2

34

4

4

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 6 3 6 3 0 0

0 0 3 2 3 0 02

0 0 6 3 6 3 0 0

0 0 3 3 2 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

II

II

II

II

v

vQ L L

Q L L L LEI

vQ L LL

Q L L L L

v

q

q

q

q

65

Matrix Assembly of Multiple Beam Elements

Element I I I

1

1

2

2

3

1 3

2 2

2 3

3 4

2 2

4 4

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 02

0 0 0 0 6 3 6 3

0 0 0 0 3 2 3

0 0 0 0 6 3 6 3

0 0 0 0 3 3 2

III

III

III

III

v

v

EI

Q vL LL

Q L L L L

Q vL L

Q L L L L

q

q

q

q

4

4

3

3

2

2

1

1

22

2222

2222

22

3

4

4

3

3

2

2

1

1

2330000

36360000

32233300

3633663600

00322333

0036336636

0000323

00003636

2

q

q

q

q

v

v

v

v

LLLL

LL

LLLLLLLL

LLLL

LLLLLLLL

LLLL

LLLL

LL

L

EI

M

P

M

P

M

P

M

P

66

Solution Procedures

?

?

?

0

?

0

0

0

2330000

36360000

340300

360123600

003403

003601236

0000323

00003636

2

0

0

?

0

?

?

?

4

4

3

3

2

2

1

1

22

222

222

22

3

4

4

3

3

2

2

1

1

q

q

q

q

v

v

v

v

LLLL

LL

LLLLL

LL

LLLLL

LL

LLLL

LL

L

EI

M

FP

M

P

M

P

M

P

Apply known boundary conditions

?

?

?

0

?

0

0

0

360123600

003601236

0000323

00003636

2330000

36360000

340300

003403

2

?

?

?

?

0

0

0

4

4

3

3

2

2

1

1

22

22

222

222

3

3

2

1

1

4

4

3

2

q

q

q

q

v

v

v

v

LL

LL

LLLL

LL

LLLL

LL

LLLLL

LLLLL

L

EI

P

P

M

P

M

FP

M

M

67

Solution Procedures

?

?

?

?

230

3630

34

004

2

0

0

0

4

4

3

2

22

222

22

3

4

4

3

2

q

q

q

v

LLL

LL

LLLL

LL

L

EI

M

FP

M

M

4

4

3

2

2

3

3

2

1

1

3603

0030

000

0003

2

?

?

?

?

q

q

q

v

LL

L

L

L

L

EI

P

P

M

P

?

?

?

?

0

0

0

0

360312600

003061236

0000323

00030636

2303000

36306000

340300

004303

2

?

?

?

?

0

0

0

4

4

3

2

3

2

1

1

22

22

222

222

3

3

2

1

1

4

4

3

2

q

q

q

q

v

v

v

v

LL

LL

LLLL

LL

LLLL

LL

LLLLL

LLLLL

L

EI

P

P

M

P

M

FP

M

M

68

Shear Resultant & Bending Moment Diagram

FL7

2

FL7

1

FL

F7

9

F7

3

2P

F

69

Plane Flame

Frame: combination of bar and beam

E, A, I, L Q1 , v1 Q3 , v2

Q2 , q1

P1 , u1

Q4 , q2

P2 , u2

2

2

2

1

1

1

22

2323

22

2323

4

3

2

2

1

1

460

260

6120

6120

0000

260

460

6120

6120

0000

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AE

Q

Q

P

Q

Q

P

70

Finite Element Model of an Arbitrarily Oriented Frame

q x

y

q x

y

71

Finite Element Model of an Arbitrarily Oriented Frame

2

2

2

1

1

1

22

2323

22

2323

4

3

2

2

1

1

40

620

6

0000

60

1260

12

20

640

6

0000

60

1260

12

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EI

Q

Q

P

Q

Q

P

local

global

72

Plane Frame Analysis - Example

Rigid Joint Hinge Joint

Beam II

Beam I Beam

Bar

F F F F

73

Plane Frame Analysis

2

2

2

1

1

1

22

2323

22

2323

4

3

2

2

1

1

40

620

6

0000

60

1260

12

20

640

6

0000

60

1260

12

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EI

Q

Q

P

Q

Q

P

I

I

P1 , u1

P2 , u2

Q2 , q1

Q4 , q2

Q1 , v1

Q3 , v2

74


P1 , u2

Q3 , v3

Q2 , q2 Q4 , q3

Q1 , v2

P2 , u3

4

3

3

2

2

2

22

2323

22

2323

4

3

2

2

2

1

460

260

6120

6120

0000

260

460

6120

6120

0000

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AE

Q

Q

P

Q

Q

PII

75


2

2

2

1

1

1

22

2323

22

2323

4

3

2

2

1

1

40

620

6

0000

60

1260

12

20

640

6

0000

60

1260

12

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EI

Q

Q

P

Q

Q

P

I

I

4

3

3

2

2

2

22

2323

22

2323

4

3

2

2

2

1

460

260

6120

6120

0000

260

460

6120

6120

0000

q

q

v

u

v

u

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AE

Q

Q

P

Q

Q

PII

76


FL8

1

F2

1

F2

1

FL8

3

77

UNIT- 3 Finite element analysis constant

strain triangle

78

Finite element formulation for 2D: Step 1: Divide the body into finite elements connected to each other through special points (“nodes”)

x

y

Su

ST u

v

x

px

py

Element ‘e’

3

2

1

4

y

x v

u

1

2

3

4

u1

u2

u3

u4

v4

v3

v2

v1

4

4

3

3

2

2

1

1

v

u

v

u

v

u

v

u

d

79

44332211

44332211

vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,v

u y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u

4

4

3

3

2

2

1

1

4321

4321

v

u

v

u

v

u

v

u

N0N0N0N0

0N0N0N0N

y)(x,v

y)(x,uu

dNu

80

TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT

...... vy)(x,N

uy)(x,Ny)(x,vy)(x,u

vy)(x,N

vy)(x,N

v y)(x,N

vy)(x,Ny)(x,v

u y)(x,N

u y)(x,N

u y)(x,N

u y)(x,Ny)(x,u

11

11

xy

44

3

3

22

11

y

44

3

3

22

11

x

xyxy

yyyyy

xxxxx

Approximation of the strain in element ‘e’

81

4

4

3

3

2

2

1

1

B

44332211

4321

4321

xy

v

u

v

u

v

u

v

u

y)(x,Ny)(x,Ny)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N

y)(x,N0

y)(x,N0

y)(x,N0

y)(x,N0

0y)(x,N

0y)(x,N

0 y)(x,N

0y)(x,N

xyxyxyxy

yyyy

xxxx

y

x

dBε

82

Displacement approximation in terms of shape functions

Strain approximation in terms of strain-displacement matrix

Stress approximation

Summary: For each element

Element stiffness matrix

Element nodal load vector

dNu

dBD

dBε

eV

k dVBDBT

S

eT

b

e

f

SS

T

f

V

TdSTdVXf NN

83

Constant Strain Triangle (CST) : Simplest 2D finite element

• 3 nodes per element • 2 dofs per node (each node can move in x- and y- directions) • Hence 6 dofs per element

x

y

u3

v3

v1

u1

u2

v2

2

3

1

(x,y)

v u

(x1,y1)

(x2,y2)

(x3,y3)

84

166212 dNu

3

3

2

2

1

1

321

321

v

u

v

u

v

u

N0N0N0

0N0N0N

y)(x,v

y)(x,uu

The displacement approximation in terms of shape functions is

321

321

N0N0N0

0N0N0NN

1 1 2 2 3 3u (x,y) u u uN N N

1 1 2 2 3 3v(x,y) v v vN N N

85

Formula for the shape functions are

A

ycxbaN

A

ycxbaN

A

ycxbaN

2

2

2

333

3

2222

1111

12321312213

31213231132

23132123321

33

22

11

x1

x1

x1

det2

1

xxcyybyxyxa

xxcyybyxyxa

xxcyybyxyxa

y

y

y

triangleofareaA

where

x

y

u3

v3

v1

u1

u2

v2

2

3

1

(x,y)

v u

(x1,y1)

(x2,y2)

(x3,y3)

86

Properties of the shape functions:

1. The shape functions N1, N2 and N3 are linear functions of x and y

x

y

2

3

1

1

N1

2

3

1

N2

1 2

3

1

1

N3

nodesotherat

inodeat

0

''1N i

87

2. At every point in the domain

yy

xx

i

i

3

1i

i

3

1i

i

3

1i

i

N

N

1N

88

3. Geometric interpretation of the shape functions At any point P(x,y) that the shape functions are evaluated,

x

y

2

3

1 P (x,y)

A1

A3

A2

A

A

A

A

A

A

3

3

22

11

N

N

N

89

Approximation of the strains

xy

u

v

u v

x

y

x

Bdy

y x

332211

321

321

332211

321

321

000

000

2

1

y)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N

y)(x,N0

y)(x,N0

y)(x,N0

0y)(x,N

0 y)(x,N

0y)(x,N

bcbcbc

ccc

bbb

A

xyxyxy

yyy

xxx

B

90

Element stresses (constant inside each element)

dBD

Inside each element, all components of strain are constant: hence the name Constant Strain Triangle

91

IMPORTANT NOTE: 1. The displacement field is continuous across element boundaries 2. The strains and stresses are NOT continuous across element boundaries

92

Element stiffness matrix

eV

k dVBDBT

AtkeV

BDBdVBDBTT

t=thickness of the element A=surface area of the element

Since B is constant

t

A

93

S

eT

b

e

f

SS

T

f

V

TdSTdVXf NN

Element nodal load vector

94

Element nodal load vector due to body forces

ee A

T

V

T

bdAXtdVXf NN

e

e

e

e

e

e

Ab

Aa

Ab

Aa

Ab

Aa

yb

xb

yb

xb

yb

xb

b

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

f

f

f

f

f

f

f

3

3

2

2

1

1

3

3

2

2

1

1

x

y

fb3x

fb3y

fb1y

fb1x

fb2x

fb2y

2

3

1

(x,y)

Xb Xa

95

EXAMPLE: If Xa=1 and Xb=0

03

03

03

0

0

0

3

2

1

3

3

2

2

1

1

3

3

2

2

1

1

tA

tA

tA

dANt

dANt

dANt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

f

f

f

f

f

f

f

e

e

e

e

e

e

e

e

e

A

A

A

Ab

Aa

Ab

Aa

Ab

Aa

yb

xb

yb

xb

yb

xb

b

96

Element nodal load vector due to traction

e

TSS

T

SdSTf N

EXAMPLE:

x

y

fS3x

fS3y

fS1y

fS1x

2

3

1

e

lS

along

T

SdSTtf

31 31N

97

Element nodal load vector due to traction

EXAMPLE:

x

y

fS3x

2

3 1

e

lS

along

T

SdSTtf

32 32N

fS3y

fS2x

fS2y

(2,0)

(2,2)

(0,0)

0

1ST

tt

dyNtfex l alongS

122

1

)1(32

2 322

0

0

3

3

2

y

x

y

S

S

S

f

tf

f

Similarly, compute

1

2

98

Recommendations for use of CST 1. Use in areas where strain gradients are small 2. Use in mesh transition areas (fine mesh to coarse mesh) 3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners) 4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.

99

Example

x

y

El 1

El 2

1

2 3

4

300 psi 1000 lb

3 in

2 in Thickness (t) = 0.5 in E= 30×106 psi n=0.25

(a) Compute the unknown nodal displacements. (b) Compute the stresses in the two elements.

100

Realize that this is a plane stress problem and therefore we need to use

psiE

D 7

210

2.100

02.38.0

08.02.3

2

100

01

01

1

nn

n

n

Step 1: Node-element connectivity chart

ELEMENT Node 1 Node 2 Node 3 Area

(sqin)

1 1 2 4 3

2 3 4 2 3

Node x y

1 3 0

2 3 2

3 0 2

4 0 0

Nodal coordinates

101

Step 2: Compute strain-displacement matrices for the elements

332211

321

321

000

000

2

1

bcbcbc

ccc

bbb

AB

Recall

123312231

213132321

xxcxxcxxc

yybyybyyb

with

For Element #1:

1(1)

2(2)

4(3)

(local numbers within brackets)

0;3;3

0;2;0

321

321

xxx

yyy

Hence

033

202

321

321

ccc

bbb

200323

003030

020002

6

1)1(B

Therefore

For Element #2:

200323

003030

020002

6

1)2(B

102

Step 3: Compute element stiffness matrices

7

)1(T)1()1(T)1()1(

10

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

Atk

u1 u2 u4 v4 v2 v1

103

7

)2(T)2()2(T)2()2(

10

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

Atk

u3 u4 u2 v2 v4 v3

104

Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom

014433 vvuvuNotice that

Hence we need to calculate only a small (3x3) stiffness matrix

710

4.102.0

0983.045.0

2.045.0983.0

K

u1

u2 v2

u1 u2

v2

105

Step 5: Compute consistent nodal loads

y

x

x

f

f

f

f

2

2

1

yf2

0

0

ySy ff2

10002

The consistent nodal load due to traction on the edge 3-2

lbx

dxx

dxN

tdxNf

x

x

xS y

2252

950

250

3150

)5.0)(300(

)300(

3

0

2

3

0

3

0 233

3

0 2332

3 2

3232

xN

106

lb

ffySy

1225

100022

Hence

Step 6: Solve the system equations to obtain the unknown nodal loads

fdK

1225

0

0

4.102.0

0983.045.0

2.045.0983.0

10

2

2

1

7

v

u

u

Solve to get

in

in

in

v

u

u

4

4

4

2

2

1

109084.0

101069.0

102337.0

107

Step 7: Compute the stresses in the elements

)1()1()1(dBD

With

00109084.0101069.00102337.0

d

444

442211

)1(

vuvuvuT

Calculate

psi

1.76

1.1391

1.114)1(

In Element #1

108

)2()2()2(dBD

With

44

224433

)2(

109084.0101069.00000

d

vuvuvuT

Calculate

psi

35.363

52.28

1.114)2(

In Element #2

Notice that the stresses are constant in each element

109

UNIT - 4

Heat transfer analysis

110

Thermal Convection

( )sq h T T

Newton’s Law of Cooling

2: convective heat transfer coefficient ( )oh W m C

111

Thermal Conduction in 1-D

Dirichlet BC:

Boundary conditions:

Natural BC:

Mixed BC:

112

Weak Formulation of 1-D Heat Conduction (Steady State Analysis)

• Governing Equation of 1-D Heat Conduction -----

( )( ) ( ) ( ) 0

d dT xx A x AQ x

dx dx

0<x<L

• Weighted Integral Formulation -----

• Weak Form from Integration-by-Parts -----

0

( )0 ( ) ( ) ( ) ( )

Ld dT x

w x x A x AQ x dxdx dx

0 0

0

LLdw dT dT

A wAQ dx w Adx dx dx

113

Formulation for 1-D Linear Element

Let 1 1 2 2(x) (x) (x)T T T

f2

x1 x2

1 2

T1

x

T2

f1

2 11 2( ) , ( )

x x x xx x

l l

x2 x1

1T1

2T2

2

2

1

1 )( ,)(x

TAxf

x

TAxf

114


Let w(x)= i (x), i = 1, 2

2 2

1 1

2

2 2 1 1

1

0 ( ) ( )

x x

jij i i i

j x x

ddT A dx AQ dx x f x f

dx dx

1122

2

1

)()( fxfxQTK iii

j

jij

2

1

2212

1211

2

1

2

1

T

T

KK

KK

Q

Q

f

f

2 2

1 21 1

1 2 , Q , f , f

x x

jiij i i

x xx x

dd dT dTwhere K A dx AQ dx A A

dx dx dx dx

115

Element Equations of 1-D Linear Element

2

1

2

1

2

1

11

11

T

T

L

A

Q

Q

f

f

2

1 21

1 2 Q , f , f

x

i i

x x x xx

dT dTwhere AQ dx A A

dx dx

f2

x1 x2

1 2

T1

x

T2

f1

116

1-D Heat Conduction - Example

A composite wall consists of three materials, as shown in the figure below. The inside wall temperature is 200oC and the outside air temperature is 50oC with a convection coefficient of h = 10 W(m2.K). Find the temperature along the composite wall.

t1 t2 t3

0 200oT C 50oT C

x

12 3

1 2 3

1 2 3

70 , 40 , 20

2 , 2.5 , 4

W m K W m K W m K

t cm t cm t cm

117

Thermal Conduction and Convection- Fin

Objective: to enhance heat transfer

dx

t

x

w

2 ( )2 ( ) 2 ( )loss

c c

h T T w th T T dx w h T T dx tQ

A dx A

Governing equation for 1-D heat transfer in thin fin

0c c

d dTA A Q

dx dx

0c c

d dTA Ph T T A Q

dx dx

2P w t where

118

Fin - Weak Formulation (Steady State Analysis)

• Governing Equation of 1-D Heat Conduction -----

( )

( ) ( ) 0d dT x

x A x Ph T T AQdx dx

0<x<L

• Weighted Integral Formulation -----

• Weak Form from Integration-by-Parts -----

0

( )0 ( ) ( ) ( ) ( ) ( )

Ld dT x

w x x A x Ph T T AQ x dxdx dx

0 0

0 ( )

LLdw dT dT

A wPh T T wAQ dx w Adx dx dx

119


Let w(x)= i (x), i = 1, 2

2 2

1 1

2

1

2 2 1 1

0

( ) ( )

x x

jij i j i

j x x

i i

ddT A Ph dx AQ PhT dx

dx dx

x f x f

1122

2

1

)()( fxfxQTK iii

j

jij

2

1

2212

1211

2

1

2

1

T

T

KK

KK

Q

Q

f

f

2 2

1 1

1 2

1 2

, Q ,

,

x x

jiij i j i i

x x

x x x x

ddwhere K A Ph dx AQ PhT dx

dx dx

dT dTf A f A

dx dx

120

Element Equations of 1-D Linear Element

1 1 1

2 2 2

1 1 2 1

1 1 1 26

f Q TA Phl

f Q TL

f2

x=0 x=L

1 2

T1

x

T2

f1

2

1 21

1 2 Q , ,

x

i i

x x x xx

dT dTwhere AQ PhT dx f A f A

dx dx

121

Time-Dependent Problems

122

Time-Dependent Problems

In general,

Key question: How to choose approximate functions?

Two approaches:

txutxu jj ,,

txu ,

xtutxu jj ,

123

Model Problem I – Transient Heat Conduction

Weak form:

txfx

ua

xt

uc ,

)()(0 2211

2

1

xwQxwQdxwft

ucw

x

u

x

wa

x

x

;

21

21

xx dx

duaQ

dx

duaQ

124

Transient Heat Conduction

let:

)()(0 2211

2

1

xwQxwQdxwft

ucw

x

u

x

wa

x

x

n

j

jj xtutxu1

, and xw i

FuMuK

2

1

x

x

jiij dx

xxaK

2

1

x

x

jiij dxcM

i

x

x

ii QfdxF 2

1

ODE!

125

Time Approximation – First Order ODE

tfbudt

dua Tt 0 00 uu

Forward difference approximation - explicit

Backward difference approximation - implicit

1k k k k

tu u f bu

a

1k k k k

tu u f bu

a b t

126

Stability Requirment

max21

2

critt

QuMK where

Note: One must use the same discretization for solving the eigenvalue problem.

127

Transient Heat Conduction - Example

02

2

x

u

t

u10 x

0,0 tu 0,1

t

t

u

0.10, xu

0t

128


129


130


131


132


133

UNIT – 5

Dynamic Analysis

134

Axi-symmetric Analysis

Cylindrical coordinates:

zzryrx ;sin ;cos qq

zr , , q

• quantities depend on r and z only • 3-D problem 2-D problem

135

Axi-symmetric Analysis

136

Axi-symmetric Analysis – Single-Variable Problem

0),(),(),(1

002211

zrfua

z

zrua

zr

zrura

rr

Weak form:

dswq

rdrdzzrwfwuaz

ua

z

w

r

ua

r

w

e

e

n

),(0 002211

where zrn n

z

zruan

r

zruaq

),(),(2211

137

Finite Element Model – Single-Variable Problem

j j

j

u u

Ritz method:

e

i

e

i

e

j

n

j

e

ij QfuK 1

where ( , ) ( , )j jr z x y

iw

Weak form

11 22 00

e

j je i iij i jK a a a rdrdz

r r z z

where

e

e

i if frdrdz

e

e

i i nQ q ds

138

Single-Variable Problem – Heat Transfer

Heat Transfer:

Weak form

1 ( , ) ( , )( , ) 0

T r z T r zrk k f r z

r r r z z

0 ( , )

e

e

n

w T w Tk k wf r z rdrdz

r r z z

wq ds

( , ) ( , )n r z

T r z T r zq k n k n

r z

where

139

3-Node Axi-symmetric Element

1 1 2 2 3 3( , )T r z T T T

1

2

3

2 3 3 2

1 2 3

3 2

1

2 e

r z r zr z

z zA

r r

3 1 1 3

2 3 1

1 3

1

2 e

r z r zr z

z zA

r r

1 2 2 1

3 1 2

2 1

1

2 e

r z r zr z

z zA

r r

140

4-Node Axi-symmetric Element

1 1 2 2 3 3 4 4( , )T r z T T T T

1 2

3 4

a

b

h

x

r

z

1 2

3 4

1 1 1

1

a b a b

a b a b

x h x h

x h x h

141

Single-Variable Problem – Example

Step 1: Discretization

Step 2: Element equation

e

j je i iijK rdrdz

r r z z

e

e

i if frdrdz

e

e

i i nQ q ds

142

Review of CST Element

• Constant Strain Triangle (CST) - easiest and simplest finite element – Displacement field in terms of generalized coordinates

– Resulting strain field is

– Strains do not vary within the element. Hence, the name constant strain triangle (CST) • Other elements are not so lucky. • Can also be called linear triangle because displacement field is

linear in x and y - sides remain straight.

143

Constant Strain Triangle

• The strain field from the shape functions looks like:

– Where, xi and yi are nodal coordinates (i=1, 2, 3) – xij = xi - xj and yij=yi - yj – 2A is twice the area of the triangle, 2A = x21y31-x31y21

• Node numbering is arbitrary except that the sequence 123 must go clockwise around the element if A is to be positive.

144

Constant Strain Triangle

• Stiffness matrix for element k =BTEB tA

• The CST gives good results in regions of the FE model where there is little strain gradient

– Otherwise it does not work well.

145

Linear Strain Triangle

• Changes the shape functions and results in quadratic displacement distributions and linear strain distributions within the element.

146

Linear Strain Triangle

• Will this element work better for the problem?

147

Example Problem

• Consider the problem we were looking at:

5 in.

1 in.

0.1 in.

I 0.113 /12 0.008333 in4

M c

I1 0.5

0.008333 60 ksi

E 0.00207

ML2

2EI

25

2 29000 0.008333 0.0517 in.

1k

1k

148

Bilinear Quadratic

• The Q4 element is a quadrilateral element that has four nodes. In terms of generalized coordinates, its displacement field is:

149

Bilinear Quadratic • Shape functions and strain-displacement

matrix

150

Example Problem

• Consider the problem we were looking at:

5 in.

1 in.

0.1 in.

.in0345.0008333.0290003

1252.0

EI3

PL

00207.0E

ksi60008333.0

5.01

I

cM

in008333.012/11.0I

3

43

0.1k

0.1k

151

Quadratic Quadrilateral Element

• The 8 noded quadratic quadrilateral element uses quadratic functions for the displacements

152


• Shape function examples:

• Strain distribution within the element

153


• Should we try to use this element to solve our problem?

• Or try fixing the Q4 element for our purposes.

– Hmm… tough choice.

154

Isoparametric Elements and Solution

• Biggest breakthrough in the implementation of the finite element method is the development of an isoparametric element with capabilities to model structure (problem) geometries of any shape and size.

• The whole idea works on mapping. – The element in the real structure is mapped to an

‘imaginary’ element in an ideal coordinate system – The solution to the stress analysis problem is easy and

known for the ‘imaginary’ element – These solutions are mapped back to the element in the

real structure. – All the loads and boundary conditions are also mapped

from the real to the ‘imaginary’ element in this approach

155

Isoparametric Element

1 2

3 4

(x1, y1) (x2, y2)

(x3, y3) (x4, y4)

X, u

Y,v

(-1, 1)

x

h

2 (1, -1)

1 (-1, -1)

4 3 (1, 1)

156

Isoparametric element

• The mapping functions are quite simple:

X

Y

N1 N2 N3 N4 0 0 0 0

0 0 0 0 N1 N2 N3 N4

x1

x 2

x3

x4

y1

y2

y3

y4

N1 1

4(1x )(1h)

N2 1

4(1 x )(1h)

N3 1

4(1 x )(1 h)

N4 1

4(1x )(1 h)

Basically, the x and y coordinates of any point in the element are interpolations of the nodal (corner) coordinates.

From the Q4 element, the bilinear shape functions are borrowed to be used as the interpolation functions. They readily satisfy the boundary values too. 157

Isoparametric element

• Nodal shape functions for displacements

u

v

N1 N2 N3 N4 0 0 0 0

0 0 0 0 N1 N2 N3 N4

u1

u2

u3

u4

v1

v2

v3

v4

N1 1

4(1x )(1h)

N2 1

4(1 x )(1h)

N3 1

4(1 x )(1 h)

N4 1

4(1x )(1 h) 158

• The displacement strain relationships:

x u

X

u

x

x

X

u

h

h

X

y v

Y

v

x

x

Y

v

h

h

Y

x

yxy

u

Xv

Yu

Y

v

X

x

X

h

X0 0

0 0x

Y

h

Yx

Y

h

Y

x

X

h

X

u

xu

hv

xv

h

But,it is too difficult to obtainx

Xand

h

X159

Isoparametric Element

u

x

u

X

X

x

u

Y

Y

x

u

h

u

X

X

h

u

Y

Y

h

u

xu

h

X

x

Y

xX

h

Y

h

u

Xu

Y

It is easier to obtainX

xand

Y

x

J

X

x

Y

xX

h

Y

h

Jacobian

defines coordinate transformation

Hence we will do it another way

X

x

N ixX i

Y

x

N ixYi

X

h

N ihX i

Y

h

N ihYi

u

Xu

Y

J 1

u

xu

h

160

Gauss Quadrature

• The mapping approach requires us to be able to evaluate the integrations within the domain (-1…1) of the functions shown.

• Integration can be done analytically by using closed-form formulas from a table of integrals (Nah..) – Or numerical integration can be performed

• Gauss quadrature is the more common form of numerical integration - better suited for numerical analysis and finite element method.

• It evaluated the integral of a function as a sum of a finite number of terms

I dx becomes I Wiii1

n

1

1

161

Gauss Quadrature

• Wi is the ‘weight’ and i is the value of f(x=i)

162

Numerical Integration Calculate: dxxfI

b

a

• Newton – Cotes integration

• Trapezoidal rule – 1st order Newton-Cotes integration

• Trapezoidal rule – multiple application

)()()(

)()()( 1 axab

afbfafxfxf

b

a

b

a

bfafabdxxfdxxfI

2

)()()()()( 1

n

n

n x

x

x

x

b

a

x

x

x

x

n dxxfdxxfdxxfdxxfdxxfI

1

2

10

1

0

)()()( )()(

1

1

)()(2)(2

n

i

i bfxfafh

I

163


b

a

• Newton – Cotes integration

• Simpson 1/3 rule – 2nd order Newton-Cotes integration

)())((

))(()(

))((

))(()(

))((

))(()()( 2

1202

101

2101

200

2010

212 xf

xxxx

xxxxxf

xxxx

xxxxxf

xxxx

xxxxxfxf

b

a

b

a

xfxfxfxxdxxfdxxfI

6

)()(4)()()()( 210

022

164


b

a

• Gaussian Quadrature

)(2

)()(

2

)(

2

)()()(

bfab

afab

bfafabI

)()( 1100 xfcxfcI

Trapezoidal Rule: Gaussian Quadrature:

Choose according to certain criteria 1010 ,,, xxcc

165


b

a

• Gaussian Quadrature

• 2pt Gaussian Quadrature

•3pt Gaussian Quadrature

3

1

3

11

1

ffdxxfI

77.055.0089.077.055.0

1

1

fffdxxfI

111100

1

1

nn xfcxfcxfcdxxfI

ab

axx

)(21~Let:

xdxabbafabdxxf

b

a

~~)(2

1)(

2

1)(

2

1)(

1

1

166

power point presentation on finite element methods...finite element methods iii b tech ii semester...

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