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1

Finite Element Method

FEM FOR 2D SOLIDS

for readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 7:

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Finite Element Method by G. R. Liu and S. S. Quek2

CONTENTS INTRODUCTION

LINEAR TRIANGULAR ELEMENTS Field variable interpolation

Shape functions construction

Using area coordinates Strain matrix

Element matrices

LINEAR RECTANGULAR ELEMENTS


Strain matrix

Element matrices

Gauss integration

Evaluation ofme

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CONTENTS

LINEAR QUADRILATERAL ELEMENTS

Coordinate mapping

Strain matrix

Element matrices

Remarks

HIGHER ORDER ELEMENTS

COMMENTS (GAUSS INTEGRATION)

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INTRODUCTION

2D solid elements are applicable for the analysisof plane strain and plane stress problems.

A 2D solid element can have a triangular,rectangular or quadrilateral shape with straight orcurved edges.

A 2D solid element can deform only in the plane

of the 2D solid. At any point, there are two components in the x

and y directions for the displacement as well asforces.

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INTRODUCTION

For plane strain problems, the thickness of the

element is unit, but for plane stress problems, the

actual thickness must be used. In this course, it is assumed that the element has a

uniform thickness h.

Formulating 2D elements with a given variation ofthickness is also straightforward, as the procedure

is the same as that for a uniform element.

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2D solids plane stress and plane strain

Plane stress Plane strain

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7/56Finite Element Method by G. R. Liu and S. S. Quek7

LINEAR TRIANGULAR

ELEMENTS Less accurate than quadrilateral elements

Used by most mesh generators for complex

geometry A linear triangular element:

x, u

y, v

1 (x1, y1)

(u1, v1)

2 (x2, y2)

(u2, v2)

3 (x3, y3)

(u3, v3)

sxsy

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Field variable interpolation

( , ) ( , )he

x y x yU N d

3nodeatntsdisplaceme



3

3

2

2

1

1

v

u

v

u

v

u

ed

31 2

31 2

Node 2Node 1 Node 3

00 000 0

NN N

NN N

N

where

x, u

y, v

1 (x1, y1)

(u1, v1)

2 (x2, y2)

(u2, v2)

3 (x3, y3)

(u3, v3)

Asx

sy

(Shape functions)

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1 1 1 1N a b x c y

2 2 2 2N a b x c y

3 3 3 3

N a b x c y

i i i iN a b x c y

Assume,

i= 1, 2, 3

1 T

T

i

i i

i

a

N x y b

c

p

p

or

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Delta function property:

1 for( , )

0 for

i j j

i jN x y

i j

1 1 1

1 2 2

1 3 3

( , ) 1

( , ) 0

( , ) 0

N x y

N x y

N x y

Therefore, 1 1 1 1 1 1 1 1

1 2 2 1 1 2 1 2

1 3 3 1 1 3 1 3

( , ) 1

( , ) 0

( , ) 0

N x y a b x c y

N x y a b x c y

N x y a b x c y

Solving, 2 3 3 2 2 3 3 21 1 1, ,

2 2 2e e e

x y x y y y x xa b c

A A A

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1 1

2 2 2 3 3 2 2 3 1 3 2 1

3 3

11 1 1

1 [( ) ( ) ( ) ]2 2 2

1

e

x y

A x y x y x y y y x x x y

x y

P

Area of triangle Moment matrix

Substitute a1, b1 and c1 back intoN1= a1+ b1x + c1y:

1 2 3 2 3 2 2

1[( )( ) ( )( )]

2 eN y y x x x x y y

A

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Similarly,

2 1 1

2 2 2

2 3 3

( , ) 0

( , ) 1

( , ) 0

N x y

N x y

N x y

2 3 1 1 3 3 1 1 3

3 1 3 1 3 3

1[( ) ( ) ( ) ]

2

1 [( )( ) ( )( )]2

e

e

N x y x y y y x x x yA

y y x x x x y yA

3 1 1

3 2 2

3 3 3

( , ) 0

( , ) 0

( , ) 1

N x y

N x y

N x y

3 1 2 1 1 1 2 2 1

1 2 1 2 1 1

1[( ) ( ) ( ) ]

2

1 [( )( ) ( )( )]2

e

e

N x y x y y y x x x yA

y y x x x x y yA

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i i i iN a b x c y

1( )2

1( )

2

1 ( )2

i j k k j

e

i j k

e

i k j

e

a x y x yA

b y yA

c x xA

where

i

jk

i= 1, 2, 3

J, kdetermined from cyclic

permutationi = 1, 2

j = 2, 3k= 3, 1

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Using area coordinates

Alternative method of constructing shape

functions

i,1

, 2

k, 3

x

P

A1

1 2 2 2 3 3 2 2 3 3 2

3 3

1

1 11 [( ) ( ) ( ) ]2 2

1

x y

A x y x y x y y y x x x y

x y

1

1

e

AL

A

2-3-P:

Similarly, 3-1-P A2

1-2-P A3

22

e

ALA

3

3

e

AL

A

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Using area coordinates

1 2 31L L L Partitions of unity:

3 2 2 32 2

1 2 3 1e e e e

A A A AA AL L L

A A A A

Delta function property: e.g.L1 = 0 at if P at nodes 2 or 3

Therefore, 1 1 2 2 3 3, ,N L N L N L

( , ) ( , )h

ex y x yU N d

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Strain matrix

xx

yy

xy

ux

v

y

u vy x

LU where

0

0

x

y

y x

L

ee BdLNdLU

0

0

x

y

y x

B LN N

1 2 3

1 2 3

1 1 2 2 3 3

0 0 0

0 0 0

a a a

b b b

b a b a b a

B

(constant strain element)

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17

Element matrices

0d ( d ) d d

e e e

hT T T

e

V A A

V z A h A k B cB B cB B cB

Constant matrix

Te ehAk B cB

0d d d d

e e e

hT T T

eV A AV x A h A

m N N N N N N

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18

Element matrices

1 1 1 2 1 3

1 1 1 2 1 3

2 1 2 2 2 3

2 1 2 2 2 3

3 1 3 2 3 3

3 1 3 2 3 3

0 0 0

0 0 0

0 0 0 d0 0 0

0 0 0

0 0 0

e

e

A

N N N N N N

N N N N N N

N N N N N Nh AN N N N N N

N N N N N N

N N N N N N

m

For elements with uniform density and thickness,

Apnm

pnmALLL

pn

A

m 2)!2(

!!!d321

Eisenberg and Malvern (1973):

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19

Element matrices

2

02.

102

0102

10102

010102

12

sy

hAe

m

x, u

, v

1 (x1, y1)

(u1, v1)

2 (x2, y2)

(u2, v

2)

3 (x3, y3)

(u3, v3)

Asx

sy

l

f

f

lsy

sx

e d][32

T

Nf

y

x

y

x

e

f

f

f

fl

0

0

2

132fUniform distributed load:

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20

LINEAR RECTANGULAR

ELEMENTS

Non-constant strain matrix More accurate representation of stress and strain

Regular shape makes formulation easy

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21


x, u

y, v

1 (x1,y1)(u1, v1)

2 (x2,y2)(u2, v2)

3 (x3,y3)(u3, v3)

2a

sy

sx

4 (x4,y4)(u4, v4)

2b

Consider a rectangular element

1

1

2

2

3

3

4

4

displacements at node 1




e

u

v

u

u

u

u

u

u

d

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22


x, u

y, v

1 (x1,y1)

(u1, v1)

2 (x2,y2)

(u2, v2)

3 (x3,y3)

(u3, v3)

2a

sy

sx

4 (x4,y4)

(u4, v4)

2b

1 (1, 1)(u1, v1)

2 (1, 1)(u2, v2)

3 (1, +1)

(u3, v3)

2a

4 (1, +1)(u4, v4)

2b

, byax

( , ) ( , )hex y x yU N d

31 2 4

31 2 4

Node 2 Node 3Node 1 Node 4

00 0 0

00 0 0

NN N N

NN N N

Nwhere

(Interpolation)

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)1)(1(

)1)(1(

)1)(1(

)1)(1(

4

14

4

13

4

12

4

11

N

N

NN

1 13 4at node 11

113 4at node 2

1

113 4at node 31

11

3 4at node 4 1

(1 )(1 ) 0

(1 )(1 ) 0

(1 )(1 ) 1

(1 )(1 ) 0

N

N

N

N

Delta function

property

4

1 2 3 4

1

14

14

[(1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 )]

[2(1 ) 2(1 )] 1

i

i

N N N N N

Partition of

unity

)1)(1(4

1 jjjN

1 (1, 1)(u1, v1)

2 (1, 1)(u2, v2)

3 (1, +1)

(u3, v3)

2a

4 (1, +1)(u4, v4)

2b

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Strain matrix

abababab

bbbb

aaaa

11111111

1111

1111

0000

0000

LNB

Note: No longer a constant matrix!

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Element matrices

, byax dxdy = ab dd

Therefore,

ddd T1

1

1

1

TcBBcBBk habAh

A

e

dddddd1

1

1

10

NNNNNNNNmTT

A

T

A

hT

V

e abhAhAxV

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Element matrices

lf

f

lsy

sx

e d][32

T

Nf

x, u

, v

1 (x1,y1)

(u1, v1)

2 (x2,y2)

(u2, v2)

3 (x3,y3)

(u3, v3)

2a

sy

sx

4 (x4,y4)

(u4, v4)

2b

For uniformly distributed load,

0

0

0

0

y

x

y

x

e

f

f

f

f

bf

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Gauss integration

For evaluation of integrals in ke and me (in practice)

In 1 direction: )()d(1

1

1 jj

m

j

fwfI

m gauss points gives exact solution of

polynomial integrand ofn = 2m - 1

1 1

1 11 1

( , )d d ( , )yxnn

i j i j

i j

I f ww f

In 2 directions:

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Gauss integrationm j wj Accuracy n1 0 2 1

2 -1/3, 1/3 1, 1 3

3 -0.6, 0, 0.6 5/9, 8/9, 5/9 5

4 -0.861136, -0.339981,

0.339981, 0.861136

0.347855, 0.652145,

0.652145, 0.347855

7

5 -0.906180, -0.538469, 0,

0.538469, 0.906180

0.236927, 0.478629,

0.568889, 0.478629,

0.236927

9

6 -0.932470, -0.661209,

-0.238619, 0.238619,

0.661209, 0.932470

0.171324, 0.360762,

0.467914, 0.467914,

0.360762, 0.171324

11

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Evaluation of me

404.

204

0204

10204

010204

2010204

02010204

9

sy

habe

m

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Evaluation of me

E.g.

)1)(1(4

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

16

31

31

1

1

1

1

1

1

1

1

jiji

jiji

jiij

hab

ddhab

ddNNhabm

94)111)(111(43

1

3

133

habhab

m

Note: In practice, Gauss integration is often used

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LINEAR QUADRILATERAL

ELEMENTS Rectangular elements have limited application

Quadrilateral elements with unparallel edges are

more useful Irregular shape requires coordinate mapping

before using Gauss integration

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Coordinate mapping

2 (x2,y2)

x1 (1, 1) 2 (1, 1)

3 (1, +1)4 (1, +1)

3 (x3,y3)4 (x4,y4)

1 (x1,y1)

Physical coordinates Natural coordinates

( , ) ( , )he

U N d (Interpolation of displacements)

( , ) ( , ) e X N x (Interpolation of coordinates)

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Coordinate mapping

( , ) ( , ) e X N x

wherex

y

X ,

1

1

2

2

3

3

4

4

coordinate at node 1




e

xy

x

y

x

y

x

y

x

)1)(1(

)1)(1(

)1)(1(

)1)(1(

41

4

41

3

41

2

41

1

N

N

N

N

ii

i

xNx ),(4

1

ii

i

yNy ),(4

1

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Coordinate mapping

Substitute 1 into iii

xNx ),(4

1

2 (x2,y2)

x1 (1, 1) 2 (1, 1)

3 (1, +1)4 (1, +1)

3 (x3,y3)4 (x4,y4)

1 (x1,y1)

321

221

321

221

)1()1(

)1()1(

yyy

xxx

or

)()(

)()(

232

1322

1

232

1322

1

yyyyy

xxxxx

Eliminating , )()}({)(

)(322

1322

1

23

23 yyxxxyy

xxy

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Strain matrix

y

y

Nx

x

NN

y

y

Nx

x

NN

iii

iii i i

ii

N N

x

NN

y

Jor

x y

x y

Jwhere (Jacobian matrix)

1 131 2 4

2 2

3 331 2 4

4 4

x yNN N N

x y

x yNN N N

x y

JSince ( , ) ( , ) e X N x ,

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Strain matrix

1

ii

i i

NN

x

N N

y

JTherefore,

NLNB

xy

y

x

00

Replace differentials ofNiw.r.t.x andy

with differentials ofNiw.r.t. and

(Relationship between differentials of shape

functions w.r.t. physical coordinates and

differentials w.r.t. natural coordinates)

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Element matrices

Murnaghan (1951) : dA=det |J | dd

1 1T

1 1 det d de h

k B cB J

dddet

dddd

1

1

1

1

0

JNN

NNNNNNm

T

T

A

T

A

hT

V

e

h

AhAxV

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Remarks

Shape functions used for interpolating the coordinates are

the same as the shape functions used for interpolation of

the displacement field. Therefore, the element is called an

isoparametric element.

Note that the shape functions for coordinate interpolation

and displacement interpolation do not have to be the same.

Using the different shape functions for coordinate

interpolation and displacement interpolation, respectively,

will lead to the development of so-called subparametric or

superparametric elements.

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Higher order triangular elements

i (I,J,K)

(p,0,0) (0,p,0)

(0,0,p)

(p1,1,0)

L1

L3

L2

(0,p1,1)

(0,1,p1)(1,0,p1)

(2,0,p2)

nd= (p+1)(p+2)/2

I J K p Node i,

Argyris, 1968 :

1 2 3( ) ( ) ( )I J Ki I J K N l L l L l L

0 1 ( 1)

0 1 ( 1)

( )( ) ( )( )

( )( ) ( )I I I

L L L L L Ll L

L L L L L L

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Higher order triangular elements (Contd)

x, u

, v

1

2

3

4

56

1 2 2 1 1(2 1)N N N L L

4 5 6 1 24N N N L L

x, u

, v

1

2

3

45

6

78

9

10

1 2 3 1 1 1

1(3 1)(3 2)

2N N N L L L

4 9 1 2 1

9(3 1)

2N N L L L

10 1 2 327N L L L

Cubic element

Quadratic element

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Higher order rectangular elements

(0,0)

0

(n,0)

0,m (n,m)

i n,m

Lagrange type:

1 1( ) ( )

D D n m

i I J I J N N N l l

0 1 1 1

0 1 1 1

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

( )( ) ( )( ) ( )

n k k nk

k k k k k k k n

l

[Zienkiewicz et al., 2000]

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Higher order rectangular elements (Contd)

1 2

34

5

6

7

8 9

1 1

1 1 1

1 1

2 2 1

1 1

3 2 2

1 1

4 1 2

1( ) ( ) (1 ) (1 )

41

( ) ( ) (1 ) (1 )4

1( ) ( ) (1 )(1 )

4

1

( ) ( ) (1 )(1 )4

D D

D D

D D

D D

N N N

N N N

N N N

N N N

1 1

5 3 1

1 1

6 2 3

1 17 3 2

1 1

8 1 1

1 1 2 2

9 3 3

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

2

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

2

1( ) ( ) (1 )(1 )(1 )2

1( ) ( ) (1 )(1 )

2

( ) ( ) (1 )(1 )

D D

D D

D D

D D

D D

N N N

N N N

N N N

N N N

N N N

(nine node quadratic element)

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Serendipity type:

1 2

34

5

6

7

8 0

=1

=1

14

212

212

(1 )(1 )( 1) 1, 2, 3, 4

(1 )(1 ) 5, 7

(1 )(1 ) 6,8

j j j j j

j j

j j

N j

N j

N j

(eight node quadratic element)

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1 2

34

5 6

7

8

910

11

12

2 21

32

29

32

13

29

32

(1 )(1 )(9 9 10)

for corner nodes 1, 2, 3, 4

(1 )(1 )(1 9 )

for side nodes 7, 8, 11, 12 where 1 and

(1 )(1 )(1

j j j

j j j

j j

j j

N

j

N

j

N

13

9 )

for side nodes 5, 6, 9, 10 where and 1

j

j jj

(twelve node cubic element)

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ELEMENT WITH CURVED

EDGES

4

2

3

8

1

5

7

6

1

4 2

5

3

6

1

2

3

4

56

1 2

34

5

6

7

8

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COMMENTS (GAUSS

INTEGRATION) When the Gauss integration scheme is used, one has to

decide how many Gauss points should be used.

Theoretically, for a one-dimensional integral, using m

points can give the exact solution for the integral of a

polynomial integrand of up to an order of (2m1).

As a general rule of thumb, more points should be used for

a higher order of elements.

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COMMENTS (GAUSS

INTEGRATION) Using a smaller number of Gauss points tends to

counteract the over-stiff behaviour associated with thedisplacement-based method.

Displacement in an element is assumed using shapefunctions. This implies that the deformation of the elementis somehow prescribed in a fashion of the shape function.This prescription gives a constraint to the element. The so-constrained element behaves stiffer than it should. It is

often observed that higher order elements are usually softerthan lower order ones. This is because using higher orderelements gives fewer constraint to the elements.

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COMMENTS ON GAUSSINTEGRATION

Two Gauss points for linear elements, and two or three

points for quadratic elements in each direction should be

sufficient for most cases.

Most of the explicit FEM codes based on explicitformulation tend to use one-point integration to achieve the

best performance in saving CPU time.

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CASE STUDY

Side drive micro-motor

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CASE STUDY

Elastic Properties of Polysilicon

Youngs Modulus, E 169GPa

Poissons ratio, 0.262

Density, 2300kgm-3

10N/m

10N/m

10N/m

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CASE STUDY

Analysis no. 1: Von Mises stress

distribution using 24 bilinear

quadrilateral elements (41 nodes)

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CASE STUDY




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CASE STUDY




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CASE STUDY


distribution using 24 eight-nodal,

quadratic elements (105 nodes)

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CASE STUDY


distribution using 192 three-nodal,

triangular elements (129 nodes)

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CASE STUDYAnalysis

no.

Number / type of

elements

Total number

of nodes inmodel

Maximum

Von MisesStress (GPa)

124 bilinear,

quadrilateral41 0.0139

2 96 bilinear,quadrilateral

129 0.0180

3144 bilinear,


424 quadratic,


5 192 linear,

triangular129 0.0167

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