1 Finite Element Method FEM FOR 2D SOLIDS for readers of all backgrounds G. R. Liu and S. S. Quek...

1

FFinite Element Methodinite Element Method

FEM FOR 2D SOLIDS

for readers of all backgroundsfor readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 7:

2Finite Element Method by G. R. Liu and S. S. Quek

CONTENTSCONTENTS INTRODUCTION LINEAR TRIANGULAR ELEMENTS

– Field variable interpolation– Shape functions construction– Using area coordinates– Strain matrix– Element matrices

LINEAR RECTANGULAR ELEMENTS– Shape functions construction– Strain matrix– Element matrices– Gauss integration– Evaluation of me


CONTENTSCONTENTS

LINEAR QUADRILATERAL ELEMENTS– Coordinate mapping– Strain matrix– Element matrices– Remarks

HIGHER ORDER ELEMENTS COMMENTS (GAUSS INTEGRATION)


INTRODUCTIONINTRODUCTION

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

A 2D solid element can have a triangular, rectangular or quadrilateral shape with straight or curved 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 as forces.


INTRODUCTIONINTRODUCTION

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 of thickness is also straightforward, as the procedure is the same as that for a uniform element.


2D solids – plane stress and plane strain2D solids – plane stress and plane strain

fx

fy

x

y

x

y fx

z

Plane stress Plane strain


LINEAR TRIANGULAR LINEAR TRIANGULAR ELEMENTSELEMENTS

Less accurate than quadrilateral elementsUsed by most mesh generators for complex

geometryA linear triangular element:

x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

A fsx

fsy


Field variable interpolationField variable interpolation

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

3 nodeat ntsdisplaceme



3

3

2

2

1

1

v

u

v

u

v

u

ed

31 2

31 2

Node 2Node 1 Node 3

00 0

00 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)

A fsx

fsy

(Shape functions)


Shape functions constructionShape functions construction

1 1 1 1N a b x c y

2 2 2 2N a b x c y

3 3 3 3N 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



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



1 1

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

3 3

11 1 1

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

1e

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 into N1 = a1 + b1x + c1y:

1 2 3 2 3 2 2

1[( )( ) ( )( )]

2 e

N y y x x x x y yA



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



i i i iN a b x c y

1( )

2

1( )

2

1( )

2

i j k k je

i j ke

i k je

a x y x yA

b y yA

c x xA

where

i

jk

i= 1, 2, 3

J, k determined from cyclic permutation

i = 1, 2

j = 2, 3k = 3, 1


Using area coordinatesUsing area coordinates

Alternative method of constructing shape functions

i, 1 j, 2

k, 3

x

y

P

A1

1 2 2 2 3 3 2 2 3 3 2

3 3

11 1

1 [( ) ( ) ( ) ]2 2

1

x y

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

x y

11

e

AL

A

2-3-P:

Similarly, 3-1-P A2

1-2-P A3

22

e

AL

A

33

e

AL

A


Using area coordinatesUsing area coordinates

1 2 3 1L L L Partitions of unity:

3 2 2 32 21 2 3 1

e 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

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


Strain matrixStrain matrix

xx

yy

xy

u

xv

y

u v

y x

LU where

0

0

x

y

y x

L

ee BdLNdLU

0

0

x

y

y x

B LN N1 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)


Element matricesElement matrices

0d ( d ) d d

e e e

hT T Te

V A A

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

Constant matrix T

e ehAk B cB

0d d d d

e e e

hT T Te

V A A

V x A h A m N N N N N N



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 0d

0 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 A

N 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):



2

02.

102

0102

10102

010102

12

sy

hAe

m

x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

A fsx

fsy

lf

fl

sy

sxe d ][

32

T

Nf

y

x

y

xe

f

f

f

fl

0

0

2

132fUniform distributed load:


LINEAR RECTANGULAR LINEAR RECTANGULAR ELEMENTSELEMENTS

Non-constant strain matrixMore accurate representation of stress and

strainRegular shape makes formulation easy



x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

2a

fsy fsx

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



x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

2a

fsy fsx

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

N

where

(Interpolation)



)1)(1(

)1)(1(

)1)(1(

)1)(1(

41

4

41

3

41

2

41

1

N

N

N

N

113 4at node 1 1

113 4at node 2 1

113 4at node 3 1

113 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 41

14

14

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

[2(1 ) 2(1 )] 1

ii

N N N N N

Partition of unity

)1)(1(41 jjjN

1 ( 1, 1) (u1, v1)

2 (1, 1) (u2, v2)

3 (1, +1) (u3, v3)

2a

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

2b



abababab

bbbb

aaaa

11111111

1111

1111

0000

0000

LNB

Note: No longer a constant matrix!



, byax dxdy = ab dd

Therefore,

ddd T1

1

1

1

T cBBcBBk habAhA

e

dddddd1

1

1

10NNNNNNNNm TT

A

T

A

hT

V

e abhAhAxV



lf

fl

sy

sxe d ][

32

T

Nf

x, u

y, v

1 (x1, y1) (u1, v1)

2 (x2, y2) (u2, v2)

3 (x3, y3) (u3, v3)

2a

fsy fsx

4 (x4, y4) (u4, v4)

2b

For uniformly distributed load,

0

0

0

0

y

x

y

x

e

ff

f

f

bf


Gauss integrationGauss 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 of n = 2m - 1

1 1

1 11 1

( , )d d ( , )yx

nn

i j i ji j

I f w w f

In 2 directions:


Gauss integrationGauss integrationm j wj Accuracy n

1 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


Evaluation of Evaluation of mmee

4

04.

204

0204

10204

010204

2010204

02010204

9

sy

habe

m


Evaluation of Evaluation of mmee

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(

4 31

31

33

habhabm

Note: In practice, Gauss integration is often used


LINEAR QUADRILATERAL LINEAR QUADRILATERAL ELEMENTSELEMENTS

Rectangular elements have limited applicationQuadrilateral elements with unparallel edges are

more useful Irregular shape requires coordinate mapping

before using Gauss integration


Coordinate mappingCoordinate mapping

2 (x2, y2)

y

x 1 (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)



( , ) ( , ) e X N x

wherex

y

X ,

1

1

2

2

3

3

4

4

coordinate at node 1




e

x

y

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

iii

xNx ),(4

1

iii

yNy ),(4

1



Substitute 1 into iii

xNx ),(4

1

2 (x2, y2)

y

x 1 ( 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

)()(

)()(

2321

3221

2321

3221

yyyyy

xxxxx

Eliminating , )()}({)(

)(322

1322

1

23

23 yyxxxyy

xxy



y

y

Nx

x

NN

y

y

Nx

x

NN

iii

iii i i

ii

N N

xNNy

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 ,



1

ii

i i

NN

xN Ny

JTherefore,

NLNB

xy

y

x

0

0

Replace differentials of Ni w.r.t. x and y with differentials of Ni w.r.t. and

(Relationship between differentials of shape functions w.r.t. physical coordinates and differentials w.r.t. natural coordinates)



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

1 1 T

1 1det 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


RemarksRemarks

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.


HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS

Higher order triangular elements

i (I,J,K)

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

(0,0,p)

(p 1,1,0)

L1

L3

L2

(0,p 1,1)

(0,1,p 1) (1,0,p 1)

(2,0,p 2)

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

I J K p Node i,

Argyris, 1968 :

1 2 3( ) ( ) ( )I J Ki I J KN 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



Higher order triangular elements (Cont’d)

x , u

y , v

1

2

3

4

5 6

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

4 5 6 1 24N N N L L

x , u

y , v

1

2

3

4 5

6

7 8

9 1 0

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

1 0 1 2 32 7N L L L

Cubic element

Quadratic element



Higher order rectangular elements

(0,0)

0

(n,0)

(0,m) (n,m)

i(n,m)

Lagrange type:

1 1 ( ) ( )D D n mi I J I JN 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]



Higher order rectangular elements (Cont’d)

1 2

3 4

5

6

7

8 9

1 11 1 1

1 12 2 1

1 13 2 2

1 14 1 2

1( ) ( ) (1 ) (1 )

41

( ) ( ) (1 ) (1 )4

1( ) ( ) (1 )(1 )

41

( ) ( ) (1 )(1 )4

D D

D D

D D

D D

N N N

N N N

N N N

N N N

1 15 3 1

1 16 2 3

1 17 3 2

1 18 1 1

1 1 2 29 3 3

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

21

( ) ( ) (1 )(1 )(1 )21

( ) ( ) (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)




Serendipity type:

1 2

3 4

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)




1 2

3 4

5 6

7

8

9 10

11

12

2 2132

2932

13

2932

(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)


ELEMENT WITH CURVED ELEMENT WITH CURVED EDGESEDGES

4

2

3

8

1 5

7

6

1 4 2

5

3

6

1

2

3

4

5 6

1 2

3 4

5

6

7

8


COMMENTS (GAUSS COMMENTS (GAUSS INTEGRATION)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.


COMMENTS (GAUSS COMMENTS (GAUSS INTEGRATION)INTEGRATION)

Using a smaller number of Gauss points tends to counteract the over-stiff behaviour associated with the displacement-based method.

Displacement in an element is assumed using shape functions. This implies that the deformation of the element is 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 softer than lower order ones. This is because using higher order elements gives fewer constraint to the elements.


COMMENTS ON GAUSS COMMENTS ON GAUSS INTEGRATIONINTEGRATION

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 explicit formulation tend to use one-point integration to achieve the best performance in saving CPU time.


CASE STUDYCASE STUDY

Side drive micro-motor



Elastic Properties of Polysilicon

Young’s Modulus, E 169GPa

Poisson’s ratio, 0.262

Density, 2300kgm-3

10N/m

10N/m

10N/m



Analysis no. 1: Von Mises stress distribution using 24 bilinear

quadrilateral elements (41 nodes)



Analysis no. 4: Von Mises stress distribution using 24 eight-nodal, quadratic elements (105 nodes)



Analysis no. 5: Von Mises stress distribution using 192 three-nodal,

triangular elements (129 nodes)


CASE STUDYCASE STUDYAnalysis

no.Number / type of

elements

Total number of nodes in

model

Maximum Von Mises

Stress (GPa)

124 bilinear,

quadrilateral 41 0.0139

296 bilinear, quadrilateral

129 0.0180

3144 bilinear, quadrilateral

185 0.0197

424 quadratic, quadrilateral

105 0.0191

5 192 linear, triangular

129 0.0167

1 Finite Element Method FEM FOR 2D SOLIDS for readers of all backgrounds G. R. Liu and S. S. Quek...

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Transcript of 1 Finite Element Method FEM FOR 2D SOLIDS for readers of all backgrounds G. R. Liu and S. S. Quek...