slide chpt09

1

FFinite Element Methodinite Element Method

FEM FOR 3D SOLIDS

for readers of all backgroundsfor readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 9:

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

CONTENTSCONTENTS INTRODUCTION TETRAHEDRON ELEMENT

– Shape functions– Strain matrix– Element matrices

HEXAHEDRON ELEMENT– Shape functions– Strain matrix– Element matrices– Using tetrahedrons to form hexahedrons

HIGHER ORDER ELEMENTS ELEMENTS WITH CURVED SURFACES


INTRODUCTIONINTRODUCTION

For 3D solids, all the field variables are dependent of x, y and z coordinates – most general element.

The element is often known as a 3D solid element or simply a solid element.

A 3D solid element can have a tetrahedron and hexahedron shape with flat or curved surfaces.

At any node there are three components in the x, y and z directions for the displacement as well as forces.


TETRAHEDRON ELEMENTTETRAHEDRON ELEMENT

3D solid meshed with tetrahedron elements


TETRAHEDRON ELEMENTTETRAHEDRON ELEMENT

z=Z

x=Xz=Z

y=Y

w 4

v4

u4

w2

u2

u2

w 1

u1

v1

w3

u3

v3 i

j

l

k 1 =

4 =

2 =

3 =

fsy

fsz

fsx

Consider a four node tetrahedron element

1

1

1

2

2

2

3

3

3

4

4

4

node 1

node 2

node 3

node 4

e

u

v

w

u

v

w

u

v

w

u

v

w

d


Shape functionsShape functions

( , , ) ( , , )hex y z x y zU N d

1 2 3 4

1 2 3 4

1 2 3 4

node 1 node 2 node 3 node 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

N N N N

N N N N

N N N N

N

where

Use volume coordinates (Recall Area coordinates for 2D triangular element)

1234

2341 V

VL P

1=i

2=j

3=k

4=l

P

y

z

x



Similarly,1234

1234

1234

1243

1234

1342 , ,

V

VL

V

VL

V

VL PPP

Can also be viewed as ratio of distances

234 134 1231241 2 3 4

1 234 1 234 1 234 1 234

, , , P P PPd d ddL L L L

d d d d

1=i

2=j

3=k

4=l

P

y

z

x

1 4321 LLLL

since

1234123124134234 VVVVV PPPP

(Partition of unity)



jkl

iLi nodes remote theat the 0

node home at the 1

44332211

44332211

44332211

zLzLzLzLz

yLyLyLyLy

xLxLxLxLx

(Delta function property)

1 4321 LLLL

4

3

2

1

4321

4321

4321

1 1 1 11

L

L

L

L

zzzz

yyyy

xxxx

z

y

x



Therefore,

where

z

y

x

dcba

dcba

dcba

dcba

V

L

L

L

L 1

6

1

4444

3333

2222

1111

4

3

2

1

1

det , det 1

1

1 1

det 1 , det 1

1 1

j j j j j

i k k k i k k

l l l l l

j j j j

i k k i k k

l l l l

x y z y z

a x y z b y z

x y z y z

y z y z

c y z d y z

y z y z

(Adjoint matrix)

(Cofactors)

i

j

k

l

i= 1,2

j = 2,3

k = 3,4

l = 4,1



l

k

j

i

l

k

j

i

l

k

j

i

z

z

z

z

y

y

y

y

x

x

x

x

V

1

1

1

1

det6

1(Volume of tetrahedron)

)(6

1zdycxba

VLN iiiiii Therefore,


Strain matrixStrain matrix

Since, ( , , ) ( , , )hex y z x y zU N d

Therefore, ee BdLNdLU where NLNB

0

0

0

00

00

00

xy

xz

yz

z

y

x

44

44

44

4

4

4

33

33

33

3

3

3

22

22

22

2

2

2

11

11

11

1

1

1

0

0

0

00

00

00

0

0

0

00

00

00

0

0

0

00

00

00

0

0

0

00

00

00

2

1

bd

cd

bc

d

c

b

bd

cd

bc

d

c

b

bd

cd

bc

d

c

b

bd

cd

bc

d

c

b

VB

(Constant strain element)


Element matricesElement matrices

e

T Te eV

dV V k B cB B cB

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

d de e

Te

V V

V V

N N N N

N N N Nm N N

N N N N

N N N N

where

ji

ji

ji

ij

NN

NN

NN

00

00

00

N



1 2 3 4

! ! ! !d 6

( 3)!e

m n p qeV

m n p qL L L L V V

m n p q

Eisenberg and Malvern [1973] :

2 0 0 1 0 0 1 0 0 1 0 0

2 0 0 1 0 0 1 0 0 1 0

2 0 0 1 0 0 1 0 0 1

2 0 0 1 0 0 1 0 0

2 0 0 1 0 0 1 0

2 0 0 1 0 0 1

2 0 0 1 0 020

2 0 0 1 0

2 0 0 1

. 2 0 0

2 0

2

ee

V

sy

m



Alternative method for evaluating me: special natural coordinate system

z

x z = Z

y

i

j

l

k

1 =

4 =

2 =

3 =

= 0

= 1

= 1

= c o n s ta n t

P

Q



z

x z = Z

y

i

j

l

k

1 =

4 =

2 =

3 =

= 0

= 0

= 1

= c o n s ta n t

P



z

x z=Z

y

i

j

l

k

1 =

4 =

2 =

3 =

=1

=1

=1

=0

=constant

P

Q R



z

x z=Z

y

i

j

l

k

1 =

4 =

2 =

3 =

=0 =0 =1

=1 =0 =1

=1 =1 =1

=0

=constant

P [xP(x3

x2)+x2, yP(y3

y2)+y2,0]

O

B

B [xB(xP

x1)+x1, yB[(yP

y1)y1],0]

O [x=(1 )(x4 xB)xB, y=(1 )(y4

yB)yB, z=(1 )z4]

=constant

=constant

0

)(

)(

223

223

P

P

P

z

yyyy

xxxx

0

)()()(

)()()(

1122311

1122311

B

PB

PB

z

yyyyyyyyy

xxxxxxxxx

4

321214444

321214444

)1(

)()()()(

)()()()(

zz

yyyyyyyyyyy

xxxxxxxxxxx

B

B

)1(

)1(

)1(

4

3

2

1

N

N

N

N



Jacobian:

z

y

x

z

y

x

z

y

x

J

2

4

312141313121

312141313121

6

0 0

]det[

V

z

yyyyyy

xxxxxx

J

1 1 1

0 0 0d det d d d

e

T Te

V

V m N N N N [J]

11 12 13 14

1 1 1 21 22 23 242

0 0 031 32 33 34

41 42 43 44

6 d d de eV

N N N N

N N N Nm

N N N N

N N N N



l

f

f

f

l

sz

sy

sx

e d ][43

T

Nf

z=Z

x=Xz=Z

y=Y

w 4

v4

u4

w2

u2

u2

w 1

u1

v1

w3

u3

v3 i

j

l

k 1 =

4 =

2 =

3 =

fsy

fsz

fsx

For uniformly distributed load:

13

13

13

13

13

13

432

1

0

0

0

0

0

0

f

sz

sy

sx

sz

sy

sx

e

f

f

ff

f

f

l


HEXAHEDRON ELEMENTHEXAHEDRON ELEMENT

3D solid meshed with hexahedron elements

P P’

P’’ P’’’



eNdU

1

2

3

4

5

6

7

8

displacement components at node 1




displacement co

e

e

e

ee

e

e

e

e

d

d

d

dd

d

d

d

d

mponents at node 5




1

1

1

( 1, 2, ,8) ei

u

v i

w

d

17

5 8

6 4

2

0

z

y

x

3

0

fsz

fsyfsx

87654321 NNNNNNNNN

)8,,2,1(

00

00

00

i

N

N

N

i

i

i

iN



4(-1, 1, -1)

(1, -1, 1)6

(1, -1, -1)2

1 7

5 8

6 4

2 0

z

y

x

3

0

fsz

fsy fsx

8(-1, 1, 1)

7 (1, 1, 1)

(-1, -1, 1)5

(-1, -1, -1)1

3(1, 1, -1)

iii

iii

iii

zNz

yNy

xNx

),,(

),,(

),,(

8

1

8

1

8

1

)1)(1)(1(

8

1iiiiN

(Tri-linear functions)



87654321 BBBBBBBBB

whereby

0

0

0

00

00

00

xNyN

xNzN

yNzN

zN

yN

xN

ii

ii

ii

i

i

i

ii LNB

Note: Shape functions are expressed in natural coordinates – chain rule of differentiation

ee BdLNdLU



z

z

Ny

y

Nx

x

NN

z

z

Ny

y

Nx

x

NN

z

z

Ny

y

Nx

x

NN

iiii

iiii

iiii

Chain rule of differentiation

z

Ny

Nx

N

N

N

N

i

i

i

i

i

i

J

where

z

z

z

y

y

y

x

x

x

J


Strain matrixStrain matrix8 8 8

1 1 1

( , , ) , ( , , ) , ( , , )i i i i i ii i i

x N x y N y z N z

Since,

or

8

1

8

1

8

1

8

1

8

1

8

1

8

1

8

1

8

1

i

ii

i

ii

i

ii

i

ii

i

ii

i

ii

i

ii

i

ii

i

ii

Nz

Nz

Nz

Ny

Ny

Ny

Nx

Nx

Nx

J

1 1 1

2 2 23 5 6 7 81 2 4

3 3 3

4 4 43 5 6 7 81 2 4

5 5 5

6 6 61 2 3 4 5 6 7 8

7 7 7

8 8 8

x y z

x y zN N N N NN N Nx y z

x y zN N N N NN N Nx y z

x y zN N N N N N N N

x y z

x y z

J



i

i

i

i

i

i

N

N

N

z

Ny

Nx

N

1J

0

0

0

00

00

00

xNyN

xNzN

yNzN

zN

yN

xN

ii

ii

ii

i

i

i

ii LNB

Used to replace derivatives w.r.t. x, y, z with derivatives w.r.t. , ,



1 1 1T T

1 1 1d det[ ]d d d

e

e

V

A

k B cB B cB J

Gauss integration: ),,(d)d,(1 1 1

1

1

1

1

1

1 jjikji

n

i

m

j

l

k

fwwwfI

1 1 1

1 1 1d det d d d

e

T Te

V

V

m N N N N [J]



For rectangular hexahedron:

det eabc V [J]

88

7877

686766

58575655

4847464544

282726252433

28272625242322

1817161514131211

.

m

mm

mmm

mmmm

mmmmm

mmmmmm

mmmmmmm

mmmmmmmm

m

sy

e



(Cont’d)

where

ddd

00

00

00

ddd

00

00

00

00

00

00

ddd

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

ji

ji

ji

j

j

j

i

i

i

jiij

NN

NN

NN

abc

N

N

N

N

N

N

abc

abc NNm



(Cont’d)

or

ij

ij

ij

ij

m

m

m

00

00

00

m

where

)1)(1)(1(8

d)1)(1(d)1)(1(d)1)(1(64

ddd

31

31

31

1

1

1

1

1

1

1

1

1

1

jijiji

jijiji

jiij

hab

abc

NNabcm



(Cont’d)

E.g.216

8)111)(111)(111(8 3

131

31

33

abcabcm

216

1216

2216

4

216

8

46352817

184538276857473625162413

483726155814786756342312

8877665544332211

abcmmmm

abcmmmmmmmmmmmm

abcmmmmmmmmmmmm

abc

mmmmmmmm



(Cont’d)

8

48.

248

4248

42128

242148

1242248

21244248

216

sy

abcex

m

Note: For x direction only

(Rectangular hexahedron)



l

f

f

f

l

sz

sy

sx

e d ][43

T

Nf

17

5 8

6 4

2

0

z

y

x

3

0

fsz

fsyfsx

13

13

13

13

13

13

432

1

0

0

0

0

0

0

f

sz

sy

sx

sz

sy

sx

e

f

f

ff

f

f

l

For uniformly distributed load:


Using tetrahedrons to form hexahedronsUsing tetrahedrons to form hexahedrons

Hexahedrons can be made up of several tetrahedrons

1

5

6

8 1 4

3

8

1

2 3

4

5

7

8

3

1 6

8

6

3

2

1

6

3

6 7

8 Hexahedron made up of 5 tetrahedrons:


Using tetrahedrons to form hexahedronsUsing tetrahedrons to form hexahedrons

1

2 3

4

5

7

8

6

1

2

4

5 8

6

2 3

7

8

6 4

1 4

5

6

1

2

4 6

5 8

6 4

Break into three

Hexahedron made up of six tetrahedrons:

Element matrices can be obtained by assembly of tetrahedron elements


HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS

Tetrahedron elements

1

9

8

7 10

2

5

6

3

4

5 2 3

6 1 3

7 1 2

8 1 4

9 2 4

10 3 4

(2 -1) for corner nodes 1,2,3,4

4

4

4 for mid-edge nodes

4

4

4

i i iN L L i

N L L

N L L

N L L

N L L

N L L

N L L

10 nodes, quadratic:



Tetrahedron elements (Cont’d)20 nodes, cubic:

12

9 95 1 1 3 11 1 1 42 2

9 96 3 1 3 12 4 1 42 2

9 97 1 1 2 13 22 2

98 2 1 22

99 2 2 32

910 3 2 32

(3 1)(3 2) for corner nodes 1,2,3,4

(3 1) (3 1)

(3 1) (3 1)

(3 1) (3 1)

(3 1)

(3 1)

(3 1)

i i i iN L L L i

N L L L N L L L

N L L L N L L L

N L L L N L L

N L L L

N L L L

N L L L

2 4

914 4 2 42

915 3 3 42

916 4 3 42

17 2 3 4

18 1 2 3

19 1 3 4

20 1 2 4

for edge nodes(3 1)

(3 1)

(3 1)

27

27 for center surface nodes

27

27

L

N L L L

N L L L

N L L L

N L L L

N L L L

N L L L

N L L L

1

13 12

7

15

2

9

6 3

4

5

8

10

11

14

16

17

18

195

20



Brick elements

Lagrange type:

i(I,J,K)

(0,0,0)

(n,m,p)

(n,0,0)

(n,m,0)

(nd=(n+1)(m+1)(p+1) nodes)

1 1 1 ( ) ( ) ( )D D D n m pi I J K I J KN N N N l l l

0 1 1 1

0 1 1 1

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

( )( ) ( )( ) ( )n k k nk

k k k k k k k n

l

where


HIGHER ORDER HIGHER ORDER ELEMENTSELEMENTS

Brick elements (Cont’d)

Serendipity type elements:

4(-1, 1, -1)

(1, -1, 1)6

(1, -1, -1)2

8(-1, 1, 1)

7 (1, 1, 1)

(-1, -1, 1)5

(-1,-1,-1)1

3(1, 1, -1)

9(1,0,-1)

10(0,1,-1)

11(-1,0,-1) 12(0-1,-1)

13 143

15

16

17 18

19 20

18

214

214

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

for corner nodes 1, , 8

(1 )(1 )(1 ) for mid-side nodes 10,12,14,16

(1 )(1

j j j j j j i

j j j

j

N

j

N j

N

214

)(1 ) for mid-side nodes 9,11,13,15

(1 )(1 )(1 ) for mid-side nodes 17,18,19,20

j j

j j j

j

N j

20 nodes, tri-quadratic:



Brick elements (Cont’d)

2 2 2164

2964

13

2964

(1 )(1 )(1 )(9 9 9 19)

for corner nodes 1, , 8

(1 )(1 9 )(1 )(1 )

for side nodes with , 1 and 1

(1 )(1 9

j j j j

j j j j

j j j

j

N

j

N

N

13

2964

13

)(1 )(1 )


(1 )(1 9 )(1 )(1 )


j j j

j j j

j j j j

j j j

N

32 nodes, tri-cubic:


ELEMENTS WITH CURVED ELEMENTS WITH CURVED SURFACESSURFACES

1

4

9 8

7 10

2 5

6 3

7 18

16

12 15

14 11

13

5 17 19

20

6

10 9

8

2

1

4 3

9 8

7 10

2

5

6 3

1

4

13 7 18 16

12 15

14 11

5 17 19

20

6

10

9

8

2

1 4

3


CASE STUDYCASE STUDY

Stress and strain analysis of a quantum dot heterostructure

Material E (Gpa)

GaAs 86.96 0.31

InAs 51.42 0.35

GaAs substrate

GaAs cap layer

InAs wetting layer

InAs quantum dot


CASE STUDYCASE STUDY30 nm

30 nm

slide chpt09

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