View
11
Download
1
Category
Tags:
Preview:
DESCRIPTION
Chpt06-FEM for 3D Solids
Citation preview
FFinite Element Methodinite Element Method
FEM FOR 3D SOLIDS
A Practical CourseA Practical Course
CHAPTER 6:
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 CASE STUDY
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 3-D solid element can have a tetrahedron and hexahedron shape with flat or curved surfaces.
At any node there are three components in 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
w4
v4
u4
w2
u2
u2
w1
u1
v1
w3
u3
v3 i
j
l
k 1 =
4 =
2 =
3 =
fsy
fsz
fsx
Consider a 4 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
Shape functionsShape functions
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)
Shape functionsShape functions
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
Shape functionsShape functions
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
Shape functionsShape functions
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
(Constant strain element)
31 2 4
31 2 4
31 2 4
3 31 1 2 2 4 4
3 31 1 2 2 3 4
3 31 1 2 2 4 4
0 00 0 0 0 0 0
0 00 0 0 0 0 0
0 00 0 0 0 0 0100 0 02
00 0 0
00 0 0
bb b b
cc c c
dd d d
d cd c d c d cV
d bd b d b d b
c bc b c b c b
B
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
Element matricesElement matrices
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
Element matricesElement matrices
Alternative method for evaluating me: special natural coordinate system
z
x
y
i
j
l
k
1 =
4 =
2 =
3 =
=0
=1
=1
=constant
P
Q
Element matricesElement matrices
z
x
y
i
j
l
k
1 =
4 =
2 =
3 =
=0
=0
=1
=constant
P
Element matricesElement matrices
z
x
y
i
j
l
k
1 =
4 =
2 =
3 =
=1
=1
=1
=0
=constant
P
Q R
Element matricesElement matrices3 2 2
3 2 2
3 2 2
( )
( )
( )
P
P
P
x x x x
y y y y
z z z z
1 1 3 2 2 1 1
1 1 3 2 2 1 1
1 1 3 2 2 1 1
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
B P
B P
B P
x x x x x x x x x
y y y y y y y y y
z z z z z z z z z
4 4 4 4 1 2 1 2 3
4 4 4 4 1 2 1 2 3
4 4 4 4 1 2 1 2 3
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
B
B
B
x x x x x x x x x x x
y y y y y y y y y y y
z z z z z z z z z z z
1
2
3
4
(1 )
(1 )
(1 )
N
N
N
N
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,zP(z3 z2)+z2]
O
B
B [xB(xP x1)+x1, yB(yP y1)y1, zB(zP y1)z1]
O [x=(1 )(x4 xB)xB, y=(1 )(y4 yB)yB, z=(1 )(z4 zB)zB]
=constant
= constant
Element matricesElement matrices
Jacobian:
z
y
x
z
y
x
z
y
x
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
21 31 31 41 21 312
21 31 31 41 21 31
21 31 31 41 21 31
det[ ] 6
x x x x x x
y y y y y y V
z z z z z z
J
Element matricesElement matrices
T
2 3[ ] d
sx
e syl
sz
f
f l
f
f N
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:
3 1
2 3
3 1
1
2
sx
sy
sz
e
sx
sy
sz
f
f
fl
f
f
f
0
f
0
HEXAHEDRON ELEMENTHEXAHEDRON ELEMENT
3D solid meshed with hexahedron elements
P P’
P’’ P’’’
Shape functionsShape functions
eNdU
1
2
3
4
5
6
7
8
displacement components at node 1
displacement components at node 2
displacement components at node 3
displacement components at node 4
displacement co
e
e
e
ee
e
e
e
e
d
d
d
dd
d
d
d
d
mponents at node 5
displacement components at node 6
displacement components at node 7
displacement components at node 8
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
Shape functionsShape functions
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)
Strain matrixStrain matrix
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
Strain matrixStrain matrix
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
Strain matrixStrain matrix
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. , ,
Element matricesElement matrices
1 1 1T T
1 1 1d det[ ]d d d
e
e
V
V
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]
Element matricesElement matrices
For rectangular hexahedron:
det / 8eabc V [J]
11 12 13 14 15 16 17 18
22 23 24 25 26 27 28
33 34 35 36 37 38
44 45 46 47 48
55 56 57 58
66 67 68
77 78
88
.
e
sy
m m m m m m m m
m m m m m m m
m m m m m m
m m m m mm
m m m m
m m m
m m
m
Element matricesElement matrices
(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
Element matricesElement matrices
(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
Element matricesElement matrices
(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
Element matricesElement matrices
(Cont’d)
8
48.
248
4248
42128
242148
1242248
21244248
216
sy
abcex
m
Note: For x direction only
(Rectangular hexahedron)
Element matricesElement matrices
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 6 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:
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
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
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
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:
HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS
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 )
for side nodes with , 1 and 1
(1 )(1 9 )(1 )(1 )
for side nodes with , 1 and 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 STUDY
CASE STUDYCASE STUDY30 nm
30 nm
CASE STUDYCASE STUDY
CASE STUDYCASE STUDY
Recommended