Finite Element Method
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Transcript Finite Element Method
Finite Element Method
A Practical Course
CHAPTER 6:
FEM FOR 3D SOLIDS
CONTENTS
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
INTRODUCTION
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 ELEMENT
3D solid meshed with tetrahedron elements
TETRAHEDRON ELEMENT
Consider a 4 node tetrahedron element
u1
v
1
w1
u2
v2
w
de 2
u3
v3
w3
u
4
v4
w
4
w4
node 1
4=
node 2
w1
v4
w3
u4
v3
1=
node 3
node 4
l
z=Z
i
fsz
u1
w2
fsy
fsx
y=Y
x=X
z =
Z
3=
u3
v1
2=
u2
u2
j
k
Shape functions
U h ( x, y, z ) N( x, y, z )de
node 1
where
N1
N 0
0
node 2
node 3
node 4
0
0
N2
0
0
N3
0
0
N4
0
N1
0
0
N2
0
0
N3
0
0
N4
0
N1
0
0
N2
0
0
N3
0
0
0
0
N 4
Use volume coordinates (Recall Area coordinates for
2D triangular element)
VP 234
L1
V1234
1=i
P
4=l
3=k
z
x
y
2=j
Shape functions
Similarly, L2
VP134
VP123
VP124
, L3
, L4
V1234
V1234
V1234
Can also be viewed as ratio of distances
d
d
d
d
L1 P 234 , L2 P 134 , L3 P 124 , L4 P 123
d1 234
d1234
d1234
d1234
L1 L2 L3 L4 1 (Partition of unity)
1=i
since
VP 234 VP134 VP124 VP123 V1234
4=l
P
3=k
z
x
y
2=j
Shape functions
1
Li
0
at the home node i
(Delta function property)
at the the remote nodes jkl
x L1 x1 L2 x 2 L3 x3 L4 x 4
y L1 y1 L2 y 2 L3 y 3 L4 y 4
z L1 z1 L2 z 2 L3 z 3 L4 z 4
1 1 1 1
x x x x
1
2
3
y y1 y 2 y 3
z z1 z 2 z 3
L1 L2 L3 L4 1
1 L1
x 4 L2
y 4 L3
z 4 L4
Shape functions
Therefore,
L1
a1
L
2 1 a 2
L3 6V a3
L4
a 4
(Adjoint matrix)
b1
c1
b2
c2
b3
c3
b4
c4
d1 1
d 2 x
l = 4,1
d3 y
l
d 4 z
(Cofactors)
where
xj yj z j
1
ai det xk yk zk ,
bi det 1
xl yl zl
1
zj
yj 1
yj
ci det yk 1
zk , d i det yk
yl 1
yl
zl
i= 1,2
i
j
k
k = 3,4
yj
yk
yl
zj
zk
zl
zj
zk
zl
1
1
1
j = 2,3
Shape functions
1
V det
6
1 xi y i z i
1 x j y j z j
1 xk y k z k
1 xl y l z l
Therefore, N i Li
(Volume of tetrahedron)
1
(ai bi x ci y d i z )
6V
Strain matrix
0
0
x
0
y
0
0
0
z
LNd e Bd e where B LN 0 z y N
z
0
x
0
y x
0 0 b3 0 0 b4 0 0
c2 0 0 c3 0 0 c4 0
0 d 2 0 0 d3 0 0 d 4
d 2 c2 0 d3 c3 0 d 4 c4
0 b2 d3 0 b3 d3 0 b4
b2 0 c3 b3 0 c4 b4 0
Since, U h ( x, y, z ) N( x, y, z )de
Therefore, LU
b1 0 0 b2
0 c 0 0
1
1 0 0 d1 0
B
2V 0 d1 c1 0
d1 0 b1 d 2
c1 b1 0 c2
(Constant strain element)
Element matrices
k e BT cBdV Ve BT cB
Ve
N11
N
me NT NdV 21
N31
Ve
Ve
N 41
where
Ni N j
N ij 0
0
0
Ni N j
0
N12
N13
N 22
N 23
N32
N33
N 42
N 43
0
0
N i N j
N14
N 24
dV
N34
N 44
Element matrices
Eisenberg and Malvern, 1973 :
m !n ! p !q !
Ve L L L L dV (m n p q 3)! 6Ve
m n p q
1 2 3 4
2 0 0
2 0
2
V
m e e
20
sy.
1 0 0 1 0 0 1 0 0
0 1 0 0 1 0 0 1 0
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 0
2 0 0 1 0
2 0 0 1
2 0 0
2 0
2
Element matrices
Alternative method for evaluating me: special natural
coordinate system
4=
l
=constant
3= k
1=
i
Q
=0
z
P
y
x
2= j
=1
=1
Element matrices
4=
l
=constant
3= k
1=
=1
i
=0
P
z
y
x
2= j
=0
Element matrices
4=
l =0
=constant
R
1=
Q
=1
i
=1
P
z
y
x
3= k
2= j
=1
Element matrices
xP ( x3 x2 ) x2
xB ( xP x1 ) x1 ( x3 x2 ) ( x2 x1 ) x1
yP ( y3 y2 ) y2
yB ( yP y1 ) y1 ( y3 y2 ) ( y2 y1 ) y1
z P ( z3 z2 ) z2
zB ( zP z1 ) z1 ( z3 z2 ) ( z2 z1 ) z1
x x4 ( x4 xB ) x4 ( x4 x1 ) ( x2 x1 ) ( x2 x3 )
y y4 ( y4 yB ) y4 ( y4 y1 ) ( y2 y1 ) ( y2 y3 )
z z4 ( z4 z B ) z4 ( z4 z1 ) ( z2 z1 ) ( z2 z3 )
4=
l =0
= constant
N1 (1 )
N 2 (1 )
3= k
=1
=1
=1
O
1=
N 3
N 4 (1 )
z
i
=0
=0
=1
B
P [xP(x3x2)+x2, yP(y3y2)+y2,zP(z3z2)+z2]
B [xB(xPx1)+x1, yB(yPy1)y1, zB(zPy1)z1]
=constant
y
x
z=Z
=1
=0
=1
2= j
= constant
O [x=(1)(x4xB)xB, y=(1)(y4yB)yB, z=(1)(z4zB)zB]
Element matrices
x
y
J
z
Jacobian:
x
y
z
x
y
z
x21 x31 x31 x41 x21 x31
det[J ] y21 y31 y31 y41 y21 y31 6V 2
z21 z31 z31 z41 z21 z31
me N NdV
T
1 1 1
0 0 0
NT Ndet[J]d dd
Ve
N11
N
1 1 1
me 6Ve 2 21
0 0 0
N31
N 41
N12
N13
N 22
N 23
N32
N33
N 42
N 43
N14
N 24
d d d
N34
N 44
Element matrices
f sx
fe [N]T f sy dl
2 3
l
f
sz
For uniformly distributed load:
w4
4=
w1
l
v4
w3
u4
v3
1=
z=Z
i
u3
v1
fsz
u1
w2
fsy
fsx
y=Y
x=X
z=Z
3=
2=
u2
u2
j
k
031
f
sx
f sy
f sz
1
f e l2 3
2
f
sx
f sy
f sz
0
31
HEXAHEDRON ELEMENT
3D solid meshed with hexahedron elements
P’
P
P’’’
P’’
Shape functions
5
8
U Nd e
d e1 displacement components at node 1
d displacement components at node 2
e2
d e3 displacement components at node 3
d displacement components at node 4
de e4
d e5 displacement components at node 5
d e 6 displacement components at node 6
d e 7 displacement components at node 7
d displacement components at node 8
e8
u1
d ei v1
w
1
(i 1, 2,
N N1 N 2
N3
N4
N5
N6
4
7
1
0
z
fsx
2
fsy
0
y
3
x
Ni
N i 0
0
,8)
fsz
6
N7
0
Ni
0
N8
0
0
N i
(i 1,2,,8)
Shape functions
(-1, -1, 1)5
5
8(-1, 1, 1)
8
(1, -1, 1)6
fsz
6
7 (1, 1, 1)
4
7
1
z
(-1, -1, -1)1
0
2
fsx
y
3
0
4(-1, 1, -1)
fsy
(1, -1, -1)2
3(1, 1, -1)
x
8
x N i ( , , ) xi
i 1
8
y N i ( , , ) y i
1
N i (1 i )(1 i )(1 i )
8
i 1
8
z N i ( , , ) z i
i 1
(Tri-linear functions)
Strain matrix
LU LNd e Bde
B B1 B 2
B3
B4
B5
B6
B7
B8
whereby
0
0
N i x
0
N
y
0
i
0
0
N i z
B i LN i
0
N
z
N
y
i
i
N i z
0
N i x
0
N i y N i x
Note: Shape functions are expressed in natural
coordinates – chain rule of differentiation
Strain matrix
N i N i
x
N i N i
x
N i N i
x
x N i
y
x N i
y
x N i
y
y N i z
z
y N i z
z
y N i z
z
N i
N i
x
N
N
i
i
J
y
N
N
i
i
z
where
Chain rule of
differentiation
x
x
J
x
y
y
y
z
z
z
Strain matrix
Since,
8
8
8
i 1
i 1
i 1
x Ni ( ,, ) xi , y Ni ( ,, ) yi , z Ni ( ,, ) zi
N1
N
J 1
N1
N 2
N 2
N 2
xi
i 81
J xi
i 1
8
xi
i 1
8
or
N 3
N 3
N 3
N i
N i
N i
N 4
N 4
N 4
N 5
N 5
N 5
N i
i 1
8
N i
y
i
i 1
8
N i
y
i
i 1
8
yi
N 6
N 6
N 6
N 7
N 7
N 7
N i
i 1
8
N i
z
i
i 1
8
N i
z
i
i 1
8
zi
x1
N8 x2
x3
N8 x4
x5
N8 x6
x7
x8
y1
y2
y3
y4
y5
y6
y7
y8
z1
z2
z3
z4
z5
z6
z7
z8
Strain matrix
N i
N i
x
N
N
i
1
i
J
y
N i
N i
z
Used to replace derivatives
w.r.t. x, y, z with derivatives
w.r.t. , ,
0
0
N i x
0
N
y
0
i
0
0
N i z
B i LN i
0
N
z
N
y
i
i
N i z
0
N i x
N
y
N
x
0
i
i
Element matrices
k e B cBdV
T
1
1
1
1
1
1
BTcB det[J ]d dd
Ve
1 1 1
n
m
l
Gauss integration: I 1 1 1 f ( , )dd wi w j wk f ( i , j , j )
i 1 j 1 k 1
me N NdV
T
1
1
1
1 1 1
Ve
NT Ndet[J]d dd
Element matrices
For rectangular hexahedron:
det[J] abc Ve / 8
m11 m12
m 22
me
sy.
m13
m14
m15
m16
m17
m 23 m 24
m 25
m 26
m 27
m33
m34
m35
m 36
m 37
m 44
m 45
m 46
m 47
m 55
m 56
m 57
m 66
m 67
m 77
m18
m 28
m38
m 48
m58
m 68
m 78
m88
Element matrices
(Cont’d)
1
1
1
where m ij 1 1 1 abcN i N j ddd
1
abc
1
1
1 1 1
1
abc
1
1
1 1 1
Ni
0
0
0
Ni
Ni N j
0
0
0
0 N j
0 0
N i 0
0
Ni N j
0
0
Nj
0
0
0 ddd
N j
0
0 ddd
N i N j
Element matrices
(Cont’d)
or
mij
m ij 0
0
where
0
0
mij
0
mij
0
mij abc
1 1
1 1
abc
1
N i N j ddd
1
1
64 1
hab
(1 13 i j )(1 13 i j )(1 13 i j )
8
1
(1 i )(1 j )d (1 i )(1 j )d (1 i )(1 j )d
1
Element matrices
(Cont’d)
E.g. m33
abc
8
(1 1 1)(1 1 1)(1 1 1) 8
1
3
1
3
m11 m22 m33 m44 m55 m66 m77 m88
1
3
m13 m24 m16 m25 m36 m47 m57 m68 m27
1abc
216
216
8 abc
216
4 abc
216
2 abc
m38 m45 m18
216
m12 m23 m34 m56 m67 m78 m14 m58 m15 m26 m37 m48
m17 m28 m35 m46
abc
Element matrices
(Cont’d)
m ex
8 4 2 4 4 2 1 2
8
4
2
2
4
2
1
8 4 1 2 4 2
8 2 1 2 4
abc
8 4 2 4
216
8 4 2
sy.
8 4
8
(Rectangular hexahedron)
Note: For x direction only
Element matrices
f sx
f e [N]T f sy dl
3 4
l
f
sz
For
uniformly
distributed
load:
031
0
31
f sx
f
sy
f sz
f
1
sx
f e l 3 4
2
f sy
f sz
0
31
031
0
31
031
5
8
fsz
6
4
7
1
0
z
2
fsx
0
y
x
3
fsy
Using tetrahedrons to form hexahedrons
Hexahedrons can be made up of several
tetrahedrons
8
5
4
8
1
1
Hexahedron
made up of 5
tetrahedrons:
5
6
8
3
8
4
1
6
7
8
7
6
2
3
3
1
6
6
1
3
3
2
Using tetrahedrons to form hexahedrons
Element matrices can
be obtained by
assembly of
tetrahedron elements
5
8
4
1
6
7
8
5
6
1
2
8
3
4
6
4
2
2
Hexahedron
made up of 6
tetrahedrons:
6
5
8
4
1
2
3
Break into three
5
1
7
4
6
6
4
HIGHER ORDER ELEMENTS
Tetrahedron elements
10 nodes, quadratic:
4
N i (2 Li -1)Li
N 5 4 L2 L3
N 6 4 L1 L3
N 7 4 L1 L2
N8 4 L1 L4
N 9 4 L2 L4
N10 4 L3 L4
for corner nodes i 1,2,3,4
for mid-edge nodes
9
8
2
7
10
1
5
6
3
HIGHER ORDER ELEMENTS
Tetrahedron elements (Cont’d)
4
20 nodes, cubic:
N i 12 (3Li 1)(3Li 2)Li
N 5 92 (3L1 1)L1 L3
for corner nodes i 1,2,3,4
N11 92 (3L1 1)L1 L4
N 6 92 (3L3 1)L1 L3 N12 92 (3L4 1)L1 L4
N 7 92 (3L1 1)L1 L2 N13 92 (3L2 1)L2 L4
N8 92 (3L2 1)L1 L2 N14 92 (3L4 1)L2 L4
N 9 92 (3L2 1)L2 L3 N15 92 (3L3 1)L3 L4
N10 92 (3L3 1)L2 L3 N16 92 (3L4 1)L3 L4
N17 27 L2 L3 L4
N18 27 L1 L2 L3
N19 27 L1 L3 L4
N 20 27 L1 L2 L4
for center surface nodes
14
12
20
13
16
11
2
8
7
17
9
19
1
5 18 15
for edge nodes
10
5
6
3
HIGHER ORDER ELEMENTS
Brick elements
nd=(n+1)(m+1)(p+1) nodes
(n,m,p)
Lagrange type:
Ni N N N
1D
I
where
lkn ( )
1D
J
1D
K
(n,m,0)
l ( )l ( )l ( )
n
I
m
J
p
K
i(I,J,K)
(0,0,0)
( 0 )( 1 ) ( k 1 )( k 1 ) ( n )
( k 0 )( k 1 ) ( k k 1 )( k k 1 ) ( k n )
(n,0,0)
HIGHER ORDER ELEMENTS
15
(-1, -1, 1)5
Brick elements (Cont’d)
Serendipity type elements:
8(-1, 1, 1)
16
13
(1, -1, 1)6
20
17
14
3
19
7 (1, 1, 1)
18
(-1,-1,-1)1
11(-1,0,-1)
4(-1, 1, -1)
12(0-1,-1)
20 nodes, tri-quadratic:
10(0,1,-1)
(1, -1, -1)2
N j 18 (1 j )(1 j )(1 j )( j j i 2)
for corner nodes j 1,
9(1,0,-1)
,8
N j 14 (1 2 )(1 j )(1 j )
for mid-side nodes j 10,12,14,16
N j 14 (1 2 )(1 j )(1 j )
for mid-side nodes j 9,11,13,15
N j 14 (1 2 )(1 j )(1 j )
for mid-side nodes j 17,18,19, 20
3(1, 1, -1)
HIGHER ORDER ELEMENTS
Brick elements (Cont’d)
32 nodes, tri-cubic:
Nj
1
64
(1 j )(1 j )(1 j )(9 2 9 2 9 2 19)
for corner nodes j 1,
Nj
9
64
,8
(1 2 )(1 9 j )(1 j )(1 j )
for side nodes with j 13 , j 1 and j 1
Nj
9
64
(1 2 )(1 9 j )(1 j )(1 j )
for side nodes with j 13 , j 1 and j 1
Nj
9
64
(1 2 )(1 9 j )(1 j )(1 j )
for side nodes with j 13 , j 1 and j 1
ELEMENTS WITH CURVED
SURFACES
4
4
9
8
9
8
10
7
2
6
10
1
5
1
2
7
5
6
3
3
10
2
2
9
1
13
5
9
10
14 11
12
4
6
18
17
16
19
20
8
3
15
7
14
1
13
3
11
12
4
6
16
15
7
18
17
5
20
19
8
CASE STUDY
Stress and strain analysis of a quantum dot
heterostructure
E (Gpa)
GaAs
86.96
0.31
InAs
51.42
0.35
Material
GaAs cap layer
InAs wetting
layer
GaAs substrate
InAs quantum dot
CASE STUDY
CASE STUDY
30 nm
30 nm
CASE STUDY
CASE STUDY