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
N1
0
0
N2
0
0
N2
0
0
N3
0
0
N3
0
0
N4
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
d1234
d1 234
d1 234
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
b3
c2
c3
b4
c4
d1  1 
d 2   x 
  l = 4,1
d 3  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  , di   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
m e   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
N32
N 23
N33
N 42
N 43
0 

0 
N i N j 
N14 
N 24 
dV

N 34

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
2
0
1
0
0
2
0
0
1
0
0
2
1
0
0
1
0
0
2
0
1
0
0
1
0
0
2
0
0
1
0
0
1
0
0
2
1
0
0
1
0
0
1
0
0
2
0
1
0
0
1
0
0
1
0
0
2
0
0 
1

0
0

1
0

0
1

0

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   ( z P  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(x3x2)+x2, yP(y3y2)+y2,zP(z3z2)+z2]
B [xB(xPx1)+x1, yB(yPy1)y1, zB(zPy1)z1]
=constant
y
x
z=Z
=1
=0
=1
2= j
= constant
O [x=(1)(x4xB)xB, y=(1)(y4yB)yB, z=(1)(z4zB)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
m e   N NdV  
T
1 1 1

0 0 0
NT Ndet[J]d dd
Ve
 N11
N
1 1 1
m e  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
031 


f
 sx 
 f sy 
 
 f sz 
1
f e  l2  3 

2
f
 sx 
 f sy 
 
 f sz 
 0 
 31 
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   e 4 
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 B 4
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


 x i
 i 81
J   x i
 i 1
8
 x i
 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

N5

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 dd
Ve
1 1 1
n
m
l
Gauss integration: I  1 1 1 f ( , )dd   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 dd
Element matrices
For rectangular hexahedron:
det[J]  abc  Ve / 8
m11 m12

m 22



me  




sy.


m13 m14
m 23 m 24
m15
m 25
m16
m 26
m17
m 27
m33
m35
m 45
m36
m 46
m37
m 47
m55
m56
m57
m 66
m 67
m 77
m34
m 44
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 ddd
1
 abc 
1
1

1 1 1
1
 abc 
1
1

1 1 1
Ni
0

 0
0
Ni
0
Ni N j

 0
 0

0 N j

0   0
N i   0
0
Ni N j
0
0
Nj
0
0 

0  ddd
N j 
0 

0  ddd
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 ddd
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
1abc
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:
031 
0 
 31 
 f sx 
 f 
 sy 
 f sz 
 f 
1
 sx 
f e  l 3 4  
2
 f sy 
 f sz 
0 
 31 
031 




0
 31 
031 
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
for corner nodes i  1,2,3,4
N 5  92 (3L1  1)L1 L3 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
N8  92 (3L2  1)L1 L2
N 9  92 (3L2  1)L2 L3
N10  92 (3L3  1)L2 L3
N17  27 L2 L3 L4
N18  27 L1 L2 L3
N19  27 L1 L3 L4
N 20  27 L1 L2 L4
N13  92 (3L2  1)L2 L4
N14  92 (3L4  1)L2 L4
N15  92 (3L3  1)L3 L4
N16  92 (3L4  1)L3 L4
14
12
20
13
16
11

2
8

7
17

9
19

1
5 18 15
 for edge nodes
10
5


6
3





 for center surface nodes


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