Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
Chapter 6. Solid Elements for 3-D Problems
I. 3-D Elasticity Theory
Stress State:
y
F
x
z
y , v
σ y
τ yx
τ yz
τ xy
τ zy
σ x
τ zx
τ
σ
xz
x, u
z
z, w
© 1999 Yijun Liu, University of Cincinnati 138
Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
σ
x
σ
y
ó ={σ } σ
z
=
,
or
[σ ij]
)
1
(
τ xy
τ
yz
τ
zx
Strains:
ε x
ε
y
å = {
ε }
ε z
=
,
or
[ε ij]
(2)
γ xy
γ yz
γ
zx
Stress-strain relation:
1 − v
v
v
0
0
0
σ
ε
x
x
v
1 − v
v
0
0
0
σ
ε
y
v
v
1 − v
0
0
0
y
σ
1 2 v
ε
z
E
−
z
=
0
0
0
0
0
τ
1
(
v 1
)(
2 v
2
)
γ
xy
+
−
1 − 2 v
xy
τ
0
0
0
0
0
γ
yz
2
yz
τ
1 2 v
γ
zx
−
0
0
0
0
0
zx
2
o r
ó = Eå
( )
3
© 1999 Yijun Liu, University of Cincinnati 139
Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
Displacement:
u( x, y, z) u
1
u =
v( x, y, z)
= u
(4)
2
(
w x, y, z)
u
3
Strain-Displacement Relation:
u
∂
v
∂
w
∂
ε =
,
ε =
, ε =
,
x
x
y
∂
y
z
∂
z
∂
v
∂
u
∂
w
∂
v
∂
u
∂
w
∂
γ =
+
, γ
=
+
, γ
=
+
( )
5
xy
x
∂
y
yz
∂
y
∂
z
xz
∂
z
∂
x
∂
or
1
u
∂
u
∂
j
i
ε =
+
,
i j
=
ij
( ,
,
1 ,
2 )
3
2
x
∂
x
∂
j
i
simply,
or
1
ε = u + u
ij
( i, j j, i ) (
notation)
tensor
2
© 1999 Yijun Liu, University of Cincinnati 140
Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
Equilibrium Equations:
∂σ
∂τ
τ
xy
x
∂
+
+
xz
+ fx = 0 ,
∂ x
∂ y
∂ z
∂τ
σ
τ
yx
∂ y
∂
+
+
yz
+ f y = 0 ,
(6)
∂ x
∂ y
∂ z
∂τ
∂τ
σ
zy
zx
∂
+
+
z
+ fz = 0 ,
∂ x
∂ y
∂ z
or
σ
f
ij j +
i = 0
,
Boundary Conditions (BC’s):
u = u , on Γ ( specified nt
displaceme
)
i
i
u
t = t , on Γ ( specified traction ) (7)
i
i
σ
(
traction t =σ n ) i
ij
j
p
n
Γσ
Γ( = Γ u + Γ )
σ
Γ u
Stress Analysis:
Solving equations in (6) under the BC’s in (7).
© 1999 Yijun Liu, University of Cincinnati 141
Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
II. Finite Element Formulation
Displacement Field:
N
u =
N u
∑ i i
i =1
N
=
v
N v
∑
(8)
i
i
i =1
N
w =
N w
∑ i i
i =1
Nodal values
In matrix form:
u 1
v
1
u
N
0
0
N
0
0
L
w
1
2
1
v
= 0 N 0 0 N 0
L
u
1
2
2
( )
9
w
0 0 N 0 0
N
L
v
N
(3× )
1
1
2
(3 3
×
)
2
w
2
M
(3 N× )1
or
u = N d
Using relations (5) and (8), we can derive the strain vector ε =B d
(6× 1) (6× 3N)× (3N× 1)
© 1999 Yijun Liu, University of Cincinnati 142
Lecture Notes: Introduction to Finite Element Method Chapter 6. Solid Elements
Stiffness Matrix:
k =
T
B E B dv
10
(
)
∫ v
(3×N) (3N× 6)× (6× 6)× (6× 3N) Numerical quadratures are often needed to evaluate the above integration.
Rigid-body motions for 3-D bodies (6 components): 3 translations, 3 rotations.
These rigid-body motions (singularity of the system of equations) must be removed from the FEA model to ensure the quality of the analysis.
© 1999 Yijun Liu, University of Cincinnati 143