Civil Engineering Reference
In-Depth Information
and [
B
]
=
[
A
][
S
], leading to,
∂N
1
∂x
∂N
2
∂x
∂N
3
∂x
∂N
4
∂x
0
0
0
0
∂N
1
∂y
∂N
2
∂y
∂N
3
∂y
∂N
4
∂y
[
B
]
=
(2.68)
0
0
0
0
∂N
1
∂y
∂N
1
∂x
∂N
2
∂y
∂N
2
∂x
∂N
3
∂y
∂N
3
∂x
∂N
4
∂y
∂N
4
∂x
Thus, we have again for this element
[
B
]
T
[
D
][
B
] d
x
d
y
[
k
m
]
=
(2.69)
which is the form in which it will be computed in Chapter 3.
Exactly the same expression holds in the case of plane strain, but the elastic [
D
]matrix
becomes (Timoshenko and Goodier, 1982), for unit thickness,
ν
1
0
1
−
ν
E(
1
−
ν)
ν
1
0
[
D
]
=
(2.70)
1
−
ν
(
1
+
ν)(
1
−
2
ν)
1
−
2
ν
0
0
−
ν)
2
(
1
2.11 Plane element mass matrix
2
where
ρ
is the mass of the element per unit volume. For an element of unit thickness this
leads, in exactly the same way as in (2.15), to the element mass matrix which has terms
given by
2
2
and
2
When inertia is significant (2.57) are supplemented by forces
−
ρ∂
u/∂t
−
ρ∂
v/∂t
[
N
]
T
[
N
] d
x
d
y
[
m
m
]
=
ρ
(2.71)
and hence to an eigenvalue equation the same as (2.19).
Evaluation of the first term in the plane element mass matrix as illustrated in Fig-
ure 2.5(b) yields,
ρab
9
m
m
1
,
1
=
(2.72)
2.12 Axisymmetric stress and strain
Solids of revolution subjected to axisymmetric loading possess only two independent com-
ponents of displacement and can be analysed as if they were two-dimensional. For example,
