Civil Engineering Reference
In-Depth Information
where
f
x
1
is the force in the
x
-direction at node 1 etc. Equation (2.11) represents the rod
element stiffness relationship, which in matrix notation becomes,
[
k
m
]
{
u
} = {
f
}
(2.12)
where [
k
m
] is the “element stiffness matrix”,
{
u
}
is the element nodal “displacements
vector”, and
{
f
}
is the element nodal “forces vector”.
2.2.2 Rod mass element
Consider now the case of an unrestrained rod in free longitudinal vibration. Figure 2.1(c)
shows the equilibrium of a segment in which the body force is now given by Newton's
law as mass times acceleration. If the mass per unit volume is
ρ
, the partial differential
equation becomes,
2
2
EA
∂
u
∂x
−
ρA
∂
u
=
0
(2.13)
2
∂t
2
On discretising
u
in space by finite elements as before, the first term in (2.13) clearly
leads again to [
k
m
]. The second term takes the form
L
N
1
N
2
u
1
u
2
d
2
d
t
−
ρA
[
N
1
N
2
] d
x
(2.14)
2
0
and assuming that
ρA
is not a function of
x
,
N
1
N
1
d
x
u
1
u
2
L
d
2
d
t
N
1
N
2
−
ρA
(2.15)
N
2
N
1
N
2
N
2
2
0
Evaluation of integrals yields
21
12
d
2
d
t
u
1
u
2
ρAL
6
−
(2.16)
2
or in matrix notation
[
m
m
]
d
2
u
d
t
−
2
where [
m
m
] is the “element mass matrix”. Thus the full matrix statement of equation
(2.13) is
[
m
m
]
d
2
u
d
t
[
k
m
]
{
u
} +
= {
0
}
(2.17)
2
which is a set of ordinary differential equations.
Note that [
m
m
] formed in this manner is the “consistent” mass matrix and differs from
the “lumped” equivalent which would lead to
ρAL/
2 terms on the diagonal with zeros
off-diagonal.
