Civil Engineering Reference
In-Depth Information
EXAMPLE 10.8
Again consider the 3 degrees-of-freedom system of Example 10.3 and determine the
steady state response when a downward force
F
=
F
0
sin
f
t
is applied to mass 2.
■
Solution
For the given conditions, the applied nodal force vector is
0
F
0
sin
f
t
0
{
F
(
t
)
} =
and the generalized forces are
=
0
.
2956
0
.
4209
0
.
6064
0
F
0
sin
f
t
0
0
.
4209
0
.
5618
−
0
.
7124
1
√
m
F
0
sin
f
t
√
m
[
A
]
T
{
F
}=
0
.
6575
0
.
5618
−
0
.
3550
0
.
6930
−
0
.
7124
0
.
0782
The equations of motion for the generalized coordinates are then
0
.
4209
F
0
sin
f
t
√
m
2
1
p
1
=
p
1
+
0
.
5618
F
0
sin
f
t
√
m
2
2
p
2
=
p
2
+
−
0
.
7124
F
0
sin
f
t
√
m
2
3
p
3
=
p
3
+
for which the solutions are
0
.
4209
F
0
sin
f
t
f
√
m
p
1
(
t
)
=
2
1
2
−
0
.
5618
F
0
sin
f
t
f
√
m
p
2
(
t
)
=
2
2
2
−
−
0
.
7124
F
0
sin
f
t
p
3
(
t
)
=
f
√
m
2
3
2
−
The actual displacements,
x
(
t
)
=
q
(
t
)
in this case, are obtained by application of Equa-
tion 10.112:
0
.
4209
2
f
0
.
5618
2
1
−
0
.
2956
0
.
6575
0
.
6930
1
√
m
F
0
sin
f
t
√
m
{
x
}=
[
A
]
{
p
}=
0
.
4209
0
.
5618
−
0
.
7124
2
f
−
0
.
7124
2
2
−
0
.
6064
−
0
.
3550
0
.
0782
2
3
−
2
f
