Civil Engineering Reference
In-Depth Information
Therefore, the free-vibration response of the 2 degree-of-freedom system is
given by
A
(1
2
sin(
A
(2
2
sin(
=
1
t
+
1
)
+
2
t
+
2
)
U
2
(
t
)
(10.45)
2
A
(1
2
5
A
(2
2
U
3
(
t
)
=
sin(
1
t
+
1
)
−
0
.
sin(
2
+
2
)
and we note the four unknown constants in the solution; specifically, these are the
amplitudes
A
(1
2
,
A
(2
2
and the phase angles
1
and
2
. Evaluation of the constants
is illustrated in a subsequent example.
Depending on the reader's mathematical background, the analysis of the
2 degree-of-freedom vibration problem may be recognized as an
eigenvalue
problem
[1]. The computed natural circular frequencies are the eigenvalues of
the problem and the amplitude ratios represent the eigenvectors of the problem.
Equation 10.45 represents the response of the system in terms of the natural
modes of vibration. Such a solution is often referred to as being obtained by
modal superposition
or simply
modal analysis
. To represent the complete solu-
tion for the system, we use the matrix notation
U
2
(
t
)
U
3
(
t
)
A
(1
2
2
A
(1
2
sin(
sin(
A
(2
2
=
1
t
+
1
)
+
2
t
+
2
)
(10.46)
5
A
(2
2
−
0
.
which shows that the modes interact to produce the overall motion of the system.
EXAMPLE 10.2
Given the system of Figure 10.4 with
k
=
40
lb/in. and
mg
=
W
=
20
lb, determine
(a) The natural frequencies of the system.
(b) The free response, if the initial conditions are
U
2
(
t
=
0)
=
U
3
(
t
U
2
(
t
=
0)
=
1in
.
U
3
(
t
=
0)
=
0
.
5in
.
=
0)
=
0
These initial conditions are specified in reference to the equilibrium position of the
system, so the computed displacement functions do not include the effect of gravity.
■
Solution
Per Equation 10.41, the natural circular frequencies are
40
20
/
g
=
k
m
=
40(386
.
4)
1
=
=
27
.
8 rad /sec
20
6(40)
20
/
g
=
6
k
m
=
6(40)(386
.
4)
2
=
=
68
.
1 rad /sec
20
The free-vibration response is given by Equation 10.35 as
U
2
(
t
)
=
A
(1
2
sin(27
.
8
t
+
1
)
+
A
(2
2
sin(68
.
1
t
+
2
)
U
3
(
t
)
=
2
A
(1
2
sin(27
.
8
t
+
1
)
−
0
.
5
A
(2
2
sin(68
.
1
t
+
2
)
