274
7 Forced Vibration Analysis of Multiple Degrees of Freedom System
Fig. 7.4 Example 7.4
Converting the damping matrix into diagonal form suggests that a reasonable
percentage of critical damping is assumed in each mode. It is not only convenient,
but also physically reasonable to assume damping in this form, than trying to evaluate
the elements of [C] matrix.
The final values of the displacements are obtained by substituting the solution of
ξ r from Eq. (7.27) into Eq. (7.21).
Example 7.4 For the two-storeyed frame with viscous damping shown in Fig. 7.4,
compute the displacement at the top storey.
The natural frequencies and mode shapes for the free undamped vibration are to
be obtained.
[M] =
1.0 0
0 1.0
and [K ] =
1500 −500
−500 500
The free vibration equation is
500
3 −1
−1 1
− p
2
1 0
0 1
φ 1
φ 2
=
0
0
The frequencies and mode shapes are found to be
p 1 = 17.11 and p 2 = 41.32
[] =
1.000 1.000
2.414 −0.414
[]
T [M][] =
1.000 1.000
2.414 −0.414
T
1 0
0 1
1.000 1.000
2.414 −0.414
=
6.827 0
0 1.171
7 Forced Vibration Analysis of Multiple Degrees of Freedom System
Fig. 7.4 Example 7.4
Converting the damping matrix into diagonal form suggests that a reasonable
percentage of critical damping is assumed in each mode. It is not only convenient,
but also physically reasonable to assume damping in this form, than trying to evaluate
the elements of [C] matrix.
The final values of the displacements are obtained by substituting the solution of
ξ r from Eq. (7.27) into Eq. (7.21).
Example 7.4 For the two-storeyed frame with viscous damping shown in Fig. 7.4,
compute the displacement at the top storey.
The natural frequencies and mode shapes for the free undamped vibration are to
be obtained.
[M] =
1.0 0
0 1.0
and [K ] =
1500 −500
−500 500
The free vibration equation is
500
3 −1
−1 1
− p
2
1 0
0 1
φ 1
φ 2
=
0
0
The frequencies and mode shapes are found to be
p 1 = 17.11 and p 2 = 41.32
[] =
1.000 1.000
2.414 −0.414
[]
T [M][] =
1.000 1.000
2.414 −0.414
T
1 0
0 1
1.000 1.000
2.414 −0.414
=
6.827 0
0 1.171