194
6 Free Vibration of Multiple Degrees of Freedom System
Hence, c 1 =
1
6.9788
m
m
.
Therefore,c 1
φ
(1)
=
⎧
⎨
⎩
0.3785
0.4652
0.4563
⎫
⎬
⎭
m
m
.
Following the above steps, we can show that
c 2
φ
(2)
=
⎧
⎨
⎩
0.1010
− 0.4169
0.4321
⎫
⎬
⎭
m
m
and
c 3
φ
(3)
=
⎧
⎨
⎩
0.6680
0.0987
− 0.3237
⎫
⎬
⎭
m
m
The modal matrix then is
[] =
m
m
⎡
⎣
0.3785 0.1010 0.6680
0.4652 − 0.4169 0.0987
0.4563 0.4321 − 0.3237
⎤
⎦
(a)
The results can now be checked by substituting [] of Eq. (a)
[]
T
[M] [] = m [I ]
Example 6.2 For two degrees of freedom system, the modal vectors are given by
[] =
1.000 1.000
1.618 − 0.618
The initial conditions at t = 0 are
(x 1 ) 0 = − (x 2 ) 0 = x 0
and
( ˙
x 1 ) 0 = ν,
( ˙
x 2 ) 0 = 0
The natural frequencies of the system are p 1 and p 2 . Determine the equations of
displacements in free vibration.
On the basis of principal coordinates chosen as per Eq. (6.29), the two uncoupled
equations in terms of the principal coordinates are as follows [Eq. (6.38)]
6 Free Vibration of Multiple Degrees of Freedom System
Hence, c 1 =
1
6.9788
m
m
.
Therefore,c 1
φ
(1)
=
⎧
⎨
⎩
0.3785
0.4652
0.4563
⎫
⎬
⎭
m
m
.
Following the above steps, we can show that
c 2
φ
(2)
=
⎧
⎨
⎩
0.1010
− 0.4169
0.4321
⎫
⎬
⎭
m
m
and
c 3
φ
(3)
=
⎧
⎨
⎩
0.6680
0.0987
− 0.3237
⎫
⎬
⎭
m
m
The modal matrix then is
[] =
m
m
⎡
⎣
0.3785 0.1010 0.6680
0.4652 − 0.4169 0.0987
0.4563 0.4321 − 0.3237
⎤
⎦
(a)
The results can now be checked by substituting [] of Eq. (a)
[]
T
[M] [] = m [I ]
Example 6.2 For two degrees of freedom system, the modal vectors are given by
[] =
1.000 1.000
1.618 − 0.618
The initial conditions at t = 0 are
(x 1 ) 0 = − (x 2 ) 0 = x 0
and
( ˙
x 1 ) 0 = ν,
( ˙
x 2 ) 0 = 0
The natural frequencies of the system are p 1 and p 2 . Determine the equations of
displacements in free vibration.
On the basis of principal coordinates chosen as per Eq. (6.29), the two uncoupled
equations in terms of the principal coordinates are as follows [Eq. (6.38)]