242
6 Free Vibration of Multiple Degrees of Freedom System
Fig. 6.22 Example 6.14
Now, if the displacements corresponding to internal degrees of freedom are to be
retracked, then
{x} i = [K ]
−1
ii {F} i − [K ]
− 1
ii [K ] ir {x} i = [K ]
− 1
ii {F} i + [W ] {x} r (6.153)
Similarly, the velocities and accelerations if needed to be retracked is given by
{ ˙
x} i = − [K ]
− 1
ii [K ] ir { ˙
x} r = [W ] { ˙
x} r
(6.154)
and
{ ¨
x} i = − [K ]
−1
ii [K ] ir { ¨
x} r = [W ] { ¨
x} r
(6.155)
If there is no damping and no load, then Eq. (6.150) reduces to
[M] red { ¨
x} r + [K ] red {x} r = {0}
(6.156)
Equation (6.156) indicates an eigenvalue problem.
Example 6.14 Use the reduction method to eliminate x 2 from the equations of motion
of the system shown in Fig. 6.22
The equation of motion for free vibration is given by
⎡
⎣
m 0 0
0 m 0
0 0 m
⎤
⎦
⎧
⎨
⎩
¨
x 1
¨
x 2
¨
x 3
⎫
⎬
⎭
+
⎡
⎣
2k − k 0
− k 2k − k
0 − k 2k
⎤
⎦
⎧
⎨
⎩
x 1
x 2
x 3
⎫
⎬
⎭
=
⎧
⎨
⎩
0
0
0
⎫
⎬
⎭
From the second equation of motion, we get after neglecting the inertia term
− kx 1 + 2kx 2 − kx 3 = 0
which yields
x 2 =
1
2k
(kx 1 + kx 3 )
Therefore,