224
6 Free Vibration of Multiple Degrees of Freedom System
forces involved in a normal mode are the inertia forces. Total deflection at any point
on the beam will be considered to be same as the deflection at that point due to
individual inertial force. If the deflections at 1, 2 and 3 are y 1 , y 2 and y 3 , then the
following equations can be written on this basis.
y 1 = − f 11 m 1 ¨
y 1 − f 12 m 2 ¨
y 2 − f 13 m 3 ¨
y 3
y 2 = − f 21 m 1 ¨
y 1 − f 22 m 2 ¨
y 2 − f 23 m 3 ¨
y 3
y 3 = − f 31 m 1 ¨
y 1 − f 32 m 2 ¨
y 2 − f 33 m 3 ¨
y 3
⎫
⎬
⎭
(a)
where f 11 , f 12 , etc., are influence or flexibility coefficients.
Equation (a) can be written in matrix form as follows:
⎧
⎨
⎩
y 1
y 2
y 3
⎫
⎬
⎭
= −
⎡
⎣
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
⎤
⎦
⎡
⎣
m 1 0 0
0 m 2 0
0 0 m 3
⎤
⎦
⎧
⎨
⎩
¨
y 1
¨
y 2
¨
y 3
⎫
⎬
⎭
(b)
or {y} = − [F] [M] { ¨
y}.
or − [D] { ¨
y} = {y}
where [D] = [F] [M].
The following solution is assumed
{y} = ae
i p t
{φ}
(c)
Substituting {y} and { ¨
y} from Eq. (c) into Eq. (b), we get
[D] {φ} =
1
p 2 {φ} = λ {φ}
(d)
where λ =
1
p 2 .
Now, the elements of [F] and [Ml matrices are to be generated. The mass matrix
for this lumped case is a diagonal matrix, which is given below:
[M] =
⎡
⎣
M
4
0 0
0
M
4
0
0 0
M
4
⎤
⎦ =
M
4
⎡
⎣
1 0 0
0 1 0
0 0 1
⎤
⎦
The elements of [F] matrix are as follows:
f 11 = f 33 =
3L
3
256E I
; f 13 = f 31 =
2.33 L
3
256 E I
; f 22 =
5.33 L
3
256 E I
;