222
6 Free Vibration of Multiple Degrees of Freedom System
Substituting the values of the masses and those of the first mode shapes, we get
φ
(3)
1
= − 0.759φ
(3)
2
− 0.336φ
(3)
3
(g)
Similarly, writing the orthogonality relationship connecting the second mode and
the third mode and substituting the values of the masses and those due to the second
mode, we obtain
φ
(3)
1
= 0.804 φ
(3)
2
+ 1.156 φ
(3)
3
(h)
Solving Eqs. (g) and (h), we get
φ
(3)
2
= − 0.956φ
(3)
3
(i)
Similarly from Eqs. (g) and (i), we get
φ
(3)
1
= 0.389φ
(3)
3
(j)
Substituting φ
(3)
1 and φ
(3)
2 from Eqs. (i) and (j) into the last equation given by Eq.
(d), we get
m
6k
× 2 [0.389 − 0.956 + 1] φ
(3)
3
= λ 3 φ
(3)
3
or
0.867
m
6k
φ
(3)
3
= λ 3 φ
(3)
3
or
λ 3 =
1
p
2
3
=
0.864m
6k
or
p 3 = 2.635
k
m
The third mode shape is given by
{φ
(3)
} =
⎧
⎨
⎩
0.389
− 0.956
1.000
⎫
⎬
⎭