220
6 Free Vibration of Multiple Degrees of Freedom System
and the first mode shape is given by
{φ
(1)
} =
⎧
⎨
⎩
1.00
0.759
0.336
⎫
⎬
⎭
Next, the second natural frequency and the second mode shape are to be calculated. Considering the first and the second modes, the following is the orthogonality
relationship.
m 1 φ
(1)
1 φ
(2)
1
+ m 2 φ
(1)
2 φ
(2)
2
+ m 3 φ
(1)
3 φ
(2)
3
= 0
( a )
Substituting the values of φ
(1)
1 , φ
(1)
2 and φ
(1)
3 of the first mode in Eq. (a) and the
values of the masses, we get
φ
(2)
1
= − 0.759φ
(2)
2
− 0.336φ
(2)
3
(b)
Writing Eq. (6.13) in expanded form and omitting superscript 2 for convenience,
we get
8φ 1 + 5φ 2 + 2φ 3 = λφ 1 ×
6k
m
5φ 1 + 5φ 2 + 2φ 3 = λφ 2 ×
6k
m
2φ 1 + 2φ 2 + 2φ 3 = λφ 3 ×
6k
m
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(c)
Substituting the values of φ 1 from Eq. (b) into Eq. (c) and arranging different
terms
1.072 φ 2 + 0.688φ 3 = − λ (0.759φ 2 + 0.336φ 3 ) ×
6k
m
1.2 φ 2 + 0.32 φ 3 = λφ 2 ×
6k
m
0.48φ 2 + 1.33φ 3 = λφ 3 ×
6k
m
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(d)
Considering the last two equations given by Eq. (d) and writing them in matrix
form, we get
m
6k
1.2 0.32
0.48 1.33
φ 2
φ 3
= λ
φ 2
φ 3