250
6 Free Vibration of Multiple Degrees of Freedom System
Equation (6.196) becomes
[ [ ]
1
B1 []
1
B2 ]
{ξ N 1 }
1
{ξ N 2 }
1
= [ ]
2
B {ξ N }
2
(6.198)
From Eq. (6.198), we get
{ξ N 1 }
1
= − ( [ ]
1
B1 )
− 1
[ ]
1
B2 {ξ N 2 }
1
+ ( []
1
B1 )
− 1
[]
2
B {ξ N }
2
(6.199)
Therefore,
{ξ } =
{ξ N }
1
{ξ N }
2
=
⎧
⎨
⎩
{ξ N 1 }
1
{ξ N 2 }
1
{ξ N }
2
⎫
⎬
⎭
= [T ] C {ξ }
(6.200)
where
{ξ } =
{ξ N 2 }
1
{ξ N }
2
(6.201)
and
[T ] C =
⎡
⎣
− ( []
1
B1 )
1
[]
1
B2 ( []
1
B1 )
−1
[]
2
B
[I ]
0
[0]
[ I ]
⎤
⎦
(6.202)
Substituting Eq. (6.200) into Eq. (6.177) and then combining with Eqs. (6.193)
and (6.194) yields
T =
1
2
{ ˙
ξ }
T
[M] R [ ˙
ξ }
(6.203)
U =
1
2
{ξ }
T
[K ] R {ξ }
(6.204)
where
[M] R = [T ]
T
C [T C ]
(6.205)
and
[K ] R = [T ]
T
C [K ] [T ] C
(6.206)