484
12 Finite Element Method in Vibration Analysis
Therefore
{F
e
R } = − {F
e + 1
L
} = − e
μ
{F
e
L }
(12.185)
The relationships of Eqs. (12.182) and (12.185) give rise to the following
transformations.
{δ
e
} =
⎧
⎨
⎩
{δ
e
L }
{δ
e
I }
{δ
e
R }
⎫
⎬
⎭
=
⎡
⎣
[I ] [0]
[0] [I ]
e
μ
[I ] [0]
⎤
⎦
{δ
e
L }
{δ
e
I }
= [W ] {δ
e }
(12.186)
and
{F
e
} =
⎧
⎨
⎩
{F
e
L }
{F
e
I }
{F
e
R }
⎫
⎬
⎭
=
⎡
⎣
[I ] [0]
[0] [I ]
e
μ
[I ] [0]
⎤
⎦
{F
e
L }
{F
e
I }
(12.187)
when the transformation Eqs. (12.186) and (12.187) are applied to Eq. (12.182), the
unknown boundary forces can be eliminated by premultiplying by the matrix [W ]
H .
H is used to denote taking the complex conjugate of the matrix and then transposing
it. Therefore,
[W ]
H
=
[I ] [0] e
μ
[I ]
0 [I ] [0]
(12.188)
Taking {F
e
I } = {0} results in the equation
[ [K (μ)] − ω
2
[M (μ) ] ]
{δ
e
L }
{δ
e
I }
= {0}
(12.189)
where
[K (μ)] = [W ]
H
[K
e
] [W ]
(12.190)
and
[M (μ)] = [W ]
H
[M
e
] [W ]
(12.191)
Both [K (μ)] and [M (μ)] are Hermitian matrices, since
( [W ]
H
[K
e
] [W ] )
H
= [W ]
H
[K
e
] [W ]
(12.192)
and