11.5 The Flexibility Matrix Method for the Analysis of Coupled …
437
Using the relations given in Eqs. (11.29), (11.34) can be written as
−Qy 0 − Rθ 0 − U α 0 = ω
2
[C 1 y 1 + · · · + C n y n + D 1 α 1 + · · · + D n α n ] (11.37)
where
R =
n
i=1
m i s i x i and U =
n
i=1
I i
Next, multiplying first of Eq. (11.26) by m 1 x 1 , the second by m 2 x 2 and so on and
similarly, multiplying first of Eq. (11.27) by m 1 s 1 x 1 , the second by m 2 s 2 x 2 and so on,
the steps similar to those are followed for obtaining Eqs. (11.37), and the following
equation results
−Py 0 − V θ 0 − Rα 0 = ω
2
[E 1 y 1 + · · · + E n y n + G 1 α 1 + · · · + G n α n ] (11.38)
where
E n =
n
i=1
(m n m i x i a in + m i m n s i s n x n b in )
(11.39)
G n =
n
i=1
(m i m n x i s n a in + m i s i x i I n b in )
(11.40)
V =
n
i=1
m i x
2
i
(11.41)
Solving Eqs. (11.33), (11.37) and (11.38), the values of y 0 , θ 0 and α 0 can be
obtained and they are expressed in terms of y 1 , y 2 , . . . , y n , α 1 , α 2 , . . . α n . Substitution
of these values of y 0 , θ 0 and α 0 into Eqs. (11.26) and (11.27) results in an equation
of the following form:
y 1 = ω
2
[k 11 y 1 + · · · + k 1n y n + h 11 α 1 + · · · + h 1n α n ]
y 2 = ω
2
[k 21 y 1 + · · · + k 2n y n + h 21 α 1 + · · · + h 2n α n ]
. . .
y n = ω
2
[k n1 y 1 + · · · + k nn y n + h n1 α 1 + · · · + h nn α n ]
α 1 = ω
2
[ p 11 y 1 + · · · + p 1n y n + q 11 α 1 + · · · + q 1n α n ]
α 2 = ω
2
[ p 21 y 1 + · · · + p 2n y n + q 21 α 1 + · · · + q 2n α n ]
. . .
α n = ω
2
[ p n1 y 1 + · · · + p nn y n + q n1 α 1 + · · · + q nn α n ]
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(11.42)