344
8 Free Vibration Analysis of Continuous Systems
λ
2 ∂
2 y s
∂ x 2 =
∂
2 y s
∂ t 2
(8.156)
where
λ
2
=
μ G
ρ
(8.157)
Equation (8.156) is of the same form as Eq. (8.43), derived for free longitudinal
vibration of bars. The general solution of Eq. (8.156) is
y s (x, t) = (C 1 sin αx + C 2 cos αx)(C 3 sin pt + C 4 cos pt)
(8.158)
where C 1 and C 2 are constants, which depend on the end conditions of the beam. C 3
and C 4 are also constants, but they depend on the initial conditions.
α
2
=
p
2
λ 2
(8.159)
For a cantilever uniform shear beam, the end conditions are
and
at x = 0, y s = 0
at x = l, ∂y s /∂ x = 0
(8.160)
Substituting Y s from Eq. (8.158) into the conditions given by Eq. (8.160), we get
C 2 = 0
(8.161)
and the frequency equation becomes
cos αL = 0
(8.162)
Solution of Eq. (8.162) gives
α =
(2n − 1)
2L
π, n = 1, 2, 3, . . .
(8.163)
Substituting the value of α from Eq. (8.159) into Eq. (8.163) gives
p =
(2n − 1)
2L
π
μG
ρ
, n = 1, 2, 3, . . .
(8.164)