8.6 Free Flexural Vibration of the Simply Supported Beam
325
At
At x = L , Y = 0 and
d
2 Y
dx 2 = 0
(8.86b)
Applying the end conditions given by Eq. (8.86a), we get
C 2 + C 4 = 0
−λ
2 C 2 + λ
2 C 4 = 0
(8.87)
Solving Eq. (8.87), we get
C 2 = C 4 = 0
(8.88)
Substituting the end conditions given by Eq. (8.86b), we get
C 1 sin λL + C 3 sinh λL = 0
λ
2
(−C 1 sin λL + C 3 sinh λL) = 0
(8.89)
which gives
and
C 3 sinh λL = 0
C 1 sin λL = 0
(8.90)
As λ is not zero, sinh λL = 0 for all values of λ. Therefore, from Eq. (8.90)
C 3 = 0
(8.91)
If we assume C 1 = 0 in Eq. (8.89), then there cannot be any vibration of the
system. Therefore,
sin λL = 0
(8.92)
which leads to
λL = r π, r = 1, 2, 3, . . .
(8.93)
or
λ =
r π
L
(8.94)