10.7 Dynamic Stiffness Matrix of a Bar Subjected to Axial …
415
Fig. 10.12 A beam segment
Proceeding as per Article 8.4, the equation of motion is given as
∂
2
∂ x 2
E I
∂
2 y
∂ x 2
+ N
∂
2 y
∂ x 2 + m
∂
2 y
∂t 2 = p
(10.64)
For free vibration of a uniform beam, Eq. (10.64) reduces to
E I
∂
4 y
∂ x 4 + N
∂
2 y
∂ x 2 + m
∂
2 y
∂t 2 = 0
(10.65)
Assuming a solution of the following form
y(x, t) = Y (x)q(t)
(10.66)
and substituting y from Eq. (10.66) into Eq. (10.65), we get
E I Y
iv
(x)
Y (x)
+
N Y
(x)
Y (x)
= −m
¨
q(t)
q(t)
= C
(10.67)
From Eq. (10.67), we obtain the following two equations
¨
q(t) + p
2 q(t) = 0
(10.68)
and
E I Y
iv
+ N Y
− mp
2 Y = 0
(10.69)
Solution of Eq. (10.68) is
q(t) = A cos pt + B sin pt
(10.70)
Equation (10.69) is written as