9.5 Forced Transverse Vibration of Uniform Damped Beam
383
L
0
m P(x, t) Y k (x) dx =
∞
j=1
P j (t)
L
0
Y j (x) Y k (x) dx
(9.58)
While writing Eq. (9.58), orthogonality properties of modes have been used on
the right hand side of the equation. The integral on the right-hand side of the equation
yields
L
0
m P(x, t) Y k (x) dx = P j (t)
(9.59)
Applying the same procedure for response
y(x, t) =
∞
j=1
y j (t) Y j (x)
(9.60)
and then carrying out multiplication and integration as above
L
0
m y(x, t) Y j (x) dx = y j (t)
(9.61)
Substituting Eq. (9.57) and (9.60) into Eq. (9.56) yields
∞
j=1
E I y j (t)
d
4 Y j
dx 4 + c
d y j
dt
Y j + m
d
2 y j
dt 2 Y j
=
∞
j=1
P j (t) Y j (x)
(9.62)
Combining Eq. (8.82) and (9.62) yields
∞
j=1
p
2 my j + c ˙
y j + m ¨
y j − P j (t)
Y j (x) = 0
(9.63)
Since Y j (x) cannot be zero, the terms within the parenthesis must vanish,
¨
y j +
c
m
˙
y j + p
2
j y j =
1
m
P j (t)
(9.64)
Equation (9.64) can be written as
¨
y j + 2 ζ j p j ˙
y j + p
2
j y j =
1
m
P j (t)
(9.65)
Solution of the convolution integral of Eq. (9.65) [using Eq. (3.68)] is