9.2 Forced Axial Vibration of Bars
373
P r (t) =
L
0 PU r dx
L
0 U 2
r dx
(9.9)
In this case, P has been assumed to be constant and independent of time and acting
at x = L. Combining Eqs. (9.5), (9.6) and (9.9), we get
ρ AU r ( ¨
ξ r + p
2
r ξ r ) =
U r
L
0 P U r dx
L
0 U 2
r dx
(9.10)
Therefore, rth equation of Eq. (9.10) becomes
¨
ξ r + p
2
r ξ r =
1
ρ A
L
0 P U r dx
L
0 U 2
r dx
(9.11)
From Eqs. (8.62) and (8.63), we know
U r = sin
(2r − 1)π x
2L
(9.12)
and
P r =
(2r − 1)π
2L
E
ρ
(9.13)
Now
L
0
PU r dx =
L
0
P (x = L) sin
(2r − 1)
2L
π x dx = P. sin
(2r − 1) π
2
(9.14)
and
L
0
U
2
r dx =
L
0
sin
2 (2r − 1)
2L
π x dx =
L
2
(9.15)
Combining Eqs. (9.11), (9.13), (9.14) and (9.15), we get
¨
ξ r +
(2r − 1)
2
π
2
4L 2
E
ρ
ξ r =
2P
ρ AL
sin
(2r − 1)
2
π
(9.16)
We know