3.11 Response of Structures Due to Transient Vibration
93
A = 0 and B = −
F 0
pkt 1
Substituting the values of A and B in Eq. (3.81) yields
x t =
F 0
kpt 1
[ pt − sin pt]
(3.82)
At t = t 1 , Eq. (3.82) gives
x t 1 =
F 0
kpt 1
[ pt 1 − sin pt 1 ]
(3.83)
Differentiation of Eq. (3.82) with respect to t and substitution of t = t I yields
˙
x t 1 =
F 0
kpt 1
[ p − p cos pt 1 ]
(3.84)
(ii) For t > t 1, F (t) = F 0
Equation of motion is
m ¨
x + kx = F 0
(3.85)
Solution of Eq. (3.85) is
x =
F 0
k
+ A cos pt
+ B sin pt
(3.86)
t
is the parameter measured from the instant t = t 1 . At t
= 0, x = x t 1 and ˙
x =
˙
x t 1 .
Substituting initial conditions and determining the values of constants A and B of
Eq. (3.86) yields
x =
F 0
k
−
F 0
kpt 1
( sin pt 1 − pt 1 ) cos pt
+
F 0
kpt 1
[1 − cos pt 1 ] sin pt
(3.87)
or
x =
F 0
k
1 −
2
pt 1
sin
pt 1
2
cos p
t
+
t 1
2
The plot of response against time is shown in Fig. 3.22.