24
2 Single Degree of Freedom (SDOF) Systems
˙
x p (t) = −ζ ω e
−ζ ω t
A cos(ω d t) + B sin(ω d t)
+ ω d e
−ζ ω t
− A sin(ω d t) + B cos(ω d t)
+
d
dt
t
0
h(t − u) f (u) du
and (see Appendix B)
d
dt
t
0
h(t − u) f (u) du =
t
0
∂h(t − u)
∂t
f (u) du + h(0) f (t)
so that ˙
x(0) = −A + ω d B = ˙
x 0 = 0.
Example 2.8 Suppose that the oscillator in the previous example is subjected to a
constant force during the time interval [0, τ ], i.e., f (u) = q 1(u ≤ τ ), u ≥ 0,
where the indictor function 1(A) = 1 if A is true and zero otherwise. The particular
solution by the Duhamel integral is
x p (t) =
t
0
h(t − u) q 1(u ≤ τ ) du = q
min{t,τ }
0
h(t − u) du,
so that it is available analytically by using the integral I (c, b; u) above.
Example 2.9 Suppose the oscillator of the previous example is used to model a
building which is subjected to a seismic ground acceleration a(t). The equation of
motion given by Eq. 2.7 takes the form m
¨
x(t) + a(t)
= −k x(t) − c ˙
x(t) since
the acceleration of the mass has two components, one caused by the motion of the
support and one caused by the system deformation. With the notations of Eq. 2.8 we
have
¨
x(t) + 2 ζ ω ˙
x(t) + ω
2 x(t) = −a(t),
(2.33)
so that the solution of Eq. 2.31 holds provided that −a(t) is substituted for f (t)/m.
We have
x(t) = e
−ζ ω t
A cos(ω d t) + B sin(ω d t)
−
t
0
h a (t − u) a(u) du,
(2.34)
where
h a (t − u) =
1
ω d
e
−ζ ω (t−u) sin
ω d (t − u)
(2.35)
denotes the unit impulse response function for the input ground acceleration. For
zero initial conditions, A = 0 since x(0) = A and B = 0 since