5.2 Shear Beams
129
l
0
v 0 (x) ϕ r (x) dx =
∞
n=1
A n
l
0
ϕ n (x) ϕ r (x) dx =
l
2
A r and
l
0
˙
v 0 (x) ϕ r (x) dx =
∞
n=1
ω n B n
l
0
ϕ n (x) ϕ r (x) dx =
l
2
ω r B r , p = 1, 2, . . . ,
(5.36)
by using the orthogonality condition of Eq. 5.31. The general solution of Eq. 5.34 is
q r (t) =
2
l
l
0
v 0 (x) ϕ r (x) dx
cos(ω r t) +
2
ω r l
l
0
˙
v 0 (x) ϕ r (x) dx
sin(ω r t)
+
t
0
h r (t − u) f r (u) du,
(5.37)
so that
v(x, t) =
∞
n=1
sin
(2 n − 1) π
2 l
x
2
l
l
0
v 0 (x) ϕ n (x) dx
cos(ω n t)
+
2
ω n l
l
0
˙
v 0 (x) ϕ n (x) dx
sin(ω n t) +
t
0
h n (t − u) f n (u) du
.
(5.38)
5.2.4 Free Vibration
The free vibration solution can be obtained from the previous subsection by setting
f (x, t) = 0. For completeness, we outline the steps for finding the free vibration
solution directly. The results are for the shear beam of Example 4.60.
Previous arguments on Eqs. 5.33 to 5.34 with f (x, t) = 0 give
¨
q n (t) + ω
2
r q n (t) = 0, n = 1, 2, . . . ,
(5.39)
which is satisfied by
q n (t) = A n cos(ω n t) + B n sin(ω n t), n = 1, 2, . . . ,
(5.40)
where the constants {A n } and {B n } result from the initial conditions v(x, 0) = v 0 (x)
and ˙
v(x, 0) = ˙
v 0 (x) by using the representation of v(x, t) given by Eq. 5.32 and the
orthogonality of the modal shapes. The free vibration solution is