Page 69 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)

66 Chapter 1 Fundamentals oF Vibration
A vertical displacement x of point B will cause the spring k 2 (boom) to deform by an amount
x 2 = x cos 45 and the spring k 1 (cable) to deform by an amount x 1 = x cos 190 - u2. The length
of the cable FB, l 1 , is given by Fig. 1.32(b):
2
2
2
l 1 = 3 + 10 - 21321102 cos 135 = 151.426, l 1 = 12.3055 m
The angle u satisfies the relation
2
2
2
l 1 + 3 - 21l 1 2132 cos u = 10 , cos u = 0.8184, u = 35.0736
The total potential energy (U) stored in the springs k 1 and k 2 can be expressed, using Eq. (1.2) as
1 1
2
U = k 1 3x cos 190 - u24 + k 2 3x cos 45 4 2 (E.1)
2 2
where
-6 9
A 1 E 1 1100 * 10 21207 * 10 2
6
k 1 = = = 1.6822 * 10 N>m
l 1 12.3055
and
-6 9
A 2 E 2 12500 * 10 21207 * 10 2
7
k 2 = = = 5.1750 * 10 N>m
l 2 10
Since the equivalent spring in the vertical direction undergoes a deformation x, the potential energy
of the equivalent spring 1U eq 2 is given by
1
U eq = k eq x 2 (E.2)
2
By setting U = U eq , we obtain the equivalent spring constant of the system as
6
2
2
2
2
k eq = k 1 sin u + k 2 cos 45 = k 1 sin 35.0736 + k 2 cos 45 = 26.4304 * 10 N>m
■
equivalent k of a rigid bar Connected by springs
example 1.9
A hinged rigid bar of length l is connected by two springs of stiffnesses k 1 and k 2 and is subjected to
a force F as shown in Fig. 1.33(a). Assuming that the angular displacement of the bar 1u2 is small,
find the equivalent spring constant of the system that relates the applied force F to the resulting dis-
placement x.

Solution: For a small angular displacement of the rigid bar 1u2, the points of attachment of springs
k 1 and k 2 (A and B) and the point of application (C) of the force F undergo the linear or horizontal
displacements l 1 sin u, l 2 sin u, and l sin u, respectively. Since u is small, the horizontal displace-
ments of points A, B, and C can be approximated as x 1 = l 1 u, x 2 = l 2 u, and x = lu, respectively.
The reactions of the springs, k 1 x 1 and k 2 x 2 , will be as indicated in Fig. 1.33(b). The equivalent spring
constant of the system 1k eq 2 referred to the point of application of the force F can be determined by
considering the moment equilibrium of the forces about the hinge point O:
k 1 x 1 1l 1 2 + k 2 x 2 1l 2 2 = F1l2
or
x 1 l 1 x 2 l 2
F = k 1 ¢ ≤ + k 2 ¢ ≤ (E.1)
l l