Page 65 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)

62 Chapter 1 Fundamentals oF Vibration
equivalent k of a suspension system
example 1.5
Figure 1.29 shows the suspension system of a freight truck with a parallel-spring arrangement.
Find the equivalent spring constant of the suspension if each of the three helical springs is made of
9
2
steel with a shear modulus G = 80 * 10 N>m and has five effective turns, mean coil diameter
D = 20 cm, and wire diameter d = 2 cm.
Solution: The stiffness of each helical spring is given by
9
4
4
d G 10.022 180 * 10 2
k = = = 40,000.0 N>m
3
3
8D n 810.22 152
(See inside front cover of this book for the formula.)
Since the three springs are identical and parallel, the equivalent spring constant of the suspen-
sion system is given by
k eq = 3k = 3140,000.02 = 120,000.0 N>m

FiGure 1.29 Parallel arrangement of springs in a freight truck. (Courtesy of Columbus
Castings Company.)
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torsional spring Constant of a propeller shaft
example 1.6
Determine the torsional spring constant of the steel propeller shaft shown in Fig. 1.30.

Solution: We need to consider the segments 12 and 23 of the shaft as springs in combination. From
Fig. 1.30 the torque induced at any cross section of the shaft (such as AA or BB) can be seen to be
equal to the torque applied at the propeller, T. Hence the elasticities (springs) corresponding to the