Page 73 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)
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70     Chapter 1   Fundamentals oF Vibration

                                                    x 1    x 2    x 3                 x eq    x 1




                                        Pivot point  m 1  m 2     m 3        Pivot point  m eq
                                   O                                    O
                                                     A      B      C                      A     B       C
                                             l 1                                  l 1
                                                l 2
                                                    l 3
                                                    (a)                                  (b)

                                   FiGure 1.36  Translational masses connected by a rigid bar.

                                   the various floor levels represent the mass elements, and the elasticities of the vertical
                                   members denote the spring elements.


                1.8.1              In many practical applications, several masses appear in combination. For a simple analy-
                Combination        sis, we can replace these masses by a single equivalent mass, as indicated below [1.27].
                of masses
                                   Case 1: Translational Masses Connected by a Rigid Bar.  Let the masses be attached
                                   to a rigid bar that is pivoted at one end, as shown in Fig. 1.36(a). The equivalent mass can
                                   be assumed to be located at any point along the bar. To be specific, we assume the location
                                                                                                         #
                                                                                               #
                                   of the equivalent mass to be that of mass m . The velocities of masses m 1x 2 and m 1x 2
                                                                                             2
                                                                      1
                                                                                                          3
                                                                                                       3
                                                                                                2
                                                                               #
                                   can be expressed in terms of the velocity of mass m 1x 2, by assuming small angular dis-
                                                                             1
                                                                               1
                                   placements for the bar, as
                                                              #    l #   #   l #
                                                                   2
                                                                              3
                                                              x =  l 1  x , x =  l 1  x 1             (1.18)
                                                                         3
                                                                     1
                                                               2
                                   and
                                                                    #    #
                                                                    x eq  = x 1                       (1.19)
                                   By equating the kinetic energy of the three-mass system to that of the equivalent mass
                                   system, we obtain
                                                       1   #    1   #   1   #    1   #
                                                            2
                                                                    2
                                                                             2
                                                         m  x +   m  x +   m  x =   m  x 2            (1.20)
                                                       2  1 1   2  2 2  2  3 3   2  eq eq
                                   This equation gives, in view of Eqs. (1.18) and (1.19):
                                                                      l  2      l  2
                                                                                3
                                                                      2
                                                         m   = m + ¢ ≤ m + ¢ ≤ m                      (1.21)
                                                           eq
                                                                1
                                                                      l 1  2    l 1  3
                                   It can be seen that the equivalent mass of a system composed of several masses (each mov-
                                   ing at a different velocity) can be thought of as the imaginary mass which, while moving
                                   with a specified velocity v, will have the same kinetic energy as that of the system.
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