Page 63 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)
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60     Chapter 1   Fundamentals oF Vibration
                1.7.4              In many practical applications, several linear springs are used in combination.  These
                Combination        springs can be combined into a single equivalent spring as indicated below.
                of springs
                                   Case 1: Springs in Parallel.  To derive an expression for the equivalent spring constant
                                   of springs connected in parallel, consider the two springs shown in Fig. 1.27(a). When a
                                   load W is applied, the system undergoes a static deflection d  as shown in Fig. 1.27(b).
                                                                                     st
                                   Then the free-body diagram, shown in Fig. 1.27(c), gives the equilibrium equation
                                                                W = k d + k d                          (1.8)
                                                                            2 st
                                                                     1 st
                                   If k  denotes the equivalent spring constant of the combination of the two springs, then for
                                      eq
                                   the same static deflection d , we have
                                                         st
                                                                   W = k d                             (1.9)
                                                                        eq st
                                   Equations (1.8) and (1.9) give
                                                                 k eq  = k + k 2                      (1.10)
                                                                        1
                                   In general, if we have n springs with spring constants k , k , c, k  in parallel, then the
                                                                                1
                                                                                          n
                                                                                   2
                                   equivalent spring constant k  can be obtained:
                                                          eq
                                                            k eq  = k + k + g + k n                   (1.11)
                                                                       2
                                                                   1
                                   Case 2: Springs in Series.  Next we derive an expression for the equivalent spring con-
                                   stant of springs connected in series by considering the two springs shown in Fig. 1.28(a).
                                   Under the action of a load W, springs 1 and 2 undergo elongations d  and d , respectively,
                                                                                          1
                                                                                               2
                                   as shown in Fig. 1.28(b). The total elongation (or static deflection) of the system, d , is
                                                                                                        st
                                   given by
                                                                 d = d + d  2                         (1.12)
                                                                  st
                                                                        1
                                   Since both springs are subjected to the same force W, we have the equilibrium shown in
                                   Fig. 1.28(c):
                                                                   W = k d
                                                                         1 1
                                                                   W = k d                            (1.13)
                                                                         2 2
                                                                    k d  k d
                                                                     1  st
                                                                          2  st
                                   k 1       k 2
                                                   k 1       k 2  k 1       k 2

                                               d   st


                                                        W              W
                                        (a)             (b)            (c)

                                   FiGure 1.27  Springs in parallel.
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