Page 353 - Cam Design Handbook
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THB11  9/19/03  7:33 PM  Page 341

                                    CAM SYSTEM MODELING                    341

                                             1
                                         PE =  Kx  2
                                             2
            while the potential energy of the equivalent system is
                                             1
                                        PE =  K x .
                                                 2
                                             2  eq  f
            Since for small angles, the relationship between x and x f is
                                           x  x  f
                                            =   .
                                          l   l
                                           1  2
            Using this expression for x above gives
                                                 2
                                               l Ê ˆ
                                               1
                                        K =  KÁ ˜ .
                                         eq    l Ë ¯
                                               2
            When l 1 > l 2, the spring appears stiffer than its physical stiffness, while for l 1 < l 2, the
            equivalent stiffness is less than the physical stiffness of the spring because of the mechan-
            ical advantage.
            11.4.5 Massive Springs

            Up to this point, the springs considered have been ideal springs and have had no mass
            associated  with  them.  In  practice,  it  is  often  acceptable  to  model  real  springs  as  ideal
            springs if the mass of other elements in the system dominates the mass of the springs. This
            is not always the case, however, and real springs may need to be modeled as a combina-
            tion  of  an  ideal  spring  and  a  lumped  mass.  The  equivalent  mass  of  a  massive  spring
            (Fig. 11.17) can be derived by applying the concepts used in the previous section.
               Since the spring is a continuum, the kinetic energy is given by

                                          1
                                                ()
                                               2
                                      KE =  Ú  v x dm.                   (11.55)
                                          2
                                           spring
            If the mass is evenly distributed along the length of the spring,
                                          V


                                                            V
                                                             l





                                  X
                                          L
                            FIGURE 11.17.  Spring with mass.
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