Page 365 - Cam Design Handbook
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THB11  9/19/03  7:34 PM  Page 353

                                    CAM SYSTEM MODELING                    353

            pushrod (the next lowest spring rate) as a flexible element. The model then takes the shape
            of  Fig.  11.25a,  where  the  mass  of  the  pushrod  has  been  divided  equally  between
            the elements on either side of it, the lifter and the rocker arm (Chen, 1982; Hanachi and
            Freudenstein, 1986). The parameter values then become
                      1
              m =  m +  m pr
                1
                   l
                      2
                                        0327 lbm
                 = .      + ( .  ) .  = .
                  0270 lbm
                            05 0114 lbm
                  1     J          m
                                2
              m =   m +  ra  +  mr +  cs  r ra 2
                              vs ra
                     pr
               2
                         2
                  2     l ra pr    3
                         ,
                                       ◊
                               0132 lbm in 2             0153 lbm
                                                          .
                                .
                                                      2
                                           0251 lbm
                                                                       .
                   05 0114 lbm
                 = ( .  ) .  +            + .      (1.7 ) +     ( 17) =.  2  110 lbm
                                (0.875 in ) 2               3
                                    K =  K =  3010 lb in
                                                  5
                                               ¥
                                             .
                                     1   pr
                                    K =  K =  660 lb in
                                     2   eq
               Since  this  is  exactly  the  same  form  as  the  suspension  model  discussed  earlier,  the
            natural frequencies of this two-mass system can be calculated and occur at 67Hz (as pre-
            dicted by the one degree of freedom model) and 3400Hz. This suggests two things. First,
            the next natural frequency is significantly higher than the first so there is ample justifica-
            tion for using a single-mass model for basic analysis. Secondly, while the single mass
                                                   K                b
                                       K 2          eq              eq
                                              X , V l
                                               l
                                   m 2                    M eq
                                               (b) Single mass with damping.
                                       K 1




                    X , V r
                     l
                                   m 1

                       (a) Two-mass model.
                     FIGURE 11.25.  Other simple models of the valve-gear system.
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