Page 271 - Cam Design Handbook
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THB9  9/19/03  7:26 PM  Page 259

                                 CAM MATERIALS AND LUBRICATION             259



















                 (a) Without deflection.        (b) With deflection.
                FIGURE 9.5.  Stress distribution on cam—spherical-faced follower.




            that the same stress values exist, which are shifted from the deflected side of the cam. This
            is the primary advantage of a spherical follower as compared to the serious stress increase
            exhibited with the cylindrical roller or flat-faced follower. The spherical radii used are
            from 10 in to 300 in.

            EXAMPLE The nose of a cam has a radius of curvature of 1/4 in. At 1200rpm, the trans-
            lating roller-follower has an axial force at this point of 132lb, with a pressure angle of
            20° and a roller-follower diameter of 3/4 in. The cam thickness is 3/4 in. Both the cam
            and the follower are made of steel. The cam shaft and bearing supports are relatively rigid.
            Find the maximum compressive stress on the cam nose at this speed.
            Solution The force distribution gives the normal force on the cam surface
                                          132
                                      P =     =1401b.
                                         cos  20
            From Eq. (9.11), the maximum compressive stress at this point
                                                 12
                                 È   Ê 1   1  ˆ  ˘
                                 Í  P Á Ë  r  +  r ¯ ˜  ˙
                       s   =  . 0564 Í  e  f    ˙
                         max                  2
                                 Í  Ê1 -  m  2  1 -  m ˆ ˙
                                              f
                                  t    c  +
                                 Í h Á         ˜ ˙
                                 Î  Ë E e   E  f  ¯  ˚
                                                     12
                                 È       Ê     ˆ    ˘
                                 Í       Á  1  1  ˜  ˙
                                 Í    140 Á 3  +  1  ˜  ˙
                                 Í       Á Ë   ˜ ¯  ˙
                       s  max  =  . 0564 Í  8  4    ˙  = 801 8001,  bin 2
                                      -
                                              -
                                        .
                                                .
                                 Í Í  3  Ê 103 2  103 2  ˆ ˙
                                 Í  4  Á Ë 3010  6  +  3010  6  ˜ ¯ ˙
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