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                 378    Mechanical Engineering Design



                                     This is low. We will look up data for a specific retaining ring to obtain K f more
                                     accurately. With a quick online search of a retaining ring specification using the
                                     website www.globalspec.com, appropriate groove specifications for a retaining ring
                                     for a shaft diameter of 1.625 in are obtained as follows: width, a = 0.068 in;
                                     depth, t = 0.048 in; and corner radius at bottom of groove, r = 0.01 in. From
                                     Fig. A–15–16, with r/t = 0.01/0.048 = 0.208, and a/t = 0.068/0.048 = 1.42
                                                    K t = 4.3, q = 0.65 (Fig. 6–20)
                                                    K f = 1 + 0.65(4.3 − 1) = 3.15
                                                         32K f M a  32(3.15)(2398)
                                                    σ a =        =               = 17 930 psi
                                                           πd 3       π(1.625) 3
                                                         S e  33 300
                                                    n f =   =       = 1.86
                                                         σ a  17 930
                                     Quickly check if point M might be critical. Only bending is present, and the moment
                                     is small, but the diameter is small and the stress concentration is high for a sharp
                                     fillet required for a bearing. From the moment diagram,
                                     M a = 959 lbf · in, and M m = T m = T a = 0.
                                        Estimate K t = 2.7 from Table 7–1, d = 1.0 in, and fillet radius r to fit a typical
                                     bearing.
                                                    r/d = 0.02, r = 0.02(1) = 0.02
                                                      q = 0.7 (Fig. 6–20)
                                                     K f = 1 + (0.7)(2.7 − 1) = 2.19
                                                          32K f M a  32(2.19)(959)
                                                     σ a =        =              = 21 390 psi
                                                            πd 3        π(1) 3
                                                          S e  33 300
                                                     n f =  =        = 1.56
                                                          σ a  21 390
                                     Should be OK. Close enough to recheck after bearing is selected.
                                        With the diameters specified for the critical locations, fill in trial values for
                                     the rest of the diameters, taking into account typical shoulder heights for
                                     bearing and gear support.
                                                                D 1 = D 7 = 1.0in
                                                                D 2 = D 6 = 1.4in
                                                                D 3 = D 5 = 1.625 in
                                                                D 4 = 2.0in

                                     The bending moments are much less on the left end of shaft, so D 1 , D 2 , and D 3
                                     could be smaller. However, unless weight is an issue, there is little advantage to
                                     requiring more material removal. Also, the extra rigidity may be needed to keep
                                     deflections small.
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