Page 291 - Analysis and Design of Machine Elements
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Shafts
                          deflection y and slope   . The equivalent diameter of a stepped shaft can be obtained  269
                          from [4]
                                    √
                                    √ L
                                    √
                                 v
                               d = √   z                                                    (10.13)
                                    4 √ ∑ l i
                                         d 4
                                      i=1  i
                          where
                          l , d –length and diameter of the ith segment in a stepped shaft;
                           i  i
                          L    –calculated length of stepped shaft. When a load is applied between bearings,
                               L = l; for an end loaded cantilever beam, L = l + K,where l is the span and K is
                               the cantilever length;
                          z    –number of segments in the stepped shaft;
                            The calculation of slope and deflection are based on successive integration of differ-
                          ential equations for the elastic beam deflection curve expressed as [5, 6]
                                 2
                                d y   M
                                   =                                                        (10.14)
                                dx 2  EI
                                   dy      M
                                  =   =      dx                                             (10.15)
                                   dx   ∫ EI
                                   d        M
                               y =    =        dxdx                                         (10.16)
                                   dx   ∫∫ EI
                            The integration may be performed either analytically, graphically or numerically. For

                          a shaft of given length and loading, the bending deflection is inversely proportional to
                          the product EI, as indicated in Eq. (10.16). Therefore, the effective way to increase the
                          rigidity of a shaft is to increase the diameter of shaft.
                            The bending rigidity criteria are then
                               y ≤ [y]                                                      (10.17)
                                  ≤ [  ]                                                    (10.18)
                            The allowable misalignment of a shaft is determined by the requirements of mounted
                          specific gears or bearings by checking gear or bearing catalogues. As a rough guideline,
                          the allowable bending deflections for a transmission shaft is (0.0003–0.0005)l,where l is
                          span between bearings. When a shaft supports gears, the allowable bending deflections
                          is selected as (0.01–0.03)m ,where m is the normal module of mounted gears. The
                                                          n
                                                 n
                          allowable slopes should be checked at locations where gears and bearings are mounted.
                          At the cross section where a gear is mounted, the allowable slopes should be within
                          0.001–0.002 rad. The allowable slopes for the shaft where a deep groove ball bearing, a
                          cylindrical roller bearing and a tapered roller bearing, is mounted should be less than
                          0.005, 0.0025 and 0.0016 rad, respectively [7].

                          10.3.2.2  Torsional Deflections
                          Torsional rigidity is less important unless in some special applications. If a shaft has
                          uniform diameter over its whole length, the unit length angular deflection may be readily
                          calculated from formula in Mechanics of Materials [5], repeated here as
                                   180   T      3          4 T
                                  =    ×    × 10 = 5.73 × 10                                (10.19)
                                         GJ                 GJ
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