Page 231 - Analysis and Design of Machine Elements
P. 231

Gear Drives
                             is difference between the trial load factor K and K, the trial values of pitch diameter  209
                                                                t
                             d (or module m ) can be modified by
                              1t
                                           nt
                                         √
                                   d = d 1t  3  K∕K t                                        (8.88)
                                    1
                             or
                                           √
                                   m = m nt  3  K∕K t                                        (8.89)
                                    n
                          The results from both strength analyses are compared and evaluated before the final
                            design decision can be made.
                          8.6.6.4  Geometrical Calculation
                          Because of the requirement of standardization and manufacturing process, variables
                          in gear transmission must be adjusted after the initial design. Basic rules are listed in
                          Table 8.7.
                            Variables in gear transmission are constrained by each other. On the one hand, they
                          have to meet the requirements listed in Table 8.7; on the other hand, they have to satisfy
                          geometrical constraints for proper meshing. A reasonable compromise between rele-
                          vant variables must be reached during design process.
                            For example, for a soft-tooth-surface helical gear in an enclosed gearing with a speed
                          ratio i = 3.243, assume the module is obtained by
                                    √
                                              2
                                       2KT cos     Y Y Y Y
                                                   Fa Sa      
                                          1
                                     3
                               m ≥          2   ⋅           = 2.31
                                 n
                                            z        [   ]
                                                      F
                                           d 1
                            Astandardmoduleisthenselectedas m      n  = 2.5 mm. Select z 1  = 24, then
                                                                              ∘
                          z = iz = 3.243 × 24 = 77.832, choose z = 78. Assuming    = 15 , the centre distance is
                                                          2
                           2
                               1
                                   m (z + z )
                                     n
                                           2
                                       1
                               a =           = 131.99
                                     2cos   
                            Round off the centre distance to 130 mm for the convenience of assembly and the helix
                          angleisconsequentlychanged to
                                         m (z + z )        2.5 ×(24 + 78)      ∘
                                                 2
                                             1
                                          n
                                  = arccos         = arccos             = 11.275
                                            2a                2 × 130
                                                          ∘     ∘
                            If    is not within the range from 8 to 20 ,select z and z again and repeat this
                                                                        1
                                                                              2
                          calculation.
                          Table 8.7 Basic rules for design variable selection.
                           Variables          Requirements   Variables      Requirements
                           Module m           Standard value  Centre distance a  0or5at digit
                                  n
                           Number of teeth z , z  Integer    Face width b   Integer
                                        1  2
                           Pitch diameter d   Calculated value  Helix angle     Calculated value
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