Page 421 - Analysis and Design of Machine Elements
P. 421

FD∕2                                                 Springs  399
                                     =                                                       (14.8)
                                   T
                                         3
                                         d ∕16
                             The torsional shear stress is inversely proportional to the cube of wire diameter. This
                             implies the great effect of the variation of wire diameter on the spring performance.
                             The transverse shear stress induced by shear force S, which is equivalent to F,is
                             (Figure 14.12c)
                                        F
                                     =                                                       (14.9)
                                   S     2
                                        d ∕4
                             The torsional shear stress    and transverse shear stress    oppose at the outer coil
                                                   T                         S
                             radius but coincide at the inner coil radius. The maximum combined shear stress in
                             the coil wire, which occurs at the inner surface of wire, may be computed by super-
                             position of the torsional shear stress and transverse shear stress as
                                               FD∕2      F     8FD  (    1  )  8FC
                                     =    +    =     +       =      1 +     ≈               (14.10)
                                           S
                                      T
                                                         2
                                                3
                                                d ∕16    d ∕4    d 3    2C       d 2
                             where C = D/d is spring index, usually within the range of 4–12. It is a measure of
                             coil curvature.
                             The maximum combined shear stress in the coil wire expressed in Eq. (14.10) is
                             for springs under static or low cycle loading and is indicated by the dashed line in
                             Figure 14.12d. However, for a spring subjected to high cycle of loading, the actual
                             stress at wire cross section is slightly higher on the inside of coil due to the influence

                             of pitch angle and coil curvature, as shown by the solid line in Figure 14.12d. The
                             Wahl factor, K , is introduced to account for the curvature of wire. For a circular
                                         w
                             cross section spring wire, the Wahl factor is expressed as:
                                       4C − 1   0.615
                                  K ≈        +                                              (14.11)
                                   w
                                       4C − 4    C
                             Since spring index C is within 4–12, Wahl factor K is within the range of 1.1–1.4. A
                                                                      w
                             midrange value of 1.2 is often selected as an initial estimate for K in spring design
                                                                                   w
                             calculations.
                          2) Static strength analysis
                             Considering the Wahl factor and strength conditions at the inner radius of coil
                             wire cross section, from Eq. (14.10), the shear stress and shear strength can then be
                             expressed as

                                               8FC
                                     = K    = K w  ≤ [  ]                                   (14.12)
                                       w T
                                                 d 2
                             The wire diameter is designed by
                                      √            √
                                        8K CF        K CF
                                          w
                                                       w
                                  d ≥          = 1.6                                        (14.13)
                                           [  ]        [  ]
                             where the allowable shear stress [  ] can be found in Table 14.3.
                          3) Fatigue strength analysis
                             Springs are almost always subject to fatigue loading. If the number of load cycles
                             is less than 1000, or the load is constant during operation, static strength analysis
   416   417   418   419   420   421   422   423   424   425   426