Page 397 - Marks Calculation for Machine Design
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P1: Naresh
                          January 4, 2005
                                      15:28
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                 Brown˙C09
                              U.S. Customary  MACHINE ENERGY      SI/Metric       379
                    Example 6. Suppose the two helical springs  Example 6. Two helical springs are used in
                    in Example 5 are used in parallel as shown in  series as shown in Fig. 9.6(a). Calculate the
                    Fig. 9.6(b). Calculate the equivalent spring rate  equivalent spring rate (k eq ), where
                    (k eq ), where
                     k 1 = 30 lb/in                     k 1 = 5,400 N/m
                     k 2 = 60 lb/in                     k 2 = 10,800 N/m
                    solution                           solution
                    Step 1. Using Eq. (9.32), determine the equiv-  Step 1. Using Eq. (9.32), determine the equiv-
                    alent spring rate (k eq ) as       alent spring rate (k eq ) as
                          k eq = k 1 + k 2                k eq = k 1 + k 2
                            = (30 lb/in) + (60 lb/in)       = (5,400 N/m) + (10,800 N/m)
                            = 90 lb/in                      = 16,200 N/m
                     Theequivalentspringrate(k eq )isgreaterthan  Theequivalentspringrate(k eq )isgreaterthan
                    either of the two individual spring rates. This is  either of the two individual spring rates. This is
                    because both springs are working together in  because both springs are working together in
                    the system.                        the system.
                     Note that these two springs will change  Note that these two springs will change
                    lengths at different rates, therefore the system  lengths at different rates, therefore the system
                    may rotate to accommodate this difference.  may rotate to accommodate this difference.



                    9.2.5 Extension Springs
                    Extensions springs are helical springs loaded in tension. To provide a way to connect these
                    springs into a mechanical system, a hook is usually fashioned from additional coils at each
                    end. The stress concentrations these hooks produce must be considered in the design.
                      Hooks come in many designs, however all hooks follow the pattern in Fig. 9.7, where the
                    ratio of the mean radius (r m ) to the inside radius (r i ) of the hook is the stress-concentration
                    factor (K) given in Eq. (9.33) as
                                                      d
                                                  r i +
                                             r m      2      d
                                         K =    =       = 1 +                   (9.33)
                                             r i    r i      2r i
                    where as the wire diameter (d) increases, or the inside radius (r i ) decreases, or both, the
                    stress-concentration factor (K) increases.



                                                           Hook end
                                      d
                                                 r i

                                                   r m


                                     FIGURE 9.7  Extension spring hook geometry.
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