Page 398 - Marks Calculation for Machine Design
P. 398

P1: Naresh
                          January 4, 2005
                                      15:28
        Brown.cls
                 Brown˙C09
                                           APPLICATION TO MACHINES
                  380
                    The free, or unstretched, length (L o ) of an extension spring is the body length (L B ) plus
                  two times the hook distance (L hook ), given in Eq. (9.34) as
                                            L o = L B + 2L hook                (9.34)
                  where the body length (LB) is given by Eq. (9.35) as
                                             L B = (N a + 1)d                  (9.35)
                    The presence of the stress-concentration factor (K) given in Eq. (9.33) prevents the hooks
                  from being designed as strong as the main coils of the spring.
                            U.S. Customary                       SI/Metric
                  Example 7. Suppose circular hooks are added  Example 7. Suppose circular hooks are added
                  to the ends of the cylindrical helical spring  to the ends of the cylindrical helical spring
                  designed in Example 2. Determine the stress-  designed in Example 2. Determine the stress-
                  concentration factor (K) for the design, where  concentration factor (K) for the design, where
                    D = 1 in (given in Example 2)      D = 0.025 m (given in Example 2)
                    d = 0.11 in (determined in Example 2)  d = 0.0027 m (determined in Example 2)
                    r i = (D − d)/2 = 0.445 in         r i = (D − d)/2 = 0.01115 m
                  solution                           solution
                  Step 1. Using Eq. (9.33), determine the stress-  Step 1. Using Eq. (9.33), determine the stress-
                  concentration factor (K) as        concentration factor (K) as
                              d       0.11 in                   d       0.0027 m
                       K = 1 +  = 1 +                   K = 1 +   = 1 +
                             2r i    2(0.445 in)               2r i   2(0.01115 m)
                        = 1 + (0.124) = 1.12              = 1 + (0.121) = 1.12
                    This means the stress at the hook ends are a  This means the stress at the hook ends are a
                  little over 12 percent greater than the stress in  little over 12 percent greater than the stress in
                  main coils.                        main coils.



                  9.2.6 Compression Springs
                  As the name implies, compression springs are helical springs loaded in compression. There
                  are four main types of ends for compression springs: (1) plain, (2) squared, (3) plain and
                  ground, and (4) squared and ground. A spring with plain ends has an uninterrupted helix
                  angle at its ends, whereas a spring with squared ends has the helix angle flattened to zero at
                  its ends. For both plain and squared types, ends that are ground flush improve load transfer,
                  particularly with squared and ground ends.
                    Additional coils must be added to the design of a helical spring if the ends are not plain.
                  Table 9.1 gives a summary of the additional coils needed for each type. In Table 9.1, a term
                  appears denoted (p) for pitch. For a cylindrical helical spring with plain ends, the pitch (p)
                  is defined as
                                                  L o − d
                                              p =                              (9.36)
                                                   N a
                  where the units of pitch are length per number of active coils. The pitch (p) of a helical
                  spring is used to determine its free length.
   393   394   395   396   397   398   399   400   401   402   403