Page 62 - Analysis and Design of Machine Elements
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Analysis and Design of Machine Elements
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                       Figure 2.7 Endurance limits for three typical loading cases.

                       limit at stress ratio r. If the working stress is located within the area of ΔOA F ,the
                                                                                         1 1
                       endurance limit can be found by solving two line equations of OM and A F to find out
                                                                             1
                                                                                   1 1
                       the intersection, by
                                     ′
                             ⎧     a     ae
                             ⎪   =   ′
                                
                             ⎨ m       me                                                 (2.27)
                                       ′
                             ⎪    = K    +       ′
                               −1       ae     me
                             ⎩
                         We can obtain the coordinates of intersection point M (   ′ me ,     ′ ae ), and then the
                                                                        1
                       fatigue strength is expressed as
                                              ′
                                          ′
                                     lim     +    me      −1
                                          ae
                             S =      =          =             ≥ [S]                      (2.28)
                              ca
                                     max     max   K    +      
                                                       a
                                                              m
                         A practical example of a constant mean stress is the stresses in a loaded vibrating
                       spring. Points on the vertical lines through the working point of M (or N), that is, MM
                                                                                              2
                       (or NN ), represent the same mean stress. Similarly, the fatigue strength for constant
                              2
                       mean stress is
                                     −1  +(K −    )   m
                                           
                                               
                             S =                  ≥ [S]                                   (2.29)
                              ca
                                     K (   +    )
                                         m
                                             a
                                        
                         A typical example of constant minimum stress is the stresses in a preloaded bolt. Since
                                                ∘
                           =    −   = Const., the 45 lines through point M (or N), that is, MM (or NN ), rep-
                         min  m   a                                               3      3
                       resent the stresses that have the same minimum stress. Similarly, the fatigue strength is
                                  2   −1  +(K −    )   min
                                                 
                                            
                             S =                     ≥ [S]                                (2.30)
                              ca
                                  (K +    )(2   +     )
                                             a   min
                         In these three cases, if the intersection is on line EF , the element will yield rather than
                                                                   1
                       fatigue. The strength is then calculated by
                                       s
                             S =         ≥ [S]                                            (2.31)
                              ca
                                     +    a
                                   m
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