Page 312 - Marks Calculation for Machine Design
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                  Step 12. Calculate the mean bending stress (σ m ) and the alternating bending stress (σ a )
                  as      January 4, 2005  15:4  STRENGTH OF MACHINES
                                     M m c  (10 in · lb)(0.03125 in)
                                σ m =    =                   = 30.6 kpsi
                                      I       1.02 × 10 −5  in 4
                                     M a c  (4in · lb)(0.03125 in)
                                σ a =    =                  = 12.3 kpsi
                                      I      1.02 × 10 −5  in 4
                  Step13. Plotthemeanbendingstress (σ m )andalternatingbendingstress(σ a )fromstep12,
                  the given ultimate tensile strength (S ut ), and the endurance limit (S e ) calculated in step 8
                  in a Goodman diagram like that shown in Fig. 7.17.

                            )
                         (s a
                                                            Scale: 2 kpsi × 2 kpsi
                         32.7  S e
                          30                     Calculated stresses
                          20                                   Goodman line
                                    s m
                         12.3
                          10
                                            s a
                                                                     S ut
                           0
                                                                          (s m )
                             0   10   20   30  40   50   60  70   80 85  90
                                          30.6
                        FIGURE 7.17  Goodman diagram for Example 1 (U.S. Customary).


                  Step 14. To answer question (a), calculate the factor-of-safety (n) using Eq. (7.25), which
                  represents the distance (d) in Fig. 7.12.

                         1   σ a  σ m  12.3 kpsi  30.6 kpsi
                          =    +    =         +        = (0.376) + (0.360) = 0.736
                         n   S e  S ut  32.7 kpsi  85 kpsi
                              1
                         n =      = 1.36
                             0.736
                  Step 15. To answer question (b), calculate the factor-of-safety (n m ) using Eq. (7.27), which
                  represents the distance (d m ) in Fig. 7.13.


                                      σ m                30.6 kpsi
                               S e 1 −      (32.7 kpsi) 1 −
                        σ a | σ m     S ut                85 kpsi   (32.7 kpsi)(0.640)
                   n m =     =            =                       =
                         σ a       σ a             12.3 kpsi            12.3 kpsi
                        20.93 kpsi
                      =         = 1.70
                        12.3 kpsi
                                           ) was substituted from Eq. (7.28).
                  where the alternating stress (σ a | σ m
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