Page 334 - Marks Calculation for Machine Design
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P1: Shashi
                          January 4, 2005
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                                           STRENGTH OF MACHINES
                  Step 1F. Using the test specimen endurance limit (S ) found in step 1E, and the modifying
                                                        e
                  factors found in steps 1A through 1D, calculate the endurance limit (S e ) for the solid shaft
                  using the Marin equation for combined loading in Eq. (7.35) as

                              S e = k a k b (1)k d S = (0.86)(0.87)(1)(1.020)(264.6MPa)
                                          e
                                = (0.763)(264.6MPa) = 202 MPa
                  Step 2. The normal and shear stresses are given and displayed in Fig. 7.32.
                                            0
                                                t = 56 MPa
                                 s axial              s axial  = 70 MPa
                                 s bending            s bending  = ± 140 MPa
                                      t
                                            0
                               FIGURE 7.32  Plane stress element for Example 4 (SI/metric).

                  Step 3. Calculate the maximum normal stress (σ max ) and the minimum normal stress
                  (σ min ) as
                           σ max = σ axial + σ bending = (70 MPa) + (140 MPa) = 210 MPa

                           σ min = σ axial − σ bending = (70 MPa) − (140 MPa) =−70 MPa

                  Step 4A. Calculate the mean normal stress (σ m ) and the alternating normal stress (σ a ) as
                            σ max + σ min  (210 MPa) + (−70 MPa)  140 MPa
                        σ m =         =                     =        = 70 MPa
                                2                2               2
                            σ max − σ min  (210 MPa) − (−70 MPa)  280 MPa
                        σ a =         =                     =        = 140 MPa
                                2                2               2
                  Step 4B. As the shear stress due to the torque is constant, the mean shear stress (τ m ) and
                  alternating shear stress (τ a ) are

                                             τ m = 56 MPa
                                              τ a = 0MPa

                  Step 5. Multiply the alternating normal stress (σ a ) by the reduced stress concentration
                  factor (K f ) to give

                                      σ a = (1.15)(140 MPa) = 161 MPa
                  Step 6. There are no alternating axial stresses, so proceed to Step 7.
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