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                 332   Mechanical Engineering Design
                  Table 6–9
                                           Material Class              φ 0.30
                  Comparison of            Wrought steels              0.50
                  Approximate Values of    Cast steels                 0.40
                  Mean Fatigue Ratio for   Powdered steels             0.38
                  Some Classes of Metals   Gray cast iron              0.35
                                           Malleable cast iron         0.40
                                           Normalized nodular cast iron  0.33



                  Table 6–10                                        k a   aS ut LN(1, C)
                                                                            b
                  Parameters in Marin                                     a                       Coefficient of
                                           Surface Finish          kpsi       MPa         b        Variation, C
                  Surface Condition Factor
                                           Ground*                  1.34       1.58      −0.086       0.120
                                           Machined or Cold-rolled   2.67      4.45      −0.265       0.058
                                           Hot-rolled              14.5       58.1       −0.719       0.110
                                           As-forged               39.8      271         −0.995       0.145

                                          *Due to the wide scatter in ground surface data, an alternate function is k a = 0.878LN(1, 0.120).
                                          Note: S ut in kpsi or MPa.




                       EXAMPLE 6–16       A steel has a mean ultimate strength of 520 MPa and a machined surface. Estimate k a .

                                Solution  From Table 6–10,
                                                           k a = 4.45(520) −0.265 LN(1, 0.058)

                                                           ¯
                                                           k a = 4.45(520) −0.265 (1) = 0.848
                                                          ˆ σ ka = Ck a = (0.058)4.45(520) −0.265  = 0.049
                                                                 ¯
                                Answer    so k a = LN(0.848, 0.049).





                                              The load factor k c for axial and torsional loading is given by
                                                                            −0.0778
                                                               (k c ) axial = 1.23S ¯ ut  LN(1, 0.125)     (6–73)
                                                                             0.125
                                                              (k c ) torsion = 0.328S ¯ ut  LN(1, 0.125)   (6–74)
                                          where S ut is in kpsi. There are fewer data to study for axial fatigue. Equation (6–73) was
                                                ¯
                                          deduced from the data of Landgraf and of Grover, Gordon, and Jackson (as cited earlier).
                                              Torsional data are sparser, and Eq. (6–74) is deduced from data in Grover et al.
                                          Notice the mild sensitivity to strength in the axial and torsional load factor, so k c in
                                          these cases is not constant. Average values are shown in the last column of Table 6–11,
                                          and as footnotes to Tables 6–12 and 6–13. Table 6–14 shows the influence of material
                                          classes on the load factor k c . Distortion energy theory predicts (k c ) torsion = 0.577 for
                                          materials to which the distortion-energy theory applies. For bending, k c = LN(1, 0).
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