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                 496   Mechanical Engineering Design
                                  9–7     Fatigue Loading

                                          The conventional methods will be provided here. In fatigue, the Gerber criterion is best;
                                          however, you will find that the Goodman criterion is in common use. For the surface
                                          factor of Eq. 6–19, an as-forged surface should always be assumed for weldments
                                          unless a superior finish is specified and obtained.
                                              Some examples of fatigue loading of welded joints follow.




                        EXAMPLE 9–5       The 1018 steel strap of Fig. 9–21 has a 1000 lbf, completely reversed load applied.
                                          Determine the factor of safety of the weldment for infinite life.

                                Solution  From Table A–20 for the 1018 attachment metal the strengths are S ut = 58 kpsi and
                                          S y = 32 kpsi. For the E6010 electrode, from Table 9–3 S ut = 62 kpsi and S y = 50 kpsi.
                                          The fatigue stress-concentration factor, from Table 9–5, is K fs = 2.7. From Table 6–2,
                                          p. 288, k a = 39.9(58) −0.995  = 0.702. For case 2 of Table 9–5, the shear area is:
                                                               A = 1.414(0.375)(2) = 1.061 in 2
                                          For a uniform shear stress on the throat, k b = 1.
                                              From Eq. (6–26), p. 290, for torsion (shear),
                                                               k c = 0.59  k d = k e = k f = 1
                                          From Eqs. (6–8), p. 282, and (6–18), p. 287,
                                                         S se = 0.702(1)0.59(1)(1)(1)0.5(58) = 12.0 kpsi
                                          From Table 9–5, K fs = 2.7. Only primary shear is present. So, with F a = 1000 lbf and
                                          F m = 0
                                                            K fs F a  2.7(1000)
                                                                             = 2545 psi   τ = 0 psi

                                                       τ =        =
                                                        a                                 m
                                                              A       1.061
                  Figure 9–21
                                                1018
                                                                          E6010
                                                        2 in
                                                                       3  in
                                                                       8
                                          2 in
                                                                       4- × 7.25-in channel
                                                 1  in
                                                 2

                                                       1018






                                                          1000 lbf
                                                          completely
                                                          reversed
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