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                                                                                     Failures Resulting from Static Loading  237
                                                                                        8
                                               which can more closely represent some of the data. However, this introduces a nonlin-
                                               ear equation for the sake of a minor correction, and will not be presented here.



                             EXAMPLE 5–5       Consider the wrench in Ex. 5–3, Fig. 5–16, as made of cast iron, machined to dimen-
                                               sion. The force F required to fracture this part can be regarded as the strength of the
                                               component part. If the material is ASTM grade 30 cast iron, find the force F with
                                               (a) Coulomb-Mohr failure model.
                                               (b) Modified Mohr failure model.

                                    Solution   We assume that the lever DC is strong enough, and not part of the problem. Since grade
                                               30 cast iron is a brittle material and cast iron, the stress-concentration factors K t and K ts
                                               are set to unity. From Table A–24, the tensile ultimate strength is 31 kpsi and the com-
                                               pressive ultimate strength is 109 kpsi. The stress element at A on the top surface will be
                                               subjected to a tensile bending stress and a torsional stress. This location, on the 1-in-
                                               diameter section fillet, is the weakest location, and it governs the strength of the assem-
                                               bly. The normal stress σ x and the shear stress at A are given by

                                                                     M       32M      32(14F)
                                                              σ x = K t  = K t    = (1)       = 142.6F
                                                                     I/c     πd 3       π(1) 3
                                                                     Tr       16T      16(15F)
                                                             τ xy = K ts  = K ts  = (1)       = 76.4F
                                                                      J       πd 3      π(1) 3
                                               From Eq. (3–13) the nonzero principal stresses σ A and σ B are

                                                           142.6F + 0    	 142.6F − 0  
 2
                                                                                               2
                                                  σ A ,σ B =         ±                + (76.4F) = 175.8F, −33.2F
                                                               2              2
                                               This puts us in the fourth-quadrant of the σ A ,σ B plane.
                                               (a) For BCM, Eq. (5–31b) applies with n = 1 for failure.
                                                                            175.8F    (−33.2F)
                                                                 σ A   σ B
                                                                    −     =        −          = 1
                                                                                            3
                                                                                 3
                                                                 S ut  S uc  31(10 )  109(10 )
                                               Solving for F yields
                                     Answer                                  F = 167 lbf

                                               (b) For MM, the slope of the load line is  |σ B /σ A |= 33.2/175.8 = 0.189 < 1.
                                               Obviously, Eq. (5–32a) applies.
                                                                               175.8F
                                                                          σ A
                                                                             =        = 1
                                                                                    3
                                                                          S ut  31(10 )
                                     Answer                                  F = 176 lbf
                                               As one would expect from inspection of Fig. 5–19, Coulomb-Mohr is more conservative.





                                               8 See J. E. Shigley, C. R. Mischke, R. G. Budynas, Mechanical Engineering Design, 7th ed., McGraw-Hill,
                                               New York, 2004, p. 275.
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