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                 338   Mechanical Engineering Design

                                          (b) The rotary endurance tests are described by S = 40LN(1, 0.05) kpsi whose mean
                                                                                  e
                                                                                                    ¯
                                          is less than the predicted mean in part a. The mean endurance strength S e is
                                                            S e = 0.816(0.869)40 = 28.4 kpsi
                                                            ¯
                                                                                    2 1/2
                                                                             2
                                                                     2
                                                          C Se = (0.058 + 0.125 + 0.05 )  = 0.147
                                          so the endurance strength can be expressed as  S e = 28.3LN(1, 0.147) kpsi. From
                                          Eq. (5–43),
                                                                  ⎛                 ⎞

                                                                     28.4  1 + 0.156 2
                                                                 ln  ⎝              ⎠
                                                                    10.56  1 + 0.147 2
                                                                                        =−4.65
                                                         z =−&             2         2
                                                                ln[(1 + 0.147 )(1 + 0.156 )]
                                          Using Table A–10, we see the probability of failure p f =  (−4.65) = 0.000 001 71,
                                          and
                                                             R = 1 − 0.000 001 71 = 0.999 998 29

                                          an increase! The reduction in the probability of failure is (0.000 001 71 − 0.000
                                          006 35)/0.000 006 35 =−0.73, a reduction of 73 percent. We are analyzing an existing
                                          design, so in part (a) the factor of safety was ¯n = S/¯σ = 31.4/10.56 = 2.97. In part (b)
                                                                                  ¯
                                          ¯ n = 28.4/10.56 = 2.69,a decrease. This example gives you the opportunity to see the role
                                          of the design factor. Given knowledge of S, C S, ¯σ, C σ , and reliability (through z), the mean
                                                                          ¯
                                          factor of safety (as a design factor) separates S and ¯σ so that the reliability goal is achieved.
                                                                             ¯
                                          Knowing ¯n alone says nothing about the probability of failure. Looking at ¯n = 2.97 and
                                          ¯ n = 2.69 says nothing about the respective probabilities of failure. The tests did not reduce
                                          S e significantly, but reduced the variation C S such that the reliability was increased.
                                          ¯
                                                                                                    ¯
                                              When a mean design factor (or mean factor of safety) defined as S e /¯σ is said to
                                          be silent on matters of frequency of failures, it means that a scalar factor of safety
                                          by itself does not offer any information about probability of failure. Nevertheless,
                                          some engineers let the factor of safety speak up, and they can be wrong in their
                                          conclusions.







                                              As revealing as Ex. 6–19 is concerning the meaning (and lack of meaning) of a
                                          design factor or factor of safety, let us remember that the rotary testing associated with
                                          part (b) changed nothing about the part, but only our knowledge about the part. The
                                          mean endurance limit was 40 kpsi all the time, and our adequacy assessment had to
                                          move with what was known.

                                          Fluctuating Stresses
                                          Deterministic failure curves that lie among the data are candidates for regression mod-
                                          els. Included among these are the Gerber and ASME-elliptic for ductile materials, and,
                                          for brittle materials, Smith-Dolan models, which use mean values in their presentation.
                                          Just as the deterministic failure curves are located by endurance strength and ultimate
                                          tensile (or yield) strength, so too are stochastic failure curves located by S e and by S ut
                                          or S y . Figure 6–32, p. 320, shows a parabolic Gerber mean curve. We also need to
                                          establish a contour located one standard deviation from the mean. Since stochastic
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