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                 292   Mechanical Engineering Design
                                          from Table 6–4 or Eq. (6–27) and proceed as usual. If the rotating-beam endurance limit
                                          is not given, then compute it using Eq. (6–8) and the temperature-corrected tensile
                                          strength obtained by using the factor from Table 6–4. Then use k d = 1.




                        EXAMPLE 6–5       A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F
                                          in service. Estimate the Marin temperature modification factor and (S e ) 450 if
                                                                                                      ◦
                                          (a) The room-temperature endurance limit by test is (S ) 70 = 39.0 kpsi.

                                                                                         ◦
                                                                                      e
                                          (b) Only the tensile strength at room temperature is known.
                                Solution  (a) First, from Eq. (6–27),
                                                                                          −5
                                                                                                 2
                                                                         −3
                                                       k d = 0.975 + 0.432(10 )(450) − 0.115(10 )(450 )
                                                                     −8
                                                                            3
                                                                                              4
                                                            + 0.104(10 )(450 ) − 0.595(10 −12 )(450 ) = 1.007
                                          Thus,
                                Answer                    (S e ) 450 = k d (S ) 70 = 1.007(39.0) = 39.3 kpsi

                                                               ◦
                                                                         ◦
                                                                      e
                                          (b) Interpolating from Table 6–4 gives
                                                                                     450 − 400
                                                    (S T /S RT ) 450 = 1.018 + (0.995 − 1.018)  = 1.007
                                                              ◦
                                                                                     500 − 400
                                          Thus, the tensile strength at 450°F is estimated as
                                                     (S ut ) 450 = (S T /S RT ) 450 (S ut ) 70 = 1.007(70) = 70.5 kpsi
                                                                        ◦
                                                                               ◦
                                                           ◦
                                          From Eq. (6–8) then,
                                Answer                    (S e ) 450 = 0.5(S ut ) 450 = 0.5(70.5) = 35.2 kpsi
                                                               ◦
                                                                           ◦
                                          Part a gives the better estimate due to actual testing of the particular material.

                                          Reliability Factor k e
                                          The discussion presented here accounts for the scatter of data such as shown in
                                                                                               .
                                          Fig. 6–17 where the mean endurance limit is shown to be S /S ut = 0.5, or as given by

                                                                                          e
                                          Eq. (6–8). Most endurance strength data are reported as mean values. Data presented
                                                              19
                                          by Haugen and Wirching show standard deviations of endurance strengths of less than
                                          8 percent. Thus the reliability modification factor to account for this can be written as
                                                                                                           (6–29)
                                                                      k e = 1 − 0.08 z a
                                          where z a is defined by Eq. (20–16) and values for any desired reliability can be deter-
                                          mined from Table A–10. Table 6–5 gives reliability factors for some standard specified
                                          reliabilities.
                                              For a more comprehensive approach to reliability, see Sec. 6–17.



                                          19 E. B. Haugen and P. H. Wirsching, “Probabilistic Design,” Machine Design, vol. 47, no. 12, 1975,
                                          pp. 10–14.
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