Page 262 - Marks Calculation for Machine Design
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P1: Shibu
                          January 4, 2005
                                      14:56
        Brown.cls
                 Brown˙C06
                  244
                                         s 2  STRENGTH OF MACHINES
                          Boundary of allowable                   Biaxial where
                                                                  s  = s  > 0
                             combinations                          1   2
                                                            3
                                         84
                                                                   Biaxial where
                                                                   s  = 2s  > 0
                                                                        2
                                                                    1
                                                            2
                             –84
                                                                 s 1
                                                       84
                                                                Uniaxial where
                                                                   > 0, s  = 0
                                                                s 1
                                                          5            2
                                                           4
                                             –84
                                             Pure shear where
                                                > 0, s  = –s
                                             s 1   2    1
                                                            Scale: 7 MPa ¥ 7 MPa
                         FIGURE 6.9  Principal stress combinations in Example 1 (SI/metric).

                    Example 5 (Sec. 5.2):

                                         1/2
                            2    2                  2       2          1/2

                           σ + σ − σ 1 σ 2   1  ((83) + (−33) − (83)(−33))
                            1   2
                                          =   =
                                 S y         n              84
                          1   (6,889 + 1,089 + 2,739) 1/2  (10,717) 1/2  103.52
                            =                      =          =       = 1.23
                          n             84              84        84
                                1
                           n =    = 0.81 (unsafe)
                              1.23
                    Example 2 (Sec. 5.3):

                                            1/2
                                2   2                  2     2        1/2

                              σ + σ − σ 1 σ 2   1   ((84) + (0) − (84)(0))
                               1    2
                                              =   =
                                    S y         n            84
                              1   (7,056 + 0 + 0) 1/2  (7,056) 1/2  84
                               =                =         =    = 1.0
                              n         84           84     84
                                  1
                              n =    = 1.0 (okay, but marginal)
                                  1.0
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