Page 169 - Marks Calculation for Machine Design
P. 169

P1: Rakesh
                                      14:16
                          January 4, 2005
        Brown.cls
                 Brown˙C03
                              U.S. Customary  ADVANCED LOADINGS   SI/Metric       151
                    Step 4. Calculate the maximum radial stress  Step 4. Calculate the maximum radial stress
                    (σ r ) and the tangential stress (σ t ) at the radius  (σ r ) and the tangential stress (σ t ) at the radius
                    √                                  √
                    ( r i r o ), using Eqs. (3.45) and (3.46).  ( r i r o ), using Eqs. (3.45) and (3.46).
                                        2                                  2
                            3 + ν    r i                       3 + ν   r i

                     σ r max  = σ o  1 +                σ r max  = σ o  1 +
                              8      r o                        8      r o
                                               2                                     2

                                3 + 0.3   1in                        3 + 0.3  0.025 m
                         = (10 kpsi)   1 +                 = (86.5MPa)     1 +
                                  8       36 in                        8       1m
                                 3.3                                 3.3
                         = (10 kpsi)  (1 + 0.0278) 2       = (86.5MPa)  (1 + 0.025) 2
                                 8                                   8
                         = (10 kpsi)(0.4125)(1.0278) 2     = (86.5MPa)(0.4125)(1.025) 2
                         = 4.36 kpsi                       = 37.5MPa

                                               
                                 2
                     √      3 + ν            r  2       √      3 + ν    1 − ν r i  r
                      r i r o        1 − ν r i  i        r i r o    1 + 2       i
                    σ t  = σ o   1 + 2     +  2        σ t  = σ o  8          +  r  2
                              8      3 + ν r o  r  o                    3 + ν r o  o
                                3 + 0.3                              3 + 0.3
                         = (10 kpsi)                       = (86.5MPa)
                                  8                                    8

                                                2                 1− 0.3 0.025 m  (0.025 m) 2
                                 1 − 0.3 1in  (1in)
                          × 1 + 2         +                  × 1+ 2         +     2
                                 3 + 0.3 36 in  (36 in) 2         3+ 0.3  1m   (1m)
                                 3.3                                 3.3
                         = (10 kpsi)                       = (86.5MPa)
                                 8                                   8

                                 0.7 1   1
                                                                    0.7 0.025  0.000625
                          × 1 + 2     +                      × 1 + 2      +
                                 3.3 36  1296                       3.3  1     1
                         = (10 kpsi)(0.4125)               = (86.5MPa)(0.4125)
                          × (1 + 0.012 + 0.00077)            × (1 + 0.0106 + 0.000625)
                         = 4.18 kpsi                       = 36.1MPa
                    Step 5. Using Eq. (3.47), calculate the factor-  Step 5. Using Eq. (3.47), calculate the factor-
                    of-safety (n) at the inside radius (r i ) where the  of-safety (n) at the inside radius (r i ) where the
                    principal stress (σ 1 ) is the maximum tangential  principal stress (σ 1 ) is the maximum tangential
                    stress found in Step 3 and the principal stress  stress found in Step 3 and the principal stress
                    (σ 2 ) is zero.                    (σ 2 ) is zero.
                                       1/2                                1/2

                            2   2                              2  2
                           σ + σ − σ 1 σ 2  1                 σ + σ − σ 1 σ 2  1
                            1  2                              1   2
                                         =                                  =
                                S y        n                       S y       n
                                                               2
                            2
                                 2
                                                                   2
                        ((8.25) + (0) − (8.25)(0)) 1/2  1  ((71.4) + (0) − (71.4)(0)) 1/2  1
                                             =                                 =
                                 50           n                     50           n
                            2 1/2
                                                               2 1/2
                        ((8.25) )  8.25  1                ((71.4) )  71.4  1
                                =    =                             =    =
                           50     50   n                     50      350  n
                           50                                350
                       n =    = 6.1                       n =    = 4.9
                          8.25                               71.4
                     Clearly the design is safe.        Clearly the design is safe.
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