Page 347 - Marks Calculation for Machine Design
P. 347

P1: Sanjay
                                      15:14
                          January 4, 2005
        Brown.cls
                 Brown˙C08
                              U.S. Customary  MACHINE ASSEMBLY    SI/Metric       329
                    where                              where
                       C 1 = (2 t tan α + D − d)(D + d)   C 1 = (2 t tan α + D − d)(D + d)
                                         ∗
                                     ◦
                         = (2 t middle tan 30 + D − d bolt )  = (2 t middle tan 30 + D − d bolt )
                                                                       ◦
                                                                           ∗
                              ∗
                                                                 ∗
                           ×(D + d bolt )                     ×(D + d bolt )

                             (2)(0.125 in)(0.577)              (2)(0.0025 m)(0.577)
                         =                                  =
                             +(1.62 in) − (0.5in)              +(0.041 m) − (0.012 m)
                           ×((1.62 in) + (0.5in))             ×((0.041 m) + (0.012 m))
                         = (1.26 in)(2.12 in)               = (0.032 m)(0.053 m)
                         = 2.68 in 2                        = 0.00170 m 2
                    and                                and
                       C 2 = (2 t tan α + D + d)(D − d)   C 2 = (2 t tan α + D + d)(D − d)
                                     ◦
                                         ∗
                                                                       ◦
                                                                           ∗
                         = (2 t middle tan 30 + D + d bolt )  = (2 t middle tan 30 + D + d bolt )
                              ∗
                                                                 ∗
                           ×(D − d bolt )                     ×(D − d bolt )

                             (2)(0.125 in)(0.577)              (2)(0.0025 m)(0.577)
                         =                                  =
                             +(1.62 in) + (0.5in)              +(0.041 m) + (0.012 m)
                           ×((1.62 in) − (0.5in))             ×((0.041 m) − (0.012 m))
                         = (2.26 in)(1.12 in)               = (0.056 m)(0.029 m)
                         = 2.54 in 2                        = 0.00162 m 2
                    Therefore,                         Therefore,

                                        lb                                 N
                            (1.814) 16 × 10 6  (0.5in)        (1.814) 1.1 × 10 11  (.012 m)
                                        in 2                               m 2
                     k middle =           	            k middle =            2
                                    2.68 in 2                         0.00170 m
                                 ln                                ln        2
                                    2.54 in 2                         0.00162 m
                                                                    9
                                  7
                            1.45 × 10 lb/in                =  2.39 × 10 N/m
                          =                                      0.048
                               0.054                                10
                                  8
                          = 2.69 × 10 lb/in                = 4.98 × 10  N/m
                    Step 4. The remaining thickness of the cast  Step 4. The remaining thickness of the cast
                    iron member is the thickness (t 2 ) minus the third  iron member is the thickness (t 2 ) minus the third
                    thickness (t middle ), which from Eq. (8.19) is half  thickness (t middle ), which from Eq. (8.19) is half
                    the grip length (L grip /2). Substitute this remain-  the grip length (L grip /2). Substitute this remain-
                    ing thickness (L grip /2), the given bolt diameter  ing thickness (L grip /2), the given bolt diameter
                    (d bolt ), and the modulus of elasticity (E 2 ) in  (d bolt ), and the modulus of elasticity (E 2 ) in
                    Eq. (8.22) to find the stiffness (k 2 ) as  Eq. (8.22) to find the stiffness (k 2 ) as
                                (π tan 30 ) Ed                     (π tan 30 ) Ed
                                                                        ◦
                                      ◦
                        k 2 =                              k 2 =
                                                                         ◦
                                      ◦
                                2 t tan 30 + 0.5 d                 2 t tan 30 + 0.5 d
                             ln 5                              ln 5
                                2 t tan 30 + 2.5 d                 2 t tan 30 + 2.5 d
                                      ◦
                                                                         ◦
                                  ◦                                  ◦
                             (π tan 30 ) E 2 d bolt            (π tan 30 ) E 2 d bolt
                          =                                  =
                                ln 5  C 1                         ln 5  C 1
                                   C 2                                C 2
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