Page 104 - Marks Calculation for Machine Design
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P1: Sanjay
                          January 4, 2005
                                      16:18
        Brown.cls
                 Brown˙C02
                                           STRENGTH OF MACHINES
                  86
                    Note that the deflection determined in the previous example was positive or upward. For
                  distances closer to the left support, the deflection will be negative or downward. The location
                  of the transition point, meaning the point of zero deflection, is not a simple expression, and
                  depends on the relative values of the length (L) between the supports and the length of the
                  overhang (a).
                    Unless the length of the overhang (a) is very short compared to the length (L) between
                  the supports, the maximum downward deflection occurs at the tip of the overhang, a distance
                  (x 1 )isequalto(a)fromtherollersupport.Substituting(a)forthedistance(x 1 )inEq.(2.47b)
                  gives the tip deflection (  Tip ) as
                                            w     4    3     3
                                       Tip =   (3 a + 4 a L − aL ) ↑           (2.48)
                                           24 EI
                            U.S. Customary                       SI/Metric
                  Example 5. Calculate the maximum down-  Example 5. Calculate the maximum down-
                  ward deflection (  Tip ) for the beam configu-  ward deflection (  Tip ) for the beam configu-
                  ration in Example 4, where         ration in Example 4, where
                    w = 113 lb/ft                      w = 1,540 N/m
                    L = 3 ft, a = 1ft                  L = 1m, a = 0.3 m
                                                                 6
                             6
                    EI = 2.5 × 10 lb · ft 2            EI = 1.02 × 10 N · m 2
                  solution                           solution
                  The maximum downward deflection (  Tip ) is  The maximum downward deflection (  Tip ) is
                  given by Eq. (2.48).               given by Eq. (2.48).
                           w    4    3     3                 w    4    3    3
                       Tip =  (3 a + 4 a L − aL ) ↑     Tip =  (3 a + 4 a L − aL ) ↑
                          24 EI                             24 EI
                              (113 lb/ft)                      (1,540 N/m)
                        =                                 =
                                       2
                                                                         2
                                                                     6
                                  6
                          24 (2.5 × 10 lb · ft )            24 (1.02 × 10 N · m )
                                                                   4
                                                                           3
                                4
                                       3
                          ×[3(1ft) + 4(1ft) (3ft)           ×[3(0.3m) + 4(0.3m) (1m)
                                  3
                                                                     3
                          −(1ft)(3ft) ]                     −(0.3m)(1m) ]
                             113 lb/ft                        1,540 N/m
                        =                                 =
                                7
                                                                  7
                          6.0 × 10 lb · ft 2                2.45 × 10 N · m 2
                                      4
                                                                              4
                          ×[(3 + 12 − 27) ft ]              ×[(0.0243 + 0.108 − 0.3) m ]

                                    1                                 1
                                                                                4
                                            4
                        =  1.88 × 10 −6  (−12 ft )        =  6.29 × 10 −5  (−0.1677 m )
                                    ft 3                             m 3
                                     12 in                            100 cm
                        =−0.000023 ft ×  ↑                =−0.00001 m ×    ↑
                                      ft                               m
                        = 0.00027 in ↓                    = 0.001 cm ↓
                  2.2.9 Double Overhang: Concentrated
                  Forces at Free Ends
                  The simply-supported beam in Fig. 2.65 has double overhangs with concentrated forces,
                  each of magnitude (F), acting directly downward at the free ends: points A and D. The
                  distance between the supports is labeled (L), and the length of each overhang is labeled
                  (a). Therefore, the total length of the beam, measured from the left end, is (L + 2a).
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