Page 64 - Marks Calculation for Machine Design
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P1: Sanjay
                                      16:18
                          January 4, 2005
        Brown.cls
                 Brown˙C02
                                           STRENGTH OF MACHINES
                  46
                    If the distance (a) is greater than the distance (b), then the maximum deflection (  max )
                  caused by this loading configuration is given by Eq. (2.10),
                                     Fb    a (L + b)    3/2       a(L + b)
                               max =                    at  x =                (2.10)
                                    3 EIL    3                      3
                  located at a point to the left of where the force (F) acts. It is clear that the distance (x) for
                  the maximum deflection is not the place where the force (F) acts.
                    If the distance (a) is less than the distance (b), then consider a mirror image of the beam
                  where the distances (a) and (b) swap values.
                    The deflection (  a ) at the point where the force (F) acts, the distance (a), is given by
                  Eq. (2.11),
                                               2 2
                                             Fa b
                                          a =        at  x = a                 (2.11)
                                             3 EIL
                  where it is not entirely obvious that the deflection (  a ) is less than the maximum deflection
                  (  max ).
                            U.S. Customary                       SI/Metric
                  Example 4. Calculate the deflection ( ) of  Example 4. Calculate the deflection ( ) of
                  a simply-supported beam with a concentrated  a simply-supported beam with a concentrated
                  force (F) at an intermediate point a distance  force (F) at an intermediate point a distance
                  (x) of (3L/8), where               (x) of (3L/8), where
                    F = 10 kip = 10,000 lb             F = 45 kN = 45,000 N
                    L = 8 ft, a = 6 ft, b = 2ft        L = 3m, a = 2m, b = 1m
                                                                9
                                                                    2
                                 2
                             6
                    E = 30 × 10 lb/in. (steel)         E = 207 × 10 N/m (steel)
                    I = 4in 4                          I = 201 cm 4
                  solution                           solution
                  Step 1. Determine the distance (x).  Step 1. Determine the distance (x).
                            3L   3 (8ft)                     3L   3 (3m)
                         x =   =      = 3ft               x =   =      = 1.125 m
                             8     8                          8     8
                  Step 2. Calculate the stiffness (EI).  Step 2. Calculate the stiffness (EI).
                                 6
                                     2
                                         4
                                                                              4
                                                                       2
                                                                   9
                       EI = (30 × 10 lb/in )(4in )       EI = (207 × 10 N/m )(201 cm )
                                      2
                                 8
                           1.2 × 10 lb · in × 1ft 2             1m 4
                         =                                   ×
                                 144 in 2                     (100 cm) 4
                                  5
                                                                   5
                         = 8.33 × 10 lb · ft 2             = 4.16 × 10 N · m 2
                  Step 3. Determine the deflection ( ) from  Step 3. Determine the deflection ( ) from
                  Eq. (2.9a).                        Eq. (2.9a).
                            Fbx  2   2   2                   Fbx   2  2   2
                        =       (L − b − x )             =       (L − b − x )
                          6 (EI) L                         6 (EI) L
                            (10,000 lb)(2ft)(3ft)          (45,000 N)(1m)(1.125 m)
                        =                                =
                                      2
                                                                   5
                                  5
                                                                        2
                          6(8.33 × 10 lb · ft )(8ft)       6 (4.16 × 10 N · m )(3m)
                                                                      2
                                                                               2
                                                                2
                                     2
                                           2
                               2
                          × [(8ft) − (2ft) − (3ft) ]       ×[(3m) − (1m) − (1.125 m) ]
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