Page 179 - Shigley's Mechanical Engineering Design
P. 179

bud29281_ch04_147-211.qxd  11/27/09  2:55PM  Page 154 ntt 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles:







                 154    Mechanical Engineering Design
                                          the results algebraically. Superposition may be applied provided: (1) each effect is
                                          linearly related to the load that produces it, (2) a load does not create a condition that
                                          affects the result of another load, and (3) the deformations resulting from any spe-
                                          cific load are not large enough to appreciably alter the geometric relations of the
                                          parts of the structural system.
                                              The following examples are illustrations of the use of superposition.



                        EXAMPLE 4–2       Consider the uniformly loaded beam with a concentrated force as shown in Fig. 4–3.
                                          Using superposition, determine the reactions and the deflection as a function of x.

                                Solution  Considering each load state separately, we can superpose beams 6 and 7 of Table A–9.
                                          For the reactions we find
                                                                            Fb   wl
                                Answer                                 R 1 =   +
                                                                             l    2
                                                                            Fa   wl
                                Answer                                 R 2 =   +
                                                                             l    2
                                              The loading of beam 6 is discontinuous and separate deflection equations are given
                                          for regions AB and BC. Beam 7 loading is not discontinuous so there is only one equa-
                                          tion. Superposition yields

                                                           Fbx                wx
                                                                                          3
                                                                                      2
                                                                 2
                                                                         2
                                                                                              3
                                                                     2
                                Answer               y AB =    (x + b − l ) +     (2lx − x − l )
                                                           6EIl              24EI
                                                           Fa(l − x)  2  2          wx     2   3   3
                                Answer               y BC =        (x + a − 2lx) +      (2lx − x − l )
                                                             6EIl                  24EI
                  Figure 4–3               y
                                                       l
                                                    F
                                               a           b
                                                       w
                                                                    C
                                          A                            x
                                                   B
                                            R 1                     R 2


                                              If the maximum deflection of a beam is desired, it will occur either where the slope
                                          is zero or at the end of the overhang if the beam has a free end. In the previous example,
                                          there is no overhang, so setting dy/dx = 0 will yield the equation for x that locates
                                          where the maximum deflection occurs. In the example there are two equations for y
                                          where only one will yield a solution. If a = l/2, the maximum deflection would obvi-
                                          ously occur at x = l/2 because of symmetry. However, if a < l/2, where would the
                                          maximum deflection occur? It can be shown that as F moves toward the left support,
                                          the maximum deflection moves toward the left support also, but not as much as F (see
                                          Prob. 4–34). Thus, we would set dy BC /dx = 0 and solve for x. If a > l/2, then we
                                          would set dy AB dx = 0. For more complicated problems, plotting the equations using
                                          numerical data is the simplest approach to finding the maximum deflection.
   174   175   176   177   178   179   180   181   182   183   184