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4.6 Application to deflection problems  83


                                         w/un i t length        Pf   .
                                                                  -1

                                                                    AT
                                                                  t
                                                 L                   I

               Fig. 4.10  Deflection of a uniformly loaded cantilever by the method of complementary energy.


               length (see Fig. 4.10). First we apply a fictitious load Pf at the point where the deflec-
               tion is required. The total complementary energy of the system is
                                                           c
                                   C = IL 1;  dt'dM - ATP, -  Awdz

               where the symbols take their previous meanings and A is the vertical deflection of any
               point on the beam. Then

                                                                                 (4.20)

               As before
                                                   M
                                              dt' = - dz
                                                  EI
               but
                                                 W2
                                       M  = PfZ + - (Pf = 0)
                                                  2
               Hence




               Substituting in Eq. (4.20) for de, M  and dM/dPf, and remembering that Pf = 0, we
               have




               giving
                                                    wL4
                                              AT  =-
                                                    8 EI
                 It will be noted that here, unlike the method for the solution of the pin-jointed
               framework, the fictitious load is applied to the loaded beam. There is, however, no
               arithmetical advantage to  be  gained by  the former approach  although  the  result
              would obviously be the same since M  would equal w2/2 and aM/aPf would have
               the value z.
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