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                 178    Mechanical Engineering Design
                                           3 Since  M 1 is the redundant reaction at  O, write the equation for the angular
                                              deflection at point O. From Castigliano’s theorem this is

                                                                              ∂U
                                                                        θ O =                                 (6)
                                                                             ∂M 1
                                          We can apply Eq. (4–31), using the variable x as shown in Fig. 4–16b. However, sim-
                                          pler terms can be found by using a variable ˆx that starts at B and is positive to the left.
                                          With this and the expression for R 2 from Eq. (5) the moment equations are

                                                              F    M 1                           l

                                                                                             x
                                                        M =     −      ˆ x               0 ≤ˆ ≤               (7)
                                                              2    l                             2

                                                              F    M 1            l      l
                                                                                             x
                                                        M =     −      ˆ x − F x −         ≤ˆ ≤ l             (8)
                                                                              ˆ
                                                              2    l              2      2
                                          For both equations
                                                                        ∂M      ˆ x
                                                                            =−                                (9)
                                                                        ∂M 1    l
                                          Substituting Eqs. (7) to (9) in Eq. (6), using the form of Eq. (4–31) where F i = M 1 , gives


                                                    ∂U     1       l/2     F  M 1       ˆ x        l      F  M 1
                                                                                       x
                                              θ O =     =              −      ˆ x −  d ˆ +        −      ˆ x
                                                   ∂M 1   EI    0    2    l       l        l/2  2    l
                                                                        l      ˆ x

                                                                                    x
                                                                    ˆ
                                                               − F x −       −    d ˆ = 0
                                                                        2      l
                                          Canceling 1/EIl, and combining the first two integrals, simplifies this quite readily to
                                                          F   M 1      2               l
                                                                      l          l
                                                                                            x
                                                                         x
                                                            −         ˆ x d ˆ− F   ˆ x −  ˆ xd ˆ = 0
                                                          2    l    0          l/2    2
                                          Integrating gives
                                                                               3
                                                                                               2
                                                      F    M 1  l   F  3    l       Fl  2    l
                                                                3
                                                        −         −    l −       +     l −       = 0
                                                      2    l   3    3       2       4       2
                                          which reduces to
                                                                              3Fl
                                                                        M 1 =                               (10)
                                                                              16
                                           4   Substituting Eq. (10) into (5) results in

                                                                       11F          5F
                                                                  R 1 =        R 2 =                        (11)
                                                                        16          16
                                          which again agrees with beam 11 of Table A–9.
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