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                 164    Mechanical Engineering Design



                        EXAMPLE 4–8       A cantilever beam with a round cross section has a concentrated load F at the end, as
                                          shown in Fig. 4–9a. Find the strain energy in the beam.

                                                                                   x
                  Figure 4–9                         l                       F
                                              F
                                          y
                                           max
                                                                                                M
                                                                                           V
                                                     (a)                           (b)

                                Solution  To determine what forms of strain energy are involved with the deflection of the beam, we
                                          break into the beam and draw a free-body diagram to see the forces and moments being
                                          carried within the beam. Figure 4–9b shows such a diagram in which the transverse shear
                                          is V =−F, and the bending moment is M =−Fx. The variable x is simply a variable of
                                          integration and can be defined to be measured from any convenient point. The same results
                                          will be obtained from a free-body diagram of the right-hand portion of the beam with x
                                          measured from the wall. Using the free end of the beam usually results in reduced effort
                                          since the ground reaction forces do not need to be determined.
                                              For the transverse shear, using Eq. (4–24) with the correction factor C = 1.11 from
                                          Table 4–2, and noting that V is constant through the length of the beam,
                                                                                     2
                                                                            2
                                                                         CV l   1.11F l
                                                                  U shear =   =
                                                                         2AG      2AG
                                          For the bending, since M is a function of x, Eq. (4–23) gives
                                                                    2            l            2 3
                                                                  M dx      1          2     F l
                                                         U bend =       =        (−Fx) dx =
                                                                   2EI    2EI  0             6EI
                                          The total strain energy is

                                                                                 2 3
                                                                                           2
                                                                                F l   1.11F l
                                Answer                       U = U bend + U shear =  +
                                                                                6EI     2AG
                                          Note, except for very short beams, the shear term (of order l) is typically small com-
                                                                       3
                                          pared to the bending term (of order l ). This will be demonstrated in the next example.







                                  4–8     Castigliano’s Theorem
                                          A most unusual, powerful, and often surprisingly simple approach to deflection analysis
                                          is afforded by an energy method called Castigliano’s theorem. It is a unique way of ana-
                                          lyzing deflections and is even useful for finding the reactions of indeterminate structures.
                                          Castigliano’s theorem states that when forces act on elastic systems subject to small dis-
                                          placements, the displacement corresponding to any force, in the direction of the force, is
                                          equal to the partial derivative of the total strain energy with respect to that force. The
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