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                                                                                              Deflection and Stiffness  173
                                               For section BC, with the variable of integration θ defined as shown in Fig. 4–13c, sum-
                                               ming moments about the break gives the moment equation for section BC.

                                                                   M BC = Q(R + R sin θ) + PR sin θ               (4)
                                                                      ∂M BC /∂Q = R(1 + sin θ)                    (5)

                                               From Eq. (4–41), inserting Eqs. (4) and (5) and setting Q = 0, we get

                                                            π/2                           π/2
                                                               1       ∂M BC         R
                                                 (δ D ) BC =       M BC      Rdθ =          (PR sin θ)[R(1 + sin θ)] dx
                                                          0   EI        ∂Q          EI   0
                                                         PR 3     π                                               (6)
                                                       =       1 +
                                                          EI       4
                                               Noting that the break in section CD contains nothing but Q, and after setting Q = 0, we
                                               can conclude that there is no actual strain energy contribution in this section.
                                               Combining terms from Eqs. (3) and (6) to get the total vertical deflection at D,

                                                                            P    2   3  2   1  3  PR  3    π
                                                     δ D = (δ D ) AB + (δ D ) BC =  (2R l + l R + l ) +  1 +
                                                                           EI        2      3      EI      4
                                                                                                                  (7)
                                                           P        3     2        2       3
                                                        =    (1.785R + 2R l + 1.5 Rl + 0.333l )
                                                          EI
                                                                               4
                                               Substituting values, and noting I = πd /64, and E = 207 GPa for steel, we get
                                                             1
                                                                                          2
                                                                                 3
                                     Answer    δ D =                    [1.785(0.05 ) + 2(0.05 )0.04
                                                                  4
                                                          9
                                                    207(10 )[π(0.002 )/64]
                                                                                                      2
                                                                                                                  3
                                                                                         + 1.5(0.05)0.04 + 0.333(0.04 )]
                                                          −3
                                                 = 3.47(10 ) m = 3.47 mm





                            EXAMPLE 4–13       Deflection in a Variable-Cross-Section Punch-Press Frame
                                               The general result expressed in Eq. (4–39),

                                                                        π FR 2   π FR   πCF R
                                                                     δ =      −       +
                                                                         2AeE    2AE     2AG
                                               is useful in sections that are uniform and in which the centroidal locus is circular. The
                                               bending moment is largest where the material is farthest from the load axis.
                                               Strengthening requires a larger second area moment I. A variable-depth cross section is
                                               attractive, but it makes the integration to a closed form very difficult. However, if you
                                               are seeking results, numerical integration with computer assistance is helpful.
                                                  Consider the steel C frame depicted in Fig. 4–14a in which the centroidal radius is
                                               32 in, the cross section at the ends is 2in × 2in, and the depth varies sinusoidally with
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