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                 174    Mechanical Engineering Design
                  Figure 4–14

                  (a) A steel punch press has a
                  C frame with a varying-depth
                                          32-in R            1000 lbf
                  rectangular cross section
                  depicted. The cross section
                                                                                                  1000 lbf
                  varies sinusoidally from
                               ◦
                  2in × 2 in at θ = 0 to
                               ◦
                  2in × 6 in at θ = 90 , and
                                     ◦
                  back to 2in × 2 in at θ = 180 .
                  Of immediate interest to the
                  designer is the deflection in the
                  load axis direction under the              1000 lbf
                  load. (b) Finite-element model.


                                                     (a)                            (b)
                                                                                                           6
                                          an amplitude of 2 in. The load is 1000 lbf. It follows that C = 1.2, G = 11.5(10 ) psi,
                                                   6
                                          E = 30(10 ) psi. The outer and inner radii are
                                                 R out = 33 + 2 sin θ  R in = 31 − 2 sin θ
                                          The remaining geometrical terms are

                                                   h = R out − R in = 2(1 + 2 sin θ) -
                                                   A = bh = 4(1 + 2 sin θ)
                                                           h               2(1 + 2 sin θ)
                                                  r n =          =                             1
                                                       ln(R out /R in )  ln[(33 + 2 sin θ)/(31 − 2 sin θ)]
                                                   e = R − r n = 32 − r n
                                          Note that
                                                          M = FR sin θ         ∂M/∂F = R sin θ
                                                          F θ = F sin θ       ∂F θ /∂F = sin θ
                                                                     2
                                                                                               2
                                                                 2
                                                        MF θ = F R sin θ     ∂MF θ /∂F = 2FR sin θ
                                                          F r = F cos θ        ∂F r /∂F = cos θ
                                          Substitution of the terms into Eq. (4–38) yields three integrals
                                                                                                              (1)
                                                                      δ = I 1 + I 2 + I 3
                                          where the integrals are
                                                                                     2
                                                                     π            sin θ dθ
                                                               −3
                                                   I 1 = 8.5333(10 )            ⎡                    ⎤        (2)
                                                                   0
                                                                                        2(1 + 2 sin θ)
                                                                                ⎢                    ⎥
                                                                      (1 + 2 sin θ) 32 −              ⎥
                                                                                ⎢
                                                                                ⎣        33 + 2 sin θ  ⎦
                                                                                      ln
                                                                                         31 − 2 sin θ
                                                                          2
                                                                       π  sin θ dθ
                                                                 −4
                                                   I 2 =−2.6667(10 )                                          (3)
                                                                     0 1 + 2 sin θ
                                                                         2
                                                                     π  cos θ dθ
                                                               −4
                                                   I 3 = 8.3478(10 )                                          (4)
                                                                   0 1 + 2 sin θ
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