15_ch 10 Mechanical Design budynas_SM_ch10

15_ch 10 Mechanical Design budynas_SM_ch10 -...

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Chapter 10 275 10-22 For a coil radius given by: R = R 1 + R 2 R 1 2 π N θ The torsion of a section is T = PR where dL = Rd θ δ p = U P = 1 GJ ± T T P dL = 1 GJ ± 2 π N 0 PR 3 d θ = P GJ ± 2 π N 0 ² R 1 + R 2 R 1 2 π N θ ³ 3 d θ = P GJ ² 1 4 ³² 2 π N R 2 R 1 ³ ´ ² R 1 + R 2 R 1 2 π N θ ³ 4 µ¶ 2 π N 0 = π PN 2 GJ ( R 2 R 1 ) ( R 4 2 R 4 1 ) = π PN 2 GJ ( R 1 + R 2 ) ( R 2 1 + R 2 2 ) J = π 32 d 4 δ p = 16 PN Gd 4 ( R 1 + R 2 ) ( R 2 1 + R 2 2 ) k = P δ p = d 4 G 16 N ( R 1 + R 2 ) ( R 2 1 + R 2 2 ) Ans . 10-23 For a food service machinery application select A313 Stainless wire. G = 10(10 6 ) psi Note that for 0 . 013 d 0 . 10 in A = 169, m = 0 . 146 0 . 10 < d 0 . 20 in A = 128, m = 0 . 263 F a = 18 4 2 = 7 lbf, F m = 18 + 4 2 = 11 lbf, r = 7 / 11 Try d = 0 . 080 in, S ut = 169 (0 . 08) 0 . 146 = 244 . 4 kpsi S su = 0 . 67 S ut = 163 . 7 kpsi, S sy = 0 . 35 S ut = 85 . 5 kpsi Try unpeened using Zimmerli’s endurance data: S sa = 35 kpsi, S sm = 55 kpsi Gerber: S se = S sa 1 ( S sm / S su ) 2 = 35 1 (55 / 163 . 7) 2 = 39 . 5 kpsi S sa = (7 /
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