7_ch 14 Mechanical Design budynas_SM_ch14

# 7_ch 14 Mechanical Design budynas_SM_ch14 - 6 psi Eq(14-13...

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Chapter 14 355 l = 2 . 25 P d = 2 . 25 12 = 0 . 1875 in x = 3 Y P 2 P = 3(0 . 303) 2(12) = 0 . 0379 in t = ± 4(0 . 1875)(0 . 0379) = 0 . 1686 in d e = 0 . 808 ± 0 . 875(0 . 1686) = 0 . 310 in k b = ² 0 . 310 0 . 30 ³ 0 . 107 = 0 . 996 k c = k d = k e = 1, k f 1 = 1 . 66 (see Ex. 14-2) r f = 0 . 300 12 = 0 . 025 in (see Ex. 14-2) r d = r f t = 0 . 025 0 . 1686 = 0 . 148 Approximate D / d =∞ with D / d = 3; from Fig. A-15-6, K t = 1 . 68 . From Fig. 6-20, with S ut = 76 kpsi and r = 0.025 in, q = 0.62. From Eq. (6-32) K f = 1 + 0.62(1.68 1) = 1.42 Miscellaneous-Effects Factor: k f = k f 1 k f 2 = 1 . 65 ² 1 1 . 323 ³ = 1 . 247 S e = 0 . 857(0 . 996)(1)(1)(1)(1 . 247)(38 000) = 40 450 psi σ all = 40 770 2 . 25 = 18 120 psi W t = FY P σ all K v P d = 0 . 875(0 . 303)(18 120) 1 . 162(12) = 345 lbf H = 345(194 . 8) 33 000 = 2 . 04 hp Ans. Wear ν 1 = ν 2 = 0 . 292, E 1 = E 2 = 30(10
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Unformatted text preview: 6 ) psi Eq. (14-13): C p = 1 2 π ² 1 − . 292 2 30(10 6 ) ³ = 2285 ± psi r 1 = d P 2 sin φ = 1 . 417 2 sin 20° = . 242 in r 2 = d G 2 sin φ = 2 . 500 2 sin 20° = . 428 1 r 1 + 1 r 2 = 1 . 242 + 1 . 428 = 6 . 469 in − 1 Eq. (14-12): Eq. (7-17): Eq. (14-3): Eq. ( b ), p. 717: Eq. (6-25): Eq. (6-20):...
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