12_ch 14 Mechanical Design budynas_SM_ch14

# 12_ch 14 Mechanical Design budynas_SM_ch14 - In terms of...

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Estimate D / d =∞ by setting D / d = 3, K t = 1 . 68 . From Fig. 6-20, q = 0.86, and Eq. (6-32) K f = 1 + 0 . 86(1 . 68 1) = 1 . 58 k f 2 = 1 K f = 1 1 . 58 = 0 . 633 k f = k f 1 k f 2 = 1 . 66(0 . 633) = 1 . 051 S e = 0 . 766(0 . 919)(1)(1)(1)(1 . 051)(58) = 42 . 9 kpsi σ all = S e n d = 42 . 9 2 = 21 . 5 kpsi W t = FY P σ all K v P d = 2(0 . 303)(21 500) 1 . 692(6) = 1283 lbf H = W t V 33 000 = 1283(830 . 7) 33 000 = 32 . 3hp Ans. (b) Pinion fatigue Wear From Table A-5 for steel: ν = 0 . 292, E = 30(10 6 ) psi Eq. (14-13) or Table 14-8: C p = ± 1 2 π [(1 0 . 292 2 ) / 30(10 6 )] ² 1 / 2 = 2285 ³ psi In preparation for Eq. (14-14): Eq. (14-12): r 1 = d P 2 sin φ = 2 . 833 2 sin 20 = 0 . 485 in r 2 = d G 2 sin φ = 8 . 500 2 sin 20 = 1 . 454 in ´ 1 r 1 + 1 r 2 µ = 1 0 . 485 + 1 1 . 454 = 2 . 750 in Eq. (6-68): ( S C ) 10 8 = 0 . 4 H B 10 kpsi
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Unformatted text preview: In terms of gear notation σ C = [0 . 4(232) − 10]10 3 = 82 800 psi We will introduce the design factor of n d = 2 and because it is a contact stress apply it to the load W t by dividing by √ 2. σ C ,all = − σ c √ 2 = − 82 800 √ 2 = − 58 548 psi 360 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design...
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