9_ch 14 Mechanical Design budynas_SM_ch14

9_ch 14 Mechanical Design budynas_SM_ch14 - V = d P n 12 =...

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Chapter 14 357 Wear C p = 2285 ± psi r 1 = (6 . 667 / 2) sin 20° = 1 . 140 in r 2 = (33 . 333 / 2) sin 20° = 5 . 700 in S C = [0 . 4(262) 10](10 3 ) = 94 800 psi σ C ,all =− S C / n d =− 94 800 / 1 . 5 =− 77 404 psi W t = ² σ C ,all C p ³ 2 F cos φ K v 1 1 / r 1 + 1 / r 2 = ² 77 404 2300 ³ 2 ² 2 . 5 cos 20° 2 . 266 ³² 1 1 / 1 . 140 + 1 / 5 . 700 ³ = 1115 lbf H = W t V 33 000 = 1115(1519) 33 000 = 51 . 3hp Ans. For 10 8 cycles (revolutions of the pinion), the power based on wear is 51.3 hp. Rating power–pinion controls H 1 = 144 hp H 2 = 51 . 3hp H rated = min(144, 51 . 3) = 51 . 3hp Ans. 14-17 Given: φ = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, N P = 16 milled teeth, N G = 30 T , S ut = 900 MPa, H B = 260, n d = 3, Y P = 0 . 296, and Y G = 0 . 359 . Pinion bending d P = mN P = 6(16) = 96 mm d G = 6(30) = 180 mm
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Unformatted text preview: V = d P n 12 = (96)(1145)(10 3 )(12) (12)(60) = 5 . 76 m/s Eq. (14-6 b ): K v = 6 . 1 + 5 . 76 6 . 1 = 1 . 944 S e = . 5(900) = 450 MPa a = 4 . 45, b = . 265 k a = 4 . 51(900) . 265 = . 744 l = 2 . 25 m = 2 . 25(6) = 13 . 5 mm x = 3 Ym / 2 = 3(0 . 296)6 / 2 = 2 . 664 mm t = 4 lx = 4(13 . 5)(2 . 664) = 12 . 0 mm d e = . 808 75(12 . 0) = 24 . 23 mm Eq. (14-13): Eq. (14-12): Eq. (6-68):...
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