11_ch 14 Mechanical Design budynas_SM_ch14

11_ch 14 Mechanical Design budynas_SM_ch14 - 337 = 663 in...

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Chapter 14 359 14-18 Preliminaries: N P = 17, N G = 51 d P = N P d = 17 6 = 2 . 833 in d G = 51 6 = 8 . 500 in V = π d P n / 12 = π (2 . 833)(1120) / 12 = 830 . 7 ft/min Eq. (14-4 b ): K v = (1200 + 830 . 7) / 1200 = 1 . 692 σ all = S y n d = 90 000 2 = 45 000 psi Table 14-2: Y P = 0 . 303, Y G = 0 . 410 Eq. (14-7): W t = FY P σ all K v P d = 2(0 . 303)(45 000) 1 . 692(6) = 2686 lbf H = W t V 33 000 = 2686(830 . 7) 33 000 = 67 . 6hp Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue Bending Eq. (2-17): S ut . = 0 . 5 H B = 0 . 5(232) = 116 kpsi Eq. (6-8): S ± e = 0 . 5 S ut = 0 . 5(116) = 58 kpsi Eq. (6-19): a = 2 . 70, b =− 0 . 265, k a = 2 . 70(116) 0 . 265 = 0 . 766 Table 13-1: l = 1 P d + 1 . 25 P d = 2 . 25 P d = 2 . 25 6 = 0 . 375 in Eq. (14-3): x = 3 Y P 2 P d = 3(0 . 303) 2(6) = 0 . 0758 Eq. ( b ), p. 717: t = 4 lx = ± 4(0 . 375)(0 . 0758) = 0 . 337 in Eq. (6-25): d e = 0 . 808 Ft = 0 . 808 ± 2(0
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Unformatted text preview: . 337) = . 663 in Eq. (6-20): k b = ² . 663 . 30 ³ − . 107 = . 919 k c = k d = k e = 1 . Assess two components contributing to k f . First, based upon one-way bending and the Gerber failure criterion, k f 1 = 1 . 66 (see Ex. 14-2). Second, due to stress-concentration, r f = . 300 P d = . 300 6 = . 050 in (see Ex. 14-2) Fig. A-15-6: r d = r f t = . 05 . 338 = . 148...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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