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13_ch 14 Mechanical Design budynas_SM_ch14

# 13_ch 14 Mechanical Design budynas_SM_ch14 -...

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Chapter 14 361 Solve Eq. (14-14) for W t : W t = 58 548 2285 2 2 cos 20 1 . 692(2 . 750) = 265 lbf H all = 265(830 . 7) 33 000 = 6 . 67 hp Ans. For 10 8 cycles (turns of pinion), the allowable power is 6.67 hp. (c) Gear fatigue due to bending and wear Bending Eq. (14-3): x = 3 Y G 2 P d = 3(0 . 4103) 2(6) = 0 . 1026 in Eq. ( b ), p. 717: t = 4(0 . 375)(0 . 1026) = 0 . 392 in Eq. (6-25): d e = 0 . 808 2(0 . 392) = 0 . 715 in Eq. (6-20): k b = 0 . 715 0 . 30 0 . 107 = 0 . 911 k c = k d = k e = 1 r d = r f t = 0 . 050 0 . 392 = 0 . 128 Approximate D / d = ∞ by setting D / d = 3 for Fig. A-15-6; K t = 1 . 80 . Use K f = 1.80. k f 2 = 1 1 . 80 = 0 . 556, k f = 1 . 66(0 . 556) = 0 . 923 S e = 0 . 766(0 . 911)(1)(1)(1)(0 . 923)(58) = 37 . 36 kpsi σ all = S e n d = 37 . 36 2 = 18 . 68 kpsi W t = FY G σ all K v P d = 2(0 . 4103)(18 680) 1 . 692(6)
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