13_ch 14 Mechanical Design budynas_SM_ch14

13_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 12/05/2006 17:39 Page 361 FIRST PAGES 361 Chapter 14 Solve Eq. (14-14) for W t: Wt = Hall = −58 548 2285 2 2 cos 20◦ = 265 lbf 1.692(2.750) 265(830.7) = 6.67 hp 33 000 Ans. For 108 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(0.4103) 3YG = = 0.1026 in 2 Pd 2(6) Eq. (b), p. 717: t= 4(0.375)(0.1026) = 0.392 in Eq. (6-25): de = 0.808 2(0.392) = 0.715 in Eq. (6-20): kb = 0.715 0.30 −0.107 = 0.911 kc = kd = ke = 1 rf r 0.050 = = = 0.128 d t 0.392 Approximate D /d = ∞ by setting D /d = 3 for Fig. A-15-6; K t = 1.80. Use K f = 1.80. kf 2 = 1 = 0.556, k f = 1.66(0.556) = 0.923 1.80 Se = 0.766(0.911)(1)(1)(1)(0.923)(58) = 37.36 kpsi σall = Se 37.36 = = 18.68 kpsi nd 2 Wt = FYG σall 2(0.4103)(18 680) = = 1510 lbf K v − Pd 1.692(6) Hall = 1510(830.7) = 38.0 hp 33 000 Ans. The gear is thus stronger than the pinion in bending. Wear Since the material of the pinion and the gear are the same, and the contact stresses are the same, the allowable power transmission of both is the same. Thus, Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for 108 /3 revolutions. ...
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