10_ch 14 Mechanical Design budynas_SM_ch14

10_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 358 12/05/2006 17:39 Page 358 FIRST PAGES Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design kb = 24.23 7.62 −0.107 = 0.884 kc = kd = ke = 1 r f = 0.300m = 0.300(6) = 1.8 mm From Fig. A-15-6 for r /d = r f / t = 1.8/12 = 0.15, K t = 1.68. Figure 6-20, q = 0.86; Eq. (6-32), K f = 1 + 0.86(1.68 − 1) = 1.58 k f 1 = 1.66 (Gerber failure criterion) k f 2 = 1/ K f = 1/1.537 = 0.651 k f = k f 1 k f 2 = 1.66(0.651) = 1.08 Se = 0.744(0.884)(1)(1)(1)(1.08)(450) = 319.6 MPa σall = Eq. (14-8): Se 319.6 = = 245.8 MPa nd 1.3 Wt = FY m σall 75(0.296)(6)(245.8) = = 16 840 N Kv 1.944 H= 16 840(96/2)(1145) Tn = = 96.9 kW 9.55 9.55(106 ) Ans . Wear : Pinion and gear r1 = (96/2) sin 20◦ = 16.42 mm Eq. (14-12): r2 = (180/2) sin 20◦ = 30.78 mm Eq. (14-13), with E = 207(103) MPa and ν = 0.292, gives √ 1 = 190 MPa Cp = 2π (1 − 0.2922 ) /(207 × 103 ) Eq. (6-68): SC = 6.89[0.4(260) − 10] = 647.7 MPa SC 647.7 σC , all = − √ = − √ = −568 MPa n 1.3 W= = Eq. (14-14): σC , all Cp −568 191 t 2 2 F cos φ 1 K v 1/ r1 + 1/ r2 75 cos 20◦ 1.944 1 1/16.42 + 1/30.78 = 3433 N W t dP 3433(96) = = 164 784 N · mm = 164.8 N · m 2 2 164.8(1145) Tn = = 19 758.7 W = 19.8 kW Ans . H= 9.55 9.55 Thus, wear controls the gearset power rating; H = 19.8 kW. Ans . T= ...
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