2_ch 15 Mechanical Design budynas_SM_ch15

# 2_ch 15 Mechanical Design budynas_SM_ch15 - 1(1(1 118 = 120...

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380 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design W t G = 10 930(1 . 25)(1)(0 . 216) 6(1)(1 . 374)(0 . 5222)(1 . 106) = 620 lbf H 2 = 620(785 . 3) 33 000 = 14 . 8hp Ans. The gear controls the bending rating. 15-2 Refer to Prob. 15-1 for the gearset speciﬁcations. Wear Fig. 15-12: s ac = 341(300) + 23 620 = 125 920 psi For the pinion, C H = 1 . From Prob. 15-1, C R = 1 . 118. Thus, from Eq. (15-2): ( σ c ,all ) P = s ac ( C L ) P C H S H K T C R ( σ c ,all ) P = 125 920(1)(1) 1(1)(1 . 118) = 112 630 psi For the gear, from Eq. (15-16), B 1 = 0 . 008 98(300 / 300) 0 . 008 29 = 0 . 000 69 C H = 1 + 0 . 000 69(3 1) = 1 . 001 38 And Prob. 15-1, ( C L ) G = 1 . 0685 . Equation (15-2) thus gives ( σ c ,all ) G = s ac ( C L ) G C H S H K T C R ( σ c ,all ) G = 125 920(1 . 0685)(1 . 001 38)
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Unformatted text preview: 1(1)(1 . 118) = 120 511 psi For steel: C p = 2290 ± psi Eq. (15-9): C s = . 125(1 . 25) + . 4375 = . 593 75 Fig. 15-6: I = . 083 Eq. (15-12): C xc = 2 Eq. (15-1): W t P = ² ( σ c ,all ) P C p ³ 2 Fd P I K o K v K m C s C xc = ² 112 630 2290 ³ 2 ´ 1 . 25(3 . 333)(0 . 083) 1(1 . 374)(1 . 106)(0 . 5937)(2) µ = 464 lbf H 3 = 464(785 . 3) 33 000 = 11 . 0 hp W t G = ² 120 511 2290 ³ 2 ´ 1 . 25(3 . 333)(0 . 083) 1(1 . 374)(1 . 106)(0 . 593 75)(2) µ = 531 lbf H 4 = 531(785 . 3) 33 000 = 12 . 6 hp...
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