6_ch 14 Mechanical Design budynas_SM_ch14

# 6_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 354 12/05/2006 17:39 Page 354 FIRST PAGES Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 14-13 24 48 = 4.8 in, dG = = 9.6 in 5 5 π (4.8)(50) = = 62.83 ft/min 12 600 + 62.83 = = 1.105 600 63 025 H = = 525.2 H 50(4.8/2) √ = 1960 psi (see note in Prob. 14-11 solution) dP = V Kv Eq. (14-4a): Wt Cp Table 14-8: Eq. (14-12): Eq. (14-14): 4.8 sin 20◦ = 0.821 in, r2 = 2r1 = 1.642 in 2 1/2 1 1.105(525.2 H ) 1 3 −100(10 ) = −1960 + 2.5 cos 20◦ 0.821 1.642 r1 = H = 5.77 hp Ans. 14-14 d P = 4(20) = 80 mm, dG = 4(32) = 128 mm π (80)(10−3 )(1000) = 4.189 m/s 60 3.05 + 4.189 Kv = = 2.373 3.05 V= Eq. (14-6 a): 60(10)(103 ) = 2387 N π (80)(10−3 )(1000) √ C p = 163 MPa (see note in Prob. 14-11 solution) Wt = Table 14-8: Eq. (14-12): Eq. (14-14): r1 = 80 sin 20° = 13.68 mm, 2 σC = −163 2.373(2387) 50 cos 20° r2 = 128 sin 20° = 21.89 mm 2 1 1 + 13.68 21.89 1/2 = −617 MPa 14-15 The pinion controls the design. Eq. (6-8): Y P = 0.303, YG = 0.359 17 30 dP = = 1.417 in, dG = = 2.500 in 12 12 π dP n π (1.417)(525) V= = = 194.8 ft/min 12 12 1200 + 194.8 = 1.162 Kv = 1200 Se = 0.5(76) = 38 kpsi Eq. (6-19): ka = 2.70(76) −0.265 = 0.857 Bending Eq. (14-4b): Ans . ...
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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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