8_ch 11 Mechanical Design budynas_SM_ch11

# 8_ch 11 Mechanical Design budynas_SM_ch11 - 82 1(16 −...

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296 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 11-16: f T = 0 . 8 Fig. 11-17: f V = 1 . 07 Thus, a 3 l = f T f V = 0 . 8(1 . 07) = 0 . 856 Individual reliability: R i = 0 . 9 = 0 . 95 Eq. (11-17): ( C 10 ) A = 1 . 4(3 . 29) ± 40 000(400)(60) 4 . 48(0 . 856)(1 0 . 95) 2 / 3 (90)(10 6 ) ² 0 . 3 = 11 . 40 kN ( C 10 ) B = 1 . 4(4 . 873) ± 40 000(400)(60) 4 . 48(0 . 856)(1 0 . 95) 2 / 3 (90)(10 6 ) ² 0 . 3 = 16 . 88 kN From Fig. 11-15, choose cone 32 305 and cup 32 305 which provide F r = 17 . 4kN and K = 1 . 95 . With K = 1 . 95 for both bearings, a second trial validates the choice of cone 32 305 and cup 32 305. Ans. 11-13 R = 0 . 95 = 0 . 975 T = 240(12)(cos 20 ) = 2706 lbf · in F = 2706 6 cos 25 = 498 lbf In xy -plane: ³ M O =−
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Unformatted text preview: 82 . 1(16) − 210(30) + 42 R y C = R y C = 181 lbf R y O = 82 + 210 − 181 = 111 lbf In xz-plane: ³ M O = 226(16) − 452(30) − 42 R z c = R z C = − 237 lbf R z O = 226 − 451 + 237 = 12 lbf R O = (111 2 + 12 2 ) 1 / 2 = 112 lbf Ans. R C = (181 2 + 237 2 ) 1 / 2 = 298 lbf Ans. F eO = 1 . 2(112) = 134 . 4 lbf F eC = 1 . 2(298) = 357 . 6 lbf x D = 40 000(200)(60) 10 6 = 480 z 14" 16" 12" R z O R z C R y O A B C R y C O 451 210 226 T T 82.1 x y...
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