6_ch 11 Mechanical Design budynas_SM_ch11

6_ch 11 Mechanical Design budynas_SM_ch11 - 1 / 1 . 483 ²...

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294 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design x D = 30 000(300)(60 / 10 6 ) = 540 C 10 = 496 ± 540 0 . 02 + 4 . 439[ln(1 / 0 . 96)] 1 / 1 . 483 ² 1 / 3 = 4980 lbf = 22 . 16 kN A 02-35 bearing will do. Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O . Check combined reliability. Ans. 11-10 For a combined reliability goal of 0.90, use 0 . 90 = 0 . 95 for the individual bearings. x 0 = 50 000(480)(60) 10 6 = 1440 The resultant of the given forces are R O = [( 387) 2 + 467 2 ] 1 / 2 = 607 lbf and R B = [316 2 + ( 1615) 2 ] 1 / 2 = 1646 lbf. At O : F e = 1 . 4(607) = 850 lbf Ball: C 10 = 850 ± 1440 0 . 02 + 4 . 439[ln(1 / 0 . 95)]
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Unformatted text preview: 1 / 1 . 483 ² 1 / 3 = 11 262 lbf or 50.1 kN Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kN. Ans. At B : F e = 1 . 4(1646) = 2304 lbf Roller: C 10 = 2304 ± 1440 . 02 + 4 . 439[ln(1 / . 95)] 1 / 1 . 483 ² 3 / 10 = 23 576 lbf or 104.9 kN Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series roller has the same bore as the 02-series ball. Ans. z 20 16 10 O F A R O R B B A C y x F C 20 ±...
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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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