4_ch 11 Mechanical Design budynas_SM_ch11

4_ch 11 Mechanical Design budynas_SM_ch11 - = 11 . 12 270 ....

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292 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 11-7 From Prob. 11-6, x D = 270 and the final value of F e is 10 . 60 kN. C 10 = 10 . 6 ± 270 0 . 02 + 4 . 439[ln(1 / 0 . 96)] 1 / 1 . 483 ² 1 / 3 = 84 . 47 kN Table 11-2: Choose a deep-groove ball bearing, based upon C 10 load ratings. Trial #1: Tentatively select a 02-90 mm. C 10 = 95 . 6, C 0 = 62 kN F a C 0 = 4 62 = 0 . 0645 From Table 11-1, interpolate for Y 2 . F a / C 0 Y 2 0.056 1.71 0.0645 Y 2 0.070 1.63 Y 2 1 . 71 1 . 63 1 . 71 = 0 . 0645 0 . 056 0 . 070 0 . 056 = 0 . 607 Y 2 = 1 . 71 + 0 . 607(1 . 63 1 . 71) = 1 . 661 F e = 0 . 56(8) + 1 . 661(4) = 11 . 12 kN C 10
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Unformatted text preview: = 11 . 12 270 . 02 + 4 . 439[ln(1 / . 96)] 1 / 1 . 483 1 / 3 = 88 . 61 kN < 95 . 6 kN Bearing is OK. Decision: Specify a deep-groove 02-90 mm ball bearing. Ans. 11-8 For the straight cylindrical roller bearing specied with a service factor of 1, R = 0.90 and F r = 12 kN x D = 4000(750)(60) 10 6 = 180 C 10 = 12 180 1 3 / 10 = 57 . 0 kN 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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