10_ch 11 Mechanical Design budynas_SM_ch11

10_ch 11 Mechanical Design budynas_SM_ch11 - (Why Ans 11-15...

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298 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Table 11-1: Y 2 = 1 . 98 F e = 0 . 56(0 . 957) + 1 . 98(2 . 47) = 5 . 43 kN F D = 1 . 3(5 . 43) = 7 . 05 kN C 10 = 7 . 05 ± 900 0 . 02 + 4 . 439[ln(1 / 0 . 99)] 1 / 1 . 483 ² 1 / 3 = 113 kN < 121 kN O.K. Select a 02-95 mm angular-contact ball bearing. Ans. Bearing at B (Roller): Any bearing will do since R = 1 . Let’s prove it. From Eq. (11-18) when ³ a f F D C 10 ´ 3 x D < x 0 R = 1 The smallest 02-series roller has a C 10 = 16 . 8kN for a basic load rating. ³ 0 . 427 16 . 8 ´ 3 (900) < ? > 0 . 02 0 . 0148 < 0 . 02 R = 1 Spotting this early avoided rework from 0 . 99 = 0 . 995 . Any 02-series roller bearing will do. Same bore or outside diameter is a common choice.
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Unformatted text preview: (Why?) Ans. 11-15 Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken: b = 1 . 5, θ = 4 . 48 . We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB . In this case, B is to the right of A . For F = 18 kN, ( x ) 1 = 115(2000)(16) 10 6 = 13 . 8 This establishes point 1 on the R = . 90 line. 1 1 2 10 18 1 2 39.6 100 1 10 13.8 72 1 100 x 2 log x F A B log F R ± 0.90 R ± 0.20...
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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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