11_ch 11 Mechanical Design budynas_SM_ch11

11_ch 11 Mechanical Design budynas_SM_ch11 - (0 99 1 6 =...

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Chapter 11 299 The R = 0 . 20 locus is above and parallel to the R = 0 . 90 locus. For the two-parameter Weibull distribution, x 0 = 0 and points A and B are related by [see Eq. (20-25)]: x A = θ [ln(1 / 0 . 90)] 1 / b (1) x B = θ [ln(1 / 0 . 20)] 1 / b and x B / x A is in the same ratio as 600 / 115 . Eliminating θ b = ln[ln(1 / 0 . 20) / ln(1 / 0 . 90)] ln(600 / 115) = 1 . 65 Ans. Solving for θ in Eq. (1) θ = x A [ln(1 / R A )] 1 / 1 . 65 = 1 [ln(1 / 0 . 90)] 1 / 1 . 65 = 3 . 91 Ans. Therefore, for the data at hand, R = exp ± ² x 3 . 91 ³ 1 . 65 ´ Check R at point B : x B = (600 / 115) = 5 . 217 R = exp ± ² 5 . 217 3 . 91 ³ 1 . 65 ´ = 0 . 20 Note also, for point 2 on the R = 0 . 20 line. log(5 . 217) log(1) = log( x m ) 2 log(13 . 8) ( x m ) 2 = 72 11-16 This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of
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Unformatted text preview: (0 . 99) 1 / 6 = . 9983 . Shaft a F r A = (239 2 + 111 2 ) 1 / 2 = 264 lbf or 1.175 kN F r B = (502 2 + 1075 2 ) 1 / 2 = 1186 lbf or 5.28 kN Thus the bearing at B controls x D = 10 000(1200)(60) 10 6 = 720 . 02 + 4 . 439[ln(1 / . 9983)] 1 / 1 . 483 = . 080 26 C 10 = 1 . 2(5 . 2) ² 720 . 080 26 ³ . 3 = 97 . 2 kN Select either a 02-80 mm with C 10 = 106 kN or a 03-55 mm with C 10 = 102 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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