12_ch 11 Mechanical Design budynas_SM_ch11

12_ch 11 Mechanical Design budynas_SM_ch11 - G ( F = 18 kN,...

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300 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Shaft b F r C = (874 2 + 2274 2 ) 1 / 2 = 2436 lbf or 10.84 kN F r D = (393 2 + 657 2 ) 1 / 2 = 766 lbf or 3.41 kN The bearing at C controls x D = 10 000(240)(60) 10 6 = 144 C 10 = 1 . 2(10 . 84) ± 144 0 . 0826 ² 0 . 3 = 122 kN Select either a 02-90 mm with C 10 = 142 kN or a 03-60 mm with C 10 = 123 kN. Ans. Shaft c F r E = (1113 2 + 2385 2 ) 1 / 2 = 2632 lbf or 11.71 kN F r F = (417 2 + 895 2 ) 1 / 2 = 987 lbf or 4.39 kN The bearing at E controls x D = 10 000(80)(60 / 10 6 ) = 48 C 10 = 1 . 2(11 . 71) ± 48 0 . 0826 ² 0 . 3 = 94 . 8kN Select a 02-80 mm with C 10 = 106 kN or a 03-60 mm with C 10 = 123 kN. Ans. 11-17 The horizontal separation of the R = 0 . 90 loci in a log F -log x plot such as Fig. 11-5 will be demonstrated. We refer to the solution of Prob. 11-15 to plot point
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Unformatted text preview: G ( F = 18 kN, x G = 13 . 8) . We know that ( C 10 ) 1 = 39 . 6 kN, x 1 = 1 . This establishes the unim-proved steel R = . 90 locus, line AG . For the improved steel ( x m ) 1 = 360(2000)(60) 10 6 = 43 . 2 We plot point G ( F = 18 kN, x G = 43 . 2), and draw the R = . 90 locus A m G parallel to AG 1 1 2 10 18 G G 39.6 55.8 100 1 10 13.8 1 100 2 x log x F A A m Improved steel log F Unimproved steel 43.2 R 0.90 R 0.90 1 3 1 3...
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