2_ch 11 Mechanical Design budynas_SM_ch11

# 2_ch 11 Mechanical Design budynas_SM_ch11 - plications,...

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290 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 11-3 For the straight-Roller 03-series bearing selection, x D = 1440 rating lives from Prob. 11-2 solution. F D = 1 . 4(1650) = 2310 lbf = 10 . 279 kN C 10 = 10 . 279 ± 1440 1 ² 3 / 10 = 91 . 1kN Table 11-3: Select a 03-55 mm with C 10 = 102 kN. Ans. Using Eq. (11-18), R = exp ³ ´ 1440(10 . 28 / 102) 10 / 3 0 . 02 4 . 439 µ 1 . 483 = 0 . 942 Ans. 11-4 We can choose a reliability goal of 0 . 90 = 0 . 95 for each bearing. We make the selec- tions, ﬁnd the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R 1 . Then set the relia- bility goal of the second as R 2 = 0 . 90 R 1 or vice versa. This gives three pairs of selections to compare in terms of cost, geometry im-
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Unformatted text preview: plications, etc. 11-5 Establish a reliability goal of . 90 = . 95 for each bearing. For a 02-series angular con-tact ball bearing, C 10 = 854 1440 . 02 + 4 . 439[ln(1 / . 95)] 1 / 1 . 483 1 / 3 = 11 315 lbf = 50 . 4 kN Select a 02-60 mm angular-contact bearing with C 10 = 55 . 9 kN. R A = exp 1440(3 . 8 / 55 . 9) 3 . 02 4 . 439 1 . 483 = . 969 For a 03-series straight-roller bearing, C 10 = 10 . 279 1440 . 02 + 4 . 439[ln(1 / . 95)] 1 / 1 . 483 3 / 10 = 105 . 2 kN Select a 03-60 mm straight-roller bearing with C 10 = 123 kN. R B = exp 1440(10 . 28 / 123) 10 / 3 . 02 4 . 439 1 . 483 = . 977...
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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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