shi20396_ch11 - Chapter 11 11-1 For the deep-groove...

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Chapter 11 11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is x D = 30 000(300)(60) 10 6 = 540 Ans. The design radial load F D is F D = 1 . 2(1 . 898) = 2 . 278 kN From Eq. (11-6), C 10 = 2 . 278 540 0 . 02 + 4 . 439[ln(1 / 0 . 9)] 1 / 1 . 483 1 / 3 = 18 . 59 kN Ans. Table 11-2: Choose a 02-30 mm with C 10 = 19 . 5 kN. Ans. Eq. (11-18): R = exp 540(2 . 278 / 19 . 5) 3 0 . 02 4 . 439 1 . 483 = 0 . 919 Ans. 11-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is x D = 50 000(480)(60) 10 6 = 1440 The design load is radial and equal to F D = 1 . 4(610) = 854 lbf = 3 . 80 kN Eq. (11-6): C 10 = 854 1440 0 . 02 + 4 . 439[ln(1 / 0 . 9)] 1 / 1 . 483 1 / 3 = 9665 lbf = 43 . 0 kN Table 11-2: Select a 02-55 mm with C 10 = 46 . 2 kN. Ans. Using Eq. (11-18), R = exp 1440(3 . 8 / 46 . 2) 3 0 . 02 4 . 439 1 . 483 = 0 . 927 Ans.
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298 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 3 / 10 = 91 . 1 kN Table 11-3: Select a 03-55 mm with C 10 = 102 kN. Ans. Using Eq. (11-18), R = exp 1440(10 . 28 / 102) 10 / 3 0 . 02 4 . 439 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- plications, etc. 11-5 Establish a reliability goal of 0 . 90 = 0 . 95 for each bearing. For a 02-series angular con- tact ball bearing, C 10 = 854 1440 0 . 02 + 4 . 439[ln(1 / 0 . 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 0 . 02 4 . 439 1 . 483 = 0 . 969 For a 03-series straight-roller bearing, C 10 = 10 . 279 1440 0 . 02 + 4 . 439[ln(1 / 0 . 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 0 . 02 4 . 439 1 . 483 = 0 . 977
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Chapter 11 299 Form a table of existing reliabilities R goal R A R B 0.912 0.90 0.927 0.941 0.872 0.95 0.969 0.977 0.947 0.906 The possible products in the body of the table are displayed to the right of the table. One, 0.872, is predictably less than the overall reliability goal. The remaining three are the choices for a combined reliability goal of 0.90. Choices can be compared for the cost of bearings, outside diameter considerations, bore implications for shaft modifications and housing modifications. The point is that the designer has choices. Discover them before making the selection decision. Did the answer to Prob. 11-4 uncover the possibilities? To reduce the work to fill in the body of the table above, a computer program can be helpful. 11-6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For F r = 8 kN and F a = 4 kN x D = 5000(900)(60) 10 6 = 270 Eq. (11-5): C 10 = 8 270 0 . 02 + 4 . 439[ln(1 / 0 . 90)] 1 / 1 . 483 1 / 3 = 51 . 8 kN Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with C 0 = 37 . 5 kN.
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