Shegley solver del 11 en adelante

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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 xD = The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (11-6), C10 540 = 2.278 0.02 + 4.439[ln(1/0.9)]1/1.483 = 18.59 kN Ans.
1/3

30 000(300)(60) = 540 106Ans.

Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN. Ans. Eq. (11-18): 540(2.278/19.5) 3 − 0.02 R = exp − 4.439 = 0.919 Ans. 11-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is xD = 50 000(480)(60) = 1440 106
1.483

The design load is radial and equal to FD = 1.4(610) = 854 lbf = 3.80 kN Eq. (11-6): C10 = 854 1440 0.02 + 4.439[ln(1/0.9)]1/1.4831/3

= 9665 lbf = 43.0 kN Table 11-2: Select a 02-55 mm with C10 = 46.2 kN. Ans. Using Eq. (11-18), 1440(3.8/46.2) 3 − 0.02 R = exp − 4.439 = 0.927 Ans.
1.483

290
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

11-3

For the straight-Roller03-series bearing selection, x D = 1440 rating lives from Prob. 11-2 solution. FD = 1.4(1650) = 2310 lbf = 10.279 kN C10 = 10.279 1440 1
3/10

= 91.1 kN

Table 11-3: Select a 03-55 mm with C10 = 102 kN. Ans. Using Eq. (11-18), 1440(10.28/102) 10/3 − 0.02 R = exp − 4.439 11-4
1.483

= 0.942 Ans.

√ We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the selections, ﬁnd theexisting 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 R1 . Then set the reliability goal of the second as R2 = 0.90 R1

or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. 11-5Establish a reliability goal of tact ball bearing, √ 0.90 = 0.95 for each bearing. For a 02-series angular con1440 0.02 + 4.439[ln(1/0.95)]1/1.483
1/3

C10 = 854

= 11 315 lbf = 50.4 kN Select a 02-60 mm angular-contact bearing with C10 = 55.9 kN. 1440(3.8/55.9) 3 − 0.02 R A = exp − 4.439 For a 03-series straight-roller bearing, C10 = 10.279 1440 0.02 + 4.439[ln(1/0.95)]1/1.483
3/10 1.483

=0.969

= 105.2 kN

Select a 03-60 mm straight-roller bearing with C10 = 123 kN. 1440(10.28/123) 10/3 − 0.02 R B = exp − 4.439
1.483

= 0.977

Chapter 11

291

The overall reliability is R = 0.969(0.977) = 0.947, which exceeds the goal. Note, using R A from this problem, and R B from Prob.11-3, R = 0.969(0.942) = 0.913, which still exceeds the goal. Likewise, using R B from this problem, and R A from Prob. 11-2, R = 0.927(0.977) = 0.906. 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? 11-6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For Fr = 8kN and Fa = 4 kN xD = Eq. (11-5): C10 = 8 270 0.02 + 4.439[ln(1/0.90)]1/1.483
1/3

5000(900)(60) = 270 106 = 51.8 kN

Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with C0 = 37.5 kN. Fa 4 = = 0.107 C0 37.5 Table 11-1: Fa /(V Fr ) = 0.5 > e X 2 = 0.56, Eq. (11-9): Fe = 0.56(1)(8) + 1.46(4) = 10.32 kN Eq. (11-6): For R = 0.90, C10 = 10.32 270 1
1/3

Y2 =1.46

= 66.7 kN > 61.8 kN

Trial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0. Check: Fa 4 = = 0.089 C0 45 Table 11-1: X 2 = 0.56, Y2 = 1.53 Fe = 0.56(8) + 1.53(4) = 10.60 kN Eq. (11-6): C10 = 10.60 ∴ Selection stands. Decision: Specify a 02-80 mm deep-groove ball bearing. Ans. 270 1
1/3

= 68.51 kN < 70.2 kN

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