3_ch 11 Mechanical Design budynas_SM_ch11

3_ch 11 Mechanical Design budynas_SM_ch11 - F a C = 4 37 5...

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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 de- cision. 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 F r = 8kN and F a = 4kN 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 ² 1 / 3 = 51 . 8kN Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with C 0 = 37 . 5kN.
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Unformatted text preview: F a C = 4 37 . 5 = . 107 Table 11-1: F a / ( V F r ) = . 5 > e X 2 = . 56, Y 2 = 1 . 46 Eq. (11-9): F e = . 56(1)(8) + 1 . 46(4) = 10 . 32 kN Eq. (11-6): For R = . 90, C 10 = 10 . 32 ³ 270 1 ´ 1 / 3 = 66 . 7 kN > 61 . 8 kN Trial #2: From Table 11-2 choose a 02-80 mm having C 10 = 70 . 2 and C = 45 . . Check: F a C = 4 45 = . 089 Table 11-1: X 2 = . 56, Y 2 = 1 . 53 F e = . 56(8) + 1 . 53(4) = 10 . 60 kN Eq. (11-6): C 10 = 10 . 60 ³ 270 1 ´ 1 / 3 = 68 . 51 kN < 70 . 2 kN ∴ Selection stands. Decision: Specify a 02-80 mm deep-groove ball bearing. Ans....
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