9_ch 11 Mechanical Design budynas_SM_ch11

9_ch 11 Mechanical Design budynas_SM_ch11 -...

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Unformatted text preview: budynas_SM_ch11.qxd 12/04/2006 15:25 FIRST PAGES Page 297 297 Chapter 11 ( C10 ) O 480 = 134.4 0.02 + 4.439[ln(1/0.975)]1/1.483 = 1438 lbf or 6.398 kN 480 ( C10 ) C = 357.6 0.02 + 4.439[ln(1/0.975)]1/1.483 1/3 1/3 = 3825 lbf or 17.02 kN Bearing at O: Choose a deep-groove 02-12 mm. Ans. Bearing at C: Choose a deep-groove 02-30 mm. Ans. There may be an advantage to the identical 02-30 mm bearings in a gear-reduction unit. 11-14 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (Roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust is heavily loaded compared to the other bearing. The second bearing is thus oversized and does not contribute √ measurably to the chance of failure. This is predictable. The reliability goal is not 0.99, but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort. Bearing at A (Ball) Fr = (362 + 2122 ) 1/2 = 215 lbf = 0.957 kN Fa = 555 lbf = 2.47 kN Trial #1: Tentatively select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN . Fa 2.47 = = 0.0392 C0 63.0 xD = 25 000(600)(60) = 900 106 Table 11-1: X 2 = 0.56, Y2 = 1.88 Fe = 0.56(0.957) + 1.88(2.47) = 5.18 kN FD = f A Fe = 1.3(5.18) = 6.73 kN C10 900 = 6.73 0.02 + 4.439[ln(1/0.99)]1/1.483 1/3 = 107.7 kN > 90.4 kN Trial #2: Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN. Fa 2.47 = = 0.029 C0 85 ...
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