Unformatted text preview: . 3) 3 (10 6 ) = 8 . 365(10 9 ) At a load of 18 kN, life L 1 is given by: L 1 = K F a 1 = 8 . 365(10 9 ) 18 3 = 1 . 434(10 6 ) rev For a load of 30 kN, life L 2 is: L 2 = 8 . 365(10 9 ) 30 3 = . 310(10 6 ) rev In this case, Eq. (7-57) – the Palmgren-Miner cycle ratio summation rule – can be ex-pressed as l 1 L 1 + l 2 L 2 = 1 Substituting, 200 000 1 . 434(10 6 ) + l 2 . 310(10 6 ) = 1 l 2 = . 267(10 6 ) rev Ans. 11-19 Total life in revolutions Let: l = total turns f 1 = fraction of turns at F 1 f 2 = fraction of turns at F 2...
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- Fall '11
- Shingley
- Ball bearing
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