13_ch 11 Mechanical Design budynas_SM_ch11

# 13_ch 11 Mechanical Design budynas_SM_ch11 - . 3) 3 (10 6 )...

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Chapter 11 301 We can calculate ( C 10 ) m by similar triangles. log( C 10 ) m log 18 log 43 . 2 log 1 = log 39 . 6 log 18 log 13 . 8 log 1 log( C 10 ) m = log 43 . 2 log 13 . 8 log ± 39 . 6 18 ² + log 18 ( C 10 ) m = 55 . 8kN The usefulness of this plot is evident. The improvement is 43 . 2 / 13 . 8 = 3 . 13 fold in life. This result is also available by ( L 10 ) m / ( L 10 ) 1 as 360 / 115 or 3.13 fold, but the plot shows the improvement is for all loading. Thus, the manufacturer’s assertion that there is at least a 3-fold increase in life has been demonstrated by the sample data given. Ans. 11-18 Express Eq. (11-1) as F a 1 L 1 = C a 10 L 10 = K For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C 10 = 20 . 3kN. K = (20
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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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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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