11_ch 15 Mechanical Design budynas_SM_ch15

# 11_ch 15 Mechanical Design budynas_SM_ch15 - . 71) + . 4375...

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Chapter 15 389 Analyze for 10 9 pinion cycles at 0.999 reliability Eq. (15-19): K R = 0 . 50 0 . 25 log(1 0 . 999) = 1 . 25 C R = ± K R = 1 . 25 = 1 . 118 Bending Pinion: Eq. (15-23): ( s at ) P = 44(300) + 2100 = 15 300 psi Eq. (15-4): ( s w t ) P = 15 300(0 . 862) 1(1)(1 . 25) = 10 551 psi Eq. (15-3): W t = ( s w t ) P FK x J P P d K o K v K s K m = 10 551(0 . 71)(1)(0 . 241) 10(1)(1 . 412)(0 . 508)(1 . 252) = 201 lbf H 1 = 201(628 . 3) 33 000 = 3 . 8 hp Gear: ( s at ) G = 15 300 psi Eq. (15-4): ( s w t ) G = 15 300(0 . 881) 1(1)(1 . 25) = 10 783 psi Eq. (15-3): W t = 10 783(0 . 71)(1)(0 . 201) 10(1)(1 . 412)(0 . 508)(1 . 252) = 171 . 4 lbf H 2 = 171 . 4(628 . 3) 33 000 = 3 . 3hp Wear Pinion: ( C H ) G = 1, I = 0 . 078, C p = 2290 ± psi, C xc = 2 C s = 0 . 125(0
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Unformatted text preview: . 71) + . 4375 = . 526 25 Eq. (15-22): ( s ac ) P = 341(300) + 23 620 = 125 920 psi ( σ c ,all ) P = 125 920(1)(1) 1(1)(1 . 118) = 112 630 psi Eq. (15-1): W t = ² ( σ c ,all ) P C p ³ 2 Fd P I K o K v K m C s C xc = ´ 112 630 2290 µ 2 ² . 71(2 . 000)(0 . 078) 1(1 . 412)(1 . 252)(0 . 526 25)(2) ³ = 144 . 0 lbf H 3 = 144(628 . 3) 33 000 = 2 . 7 hp...
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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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