22_ch 14 Mechanical Design budynas_SM_ch14

22_ch 14 Mechanical Design budynas_SM_ch14 -...

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Eq. (14-17): ( σ all ) P = 32 125(0 . 832) 1(1)(1) = 26 728 psi B = 0 . 25 ( 12 6 ) 2 / 3 = 0 . 8255 A = 50 + 56 ( 1 0 . 8255 ) = 59 . 77 K v = ± 59 . 77 + 1649 59 . 77 ² 0 . 8255 = 1 . 534 K s = 1, C m = 1 C mc = F 10 d 0 . 0375 + 0 . 0125 F = 3 . 25 10(5 . 5) 0 . 0375 + 0 . 0125(3 . 25) = 0 . 0622 C ma = 0 . 127 + 0 . 0158(3 . 25) 0 . 093(10 4 )(3 . 25 2 ) = 0 . 178 C e = 1 K m = C mf = 1 + (1)[0 . 0622(1) + 0 . 178(1)] = 1 . 240 K B = 1, K T = 1 Eq. (14-15): W t 1 = 26 728(3 . 25)(0 . 345) 1 . 25(1 . 534)(1)(4)(1 . 240) = 3151 lbf H 1 = 3151(1649) 33 000 = 157 . 5hp Gear bending By similar reasoning, W t 2 = 3861 lbf and H 2 = 192 . 9hp Pinion wear m G = 60 / 22 = 2 . 727 I = cos 20 sin 20 2 ³ 2 . 727 1 + 2 . 727 ´ = 0 . 1176 0 . 99 ( S c ) 10 7 = 322(250) + 29 100 = 109 600 psi ( Z N ) P = 2 . 466[3(10 9 )] 0 . 056 = 0 . 727 ( Z N ) G = 2 . 466[3(10 9 ) / 2 . 727] 0 . 056 = 0 . 769 ( σ c ,all ) P = 109 600(0
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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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