23_ch 14 Mechanical Design budynas_SM_ch14

23_ch 14 Mechanical Design budynas_SM_ch14 - 1.18 where the...

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Chapter 14 371 Gear wear Similarly, W t 4 = 1182 lbf, H 4 = 59 . 0hp Rating H rated = min( H 1 , H 2 , H 3 , H 4 ) = min(157 . 5, 192 . 9, 53, 59) = 53 hp Ans . Note differing capacities. Can these be equalized? 14-25 From Prob. 14-24: W t 1 = 3151 lbf, W t 2 = 3861 lbf, W t 3 = 1061 lbf, W t 4 = 1182 lbf W t = 33 000 K o H V = 33 000(1 . 25)(40) 1649 = 1000 lbf Pinion bending : The factor of safety, based on load and stress, is ( S F ) P = W t 1 1000 = 3151 1000 = 3 . 15 Gear bending based on load and stress ( S F ) G = W t 2 1000 = 3861 1000 = 3 . 86 Pinion wear based on load: n 3 = W t 3 1000 = 1061 1000 = 1 . 06 based on stress: ( S H ) P = 1 . 06 = 1 . 03 Gear wear based on load: n 4 = W t 4 1000 = 1182 1000 = 1 . 18 based on stress: ( S H ) G = 1 . 18 = 1 . 09 Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06,
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Unformatted text preview: 1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors ( S F ) P , ( S F ) G , ( S H ) P , ( S H ) G are 3 . 15, 3 . 86, 1 . 03, 1 . 09 or 3 . 15, 3 . 86, 1 . 06 1 / 2 , 1 . 18 1 / 2 and the threat is again from pinion wear. Depending on the magnitude of the numbers, using S F and S H as dened by AGMA, does not necessarily lead to the same conclusion concerning threat. Therefore be cautious....
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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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