26_ch 17 Mechanical Design budynas_SM_ch17

26_ch 17 Mechanical Design budynas_SM_ch17 - 0.423 C ± 2...

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Chapter 17 445 17-28 Follow the decision set outlined in Prob. 17-27 solution. We will form two tables, the first for a 15 000 h life goal, and a second for a 50 000 h life goal. The comparison is useful. Function : H nom = 50 hp at n = 1800 rev/min, n pump = 900 rev/min m G = 1800 / 900 = 2, K s = 1 . 2 life = 15 000 h, then repeat with life = 50 000 h Design factor : n d = 1 . 1 Sprockets : N 1 = 19 teeth, N 2 = 38 teeth Table 17-22 (post extreme): K 1 = ± N 1 17 ² 1 . 5 = ± 19 17 ² 1 . 5 = 1 . 18 Table 17-23: K 2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0 Decision variables for 15 000 h life goal : H ± tab = K s n d H nom K 1 K 2 = 1 . 2(1 . 1)(50) 1 . 18 K 2 = 55 . 9 K 2 (1) n fs = K 1 K 2 H tab K s H nom = 1 . 18 K 2 H tab 1 . 2(50) = 0 . 0197 K 2 H tab Form a table for a 15 000 h life goal using these equations. K 2 H ± tab Chain # H tab n fs Lub 1 1.0 55.90 120 21.6
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Unformatted text preview: 0.423 C ± 2 1.7 32.90 120 21.6 0.923 C ± 3 2.5 22.40 120 21.6 1.064 C ± 4 3.3 16.90 120 21.6 1.404 C ± 5 3.9 14.30 80 15.6 1.106 C ± 6 4.6 12.20 60 12.4 1.126 C ± 8 6.0 9.32 60 12.4 1.416 C ± There are 4 possibilities where n f s ≥ 1 . 1 Decision variables for 50 000h life goal From Eq. (17-40), the power-life tradeoff is: ( H ± tab ) 2 . 5 15 000 = ( H ±± tab ) 2 . 5 50 000 H ±± tab = ³ 15 000 50 000 ( H ± tab ) 2 . 5 ´ 1 / 2 . 5 = . 618 H ± tab Substituting from (1), H ±± tab = . 618 ± 55 . 9 K 2 ² = 34 . 5 K 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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