20_ch 16 Mechanical Design budynas_SM_ch16

20_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 415 (b) For R = constant = nm , I e = I M + I P + n 2 I P + I P n 2 + R 2 I P n 4 + I L R 2 Ans. (c) For R = 10, I e n = 0 + 0 + 2 n (1) 2(1) n 3 4(10 2 )(1) n 5 + 0 = 0 n 6 n 2 200 = 0 From which n * = 2 . 430 Ans. m * = 10 2 . 430 = 4 . 115 Ans. Notice that n *and m * are independent of I L . 16-28 From Prob. 16-27, I e = I M + I P + n 2 I P + I P n 2 + R 2 I P n 4 + I L R 2 = 10 + 1 + n 2 + 1 n 2 + 100(1) n 4 + 100 10 2 = 10 + 1 + n 2 + 1 n 2 + 100 n 4 + 1 nI e 1.00 114.00 1.50 34.40 2.00 22.50 2.43 20.90 3.00 22.30 4.00 28.50 5.00 37.20 6.00 48.10 7.00 61.10 8.00 76.00 9.00 93.00 10.00 112.02 Optimizing the partitioning of a double reduction lowered the gear-train inertia to
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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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