9_ch 13 Mechanical Design budynas_SM_ch13

9_ch 13 Mechanical - shi20396_ch13.qxd 12:16 PM Page 341 341-22 N5 = 12 2(16 2(12 = 68 teeth Ans Let gear 2 be first n F = n 2 = 320 rev/min Let

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Unformatted text preview: shi20396_ch13.qxd 8/29/03 12:16 PM Page 341 341 Chapter 13 13-22 N5 = 12 + 2(16) + 2(12) = 68 teeth Ans. Let gear 2 be first, n F = n 2 = 320 rev/min. Let gear 5 be last, n L = n 5 = 0 3 nL − n A 12 16 12 = , e= 16 12 68 17 nF − nA 17 320 − n A = (0 − n A ) 3 3 n A = − (320) = −68.57 rev/min 14 The negative sign indicates opposite of n 2 ∴ n A = 68.57 rev/min cw Ans. e= Alternatively, v 0 nA(N2 2N3 N4) 2nA(N2 nA(N2 v n 2 N2 nA = N3) 2N3 N4) n2N2 2nA(N2 2N3 N4) 2nA(N2 N3) 2nA(N2 N3) n2N2 320(12) n 2 N2 = 2( N3 + N4 ) 2(16 + 12) = 68.57 rev/min cw Ans. 13-23 Let n F = n 2 then n L = n 7 = 0. 24 18 8 36 =− 18 30 54 15 n L − n5 8 e= =− n F − n5 15 8 15 0−5 =− ⇒ n 2 = 5 + (5) = 14.375 turns in same direction n2 − 5 15 8 e=− 13-24 (a) Let n F = n 2 = 0, then n L = n 5 . e= 99 100 101 100 = 9999 , 10 000 e= nL − n A nL − n A = nF − nA 0 − nA n L − n A = −en A n L = n A ( −e + 1) 1 9999 nL = = 0.0001 Ans. =1−e =1− nA 10 000 10 000 101 N4 = = 10.1 in P 10 100 d5 = = 10 in 10 10 d5 dhousing > 2 d4 + = 2 10.1 + 2 2 (b) d4 = = 30.2 in Ans. ...
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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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