10_ch 13 Mechanical Design budynas_SM_ch13

10_ch 13 Mechanical Design budynas_SM_ch13 - 2 N 2 N 3 + N...

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342 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-25 n 2 = n b = n F , n A = n a , n L = n 5 = 0 e =− 21 444 = n L n A n F n A 21 444 ( n F n A ) = 0 n A With shaft b as input 21 444 n F + 21 444 n A + 444 444 n A = 0 n A n F = n a n b = 21 465 n a = 21 465 n b ,i n the same direction as shaft b , the input. Ans . Alternatively, v A N 4 = n 2 N 2 N 3 + N 4 v A = n 2 N 2 N 4 N 3 + N 4 n a = n A = v A N 2 + N 3 = n 2 N 2 N 4 ( N 2 + N 3 )( N 3 + N 4 ) = 18(21)( n b ) (18 + 72)(72 + 21) = 21 465 n b in the same direction as b Ans . 13-26 n F = n 2 = n a , n L = n 6 = 0 e =− 24 18 µ 22 64 =− 11 24 , e = n L n A n F n A = 0 n b n a n b 11 24 = 0 n b n a n b n b n a = 11 35 Ans . Yes, both shafts rotate in the same direction. Ans. Alternatively, v A N 5 = n
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Unformatted text preview: 2 N 2 N 3 + N 5 = N 2 N 3 + N 5 n a , v A = N 2 N 5 N 3 + N 5 n a n A = n b = v A N 2 + N 3 = N 2 N 5 ( N 2 + N 3 )( N 3 + N 5 ) n a n b n a = 24(22) (24 + 18)(22 + 18) = 11 35 Ans . n b rotates ccw ∴ Yes Ans . 13-27 n 2 = n F = 0, n L = n 5 = n b , n A = n a e = + 20 24 µ 20 24 ¶ = 25 36 3 5 v 5 v A n 2 N 2 2 3 4 2 v 5 v A n 2 N 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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