36_ch 03 Mechanical Design budynas_SM_ch03

36_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 49 l = π 180 JG θ T = π 180 ± π 32 d 4 G θ ( π/ 16) τ d 3 ² = π 360 dG θ τ = π 360 (0 . 015)(79 . 3)(10 9 )(30) 110(10 6 ) = 2 . 83 m Ans. 3-40 d = 3 in, replaced by 3 in hollow with t = 1 / 4 in (a) T solid = π 16 τ (3 3 ) T hollow = π 32 τ (3 4 2 . 5 4 ) 1 . 5 % ± T = ( π/ 16)(3 3 ) ( π/ 32) [(3 4 2 . 5 4 ) / 1 . 5] ( π/ 16)(3 3 ) (100) = 48 . 2% Ans. (b) W solid = kd 2 = k (3 2 ) , W hollow = k (3 2 2 . 5 2 ) % ± W = k (3 2 ) k (3 2 2 . 5 2 ) k (3 2 ) (100) = 69 . 4% Ans. 3-41 T = 5400 N · m, τ all = 150 MPa (a) τ = Tc J 150(10 6 ) = 5400( d / 2) ( π/ 32)[ d 4 (0 . 75 d ) 4 ] = 4 . 023(10 4 ) d 3 d = ³ 4 . 023(10 4 ) 150(10 6 ) ´ 1 / 3 = 6 . 45(10 2 ) m = 64.5 mm From Table A-17, the next preferred size is d = 80 mm; ID = 60 mm Ans. (b) J = π 32 (0 . 08 4 0 . 06 4 ) = 2 . 749(10 6 )mm 4 τ i = 5400(0 . 030)
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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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