16_ch 16 Mechanical Design budynas_SM_ch16

16_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 411 Eq. (16-33): F = ( θ 2 θ 1 ) p a r i ( r o r i ) Thus, T fFD = (1 / 2)( θ 2 θ 1 ) fp a r i ( r 2 o r 2 i ) f ( θ 2 θ 1 ) p a r i ( r o r i )( D ) = r o + r i 2 D = D / 2 + d / 2 2 D = 1 4 ± 1 + d D ² O.K. Ans. Uniform pressure Eq. (16-38): T = 1 3 ( θ 2 θ 1 ) fp a ( r 3 o r 3 i ) Eq. (16-37): F = 1 2 ( θ 2 θ 1 ) p a ( r 2 o r 2 i ) T fFD = (1 / 3)( θ 2 θ 1 ) fp a ( r 3 o r 3 i ) (1 / 2) f ( θ 2 θ 1 ) p a ( r 2 o r 2 i ) D = 2 3 ³ ( D / 2) 3 ( d / 2) 3 [( D / 2) 2 ( d / 2) 2 D ] ´ = 2( D / 2) 3 (1 ( d / D ) 3 ) 3( D / 2) 2 [1 ( d / D ) 2 ] D = 1 3 µ 1 ( d / D ) 3 1 ( d / D ) 2 O . K . Ans . 16-21 ω = 2 π n / 60 = 2 π 500 / 60 = 52 . 4 rad/s T = H ω = 2(10 3 ) 52 . 4 = 38 . 2N · m Key : F = T r = 38 . 2 12 = 3 . 18 kN Average shear stress in key is τ = 3 . 18(10 3 )
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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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