13_ch 16 Mechanical Design budynas_SM_ch16

13_ch 16 Mechanical Design budynas_SM_ch16 - = F f 4 ( D +...

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408 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) The torque ratio is 150(12) / 100 or 18-fold. P 2 = 349 18 = 19 . 4 lbf P 1 = 2 . 25 P 2 = 2 . 25(19 . 4) = 43 . 6 lbf p = 89 . 6 18 = 4 . 98 psi Ans . Comment: As the torque opposed by the locked brake increases, P 2 and P 1 increase (although ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be provided by a shear key. 16-16 (a) From Eq. (16-23), since F = π p a d 2 ( D d ) then p a = 2 F π d ( D d ) and it follows that p a = 2(5000) π (225)(300 225) = 0 . 189 N/mm 2 or 189 000 N/m 2 or 189 kPa Ans . T
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Unformatted text preview: = F f 4 ( D + d ) = 5000(0 . 25) 4 (300 + 225) = 164 043 N mm or 164 N m Ans . (b) From Eq. (16-26), F = p a 4 ( D 2 d 2 ) p a = 4 F ( D 2 d 2 ) = 4(5000) (300 2 225 2 ) = . 162 N/mm 2 = 162 kPa Ans . From Eq. (16-27), T = 12 f p a ( D 3 d 3 ) = 12 (0 . 25)(162)(10 3 )(300 3 225 3 )(10 3 ) 3 = 166 N m Ans . 16-17 (a) Eq. (16-23): F = p a d 2 ( D d ) = (120)(4) 2 (6 . 5 4) = 1885 lbf 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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