15_ch 16 Mechanical Design budynas_SM_ch16

15_ch 16 Mechanical Design budynas_SM_ch16 - = (0 . 26) p a...

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410 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Multiply through by D 0 . 45 D d 0 . 80 D 0 . 45(6 . 5) d 0 . 80(6 . 5) 2 . 925 d 5 . 2in ± d D ² = d / D = 1 3 = 0 . 577 which lies within the common range of clutches. Yes. Ans. 16-19 Given: d = 0 . 306 m, l = 0 . 060 m, T = 0 . 200 kN · m, D = 0 . 330 m, f = 0 . 26 . α = tan 1 ± 12 60 ² = 11 . 31° Uniform wear Eq. (16-45): 0 . 200 = π (0 . 26)(0 . 306) p a 8 sin 11 . 31° (0 . 330 2 0 . 306 2 ) = 0 . 002 432 p a p a = 0 . 200 0 . 002 432 = 82 . 2kPa Ans. Eq. (16-44): F = π p a d 2 ( D d ) = π (82 . 2)(0 . 306) 2 (0 . 330 0 . 306) = 0 . 949 kN Ans. Uniform pressure Eq. (16-48): 0 . 200
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Unformatted text preview: = (0 . 26) p a 12 sin 11 . 31 (0 . 330 3 . 306 3 ) = . 002 53 p a p a = . 200 . 002 53 = 79 . 1 kPa Ans. Eq. (16-47): F = p a 4 ( D 2 d 2 ) = (79 . 1) 4 (0 . 330 2 . 306 2 ) = . 948 kN Ans. 16-20 Uniform wear Eq. (16-34): T = 1 2 ( 2 1 ) f p a r i ( r 2 o r 2 i ) 165 153 12 60 Not to scale C L...
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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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