10_ch 16 Mechanical Design budynas_SM_ch16

10_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 405 16-12 Given: D = 300 mm, f = 0 . 28, b = 80 mm, φ = 270°, P 1 = 7600 N . f φ = 0 . 28( π )(270 / 180 ) = 1 . 319 P 2 = P 1 exp( f φ ) = 7600 exp( 1 . 319) = 2032 N p a = 2 P 1 bD = 2(7600) 80(300) = 0 . 6333 N/mm 2 or 633 kPa Ans. T = ( P 1 P 2 ) D 2 = (7600 2032) 300 2 = 835 200 N · mm or 835.2 N · m Ans. 16-13 α = cos 1 ± 125 200 ² = 51 . 32° φ = 270° 51 . 32° = 218 . f φ = 0 . 30(218 . 7) π 180° = 1 . 145 P 2 = (125 + 275) F 125 = (125 + 275)400 125 = 1280 N Ans. P 1 = P 2 exp( f φ ) = 1280 exp(1 . 145) = 4022 N T = ( P 1 P 2 ) D 2 = (4022 1280) 250
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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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