2_ch 14 Mechanical Design budynas_SM_ch14

# 2_ch 14 Mechanical Design budynas_SM_ch14 - From Table A-17...

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350 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 14-4 d = 5(15) = 75 mm, Y = 0 . 290 V = π (75)(10 3 )(200) 60 = 0 . 7854 m/s Assume steel and apply Eq. (14-6 b ): K v = 6 . 1 + 0 . 7854 6 . 1 = 1 . 129 W t = 60 H π dn = 60(5)(10 3 ) π (75)(10 3 )(200) = 6366 N Eq. (14-8): σ = K v W t FmY = 1 . 129(6366) 60(5)(0 . 290) = 82 . 6MPa Ans. 14-5 d = 1(16) = 16 mm, Y = 0 . 296 V = π (16)(10 3 )(400) 60 = 0 . 335 m/s Assume steel and apply Eq. (14-6 b ): K v = 6 . 1 + 0 . 335 6 . 1 = 1 . 055 W t = 60 H π dn = 60(0 . 15)(10 3 ) π (16)(10 3 )(400) = 447 . 6N Eq. (14-8): F = K v W t σ mY = 1 . 055(447 . 6) 150(1)(0 . 296) = 10 . 6mm
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Unformatted text preview: From Table A-17, use F = 11 mm Ans. 14-6 d = 1 . 5(17) = 25 . 5 mm, Y = . 303 V = π (25 . 5)(10 − 3 )(400) 60 = . 534 m/s Eq. (14-6 b ): K v = 6 . 1 + . 534 6 . 1 = 1 . 088 W t = 60 H π dn = 60(0 . 25)(10 3 ) π (25 . 5)(10 − 3 )(400) = 468 N Eq. (14-8): F = K v W t σ mY = 1 . 088(468) 75(1 . 5)(0 . 303) = 14 . 9 mm Use F = 15 mm 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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