7_ch 08 Mechanical Design budynas_SM_ch08

# 7_ch 08 Mechanical Design budynas_SM_ch08 - 8) = 2 . 4...

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210 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design | δ b |+| δ m |= N t N F i k b + F i k m = N t N N t = NF i ± 1 k b + 1 k m ² = ³ k b + k m k b k m ´ F i N = θ 360 Ans . As a check invert Prob. 8-15. What Turn-of-Nut will induce F i = 11 808 lbf? N t = 16(11 808) ³ 1 1 . 02(10 6 ) + 1 1 . 27(10 6 ) ´ = 0 . 334 turns . = 1 / 3 turn (checks) The relationship between the Turn-of-Nut method and the Torque Wrench method is as follows. N t = ³ k b + k m k b k m ´ F i N (Turn-of-Nut) T = KF i d (Torque Wrench) Eliminate F i N t = ³ k b + k m k b k m ´ NT Kd = θ 360 Ans . 8-17 (a) From Ex. 8-4, F i = 14 . 4 kip, k b = 5 . 21(10 6 ) lbf/in, k m = 8 . 95(10 6 ) lbf/in Eq. (8-27): T = kF i d = 0 . 2(14 . 4)(10 3 )(5 / 8) = 1800 lbf · in Ans . From Prob. 8-16, t = NF i ³ 1 k b + 1 k m ´ = 16(14 . 4)(10 3 ) ± 1 5 . 21(10 6 ) + 1 8 . 95(10 6 ) ² = 0 . 132 turns = 47 . 5 Ans . Bolt group is (1 . 5) / (5 /
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Unformatted text preview: 8) = 2 . 4 diameters. Answer is lower than RB&W recommendations. (b) From Ex. 8-5, F i = 14 . 4 kip, k b = 6 . 78 Mlbf/in, and k m = 17 . 4 Mlbf/in T = . 2(14 . 4)(10 3 )(5 / 8) = 1800 lbf · in Ans . t = 11(14 . 4)(10 3 ) ± 1 6 . 78(10 6 ) + 1 17 . 4(10 6 ) ² = . 0325 = 11 . 7 ◦ Ans . Again lower than RB&W. 8-18 From Eq. (8-22) for the conical frusta, with d / l = . 5 k m Ed µ µ µ µ ( d / l ) = . 5 = . 5774 π 2 ln { 5[0 . 5774 + . 5(0 . 5)] / [0 . 5774 + 2 . 5(0 . 5)] } = 1 . 11...
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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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