41_ch 04 Mechanical Design budynas_SM_ch04

41_ch 04 Mechanical Design budynas_SM_ch04 -...

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110 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design k = 0 . 875 4 = 0 . 2188 in l / k = 18 0 . 2188 = 82 . 3 try Johnson P cr = π 4 (0 . 875 2 ) ± 37 . 5(10 3 ) ² 37 . 5(10 3 ) 2 π 82 . 3 ³ 2 1 1(30)(10 6 ) ´ = 17 714 lbf Use d = 0 . 875 in Ans. (c) n ( a ) = 20 440 5655 = 3 . 61 Ans . n ( b ) = 17 714 5655 = 3 . 13 Ans . 4-78 4 F sin θ = 3920 F = 3920 4 sin θ In range of operation, F is maximum when θ = 15 F max = 3920 4 sin 15 = 3786 N per bar P cr = n d F max = 2 . 5(3786) = 9465 N l = 300 mm, h = 25 mm Try b = 5 mm: out of plane k = (5 / 12) = 1 . 443 mm l k = 300 1 . 443 = 207 . 8
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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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