11_ch 08 Mechanical Design budynas_SM_ch08

11_ch 08 Mechanical Design budynas_SM_ch08 - 1 25 = 500 in...

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214 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From Eq. (8-20): Top frustum: D = 18, t = 13, E = 207 GPa k 1 = 5316 MN/m Mid-frustum: t = 7, E = 207 GPa, D = 24 . 9mm k 2 = 15 620 MN/m Bottom frustum: D = 18, t = 6, E = 100 GPa k 3 = 3887 MN/m k m = 1 (1 / 5316) + (1 / 55 620) + (1 / 3887) = 2158 MN/m Ans . C = 744 744 + 2158 = 0 . 256 Ans . From Prob. 8-22, F i = 37 . 9kN n = S p A t F i CP = 600(0 . 0843) 37 . 9 0 . 256(8 . 48) = 5 . 84 Ans . 8-24 Calculation of bolt stiffness: H = 7 / 16 in L T = 2(1 / 2) + 1 / 4 = 11 / 4in l = 1 / 2 + 5 / 8 + 0 . 095 = 1 . 22 in L > 1 . 125 + 7 / 16 + 0 . 095 = 1 . 66 in Use L = 1 . 75 in l d = L L T = 1 . 75
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Unformatted text preview: 1 . 25 = . 500 in l t = 1 . 125 + . 095 − . 500 = . 72 in A d = π (0 . 50 2 ) / 4 = . 1963 in 2 A t = . 1419 in 2 (UNC) k t = A t E l t = . 1419(30) . 72 = 5 . 9125 Mlbf/in k d = A d E l d = . 1963(30) . 500 = 11 . 778 Mlbf/in k b = 1 (1 / 5 . 9125) + (1 / 11 . 778) = 3 . 936 Mlbf/in Ans . 5 8 1 2 0.095 3 4 1.454 1.327 3 4 0.860 1.22 0.61...
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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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