12_ch 08 Mechanical Design budynas_SM_ch08

# 12_ch 08 Mechanical Design budynas_SM_ch08 - E = 70 Mpsi, d...

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Chapter 8 215 Member stiffness for four frusta and joint constant C using Eqs. (8-20) and ( e ). Top frustum: D = 0 . 75, t = 0 . 5, d = 0 . 5, E = 30 k 1 = 33 . 30 Mlbf/in 2nd frustum: D = 1 . 327, t = 0 . 11, d = 0 . 5, E = 14 . 5 k 2 = 173 . 8 Mlbf/in 3rd frustum: D = 0 . 860, t = 0 . 515, E = 14 . 5 k 3 = 21 . 47 Mlbf/in Fourth frustum: D = 0 . 75, t = 0 . 095, d = 0 . 5, E = 30 k 4 = 97 . 27 Mlbf/in k m = ± 4 ² i = 1 1 / k i ³ 1 = 10 . 79 Mlbf/in Ans . C = 3 . 94 / (3 . 94 + 10 . 79) = 0 . 267 Ans . 8-25 k b = A t E l = 0 . 1419(30) 0 . 845 = 5 . 04 Mlbf/in Ans . From Fig. 8-21, h = 1 2 + 0 . 095 = 0 . 595 in l = h + d 2 = 0 . 595 + 0 . 5 2 = 0 . 845 D 1 = 0 . 75 + 0 . 845 tan 30 = 1 . 238 in l / 2 = 0 . 845 / 2 = 0 . 4225 in From Eq. (8-20): Frustum 1: D = 0 . 75, t = 0 . 4225 in, d = 0 . 5in, E = 30 Mpsi k 1 = 36 . 14 Mlbf/in Frustum 2: D = 1 . 018 in, t = 0 . 1725 in,
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Unformatted text preview: E = 70 Mpsi, d = . 5 in k 2 = 134 . 6 Mlbf/in Frustum 3: D = . 75, t = . 25 in, d = . 5 in, E = 14 . 5 Mpsi k 3 = 23 . 49 Mlbf/in k m = 1 (1 / 36 . 14) + (1 / 134 . 6) + (1 / 23 . 49) = 12 . 87 Mlbf/in Ans . C = 5 . 04 5 . 04 + 12 . 87 = . 281 Ans . 0.095&quot; 0.1725&quot; 0.25&quot; 0.595&quot; 0.5&quot; 0.625&quot; 0.4225&quot; 0.845&quot; 0.75&quot; 1.018&quot; 1.238&quot; Steel Cast iron...
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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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