32_ch 17 Mechanical Design budynas_SM_ch17

32_ch 17 Mechanical Design budynas_SM_ch17 -...

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Unformatted text preview: budynas_SM_ch17.qxd 12/06/2006 17:29 Page 451 FIRST PAGES 451 Chapter 17 (b) Try m = 4 strands Ft = 8 + 3.2d 2 4 1+ 2 32.2 = 2.12 + 3.4d 2 kip Ff = 12.1d kip 12.1d n= 2.12 + 3.4d 2 d, in n 0.5000 0.5625 0.6250 0.7500 0.8750 1.0000 2.037 2.130 2.193 2.250 ← maximum n Ans. 2.242 2.192 Comparing tables, multiple ropes supporting the load increases the factor of safety, and reduces the corresponding wire rope diameter, a useful perspective. 17-32 n= ad b/ m + cd 2 dn ( b/ m + cd 2 ) a − ad (2cd ) =0 = dd ( b/ m + cd 2 ) 2 From which d* = b mc Ans. √ a a b/( mc) = n* = ( b/ m ) + c[b/( mc)] 2 m bc Ans. These results agree closely with Prob. 17-31 solution. The small differences are due to rounding in Prob. 17-31. 17-33 From Prob. 17-32 solution: n1 = ad b/ m + cd 2 Solve the above equation for m m= b ad / n 1 − cd 2 dm [( ad / n 1 ) − ad 2 ](0) − b[( a / n 1 ) − 2cd ] =0= ad [( ad / n 1 ) − cd 2 ]2 (1) ...
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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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