52_ch 03 Mechanical Design budynas_SM_ch03

52_ch 03 Mechanical Design budynas_SM_ch03 - 3 . 5 2 + . 75...

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Chapter 3 65 For ± dA / r , add the results of the following equation for each of the four rectangles. ² r o r i bdr r = b ln r o r i , b = width ² dA r = 0 . 375 ln 1 . 8125 1 + 1 . 25 ln 2 . 1875 1 . 8125 + 1 . 25 ln 3 . 6875 3 . 3125 + 0 . 375 ln 4 . 5 3 . 6875 = 0 . 666 810 6 r n = 1 . 546 875 0 . 666 810 6 = 2 . 3198 in e = r c r n = 2 . 75 2 . 3198 = 0 . 4302 in c i = r n r i = 2 . 320 1 = 1 . 320 in c o = r o r n = 4 . 5 2 . 320 = 2 . 180 in Shear stress due to 206 lbf force is zero at inner and outer surfaces. σ i =− 142 1 . 547 + 2000(1 . 32) 1 . 547(0 . 4302)(1) = 3875 psi Ans. σ o =− 142 1 . 547 2000(2 . 18) 1 . 547(0 . 4302)(4 . 5) =− 1548 psi Ans. 3-77 A = (6 2 1)(0 . 75) = 2 . 25 in 2 r c = 6 + 2 2 = 4in Similar to Prob. 3-76, ² dA r = 0 . 75 ln
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Unformatted text preview: 3 . 5 2 + . 75 ln 6 4 . 5 = . 635 473 4 in r n = A ( d A / r ) = 2 . 25 . 635 473 4 = 3 . 5407 in e = 4 3 . 5407 = . 4593 in i = 5000 2 . 25 + 20 000(3 . 5407 2) 2 . 25(0 . 4593)(2) = 17 130 psi Ans . o = 5000 2 . 25 20 000(6 3 . 5407) 2 . 25(0 . 4593)(6) = 5710 psi Ans . 3-78 A = r o r i b dr = 6 2 2 r dr = 2 ln 6 2 = 2 . 197 225 in 2...
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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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