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PAGE #3 - = ~ (c) The strain-at-fracture (plastic) is 8 f =...

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~5" Professor Meletis Spring 2004 Louisiana State U niversity Mechanical Engineering Department ME 2733 Materials of Engineering Homework Assignment #4 Solution * 1, 12000/0.1 0.1/10 120,000 0.01 (a) E = ~ = = 12 x 106 psi ( slope in the elastic region) = (b) The yield strength corresponds to plastic strain of 0.002. So, 0.002 = .!.- ~ x = 0.02in .The 10 extension x corresponding to 0.002 plastic strain is 0.02 in. 0- y = 17,000/0.1 = 17~ o-uts = 21,000/0.1
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Unformatted text preview: = ~ (c) The strain-at-fracture (plastic) is 8 f = 0.43/10 = 9.043 The elastic strain just before fracture is the total strain minus the plastic strain: 0.54- 0.43 = 0.011 s = e 10 (d) At 20,000 lbs the total extension is -0.23 in that is composed of 0.06 in plastic extension and 0.17 in elastic extension. Thus, the plastic strain is fp = 0.06/10 = Q,QQ.Q and the elastic strain is: fe = 0.17/10 = Q.Ql1...
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This note was uploaded on 01/22/2012 for the course ME 2733 taught by Professor Meng during the Fall '10 term at LSU.

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