hw2sol - HW#2 Solution 1(a 90000 80000 70000 60000 50000...

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HW#2 Solution 1. (a) 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 0 0.05 0.1 0.15 0.2 0.25 0.3 True Eng () y y y TS b 40,072 40,072 E= 22,262 ksi 0.0018 c From the graph, 76,440 psi d In the plastic range. ln =lnK+nln From the ln vs. ln plot (see right) n = 0.135 σ ε σε == = 0.135 76,440=K 0.085 K=106,622 psi. log strain hardening y = 0.1352x + 11.425 R 2 = 0.9276 10.4 10.6 10.8 11 11.2 11.4 -8 -6 -4 -2 0 Series1 Linear (Series1)
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2. () yy 2 y 2 6 n 0.15 uu a Resilience 1 = 2 1 = 2E 1 25,000 = 210 10 =31.25 psi (b) ductility at necking point = 50,000=60,000 0.296 σε σ εε × Κ ⇒= 0.296 u o o o oo u o 0.35 fo o f 0 ln e 1.34 1.34 e = 0.34 If you use fracture point e 1.419 0.419 c toughness = d f ε == = ∴= ⋅= A AA A A A A A n+1 f 1.15 n+1 60,000 = (0.35) 15,600 psi 1.15 Κ =
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3 () a Prob. 2.54 HRA = 100 500 t =100-500 0.06 =70 It is unacceptable since 70<75.
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This note was uploaded on 12/25/2011 for the course ME 300363 taught by Professor Staff during the Fall '10 term at Purdue.

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hw2sol - HW#2 Solution 1(a 90000 80000 70000 60000 50000...

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