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Homework 7 - solutions

# Homework 7 - solutions - MAT 33 Fall 2008 Due 1 Homework#7...

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MAT 33 – Fall 2008 Homework #7 – Solutions Due: 10/27/08 1.) For a brass alloy under tension, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking. Engineering Stress (MPa) Engineering Strain 315 0.105 340 0.220 (i) Calculate the true stress, σ T and true strain and ε T for both of these loading situations (ii) The true stress and true strain are related by σ T = K ε T n . Calculate the strain hardening exponent, n, and the constant K for this material. (iii) Calculate the engineering stress necessary to produce an engineering strain of 0.32 in this material. (i) Since σ T = σ (1 + ε ) then σ T 1 = (315 MPa)(1+ 0.105)= 348 MPa σ T 2 = (340 MPa)(1+ 0.220)= 415 MPa Similarly for strains, since ε T = ln(1 + ε ) then ε T 1 = ln (1 + 0.105)= 0.09985 ε T 2 = ln (1 + 0.220)= 0.19885 (ii) Taking logarithms of σ T = K ε T n we get log σ T = log K + n log ε T which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and n as unknowns. Thus

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log (348) = log K + n log(0.09985) ) log (415) = log K + n log(0.19885) ) Solving for these two expressions yields K = 628 MPa and n = 0.256. (iii) Converting ε = 0.32 to true strain 0.278 = 0.32) + (1 ln = T ε The corresponding σ T to give this value of ε T [using σ T = K ε T n ] is just MPa 453 = ) MPa)(0.278 (628 = K = 0.256 n T T ε σ Now converting this σ T to an engineering stress MPa 343 = 32 . 0 1 453
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Homework 7 - solutions - MAT 33 Fall 2008 Due 1 Homework#7...

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