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19 April, 11:04

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking. On the basis of this information, what engineering stress (in MPa) is necessary to produce an engineering plastic strain of 0.250?

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  1. 19 April, 11:12
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    The missing stresses in the question:

    Engineering Engineering strain

    Stress (MpA)

    235 0.194

    250 0.296

    Answer:

    σ = 245MPa

    Explanation:

    For this problem, we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n may be determined. Since σT = σ (1 + ε)

    So,

    σT1 = (235 MPa) (1 + 0.94) = 280 MPa

    σT2 = (250 MPa) (1 + 0.296) = 324MPa

    Similarly for strains, since;

    εT = ln (1 + ε)

    Thus;

    ε T1 = ln (1 + 0.194) = 0.177

    εT2 = ln (1 + 0.296) = 0.259

    We know that true stress;

    σε = (σT/K) ^ (1/n)

    So taking log of both sides, to get;

    log σT = log K + n log εT

    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,

    log (280) = log K + n log (0.177)

    and

    log (324) = log K + n log (0.259)

    Solving both log equations simultaneously, we get;

    K = 543 MPa and n = 0.383

    Now converting, ε to true strain;

    ε T = ln (1 + 0.25) = 0.223

    From the true stress equation earlier quoted, we'll make σT the subject and get;

    σT = (εT^n) x k

    σT = 543 x (0.223^ (0.383)) = 306 MPa

    Now converting this value of σT

    to an engineering stress using the first equation quoted in this answer, gives;

    σ = σT / (1 + ε) = 306 / (1+0.25) = 245MPa
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