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1 July, 11:08

A tensile test is performed on a metal specimen, and it is found that a true plastic strain given is produced when a true stress of listed is applied; for the same metal, the value of K in the following equation: is 860 MPa (125,000 psi). Calculate the value of n.

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  1. 1 July, 14:29
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    The value of the applied true stress is missing and this is the full question indicating it's value;

    A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.2 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the value of n

    Answer:

    n=0.25

    Explanation:

    True plastic strain (εt) = 0.20

    σT = 575 MPa

    K = 860 MPa

    Using the formula σT = K (εt) ^n

    We solve for n;

    575 MPa = {860 MPa (0.20) } ^n

    Take log of each side

    log (575) = log860 - nlog (0.20)

    2.76 = 2.93 - n (-0.67)

    Simplifying this, we get n to be approximately = - 0. 25.

    Ignoring the negative sign, n=0.25
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