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24 December, 06:50

A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.22 is produced when a true stress of 572 MPa (82960 psi) is applied; for the same metal, the value of K in the equation is 860 MPa (124700 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87020 psi).

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  1. 24 December, 07:03
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    True strain; ε' = 0.26

    Explanation:

    True strain is given by;

    ε' = (σ'/k) ^ (1/n)

    Where;

    ε' = true strain

    σ' = true stress

    k = strength coefficient

    n = the strain-hardening exponent

    We are given;

    σ' = 572 MPa

    k = 860 MPa

    ε' = 0.22

    Now, let's find the unknown 'n'

    ε' = (σ'/k) ^ (1/n)

    Thus,

    Raise both sides to the power of n;

    ε'ⁿ = (σ'/k)

    So, n log ε' = log σ' - log k

    n = (log σ' - log k) / log ε'

    n = (log 572 - log 860) / log 0.22

    n = 0.2693

    Now, the second part of the question gives a new condition which is;

    true stress (σ') = 600 MPa

    Thus, plugging this into the first equation quoted;

    ε' = (σ'/k) ^ (1/n)

    ε' = (600/860) ^ (1/0.2693) = 0.2627 ≈ 0.26
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