Ask Question
28 December, 09:25

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).

+2
Answers (1)
  1. 28 December, 13:00
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers