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).

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

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: