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.

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

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