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
Step-by-step 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