Answer:
The strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex]
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find [tex](K)[/tex] and the [tex](n)[/tex].
[tex]\sigma =K\epsilon^n[/tex]
We will plug the values in the formula.
[tex]250=K\times (0.12)^n\\350=K\times (0.26)^n[/tex]
We will solve these equation.
[tex]K=\frac{250}{(0.12)^n}[/tex] plug this value in [tex]350=K\times (0.26)^n[/tex]
[tex]350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n[/tex]
Taking a natural log both sides we get.
[tex]ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435[/tex]
Now, we will find value of [tex]K[/tex]
[tex]K=\frac{250}{(0.12)^n}[/tex]
[tex]K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625[/tex]
So, the strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex].
