In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and the strain-hardening exponent in the flow curve equation.

Respuesta :

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].