Question

For some metal alloy, a true stress of 345 MPa (50,000 psi) produces a plastic true strain of 0.02. How much does a specimen of this material elongate when a true stress of 415 MPa (60,000 psi) is applied if the original length is 500 mm (20 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.

Answer 1

Answer:

ΔL = 0.02315 m = 23.15 mm

Explanation:

σ₁ = 345 MPa

ε₁ = 0.02

σ₂ = 415 MPa

L₀ = 500 mm = 0.50 m

n = 0.22

We can apply the following equation

σ = σ₀*εⁿ

then

σ₁ = σ₀*ε₁ⁿ    ⇒    σ₀ = σ₁ / ε₁ⁿ

⇒    σ₀ = 345 MPa / (0.02)∧(0.22)  

⇒    σ₀ = 815.8165 MPa

Now,

σ₂= σ₀*ε₂ⁿ    ⇒   ε₂ = (σ₂ / σ₀)∧(1 / n)

⇒   ε₂ = (415 MPa / 815.8165 MPa)∧(1 / 0.22)

⇒   ε₂ = 0.0463

we now that

ε = ΔL / L₀    ⇒    ΔL = ε₂*L₀

⇒    ΔL = 0.0463*0.50 m

⇒    ΔL = 0.02315 m = 23.15 mm

Answer 2

When a true stress of 415 MPa is applied, the elongation will be "15.85 mm".

Stress and Strain

According to the question,

True stress, r₁ = 375 MPa

                    r₂ = 415 MPa

True strain, $\varepsilon_1$ = 0.02

Strain hardening exponent, n = 0.22

Original length = 500 mm

By using Holloman's equation, we get

→        r = K$\varepsilon^n$

         r ∝ $\varepsilon^n$

or,

        $\frac{r_1}{r_2} = (\frac{\varepsilon_1}{\varepsilon_2} )^n$

By substituting the values, we get

       $\frac{375}{415} = (\frac{0.02}{\varepsilon_2})^{0.22}$

$[\frac{375}{415}]^{\frac{1}{0.22} } =\frac{0.22}{\varepsilon_2}$

         $\varepsilon_2$ = 0.31703

hence,

→      Strain = $\frac{Elongation}{500}$

By substituting the values,

    0.31703 = $\frac{Elongation}{500}$

Elongation = 0.31703 × 500

                   = 15.85 mm

Thus the response above is correct.

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