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 assume a value of 0.22 for the strain-hardening exponent, n.

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

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