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

ΔL = 0.02315 m = 23.15 mm

Step-by-step 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