Strain-hardening exponent and true stress

(a) n and K

Taking logs of the power law $\sigma_T = K\varepsilon_T^{\,n}$ for the two points:

$$n = \frac{\ln(\sigma_2/\sigma_1)}{\ln(\varepsilon_2/\varepsilon_1)} = \frac{\ln(520/450)}{\ln(0.12/0.05)} = \frac{0.1449}{0.8755} = 0.165$$ $$K = \frac{\sigma_1}{\varepsilon_1^{\,n}} = \frac{450}{(0.05)^{0.165}} = 738\ \text{MPa}$$

(b) Necking

Necking begins when the strain-hardening rate equals the stress, $d\sigma_T/d\varepsilon_T = \sigma_T$, which gives $\varepsilon_T = n$.

$$\varepsilon_{T,neck} = n = 0.165$$ $$\sigma_{T,neck} = K n^{n} = 738 \times (0.165)^{0.165} = 548\ \text{MPa}$$

Engineering UTS: $\sigma_T = \sigma(1+e)$ with $1 + e = e^{\varepsilon_T} = e^{0.165} = 1.180$, so

$$\sigma_{UTS} = \frac{548}{1.180} = 465\ \text{MPa}$$

(c) Engineering to true

$$\sigma_T = \sigma(1+e) = 400 \times 1.10 = 440\ \text{MPa}$$ $$\varepsilon_T = \ln(1+e) = \ln 1.10 = 0.0953$$

This is valid because $e = 0.10$ is less than the engineering strain at necking ($1.180 - 1 = 0.18$).

Answer: n = 0.165, K = 738 MPa; necking at ε_T = 0.165, σ_T = 548 MPa, engineering UTS = 465 MPa; (c) σ_T = 440 MPa, ε_T = 0.0953.