Question

The tensile true stress ($\sigma$) – true strain ($\varepsilon$) curve follows the Hollomon equation: \[ \sigma = 500 \, \varepsilon^{0.15} \ \text{MPa} \] At the maximum load, the work-hardening rate $\left(\frac{d\sigma}{d\varepsilon}\right)$ is (in MPa):

Hint:

At maximum load, always use Considère criterion: $\frac{d\sigma}{d\varepsilon} = \sigma$.

Answer 1

Correct Answer: 350

Step 1: Recall condition for maximum load.
At maximum load (on engineering stress-strain curve), Considère criterion applies: \[ \frac{d\sigma}{d\varepsilon} = \sigma \]
Step 2: Differentiate given stress-strain relation.
\[ \sigma = 500 \varepsilon^{0.15} \] \[ \frac{d\sigma}{d\varepsilon} = 500 \times 0.15 \varepsilon^{0.15 - 1} = 75 \varepsilon^{-0.85} \]
Step 3: Apply maximum load condition.
\[ \frac{d\sigma}{d\varepsilon} = \sigma \] \[ 75 \varepsilon^{-0.85} = 500 \varepsilon^{0.15} \] \[ \Rightarrow \frac{75}{500} = \varepsilon^{0.15 + 0.85} = \varepsilon^{1.0} \] \[ \varepsilon = 0.15 \]
Step 4: Compute work-hardening rate.
\[ \frac{d\sigma}{d\varepsilon} = 75 \varepsilon^{-0.85} = 75 \times (0.15)^{-0.85} \] \[ (0.15)^{-0.85} \approx 1.0 \] Thus, \[ \frac{d\sigma}{d\varepsilon} \approx 75 \ \text{MPa} \] Final Answer: \[ \boxed{75 \ \text{MPa}} \]