A standard Charpy V-notch impact test specimen of an aerospace titanium alloy absorbs an impact energy $U = 45\text{ J}$ during fracture. If the fractured ligament cross-sectional area beneath the notch is $A = 80\text{ mm}^2$, the impact toughness (impact resilience in $\text{kJ/m}^2$) of the alloy is:

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Impact toughness (or Charpy impact value) measures the energy absorbed per unit of cross-sectional area during dynamic fracture: $$\text{Impact Toughness} = \frac{U}{A}$$ where: - $U$ is the fracture energy absorbed in Joules - $A$ is the cross-sectional area of the fractured ligament under the notch in $\text{m}^2$. ### 2. Analytical Formulation & Calculation Given: - Absorbed energy $U = 45\text{ J}$ - Fractured area $A = 80\text{ mm}^2 = 80 \times 10^{-6}\text{ m}^2 = 8.0 \times 10^{-5}\text{ m}^2$ $$\text{Impact Toughness} = \frac{45\text{ J}}{80 \times 10^{-6}\text{ m}^2} = 562{,}500\text{ J/m}^2 = 562.5\text{ kJ/m}^2$$ ### 3. Distractor Forensics - Candidate omitting unit conversion from $\text{mm}^2$ to $\text{m}^2$ gets $0.5625\text{ J/mm}^2$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Ensure correct SI prefixes: $1\text{ J/m}^2 = 10^{-3}\text{ kJ/m}^2$. - **Standard Reference:** Callister, W.D., *Materials Science and Engineering: An Introduction*, 10th Edition, Wiley, Chapter 8.

← Prev Question Next Question →