Modulus of toughness and resilience (300 kN)

Given: $d = 10$ mm, $L = 400$ mm, $P_u = 300$ kN, $P_y = 150$ kN, elongation $\delta = 10$ mm.

Assumptions: stresses are based on the original area $A_0$ (the 2 mm diameter at maximum load is the neck diameter and does not change $A_0$). The question gives only one elongation (10 mm), so this strain is used for both moduli. The elastic energy is taken up to yield, and the modulus of toughness is taken as the approximate area under the curve (average of yield and ultimate stress times strain).

$$\begin{aligned} A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\ \sigma_y &= \frac{150\times10^3}{78.54} = 1909.86\ \text{N/mm}^2\\ \sigma_u &= \frac{300\times10^3}{78.54} = 3819.72\ \text{N/mm}^2\\ \varepsilon &= \frac{10}{400} = 0.025 \end{aligned}$$

Modulus of resilience:

$$U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1909.86)(0.025) = 23.87\ \text{N/mm}^2$$

Modulus of toughness:

$$U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1909.86+3819.72}{2}(0.025) = 71.62\ \text{N/mm}^2$$

Answer: Modulus of resilience = 23.87 N/mm² (MJ/m³); modulus of toughness ≈ 71.62 N/mm² (MJ/m³).