Modulus of toughness and resilience (250 kN)

Given: $d = 10$ mm, $L = 300$ mm, $P_u = 250$ kN, $P_y = 120$ kN, elongation $\delta = 8$ mm.

Assumptions: the 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 (8 mm), so this strain is used for both moduli. The modulus of resilience is the elastic energy per unit volume 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). The data give stresses far above those of real mild steel, but they are used as given.

$$\begin{aligned} A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\ \sigma_y &= \frac{120\times10^3}{78.54} = 1527.89\ \text{N/mm}^2\\ \sigma_u &= \frac{250\times10^3}{78.54} = 3183.10\ \text{N/mm}^2\\ \varepsilon &= \frac{8}{300} = 0.02667 \end{aligned}$$

Modulus of resilience (elastic limit taken at yield):

$$U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1527.89)(0.02667) = 20.37\ \text{N/mm}^2$$

Modulus of toughness:

$$U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1527.89+3183.10}{2}(0.02667) = 62.81\ \text{N/mm}^2$$

Answer: Modulus of resilience = 20.37 N/mm² (MJ/m³); modulus of toughness ≈ 62.81 N/mm² (MJ/m³).