Tensile test numerical, 40 mm specimen

Given: $d = 40$ mm, $L_0 = 200$ mm; in the elastic range $P = 40$ kN gives $\delta = 2$ mm; $P_y = 160$ kN; $P_u = 240$ kN; $L_f = 250$ mm.

Step 1: area and stresses

$$\begin{aligned} A_0 &= \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2\\ \sigma_y &= \frac{160\times10^3}{1256.64} = 127.32\ \text{N/mm}^2\\ \sigma_u &= \frac{240\times10^3}{1256.64} = 190.99\ \text{N/mm}^2 \end{aligned}$$

Step 2: modulus of elasticity (from the elastic reading)

$$\sigma = \frac{40\times10^3}{1256.64} = 31.83\ \text{N/mm}^2,\quad \varepsilon = \frac{2}{200} = 0.01,\quad E = \frac{31.83}{0.01} = 3183.1\ \text{N/mm}^2$$

Step 3: modulus of resilience (up to yield)

$$U_r = \frac{\sigma_y^2}{2E} = \frac{(127.32)^2}{2(3183.1)} = 2.546\ \text{N/mm}^2$$

Step 4: modulus of toughness

Strain at fracture: $\varepsilon_f = \dfrac{250-200}{200} = 0.25$. Taking the average of yield and ultimate stress over the whole strain:

$$U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{127.32+190.99}{2}(0.25) = 39.79\ \text{N/mm}^2$$

Answer: Modulus of resilience = 2.55 N/mm² (MJ/m³); modulus of toughness ≈ 39.79 N/mm² (MJ/m³).