In a tensile test on a round bar with a diameter of 18 mm, a load of 55 kN caused an extension of 0,190 mm when the original length was 202 mm. Calculate: 7.1.1 The stress in the bar, 7.1.2 The strain in the bar, 7.1.3 Young's modulus

$$d = 18 \text{ mm} = 0.018 \text{ m}$$ $$F = 55 \text{ kN} = 55 \times 10^3 \text{ N}$$ $$\Delta L = 0.190 \text{ mm} = 0.190 \times 10^{-3} \text{ m}$$ $$L_0 = 202 \text{ mm} = 0.202 \text{ m}$$

7.1.1 The stress in the bar

The bar is round, so its area is:

$$A = \frac{\pi d^2}{4}$$ $$A = \frac{\pi (0.018 \text{ m})^2}{4}$$ $$A = \frac{\pi (0.000324)}{4} \text{ m}^2$$ $$A \approx 2.54469 \times 10^{-4} \text{ m}^2$$

Stress is defined as force per unit area:

$$\sigma = \frac{F}{A}$$ $$\sigma = \frac{55 \times 10^3 \text{ N}}{2.54469 \times 10^{-4} \text{ m}^2}$$ $$\sigma \approx 216130000 \text{ Pa}$$ $$\sigma \approx 216.13 \text{ MPa}$$

The stress in the bar is $\boxed{\text{216.13 MPa}}$.

7.1.2 The strain in the bar

Strain is defined as the change in length divided by the original length:

$$\epsilon = \frac{\Delta L}{L_0}$$ $$\epsilon = \frac{0.190 \text{ mm}}{202 \text{ mm}}$$ $$\epsilon \approx 0.00094059$$

The strain in the bar is $\boxed{\text{0.00094059}}$.

7.1.3 Young's modulus

Young's modulus is the ratio of stress to strain:

$$E = \frac{\sigma}{\epsilon}$$ $$E = \frac{216.13 \times 10^6 \text{ Pa}}{0.00094059}$$ $$E \approx 229782000000 \text{ Pa}$$ $$E \approx 229.78 \text{ GPa}$$

Young's modulus is $\boxed{\text{229.78 GPa}}$.