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

Here are the solutions to Question 7:

Given:

• Diameter, $d = 18 \text{ mm}$
• Load, $F = 55 \text{ kN}$
• Extension, $\Delta L = 0.190 \text{ mm}$
• Original length, $L_0 = 202 \text{ mm}$

Step 1: Convert given values to SI units. $d = 18 mm = 0.018 m$ $F = 55 kN = 55 \times 10^3 N$ $\Delta L = 0.190 mm = 0.190 \times 10^{-3} m$ $L_0 = 202 mm = 0.202 m$

7.1.1 The stress in the bar

Step 2: Calculate the cross-sectional area ($A$) of the bar. The bar is round, so its area is: $A = \frac{\pi d^2}{4}$ $A = \frac{\pi (0.018 m)^2}{4}$ $A = \frac{\pi (0.000324)}{4} m^2$ $A \approx 2.54469 \times 10^{-4} m^2$

Step 3: Calculate the stress ($\sigma$) in the bar. Stress is defined as force per unit area: $\sigma = \frac{F}{A}$ $\sigma = \frac{55 \times 10^3 N}{2.54469 \times 10^{-4} m^2}$ $\sigma \approx 216130000 Pa$ $\sigma \approx 216.13 MPa$

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

7.1.2 The strain in the bar

Step 4: Calculate the strain ($\epsilon$) in the bar. Strain is defined as the change in length divided by the original length: $\epsilon = \frac{\Delta L}{L_0}$ $\epsilon = \frac{0.190 mm}{202 mm}$ $\epsilon \approx 0.00094059$

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

7.1.3 Young's modulus

Step 5: Calculate Young's modulus ($E$). Young's modulus is the ratio of stress to strain: $E = \frac{\sigma}{\epsilon}$ $E = \frac{216.13 \times 10^6 Pa}{0.00094059}$ $E \approx 229782000000 Pa$ $E \approx 229.78 GPa$

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