QUESTION 7: STRESS, STRAIN AND YOUNG'S MODULUS A copper rod that is 5,2 m long is subjected to a tensile load of 850 kg. The rod elongates by 1,25 mm.

Here are the solutions to Question 7:

QUESTION 7: STRESS, STRAIN AND YOUNG'S MODULUS

7.1 A copper rod that is 5,2 m long is subjected to a tensile load of 850 kg. The rod elongates by 1,25 mm.

Given:

• Original length, $L = 5.2 \text{ m}$
• Mass of load, $m = 850 \text{ kg}$
• Elongation, $\Delta L = 1.25 mm = 0.00125 \text{ m}$
• Acceleration due to gravity, $g = 9.81 m/s^2$

Assumption: The original diameter of the copper rod is $16 \text{ mm}$, as provided in the context of Question 7.2 for a gauge length section, and no other diameter is given for 7.1.

• Original diameter, $D = 16 mm = 0.016 \text{ m}$

Step 1: Calculate the tensile force ($F$). $F = mg = (850 kg)(9.81 m/s^2) = 8338.5 N$

Step 2: Calculate the cross-sectional area ($A$). $A = \frac{\pi D^2}{4} = \frac{\pi (0.016 m)^2}{4} = 0.000064\pi m^2 \approx 2.0106 \times 10^{-4} m^2$

7.1.1 Calculate the tensile stress on the rod.

Step 3: Calculate the tensile stress ($\sigma$). $\sigma = \frac{F}{A} = \frac{8338.5 N}{2.0106 \times 10^{-4} m^2} \approx 41472697 Pa$ $\sigma \approx 41.47 MPa$

The tensile stress on the rod is $\boxed{41.47 MPa}$.

7.1.2 Calculate the strain on the rod.

Step 4: Calculate the strain ($\epsilon$). $\epsilon = \frac{\Delta L}{L} = \frac{0.00125 m}{5.2 m} \approx 0.00024038$

The strain on the rod is $\boxed{0.000240}$.

7.2 The following data were recorded during a tensile test on a steel specimen:

Given:

• Original diameter, $D_0 = 16 mm = 0.016 \text{ m}$
• Gauge length, $L_0 = 80 mm = 0.080 \text{ m}$
• Gauge length at fracture, $L_f = 92.16 \text{ mm}$
• Neck diameter at fracture, $D_f = 11.52 \text{ mm}$

Step 1: Calculate the original cross-sectional area ($A_0$). $A_0 = \frac{\pi D_0^2}{4} = \frac{\pi (0.016 m)^2}{4} = 0.000064\pi m^2 \approx 2.0106 \times 10^{-4} m^2$

Step 2: Calculate Stress and Strain for each data point.

• Stress ($\sigma = \frac{Load}{A_0}$)
• Strain ($\epsilon = \frac{Extension}{L_0}$)

| Load (kN) | Load (N) | Extension (mm) | Extension (m) | Stress ($\sigma$) (MPa) | Strain ($\epsilon$) | | :-------- | :------- | :------------- | :------------ | :---------------------- | :------------------ | | 0 | 0 | 0 | 0 | 0 | 0 | | 15 | 15000 | 0.0182 | 0.0000182 | 74.60 | 0.0002275 | | 30 | 30000 | 0.0363 | 0.0000363 | 149.21 | 0.00045375 | | 45 | 45000 | 0.0547 | 0.0000547 | 223.82 | 0.00068375 | | 60 | 60000 | 0.0724 | 0.0000724 | 298.42 | 0.000905 | | 75 | 75000 | 0.0910 | 0.0000910 | 373.03 | 0.0011375 |

7.2.1 Draw a stress-strain graph for these values.

(Due to the text-based format, I cannot draw the graph. However, the table above provides the calculated stress and strain values needed to plot the graph. The graph should have strain on the x-axis and stress on the y-axis. The points should be plotted and connected to form the stress-strain curve. The tip suggests a scale of $10 mm = 10 \text{ MPa}$ for stress and $10 mm = 50 \times 10^{-6}$ for strain.)

7.2.2 Determine Young's modulus of elasticity with the aid of the graph.

Step 3: Determine Young's Modulus ($E$) from the linear elastic region. Young's Modulus is the slope of the stress-strain graph in the elastic region. We can use any two points from the linear part of the table (e.g., from Load 15 kN to Load 60 kN). Let's use the points corresponding to 15 kN and 60 kN:

• Point 1: $(\epsilon_1, \sigma_1) = (0.0002275, 74.60 MPa)$
• Point 2: $(\epsilon_2, \sigma_2) = (0.000905, 298.42 MPa)$

$E = \frac{\Delta \sigma}{\Delta \epsilon} = \frac{\sigma_2 - \sigma_1}{\epsilon_2 - \epsilon_1}$ $E = \frac{(298.42 - 74.60) \times 10^6 Pa}{(0.000905 - 0.0002275)}$ $E = \frac{223.82 \times 10^6 Pa}{0.0006775} \approx 330361623 Pa$ $E \approx 330.36 GPa$

Young's modulus of elasticity is $\boxed{330.36 GPa}$.

7.2.3 Determine the percentage reduction in area.

Step 4: Calculate the final cross-sectional area ($A_f$). $A_f = \frac{\pi D_f^2}{4} = \frac{\pi (0.01152 m)^2}{4} = 0.00010432896\pi m^2 \approx 1.0433 \times 10^{-4} m^2$

Step 5: Calculate the percentage reduction in area. $Percentage reduction in area = \frac{A_0 - A_f}{A_0} \times 100\%$ $Percentage reduction in area = \frac{2.0106 \times 10^{-4} m^2 - 1.0433 \times 10^{-4} m^2}{2.0106 \times 10^{-4} m^2} \times 100\%$ $Percentage reduction in area = \frac{0.9673 \times 10^{-4}}{2.0106 \times 10^{-4}} \times 100\% \approx 48.11\%$

The percentage reduction in area is $\boxed{48.11\%}$.

7.3 A hollow steel tube has an outside diameter of 50 mm and an inside diameter of 30 mm. The tube is 3,8 m long and is subjected to an axial tensile load of 60 kN.

Given:

• Outside diameter, $D_o = 50 mm = 0.050 \text{ m}$
• Inside diameter, $D_i = 30 mm = 0.030 \text{ m}$
• Length, $L = 3.8 \text{ m}$
• Axial tensile load, $F = 60 kN = 60000 \text{ N}$

Step 1: Calculate the cross-sectional area ($A$) of the hollow tube. $A = \frac{\pi}{4}(D_o^2 - D_i^2) = \frac{\pi}{4}((0.050 m)^2 - (0.030 m)^2)$ $A = \frac{\pi}{4}(0.0025 - 0.0009) m^2 = \frac{\pi}{4}(0.0016) m^2 = 0.0004\pi m^2 \approx 1.2566 \times 10^{-3} m^2$

7.3.1 Calculate the normal stress in the tube.

Step 2: Calculate the normal stress ($\sigma$). $\sigma = \frac{F}{A} = \frac{60000 N}{0.0004\pi m^2} \approx 47746482.9 Pa$ $\sigma \approx 47.75 MPa$

The normal stress in the tube is $\boxed{47.75 MPa}$.

7.3.2 If the extension of the tube is 0,76 mm, calculate the strain.

Given:

• Elongation, $\Delta L = 0.76 mm = 0.00076 \text{ m}$

Step 3: Calculate the strain ($\epsilon$). $\epsilon = \frac{\Delta L}{L} = \frac{0.00076 m}{3.8 m} = 0.0002$

The strain is $\boxed{0.0002}$.

7.3.3 Determine Young's modulus of the material.

Step 4: Determine Young's Modulus ($E$). $E = \frac{\sigma}{\epsilon} = \frac{47.746 \times 10^6 Pa}{0.0002} \approx 238730000000 Pa$ $E \approx 238.73 GPa$

Young's modulus of the material is $\boxed{238.73 GPa}$.