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.

$$A = \frac{\pi D^2}{4} = \frac{\pi (0.016 \text{ m})^2}{4} = 0.000064\pi \text{ m}^2 \approx 2.0106 \times 10^{-4} \text{ m}^2$$

7.1.1 Calculate the tensile stress on the rod.

$$\sigma = \frac{F}{A} = \frac{8338.5 \text{ N}}{2.0106 \times 10^{-4} \text{ m}^2} \approx 41472697 \text{ Pa}$$ $$\sigma \approx 41.47 \text{ MPa}$$

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

7.1.2 Calculate the strain on the rod.

$$\epsilon = \frac{\Delta L}{L} = \frac{0.00125 \text{ m}}{5.2 \text{ m}} \approx 0.00024038$$

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

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

Given:

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

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

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 $10 \text{ mm} = 10 \text{ MPa}$ for stress and $10 \text{ mm} = 50 \times 10^{-6}$ for strain.)

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

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

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

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

7.2.3 Determine the percentage reduction in area.

$$\text{Percentage reduction in area} = \frac{A_0 - A_f}{A_0} \times 100\%$$ $$\text{Percentage reduction in area} = \frac{2.0106 \times 10^{-4} \text{ m}^2 - 1.0433 \times 10^{-4} \text{ m}^2}{2.0106 \times 10^{-4} \text{ m}^2} \times 100\%$$ $$\text{Percentage reduction in area} = \frac{0.9673 \times 10^{-4}}{2.0106 \times 10^{-4}} \times 100\% \approx 48.11\%$$

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

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

$$A = \frac{\pi}{4}(D_o^2 - D_i^2) = \frac{\pi}{4}((0.050 \text{ m})^2 - (0.030 \text{ m})^2)$$ $$A = \frac{\pi}{4}(0.0025 - 0.0009) \text{ m}^2 = \frac{\pi}{4}(0.0016) \text{ m}^2 = 0.0004\pi \text{ m}^2 \approx 1.2566 \times 10^{-3} \text{ m}^2$$

7.3.1 Calculate the normal stress in the tube.

$$\sigma = \frac{F}{A} = \frac{60000 \text{ N}}{0.0004\pi \text{ m}^2} \approx 47746482.9 \text{ Pa}$$ $$\sigma \approx 47.75 \text{ MPa}$$

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

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

Given:

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

$$\epsilon = \frac{\Delta L}{L} = \frac{0.00076 \text{ m}}{3.8 \text{ m}} = 0.0002$$

The strain is $\boxed{\text{0.0002}}$.

7.3.3 Determine Young's modulus of the material.

$$E = \frac{\sigma}{\epsilon} = \frac{47.746 \times 10^6 \text{ Pa}}{0.0002} \approx 238730000000 \text{ Pa}$$ $$E \approx 238.73 \text{ GPa}$$

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