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. 7.1.1 Calculate the tensile stress on the rod. 7.1.2 Calculate the strain on the rod.

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QUESTION 7: STRESS, STRAIN AND YOUNG'S MODULUS

7.1.1 Calculate the tensile stress on the rod.

Step 1: Calculate the tensile force $F$. The mass $m = 850 \text{ kg}$. The acceleration due to gravity $g = 9.8 m/s^2$. $F = mg = 850 kg \times 9.8 m/s^2 = 8330 N$

Step 2: State the formula for tensile stress. Tensile stress $\sigma = \frac{F}{A}$, where $F$ is the tensile force and $A$ is the cross-sectional area of the rod. Note: The cross-sectional area of the copper rod is not provided in the question. Therefore, a numerical value for the tensile stress cannot be calculated. The tensile force is $\boxed{8330 N}$.

7.1.2 Calculate the strain on the rod.

Step 1: Identify the original length and the elongation. Original length $L = 5.2 \text{ m}$. Elongation $\Delta L = 1.25 mm = 0.00125 \text{ m}$.

Step 2: Calculate the strain $\epsilon$ using the formula $\epsilon = \frac{\Delta L}{L}$. $\epsilon = \frac{0.00125 m}{5.2 m}$ $\epsilon \approx 0.00024038$ The strain on the rod is $\boxed{0.000240}$.

7.2.1 Draw a stress strain graph for these values.

Step 1: Calculate the original cross-sectional area $A_0$. Original diameter $D_0 = 16 mm = 0.016 \text{ m}$. $A_0 = \frac{\pi D_0^2}{4} = \frac{\pi (0.016 m)^2}{4} = \frac{\pi \times 0.000256 m^2}{4} = 0.000064\pi m^2 \approx 2.0106 \times 10^{-4} m^2$

Step 2: Calculate the stress ($\sigma = \frac{Load}{A_0}$) and strain ($\epsilon = \frac{Extension}{L_0}$) for each data point. Gauge length $L_0 = 80 mm = 0.080 \text{ m}$.

| Load (kN) | Extension (mm) | Stress ($\sigma = Load/A_0$) (MPa) | Strain ($\epsilon = Extension/L_0$) | |:---------:|:--------------:|:---------------------------------------:|:---------------------------------------:| | 0 | 0 | 0 | 0 | | 15 | 0.0182 | $\frac{15 \times 10^3}{2.0106 \times 10^{-4}} \approx 74.60$ | $\frac{0.0182}{80} = 0.0002275$ | | 30 | 0.0363 | $\frac{30 \times 10^3}{2.0106 \times 10^{-4}} \approx 149.21$ | $\frac{0.0363}{80} = 0.00045375$ | | 45 | 0.0547 | $\frac{45 \times 10^3}{2.0106 \times 10^{-4}} \approx 223.91$ | $\frac{0.0547}{80} = 0.00068375$ | | 60 | 0.0724 | $\frac{60 \times 10^3}{2.0106 \times 10^{-4}} \approx 298.42$ | $\frac{0.0724}{80} = 0.000905$ | | 75 | 0.0910 | $\frac{75 \times 10^3}{2.0106 \times 10^{-4}} \approx 373.02$ | $\frac{0.0910}{80} = 0.0011375$ |

Step 3: Describe how to draw the graph. Plot the calculated stress values on the y-axis and the strain values on the x-axis. Use the given scale: $10 mm = 10 \text{ MPa}$ for the stress axis and $10 mm = 50 \times 10^{-6}$ for the strain axis. The points should form a roughly linear relationship, representing the elastic region of the material. (As an AI, I cannot physically draw the graph, but the table above provides the necessary data points for plotting.)

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

Step 1: Young's modulus $E$ is the slope of the linear portion of the stress-strain graph. We can calculate it using any point in the elastic region (e.g., the last data point before yielding, assuming the given data is within the elastic range). Using the last data point: $\sigma = 373.02 \text{ MPa}$ and $\epsilon = 0.0011375$. $E = \frac{\sigma}{\epsilon} = \frac{373.02 \times 10^6 Pa}{0.0011375}$ $E \approx 3.279 \times 10^{11} Pa = 327.9 GPa$ Young's modulus of elasticity is $\boxed{327.9 GPa}$.

7.2.3 Determine the percentage reduction in area.

Step 1: Calculate the original area $A_0$ and the final area $A_f$. Original diameter $D_0 = 16 \text{ mm}$. Neck diameter at fracture $D_f = 11.52 \text{ mm}$. $A_0 = \frac{\pi D_0^2}{4} = \frac{\pi (16 mm)^2}{4} = 64\pi mm^2$ $A_f = \frac{\pi D_f^2}{4} = \frac{\pi (11.52 mm)^2}{4} = \frac{\pi \times 132.7104 mm^2}{4} = 33.1776\pi mm^2$

Step 2: Calculate the percentage reduction in area. $Percentage reduction in area = \frac{A_0 - A_f}{A_0} \times 100\%$ $Percentage reduction in area = \frac{64\pi mm^2 - 33.1776\pi mm^2}{64\pi mm^2} \times 100\%$ $Percentage reduction in area = \frac{64 - 33.1776}{64} \times 100\%$ $Percentage reduction in area = \frac{30.8224}{64} \times 100\% = 0.4816 \times 100\% = 48.16\%$ The percentage reduction in area is $\boxed{48.16\%}$.

7.3.1 Calculate the normal stress in the tube.

Step 1: Calculate the cross-sectional area $A$ of the hollow tube. Outside diameter $D_o = 50 mm = 0.050 \text{ m}$. Inside diameter $D_i = 30 mm = 0.030 \text{ m}$. $A = \frac{\pi}{4}(D_o^2 - D_i^2) = \frac{\pi}{4}((0.050 m)^2 - (0.030 m)^2)$ $A = \frac{\pi}{4}(0.0025 m^2 - 0.0009 m^2) = \frac{\pi}{4}(0.0016 m^2) = 0.0004\pi m^2 \approx 1.2566 \times 10^{-3} m^2$

Step 2: Calculate the normal stress $\sigma$ using the formula $\sigma = \frac{F}{A}$. Axial tensile load $F = 60 kN = 60 \times 10^3 \text{ N}$. $\sigma = \frac{60 \times 10^3 N}{0.0004\pi m^2}$ $\sigma \approx 47746482.9 Pa \approx 47.75 MPa$ 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.

Step 1: Identify the original length and the extension. Original length $L = 3.8 \text{ m}$. Extension $\Delta L = 0.76 mm = 0.00076 \text{ m}$.

Step 2: Calculate the strain $\epsilon$ using the formula $\epsilon = \frac{\Delta L}{L}$. $\epsilon = \frac{0.00076 m}{3.8 m}$ $\epsilon = 0.0002$ The strain is $\boxed{0.0002}$.

7.3.3 Determine Young's modulus of the material.

Step 1: Use the calculated stress from 7.3.1 and strain from 7.3.2. Stress $\sigma = 47.75 MPa = 47.75 \times 10^6 \text{ Pa}$. Strain $\epsilon = 0.0002$.

Step 2: Calculate Young's modulus $E$ using the formula $E = \frac{\sigma}{\epsilon}$. $E = \frac{47.75 \times 10^6 Pa}{0.0002}$ $E = 238.75 \times 10^9 Pa = 238.75 GPa$ Young's modulus of the material is $\boxed{238.75 GPa}$.

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