14. (a). A tensile test was conducted on a mild steel rod The following data was obtained from the test Diameter of stcel rod=12.5m Length of the rod=200mm Extension under load of 35KN=0.28 Maximum load=58.5KN Total extension=68.7mm Diameter of rod at failure=7.35mm Determine: hat (1) The value of the Young's modulus of elasticity; ii The maximum stress; iii. The percentage reduction in area; iv.The percentage elongation. (10 marks)

Let's go through each part of the problem step-by-step:<br /><br />### i. Young's Modulus of Elasticity<br /><br />Young's modulus (E) is given by the formula:<br />\[ E = \frac{\sigma}{\varepsilon} \]<br /><br />Where:<br />- \(\sigma\) is the stress,<br />- \(\varepsilon\) is the strain.<br /><br />First, we need to calculate the stress (\(\sigma\)):<br /><br />\[ \sigma = \frac{F}{A} \]<br /><br />Where:<br />- \(F\) is the force (load),<br />- \(A\) is the cross-sectional area.<br /><br />Given:<br />- Diameter of the rod = 12.5 mm,<br />- Length of the rod = 200 mm,<br />- Maximum load = 58.5 kN.<br /><br />The cross-sectional area \(A\) is:<br />\[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{12.5 \, \text{mm}}{2}\right)^2 = \pi \left(6.25 \, \text{mm}\right)^2 = \pi \times 39.0625 \, \text{mm}^2 \approx 122.7 \, \text{mm}^2 \]<br /><br />The stress \(\sigma\) is:<br />\[ \sigma = \frac{58.5 \, \text{kN}}{122.7 \, \text{mm}^2} = \frac{58500 \, \text{N}}{122.7 \, \text{mm}^2} \approx 476.8 \, \text{MPa} \]<br /><br />Next, we calculate the strain \(\varepsilon\):<br /><br />\[ \varepsilon = \frac{\Delta L}{L_0} \]<br /><br />Where:<br />- \(\Delta L\) is the total extension,<br />- \(L_0\) is the original length.<br /><br />Given:<br />- Total extension = 68.7 mm,<br />- Original length = 200 mm.<br /><br />The strain \(\varepsilon\) is:<br />\[ \varepsilon = \frac{68.7 \, \text{mm}}{200 \, \text{mm}} = 0.3435 \]<br /><br />Now, we can calculate Young's modulus \(E\):<br /><br />\[ E = \frac{476.8 \, \text{MPa}}{0.3435} \approx 1390 \, \text{GPa} \]<br /><br />### ii. Maximum Stress<br /><br />The maximum stress is already calculated in the first part:<br />\[ \sigma = 476.8 \, \text{MPa} \]<br /><br />### iii. Percentage Reduction in Area<br /><br />The reduction in area is given by the difference between the original area and the area at failure:<br /><br />\[ \Delta A = A_0 - A_f \]<br /><br />Where:<br />- \(A_0\) is the original area,<br />- \(A_f\) is the area at failure.<br /><br />Given:<br />- Original area \(A_0 = 122.7 \, \text{mm}^2\),<br />- Diameter at failure = 7.35 mm.<br /><br />The area at failure \(A_f\) is:<br />\[ A_f = \pi \left(\frac{7.35 \, \text{mm}}{2}\right)^2 = \pi \left(3.675 \, \text{mm}\right)^2 = \pi \times 13.47 \, \text{mm}^2 \approx 42.3 \, \text{mm}^2 \]<br /><br />The reduction in area \(\Delta A\) is:<br />\[ \Delta A = 122.7 \, \text{mm}^2 - 42.3 \, \text{mm}^2 = 80.4 \, \text{mm}^2 \]<br /><br />The percentage reduction in area is:<br />\[ \text{Percentage Reduction} = \frac{\Delta A}{A_0} \times 100 = \frac{80.4 \, \text{mm}^2}{122.7 \, \text{mm}^2} \times 100 \approx 65.5\% \]<br /><br />### iv. Percentage Elongation<br /><br />The percentage elongation is given by:<br />\[ \text{Percentage Elongation} = \frac{\Delta L}{L_0} \times 100 \]<br /><br />Where:<br />- \(\Delta L\) is the total extension,<br />- \(L_0\) is the original length.<br /><br />Given:<br />- Total extension = 68.7 mm,<br />- Original length = 200 mm.<br /><br />The percentage elongation is:<br />\[ \text{Percentage Elongation} = \frac{68.7 \, \text{mm}}{200 \, \text{mm}} \times 100 \approx 34.35\% \]<br /><br />### Summary<br /><br />i. Young's Modulus of Elasticity: