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:### i. Young's Modulus of ElasticityYoung's modulus (E) is given by the formula: Where:- is the stress,- is the strain.First, we need to calculate the stress ( ): Where:- is the force (load),- is the cross-sectional area.Given:- Diameter of the rod = 12.5 mm,- Length of the rod = 200 mm,- Maximum load = 58.5 kN.The cross-sectional area is: The stress is: Next, we calculate the strain : Where:- is the total extension,- is the original length.Given:- Total extension = 68.7 mm,- Original length = 200 mm.The strain is: Now, we can calculate Young's modulus : ### ii. Maximum StressThe maximum stress is already calculated in the first part: ### iii. Percentage Reduction in AreaThe reduction in area is given by the difference between the original area and the area at failure: Where:- is the original area,- is the area at failure.Given:- Original area ,- Diameter at failure = 7.35 mm.The area at failure is: The reduction in area is: The percentage reduction in area is: ### iv. Percentage ElongationThe percentage elongation is given by: Where:- is the total extension,- is the original length.Given:- Total extension = 68.7 mm,- Original length = 200 mm.The percentage elongation is: ### Summaryi. Young's Modulus of Elasticity: