A 60 mm diameter, 0.5 m long aluminium rod is loaded axially in tension as shown.Aluminium has a modulus of elasticity of 69 GPa and a poisson’s ratio of 0.35. Determine (a)The axial strain, (b)The r

(a) The axial strain can be calculated using the formula ?=axial stress/Young's modulus. The axial stress can be calculated as force/cross-sectional area, where the cross-sectional area is ?r^2. Assuming a tensile load of 10 kN, the axial stress would be 10,000 N/(?(30 mm)^2) = 0.011 MPa. Therefore, the axial strain is ? = 0.011 MPa/69 GPa = 1.59 x 10^-7. (b) The radial strain can be calculated using the formula ?_r = -??, where ? is the Poisson's ratio and ? is the axial strain calculated in part (a). Substituting the values, we get ?_r = -0.35 x 1.59 x 10^-7 = -5.57 x 10^-8. (c) The decrease in diameter of the rod can be calculated using the formula ?d = d?_r, where d is the original diameter of the rod. Substituting the values, we get ?d = 60 mm x (-5.57 x 10^-8) = -3.34 x 10^-6 m = -3.34 ?m (negative value indicates decrease). (d) The new volume of the rod can be calculated using the formula V = ?r^2l, where r is the new radius (after considering the radial strain), and l is the original length of the rod. The new radius can be calculated as r = d/2 + ?d/2, where d/2 is the original radius. Substituting the values, we get r = 30.00000167 mm. Therefore, the new volume of the rod is V = ?(30.00000167 mm)^2(0.5 m) = 706.86 cm^3.