Please show all work!! 8) The 1-meter long, tapered bar with circular cross section illustrated below has large diameterof 0.5 m (at the top), small diameter of 0.25 m (at the bottom) and modulus E = 20 x 10⁹ Pa. It issubject to a point axial load of F =10 kN. Use the finite element method to calculate the bar'sresulting change in length. Compare your answer to the analytical solution (it's in the notes). Useat least 3 elements.TEF

The objective of the question is to calculate the change in length of a tapered bar under a given axial load using the finite element method and compare it with the analytical solution.First, we need to calculate the cross-sectional area of the bar at the top and bottom. The cross-sectional area of a circle is given by the formula A = πr^2, where r is the radius. The radius is half of the diameter, so the top area A1 = π(0.5/2)^2 = 0.196 m^2 and the bottom area A2 = π(0.25/2)^2 = 0.049 m^2.Next, we divide the bar into 3 elements, each with a length of 1/3 m. The area of each element can be approximated as the average of the areas at the two ends. So, the areas of the three elements are A1 = (0.196+0.196)/2 = 0.196 m^2, A2 = (0.196+0.049)/2 = 0.123 m^2, and A3 = (0.049+0.049)/2 = 0.049 m^2.The axial deformation of an element under load is given by the formula δ = FL/(AE), where F is the force, L is the length, A is the cross-sectional area, and E is the modulus of elasticity. Substituting the given values, we get the deformations of the three elements as δ1 = (10*10^3)*(1/3)/(0.196*20*10^9) = 0.0000085 m, δ2 = (10*10^3)*(1/3)/(0.123*20*10^9) = 0.0000136 m, and δ3 = (10*10^3)*(1/3)/(0.049*20*10^9) = 0.000034 m.The total deformation of the bar is the sum of the deformations of the three elements, which is δ = δ1 + δ2 + δ3 = 0.0000085 + 0.0000136 + 0.000034 = 0.0000561 m or 0.0561 mm.For comparison, the analytical solution for the deformation of a tapered bar under axial load is given by the formula δ = FL/(πr1r2E), where r1 and r2 are the radii at the top and bottom. Substituting the given values, we get δ = (10*10^3)*1/(π*(0.5/2)*(0.25/2)*20*10^9) = 0.000051 m or 0.051 mm. The finite element solution is slightly larger than the analytical solution, which is expected due to the approximation of the cross-sectional areas in the finite element method.The change in length of the tapered bar under the given axial load was calculated using the finite element method to be 0.0561 mm. This is slightly larger than the analytical solution of 0.051 mm, due to the approximation of the cross-sectional areas in the finite element method.