The cross-sectional area: A = (π/4) d^2 The circular rod shown is made of the steel alloy AISI 4140 OQT 900. It has a diameter of1.00 in and an initial length of 48 in. An axial tensile load of 15 000 lb. is applied during acertain operation.Compute:1. the equivalent spring constant K,2. the deformation X of the rod.The Young's modulus of the steel is known to be 30,000,000 psi.F=15 000 lbL= 48 inF=15 000 lb

The objective of the question is to calculate the equivalent spring constant and the deformation of a steel rod under a given axial tensile load. The properties of the rod and the load applied are provided.First, we need to calculate the cross-sectional area of the rod using the given formula A = (π/4) * d^2. Substituting the given diameter d = 1 in, we get A = (π/4) * (1 in)^2 = 0.7854 in^2.Next, we calculate the equivalent spring constant K. The formula for the spring constant is K = (A * E) / L, where A is the cross-sectional area, E is the Young's modulus, and L is the length of the rod. Substituting the given values, we get K = (0.7854 in^2 * 30,000,000 psi) / 48 in = 490,291 lb/in.Finally, we calculate the deformation X of the rod. The formula for deformation is X = F / K, where F is the force applied and K is the spring constant. Substituting the given values, we get X = 15,000 lb / 490,291 lb/in = 0.0306 in.The equivalent spring constant of the steel rod is 490,291 lb/in and the deformation of the rod under the given axial tensile load is 0.0306 in.

Answered: The circular rod shown is made of the…