Final answer:
The displacement of end A when a 60-kN force is applied to the A-36 steel rod is 1.2 mm, calculated by balancing the spring force with the applied force and using the spring's stiffness of 50 MN/m.
Step-by-step explanation:
The question asks about the displacement of end A when a 60-kN force is applied to a 20-mm-diameter rod made of A-36 steel, which is opposed by a spring with a stiffness of k = 50 MN/m. To solve this, we consider the spring force which is equal to the displacement of the spring (x) multiplied by the stiffness of the spring (k). As the force applied to the rod is 60 kN, and this force must be balanced by the spring force to reach equilibrium, we can set up the equation: F = kx.
To find the displacement, x, we rearrange this to x = F/k.
First, we must convert the known force into meganewtons to match the given stiffness: 60 kN = 0.060 MN. Now, we can substitute into the equation: x = 0.060 MN / 50 MN/m, which gives us: x = 0.0012 m or 1.2 mm. Therefore, the displacement of end A when the 60-kN force is applied is 1.2 mm.