Question

A 9,000-lb load is suspended from the roof in a shopping mall with a 16-ft-long solid aluminum rod. The modulus of elasticity of aluminum is 10,000,000 psi. If the maximum rod elongation must be limited to 0.50 in. and the maximum stress must be limited to 30,000 psi, determine the minimum diameter that may be used for the rod (precision to 0.00).

Answer 1

Step-by-step explanation:

First, we can calculate the stress in the rod using the formula:

stress = load / (pi * (d/2)^2)

where load = 9,000 lb and d is the diameter of the rod in inches. We want the stress to be limited to 30,000 psi, so we can rearrange the formula to solve for the minimum diameter:

d = 2 * sqrt(load / (pi * stress))

Substituting the given values, we get:

d = 2 * sqrt(9000 / (pi * 30000)) = 1.22 in. (rounded to 0.01)

Therefore, the minimum diameter that may be used for the rod is 1.22 inches.