Answer:
F_max = 184.08 lb
Explanation:
Given:
- The diameter of steel core d = 1/16 inch
- The thickness of copper shell t = 3/64 inch
- The allowable stress for steel = 60 ksi
- The allowable stress for copper = 24 ksi
Find:
- What tension can be applied without exceeding allowable limits.
Solution:
- Compute the cross sectional areas of the each material.
A_steel = pi*d^2/4
A_steel = pi*(1/16)^2 / 4
A_steel = 3.06796*10^-3 in^2
A_copper = pi*((d+2t)^2 - d^2) / 4
A_copper = pi*((5/32)^2 - (1/16)^2) / 4
A_copper = 0.016107 in^2
- We will use the allowable stresses to calculate the maximum force possible for each component of wire:
F_steel = sigma_steel * A_steel
F_steel = 60,000 * 3.06796 * 10^-3
F_steel = 184.08 lb
F_copper = sigma_copper * A_copper
F_copper = 24,000 * 0.016107
F_steel = 386.56 lb
- Hence, the maximum possible force that can be applied is:
F_max = min ( 184.08 , 386.56 )
F_max = 184.08 lb
