Question

If a steel cable is rated to take 800-lb and the steel has a yield strength of 90,000psi, what is the diameter of the cable?

Answer 1

Answer:

d = 2.69 mm

Explanation:

Assuming the cable is rated with a factor of safety of 1.

The stress on the cable is:

σ = P/A

Where

σ = normal stress

P: load

A: cross section

The section area of a circle is:

A = π/4 * d^2

Then:

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

Replacing:

$d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches$

0.106 inches = 2.69 mm