Max Tension Load

Consider a 18-8 stainless steel bolt with a 0.164 inch diameter has a tensile strength of 80,000 psi. What is the max tension force (lb) the bolt can handle before failure with a minimum factor of safety of 3?

Expand Hint
Failure by Pure Tensile:
$$$\sigma=\frac{T}{A} $$$
where $$T$$ is the tension load, $$\sigma$$ is the tensile strength, and $$A$$ is the cross sectional area of the bolt or rivet.
Hint 2
Area of a circle:
$$$A=\frac{\pi}{4}d^2$$$
where $$d$$ is the diameter.
Failure by Pure Tensile:
$$$\sigma=\frac{T}{A} $$$
where $$T$$ is the tension load, $$\sigma$$ is the tensile strength, and $$A$$ is the cross sectional area of the bolt or rivet.

Area of a circle:
$$$A=\frac{\pi}{4}d^2$$$
where $$d$$ is the diameter.

Combining the two equations to solve for the max tension force:
$$$A=\frac{T}{\sigma}=\frac{\pi}{4}d^2$$$
$$$T=\sigma\frac{\pi}{4}d^2=(80,000\frac{lb}{in^2})\frac{\pi}{4}(0.164in)^2=1,689\:lb$$$
With a factor of safety of 3, the max tensional force is:
$$$\frac{1,689lb}{3}=563\:lb$$$
563 lb