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