## Bolt Diameter

Consider a certain material with a shear strength of 50 MPa is machined into a bolt. If the bolt is subjected to a 7,000 N shear force, what is the minimum bolt diameter (to the nearest mm) needed to prevent failure?

Hint
Failure by Pure Shear:
$$\tau=\frac{F}{A}$$$where $$F$$ is the shear load 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 Shear:
$$\tau=\frac{F}{A}$$$where $$F$$ is the shear load 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 minimum bolt diameter:
$$A=\frac{F}{\tau}=\frac{\pi}{4}d^2$$$$$d=\sqrt{\frac{F(4)}{\tau \pi}}=\sqrt{\frac{(7,000N)(4)(m^2)}{(50\cdot 10^6N)\pi}}=\sqrt{\frac{28,000m^2}{157\cdot 10^6}}$$$
$$d=\sqrt{1.783\cdot 10^{-4}m^2}=0.01335\:m$$$Rounding up to the nearest mm to ensure non-failure: $$d=14\:mm$$$
14 mm