- #1

- 1

- 0

Ultimate shear strength = Force/Area

Therefore Force = USS x Area

The USS of the material = 360 N/mm^2

I assume the only cutting surface on the drill bit is the tip which is approximately a cone shape. So I worked out the surface area of a cone

A = pi*r*(r+(√(h^2+r^2)))

Then subtracted the area of the circular base of the cone because we only care about the pointy bit of the cone. I'm thinking the surface area of the pointy part of the cone would equal the area resisting shear, assuming the drill was already starting in a pre-drilled hole rather than on a flat surface.

Area resisting shear = 91.6 mm^2

Put this into our first formula. F = 360 * 91.6 = 33 KN

This gives me a linear force but the drill is rotating so the force needs to be expressed as torque. And to do that I used

Torque = Force*radius = 33000*0.005 = 165 N.m of torque required to shear through the material.

Where radius is the radius of the drill bit (5mm)

So can this process give me an approximately accurate idea of the torque needed for a drill to cut through something? Or am I way off?