# Torque for drill to shear through material

I'm attempting to figure out the torque required for a twist drill bit to shear through a material. I have no idea if this is even remotely correct but this is the process I used.

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.

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?

Bystander
Homework Helper
Gold Member
approximately accurate
33000*0.005 = 165 N.m
Area resisting shear = 91.6 mm^2
This is probably high. The material is not going to be shearing over the entire surface area, but deformation of chips could make up the difference.

Baluncore
2021 Award
A drill bit that is advanced at a fixed feed rate does two things. It shears the material, then rolls the material into a spiral, (or breaks it into chips), that will pass out through the spiral flutes of the drill.

The bottom of the cutting edge has a relief angle sufficient to prevent continued frictional contact with the material not yet displaced. The top of the cutting edge is subjected to frictional, where the swarf is curled. The hard TiC coating on drill bits is really only a benefit on top of the cutting edge for curling the swarf.

By using a lower feed rate with a closer pitch, more shearing must be done per unit length of hole, while the swarf is easier to roll into a spiral because it is thinner. The fine swarf produced will have less stability and a greater velocity in the drill flutes. By using a greater feed per turn, less shearing is being done while it is significantly harder, (a power law of feed rate), to roll the swarf or break it into chips.

The optimum feed rate will therefore be decided by the balance between the forces needed to shear and to deform the material.
Maximum RPM is decided by the temperature of the cutting edge, the drill material characteristics and the coolant used, (which may also modify friction).

For those reasons, the energy needed to drill a hole must consider shear, deformation and friction.
The torque requirement will be a function of feed rate since deformation of the sheared material plays such an important part. In my opinion, the quick and therefore economic drilling of holes may not need to consider shear if the drill is sharp. The greater proportion of the energy goes into curling the swarf.