Physics of lathe

[Downunder Bob]

Cactus Farmer The amount to work it takes to cut 4140 versus 12L14 is almost completely related to the shear strength of the material. No matter how sharp a cutter is, just ahead of the cutter is a zone that is deformed by the process - As it deforms it shears. 12L14 has a lot lower shear strength (that is why lead is added to the steel).

BTW machining heat treated and drawn 4140 is a lot easier to machine and produces a better finish. It is referreed to as 'hard turning'. you take much shallower cuts to compensate for the greater tensile strength. I usually rough turn it with the typical poor finish, heat treat and draw it to RC40, then finish the part. Depending on the application, that is good enough, or I can reharden and redraw it to the preferred characteristic.

The lead acts as a lubricant, as well as lowering the shear strength.

 

[whitmore]

On a simple lathe cut, what happens to the energy? Energy being force time distance and both are happening.
...
I guess some energy must go to shearing.

That's true; you have (at least) to break the chemical bond that holds the material together,
and the amount of energy must be at least the area of the newly created surface
times the energy per square inch that is the 'shear strength' of the material.
The Charpy shear test is intended to determine this energy.
There's also reformation of the bonds near the surface (cold working) especially
if the tool is dull... and plows rather than shearing.

That surface energy is why heavy cuts and coarse grinding is preferred for fast
material removal: it's more energy efficient, makes the least new surface.

On really tough materials, any fracture, rather than parting cleanly, takes a forked
path (forced sideways by inclusions like glass in fiberglass, or by vanadium
carbide needle crystals in tool steel) so that much additional energy can be
absorbed. Deformation like that causes local heating, so sharp hard tools
that slice clean are energy-savers; sharp abrasive grains are very effective. Accidental adhesion
with the tool causes friction heating, so cutting 'lubricant' also lessens the
energy need (it poisons the surface sites that would form weld-like bonds
with the cutting tool). The TiN coatings also deter bonding with some
sticky metals (aluminum for one). The cutting tool relief angle, and any chip-breaker
grooves, lessen the energy cost by reducing such adhesion, too.

Some alloys and some tools are just TOO weld-prone, or worse. That's why
diamond wheels are less effective than CBN on steel.

Chemistry, and physics, are complex enough that it's easier to cut-and-try than to predict.
For well-understood physics and quick cuts, use shaped-charge explosives. There,
you want prediction rather than trial-and-error.

 

[RandyM]

You have no idea what interests I may have, (aside from over the top comedic performances).

Well, obviously this topic is of no interest to you. Then maybe it best you refrain from posting until you have something of value to add.

 

[higgite]

The Turbo Encabulator post is the most “liked” post of any in this thread by two-fold. Maybe I'm wrong, but it struck me as some friendly humor for us non-physicists and physicists alike, nothing more. Let's keep the Friendly Forum friendly. Just saying.

Tom

 

[DHarris]

Whitmore, excellent explanation! Thank you

 

[Bill Kahn]

That's true; you have (at least) to break the chemical bond that holds the material together,
and the amount of energy must be at least the area of the newly created surface
times the energy per square inch that is the 'shear strength' of the material.
...

Thank you Whitmore. You made part of the puzzle simple. The energy goes, in part, to shearing and is proportional to a constant k1 (dependent on the cutter and the material), a constant k2 (dependent on the material alone), and the surface area created. Lots of other mechanisms explained in other posts on how the energy ends up as heat. And, in addition, energy is used to bend the chips. One poster said this was proportional to the cube of their thickness.

So, I now have a more refined question. If you look at the actual force on the cutter (conceptually easily measured with a force gauge, but I have now clearly learned ain't nothing easy in practice), and the linear distance there is a total amount of mechanical work done (e.g. energy expended) in some fixed time (yes, work per time is power, but I'll just think of work right now). Through multiple mechanisms much of that work becomes heat. Even the sound produced ultimately become heat. But, there seem to be two (at least) receivers of the energy which do not end up as heat:
1) The shearing used energy. (Yes, there is heating too, but the shearing itself uses energy independent of the heat). And 2) bending the chips. (Again, yes, the chips get hot, but simply bending a rod take energy independent of the rod getting warm).
So, a refined physics question...

How much of the mechanical work put into the workpiece ends up a heat? How much is in the shear? How much is in the bending?

I almost wonder if chip breakage is actually important separate from ugly surface-finish-damage-inducing sharf. If the chip didn't have to bend at all the I could cut 10 mil as easily as 1 mil. I have the sense that the radius of curvature at the end of a curly chip is less than in the middle. So, a set of broken chips of equal total length to a single long one I think would have less total curvature. And so less total energy.

Might it be the case the core mechanism enabling deeper cuts (all else being equal) is getting the chips to break so you don't have to bend them as much? Could bending the chips actually be a major use of the input mechanical energy?

Am loving this hobby--even while I muse on the physics, today I made my first internal threads (1" 20TPI). How absolutely satisfying to have made my own coupler and threaded rod.

-Bill

 

[Boswell]

I have looked at these videos several times and they are facinating. This time I noticed something I had not before. The chip is being compressed. If you look at the speed the metal is moving past the cutter, it looks to be much slower by at least half than the non-chip metal, that to me implies that it is being compressed and thus another energy sink.

 

[whitmore]

How much of the mechanical work put into the workpiece ends up a heat? How much is in the shear? How much is in the bending?

I almost wonder if chip breakage is actually important separate from ugly surface-finish-damage-inducing sharf.l

We can look at the motor power at idle, and under heavy cutting load, and get a fairly good idea
of the energy use, and that will relate somewhat to the sheared area (which we can tell from
examining the swarf). What one learns, becomes part of a SFM recommendation...
so looking at tabulated surface feet/minute for the material ought to be a starting
point (and refine the edge and coolant/lube if that recommendation isn't working).

Chip breakage may be VERY important, as it determines if the
swarf was a spring compressed on the cutting edge, or if the cold-work damage to the swarf
was high enough to produce an occasional brittle fracture. Broken chips means less friction, and probably
a cooler edge.

It isn't clear how to interpret chip breakage (or temperature, or color), though.
Chips fracture easily with a non-ductile or low tensile strength workpiece (cast iron).
In a sense, steel wool and bronze wool are useful spring-tempered products made on a lathe
in the absence of chip breakage.
So, if the chip character changes, probably consider resharpening. Otherwise, I dunno.

 

[john.oliver35]

I have a comment related to the where all of the energy ultimately ends up - this is probably not helpful to machining, but the engineer in me just had to blurt it out I believe that at the end of the day all of the energy has been dissipated as heat:

* Deformation of metal raises the temperature of atoms as they slide past each other
* flying chips hit the floor and transfer all of their kinetic energy to deformation and vibration of the colliding surfaces
* Bent chips either spring back to lower energy state or are permanently deformed, dissipating the bending energy imparted onto them as heat
* Audible noise and vibration are transferred to walls, floor, and air where they heat those bodies

In the end, every watt of electrical power that came from the outlet ends up heating the room. I guess one tiny exception would be the energy retained in residual stresses within a part. This energy would be stored as potential energy until released by a future machining operation, or as heat as the part warps!

 

[KBeitz]

Some of the energy is converted into light. Not all is in the band we can see. some is in the inferred range.