## required chucking force for horizontal lathe

## required chucking force for horizontal lathe

(OP)

What I am looking for is - if I chuck a hollow cylinder in a horizontal lathe, how much force is required for a chuck jaw (the one at the 'top', that is 12:00 if viewed from either end of the cylinder's axis) to retain the cylinder? I know that coefficient of friction factors in to this, and I think I can handle that part... the part that I do not have a handle on is how to I calculate the force that the weight of the cylinder is applying to the 'top' jaw of the chuck? In other words... if the cylinder were allowed to 'fall', provided that there is also a jaw at the bottom of the chuck, the top-most point at the chuck-end of the cylinder would describe an arc with a radius equal to the cylinder's diameter. There is force that the cylinder is applying to the top-jaw, along this arc, as it wants to pull itself out of the chuck - how much is it?

Now, one caveat - what about if it is a 3-jaw chuck? That means that the bottom-most point of constraint, the point that the part wants to rotate about, is only .86603 (sine of 60°) times the cylinder's diameter. It seems like that would increase the pull against the top jaw as the part now has more leverage than if it were constrained at the very bottom.

I will know the ID, OD, length and density of the cylinder so I should be able to calculate this - that is, if I were smart enough!

A typical scenario for me would be something like a 22" OD, 11" ID, 26" length - so with a density of .283 pounds per cubic inch for steel that is like 2,097 lbs. These parts are not actually hollow cylinders, they are flanged and I know for this particular size the weight is actually 1,094 lbs. but I am calculating it is a hollow cylinder for a factor of safety of around 2.

and by the way, is this a moment arm calculation? moment of inertia? I'd like to know what it really is called.

There are two places that I can get information on the forces applied by the machining operation, Harding and Kitagawa, they're chuck manufacturers... but they completely ignore the force applied by the part itself, and that scares the heck out of me. The chuck force required to counteract the machining looks to be around 91.5kN, if I did the calculation properly. This is within the capability of most power chucks, but it may not be safe if the weight of the part adds much to it.

Also, I might add, all of the calculations that I have seen are for front-turret lathes, so the machining force applied, at least in the case of the turning operation would actually serve to support the part and offset the force that I am concerned about. However, my machines are slant-bed lathes, so the turrets are in the rear, with the tools upside down as is the usual arrangement for slant-beds. So the weight of the part and the machining forces are added.

Thanks in advance,

Gary

Now, one caveat - what about if it is a 3-jaw chuck? That means that the bottom-most point of constraint, the point that the part wants to rotate about, is only .86603 (sine of 60°) times the cylinder's diameter. It seems like that would increase the pull against the top jaw as the part now has more leverage than if it were constrained at the very bottom.

I will know the ID, OD, length and density of the cylinder so I should be able to calculate this - that is, if I were smart enough!

A typical scenario for me would be something like a 22" OD, 11" ID, 26" length - so with a density of .283 pounds per cubic inch for steel that is like 2,097 lbs. These parts are not actually hollow cylinders, they are flanged and I know for this particular size the weight is actually 1,094 lbs. but I am calculating it is a hollow cylinder for a factor of safety of around 2.

and by the way, is this a moment arm calculation? moment of inertia? I'd like to know what it really is called.

There are two places that I can get information on the forces applied by the machining operation, Harding and Kitagawa, they're chuck manufacturers... but they completely ignore the force applied by the part itself, and that scares the heck out of me. The chuck force required to counteract the machining looks to be around 91.5kN, if I did the calculation properly. This is within the capability of most power chucks, but it may not be safe if the weight of the part adds much to it.

Also, I might add, all of the calculations that I have seen are for front-turret lathes, so the machining force applied, at least in the case of the turning operation would actually serve to support the part and offset the force that I am concerned about. However, my machines are slant-bed lathes, so the turrets are in the rear, with the tools upside down as is the usual arrangement for slant-beds. So the weight of the part and the machining forces are added.

Thanks in advance,

Gary

## RE: required chucking force for horizontal lathe

Depending on what surfaces need machining, you can also look at a faceplate operation instead of a chuck, with the workpiece

boltedin place.My 2 cents.

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.

## RE: required chucking force for horizontal lathe

"

With a rotating workpiece, as in a lathe, the inertial force of the workpiece helps to counteract the cutting forces being applied to it."More so with very large pieces where the ratio between the weight of the workpiece and volume/time rate of material removal are at a greater spread.

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.