Thunderbird336
Mechanical
- May 16, 2013
- 30
Please forgive me if my terminology is not correct, I'm here on this forum precisely because I am not well versed in this topic...
What I am looking for is - if I chuck a hollow cylinder in chuck mounted to the spindle of a horizontal lathe, how much force is required for a chuck jaw (that is 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.
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.
Thanks in advance,
Gary
What I am looking for is - if I chuck a hollow cylinder in chuck mounted to the spindle of a horizontal lathe, how much force is required for a chuck jaw (that is 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.
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.
Thanks in advance,
Gary