How to test the gripping force of a chuck?

I've been installing a hydraulic chuck on the Traub CNC lathe, and it occurred to me that to have some means of testing gripping force would be mighty handy.

I've only operated it on much reduced hydraulic pressure at the moment during testing, but even at 5 bar as opposed to it's intended 22 bar I cannot move a bar tapping axially with a copper hammer - but that's a bit of a crude test!

Problem is there are too many variables. The jaws move a maximum of 3 mm, and being bored jaws are seated on serrations and bolted on, so themselves are movable.

Only method that I can come up with at the moment is to turn up a slug of lead, or other very malleable metal, grip it with the chuck, release and examine the (presumable) deformation. Then do the same with a manual chuck of similar size and compare them.

I suppose some sort of deformable tube filled with hydraulic oil and equipped with a suitable pressure gauge is another possibility?

Any suggestions welcome.   Google has found me this clever little gizmo:

http://www.schmachtl.cz/web_get_img_data?aID=

I'd imagine there are three pistons the the jaws bear on with lip seals communicating with a central chamber that the gauge attaches to.

Now can some mathematically versed member tell me how to relate the pressure read on the gauge, to the force on the three pistons. Piston area obviously influences it.

Working in Imperial units if (say) we have a pressure of 100 lbs/sq in reading on the gauge, each piston contributes 1/3rd so 33.333.. lbs sq  in  per piston, so (to make calculation easy) if we have a piston area of 0.333.. sq in is that a force of 100 lbs on each jaw.

Or is it more complicated than that?

It's a bit like the 'Dead Weight' testers you can get to calibrate gauges I suppose, but I've never used one or seen how that they are set up  Thanks Lew for your reply.

It doesn't look to be beyond the capabilities of a Madmodder to make such a device  I suppose the issue is to size the pistons so that a standard gauge reads sensible figures. I'd imagine lip seals are available in a multitude of sizes and it's be a case of choosing one that works out sensibly mathematically and making cylinders and pistons to suit.

Bleeding all the air from it might be fun. Andrew,

Dual o-ring seals should be sufficient for most hydraulic pumps (3 ksi or lower pressure).  I often design piston parts for pyrotechnic actuators that see something in excess of 60 ksi for short periods of time that allows me to design o-ring seals (30% compression) that use a single o-ring.  In essence, you want a "gap" between the piston & cylinder of .004/.002 inches.  I have (and am still "debugging") a spreadsheet for designing both static and dynamic o-ring seal glands from (SAE, NAS, and MS) standard o-ring sizes and materials.  If you need a hand in such an effort, drop me a line.

Basically, for a male gland piston/bore seal I need: Minimum ID of Bore, % of compression desired (which comes from SAE tables for desired pressure loads), and the o-ring material you would like to use.  Everything else is (almost) automagic.  This is a back burner project that moves ahead by fits and starts (and stops). Do you know Poldi hardness measurement system? Anyway the big idea is that you strike a plunger that presses a specimen of known strengt and a hard ball againsta unknown specimen. Indentations (of the ball) are measured from both specimens and hardness is deducted from the table. Anyway, this is pretty simple method and it produces results that are not 1:1 brinnel etc. hardness.

http://www.measureshop.biz/images/products/ultra/Ultra_tisk-193-i2.jpg

Now, we are not measuring Brinnell hadness here, but only a force. Just incidently same method:
http://home.iitk.ac.in/~kamalkk/Image10.gif


Could this idea work in sort of reverse engineering manner here? If a bush that has holes squarely against the jaws. Holes accomondate one ball bearing "ball" each. In the centre there is a round rod. When jaws are closed, an indentation is made. Now. If you have a shop press, or something close to it (I would not be surpriced, if someone had a deadweight calibrator) and you would use same size ball bearing to press indetations on the same rod. When you get close to same size indentation, you have pretty close the same force than the jaw?

Right or wrong?

I'm a little handicapped mechanically (I'm electrical engineer we did not read that much ironbashing). So, if I'm completelty off I don't mind corrected.

Pekka Happy if you find that idea usefull.

That meter surely looks spiffy and it's a instant reading.

My line of though was that you probably have a meter or set value for the pressure setting. Now, if you make a table of the pressure vs. force (indentation) you can set it later easy and if something does not check you can allways fall back known rod/ball/indetion size to check system up.

I saw once a measurement station for load cells, they had a really nice system to minimize frictions and enviromental effect. one thing that I found genious was that they were rotating the cylinder piston - all friction was sliding friction, static friction would be very much bigger loss.

Pekka Beat me to it Pekka.

A simpler idea though never tried it can be to make up a self contained container thats held in the jaws that will collapse. Link this to a pressure guage just make sure you take a reading from a designated point  ie 0 psi or 34 psi or whatever.
As cylinder compresses so will the internal pressure, sure I am missing something?
To pressurise could use oil or air, its low pressure so no harm should things go pear shaped.

Standard nitrile O rings will take in excess of psi and operate for years at 300 bar in a static arrangement. On moving pistons have had several Viton pieces last 8 1/2 years running 120bar at about 140 cycles per week with just one o ring 1mm cross section. I prefer to expand up on inner bore and compress o/d with minimal width guaranteed to work from zero psi. Some specs rely on pressure against the o ring to create a seal either on I/d and face they always pack up and distort.
Unless its a 1mm cross section o ring I usually go for a compression on o/d around 0.3mm for sub 1 1/4" bores, still nice and easy to screw 1.5 to 2.4 cross section together by hand. Do you want to know the pressure exerted, or the grip? a test of the actual gripping capacity could be done by gripping the plain part of a bolt in the chuck and using a torque wrench on the hexagon. you could then work out the surface area of the chuck jaws, and extrapolate mathematically from there (don't ask how, I leave that to the wife, she has a maths degree:-) Or have I missed the point? Basically I want to check that the way I've fitted it allows it to apply the force specified by the manufacturer (Schunk).  I need to adjust the operating hydraulic pressure to get the right amount of grip without exceeding the maximum pull on the chuck (22kN)

Researching (googling) this I've come across all sorts of interesting things. For instance I found one site with a video demonstrating how the grip reduces HUGELY as the speed increases due to the centrifugal force on the jaws. Sadly I now cannot find the link ! Pekka, although I've seen that Youtube video, the one I'm thinking of isn't youtube but a manufacturers demo one. It shows a device measuring the force and remotely monitoring it by wireless transmission on a hand held display. As the operator increases the spindle rpm the display shows the gripping force almost halving at top speed given a constant operating pressure

Phil, my machine hasn't got the facility to change pressure under program or panel control - it's a 19 mm spanner on the regulator  So really I'm only wanting to check it statically. I have to say that the little 3 cylinder hydraulic device looks quite easy to make, and even accepting that it won't be calibrated it will be a very useful comparative device. Apparently as the operating ramps in the chuck degrade the grip reduces rapidly, though to be honest it probably is a 'non problem' with my low usage.

Everyone
I am involved in a workshop area where huge (up to 7 ton) rollers (at a steel plant) are reconditioned. This involves recentering after welding it up and heat treatment, followed by machining down to size.

The current cracked chuck of the recentering lathe has 4 jaws, each approximately 70mm (2.76&#;&#;) wide, 90mm (3.54&#;&#;) long (griping area). The jaws of the replacement chuck is much narrower, only 30mm (1.18&#;&#;). The machinist working for 20 odd years on the lathe refuses to use the new chuck and I must convince him otherwise.

We all know that the magnitude of the contact area between the jaw and workpiece should not affect the gripping force (friction force caused by the clamping force due to the friction coefficient between the jaw and work piece). The surface pressure obviously increases as the contact area is reduced, but the gripping friction force should remain unchanged if the clamping force stays the same. However, in the limit, if one uses 4 knife-blades (assumed to be indestructable) to clamp the workpiece, in theory the gripping force should remain unchanged, but practically there must be a limit (apart from the obvious exceeding of the work piece&#;s elastic stress limit).
Moreover, the driven rolls (most work pieces) have non-circular cross sections where only half of the jaw width is actually clamping to the workpiece, reducing the contact area further.

Can anyone advise me on the effect of the magnitude of the contact area between the jaws and work piece and the practical narrowest limit, or give any relevant information to either support the machinist&#;s view or persuade him to use the new chuck?

Regards
GJdW
You asked a simple yet complicated question. Although theoretically there should be no difference, practically it is not quite so. If the jaws of your chuck are flat then you have only one point of initial contact which eventually become an area as the clamping force increases and workpiece deforms elastically with the growing contact stress. In practice, however, the jaw&#;s clamping area is not flat but a part of a cylinder of certain diameter. Hence, two different cases are possible in this situation. First, if the diameter of your workpiece is less that that of this cylinder then you have theoretically only point contact at the highest pint (middle) of the jaw. If otherwise, you have two points contact by the edges of the jaw. What I would recommend you to do is to draw in scale (using any CAD) the jaw and the workpiece and see what kind of contact you have now and what would happen with new jaws. I suspect that you might see the real picture which helps you to make an intelligent decision.

A few works about the elastic limit at the point of contact. In the both mentioned cases you have indentation so the maximum stress should not exceed the elastic limit of your work material (if you do not want to have marks from the jaws on your workpiece). However, in the second case this limit is achieved much faster (I mean, under mush lower clamping force).


Viktor
When gripping a part the contact area must be large enough to not exceed the yield strength of either part. If the chuck is rated for the part weight then calculating the actual contact area is all that is left. If these are hard jaws then the shaft will probably be the softest part. If they are steel soft jaws then the yield should be at least 30,000 psi. A contact of 3.54 X 1.18 X 30,000 will support 125,316 lbs of weight, sufficient for this part. If the jaws are aluminum then you must do a hardness test to determine the yield strength.
The disclaimer, the jaws must be machined to the same diameter as the shaft to use the above calculation. If these are hard jaws, then you will only have line contact in one or two places and will probably mark the grip area. When a guy in the shop says there is a problem, then 9 times out of 10, in my experience anyway, there's a problem! What he says the problem is, or his proposed solution of it, is often invalid. But somewhere in there, there's likely to be a problem - its just a case of quantifying it precisely. Echoing Barry's comments, it may just be the general wimpiness of the jaws themselves that is worrying this guy. I do know that there are a lot of dead and maimed people out there because large parts came out of chucks - it's a frightening thing to see and hear! Where did I say that there were not plenty of machinists who are adept at pinpointing problems? They frequently do not have the time or inclination to do this - it's not their job, but that doesn't necessarily mean that they are not potentially adept. And where did I use the term "lowly"? I would claim to be a reasonable machinist myself, but it's a funny thing - when I'm doing machining, I don't think like an engineer - I become an artisan, picking feeds and speeeds by instinct and feel, listening to the sounds the machine and process are making ("that doesn't sound right" etc) and making instinctive adjustments. I think machining is an art to some extent, like playing a musical instrument or driving a race car, and it's not necessaily the job of such practitioners to "pinpoint" problems. After all, if this particular operator had "pinpointed" the problem, assuming it exists, we wouldn't be having this discussion.