Dealing with unbalance tolerance on fasteners in high speed coupling (statistical vector combo?) 8

[electricpete]

Here's a coupling I'm looking at:
Bendix type diaphram coupling
7600 rpm
Total Coupling weight (hubs plus spacer) - 23 pounds
Imbalance tolerance in each plane is 4*W/N = 4*23/7600 ~ 0.01 inch-ounce ~ 0.3 inch-gram

The coupling includes 12 nuts and 12 bolts in each balance plane, at a radius of approx 3" from rotational axis. The vendor states each nut is controlled to within 0.05 grams and each bolt is controlled to within 0.05 grams. The vendor instructions state that the position of the nuts and bolts does not need to be controlled. The vendor will will assemble the coupling at their facility and perform balance, then they will disassemble and ship to us with the nuts and bolts not labeled for which position they belong to. We don't plan to (don't want to) perform any balance check on site.

That is their plan. Their plan does not seem reasonable to me since the fastener tolerance might easily exceed the balance tolerance.
I haven't had a chance to talk to them yet, but I'd like to be prepared when I get my opportunity to talk to them.

The 0.3 inch gram balance tolerance when considered at a 3" radius translates to 0.1g.

If they happen to balance with a +0.05g nut and +0.05g bolt at 12:00 position and -0.05g nut and -0.05g bolt at 6:00 position, and then we reassemble with these four pieces swapped, we have change in balance of 2*0.05 at each of 4 positions at a 3" radius or 1.2 g-inch... blows away our balance tolerance of 0.3g-inch.

I realize this exact scenario is very unlikely, but it's unsettling that by just changing position of 4 out of 24 fasteners potentially exceeds our imbalance tolerance. I'd like to be able to evaluate or understand it better.


MY QUESTIONS ARE AS FOLLOWS:
Q1 - Does above approach proposed by vendor seem reasonable or unreasonable?
Q2 - If you were going to attempt to establish a balance tolerance for the nuts and bolts in this situation, how would you do it?

Here's my initial rough attempt at item Q2
If we assumed N pieces with normal weight distribution and a standard deviation of sigma = S and , then the standard deviation of the scalar sum of the weights of the N pieces by central limit theorem is N*S/sqrt(N) = sqrt(N)*S.

In our case N=24 and the deviation of the scalar sum of weights is [sqrt(24)*S] ~ 5*S.

So far we are dealing with scalar sum and I don't know how to translate it to vector sum, but it seems reasonable (conservative?) to estimate the standard deviation of the vector sum (of unbalances) as the [standard deviation of the scalar sum of weights] times the radius.
[sqrt(N) * S] * R <= Balance Tolerance
[sqrt(24) * S] * 3*inch <= 0.3 gram*inch
S <= 0.3/(3*sqrt24) ~ 0.1/5 = 0.02 gram

Accordingly I will propose that the vendor either control the tolerance of each nut and each bolt to 0.02gram or else label the position of each nut and bolt.

Does this make sense?
Is there a better way to do statistical vector sum to estimate the effects of individual piece variability on overall unbalance?


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(2B)+(2B)' ?

 

[Strong]

Forget the math and statistical analysis and worst case scenario guessing. Have the Vendor match mark each bolt, nut and hub position after balancing. There is still possibility of eccentricity/unbalance when installing coupling on actual machine.

Walt

 

[electricpete]

Thanks, I appreciate the suggestion.

That is in fact my preference (to have vendor mark position of all fasteners).

I have requested labeling position of fasteners earlier and was told vendor doesn't want to do that... they have their own proven way of doing business which they believe to be acceptable. I am working though a 3rd party and lines of communications so far are weak and messy. But I have a conference call possibly as early as tomorrow and I'd like to prepared to address/evaluate all options, including a lower acceptable tolerance. I agree we cannot allocate the entire coupling unbalance to the fasteners, and that's a tricky part of it. As a starting point I need to be able to at least prove their approach is unacceptable through a more rigorous analysis than postulating swapping of selected fasteners (figure out a statistical way to combine balance tolerances).

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(2B)+(2B)' ?

 

[Tmoose]

23 lbs X 454g/lb ≈ 10440 grams

.3 gm-in/10440 gm ≈ 0.000,03" which is the simplistic eccentricity or change in centering that would induce the full 0.3 gm-in permissible residual imbalance.

If the coupling is to be balanced on an arbor made by humans, I doubt the centering accuracy can be that good. "Proving" the balance by rotating the coupling or pieces on the arbor requires a slip fit arbor (with clearance a good bit greater than 0.00003", which may not be at all similar to the shafting of the drive and driven equipment), or a tapered arbor, but a tapered arbor will only be able to snugly engage one hub at a time.

Then of course there is probably a key's mass to compensate for in each hub.

It is usually relatively easy to "balance" a rotor in a machine to a G "0.whatever" tolerance.
Having that condition survive to the actual installation is quite another story .

Similarly even claiming to have successfully balanced a part to a G "0.whatever" tolerance, just because the machine "said so," is danged risky in my opinion, as tooling vs reality is frought with so many similar and un-anticipated vagaries.

I'm sorry e-Pete.

Dan Timberlake

 

[electricpete]

I agree it is a ridiculously tight tolerance, owing to high speed and low weight. We don't have anything else similar. That's one reason we don't want to mess with balancing at our site.

Your logic started with my unbalance tolerance m*r=M*e = (4*W/N)*[unit-related constant] and then divided by M to get e = (4/N) *[same constant]. The calculated e would depend only on machine speed (independent of W=M*g).

So for any 7600rpm piece with 4W/N tolerance, we would come to same value of e (and same conclusion about difficulties) by your logic, regardless of weight, wouldn't we?



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(2B)+(2B)' ?

 

[Tmoose]

Hi EPete,

"So for any 7600rpm piece with 4W/N tolerance, we would come to same value of e (and same conclusion about difficulties) by your logic, regardless of weight, wouldn't we? ".

Exactly. But it Tain't really MY logic, despite having read, thought, talked, and written about it a few times.
"e" is embedded in the ISO rigid rotor tolerance as well. See the attached image.

It is common when we members are gathered at the luxurious balance folks club, sitting in comfy leather chairs before the roaring fire in the magnificent field stone and polished teak fireplace , sipping drinks and smoking our pipes, to proudly swap modest tales about routinely shop balancing to no higher than Grade 0.4, and often until the balance display's lights go out.

At those otherwise enjoyable times what goes thru my mind, having run a balance machine in a specialty shop for a while, and having had the pleasure of measuring the 1X vibration of my handiwork and the handiwork of others after it was back in service, is that the "residual imbalance" written either by hand or a computer on a shop balancing report needs an asterisk leading to a couple of paragraphs of heart breaking qualifying boilerplate about the centering and thus "balance" effects of balance tooling, shaft runout, commercial shaft fits in bearing races, and even ball and roller bearing runout.

 

[CouplingGuru]

Well said Tmoose,

Abor balancing is inherently limited, It is much better to roll a rigidized coupling, but even then you will find that the biggest issue is your bore and how well it is placed inside the the hub in comparison to the geometric constraints in manufacturing. Lots of coupling manufacturers buy hubs from overseas that are cheap then re-machine the bore per order. I don't care how good of a machinist you are unless you are making soft jaws for every hub bore set-up, which I doubt. Best case you may get 0.002" run-out from hub bore to axial positioning feature like the bolt circle or pilot diameters of a coupling. This severely limits a manufacturers ability balance a coupling. I will stand by this, a customers best bet is to request components of the coupling individually balanced as well as the hardware. Typically we do 0.01 gram, which sounds challenging, but mostly is it just a sorting procedure. In some rare occasions we take a file to the end of the bolt.

A residually balanced 4W/N coupling assembly is a myth, I have seen routinely in the balance cell, where you run the balance procedure take the coupling assembly out of the machine and put it back in the machine and re-run the test and it doesn't pass. This is especially true on high speed couplings. You will get it in spec, but then take your data pic, and don't breath on it.

Don't even get me started on 4W/N balancing with keyways, headspace and exposed key stock. . . . .

When it comes to couplings we are always here to help.

http://WWW.PSCCOUPLINGS.COM

 

[desertfox]

Hi electricpete

Found this link which discusses balancing of couplings, it also talks about balancing subassemblies and match marking components:-

http://info.ringfeder.com/engineers-blog/balancing-of-shaft-couplings

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[electricpete]

Lots of good comments.

Tmoose - that is a good way to put the balance tolerances in perspective: After a certain point, any improvement in balance is swamped by inaccuracies associated with centering during balance and operation.

For the particular coupling discussed in op, the tolerance stated was "per plane". So the computed e would be twice the 0.00003", i.e. 0.00006". That's still less than a tenth of a thousandths, so probably doesn't change the conclusion significantly. Reading that value of eccentricity off the chart, it's a little above ISO 1.0.... let's say it would be ISO 1.2 if we were allowed to vary that ISO grade continuously instead of stepwise.

If we judge our realistic centering error (change centering during balance to centering during installation) it might be on the order of 0.001". We might obtain some small benefit specifying a balance grade that yields e a factor of 10 lower, let's say 0.0001" (I agree no meaningful benefit beyond that). For 7600rpm that e=0.0001 would correspond to a balance grade around G2.5. For 3600rpm that would be closer to G1.0. All the way down at 1200rpm it is about G0.4, although I don't think many people would ask for G0.4 on a 1200rpm.

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(2B)+(2B)' ?