How to find the axial load rating of a deep grove ball bearing.

I

Iron Filing

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Hello,
I was reading the ratings of bearings here: Timken Deep Groove Ball Bearings.pdf and I couldn't find out what the ratings of axial loads on them are.

Take, for example, the number 628 bearing. It's Cr, or dynamic radial load rating is 3.3 kilo newtons. But the axial rating, Ca, is missing. I don't need much axial strength, 0.5 kilo newtons static and dynamic load would be overkill for my application. My goal is to build a centrifugal mixer, like for solder paste and thermal compound.

How would I discover the axial rating of a deep grove ball bearing?

Thanks

PS: I'm a total newbie here. So if I am misunderstanding something, please tell me.
 
There is not an axial rating for deep groove ball bearings, for the dynamic load, it is based on the bearing size, clearance and the axial and radial loads ratio and then uses those factors to get an equivalent load that is then used in the life formulas. You can use the tables in the bearing catalogs to calculate it, but the easiest way is to do as Monostre suggested and use the online calculators.
 
Here is a link to the formulas on the Schaeffler website, the process should be the same for Timken bearings, but you would want to use their tables for the calculations in case the bearing internals are different.

Deep groove ball bearings | Schaeffler medias

medias.schaeffler.us medias.schaeffler.us
 
Monostre said:
the easiest method would be a calculator like SKF Bearing Select.
Click to expand...

These catalogs have always required a bit of spreadsheet work. The calculator is a good shortcut. It's all in metric, though. 1 lbf is 4.448 N.

What you're looking for is hidden in the calculation of P. P is an equivalent load, which includes a radial and axial component. The radial and axial load terms (X and Y) are available but I couldn't find them quickly.

Here's a more detailed explanation. https://www.skf.com/group/products/...rating-life/equivalent-dynamic-bearing-load-p
 
When using the online calculator, I got stuck at the point where I enter the "load case". I'm not sure what's going on, but I just get errors when I try and calculate the entered data.
I'm not sure what I should use for X, Y , or Z. I'm assuming that Fx, Fy, and Fz are forces in the x, y, and z directions measured in kilo newtons. So I entered 0.15, 0.15, and 0.1 respectively.
The errors I get are: "Right: The minimum load requirement is not met. Other calculations may not be available." and "Right: the equivalent dynamic load is invalid."
I left other values at their defaults and set the shaft to vertical alignment with 101.6mm distance between bearings. EDIT: RPM is 1600.




I did the calculations using the schaeffler website's manual method using their table for e, X, and Y values:
X * Radial load in newtons * factor (which neither was suggested nor was not suggested.) + Y (estimated) * Axial load in newtons * factor (which was neither suggested nor not suggested.)
0.56*50*3+2.2*50*2 == 304 Newtons for the value P.


Now, what do I do with P?

Thanks!
 
Last edited:
The minimum load not met error means the bearing is not loaded enough to prevent skidding, so you need to select a different bearing that will meet the minimum load. Usually that just means a smaller bearing. How did you determine the loads on the bearing?

Once you have P, you use that in the life calculation to determine how long the bearing will last under your operating conditions. If that life is good enough, you’re done. If it comes up with a short life, then you will need to go back and see if a different bearing will help, or maybe a change to your design.

This link has the life calculations to use with the P value you calculated above.

Load carrying capacity and life | Schaeffler medias

medias.schaeffler.us medias.schaeffler.us
 
Ischgl99 said:
The minimum load not met error means the bearing is not loaded enough to prevent skidding, so you need to select a different bearing that will meet the minimum load. Usually that just means a smaller bearing. How did you determine the loads on the bearing?
Click to expand...

A centrifugal anything will generate centrifugal force. This force, if it's equal on both sides, cancels each other out at the bearing interface. I calculated what a 10 gram imbalance would cause and used that as my radial loading. I cannot imagine it would be worse as I can measure the weight of anything I stick into the machine within +/- 1 gram using a cheap scale and, I can borrow a actual old school scale if needed for even more precise measurements. As you might imagine, some of the material may move more than other portions, so some imbalance should be expected. But again, I don't see more than 10 grams moving out of balance. Just in case, I used 3x as my correction factor.

Now it's possible that I would consistently have an almost perfectly balanced apparatus, like within a newton or 2, in which case, given the minimum load requirement, *what will happen to the bearing*?



Because the shaft will be vertical, all the weight of the shaft and the centrifugal mixer's arm will be on the bearings. In an ideal world, this force would be distributed across the 2 bearings I intend to use. But IDK how close things work to the ideal. The cups of the centrifugal mixer would hold up to 1kg each of material. There are 2 cups. So, that's 2 kg. Now a kg maps to roughly to 1/10th of a newton or, 10 newtons per kg. The arm will be made of steel and the shaft of aluminum, most probably, so maybe 1kg for them. Also, I need to take into account the force I exert in placing the pales of paste into the machine. I figured 1kg for that as the cups use rubber bumpers, not force of some kind, to hold the paste pales in. Then I added 1kg for safety and used a factor of 2 just in case.

So, why an 8mm shaft? I reasoned that if something went waaaay out of balance, I'd like the machine to be able to shut down and recover gracefully as opposed to bending the shaft.



Ischgl99 said:
Once you have P, you use that in the life calculation to determine how long the bearing will last under your operating conditions. If that life is good enough, you’re done. If it comes up with a short life, then you will need to go back and see if a different bearing will help, or maybe a change to your design.

This link has the life calculations to use with the P value you calculated above.
Click to expand...

(constant / ( rpm to the -1) * (Cr / P ) ** ball bearing life exponent.
(16666/(1600**−1))*(3300/304)**3 == 34109270324.6 hours.

Well, this looks promising.
 
Iron Filing said:
(constant / ( rpm to the -1) * (Cr / P ) ** ball bearing life exponent.
(16666/(1600**−1))*(3300/304)**3 == 34109270324.6 hours.

Well, this looks promising.
Click to expand...
it's not, that actually tells me the bearings are way underloaded and could fail very quickly from skidding. I just watched a video on how these things work, it looks like it rotates at a decent speed, and it looks like you are using 1,600rpm in your calculation, so skidding is a possibility at that light a loading. You will have imbalance, no matter how good you are at machining and assembling, there will be some. How much will depend on the speed and how much weight difference there is in the assembly. You need to factor in the load from what is driving this too, do you have a pulley and some sort of belt? It looks like the rotating cups are driven off of the center shaft? That needs to be factored into your load equations. Most of your axial load will be taken up by one bearing, and you usually want one bearing floating axially so that the shaft can expand from heat. For what you are doing that might be minimal, but not knowing the design, it might not be either. In either case, it is usually a good idea to have a bearing that floats axially. You can also have bearings of different sizes at each end, they do not need to be identical, the important thing is that they are sized correctly for the loads on them.

I would not use an aluminum shaft for this, it has a much lower rigidity than steel and you could cause more vibration problems by doing that. Nothing is ever perfectly balanced, so your device will have some imbalance in it, and then the mixture would too. That imbalance is going to try and bend the shaft, so stiffer is better.

I couldn't find an O&M for these mixers, I am curious how the commercial units are built internally.
 
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