Theoretical Bearing Load Question

atomarc · Aug 2, 2018

This may be a silly question but I can't quite get my poor head around it.

If I have a shaft with a bearing at each end and I apply a force of 1000 pounds evenly to the shaft (radially), will each bearing share that load and see only 500 pounds of force? Probably not, huh!

I guess it would be akin to putting each foot on a bathroom scale and expecting each scale to read 1/2 your total weight. Probably not, huh!

I have spec'd many casters over the years and the SOP is to take the total expected weight and divide it by the number of casters..hence, a 8K weight would require 4 2K casters. What I don't understand is the difference between the first two scenarios and this caster scenario

Keep in mind I only made it to the 4th grade before reform school!

Stuart

HuFlungDung · Aug 2, 2018

IIRC, bearings were rated life for 10 million revolutions or something like that. A caster wheel likely has a factor of safety built in, and the rating wouldn't be the max load ever, but would be the max load continuous. A smart designer will anticipate a permanent unequal loading and spec accordingly.

atomarc · Aug 2, 2018

Well..this here is a dumb designer. If I have a bearing rated at 500 lbf and I use two of these on a shaft that see's 1000 lbf, am I screwed or does the fact that I have split the 1000 lbf between two bearings rated for 500 lbf keep me golden.

I don't know how else to explain this, although it seems intuitive to me that applying 1000 lbf to the shaft will impose that very same load anywhere on the shaft or bearings, not 1/2 the load on each bearing..but I don't know for sure so that's why I'm asking.

Stuart

CarbideBob · Aug 2, 2018

Put one foot on one bathroom scale and the other foot on another duplicate scale.
These are your two bearings if the shaft is loaded exactly in the middle.
Each one of your feet carries half the load. Lift one foot and all load is transferred.
Another way, lift 200 lbs with one leg in the air and then with both feet on the ground ... feel any difference in your leg/foot. We call this awkward lifting in ergo stuff.
Four caster calc would be real world wrong as uneven surfaces will inevitably transfer the the load to 3 or even 2 legs.

I do so get your thinking and puzzlement but the load is shared among supports in some fashion.
The total of these supports must be the entire load. It can not be greater.
The output end can not be bigger than the input.
Bob

HuFlungDung · Aug 2, 2018

Taking the bathroom scale as an example, you'd only need one. Put one foot on and slowly shift your Center Of Gravity (taking more weight on that leg) and what happens? The scale registers more and more weight. It doesn't instantly show a transfer of half, or of all of your weight.

A shaft with two bearings equidistant from the COG will share the same radial load. Wherever the COG is offset towards one wheel or the other, the load is split proportionally.

It is why leverage works.

JCByrd24 · Aug 2, 2018

A fairly level shaft with the load fairly well in the middle will absolutely have fairly equal weight on both bearings. How close depends on how level and how well you measure the distances, this is easy to do and also fairly easy to calculate the error. Standing on two scales introduces more variables, and is like eyeballing the level of the shaft and the centering of the load.

Multi-caster problems are much trickier once you are over 3 casters per support and likely need a significant safety factor included if critical.

Air Caster Systems - Heavy Load Moving Made Easy

atomarc · Aug 2, 2018

Wow..this fact certainly makes the design of the machine much easier. The available space for the bearings rated for the total load wasn't enough..but if the bearings need only be rated for half the load their size can be scaled down.

If I actually understand what has been noted above then it's good news for me, thanks.

Stuart

CalG · Aug 2, 2018

Two bearings on a shaft do not SHARE any load.

Think about a shaft supported by two V blocks. with both ends hanging out past the V blocks by some distance.

A load on one end of the shaft can actually tilt the opposite end up off the V blocks. The load on the near to load bearing now needs to carry the full load PLUS the full weight of the shaft. The single bearing will see a load GREATER than the applied load.

Sucks, but the math is easy. Distance from load to support is all it takes. Pay attention to the "sign", that is direction of the resultant.

No a load along the shaft may need a good understanding of stiffness etc. But a radial load is all proportionality (assuming infinite stiffness of the bearing mounting.

I just came in from the shop, where I just "hung a shaft" for a small tractor implement PTO intermediate shaft.

The 22 inch long shaft, splined at both ends, is supported by flange bearings. One about 3 inches from a 1 X 10 spline and the other end
about 6 inches in adjacent to a 1X15 spline. There is NO combination of loads that I can imagine that will ever be shared in balance between the two support bearings.

Fine with me! ;-)

atomarc · Aug 2, 2018

Cal,

You're getting pretty technical for me but I think I understand what you're saying. In my case, the load is centered exactly mid distance from both bearings so I think this would qualify as "sharing", as I understand the word.

To further complicate my original post, my application involves two rolls, one directly opposite the other, both the same diameter, length and both have identical bearing arrangements. The force involved is passing through the opening between the rolls and trying to force them apart. Much like a rolling mill.

My question now is, if the force on a single roll is shared equally by both bearings then is that force also shared equally by the bearings on the opposite roll that is seeing that force as well. Would that mean that each bearing of each roll is actually seeing 1/4th of the total load.

Sort of like standing on 4 bathroom scales at once.

Stuart

Tony Quiring · Aug 2, 2018

1/4 total load if equally distributed...

But shaft bending and other issues mean bearing may need to be larger just because.

If the pinch between the rolls needs to be a measured pressure then this is usually in psi which then is used to determine bearing loads.

If area is 2 sq inches and it needs 1000 psi to function then force is 2000 pounds.

Where we "step out" of state not our pay grade is that the force being between rolls needing to be the above 2000 pounds would that be divided across the 2 rolls at 1000 pounds each or does each need something else.

Real engineering needed here and better detail needed to be better than guesses.

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Mark Rand · Aug 2, 2018

In that case, (1000 load compression between two rolls), absolutely not. Each roll is applying the same load to the opposite roll. So both see the full 1000lbs.

Tony Quiring · Aug 2, 2018

Duh on our part...been working on fire dept bulldozers in 102 heat all day...brain not 100%...

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atomarc · Aug 2, 2018

I'll attach a very crude drawing I made while attempting to solve another question that was beyond my pay grade. The sketch will show you how the last set of crush rolls are configured and give a vague idea of the process.

Stuart

CalG · Aug 2, 2018

Now you are getting into the difference between force and weight.

The load has "weight" due to gravity. Gravity is only one direction. DOWN. Two rollers means that the "MASS" of the load is supported at two points (lines really), but at some angle from "down".

Think of the difference between picking up a length of wire rope in it's middle. The force needed to lift and hold the wire rope would be equal to the "weight" due to it's size and length.

Now, think about what happens if one end of that wire rope is secured to a wall, and you want to lift the wire rope "up" off the ground by pulling the other end horizontally. Whoa, lots of force needed there. The wire rope is in TENSION.

Similarly, if a round bar is being supported by two equal rollers, the distance between the support rollers is significant. If the supporting rollers are spaced at nearly the load's diameter, the spreading forces can be HUGE.

Resolving forces is all about Trigonometry.

sfriedberg · Aug 2, 2018

If you are crushing something between two rolls, both rolls see equal forces in opposite directions from the crushing. (They see additional forces due to gravity, etc). So the two bearings at the end of each roll can share load, more-or-less equally if your crushing load is more-or-less centered and uniform.

But the right-hand (example) bearings on the two different rolls do not get to share load. The right-hand machine wall in which those right-hand bearings are mounted will see the sum of the right-hand bearing loads in tension between the two bearing mounts. So if you put the stock being crushed all the way over at the right side of the rollers, the right-hand bearing mounts will each bear the full crushing force (because you are totally imbalanced), and the right-hand machine wall will see twice the crushing force (because it sees two equal bearing loads in opposite directions).

atomarc · Aug 2, 2018

The rolls are 2 1/2" in diameter and only 4" across the face, relatively small. The wood bar is only 3" wide at its max and will always be centered exactly between the bearings. I mashed a piece of that material on my hydraulic shop press and using a 1 1/4" steel bar was able to indent it .014 with 8000 pounds of force.

I'm hoping that because the machine will have a rolling-squeezing action versus a straight push/mash action that the forces to compress it won't exceed what I saw on my hydraulic press. I can find bearings that will fit the design of the machine frame with a dynamic rating of around 4300 lbs. If I use what I've learned here I hope to double that figure and safely crush 8600 lbs as that will be shared between the two bearings.

Stuart

Milland · Aug 2, 2018

Depending on the area of the 1.25" bar actually in contact with the wood, you might have higher or lower pressures during your test over what your roller mechanism will impart on the wood, as that compression will be a function of a gradient of the arc of contact of the roller. The larger the diameter of the rollers, the closer they'll approximate a "flat" pressing action, but round rollers can't apply a uniform load from beginning of contact through passing the narrowest gap.

You'll also see some "rebound" force as the wood travels from left to right in your sketch.

Will this be a single-pass process, or are you narrowing the gap for successive passes (like a rolling mill)? And if you're close to load limits for the bearings, you could substitute roller or needle element bearings for the ball element type it sounds like you're using.

atomarc · Aug 2, 2018

Milland said:
Depending on the area of the 1.25" bar actually in contact with the wood, you might have higher or lower pressures during your test over what your roller mechanism will impart on the wood, as that compression will be a function of a gradient of the arc of contact of the roller. The larger the diameter of the rollers, the closer they'll approximate a "flat" pressing action, but round rollers can't apply a uniform load from beginning of contact through passing the narrowest gap.

You'll also see some "rebound" force as the wood travels from left to right in your sketch.

Will this be a single-pass process, or are you narrowing the gap for successive passes (like a rolling mill)? And if you're close to load limits for the bearings, you could substitute roller or needle element bearings for the ball element type it sounds like you're using.

Yes..the crushing process is carried out in .007 steps through 4 sets of rolls. I have looked extensively at needle bearings and as you note, they do withstand much more dynamic force but there are facets of them that make them problematic in this design.

Realizing now that the measured force can be divided by the two bearings on a common roll make conventional bearings more appealing.

The larger diameter of the rolls allow the bars to more easily grabbed and pulled in as their radius is greater. I have measured approximately a .001 rebound on the material but that is inconsequential as the roll assemble is fully adjustable and the target thickness is quite flexible.

Stuart

HuFlungDung · Aug 3, 2018

What does crushing wood a few thousandths do for it?

Something that small, I would just decide what size shaft will handle the bending force and get some bearings to fit that shaft. Small ball bearings aren't very expensive. You could even put pairs of them on each end should the need arise.

atomarc · Aug 3, 2018

HuFlungDung said:
What does crushing wood a few thousandths do for it?

Something that small, I would just decide what size shaft will handle the bending force and get some bearings to fit that shaft. Small ball bearings aren't very expensive. You could even put pairs of them on each end should the need arise.

Crushing the Rosewood mashes the cells top and bottom and makes the surface much harder plus it changes the acoustic quality a bit. Tapping on the wood with a bare mallet before it's crushed leaves a divot..post crush, no divot! Xylophones used bare mallets..can't have divots in the bars.

The larger diameter rolls allow the bars to enter and get grabbed easier than small rolls. The current design of the machine also requires the larger rolls for clearance issues.

Stuart

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