Tires A physics question of leverage

[jimmysisson]

Agree w Egon and others. Just imagine the axle is ten feet long rather than 6" further out than normal. Just like a breaker bar w a pipe slipped over it. More torque to break free the [bearing] nut.
Jim

 

[AxleHub]

The total load stays the same. The internal bearing forces will change.

Egon . . Your post is correct.

And now for thread participants in general: The initial topic created by the OP was (for sidehill benefit and stability) either moving the rim outward or using spacers. In either scenario we all pretty much agreed they would be similar result.

But the OP also indicated he would use both. For those who say this is was off topic . . they were wrong because the OP generated that issue as the result of the original issue. But total load against the rear axle stays the same . However going from 0 inches of extension on each side . . Then 3 or 4 inches oon each side . . And then 6 or 8 inches of extension on each side alters the load pressure points. Whether flat land or sidehill there is a shift in angular presdure of the same load. Longer distances of gap between rim and bearings shifts the load to the outer edges of the bearings and alters the pressure points on the lugs. But add in sidehill pressure changes even further increases the shift of the load weight . . . Same weight . . . But instead of lugs and bearing having the weight spread evenly across them . . the weight is shift to more weight in specific areas of the lugs and bearings.

Now 4 inches of spacing adjustment of the rims (engineered by manufacturer) on each side (left rim) and 4 inches on right side (right rim) may be calculated by maker to be acceptable (with limited sudehill use and recommended load weight). But 4 added inches of spacers (4" on left and 4" right) is no whete in the engineering calculations too.

Because an axle supports a weight does not mean it does not deflect from perfectly straight . . and sidehills change pressure points and deflection if an axle changes pressure points.

Those pressure point changes impact lugs and bearings.

Had the thread OP only addresses rim widening OR spacers . . this thread would have been completed long ago. But the OP stated whatever the answers were (rims or spacers) he was doing both and others also said they'd do both . . and so BOTH stories are on topic and both are issues to be considered imo.

If someone wants to do both . . that's their decision . . but leverage changes effect parts and longevity of parts . . and claiming there is no harm to doing both imo is not factual . . its an opinion that an engineer could determine factually as correct or incorrect . . but not us.

 

[orezok]

I believe that some people ave viewing this situation as a lever. It is not. It should be viewed as a simple beam supported at both ends by tires and with a concentrated load at the center.

 

[Egon]

Think you'd have to make that several cantilered beams with fixed ends.

 

[orezok]

Think you'd have to make that several cantilered beams with fixed ends.

I don't think so. The rigid case of the transmission makes it one continuous beam. The axles are not connected, but the assembly is a beam.

 

[Egon]

Sorry, two cantilevered beams. There is no continuity for one beam. The axle housing has independent forces acting within it.

 

[orezok]

I’ll try to explain it in a simpler example.

I want to turn my tractor over on its side so I can easily polish the undercarriage. I bring in my good friend Hulk Hogan to turn it over. Using an example of 1000# centered on the rear axle, Hulk can’t lift enough at the existing 2’ axle stub (500#) to turn it over. Hulk adds a 2’ cheater to the axle making it possible for him to exert 250# of upward force to rotate the tractor (250# x 4’=1000# of rotational torque). Shine away.

Next time I need to do this Hulk isn’t available and I can’t lift 500 or even 250#. So I add a 6’ cheater to the axle and I only have to lift 125#, which in my dreams I can do.

The bottom line is that the length of the axle has no effect on the load on the bearing at the end of the transmission case stub. It was 1000# with the factory axle, 1000# with Hulks 2’ cheater and 1000# with my 6’ cheater.

Now this is not a perfect example as because of the rigid transmission case, the “beam” would rotate on the opposite tire, not the center of the transmission as if it was a lever condition. Tractors do not have independent rear suspension.

 

[AxleHub]

Orezok, I totally disagree.

1. That axle is not a single item . . It is based on a series of bearing and differential components which is why there is such a thing as a diff lock.

2. Engineering is based on directional capacity to determine load testing and limits. The bearings are based on a vertical solidarity in order to support those axle areas that you consider "ridgid" but in reality are merely supported components.

3. Deflection is a part of everything we have contact with. Even gravity or light are subject to deflection and certainly any metal rod is a slave to deflection. Bearings support the axle and every inch the wheels/rims are moved away from those bearing surfaces adds to deflection.

4. I or others could mention bearing drag, or could mention bearing surface design, or the angular stresses that deflection has on vertical bearings. Or the fact that the frame is what supports the transmission/differential/axle assembly.

In effect in your example . . You don't want hulk hogan or a cheater bar to lift 1 side of your tractor . . you want 2 hogan s so each side can be equally lifted upward . . not side to side . . Because the axle or wheel rims or bearings or lugs are NOT designed for side force as your example implies imo.

 

[orezok]

Orezok, I totally disagree.

1. That axle is not a single item . . It is based on a series of bearing and differential components which is why there is such a thing as a diff lock.

2. Engineering is based on directional capacity to determine load testing and limits. The bearings are based on a vertical solidarity in order to support those axle areas that you consider "ridgid" but in reality are merely supported components.

3. Deflection is a part of everything we have contact with. Even gravity or light are subject to deflection and certainly any metal rod is a slave to deflection. Bearings support the axle and every inch the wheels/rims are moved away from those bearing surfaces adds to deflection.

4. I or others could mention bearing drag, or could mention bearing surface design, or the angular stresses that deflection has on vertical bearings. Or the fact that the frame is what supports the transmission/differential/axle assembly.

In effect in your example . . You don't want hulk hogan or a cheater bar to lift 1 side of your tractor . . you want 2 hogan s so each side can be equally lifted upward . . not side to side . . Because the axle or wheel rims or bearings or lugs are NOT designed for side force as your example implies imo.

Given all of that, when you run over a rock with one rear wheel the tractor lifts only one side and the whole ASSEMBLY acts like a single beam.

 

[SPYDERLK]

I’ll try to explain it in a simpler example.

I want to turn my tractor over on its side so I can easily polish the undercarriage. I bring in my good friend Hulk Hogan to turn it over. Using an example of 1000# centered on the rear axle, Hulk can’t lift enough at the existing 2’ axle stub (500#) to turn it over. Hulk adds a 2’ cheater to the axle making it possible for him to exert 250# of upward force to rotate the tractor (250# x 4’=1000# of rotational torque). Shine away.

Next time I need to do this Hulk isn’t available and I can’t lift 500 or even 250#. So I add a 6’ cheater to the axle and I only have to lift 125#, which in my dreams I can do.

The bottom line is that the length of the axle has no effect on the load on the bearing at the end of the transmission case stub. It was 1000# with the factory axle, 1000# with Hulks 2’ cheater and 1000# with my 6’ cheater.

Now this is not a perfect example as because of the rigid transmission case, the “beam” would rotate on the opposite tire, not the center of the transmission as if it was a lever condition. Tractors do not have independent rear suspension.

The flaw here is that in real life both wheels are extended outward. Not like Hulks cheater on one side only.

 

[AxleHub]

Given all of that, when you run over a rock with one rear wheel the tractor lifts only one side and the whole ASSEMBLY acts like a single beam.

That statement is not in disagreement with my post. But it does not prove your concept of the axle is a single beam or piece..

 

[Egon]

The total load stays the same. The internal bearing forces will change.

Egon, after reading other post's, you may be right on the total load but the internal forces comment may indicate your physic's acumen is challenged! As such I'd disregard most of your comment! You may have had an impulse response that preceded rational thought?

 

[SPYDERLK]

The total load stays the same. The internal bearing forces will change.

,,,,:thumbsup:,,,,leverage.

Egon, after reading other post's, you may be right on the total load but the internal forces comment may indicate your physic's acumen is challenged! As such I'd disregard most of your comment! You may have had an impulse response that preceded rational thought?

,,,,,,,,,,,When talking to yourself, try not to contradict.

 

[orezok]

Think about it. If the weight of the tractor remains the same, it takes the same amount of upward force on the outer bearing and downward force on the inner bearing to raise the tractor NO MATTER HOW LONG THE AXLE IS.

The point being that it takes less force on the tire on a longer axle to create the minimum force to raise the tractor than on a shorter axle. If that's what you are calling leverage - OK, but the bearing doesn't know about that.

 

[SPYDERLK]

Think about it. If the weight of the tractor remains the same, it takes the same amount of upward force on the outer bearing and downward force on the inner bearing to raise the tractor NO MATTER HOW LONG THE AXLE IS.

The point being that it takes less force on the tire on a longer axle to create the minimum force to raise the tractor than on a shorter axle. If that's what you are calling leverage - OK, but the bearing doesn't know about that.

Wrong. ... It would be true if only one wheel were extended. - - That would place the suspended weight [tractor] off center tho. This would result in a higher proportion of load borne by the wheel nearer that off center concentrated mass. ... The extended wheel would bear a lighter load, but due its increased lever on its bearing pair the bearing loads would be equal to that on the short side.

 

[dmac128]

My Kubota with AG tires had adjustable rims.
There is about 3-4" of adjustment in the rims.
I just bought wheel spacers that are 3".

Just on the science theory, which way would make the tractor more stable on a side hill?
Say the outer part of the tires touching the ground is 65" as a baseline with both settings.
With the "rim" sticking out past the end of axle or the end of the axle/spacer sticking out past the "centerline" of the tire vertically,
Causing a bit of cantilever. Which would be more stable in respect to the center of gravity overturning the tractor on a side slope.

Yes I know. Both is better.
Inquiring minds want to know. I know some of you will ponder this. **** OCD![/QUOTE]


Let's apply some actual numbers to this thread...

Here's my read on this. I made some very basic geometry assumptions for only one condition - 1000 lb. vertical load at the tire centerline. Generally, everything will be linear for any given load, i.e. if you double the load to 2000 lbs, the bearing loads will also double.

I looked at 3 geometry conditions; baseline, 4" adjustment, and 4" adjustment + 3" spacer. Each condition will give greater side slope stability, and will increase bearing, axle, diff. housing, etc. stresses as shown. I've treated the axle as a free body, meaning the forces and moments (bending) acting on it must balance so it remains static. This is why the bearing loads are not equal to the 1000 lb. applied load.

Bottom line; assuming the 4" adjustment is OEM approved, adding the 3" spacer will increase local stresses by about 9% FOR MY ASSUMED GEOMETRY. Your results may vary.

I'm guessing most of us guys on here do the occasional overload. Note however that this will be all the time and for all load conditions.View attachment IMG_1718.JPG

 

[SPYDERLK]

My Kubota with AG tires had adjustable rims.
There is about 3-4" of adjustment in the rims.
I just bought wheel spacers that are 3".

Just on the science theory, which way would make the tractor more stable on a side hill?
Say the outer part of the tires touching the ground is 65" as a baseline with both settings.
With the "rim" sticking out past the end of axle or the end of the axle/spacer sticking out past the "centerline" of the tire vertically,
Causing a bit of cantilever. Which would be more stable in respect to the center of gravity overturning the tractor on a side slope.

Yes I know. Both is better.
Inquiring minds want to know. I know some of you will ponder this. **** OCD!


Let's apply some actual numbers to this thread...

Here's my read on this. I made some very basic geometry assumptions for only one condition - 1000 lb. vertical load at the tire centerline. Generally, everything will be linear for any given load, i.e. if you double the load to 2000 lbs, the bearing loads will also double.

I looked at 3 geometry conditions; baseline, 4" adjustment, and 4" adjustment + 3" spacer. Each condition will give greater side slope stability, and will increase bearing, axle, diff. housing, etc. stresses as shown. I've treated the axle as a free body, meaning the forces and moments (bending) acting on it must balance so it remains static. This is why the bearing loads are not equal to the 1000 lb. applied load.

Bottom line; assuming the 4" adjustment is OEM approved, adding the 3" spacer will increase local stresses by about 9% FOR MY ASSUMED GEOMETRY. Your results may vary.

I'm guessing most of us guys on here do the occasional overload. Note however that this will be all the time and for all load conditions.View attachment 473599

You really need to know the spacing between each axle bearing pair, and the placement of wheel center wrt the outer bearing to get a good handle on % load increase. ... The proportion increase in bearing load seems it would be much higher -- ~ 50%

 

[orezok]

Wrong. ... It would be true if only one wheel were extended. - - That would place the suspended weight [tractor] off center tho. This would result in a higher proportion of load borne by the wheel nearer the center. ... The extended wheel would bear a lighter load, but due its increased lever on its bearing pair the bearing loads would be equal to that on the short side.

You're grasping at straws now. NO ONE only widens one axle.

Perhaps you could show me a diagram that displays how the force on the outer bearing increased to an amount greater than the original force with the narrower axles.

SHOW ME!

 

[orezok]

You really need to know the spacing between each axle bearing pair, and the placement of wheel center wrt the outer bearing to get a good handle on % load increase. ... The proportion increase in bearing load seems it would be much higher -- ~ 50%

The load can ONLY increase if the weight of the tractor increases. No matter how long the lever, it can never exert more force to the bearing than is required to move the bearing upward. It's simple physics.

 

[SPYDERLK]

Think about it. If the weight of the tractor remains the same, it takes the same amount of upward force on the outer bearing and downward force on the inner bearing to raise the tractor NO MATTER HOW LONG THE AXLE IS.

The point being that it takes less force on the tire on a longer axle to create the minimum force to raise the tractor than on a shorter axle. If that's what you are calling leverage - OK, but the bearing doesn't know about that.

Wrong. ... It would be true if only one wheel were extended. - - That would place the suspended weight [tractor] off center tho. This would result in a higher proportion of load borne by the wheel nearer the center. ... The extended wheel would bear a lighter load, but due its increased lever on its bearing pair the bearing loads would be equal to that on the short side.

You're grasping at straws now. NO ONE only widens one axle.

Perhaps you could show me a diagram that displays how the force on the outer bearing increased to an amount greater than the original force with the narrower axles.

SHOW ME!

That was the point as to why your Hulk example was not realistic to the real situation [in which a tractor is centered between its support points - wheel/wheel.] It was not centered for Hulk. Ref 2nd quote above.

You really need to know the spacing between each axle bearing pair, and the placement of wheel center wrt the outer bearing to get a good handle on % load increase. ... The proportion increase in bearing load seems it would be much higher -- ~ 50%

The load can ONLY increase if the weight of the tractor increases. No matter how long the lever, it can never exert more force to the bearing than is required to move the bearing upward. It's simple physics.

Again : Extending support points increases bearing load because the extended support points exert more leverage on the bearing pair.