Effect of an offset load on a bearing - (static calc)

[ozmac]

I was positive this is a simple uni style question, but I can't seem to work it out.

I have a pneumatic cylinder that has an allowable 10N of side load acting on the end of the piston rod, at an extension of 0.1m.

Now Let's say the cylinder is generating a force of P=200N.
And the load it is attached to is offset 0.03m.

I am trying to determine the effective side load generated by the cylinder force and the offset load.

The 10N allowable would give you 1N.m alowable moment at the bearing (rod bush), so if i can calculate the moment at the bearing for this example then that should be all i need.

I've tried summing up forces (Fax, Fay, Fby, P), moment equations, but i seem to end up with too many variables, or everything cancels out... The only time i got an answer was when i said the moment about point A equalled zero, which gave me a side load of 60N. But I'm pretty sure you can't say the moment about point A is zero?
But I also got that answer when i looked at it it with simple vectors: tan(alpha)=30/100=y/200-->y=60

I'm lost, having not touched this since i left uni (4 years ago) A shame really! But funny how I'm actually interested in it now.

Hopefully this image helps illustrate the scenario.

[PLAIN]http://img441.imageshack.us/img441/4512/sideload.jpg

In reality, where the rod connects at Point A is actually a conveyor which oscillates back and forward (to adjust tension in the line), however to keep it as a simple static problem this is how i want to look at it.
Please note i have not drawn the cylinder, only the piston rod and the rod bush.

If someone wants to offer further information on the distribution of the forces on the bearing surface etc, let's say the bushing is 10mm long...

Cheers

 

[AlephZero]

I agree with your 60N.

I think it is easier to see it you take moments about point B, not point A. There is a 200N reaction force on the rod in the X direction at A. There must be a 60N force in the Y direction to balance out the moment at point B.

 

[ozmac]

Yeah you've essentially done the same thing i did , except your assumption is the moment about B is zero. (I stated the moment at A was zero, so worked out the "Fby" force which gives me "Fay" due to the sum of forces equalling zero) That's the bit that's confused me, as I thought there is a moment there, yet we've stated the moments are zero!

And to confuse me more, consider the case where there is no bushing there, so it's just a force P acting on the rod (or L bracket i suppose). You don't have a reaction force Fby as the end of the rod is free to deflect! So if you have no Fby, where do you calculate your Fay from! Edit, actually that's an easy one i think, the moment at Mb would then definately be zero, so then you could work it out as you have Mb=0=Fay*0.1 - 200*0.03 --> Fay=60

so that just confuses me more...

 

[tvavanasd]

The 10N allowable would give you 1N.m alowable moment at the bearing (rod bush), so if i can calculate the moment at the bearing for this example then that should be all i need.
...
In reality, where the rod connects at Point A is actually a conveyor which oscillates back and forward (to adjust tension in the line), however to keep it as a simple static problem this is how i want to look at it.
Please note i have not drawn the cylinder, only the piston rod and the rod bush.

Cheers

Since the allowable side load on the cylinder is only 10 N (2.2 lbf), I will assume that you are using a non-guided cylinder. It is probably best to practically elminate all side load and moment transfer from the rod, by either using a rod end or an alignment coupling. Otherwise, there will be bending in the piston rod which is undesirable.
The bearing at the head end of the cylinder cannot resist bending, only radial loads. The piston inside the cylinder (some distance from the head end bearing) provides the second radial force to resist the side load.

 

[ozmac]

yes you're right it is not a guided cylinder, hence the low allowable side load, however lucky guess as it could of just been a tiny guided cylinder!

Yes it would be nice to eliminate/reduce this side load, however i cannot see how a special rod end will assist, assuming there are no missalignments in the load. Alignment coupling?

Not sure what bearing at the head end you are referring to. The entire cylinder body is firmly mounted (foot mounts). You're right that there is a second force that resists the side load, however I have the piston in the fully extended position so we will assume the rod bush and piston are side by side. I would usually call these forces lateral or axial loads rather than radial, as a radial load would cause the piston rod to rotate. But that's probably just terminology and how you choose to read the diagram.

And as a matter of fact, it is the the transfer of this side load to the bearing in the piston, rather than the rod cover bush (the usual fail point), that is the problem that I am investigating. It is a precision cylinder with metal-metal seals, and we are getting excessive piston bypass.
Other than the possibility of air contamination (likely), I am just investigating how far out of spec the cylinder is being operated due to the loading conditions/application. The numbers I have given in this topic are not the actual numbers from my investigation, just nominal figures to ensure I am going about this the right way.

So can anyone else confirm the 60N figure, as the zero moment thing that is concerning me.

Cheers

 

[tvavanasd]

yes you're right it is not a guided cylinder, hence the low allowable side load, however lucky guess as it could of just been a tiny guided cylinder!

Yes it would be nice to eliminate/reduce this side load, however i cannot see how a special rod end will assist, assuming there are no missalignments in the load. Alignment coupling?

Not sure what bearing at the head end you are referring to. The entire cylinder body is firmly mounted (foot mounts). You're right that there is a second force that resists the side load, however I have the piston in the fully extended position so we will assume the rod bush and piston are side by side. I would usually call these forces lateral or axial loads rather than radial, as a radial load would cause the piston rod to rotate. But that's probably just terminology and how you choose to read the diagram.

And as a matter of fact, it is the the transfer of this side load to the bearing in the piston, rather than the rod cover bush (the usual fail point), that is the problem that I am investigating. It is a precision cylinder with metal-metal seals, and we are getting excessive piston bypass.
Other than the possibility of air contamination (likely), I am just investigating how far out of spec the cylinder is being operated due to the loading conditions/application. The numbers I have given in this topic are not the actual numbers from my investigation, just nominal figures to ensure I am going about this the right way.

So can anyone else confirm the 60N figure, as the zero moment thing that is concerning me.

Cheers

There is always misalignment. An alignment coupler will prevent side forces and moments from being transferred to the cylinder rod (as long as the system is well designed and aligned within reason). Rod alignment couplers are analogous to shaft couplings for transferring torque except that they only transfer thrust.

The head end is that with the rod seal and bushing. There must be some distance between the 2 bearing elements, even if it is only 10 mm.

The problem the way you have defined it is statically indeterminate. Use an alignment coupler, and design the system correctly.

Alignment coupler images:
http://www.google.ca/images?hl=en&s...arch+Images&gbv=2&aq=f&aqi=&aql=&oq=&gs_rfai=

 

[ozmac]

It sounds like you know what you are talking about, but not to be rude, i think you're mistaken (or perhaps i am, but i don't think so)

A ball and socket joint does not eliminate torque or side load in this case.
All it really does is take up any mis-aligments with the load, and helps prevent the cylinder from binding. So yes it can reduce the side load, but will not eliminate it if the load is offset. As is the case.

Have you studied engineering or done free body diagrams before? Draw one with a fixed rod end, and one with an alignment coupler. Notice how they are identical!
It is physically impossible (to my knowledge) to put any sort of coupling in place to eliminate the torque. It defies physics! You could replace the rod end with a banana and a bunch of grapes, and they will still try to pass on the same forces!

I suggest you brush up on your statics before telling others how to design systems correctly ;)btw, the rod cover is not the head end. the Head end or head cover is the other end.

 

[tvavanasd]

It sounds like you know what you are talking about, but not to be rude, i think you're mistaken (or perhaps i am, but i don't think so)

A ball and socket joint does not eliminate torque or side load in this case.
All it really does is take up any mis-aligments with the load, and helps prevent the cylinder from binding. So yes it can reduce the side load, but will not eliminate it if the load is offset. As is the case.

Have you studied engineering or done free body diagrams before? Draw one with a fixed rod end, and one with an alignment coupler. Notice how they are identical!
It is physically impossible (to my knowledge) to put any sort of coupling in place to eliminate the torque. It defies physics! You could replace the rod end with a banana and a bunch of grapes, and they will still try to pass on the same forces!

I suggest you brush up on your statics before telling others how to design systems correctly ;)


btw, the rod cover is not the head end. the Head end or head cover is the other end.

An alignment coupler is not simply a ball and socket joint. Alignment couplers are designed to eliminate side and moment loads from being transferred to the rod (and to allow for lateral and angular misalignment).

An example cross section can be found at:
http://us.misumi-ec.com/us/ItemDetail/10302010340.html

Most manufactures of pneumatic cylinders supply them:
http://xdki.festo.com/xdki/data/doc_ENGB/PDF/EN/PISTON-ROD-ATTACHMENTS_EN.PDF

If the alignment of the system is off significantly, or the design is not rigid enough, an alignment coupler may not work.

Note that a ball and socket will not transfer a moment within its free range of motion.

I am familiar with FBDs. Adding an alignment coupler (or any flexible or floating device) significantly changes the analysis of a body or system of bodies.

Numatics (the cylinder manufacturer) refers to the end with the rod seal as the head end; the other is the cap end. This may not apply to other manufacturers (type “head” in the following link):
http://www.numatics.com/common/deliverables/catalogs/cylinders/aseries.pdf

 

[ozmac]

An alignment coupler is not simply a ball and socket joint. Alignment couplers are designed to eliminate side and moment loads from being transferred to the rod (and to allow for lateral and angular misalignment).

An example cross section can be found at:
http://us.misumi-ec.com/us/ItemDetail/10302010340.html

Most manufactures of pneumatic cylinders supply them:
http://xdki.festo.com/xdki/data/doc_ENGB/PDF/EN/PISTON-ROD-ATTACHMENTS_EN.PDF

If the alignment of the system is off significantly, or the design is not rigid enough, an alignment coupler may not work.

Note that a ball and socket will not transfer a moment within its free range of motion.

I am familiar with FBDs. Adding an alignment coupler (or any flexible or floating device) significantly changes the analysis of a body or system of bodies.

Numatics (the cylinder manufacturer) refers to the end with the rod seal as the head end; the other is the cap end. This may not apply to other manufacturers (type “head” in the following link):
http://www.numatics.com/common/deliverables/catalogs/cylinders/aseries.pdf

ok so you are reffering to floating joints, another type of alignment coupler.
once again though, these do not necessarily eliminate moments from being transferred to the piston rod.
Did you not draw that FBD?
The only way to eliminate the side load in this case is to have an external guide in which case you need to draw additional external forces.

The floating joint you gave an image for will allow for no more than 1-2mm of axial misaligment. For arguements sake, let's say this is enough for the example in the original post. All this is doing is transferring the load 1-2mm (or whatever it allows for) offset from the central axis of the rod. So yes it may may not appear to have side load as the force is still parrallel to the piston rod, however it is offset to the axis, and hence you have a moment at the rod bush! You can't cheat physics. Though if the offset is less than the radius of the rod, you may be in the clear, however clearly that is not the case in the example i have given.

And when you say a ball and socket joint will not transfer a moment, once again you are thinking too much about the small picture, and not the system. A ball and socket joint will transfer a force that can be drawn at one point, depending on it's angle. Unless this angle is zero degrees (in which case there is no need for a ball and socket joint), then this means the force will not be acting towards the origion of the rod bush, and hence you get a moment at the rod bush. So yes it prevents a moment at it's location, but it does not cause the moment in the system to mysteriously dissapear.

And good to know the definition of the head end is dependant on the manufacturer, explains the original confusion.

Can someone please respond to my initial post?

Cheers

 

[ozmac]

Hi, can anyone please confirm the 60N figure, preferably with a step by step explanation because the way i have done it i am sure i made an incorrect assumption regarding the moment?

Cheers

 

[ozmac]

anyone?

 

[ozmac]

someone?

I think I've drawn the diagram wrong for the loading.
As the belt is flexible, it is offering negligable guidance, so couldn't you consider Fay = 0?

soo confused.

 

[tvavanasd]

someone?

I think I've drawn the diagram wrong for the loading.
As the belt is flexible, it is offering negligable guidance, so couldn't you consider Fay = 0?

soo confused.

Your FBD is indeterminate as drawn (a fixed cantilever with a second bearing is indeterminate), and you haven't supplied enough information regarding the rest of your design to create a proper FBD. This is why you are not able to solve it simply and why you keep asking the same question.

If you don't have an external guide in your design, it likely will not work well. The vertical cantilever (0.03 m) in your second diagram suggests rigidity at its base/mount (perhaps linear bearings elsewhere providing enough support to approximate such).

So the base of the cantilever is not actually rigid...

Why not start a new post with proper drawings or pictures of your system? Give it a fresh start. I will not respond to any of your future posts on this topic (unless you request via Private Message).

Good Luck

 

[ozmac]

Hey, yep I've drawn it wrong (or in a misleading way) which explains why i had to make assumptions that disagreed with my drawing. So yes, statically indeterminate would make sense.

I won't need to re-draw it as I've worked it out.
Considering the worst case scenario, the conveyor belt offers neglibale force in the 'y' direction. This means the only forces I have is Fbx=-Fax=P, plus the moment at B. (moment at A is zero)

And yes there is no other guide, and yes it is a bad design, but no, I did not design it nor did anyone from my work. anyways...

So as is, the moment at B is therefore Fax*0.03=6N.m
So 6 times that of the allowable (calculated to be 1N.m in the O.P), hence the failure.If you were to go off the incorrect diagram i drew in the O.P, there would only be a tiny amount of sideload however that would be due to deflection in the arm, and so it would only be a very small amount (depending on material of course). I've basically drawn it as though it were a guided system, which it is not. And yeah as it is statically indeterminate, you need to look at the material properties to determine those forces...

Well we got there in the end! :)

thanks

funny how i had the correct answer all along, I just got the answer the wrong way...