rotational friction
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Posted by: Pooch ® 06/15/2009, 10:15:55 Author Profile eMail author Edit

Hello All, Can anyone offer a friction equation for a shaft riding in 2 plain bearings, with a pressure pad made of the same material as the bearings (same CoF), applying a force to a reduced or increased diameter of the shaft between the bearings? Thanks.

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: rotational friction
: rotational friction -- Pooch	Post Reply	Top of thread	Engineering Forum

Posted by: jboggs ® 06/15/2009, 12:33:19 Author Profile eMail author Edit

Sounds like homework to me. If so, the policy here is that it is best for you to do it yourself.

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Posted by: Pooch ® 06/15/2009, 14:31:28 Author Profile eMail author Edit

Well, thanks for the insult. I really am out of my realm of this matter. While I know the usual calculation for the friction of a shaft in bearing, this concept is not addressed in any of my books. Additionally, the 2 other engineers working with me don't know the answer. So, forgive us for being so ignorant, but I'm still hoping for help.

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Posted by: Pooch ® 06/15/2009, 14:11:57 Author Profile eMail author Edit

Well, thanks for the insult. I really am out of my realm of this matter. While I know the usual calculation for the friction of a shaft in bearing, this concept is not addressed in any of my books. Additionally, the 2 other engineers working with me don't know the answer. So, forgive us for being so ignorant, but I'm still hoping for help.

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Posted by: bpd ® 06/15/2009, 15:09:20 Author Profile eMail author Edit

With all due respect Pooch, it seems like a homework problem. No specific materials or applications are mentioned. The diagram looks a little contrived, and I can't figure out why in the heck you would put a pad on the top of the middle of the shaft (is it a brake, if so why make a break of the same material as bushings?). It looks like a homework problem to me also. But if you want a little help, start with the basic F,friction=CoF*F,normal. Be creative, look at relationships between velocities and forces, and algebra your way through. Good luck. Modified by bpd at Mon, Jun 15, 2009, 15:10:12

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Posted by: jboggs ® 06/15/2009, 18:08:50 Author Profile eMail author Edit

As the originator of the insult, please forgive me. Try this: Normal force*friction coefficient=friction force. Friction force*radius=friction torque.

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Posted by: Pooch ® 06/16/2009, 07:20:49 Author Profile eMail author Edit

J Boggs, No problem, I understand. That's the equation I have, but in this instance I have 2 radii. Additionally I haven't mentioned that the pressure pad is eccentric. (of no importance in the equation, but proof that the equation is very needed) I have built a test stand and derived an Excel sheet to calculate the forces, but the calcs don't match the "feel" of the test stand. After numerous iterations, I have concluded that it is primarily because I am using the equation for friction on only the eccentric radius. (It only gets worse if I use the bearing radius alone) I am convinced the equation must demonstrate more of the combined frictional forces. Also, I built a test stand that fits our lathe to test empirically the CoF, so I know I'm not too far off. Regardless, even by varying the CoF in the Excel program, the "feel" is not matched by the calcs. So. now one can see why I ask the question. Thanks

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Posted by: jboggs ® 06/16/2009, 14:44:11 Author Profile eMail author Edit

Ok. I'm not exactly sure what your setup is. What is the driver? These two frictional forces seem to be working together. Right? Each one contributes to the total frictional torque based on its own geometry and normal force, independent of the other. Each has its own normal force, its own area, its own radius of operation, and creates its own frictional torque. The total is a sum of the two. Does that help?

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Posted by: Pooch ® 06/15/2009, 16:03:23 Author Profile eMail author Edit

I haven't listed the CoF because I'm not looking for an answer, I'm looking for an equation. The reason the bearing and pad are the same is because I don't want great friction, but can't avoid it because I need high forces. We manufacture counterbalances, this is new product in R&D. Velocity is insignificant, as the attached component is moved by hand, and power loss is not important. You can check out our web site, poor as it is, at www.cbal.com Lastly, if you don't know the equation yourself, please don't treat me as a high school kid. Thanks for your understanding.

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Posted by: Kelly Bramble ® 06/15/2009, 17:44:54 Author Profile eMail author Edit

See /lubrication/laws_rolling_friction.htm Honestly, your question has been posted before and it looks like a classic statics (College Sophmore) problem.

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Posted by: Pooch ® 06/16/2009, 07:42:46 Author Profile eMail author Edit

Kelly, See my reply to J Boggs, above. Nothing classic about this problem. Pooch

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