INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS
Come Join Us!
Are you an Engineering professional? Join EngTips now!
 Talk With Other Members
 Be Notified Of Responses
To Your Posts
 Keyword Search
 OneClick Access To Your
Favorite Forums
 Automated Signatures
On Your Posts
 Best Of All, It's Free!
*EngTips's functionality depends on members receiving email. By joining you are opting in to receive email.
Donate Today!
Do you enjoy these technical forums?
Posting Guidelines
Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Coefficient of friction for plain bearings(3)

TerryR1 (Mechanical) 
11 Jan 10 14:38 
I derived this formula for calculating friction for round plain bearings:
f = pi/2 * μ * P
f = frictional force μ = Coefficient of sliding friction P = Radial force on bearing
I would expect this formula to be published in engineering books, but I'm not finding it. Can someone provide a source? Or prove me wrong? Thanks! 

If you want an equivalent force at the sliding face, the pi/2 doesn't belong. If you want a torque, you need the pi/2 and a diameter. As modified above, the equation would be valid for sliding friction at low speeds in the absence of lubrication. In situations like that, you would also usually want a bearing length and a rotational speed, so you could estimate the radial wear rate based on sliding velocity and pressure on the projected area, using a material factor you'd find under 'tribology'. If lubrication is present and rotational speed is sufficient, there's essentially no wear, and little friction, and you'd find the equations under 'hydrodynamic pressurization' or something similar. In between those extremes, well, it depends... Mike Halloran Pembroke Pines, FL, USA 

No need to reinvent the wheel (or the sleeve or whatever). As a start, Google petroff's equation ===================================== Engtips forums: The best place on the web for engineering discussions. 

TerryR1 (Mechanical) 
11 Jan 10 22:37 
I'm fairly sure the pi/2 needs to be there for round bearings, to convert radial force to normal force.
I'd prefer not to reinvent the wheel, but I can't find a good published formula for plain (round) bearing friction.
My application uses unlubed bearings on a round shaft, either sliding or turning at very low speeds (~2ft/min). No viscosity factors. Only 'simple' sliding friction.
Here is my thought process:
f = μ * Fn (Fn= normal force)
For a block sliding on flat ground, Fn=Fy (Fy=vertical force) f= μ * Fy
For a square bearing sliding on a squareshaped shaft, loaded only in y: Fn=Fy f= μ * Fy
For a square bearing sliding on a squareshaped shaft, loaded in x and y: Fn=Fx+Fy (it is NOT sqrt(Fx^2+Fy^2)!) f= μ * Fx+Fy
For a diamond bearing sliding on a diamondshaped shaft, loaded in y: Fn=Fy/sin(θ) (θ= angle of loaded face to horizontal. 90deg is flat surface, 10deg is a wedge) f= μ * Fy/sin(θ)
So the frictional force depends on the shape of the bearing.
For a round bearing (worn so the full half circle is in contact), I figure: f= pi/2 μ * P
I need to find a good way to prove that.. 

Hi TerryR1 Have a look at this link: http://www.roymech.co.uk/Useful_Tables/Tribology/Plain_Bearing%20Friction.htmlIt shows for a plain journal bearing which can be hydrostatically,hydrodynamically,or run dry that: Mr=P.f.(D/2) where Mr = frictional moment f = frictional coeff P = radial force D = bearing diameter If you transpose the formula the friction force is: P.f Also the friction force at low loads is not dependant on bearing shape but on the magnitude of the normal force clamping them together. desertfox 

TerryR1 (Mechanical) 
12 Jan 10 13:13 
Desertfox, Thanks for the link. The formula uses a "coefficient of friction of the bearing", I believe this is different than the "sliding coefficient of friction".
Maybe bearing manufacturer's published friction data has the pi/2 factor included. They probably get all their data empirically and would not know. Hmm... 

hi Terry
Its not really different if you scroll to the bottom of the page, for a dry bearing it gives 0.2 to 0.4 which is sliding friction coefficient values,
desertfox 

You won't get a bearing manufacturer or metallurgist to give you a number for the coefficient of friction of a dry bearing because there are so many variables that affect it (trust me, I've tried). The best thing to do is to assume a range (ex 0.20.4). Liquid lubricated bearings are a whole different animal; a hydrodynamic bearing is extremely reliable and has a much more predictable running friction. Of course, they work best in machines that run continuously. 

TerryR1 (Mechanical) 
14 Jan 10 9:43 

TerryR1 (Mechanical) 
18 Jan 10 14:04 
Ok so mfr measures friction on round bearings: f = P * T / r
f = coef. friction P = radial force T = measured torque to turn bearing r = bearing radius
My thoughts after talking with them: Use of the pi/2 factor depends on the wear of the bearing.
In the application where the bearing turns, it will wear such that the ID will get larger and pinch effects on the shaft will decrease. There is't really any wedging. So multiplying the CoF pi/2 is not necessary for this application.
Where the bearing doesn't turn, it will wear such that the shaft wears a groove into one part of the bearing, causing a wedge effect. To get a more accurate CoF for a wornin sliding bearing, multiply the list CoF by pi/2.
Thanks all for your responses. 

hi TerryR1
Take a look at your formua in your last post. The friction coefficient should be unitless but it seems to me you have the coefficient of friction to be in N^2 unless I have missed something.
desertfox


Quote (TerryR1):Flash3780: http://www.cipcomposites.com/pdf/Engineering_Manual.pdf p7 I've asked the manufacturer about pi/2. We will see what they say.
Just remember that there are a few variables that affect the coefficient of friction. Different surface roughnesses, loads, contact patch sizes, or speeds affect the coefficient of friction. When you start digging into it, it's a more complicated number than you think. When you come up with a number, you may wish to look at +/20% error bands and make sure that your design holds up. 

TerryR1 (Mechanical) 
18 Jan 10 18:27 
Desertfox, You are right, very bad post! μ = T / (P * r)
flash3780 Definitely. And the pi/2 is yet another factor making up to 60% difference... 

Hi TerryR1
Thanks for the confirmation, I am having trouble with your pi/2 factor, if I understand what your saying,then the shaft will wear away the bearing until it generates a groove half the diameter of the shaft? If thats the case I doubt that will happen because journal bearing are much thinner in section than the shaft,I can except you can wear a groove say segment shape but it will never reach half the shaft diameter. If I have misunderstood please straighten me out.
regards
desertfox 

TerryR1 (Mechanical) 
18 Jan 10 20:57 
Hmm say we have a 3" bearing with 0.010" initial clearance. With autocad.. it looks like the bearing would need to wear ~0.09" for it to theoretically be fullsemicircle contact. That is a lot but not unreasonable imho.
For a more reasonable wear... say it wore 0.02". That would result in 99% contact (2.97" projected length) and the "wedge factor" would be 1.42, not pi/2.
Interesting... 

Hi TerryR1 Well I can't find anything to verify or deny your pi/2 factor but my gut feeling is that it is correct. I have found this pdf which I give a link to below. If you go to page 3 on the tool bar of adobe you will see a graph of friction factor plotted against Zn/p for Hydro Dynamic action of journal bearings,however if you look to the left of point B you can see a line plotted for imperfect lubrication or a very thin film and according to the book "Machine Design" by Paul H.Black & O.Eugene Adams which as a similar graph also mentions partial metal to metal contact which is the nearest thing I have come across to your situation. pi/2 is not needed to convert radial force to normal force as the normal force is given by the propertion of shaft mass and any external load acting on the shaft in accordance with normal statics/dynamics. I am curious to know what exactly the bearing manufactures said in reference to your pi/2 factor. Its a complicated subject for sure as others have stated, it depends on a lot of other factors. Regards desertfox 

hi TerryR1 Quote:Well I can't find anything to verify or deny your pi/2 factor but my gut feeling is that it is correct
should have read:Well I can't find anything to verify or deny your pi/2 factor but my gut feeling is that it is incorrect. regards desertfox 

TerryR1 (Mechanical) 
19 Jan 10 16:07 
Slide 15 of the slideshow says "as clearance increases it decreases the coefficient of friction".
This effect is what I have been proposing: "as clearance decreases (wear), it increases the apparent CoF." I calculate that increase to theoretically be pi/2 for zero clearance.
The bearing mfr engineer agreed that it is logical that there could be some wedging effect. They don't have a way to calculate it, but they could quantify it with a lab test. If someone paid for it. =) 

Hi TerryR1 I have managed to find a pdf which gives the situation you describe in terms of a close fitting bearing which I have uploaded for your reference, it also deals with a loose fitting shaft and a shaft with a bearing that as worn evenly. At the end of the three cases it shows that the virtual coefficient of friction for a loose,wellfitted and well worn bearing, maybe this is what our looking for. regards desertfox 

TerryR1 (Mechanical) 
20 Jan 10 11:00 
Very nice find!
On page 123: f'=1.57 * f = pi/2 * f
Where f = CoF f' = adjusted CoF for tight fit of shaft in bearing
How did you find this pdf may I ask? 

Here is a thought about where the pi/2 comes from. It is simply converting between applied force straight down and normal force which is normal to the bearing surface. Assume Fapplied is straight down. Fn = Fapplied * cos(theta) where theta goes from Pi/2 to +Pi/2 over half the bearing... theta is the angle from vertical The average value of Fn is <Fn> where <Fn> = (1/pi)*int[Fapplied * cos(theta)] dtheta integrated over the range theta =Pi/2..Pi.2 [the 1/pi is the length of the averaging interval) Integral of cos(theta) is sin(theta) <Fn> = (1/pi) * Fapplied*[sin(Pi/2)  sin(Pi/2)] <Fn> = (1/pi) * Fapplied*[1  1] <Fn> = (1/pi) * Fapplied*2 <Fn> = (2/pi) * Fapplied Ffriction = <Fn> * f = (2/pi) Fapplied * f Ffriction = (f*2/pi) * Fapplied Ffriction = f' * Fapplied where f' = f*2/pi Hmmm... seems to have come out inverted? I have to go back and look a little closer. ===================================== Engtips forums: The best place on the web for engineering discussions. 

hi TerryR1
I have access to some online books at Knovel and I found it in this book:
Tribology in Machine DesignSearch WithinSearch By: Stolarski, T.A. © 1990 Elsevier
All I did was save the chapter to my comp, printed the relevant pages and scanned them to post here.
regards
desertfox 

TerryR1 (Mechanical) 
20 Jan 10 15:09 
Signing up for knovel soon... ty! 

Wouldn't Pi/2 be one half the area of the circle. Looking back at some very old calculations for determining the COG of a shaft and bearing where I used Pi/2 as !/2 the area of the shaft. I think I'm missing a page in my notebook because it's not very self evident in my written approach. 

abdul (Industrial) 
2 Mar 10 2:11 
Gentlemen,
So far, it seems that everyone is claculating the friction force, however, the formulas used are not mentioning the length of the circular sleeve bearing. Therefore, the bearing length in contact with the shaft must be multiplied by the friction force in your equations to acheive the true value of friction force or friction moment acting at each bearing.
A second factor is to recognize that all such equations assume a 100% geometrical alignment of bearing to shaft journal. Many real world machinary do not have such alignment, so you may want to include for a misalignement factor that adds to the bearing friction formula. Then you are close to modelling the real world of the shaft to bearing friction.
Thanks,
Abdul 



