Collar Friction 4

[meca]

I have a situation where I have two concentric pipes, with the outer pipe slightly larger than the inner pipe. The outer pipe is free to rotate about the axis of the pipes. There are Axial Forces (along the axis of the pipes) and radial forces being transferred from the outer pipe to the inner pipe.

My problem is to try to estimate the amount of torque required to cause the outer pipe to rotate. Any suggestions on how I can estimate? All material is carbon steel.

 

[Metalguy]

First, where doesn't your rotational torque come from? You stated you have axial and radial forces, but they don't cause torque/rotation.

I had a case recently involving a 550 ton pressure vessel that had two "rolling rings" clamped around it. There was a very slight diff. in diameters, and the vessel rotated very slightly faster than the rings (motor driven, very slowly for welding).

This shouldn't have been a problem, but the shop didn't like the idea, so they tried to lock them together. Talk about friction-driven torque! You should see the photos.

 

[desertfox]

Hi meca

I might be able to help you here but i need more information, like how long is the outer pipe and how much clearence have between the two pipes and also there outer and inner diameters, what are the value forces you mention and how fast do u want the pipe to turn.

regards desertfox

 

[meca]

The inner pipe is 16" OD, and the output Pipe is 16.124" ID. The length of the outer pipe is 12 inches.

The outer pipe is sitting on a ledge which is rigidly attached to the inner pipe. Approx. 15,000 lbs of Axial (Along pipe axis) load is transfered from the outer pipe to the ledge. There is a radial force of 60,000 lbs acting on the outer pipe.

The speed of rotation is not critical, it just needs to be able to rotate. This will not be performed very often. The rotation will occur by using a winch and a lever arm, and I want to be sure I size the winch correctly.

Thanks for your help.

 

[desertfox]

Hi meca

I am confused here I don't understand your arrangement of pipes if the outer pipe is sat on the ledge how does the inner pipe rigidly attach to the ledge or do you mean the outer pipe is attached rigidly to the inner pipe in which case both pipes would turn.
To estimate the torque I need to understand your set up so I
need all diameters of the pipes, both pipe lengths, how is the axial load transfered and where both loads axial and radial act in terms of position along the pipe.

 

[meca]

Sorry for my poor description. I have attached a link to a rough sketch to better illustrate what I am describing.

http://www.mecaconsulting.com/Downloads/Rotating Collar.pdf

The inner pipe is fixed and it is very long (approx. 20 ft). The outer pipe rotates. Both pipes are sitting vertically, their longitudinal axis is vertical. To keep the outer rotating collar from falling down, there is a ledge to support the vertical loads. The ledge is rigidly attached to the inner pipe. There are frictional forces between the rotating collar and the ledge, and the rotating collar and the inner fixed pipe.

 

[desertfox]

Hi meca

Thanks for the sketch I will get back to you asap

regards desertfox

 

[Metalguy]

Meca,

Using a coef. of friction of 0.5, I think you'll need about 25,000+ ft. lb. torque.

I get ~5,400 ft. lb. for the axial load, and 20,000 for the radial--all of this assuming it doesn't rust solid!

 

[desertfox]

Hi meca

Just to confirm I calculate the similar values as metalguy using a coeff of friction of 0.5 the method of calculation is as follows:-

f= u * p

where f is the force req to overcome friction

u = coeff of friction 0.5

p = the applied load


so f = 0.5 * 60000 = 30000lbs

f = 0.5 * 15000 = 7500lbs

add these together so you get 37500lbs and apply this to the radius of the 16" pipe ie:- 8"

therefore 37500 * 8"/12 = 25000 lbs/ft


My final comment would be that the coeff of friction 0.5 is probably a bit high the usual average is about 0.25 to about 0.35. However it makes good sense to give a good margin of safety in sizing the winch which using this coefficient of 0.5 would do.


hope this helps



regards desertfox

 

[meca]

desertfox and metalguy,

Thank you so much for the assistance. This is exactly what I have done, with a slight difference in the coefficient of friction. But I feared that I was oversimplifying the problem. Now I feel more confident in my answer.

Thanks Again!!!

 

[desertfox]

Hi meca

Glad we could help

regards desertfox

 

[Metalguy]

Desertfox,

Good full answer--here's a star for you!
(Seems to be a real shortage of these lately).

 

[desertfox]

Hi metalguy

Thanks for the star, i return the compliment.


regards desertfox

ps theres more stars in here than the galaxy

 

[meca]

I gave you both stars, because you are the wind beneath my wings.