SETTLE A FRICTION ARGUMENT? 2

[FDS2008]

I have a physics question that I can't seem to answer using textbooks.

I have a vertical pipe that will be clamped to some structural steel. We are trying to calculate the forces required to restraint the pipe based on friction. The clamp will pull the pipe towards the structural base and there will be two different contact surfaces.
1 - The clamp
2 - THe structural steel frame

The two possibilities that we are trying to settle in the design camp are as follows. The first is that the friction force will be doubled (i.e., friction on item 1 + reaction on item 2 = 2 x friction force) since there are two friction surfaces. The second one considers the friction force on the structural steel frame as a reaction of the clamping force and the pipe would opnly need to overcome the friction force of either surface for it to slip axially. This would mean that the resistance is equivalent to the clamping force only.

What do you all think?

 

[TrippL]

RB1957,

Ok, I have come to terms that my first try was wrong (haha) and that the frictional forces from both surfaces would have to be overcome to move the pipe (the frictional forces are additive).

Would the added forces be the sum of the static max friction forces (Fsmax) or the kinectic friction force of the smaller force plus the Fsmax of the larger friction force?

Even though there is no motion yet, the smaller static frictional force would have been reached and reduced to the kinetic frictional force correct?

Let me know your thoughts, done twisting my head around this one for the day.

 

[rb1957]

i don't think dynamic friction would play a part unitll the surfaces were actually slipping ... i see it as something like yielding, below the static friction both interfaces are attracting load, as one approaches slipping it'd offload to the other (that's still stable).

if we allow one side to slip relative to the other, then we'll have to include effects like the clamp rotating (slightly), so that one edge digs into the pipe (clearly increasing the allowable load), and how much deformation does the clamp (or the chnnl) do to the pipe ?

 

[FDS2008]

I'm going to throw another wrench into the original question, as I spoke to a few more engineers regarding the friciotn question and I am still not convinced about the doubling up of the friction force. I realize my last thread indicated that I was, but now I am not so sure.

The new example is that of a regular clamp (two half moons that clamp around a pipe and bolted together). Assuming one of the clamps is welded to some structural steel to support the vertical pipe (omit any bending influece to maintain an ideal situation), if you bolt up the clamps tight, will there be two friction forces?? I believe you will have one firction force equivalent to the clamping force (2P) where P is the tenson on the bolts.

What does not make sense with the double friction force argument is that by tensioning two bolts with a force of say P=1000 lbs each (total of 2000 lbs), you would get a friction force that is calculated from a normal force that is double this, or 4,000 lbs. It would seem that energy is being created here. I spoke to a hanger engineer (designer, not applicaitns engineer)and they were on board with this conclusion.

Let me know your thoughts!

 

[hydtools]

Look at a free body of the pipe and the applied clamp forces.
The force normal to the pipe surface is, in your example, 2,000lbs. In balance there are two equal and opposite normal forces acting on the pipe. The friction force due to each normal force is mu*2000. There are 2 friction faces. Therefore, the total friction force is 2*mu*2000 = mu*4000.

Ted

 

[rb1957]

how about this way to look at it ... what is the friction area ? there are two sites providing area, and i don't think the pipe'll slip before all the friction is overcome.

if you have a table (with 4 legs) standing on a floor, and you push on it, you have to overcome the friction at all 4 locations before the table will move, no? admittedly the normal force at each is 1/4, but i think the analogy works.

in your case (of 2 1/2 clamps bolted together), i think you're getting hung up on the 1,000 lbs ... each 1/2 clamp is applying a 2,000 lbs force to the pipe (think load and reaction).

byw, not sure i'd ask a designer a theoretical strnegth question ... but that clearly depends on the designer ...

 

[TrippL]

That explains it. The normal force is distributed throughout the pipe surface. Although there is multiple materials and contact surfaces. That normal force will be distributed evenly over all of the contact area. Therefore, there is 1 normal force and one coefficient of friction which would be the average of the 2 materials' static friction coef's.

 

[desertfox]

Hi FDS2008

Take a look at the file I have uploaded.
Without having 2000lbs acting on each side of the pipe you cannot satisfy equilibrium.

desertfox

 

[TrippL]

DESERTFOX,

The clamp will exert a pressure on the pipe that will be distributed evenly around the entire pipe or whatever the contact area is for the pipe. That pressure will be divided by the contact area to get the normal force in order to determine the friction force.

 

[TrippL]

This should settle everything.

http://www.apevibro.com/pdfs/Cylinder_vs_Gripping_force.PDF

 

[desertfox]

Hi Trippl

The force I have shown is the resultant force at each side and secondly the chances of getting an equally distributed load around the pipe is very unlikely.
In fact your link in your second post is in agreement with my sketch ie equal and opposite forces.

desertfox

 

[TrippL]

Desertfox,

I was not arguing with you, you are correct. However, the reason friction is not dependant on area is because the force produces a pressure which (theoretically) distributes evenly over the surface of the friction interface, in this case that will be the contact area of the clamp and structure.

 

[desertfox]

Hi Trippi

Its not a problem no need to apologise I didn't think you were arguing with me.
Friction is not independant of area for the reason you have stated see link below:-

http://www.virginia.edu/ep/SurfaceScience/friction.html

 

[TrippL]

Desertfox,

Doesn't the website you sent state that the friction force is independent of area, which is what I stated before?

 

[desertfox]

Hi Trippi

Yes it does but your reasoning was incorrect:-

[/quote Trippi the reason friction is not dependant on area is because the force produces a pressure which (theoretically) distributes evenly over the surface of the friction interface]

The above statement is not the reason friction is independant of area which is why I left you the link.

desertfox

 

[TrippL]

The force is evenly distributed as a load over the "effective" contact surface. How is that not correct reasoning? The link you sent is an explanation of the microscopic interactions between the surfaces in obtaining accurate coef's of friction. Please explain to me the correct reason why friction is independent of area.

 

[desertfox]

To understand why friction is independent on the surface area we need a microscopic view of the contact. Practical surfaces are rough, and they only touch through contact "points" or "junctions". The figure shows two rough surfaces sliding past each other. When the load increases (bottom) the asperities become flattened by elastic deformation. This increases the effective contact area, and therefore friction.

The effective contact area A' is only a small fraction of the apparent macroscopic area A. One can derive an average shear yield stress at the contacts,

t = F/A'


Perhaps you can point out in your original statement where the word "effective area" is.