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Thread: Friction Question

  1. #1
    Associate Engineer
    Join Date
    Aug 2012

    Confused Friction Question

    Good Morning Ladies and Gents,
    I was asked a question just now about the frictional force in a clamping situation:

    If you have a rod of diameter D and you clamp it with two V-Jaw clamps (situation 1) with force F pushing toward the rod on either V-Jaw, what would be an equation showing the difference in the frictional force (which resists the rod from pulling out of the clamp) between this set up and when you would use two Radius Clamps (nearly matching the rod's diameter - Situation 2)?

    My preliminary answer was that the frictional force Ff=Mu*N where Mu is the coefficient of friction and N is the normal force perpendicular to the contact surface.

    That means that in Situation 1 you have 4 contact points - I'd assume each contact point faces 1/ your total frictional force is: Ff=4*(1/2*Mu*F) = 2*Mu*F

    In Situation 2 (where you have a theoretical full circular contact around the rod) you have infinite Normal forces which are Perpendicular to the rod's surface, so I figure the friction equation goes as follows: Ff=(INTEGRAL FROM 0 TO C) x*Mu*N dx where C is the rod's circumference (Pi*D) and x is the contact surface. This means that, if the diameter were 1, your frictional force in this situation would be Ff=3.14*Mu*F

    Is this accurate or is there some other explanation for this? I am assuming the frictional force would be higher in Situation 2, but am not sure if I vary the contact force in the integral or the Normal Force being applied (and if so, how do I do that?)

    Let me know your thoughts...

    Drawing out some quick FBD's I think that the equation for the circular jaws would have the applied force of F distributed into F/C on each "portion" of the integral (whereas Situation 1 has F/2 applied). This would make the equation: Ff=(INTEGRAL FROM 0 TO C) x*Mu*F/x dx possibly? I've posted my potential solution with free body diagrams and equations, let me know any thoughts please!

    Problem Setup.jpg
    Last edited by julchak; 05-30-2013 at 11:16 AM.

  2. #2
    Technical Fellow
    Join Date
    Feb 2011
    It depends a lot on how big these things are. If very tiny then testing and measuring is the only way to get something approaching accurate results.

    If you are talking say, 1/2" diameter and up then things change a little, but it still, ultimately, is going to come down to testing.

    Your assumptions with the full circle are a little off in that the force is applied in a single plane, albeit from opposite directions. That means said force will be applied in that direction. There will be greater force acting at the horizontal center-line than at points around the circumference. The applied force will diminish (effectively) to zero at the vertical center-line.

    What are you trying to achieve? Holding by force is not the first choice for holding anything.

  3. #3
    Associate Engineer
    Join Date
    Aug 2012
    I see what you mean - I'd pretty much have to go into polar coordinates, find the equation that dictates the force as it varies with R and THETA and do a double integral with this for R and THETA to get an accurate equation of the frictional force in the circular clamp design.

    I believe this is going to be used in a wire machine, holding/stopping 18-10 ga wire as it is cut before being pushed through the next forming process - not much force should be required for these small wires. I was simply asked to give an engineer's side of why one may be better over the other so I ran these equations up real quick as a simplified model.

    I sketched up a further analysis of the V-Jaw showing its mechanical advantage based on the angle of the V (attached an image of this). Additionally I've attached an image where I quickly go over this idea of the polar coordinates going along with what you said Pinkerton, just not coming up with the actual equation.

    Is this all accurate, and am I correct in saying that the V-Jaw will in fact provide more frictional holding force?

    V-Jaw Further Analysis.pngCircular Jaw Polar Force Distribution.png

  4. #4
    Lead Engineer RWOLFEJR's Avatar
    Join Date
    Mar 2011
    Rochester Pennsylvania
    Tricky thing to compare in real world situation but not so tricky in theoretical world. Thing is... they're nothing alike.
    In the real world you need to test like Dave said...

    In the theoretical world full circle would give more contact and therefore more friction than four line points of contact. It's a geometry thing...
    Thing is in the real world you have to tolerate some variations in size and surface finishes that aren't as black and white as the theoretical world.

    So to keep a bar from pulling through your clamps you need to look at what it is you need to accomplish and deal with it somehow. There are a lot of ways to skin a cat. Less fabrication force on the work piece... More clamping force... finger trap arrangement... collets... mechanical stops or backstops... keys or pins... more clamps... longer clamps... Can you tolerate imparting some marks onto your bar? Fine knurling or saw tooth or ratchet rings to slightly dig into the part? Surface finishes of both your clamps and stock will change things a lot. Or maybe just regular maintenance of your clamps for good fit? You basically want your grip to be at least... 125% of the work being applied to the stock.

    For positioning accuracy in a two jaw clamp most use V-Jaws. For better resistance to pull through most use interference fit in full circle for two jaw clamps.

    Good luck.

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