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Dynamics taper roll

  1. Jun 20, 2012 #1
    Sorry guys,
    As I understand there is no introduction topic or what so ever so I'll do it here. Don't want to burst in without introducing myself.

    Hi I'm Andrew from the netherlands and during my websearch for a solution to my problem came across this forum.
    After finishing my mechanical engineering study came to work in the metal industry.

    At the moment I working on a project where I'm looking for some relations between forces and movements. In the picture below I have schematically drawn the current static situation.

    http://dl.dropbox.com/u/4974652/dynamic%20roll.JPG [Broken]

    When using the static formulas I can calculate the relation between the forces F & R as follows

    http://dl.dropbox.com/u/4974652/dynamic%20roll%202.JPG [Broken]

    the two formulas of interest are the yellow highlighted ones.

    Now the problem. If A and C are flanges of which A is comming out of the plan and C is going in and B is a taper roll stuck between them. How can I calculate either the force F or R when the other is know and how does that change with vaiation of the flange speeds.

    I presume the taper roll will act like a screw with a kind of pitch and will turn up. How does the speed and force of the taper roll relate to the rotational speeds of the flanges and the force F or R?

    Thank you for helping me think
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 31, 2012 #2
    hmm found the other topic. hope someone can help me
  4. Jul 31, 2012 #3
    Take one equation, solve for Fr, plug it into the other. (Note: I didn't check your work for errors).

    No. Have you ever seen a screw without threads? Yes, the "squeezing force" will try to push the taper roll upwards, but that's due to geometry, and should be covered by the static equations.

    The rotational speed and force of the taper roll will not affect F or R.
  5. Aug 1, 2012 #4
    I apreciate the reply, but I don't agree with you.

    The static friction is the maximum friction. so if static the stays in its place the static friction is larger than the outgoing force created by the flanges pressure.

    when the flanges start to turn the static friction turns in a rolling friction that is less than the static friction and when the flanges pressure is larger than the rolling friction the roll will com out with a specific force equal to the outgoing flanges pressure minus the rolling friction in the same direction.

    question is what is the rollingfriction in outgoing direction an how does it depend on the roller angle and flange rev's
  6. Aug 1, 2012 #5
    Ah, I see what you're asking now. I was over-idealizing (I believe my answer is correct if you assume no slippage and no deformation). In the real world, rolling friction will likely be less than static friction.

    But it's non-trivial, and you haven't given enough information to answer the question. Rolling resistance is largely caused by plastic deformation, and therefore dependent on the material properties, as well as rotational velocity and slippage.
  7. Aug 1, 2012 #6
    At the moment neglect slippage. so the flanges are rotating at a same speed only in opposite direction.
    when this is clear I'll go one step further and will have a velocity difference creating slippage.

    The roller as well as the flanges are made of steel.

    what I'm looking for is a formula I can program in excel so when someone provides an roller angle and flange speeds, the sheet can calculate what pressure F nees to be give to finaly get the same force R as with other angles and speeds
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