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External Torque-Force Relation of Bolt

  1. Jul 10, 2013 #1
    Hi there everybody

    I have a nice little problem which might not be so complicated after all.

    I need to push a heavy object using torque applied on a bolt. This is done manually, and the necessary torque needed is of interets. The attachment explains it quite well, I think.

    To overcome friction, the needed horisontal force to get moving is approximately 10 kN and the data of the metric bolt is as follows:

    Diameter = 30 mm
    Pitch = 1 mm (fine)

    So to be able to push the block with F, T is to be found.

    Suggestions anyone ?
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2013 #2
    My attempt is currently to use

    T = F * K * D
    T = Torque
    F = Force
    K = Friction coefficient
    D = Diameter of bolt

    But my current understanding of this formula, is that it calculates the internal force of the bolt (as to estimate whether the bolt will break) - therefore I'm not sure if it will work on calculating the external force
     
  4. Jul 11, 2013 #3
    Agree with 10.2kN. For that force, a 30mm bolt is hugely oversized, requiring too much torque. The fine pitch needs too many turns but doesn't reduce the torque.

    For comparison, an M16 10.9 pulls 100kN at usual pre-tension.

    The torque results essentially from friction at the screw:
    - At the thread. Take the radius at mean thread depth, multiply by the friction coefficient (for instance 0.12 depending on materials and lubrication).
    - At the screw's tip. You might take 2/3 of the radius, and some friction coefficient.
    - The thread's pitch contributes very little! Compare it with the circumference.

    To reduce the torque:
    - Take a diameter as small as possible, that is about M5 or M6 (or even less with special steel) as long as it doesn't buckle.
    - Lubricate with MoS2 powder (that's really dirty) or MoS2 grease.
    - Cover all parts with Ptfe-impregnated nickel. Damned efficient, clean.
    - Use a ball screw, as everyone does.

    To reduce the number of turns, have a big pitch, which usually means a trapezium-shaped profile.
     
  5. Jul 29, 2013 #4
    Great thanks for the reply and the tips on design. I have some constraints regarding changes in the design, but I thank you for the inputs, and I will consider them.

    I guess in this case that it is the friction between bolt and thread that is of greatest concern and the pitch is almost negligible. I just find it a little counter-intuitive, that it will be more easy to move the object with a small bolt than a large one (that is if the load is below bolt yield/buckling), and also that you need almost the same force to move a heavy object with a 1 mm pictch as a 3 mm pitch.
     
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