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Thread: Bolt torque vs clamping force

  1. #1
    Associate Engineer
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    Bolt torque vs clamping force

    OK I'm officially confused. The torque calculator found here: "http://www.engineersedge.com/cad-forum/posts/280.html", tells me if I apply 44 in/lb to a 1/4 bolt it will apply 880# clamping force. (to the assembly, I think)

    44in/lb because, supposedly, that's about what a burly guy can apply w bare hand to a 1.5" diameter knob. PLEASE correct me if that's wrong.

    The boss' design team says, "If you have a 1/4-20 bolt, one revolution moves the nut .05 inches. If 44 lbs is applied at a radius of 1 inch (44 inch lbs), the force has to move 2 x pi inches or about 6.28 inches. Therefore the force applied is 44 x 6.28 / .05 = 5580 lbs."

    Who is right?
    How did it go from 880 to 2.8 tons?
    Does that mean, if I have to use a spacer under this knob that will be adjusted frequently, it has to take a 2.8 ton hit ever time it's tightened to 44in/lb.
    Shouldn't the calculator referenced above, need to know the thread pitch or why not?

    Barnyard engineer jerry

  2. #2
    Administrator Kelly Bramble's Avatar
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    The equation T = .2DF is a general equation to estimate applied fastener axial loading for industry standard threads – steel material.

    The equation is imperfect as is the torque process..

    There is more accurate force to torque equations available for critical applications however; even these equations are ultimately imperfect.

    For installations that are expected to be assembled and unassembled repeatedly:
    F = .75(At)(Sp)

    For permanent installations”
    F = .9(At)(Sp)

    Where
    F = Approximate axial preload force applied
    At = tensile area of fastener
    Sp = proof load of the fastener

    Let’s rewrite the top most equation

    From
    T = .2DF

    To
    F = cDF

    Where:
    F = force
    D = diameter (Major)
    F = Axial force applied
    c= roughness approximation
    Steel and zinc plated, dry threads = .2
    Cadmium plated = .16
    Lubricated = .16 to .17

    There are equations that attempt to compensate for thread pitch and flank angle however if one needs torque and axial clamp force that accurate for industry standard fasteners they should consider alternative installation designs.

    See: Bolt Torque - Axial Force Equation and Bolt Torque - Axial Force Calculator
    Last edited by Kelly Bramble; 05-31-2014 at 08:01 AM. Reason: Added link

  3. #3
    Principle Engineer
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    "The boss' design team says, "If you have a 1/4-20 bolt, one revolution moves the nut .05 inches. If 44 lbs is applied at a radius of 1 inch (44 inch lbs), the force has to move 2 x pi inches or about 6.28 inches. Therefore the force applied is 44 x 6.28 / .05 = 5580 lbs."

    The boss' design team did a theoretical calculation that assumed a frictionless system,
    not the reality. As you discovered, friction is so dominant that 80 % of that guy's torque went into it.

    The equation you used to get 880 is based on a large number of tests over a variety of materials and screw systems and as Kelly points out must be used and adjusted according to the application.




    "

  4. #4
    Project Engineer CCR5600Design's Avatar
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    Interesting stuff....

    How do you account for bolt stretch?

  5. #5
    Technical Fellow
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    ...back-off the nut a quarter turn.

  6. #6
    Technical Fellow jboggs's Avatar
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    Actually, on a micro-scale that "bolt stretch" is the source of the clamping force. Think about it. Any force will cause some deflection, or stretching. Assuming we stay in the elastic region, without any deflection (stretching) there is no clamping force. Outside the elastic region, all bets are off.

  7. #7
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    okay, let use say you're building a torque limiter. Your torque arm drives a ball held into a groove by a spring perpendicular to the direction of toque. To correctly preload the spring it must equal the clamping force created by the torque arm. How do you figure that out. T=F**.2 (or whichever resistance) doesn't yield a correct answer. In addition there is a nonlinear motion to the ball as it leaves the groove with an arc pivoting on the edge of the groove. So if the arm has a radius of 1.5" where it contacts & drives this ball (or several balls) and I want it to disengage the ball from the groove @ 50 Lb Ft, this formula says that it generates 1000 lb of force upon the ball. I need to know what to set my spring preload at. I have taken into account that the spring load will increase as the ball moves from the groove and deflects the spring.

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