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Some Torque Q's

  1. Oct 17, 2012 #1
    so, i am fairly good in physics, but, i do have a Q.

    with a torque wrench, does the wrench move force perpendicular on the axial of the pivot point? so, with torq on a bolt, i apply 80lbs exactly 1ft away from the pivot point this means i have loaded the bolt in compression with 80lbs of force?

    2nd, anyone know of any DIY's for building a torq measuring apparatus, something that can be hooked into a laptop to capture graphs of the forces being applied. i was thinking of capturing force applied to a Omega LC304 load cell, and, also capture the range of motion applied to the wrench so we can see on a graph the sudden 'click' when the wrench reaches its set point.

    your thoughts?
  2. jcsd
  3. Oct 17, 2012 #2
    You apply the torque your torque wrench displays. That's why they have gauge readings.

    no, 80 lbs of pressure at one ft is 80 ft-lbs. 80 lbs at two ft is 160 ft-lbs.

    Perhaps you can find a sensor idea here:


    you want something that converts pressure [force] to an electrical output
  4. Oct 17, 2012 #3


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    When you apply 80 pounds exactly 1 foot away, you are applying 80 ft-lbs of torque and 80 pounds of shear onto the bolt (more accurately, the shear is acting on the threaded rod). Technically, you are applying compression to one side of the bolt, and that is transmitting it to the stud, which is bearing the load in shear. This is usually ignored however, since 80 pounds is a pretty trivial amount for the bolt to bear in compression or for the stud to bear in shear, so it's not terribly relevant.

    Also note that you have to balance forces and moments separately - if you apply 40 pounds of force 2 feet away, you're still applying 80 foot pounds of torque, but the compression on the bolt and shear on the stud is reduced to 40 pounds.

    (I think I understood what you are asking here...)
  5. Oct 17, 2012 #4
    i am not so worried about how the torque is calculated (40x2ft vs 80x1ft), i will let the click of the wrench be the point at which i need to validate. the Q is, if the wrench is applying 80ft-lbs onto the bolt which is pushing down onto the button of the LC304 load cell, how much force do i expect to see on the load cell? this is a calibration Q, etc. in my case, the head of the bolt never bottoms out, so its basically just threaded rod pushing down onto the load cell, etc.

    so if i load the threaded rod with 80lbs of shear does that equate to 80lbs on the load cell. these two forced are perpendicular to each other.

    thus far i see i can use Omega LC304 load cell and one of their DAQ items to display and data-log the load cell, but i need to have the right equation to calibrate the charts, etc.
    Last edited: Oct 17, 2012
  6. Oct 17, 2012 #5


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    Oh... is the load cell measuring force parallel to the threaded rod? If so, then the force it will measure is dependent on several things - the thread pitch of the bolt (finer threads = more force for a given torque), the material the bolt and threaded rod are made of, and even the size of the bolt. What you're trying to measure and/or calculate then is called the bolt preload - if you google that, it should give you a decent number of resources.
  7. Oct 17, 2012 #6
    thanks for the info. so, the load cell is measuring a force, like placing a 5lb weight onto a scale. in my case, the end of the bolt opposite of the bolt head will be resting on the load cell button (perhaps using a small hardening steel plate as a buffer) and as the bolt is tightened (torq'd) there will be a force applied to the load cell. i need the translation (as mathematically accurate as i can get) so i can calibrate my chart axis so it shows correct data, etc.

    the purpose is to test the "click" of the torque wrench for accuracy across various torque settings of the wrench's usable range, etc.

    another option is not a bolt at all, but rather just a bolt head that has a arm under it that sticks out and contacts the load cell button and i just capture force until the wrench reaches its setting. so in this case, if the distance from torque pivot point to the center of load cell button is exactly 2" then when i apply 80ft-lbs via the wrench what force do i expect to see on the load cell, etc. 2" is 6x less than 12", so should i expect to see 480lbs on the load cell when 80ft-lbs of torque is applied? or perhaps just a large bolt welded to a small steel i-beam and the load cell is centered at exactly 6" from the torque pivot point? is this a better way to do it? i suspect measuring at 2" would yield a more sensitive graph?

    so, i think i have some info to go by, now i have to do some homework.

    Last edited: Oct 17, 2012
  8. Oct 17, 2012 #7
    This depends on the diameter of the bolt....it is NOT 'exact'..
  9. Oct 18, 2012 #8
    so, would it be better to avoid the threaded rod math & physics and just use an arm as i described?

    and, if the force multiplies as you get closer to the pivot point, is this force the "clamping" force exerted by the bolt head? so whats the relationship of shear force and clamping force?
    Last edited: Oct 18, 2012
  10. Oct 19, 2012 #9
    so, my idea is to have a large diameter rod (~2" diam. very strong steel rod) pass through two pillow blocks, and while passing through this pawl (attached pic) will slip over the rod. rod will be machined square in the middle, and then smaller round diameter on the far end into the other pillow block. the rod on the larger side will have a 3/4" square hole machined into it since that is probably the largest square drive torq wrench this apparatus will accomodate. for smaller wrenches with smaller drives i will use drive adapters.

    less the torq (force) of the pawl due to gravity, and some factored in minor friction from the pillow block bearings, does anyone have comments about what i am proposing?

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