Announcement

Collapse
No announcement yet.

Tension calculation from deflection?

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Tension calculation from deflection?

    Suppose one has wire in tension, tension force in Newtons as the unknown.

    The wire goes over two stationary pulleys. Between them is a third pulley, that is able to move perpendicular to the wire travel. If I push this third middle roll with a known force and the wire deflects a distance X, what sort of calculation I need to find the tension in wire?

    This is probably some very elementary solution, just can't get my head around it now. Thank you very much for help!
    Amount of experience is in direct proportion to the value of broken equipment.

  • #2
    Something like this? http://structuresworkshop.com/blog/?p=1389



    IIRC, it's something like 1/2 the load divided by the cosine of the angle between the rope and the vertical. Been a while, so don't hold me to it.
    Any products mentioned in my posts have been endorsed by their manufacturer.

    Comment


    • #3
      Jaakko, when I envisioned this problem as you described it, I pictured the ends of the wire being fixed rather than weighted and movable. Which way did you mean?

      While I am sure that there is a way to calculate this, I would be tempted to place a scale in tension between one end of the wire and the anchor point.

      Tim

      Comment


      • #4
        Not really enough info to calculate an answer. Is the wire under a constant load as shown under the saw horse? If it is fixed, tottally dependent on the amount of deflection. Bob.

        Comment


        • #5
          If L is the distance between the fixed pulleys, X the vertical deflection, W the known vertical force:

          Find the angle whose tangent is L/2X. This is half the angle between the wires at the centre pulley.

          Call that half-angle "A"

          The line pull is W divided by 2cosineA.

          Example: If L=2X, tangent =1, half angle of wires is 45 degrees, 2cos 45=1.414, line pull is W/1.414 .

          Sorry, first post was wrong, above should be OK.
          Last edited by cameron; 10-05-2013, 02:11 PM.

          Comment


          • #6
            Originally posted by Jaakko Fagerlund View Post

            The wire goes over two stationary pulleys. Between them is a third pulley, that is able to move perpendicular to the wire travel.
            I think the picture fits the OP's description close enough to work with. It's going to take a LOT of tension if the cable is nearly straight between the top pulleys. Cosine approaches 0 as it gets near 90 degrees, so my equation seems to make sense.
            Any products mentioned in my posts have been endorsed by their manufacturer.

            Comment


            • #7


              Like this. The two pulleys on the left and right are fixed with a known center distance and the middle pulley is free to move in the direction of the blue arrow. Red is the wire in tension. Now, if you push (or pull) the middle wheel down with a known force, the wire deflects a certain amount. Less deflection means more tension in the wire.

              I see I got some useful responses, seems like they are correct and useful, thank you very much

              The idea is to make a tension measurement device for my wire EDM, as I want to know the tension in the wire. Preliminary idea is the one I pictured, with the middle wheel being spring loaded with a spring whose spring constant I know. This makes it a matter of seeing the deflection distance to figure out the force on the middle pulley and then just calculating the wire tension as per formulae.

              Advanced idea is to put a fixed weight on the middle pulley and a linear potentiometer fixed to it to measure the tension with a microcontroller.
              Amount of experience is in direct proportion to the value of broken equipment.

              Comment


              • #8
                Here is another way to approach the situation.
                http://www.tensitron.com/fine_wire.html
                If You scroll down there is a section discussing edm wire tension.

                Steve

                Comment


                • #9
                  Originally posted by doctor demo View Post
                  Here is another way to approach the situation.
                  http://www.tensitron.com/fine_wire.html
                  If You scroll down there is a section discussing edm wire tension.

                  Steve
                  Yup, that kind of measuring tool I'm after. But I'm building it myself, not buying unless the sensor is 50 EUR.
                  Amount of experience is in direct proportion to the value of broken equipment.

                  Comment


                  • #10
                    Could you use a load cell on one of the wire mounting points. You can get ones quite cheap now from HK...

                    I used one on a recent project married to a Microchip MCP6N11-100 preset gain instrmentation amplifier.

                    http://www.ebay.co.uk/itm/Electronic...-/380642622174

                    Rob

                    Comment


                    • #11


                      Cheers

                      .

                      Comment


                      • #12
                        Same principle as the Jobst Brandt FSA spoke tension meter.

                        Comment


                        • #13
                          Bicycle spoke tension meters and band saw tension meters work on the same principle. Buggabo is calibration, geometry calcs are straight forward. Using Barrington's approach would require knowing
                          the cross sectional area of the piece under tension and the applied force. Fine Woodworking had an article years ago showing a device using a dial indicator to measure deflection of bandsaw blades
                          and infer tension, but the article was at least 15 yrs ago.
                          Steve

                          Comment


                          • #14
                            This site is like going back to school!, not a bad idea, the simplest way is to use vectors ( a quantity that has magnitude and direction, unlike scalars that have only a magnitude), solving can be a trig exercise if you want to or a drawing exercise, each force is represented on a scale drawing, at a convenient scale, say 10Kn per cm, in a direction, the actual angle of the force is used, each drawn from a point, a triangle polygon is then drawn with each line head to tail, the gap thats left is the resultant, note that the resultant is the reverse closing line, in this way quite complex frameworks can be analysed to find the forces, redundant members and all sorts.
                            Look at triangle of forces by drawing, Bows notation, Polygon of forces, vectors by drawing, adding vectors graphically
                            Mark
                            Last edited by boslab; 10-05-2013, 08:48 PM.

                            Comment


                            • #15
                              Originally posted by sch View Post
                              Using Barrington's approach would require knowing
                              the cross sectional area of the piece under tension and the applied force.
                              I was assuming that the stiffness of EDM wire was negligible compared to the tension ???

                              i.e. not the same case as bicycle spokes or hacksaw blades, but more like a tape tensiometer.

                              Cheers

                              .

                              Comment

                              Working...
                              X