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Loads in cables.

  1. Aug 24, 2010 #1
    Hi.

    I was trying to calculate a situation where you have a piece of cable that is attached to a horizontal bar and then the two ends of the cables goes up vertically and is attached. Now I pull in the part of the cable hanging down under the horizontal bar - but I don't pull in the center line, I pull off center. What will be the loads in the cables to the left and right of the pull and what will be the vertical loads in the cables where they are attached.

    I don't think my calculation is correct - and this is 2nd try. It is ideal if T6 = T7 = ½T1 as my calculation shows, but I have a feeling that it is not correct. But I have run out of good ideas on how to make the correct calculation. I guess the main problem is that I divide T1 by ½ in the beginning. But I can't figure out how T1 is split in the 2 forces.
     

    Attached Files:

  2. jcsd
  3. Aug 24, 2010 #2
    This revised diagram should help you proceed.

    You need to realise
    That the bar is in compression to maintain equilibrium of points A and B.
    That the forces in a compression member point outwards and in a tension member point inwards.

    To solve this you need some geometric information so you can take moments to apportion the load between the wires.
     

    Attached Files:

  4. Aug 24, 2010 #3
    Yes this is exactly the system I tried to explain. Very good. Yes of course the bar is a compression bar and the wires are tension members. If I put arrows on a member (2 arrows to show compression and tension), I have allways showed compression as two arrows pointing towards each other and tension as two arrows pointing away from each other (because the force is trying to rip the tension member apart).

    But ok, that is a detail. But how to calculate the forces if force T1 is known? Do you mean, calculating moment equilibrium around one of the 3 points in the triangle (probably on of the 2 top points)?? But isn't there still too many unknown forces?

    Oh and actually the system that you and I have sketched is only part of the system. See attached.

    I pull with known force T1 (1,5 kN is just an example until I solve how to calculate it). The "T1-cable, goes up to point 4, through the cable 3-1-2 and is attached at point 1, so it can't move through this cable. Now cable 1-4 and 1-2 will be subjected to tension. In point 2 the cable continues up vertically. At point 4, the cable will provide a force downwards and inwards at the bar. The downwards part of the force will give a tension reaction in the cables passing 3 and 2. The cable 1-3 is left out of the equation (I think), because it can't take compression.

    My goal was to find a geometric system where the load T1, results in the vertical end loads "T6" and "T7", that are T6 = T7 = ½T1. - If possible when T1 is off the centerline. But to understand how to get to that goal (if possible at all), I need of course to understand the calculation of this first system (see below).

    Forcesinwiressketch.jpg
     
  5. Aug 24, 2010 #4
    Have you actually tried taking moments?

    In my diagram let L be the length of the bar and let the line of action of W be a distance d from A.

    Taking moments about A

    T4,T2,C1 all pass through A and have zero moment.
    T1 appears twice in opposite directions so exerts zero net moment

    That leaves (T3 x L) = (W x d)

    so T3 = wd/L

    does this help?
     
  6. Aug 24, 2010 #5
    No I didn't yet try to take moments, as I didn't understand exactly what you meant. I had an idea that it was what you just explained, but wasn't sure. English is not my main language, so sometimes when speaking technical stuff, I not allways understand everything exactly.

    But I will try to do after your explanation. I think it helps, yes... :-)
     
  7. Aug 24, 2010 #6
    Yes I got now the loads in T3 and T4. But it was not exactly the result I was hoping for. I mean, I can see that even with the longer bar and extra wire, I won't get T3 = T4 = ½W, when W is off center.

    I is allowed to install the bar at an angle, but I don't think it helps. What system could do that? If I wrap W around or into a "wheel" with a tangent in the center line and W going to the middle of the bar, this "wheel" will just rotate.

    Hmm, have to think more about this.
     

    Attached Files:

  8. Aug 24, 2010 #7
    Now you have T3 and T4, equilibrium and simple geometry gives you T1 , T2 and C1.

    Yes of course the equations reduce to T3 = T4 = W/2, only when W is placed halfway between A and B.

    Tilting the bar does not affect this since the moments are derived from perpendicular distances.
    I don't understand the last comment about W and a wheel.
     
  9. Aug 24, 2010 #8
    Yes with T3 and T4, equilibrium and geometry gives me T1, T2 and C1. And yes T3 = T4 = W/2 only if W is placed halfway between A and B, I can see that from the equations.

    I will explain the other things later, I have a slight idea of how maybe I can do what I'd like (I need to make another system).
     
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