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Truss Analysis - Method of Joints

  1. Nov 7, 2013 #1
    trus.png

    Ok so first I considered the entire structure (which is in equilbrium) and resolved to find out that HA = -4.8 kN, VA= -0.9 kN and VE=0.9 kN. (these answers were correct according to mark scheme) Then I considered each joint separately (in order to find the forces in each of the members). I started with A and I found that FAB = 1.5 kN and FAF = 3.6kN. I then decided to consider joint B. However, this is where I ran into a problem. When considering joint A, I calculated FAB to be 1.5kN. Since it is positive, surely this means that my initial assumption was correct and the direction of the force is indeed from A to B? So it should be as so:

    image1dld.jpg

    However, in the mark scheme they appear to have drawn the 1.5kN force in the opposite direction (so from B to A)... Why is this so? I have been getting stuck on the direction in several problems like this..

    Thanks
     
  2. jcsd
  3. Nov 7, 2013 #2

    PhanthomJay

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    It is very easy to get beaten by the plus and minus sign. You have correctly calculated the support reactions, but be sure to indicate their direction on the diagram. Va points down and Ha points left.. You will then find that Fab pulls away from joint A (tension), and thus when looking at joint B, Fab must point away from joint B, not toward it, per newton's 3rd law. Tension forces always pull away from the joints on which they act.
     
  4. Nov 7, 2013 #3
    Thanks..

    IF FAB = -1.5 kN , it would mean the member AB is under compression and hence the force exerted by the two ends would point at each joint (rather than away from it), but would the force pointing at each joint equal -1.5 kN or +1.5kN?

    Cheers !!
     
  5. Nov 7, 2013 #4

    PhanthomJay

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    You have to be very careful with the interpretation of the minus sign, because it often means different things. As in your example, you chose roght and up as positive, and determined that Ha and Va were negative, hence, acting opposite to the direction you showed on the diagram. Immediately, correct your diagram to show the proper direction if the support forces on the structure. Otherwise you will get hopelessly buried by the minus sign. Now when you look at the equilibrium of joint A, you determine that the y force component of Fab on joint A is acting up, and its x component is acting to the right, and thus the magnitude of the resultant force of Fab on joint A is the sq rt of the sum of the squares, or 1.5 N , but its direction is determined from vector addition, pointing away from the joint, or in tension. If you look at the forces in the member AB, at either end, they also pull away from the member.....tension. joint
     
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