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Baltimore Truss Analysis Problem

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the force in member JQ for the Baltimore truss where all angles are 30º, 60º, 90º or 120º.


    2. Relevant equations

    Length from A to N can be taken as 6 units.

    3. The attempt at a solution

    To use method of sections.
     

    Attached Files:

  2. jcsd
  3. Apr 2, 2012 #2
    What is your understanding of the "method of sections"? How do you think it might apply in this case?
     
  4. Apr 3, 2012 #3
    Well.. method of sections is like, we have to cut the truss into two portions. the section we take should pass along the member whose force has to be determined. Both the portions will be in equilibrium and we can use either one of the two portions to determine the force on the member.

    I have tried to solve this question.. but I am getting stuck.

    Assuming the distance from A to N as 6units, then the height of the full truss will be √3 units.

    RA (6)= 2.5 (100) + 100 (2)
    RA (6)= 450
    RA = 450 / 6
    RA = 75kN

    RN = 100 + 100 - RA
    RN = 100 + 100 - 75
    RN = 125kN

    Figure : (See image attached)

    Taking Moments at Y,

    [itex]\Sigma[/itex]MJ = 0
    Therefore, 125(2) + FXY (√3) = 0
    FXY = - 250 / √3 (tension)
    FXY = 144.5 (compression)

    [itex]\Sigma[/itex]FY = 0
    Therefore, 100 = 125 + FXJ + FQJ . (cos30)

    [itex]\Sigma[/itex]FX = 0
    Therefore, 144.5 = FHJ + FQJ . (cos60)
     

    Attached Files:

  5. Apr 4, 2012 #4
    In general, in 2-dimensional problems of equilibrium, you have available 3 equations of equilibrium, and that is what you have applied, without success. The reason here is that your chosen section intersected 4 members whose forces you didn't know, and so you need an additional equation. For example, if the first section you chose passed through WX, GQ and GH, you would be able to solve for the force in WX in particular. Then, a second section through WX, QX, QJ and HJ would have 3 unknowns and is therefore solvable. In your case you have found the force in XY correctly, but that is not immediately helpful. You could also analyse member HQ by inspection. Incidentally, the method of joints is also a specific application of the method of sections. So the method of sections is valuable. The Baltimore truss is statically determinate, but is a special case that requires a stepwise process to unravel it. There are other examples.
     
  6. Apr 6, 2012 #5
    thanks a loot :D i was able to solve it with your help :)
     
  7. Nov 13, 2012 #6
    can you finish how to solve for F(qj)
     
  8. Nov 14, 2012 #7
    If you look at the rectangular panel XYJL, and imagine the diagonal member PY removed, you will see that the panel becomes rhombic and behaves like a mechanism. The triangle JPL just stiffens the bottom chord and doesn't contribute to the shear resistance of the panel. So PY is the only member resisting the shear in that panel, namely the 125 reaction. So now you can find the force in PQ. If you understood that, you can now look at joint J, where there are 3 unknowns and not enough equations. So you have to find one of those unknowns, again indirectly. Can you try some other sections or the technique I have just suggested to unravel one of HJ, QJ or XJ?
     
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