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Truss Prob: finding the force in a member

  1. Nov 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The diagram is attached. I must solve using method of sections the force in member HG.



    2. Relevant equations
    ƩF(y) = 0, ƩF(x) = 0, ƩM = 0


    3. The attempt at a solution

    My hand-drawn pic shows the section.
    Solving vertically, ƩF(y) = 0
    F(DG)cos45 = 10
    F(DG) = 14.1kN

    Moments about E
    10(2) + 2F(DG)cos45 + 2F(HG) = 0
    Dividing by 2 and rearranging:
    F(HG) = -10 - 14.1cos45 = -20kN



    I don't think I'd have a problem if there wasn't a pulley system and I can only assume the reason I'm not getting the correct answer for F(HG) is because I'm not taking into account the effect the pulley has.

    Can anyone advise a nudge in the right direct... ?
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2012 #2

    SteamKing

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    Well, what effect does the pulley have on the truss? Can the pulley be replaced by equivalent forces at the end of the truss?
     
  4. Nov 28, 2012 #3
    Your free body diagram looks right. Now take a moment around G to find what member DE is. Next, sum the forces in x and y to find DG and GH.

    Was I helpful to you with this? I really struggled with trusses in statics and now have a really good understanding of how to solve these problems.
     
  5. Nov 29, 2012 #4
    In your free body diagram you need to identify the value of the horizontal force at the top of the pulley. What do think that is? (in theory, and in reality)
     
  6. Nov 29, 2012 #5
    I have attached my workings out to find the reactions at the supports. Originally I was taking the total external forces at face value (i.e 10kN vertically down) but now I can see because the pulley system support is not connected to the structure there are extra forces.



    Going back to the original FBD, I'll add and resolve with a horizontal force of 10kN in there, too. Ive attached my workings out.
     

    Attached Files:

  7. Nov 29, 2012 #6
    Reactions look good, but meanings of F1 F2 and F3 not well-defined.
     
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