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Calculating Forces in Truss Members using Method of Section

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data

    a) Find the reactions provided by the foundations at A and B.
    b) Use the method of Resolution at the Joints to calculate the forces in all the members.
    c) Which member will fail first and why ?

    2. Relevant equations

    ƩFx = 0
    ƩFy = 0
    ƩM = 0

    3. The attempt at a solution

    I am really just concerned with whether I have got the following correct. If I have, then I have grasped the concept if not then if someone could point me in the right direction and advise me on where to go next of what I am doing wrong I would be very grateful.

    Horizontal Reaction Forces in the X direction at point A = 0
    Vertical Reaction Forces in the Y direction at A = 4000N Upwards
    Reaction Forces about the moment A at point B = 6000 Newtons Upwards

    Forces in Member AC = 0N (0 + AC + AF = 0)
    Forces in Member AF = 4000N in Compression (4000 + AC + AF = 0)
    Forces in Member BK = 6000N in Compression (6000 + FBK.Sin(90) + FBE = 0)
    Forces in Member BE = 0N (-FBE + FBK = 0)

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 7, 2011 #2

    PhanthomJay

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    This is all good, you got them all correct:smile::approve::cool:
     
  4. Nov 7, 2011 #3
    Thanks very much!!! One last thing. I am now finding the forces in Members FC and FG. Am I right in saying:

    Forces in direction:
    -(-4000) - FC x sin(53.1) = 0 and then I just rearrange for FC = 5000 Newtons

    and now I know FC I can do the following:

    Force in direction:
    0 + (FC x cos(53.1)) + FG = 0 and again rearrange for FG = -3002 Newtons
     
  5. Nov 7, 2011 #4

    PhanthomJay

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    yes , compression or tension?
    You have a round off error, these are 3-4-5 right triangles, FG = -3000 N. What does the minus sign mean?
     
  6. Nov 7, 2011 #5
    Sorry a negative number means it is compression and a positive number means it is in tension.
     
  7. Nov 7, 2011 #6

    PhanthomJay

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    OK, you are again correct.
     
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