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Finding the Reaction Force in a Truss

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data

    EjdVTfu.png

    2. Relevant equations

    In static equilibrium the forces are = 0

    Method of Joints

    3. The attempt at a solution

    i dont know how to find the reaction forces of A and B. i know how to calculate the force of each member by working backwards from where F is applied but once I do that, how will I know the reaction force? isn't there an easier way using symmetry and do the lengths of the members matter?
     
  2. jcsd
  3. Nov 14, 2015 #2

    SteamKing

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    The same equations of static equilibrium apply here like they do for all objects.

    The first step is to draw a free body diagram of the truss as a whole. You know that there will be reactions at A and B which keep the truss from moving and/or rotating.
    Since you are given no other information, you can assume that the truss has no weight, so the reactions at A and be will depend only on the applied load F and where it is located.
     
  4. Nov 14, 2015 #3
    So:

    Ay+By =0

    Ax+Bx=0

    But how does this bring me any closer to finding out what each component actually is? Using a free body diagram ignoring the weight of the truss I get the horizontal components to be 0 and the vertical to be F/2 for Ay and 4/2 for By.
     
  5. Nov 14, 2015 #4

    SteamKing

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    Did you write any moment equations for this truss?
     
  6. Nov 14, 2015 #5
    Sorry but what is a moment equation?
     
  7. Nov 14, 2015 #6

    SteamKing

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    You must be new to statics.

    For a body to remain in equilibrium,

    Σ F = 0
    Σ M = 0

    The moment is the tendency of a body or structure to rotate about a point due to an applied load acting on the body.

    In the case of this particular truss, the applied load F tends to cause the truss to rotate about point A. As a free body, the reactions at A and B must not only keep the truss from translating in the y-direction, but they must also keep the truss from rotating about point A.

    If this is unfamiliar to you, you may require some review of the equations of equilibrium, such as is presented in a course on statics.

    This article provides a quick outline:

    http://webassign.net/question_assets/buelemphys1/chapter10/section10dash11.pdf
     
  8. Nov 14, 2015 #7
    We were taught Torque and Static Equilibrium. We're supposed to be able to do this problem with just the method of joints.

    I understand that forces in the X and Y direction equal 0 meaning no rotation but how am I supposed to figure it out? I don't have a model truss to play around with and figure it out...

    Edit: Do I do F*r?
     
  9. Nov 14, 2015 #8

    SteamKing

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    No one is stopping you.

    If you don't feel comfortable writing the equilibrium equations for the truss as a whole, then don't. There's often more than one way to solve a problem.
     
  10. Nov 14, 2015 #9
    I went back and looked through the class notes for this topic and still I don't understand what you mean.
    Can you at least get me started...?
     
  11. Nov 14, 2015 #10
    I got:
    Ay= 3F
    By=-3F

    Ax=(-3/2)F
    Bx=(5/2)F

    But I don't have an option from these so I'm guessing they are wrong?
     
  12. Nov 14, 2015 #11

    SteamKing

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    It's not clear how you got these results. Can you show some of your work?
     
  13. Nov 14, 2015 #12
    ΣFy=0
    Ay+By=F (Will just assume it is 1)
    Ay+By=1

    ΣFx=0
    Ax+Bx=0

    -------------------------

    ΣMa=0
    4Bx-12F=0
    Bx=3N

    Since Ax+Bx=0 then
    Ax=-3N

    --------------------------

    ΣMb=0
    3Ay+4*Bx-3By=0
    3Ay+12-3By
    Ay-By=-4

    Ay=1-By (From the underlined equation)
    so:
    (1-By)-By=-4
    -2By=-5
    By=5/2

    Knowing that:
    Ay-(5/2)=-4
    Ay=-3/2

    It is however, wrong. It asks for magnitude so negatives don't matter but 5/2 isn't an option.
     
  14. Nov 14, 2015 #13

    SteamKing

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    I agree with your work up to this point.
    You can write only one moment equation here. Since you wrote your moment equation about point A, you can't write another about point B.

    Apparently, you are supposed to enter the horizontal and vertical components of the reactions at A and B separately.
    Which of the choices from the list of available answers would most likely satisfy ΣFy = 0, given that the applied load is equal to F?
     
  15. Nov 14, 2015 #14
    Oh,

    Well i thought the y components of both would be F/2 as they add up nicely but the answer was wrong.
     
  16. Nov 14, 2015 #15
    Or wait, since it's separate wouldn't it just be F for each?
     
  17. Nov 14, 2015 #16

    SteamKing

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    Not if Ray + Rby - F = 0
     
  18. Nov 14, 2015 #17
    So F/2?
     
  19. Nov 14, 2015 #18

    PhanthomJay

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    Not sure what you mean. SteamKing agrees with your answer of 3F for the horizontal reaction at A. The problem asks for the magnitude of the horiz and vert reactions at A. If you think the horiz reaction at A is 3F and the vert reaction at A is F/2, then enter "ji". If however you think the horiz reaction at A is 3F and the vert reaction at A is F, enter "jf". Or if you think the horiz reaction at A is F and the vert reaction at A is F, enter "ff". Etcetera. I'm not sure what you are thinking.
     
  20. Nov 14, 2015 #19
    I know how to format my answer but he said:

    So it got me a bit confused.

    I thought the answer would be 3F and F/2 but that was wrong, and I have a limited number of attempts so I don't want to hastily guess.
     
    Last edited: Nov 14, 2015
  21. Nov 15, 2015 #20

    PhanthomJay

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    Well now the horizontal reaction magnitude at A is certainly 3F from equilibrium considerations, but the vertical reaction at A cannot be solved from statics alone, because the problem states that both supports are pinned and thus both capable of supporting vertical loads. I guess you have to assume the supports are rigid but the member AB is not. So if the vert reaction at A is not F/2 , what else might it be if you assume no vert reaction at B? Sort of an unfair question I'd say.
     
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