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Reaction force in a beam

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-03-27 at 5.21.42 pm.png
    Calculate the reaction force at A?
    2. Relevant equations
    ΣMb = 0

    3. The attempt at a solution
    As there is a pin joint at B it is possible to consider the bar only from A-B.
    From here I applied
    ΣMb = 0
    120 + 20*2 + Ay*4 = 0
    This gives Ay = -40kn
    Also since there are no forces in the x direction Ax = 0
    and Hence A = -40kn or 40n downwards
    However this is wrong. I am not sure what I am doing wrong unless it is something that the sum of the y forces don't equal 0, but I think there would be a force at By= 60kn up to combat this.
    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Mar 27, 2016 #2

    Simon Bridge

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    120kN.m is a moment about A isn't it?
    What is the reasoning that suggests that you need only consider the A-B length?
    The pin connection at B does not stop rotation about B caused by the force distribution on B-E - just prevents translation of that point.
     
  4. Mar 27, 2016 #3
    So would the FBD look like this:
    Screen Shot 2016-03-28 at 12.40.31 pm.png
    The question says consider the reactions at A and D to be rollers, so thus there is no horizontal force
     
  5. Mar 27, 2016 #4

    SteamKing

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    The pin at point B can develop a force, but it cannot sustain a moment. That's what's missing from your FBD above.
     
  6. Mar 27, 2016 #5
    So would I add a vertical and horizontal force at point B and then use the equations of equilibrium to solve?
     
  7. Mar 27, 2016 #6

    SteamKing

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    You can add a horizontal force at pin B, but it looks like all of the applied loads are vertical.
     
  8. Mar 27, 2016 #7
    So you're saying it is there, but it would just be zero. Also i'm assuming then that Ex would be 0?
    If i do that won't I end up with more unknowns than equations, in which case it would be unsolvable.
     
  9. Mar 27, 2016 #8

    SteamKing

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    That's unlikely, since there are only two supports for the entire beam.

    BTW, the beam is only resting against the wall at E. It can't develop a moment at that location, only an axial force.
     
  10. Mar 27, 2016 #9
    So would the FBD look more like this:
     

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  11. Mar 28, 2016 #10

    SteamKing

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    Yes, that's more like it.
     
  12. Mar 28, 2016 #11

    PhanthomJay

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    In my estimation, I believe that the wording in this problem is not very good. When it says "resting against a wall", I think it implies "resting atop a wall", otherwise, with the hinge at B, the beam would be unstable.

    The force from the pin at B is internal, so when drawing the FBD of the entire beam system, that reaction doesn't enter into the equilibrium equation. I would proceeded by isolating AB first, but you have to be careful with your plus and minus signs, and interpretation of them.
     
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