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Fixed Ended Beam applied to couple moment

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known dataHi, i am trying to understand how to find the reactions (Ra & Rb) and moments (Ma & Mb) of the beam shown in the sketch below. Not an exact number but in terms of Mo(couple moment) and L(length).



    2. Relevant equationsI have the answers which are Ra = -Rb = 3Mo /2L
    AND Ma = -Mb = Mo/4



    3. The attempt at a solution
    taking moments about B: Ma - Ra(L) + Mo - Mb = 0
    After that i broke the beam into 2 pieces (in the middle of the beam)
    took the elastic equation for both sides where:
    Mx = Ra(x) - Ma (left side)
    Mx =Rb (x) - Mb (right side)
    Intergrated both equations twice and found C1 and C2,and set these 2 equations equal for x= L/2 (as the delfection in the middle is 0)

    But i cant get this result

    Have i done something wrong?
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2012 #2

    SteamKing

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    It would probably be easier if you set the deflections and slopes at A and B equal to zero, and solved for the reactions. Is the deflection at x = L/2 equal to zero? Maybe, but the deflections at A and B must be zero, as should the slopes.
     
  4. Nov 20, 2012 #3
    Thank you for your response.I just tried to solve it like this but I couldnt get the result.
    Any other suggestions???
     
    Last edited: Nov 20, 2012
  5. Nov 20, 2012 #4
    any suggestions??
     
  6. Nov 20, 2012 #5

    SteamKing

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    Show your detailed calculation. There may be a mistake in your algebra.
     
  7. Nov 20, 2012 #6
    For Ra:
    φEI= (Ra (x^2 ))/2- Ma (x)+ C1 (eqn 1)

    where φ=slope, x=distance, C1 = (Ra (x^3))/6 + Ma (x^2)


    vEI= (Ra (x^3))/6 - (Ma (x^2))/2 + C1(x) + C2 (eqn 2)

    Where v=delfection,x=distance, C1 = (Ra (x^3))/6 + Ma (x^2), C2=0


    By setting (eqn 1) equal to (eqn 2) I cannot get the result I want because the result for Ra must be in terms of Ra and Mo.
    What u think?
     
  8. Nov 21, 2012 #7

    SteamKing

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    Remember, at x = 0, the slope = 0 and the deflection = 0.
    At x = L, the slope = 0 and the deflection = 0.

    You've got to use these four conditions in order to solve for the unknown reactions and fixed end moments.
     
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