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Finding the reactions at the supports, 3D model

  1. Dec 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi. See attached, I need to determine the forces in all members.


    2. Relevant equations
    Sum of forces in X, Y & Z all equal 0.
    Sum of moments in X, Y & Z all equal 0.


    3. The attempt at a solution

    I've drawn the reactions at each support, A, B and C in my 2nd attached

    ƩFx=0 ∴ 8+3-R(ax)-R(bx)-R(cx)=0
    ƩFy=0 ∴ R(ay)+R(by)+R(cy)=0
    ƩFz=0 ∴ 20+20+20-R(az)-R(bz)-R(cz)=0

    After taking moments from various places and about various axis, I get Reactions at:

    A-x= 8
    y= 0
    z=20

    B-x= ?
    y= 0
    z= 9

    C-x= ?
    y= 0
    z= 31

    All in kN. The workings are just long winded so I'll put the important bit in - the unknowns R(bx) and R(cx).

    Taking moments at A about the Z axis
    0.866R(bx)+0.866R(cx)+0.866(3)+0.5R(cy)-0.5R(by)=0

    R(cy)=R(by)=0 & dividing by 0.866
    The equation becomes:

    R(bx) + R(cx) = -3kN

    I would say 3kN is shared equally between point C and B, but I have no proof, i.e. I can't see an equation that shows R(bx)=R(cx). Can I just assume 3kN is shared equally and R(bx) = R(cx) = 1.5kN
     

    Attached Files:

  2. jcsd
  3. Dec 30, 2012 #2
    If E is a pin joint, not able to resist moment about any axis, what does this imply for the reaction components at B? You have only 6 equilibrium equations for the whole frame, but you have assumed 9 reaction components. You need three more equations. If E is a pin joint, that gives you two more equations, considering the equilibrium of EB alone. Perhaps you are too algebraic in your approach? Yes, you do in the end have to solve equations, but have you considered drawing the three orthogonal views parallel to each of the x, y, z directions, to see if some simplification or symmetry makes itself apparent? And you cannot assume that B and C equally resist the 3 kN force.
     
    Last edited: Dec 30, 2012
  4. Jan 4, 2013 #3
    Yes Im quite algebraic in my approach, but it's all logic so I have more confidence it's correct. I apprectiate your response.

    Ive had a think about why you chose a pin joint at E. The EB leg seems only suited to taking a tensile force (z component). Realistically speaking there would be a huge force at the joint if it wasn't a pin joint. I now believe the force is zero at reactions B in the x and y planes. So I can say the force at C in the x direction is 3kN.


    Now I have all 9 values I can start calculating the force in each member.

    You mention orthogonal views. I will research this a little, the book I'm using hasn't mentioned that method. Does it help finding the reactions or finding the forces in the members?
     
  5. Jan 4, 2013 #4
    You may well be correct with your reaction components. Before calculating member forces, you can check the reaction components by using 6 independent equilibrium equations. Algebraic sum of forces in 3 directions should be zero. Algebraic sum of moments about any 3 non-parallel axes should be zero. That will give you the confidence to proceed.
     
    Last edited: Jan 4, 2013
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