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Working out support reaction of a frame - Statics

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data

    For the frame shown (see attached picture) , work out the horizontal reaction at B (ie find Bx)


    2. Relevant equations

    EQUILIBRIUM EQUATIONS:

    sum of moment = 0 (CCW is positive)
    sum of vertical forces = 0 (up is positive)
    sum of horizontal forces = 0 (left to right is positive)

    3. The attempt at a solution

    I thought this was easy but for some reason everytime I equate the forces, Bx gets cancelled out.

    [sum of moments at A = 0];
    -38(0.6) + By(0.3) = 0
    By = 76 kN

    [sum of moments at B = 0];
    -Ay(0.3) + 38(0.6) = 0
    Ay = 76 kN

    [sum of moments at D = 0];
    Ax(0.6) + Bx(0.6) + By(0.3) = 0 (***)

    [sum of moments at C = 0];
    (-38)(0.2) + Ax(0.4) + Bx(0.4) = 0

    so, Ax = 19 - Bx

    substitute Ax = 19 - Bx in (***) :
    (19 - Bx)(0.6) + Bx(0.6) + By(0.3) = 0

    but By = 76 kN so:

    (19 - Bx)(0.6) + Bx(0.6) + 76(0.3) = 0

    BUT Bx just cancels out??

    I have no clue how to approach this question any other way
    Any help/guidance would be great!
    Thanks!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2012 #2
    I got it :)
    I realised that when taking the moment about C, the force Bx does not need to included.
    So Bx = -57 kN
     
  4. Oct 8, 2012 #3

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    NO. By produces a moment also.


    What you mean to say is that since member BC is a 2-force member with the force BC directed along the axis, then the force BC produces no moment about C.
     
  5. Oct 13, 2012 #4
    Thanks for that PhanthomJay! :) My exam is very very soon and I'm glad you cleared that up for me.
     
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