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Moment about a point on a beam

  1. Jan 12, 2017 #1
    1. The problem statement, all variables and given/known data
    For the moment about B ' , why there is extra θA L behind ?
    2. Relevant equations



    3. The attempt at a solution
    is that wrong ? I think there should be no θA L behind in the equation of moment about B '
     

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  2. jcsd
  3. Jan 13, 2017 #2
    The author takes moment using the conjugate beam, and at point A there is a shear force acting downward which is equal to the rotation of A in the real beam.
     
  4. Jan 13, 2017 #3
    why theta_A multiply by L , we will get moment ? what is theta_A actually ? I'm confused
     
  5. Jan 13, 2017 #4
    Theta A is the angular rotation in the real beam. Conjugate beam theory tells us that the shear force in the conjugate beam is the rotation in the real beam.

    In the conjugate beam, moment = shear force A * distance to B = Theta A * L
     
  6. Jan 13, 2017 #5
    why? Can you explain further ? Why are they equal ?
     
  7. Jan 13, 2017 #6
    review conjugate beam theory, you will find the answer there.
     
  8. Jan 13, 2017 #7
    Can you explain further ? It's not explained in my module
     
  9. Jan 13, 2017 #8
    http://www.ce.memphis.edu/3121/notes/notes_08b.pdf
    In this link , i only notice that integral of M/EI and dx = theta(angle of rotation) ... or d(theta) /dx = M/EI

    I rewrite it as M = EI(dtheta)/dx

    How could that be true ? I found that (dtheta)/dx = M/EI , M/EI is the moment diagram , am i right ?
     
    Last edited: Jan 13, 2017
  10. Jan 13, 2017 #9
  11. Jan 13, 2017 #10
    That's the 'trick' to this method. Since 'w' represents any distributed load... let it be w=M/EI , then from the equations we see that when we solve for shear in the conjugate beam, we get rotation in the real beam.
     
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