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Homework Help: Reaction at a point

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data


    Segments AB and BC of the beam ABC are pin connected a small distance to the right of joint B. Axial loads act at A and at the midspan of AB. A concentrated moment is applied at joint B.
    Find the reaction at A, B, and C.

    2. Relevant equations


    ƩM= (Fi)(di)

    3. The attempt at a solution

    I split the bar at joint B and drew two separate free-body diagrams. For the right side of the bar, I have Cy and Cx at C, Bx from the pin, and the moment about B (100 ft-lb).

    I solved:


    From there I took the moment of the entire bar about point A, and found By=-20 lb, which I don't think is correct.

    I think I'm mostly confused because there are no downward y forces in this problem, and I didn't think rollers could ever have a downward force.

  2. jcsd
  3. Sep 6, 2012 #2


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    In this problem, the roller supports are free to slide in the x direction, and thus cannot support any load in that direction, but they are quite capable of supporting vertical forces in the y direction. The applied moment is applied at the support joint B, not at the pin to the right of B. When you look at a FBD of the right section from the pin to C, you have at most just x and y forces at the pin and at C, no applied moment . By using the equilibrium equations, you can find Cy and the vert pin force rather easily.

    Then look at the left section from A to the pin. Use the equilibrium equations again on this section to solve for Ay , By, and the horiz force at the pin. Then back to the right section for Cx.
  4. Sep 6, 2012 #3
    Sorry, but how would I find any forces on the right side? Wouldn't my equilibrium equations just be:

    Fy= By+Cy=0
  5. Sep 6, 2012 #4


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    Yes, more or less, but remember you are not looking at the joint B at the support in this FBD, you are looking at the pin just to the right of B, call it B'. So your equations should be
    Sum of forces in y direction = B'y + Cy = 0, and
    Sum of moments about B' = 10Cy = 0.

    from this last equation, Cy = ???
    And then from the first, B'y = ???
    You still have B'x and Cx forces to contend with.
  6. Sep 6, 2012 #5
    So they're both just equal to zero?
  7. Sep 7, 2012 #6


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