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Homework Help: Reaction forces in a two segment beam

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    I am just concerned with part a.

    I got the correct answers doing this:

    First, FBD of segment AB:
    +cw ∑M_A=0: -136 - 6*By = 0
    By = -22.7N

    two ways to get Ay:

    +cw ∑M_B=0: -136 + 6*Ay = 0
    Ay = 22.7N

    ∑Fy=0: -22.7 + Ay = 0
    Ay = 22.7N

    FDB entire beam:
    ∑Fy=0: 22.7 - 22.7 + Cy = 0
    Cy = 0

    ∑Fx=0: -440 + 220 + Cx = 0
    Cx = 220N

    Ok, so I got the correct answers. However, here is where I get confused. If I make a FBD on segment BC I get a different answer for Cy.

    My attempt at a FBD:
    this leads to either:
    ∑Fy=0: -22.7 + Cy = 0
    Cy = 22.7N


    +cw ∑M_B=0: -136 + 3*Cy = 0
    Cy = 45.3N

    Clearly my FBD is wrong for segment BC. I assume the correct FBD would have a zero reaction at point B; this would give the correct answer (0) for Cy. In order for Cy to be zero, there must be no forces or moment at point B in the FBD, but this doesn't make sense to me.
  2. jcsd
  3. Jan 16, 2016 #2


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    Your free body diagram is wrong because when looking at that segment between the support at B and C , you must include the internal moment in the beam at B. Far simpler to take your FBD between the pin (just to the right of B ) and C. In this manner, the applied couple does not enter into your diagram.
  4. Jan 16, 2016 #3
    So my fbd is just missing an internal moment at B, and the resulting equation would be:
    +cw ∑M_B=0: -136 + 3*Cy + M_B = 0
    I can't solve directly for Cy since I don't have M_B.
    Is this correct?

    Would I not still have to include an internal moment? Wouldn't a complete diagram look something like this?
  5. Jan 16, 2016 #4


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    unlike at an externally pinned support , there can be no internal moment in the beam at a pinned connection.
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