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Statics question - change in loading

  1. Jul 5, 2014 #1


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    1. The problem statement, all variables and given/known data

    In the structure shown, the beam is pinned at point B. Point E is a roller support. The beam is loaded with a distributed load from point A to point B of 400 N/m, a 500 N·m couple at point C, and a vertical 900 N force at point D. If the distributed load and the vertical load at D are removed, and a vertically upward force of 1700 N at point F is added, what moment at point F would be necessary to keep the reaction at point E the same?

    2. Relevant equations


    3. The attempt at a solution

    Since the reaction Ey at point E is to be unchanged, we only need to calculate the change in loading. Also, the location of the new moment is irrelevant.

    Since the reaction forces at B are unknown, it is convenient to find the change in loading by summing the moments at point B.

    The distributed 400 N/m force creates an 800 N downward force 1m to the left of point B.

    The changes at point B needed to counteract the removal of the 800 N force and the 900 N force, and the addition of an upward force at point F are:

    ∑MB = (800N)(1m) - (900N)(5.5m) + (1700N)(4.5m) = 3,500N clockwise

    That's my answer, but the book has:

    -(800N)(1m) + (900N)(5.5m) + (1700N)(4.5m) = 11,800N clockwise

    I've stared at this problem for an hour now and I can't see how my calculation is incorrect.
  2. jcsd
  3. Jul 5, 2014 #2


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    Hmm, I haven't written this out, but it looks like if you leave the diagram as is, and then apply ##\sum M_B = 0##, that should get you ##E_y##.

    ##\sum F_x = 0 \Rightarrow ?##
    ##\sum F_y = 0 \Rightarrow ?##

    Should give you the rest.

    Place the new force at ##F##. Apply your equilibrium equations again, what do you get?
  4. Jul 7, 2014 #3


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    Hello Zondrina,

    Assuming clockwise moments, y-axis up and x-axis to the right...

    From the original diagram,

    [tex]\Sigma M_B=-(800N)(1m)-(500N\cdot m)+(900N)(5.5m)-(E_y)(7m)=0[/tex]
    [tex]E_y=\frac{-(800N)(1m)-(500N\cdot m)+(900N)(5.5m)}{(7m)}=521N[/tex]

    No need to calculate the sum of the forces in the x direction.

    With the distributed load and load at point D removed, and with the new 1,700 N force at point F:

    [tex]\sum M_F=(B_y)(4.5m)-(500Nm)-(E_y)(2.5m)=M_F[/tex]
    [tex]\sum F_y=B_y+1700N+E_y=0[/tex]
    [tex]\sum M_F=(-2221N)(4.5m)-(500Nm)-(521N)(2.5m)=-11797Nm\approx-12000Nm[/tex]

    I still think the book's answer of 12,000 N·m clockwise is incorrect. My assumption of clockwise moments throughout and the negative 12,000 N·m moment just obtained seems to suggest that the answer is truly 12,000 N·m counter-clockwise. Would you agree?

    Some background: I'm studying for the FE. My strengths are in electrical engineering but I've taken statics and dynamics, physics and math, etc. I've read that one gets about 3 minutes per problem on the FE, so I need to find a way to easily spot wrong answers (test-taking tricks, if you will). Some of these statics problems easily take me 5 - 10 minutes.
  5. Jul 7, 2014 #4


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    Indeed, it is out of habit I write the equations down. Although it does inform you trivially that ##B_x = 0##.

    As for the moment around ##F##. Assuming clockwise:

    ##\sum M_F = 0 \Rightarrow M_F - (2221N)(4.5m) - (500Nm) - (521N)(2.5m) = 0##

    So ##M_F ≈ 12000 N## clockwise. This is because you were asked "what moment at point F...".

    I'm in electrical too. There's no better way to check your answer than to have drawn a free body diagram imo.
    Last edited: Jul 7, 2014
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