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Direction of moment

  1. Jul 13, 2016 #1
    1. The problem statement, all variables and given/known data

    why does the moment -400Nm only act at region B and C?

    Why it doesnt act from A to C?

    http://imgur.com/a/5tfCg
    2. Relevant equations



    3. The attempt at a solution

    As we can see from figure, there's moment 400Nm on the beam at B at the top part and also the below part....
     
  2. jcsd
  3. Jul 13, 2016 #2
    notes here
     

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  4. Jul 14, 2016 #3
    is it wrong?
     
  5. Jul 14, 2016 #4

    SteamKing

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    The moment at B is called a couple. It acts on the beam as a whole. It's applied at point B, so naturally it acts there. The beam is fixed at point C, so a moment reaction will develop there to keep the beam in static equilibrium. This is why the moment diagram looks the way it does.
     
  6. Jul 14, 2016 #5
    so, the moment 400Nm at B has anticlokwise direction, it cause the beam at B to become concave downwards, at c, the moment is clockwise , so , the beam at c will also concave downward , so, at both B and C the moment is negative, am i right?
    Thus, moment anticlokwise = moment clockwise , beam in equlibrium?
     
  7. Jul 14, 2016 #6

    SteamKing

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    At B, the couple is CCW and has a positive magnitude.
    At C, the reaction must be CW, and the moment there will have a negative magnitude, to ensure that there is no net moment acting on the beam.
     
  8. Jul 14, 2016 #7
    If the moment at B is positive, then why it's - 400Nm oat B on the graph?
     
  9. Jul 14, 2016 #8

    SteamKing

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    It's not -400 N-m at B. The diagram is indicating that the reaction moment at C is -400 N-m, so that the net moment becomes zero.
     
  10. Jul 14, 2016 #9
    The
    Then it's positive 400Nm at B?
    Why the author drew the line as negative negative 400Nm from B to C?
     
  11. Jul 14, 2016 #10

    SteamKing

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    I think you're misinterpreting the diagram.

    After all, there is a lot of additional information being presented here, and some compromises must be made on presentation.
     
  12. Jul 14, 2016 #11
    Can you whether diagram is negative 400Nm from B to C? Why not positive 400Nm at B?
     
  13. Jul 14, 2016 #12
    Why can't I interprete the moment causes the beam to concave downward as negative? I ran that I was told this theory earlier in the forum.....
     
  14. Jul 14, 2016 #13

    SteamKing

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    The magnitude of the couple is +400 N-m at B.

    In order for the beam to remain in equilibrium, the reactive couple at C must be -400 N-m, which is what the diagram indicates to me.
     
  15. Jul 14, 2016 #14
    Since you said that the moment at B is positive 400Nm, then why the graph shouldn't change from positive 400Nm to negative 400Nm from B to C?
     
  16. Jul 14, 2016 #15

    SteamKing

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    If you look at the moment plot at C, it clearly indicates that the reactive moment there is -400 N-m, which, when added to the moment at B of +400 N-m, equals zero net moment.

    Your interpretation of what the plot would look like would require a net change in moment of -800 N-m, which is not supported by the loading of the beam.
     
  17. Jul 14, 2016 #16
    So, we draw the graph from C to B?
    That's very the graph stop at B - 400Nm +400Nm =0?
     
  18. Jul 14, 2016 #17
    So, we draw the graph from C to B?
    That's very the graph stop at B - 400Nm +400Nm =0?
     
  19. Jul 14, 2016 #18

    SteamKing

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    It's not clear what you mean here.

    The moment diagram doesn't plot discrete values of BM in isolation, like you apparently want to do.

    At point B, the moment goes from 0 to +400 N-m, which is why there is a jump at this location.

    The moment remains constant at +400 N-m from point B to point C, where the reactive moment of -400 N-m is added to the +400 N-m, to bring the net moment back to zero.

    You should review how shear force and bending moment values are plotted for regular beam problems.
     
  20. Jul 14, 2016 #19
    Changing from positive 400Nm to 0 from B to C, why shouldn't the graph drawn at the positive side?
    I mean in the figure, the author indicated it as negative 400Nm and at C, it become 0
    Why shouldn't the graph drawn from positive 400Nm to 0 from B to drop 'directly' to become 0 at C?
     
  21. Jul 14, 2016 #20

    SteamKing

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    The "positive" side is the side which is shown by the author on the diagram. Certain allowances must be made so that the two diagrams don't overlap, which would make things even more confusing.
    I think I have explained this diagram enough. If you want to plot a different diagram, that is your prerogative.
     
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