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External moments when assembling internal force functions

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a 6 meter long beam that is supported by a roller at 0m and a pin at 6m.

    There is a triangular distributed load that begins at 0m to 3m which is 2kN/m. (Max load is at 0m)

    There is an additional triangular distributed load that begin at 3m which mirrors the first triangular load (2kN/m, max load is at 6m)

    And there is a counter-clockwise moment acting at 0m of 18 kNm


    2. Relevant equations

    ƩFx = 0
    ƩFy = 0
    ƩMr = 0

    Vertical Reaction @ A (0m) = 6kN
    Vertical Reaction @ C (6m) = 0kN

    3. The attempt at a solution

    So my shear functions are fine but my bending functions always run into a similar problem.

    From 0m to 3m my bending function is f(x) = -(2/3)^2(x)^3(.5) + 6(x)
    From 3m to 6m my bending function is f(x) = -(.5)(2)(3)(x-1) + 6(x) - ((.5)(2/3)(x-3)^3)/3

    My method for drawing internal force diagrams involves checking shared values of each function. So in this case, I like to use 3m in each function to see if I get the same thing, since I know that the bending function is continuous.

    When I plug 3 into my first equation I get: -12 kNm
    When I plug 3 into my second equation I get: -12 kNm

    Everything checks out!

    Now, if I calculate my bending forces at 0 m, I get 0 kNm; If I calculate my bending forces at 6 m I get -18kNm.

    So basically I now have a bending force diagram that begins at 0, hits -12 kNm at 3m, finally hits -18 kNm at 6m.

    I'm looking at the solution and for some reason it's reversed. So I decided to see what I did wrong by including the external force into my functions and while I get the correct answer at 0m and 6m, my 3m answer is completely off.

    Is my solution correct? Should it be reversed? Why/Why not? (I solve left to right)
     
  2. jcsd
  3. Dec 12, 2013 #2

    SteamKing

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    You've got to include the applied couple of 18 kN-m @ x = 0 m in your bending moment function, just like you include the reaction forces at x = 0 m and at x = 6 m in your shear force functions.
     
  4. Dec 12, 2013 #3

    SteamKing

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    With these problems, make a sketch of the shear force and bending moment curves. It's much easier to construct your shear force and bending moment functions if you have a sketch.
     
  5. Dec 12, 2013 #4
    Thanks SteamKing. When I include the couple at x = 0 m, it checks out but at x = 6 m i still get -18.
     
  6. Dec 12, 2013 #5

    SteamKing

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    Why don't you post your work? That way, everything you've done should be clear.
     
  7. Dec 12, 2013 #6
    What part should I post? My functions are the same on paper as my original post.
     
  8. Dec 13, 2013 #7

    SteamKing

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    That's your problem. They shouldn't be.

    When x = 0, M = 18 kN-m
     
  9. Dec 13, 2013 #8
    This is where my confusion lies.

    I include this couple in calculation only at x=0? That I can accept but then at 6m I have to also include it to get a bending moment of 0.
     
  10. Dec 13, 2013 #9

    SteamKing

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    Look, it will be easier to construct the shear force and bending moment functions if you construct the shear force and bending moment diagrams to check yourself. The applied couple at x = 0 affects ALL of the subsequent bending moment values, just like the reaction force RA affects ALL subsequent shear force values.
     
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