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How do I find my x bar?

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Replace the distributed loading with an equivalent resultant
    force, and specify its location on the beam measured from
    point O.

    W = 3 kn/m


    2. Relevant equations

    F*xbar = torque

    3. The attempt at a solution

    I got the resultant forces:

    (1/2)*3*3 = 4.5 kn
    (1/2)*1.5*3 = 2.25 kn

    4.5 + 2.25 = 6.75 kn

    I know my general equation is:

    6.75*xbar = torque

    I want to solve for xbar.

    xbar should equal 2.5m
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2014 #2

    SteamKing

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    What's the centroidal location of a right triangle? Your distributed load can be broken up into two triangular loads.
     
  4. Oct 21, 2014 #3
    I'm unsure, could you explain it to me?
     
  5. Oct 21, 2014 #4

    SteamKing

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    If the centroid of a rectangle is located at half the length or half the depth, what is the centroid of a triangle?

    If you're still stumped, you can always google 'centroid of right triangle'
     
  6. Oct 21, 2014 #5
    If I multiply both distances by 1/3, I still do not get the proper answer.
     
  7. Oct 21, 2014 #6
    I see, so when I approach the triangle from the one side I have 1/3 so the rest must be 2/3. But how do I know which is the 1/3 side and which is the 2/3 side?
     
  8. Oct 21, 2014 #7

    SteamKing

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    If you have a right triangle with the pointy end at O, how far from O will the centroid be, 1/3 of the length of the base or 2/3 of the length of the base?
    Remember, the location of the centroid coincides with the balance point of the triangle.
     
  9. Oct 21, 2014 #8
    It would have to be 2/3.

    So, if I had a problem where I had to find, say a location at point, "A" that was directly under W. The distance would then be 1/3 for each triangle?
     
  10. Oct 21, 2014 #9

    SteamKing

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    I'm not sure what you are asking here. The choice of whether to use 1/3 or 2/3 of the length of the base as the centroid depends on the orientation of the triangle.
     
  11. Oct 21, 2014 #10
    I meant a situation like the one pictured. I would use 1/3 for each triangle, and I would use 1/2 for the rectangle, correct?
     

    Attached Files:

  12. Oct 21, 2014 #11

    SteamKing

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    1/3 of the base for each triangle would be OK, as long as the reference was taken about a vertical line running thru the support at B.
     
  13. Oct 22, 2014 #12
    I see, thanks for your help!
     
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