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Load taken by columns

  1. Mar 15, 2016 #1
    Hi everyone.


    I would really appreciate if someone would help me with determination of floor load on columns (column tributary area).

    Normally uniformly distributed load acting on a beam is halved and each half-load is allocated to one column supporting the beam. In this case however, there is an 8 m beam spanning between a column 1 and 2, uniformly distributed load of magnitude of 80kN/m runs from column 1 up to 5m along the beam. Then another UDL 10kN/m runs from that point for 3 m up the column 2. On the grid distance between columns 1 and 2 and adjacent columns is 7 m.
    What distance of load along that beam do column 1 and 2 take respectively? Reaction Force at column 1 is 281 kN and column 2 is 149 kN.

    M I look for explanation (calcs would be welcome however). The beam is simply supported. Below is a simple diagram.

    upload_2016-3-15_16-39-38.png
    Thanks a lot !
     
  2. jcsd
  3. Mar 15, 2016 #2

    haruspex

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    Not sure I understand the question. What do you mean by "distance of load"? You seem to have found the loads on the columns ("reaction force" in your post), so I don't understand what else you are after.
    Also, is the 7m spacing in the X direction relevant? I couldn't see how.
     
  4. Mar 16, 2016 #3
    Thank you for your reply.

    Yes the 7m is in X direction ( not really relevant), the 8 m beam is in Y direction. I was given a task to determine load area on each column. In X direction the distance of the imposed load that each column carries is 3.5 m(half the distance because the load is the same and uniform for all the beams along the X direction ). If the load was uniformly distributed and of the same magnitude along the beam in Y direction ( 8m) the load distribution would be halved for each column and each column would carry 4 m of that load in Y direction ( the same as in X direction). But the load in Y direction is uneven so the column 1 will not carry 4m of that load, neither will the column 2. So I need to get ' centre' or 'centre of gravity' or 'resultant force of all the forces' and then I could say that to the right of this point all the load will be carried by the column 1 and all the load to the left of that point will be carried by the column 2.
     
  5. Mar 16, 2016 #4

    haruspex

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    Ok, but in post #1 you quoted 'reaction forces' for those two columns which confused me. So I thought you must mean something else.
    But checking those numbers now, I see they are much too high for the loads you are trying to calculate, so I'm mystified as to what those two forces represent.
    Anyhow, you need to take moments to find how the load is distributed. Pick one of the columns to take moments about - either will do.
     
  6. Mar 16, 2016 #5
    Sorry, I rounded the numbers up a bit. So you are saying that I should take it from bending moment? From the elevation it looks like this upload_2016-3-16_11-23-24.png
    Should I take the point from the bending moment? Elevation view looks like this


    upload_2016-3-16_11-25-49.png
     
  7. Mar 16, 2016 #6

    haruspex

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    Originally you wrote 80kN/m etc., now you have kN/m2. I assume it's kN/m.
    You still have not explained how you got 281 and 149, so I shall assume you have been told these are the answers and you are trying to prove it.
    To find the loads on those columns you need two equations. Any two of the following will do it:
    - sum of vertical forces
    - sum of moments of vertical forces about the top of one column
    - ditto, moments about the other column
    - ditto, moments about the point where the load changes.

    To get the moment of a spread load, you could write an integral, but it should not be necessary. For this purpose, you can take each uniform load as acting at its midpoint.
     
  8. Mar 16, 2016 #7

    PhanthomJay

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    Peter

    You seem to have double posted but you are mixing up your definition of tributary area versus tributary distance. Forget about tributary distance and focus on tributary area. The center beam between columns 1 and 2 takes the floor slab load from 3.5 m either side of it. So for example where the slab load is 1.4 kN/m^2, the distributed load on that beam is 1.4 times 7 or 9.8 kN/m which you have rounded off to 10 kN/m. Same procedure for other part of the beam and you get about 80 kN/m. Did you get these by yourself or what? That is how you calculate those loads to the beam using the tributary areas. Then solve for the column reactions as haruspex has noted. Done. Don't try to come up with some tributary 'distance 'along the beam.
     
  9. Mar 17, 2016 #8
    Thank you every one for you answers. they were very helpful
     
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