Floor load on columns
Floor load on columns
(OP)
I would really appreciate if someone would help me with determination of floor load on columns (column tributary area). Normally UDL 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. How much load do column 1 and 2 take respectively? Reaction Force at column 1 is 281 kN and column 2 is 149 kN. I think in calculating take down load we are to assume that connections are pinned and the calcs results are based on that.
I also attached a diagram to this thread.
Thanks a lot !
I also attached a diagram to this thread.
Thanks a lot !






RE: Floor load on columns
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RE: Floor load on columns
RE: Floor load on columns
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RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
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' ( whatever it is called) 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.
RE: Floor load on columns
Since:
R1= 281 kN
R2= 149 kN
W= R1 + R2 = 430kN
and:
281kN / 430kN = 0.65
then:
0.65 x 8m = 5.23m from Column 1.
But, please explain, what is the purpose of determining this?
RE: Floor load on columns
RE: Floor load on columns
The purpose for it is to draw on CAD load areas allocated to every column in respect to other columns ,since the grid is irregular (the spacing between columns)
RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
RE: Floor load on columns
My calc was based upon determining what proportion of the total load was being distributed to Column 1.
I misunderstood and gave a wrong answer.
Sorry for adding to the confusion!
dozer, you are correct!
RE: Floor load on columns
We do a lot of dumb crap these days just because we can, because we have the computer power and software to do it. That may be FEA’ing or CAD’ing the hell out of something for days just becuase we can and have the time to fill. Then, we have the high self opinion to think we have produced something profound, and exact to all ten digits of the computer printed output. The crazy thing is that the structure is smarter than we are. It will act exactly the way we designed it to act, with all the flaws, and good features, we may have incorporated into our design and details. All this wether we know and understand it or not. At the same time, the structure is too dumb to know that we analyzed the hell out of it, the wrong way, with false assumptions, and now it had better act that way, or else.
The trick to good Structural Engineering is to know this, to understand basically how the structure really acts, based on how you have designed and detailed it, and then to select an analysis approach, with appropriate and reasonably conservative assumptions, as needed, to arrive at a practical, economical, buildable design. One of the first question, to inform your approach to this problem will be, how is the floor framed? Your sketch suggests 7m floor spans, onto 8m long beams which frame into columns. And, you have found the beam reactions for this condition, which are then the column loads, to be accumulate for that column, as you move down in the structure. ‘Column tributary area’ really doesn’t have much meaning in this framing situation. As you indicated in your later post, if the floor load was truly uniform over the whole floor or if this were a two-way slab system, then tributary area could be more practically applied, and half the load would go to each column. In these types of calcs., I do carry my calcs. out to more digits, not because I think I know the loads or span lengths that accurately, but rather to self check results or avoid the confusion of why the two answers aren’t the same result; the same applies to geometry and trig. calcs., for self checking; then I finally round up my answers for reporting. You should also pay attention that some loading change might give you a larger result. This is usually obvious with simple spans. But, with two-way spanning systems and with continuos beam system, skip loading, or shifting loads can draw more reaction (col. load) to one reaction location or another.
It seems to me that your boss’ question or direction was poorly phrased, maybe ill informed, and you and he/she should sit down and go over the above posts, all good and informative, so you both have a better understanding of the facts of the matter.
RE: Floor load on columns
That is interesting thought and I keep it in mind. As to computer software... I just started working in a design office and I have to say that using computer software does not upgrade skills or teach much... I might be wrong though.
Anyway thank you all and sorry for causing confusion, but as some say... There isn't such a thing as a stupid question.
RE: Floor load on columns