Floor load on columns

[Bubik]

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 !

 

[JAE]

Is this for a project or homework?

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[Bubik]

Thanks for your reply. It is for work.

 

[JAE]

Well the beam end reactions of simple span beams (with no moment connections at the ends) is really basic statics. If the beam ends have moment connections then the analysis is more involved and depends on the column lengths and other framing and would require either an extensive hand calculations (moment distribution analysis for example) or a software analysis.



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[Leftwow]

What job number should I charge to?

 

[Bubik]

Lets say the end connections are pinned ( the problem is not to analyse as I have all the tools needed, and I know how to analyse continuous beams anyway). I just wanted to know the area of load taken by each of the mentioned columns so I can draw it on CAD.

 

[Bubik]

I wanted to know how to calculated load taken by each column from that beam as the load very along the beam (take it from bending moment, resulting force on a beam or something like that).

 

[Bubik]

Thanks very much everyone!!!

 

[dozer]

I'm lost. In your OP you ask how much load each column gets (which you answer yourself also in the OP). Later you ask what is the area of the load taken by each column. It appears that the distributed load on the middle beam is a result of the given floor load, not in conjunction with the floor load. It further appears that one-way loading is assumed. Thus the tributary width of the middle beam is 7 meters. 1.4 kn/m^2 x 7 m = 9.8 kn/m, close enough to 10. 11.4 kn/m^2 x 7 m = 79.8 kn/m, pretty dang close to 80. Like JAE said, it's just simple statics from here. The reactions you got look correct to me. I don't understand what the purpose of the area of the load taken by each column is. Can you explain?

 

[Bubik]

Sorry for confusion.



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.

 

[BSVBD]

Peter,

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?

 

[dozer]

I think BSVBD's calc shows how useless this floor area thing is. I wouldn't have done it that way. I would have said what length of the 80 kN/m distributed load would it take to cause a 281 kN reaction at column 1. The answer is 281/80 = 3.512 m. Since your tributary width is 7 m the tributary area is 7 x 3.512 = 24.5875 m^2. Or done another way, reaction divided by floor load, 281/11.4 = 24.65 m^2. It doesn't come out the exactly the same because the 80 kN/m was a rounded number. Column 2 gets the balance. But what is this for?! As we just saw from two different responses, it's just an opportunity to introduce confusion. You say the reason you want to find these areas is because you were given that task. To me, that's not a very good reason, even if its your boss giving you this task. I'm not convinced the person who is asking you this knows what he or she wants.

 

[Bubik]

Thank you all for your answers.

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)

 

[Bubik]

Bu the purpose of it might be useless, but the two answers are completely different

 

[dozer]

So which answer are you going to go with?

 

[Bubik]

Good question.

 

[BSVBD]

Please forgive me!

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!

 

[dhengr]

PeterOdron:
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.

 

[Bubik]

Dhengr,
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.

 

[tempeng]

OP: Did you do your own calculations to obtain the reactions at the columns or is this a training exercise? I only ask this because out there, somewhere in your office is the loading areas used to generate the line loads acting on the central beam.