Calculate Size of Column for 10 Storey Building

[Nezero]

Homework Statement

A 10 storey building has columns spaced on a 8m by 10m grid.
a) Calculate the reaction at the base of an internal column assuming a 300mm thick concrete floor slab and a live load of 3.0kPA (ignore self weight of column and assume concrete density 2.4t/m^3

b) Assuming an allowable axial stress of 20MPa in the reinforced concrete column estimate the size of column assuming a square cross section

Homework Equations

P=F/A
Volume * density = mass

The Attempt at a Solution

Well, I'm a bit confused by the wording of the question firstly.
So there is an 10 storey building with a concrete floor slab which is 8m by 10m with 300mm thickness right. So there is one column supporting this floor slab.
Firstly I calculated the mass of the slab which is 8 * 10 * 0.3 * 2.4 = 57.6tons
So the weight of that is 57600 * 9.8 = 564.48kN
So the column must support that + the live load.
Live load = 3000 Pa.
P= F/A
F=PA
F= 3000 * 8 * 10 / 1000 = 240kN
So total reaction force = 564.48kN+ 240kN = 804.48kN
Now there are no answers so I can't check this. Is this right?
b)
So is P = 20MPa
P = F/A
F= 804.48kN
A= 804.48kN/20MPa = .040224m^2
Sqrt(A) = 0.2m
So 0.2 by .02 m?

 

[PhanthomJay]

Oh being from the States where the metric system is not used in structural design, I don't have a good feel for your numbers. Nonetheless, your calcs seem OK except for one important step you missed..the number of stories supported by the interior column. The column base must support all 10 floors, so your loads are off by a factor of 10, and that will increase the size of your column significantly.

 

[Nezero]

So does that mean that the weight of the concrete slab must be multiplied by 10 for each level?

 

[PhanthomJay]

So does that mean that the weight of the concrete slab must be multiplied by 10 for each level?

I believe the problem implies that there is a 300mm floor slab and at each level and a 3kPa live load at each level. The base of the column supports all 10 levels, wheras the upper part of the column just supports one level, and the middle part supports 5, etc;, so in theory, you could use columns of different sizes from top to bottom, the lowest being the largest, but economy might dicate just using one or 2 sizes; but in any case, you're asked to design based on the highest load, which occurs at the base.

 

[Nezero]

Whoops I just realized, I meant do I multiply by 10 because there are 10 levels. If there were 20 then it would be by 20.
So there for I would also multiply 3kPa by 10 as well.

 

[PhanthomJay]

Whoops I just realized, I meant do I multiply by 10 because there are 10 levels. If there were 20 then it would be by 20.
So there for I would also multiply 3kPa by 10 as well.

Yes, botom line here is that your total vertical compressive load at the column base is 8045kN, the column area required is thus 0.4m^2, and the column dimensions are roughly 0.6m by 0.6m.