ULS capacity of circular RC column
ULS capacity of circular RC column
(OP)
Ok, writing a spreadsheet for checking moment capacity of circular RC column at a given axial load with uniformly distributed longitudinal bars following standard flexible theory. Spreadsheet is to the New Zealand Concrete Standard NZS3101 which generally follows the same theory/methodology as the ACI318 code. For the purposes of the discussion the code doesn't really matter too much.
FYI, the differences between ACI318 & NZS3101 are that there is just a single strength reduction factor of 0.85 vs the different strength reduction factors in ACI depending on the type of failure. The α1 and β1 factors on the rectangular Whitney stress block are also ever so slightly different (probably just a metric/imperial thing), but for the purposes of the exercise can be taken as 0.85 for both.
So the crux of the question is that after writing the spreadsheet and going on and verifying it using SPColumn (Link), I get the exact same answer to a few decimal places with zero axial load for any cross sections, so all good there.
However when I factor in the axial load the answer starts to differ by a small margin (not much only a few percent). Depending on the arrangement SPColumn is generally giving me a few percent higher capacity, so the question is I think I am probably going slightly wrong somewhere, but for the life of me I cannot figure out where!
SPColumn allows a user to manually adjust the strength reduction factors, and the α1 and β1 factors. So I should be getting the exact same answer (on the assumption that the folks who write commercial software get it right 99% of the time (right?). But because SPColumn is written for ACI318 maybe there is something also going on in the background relating to the way ACI318 works out the capacity that I'm not factoring in. Perhaps SPColumn approximates the circular shape as a series of trapezoids, but then I would expect a lower capacity.
General procedure I'm following is outlined as follows:-
1. determine bar center line set out based on cover, column diameter, spiral diameter & longitudinal bar diameter
2. from bar set out determine depth of bar from the extreme compression fiber (i.e. say the top of the section)
3. Assemble a series of equations to determine the stress in relation to the distance from the neutral axis depth 'c' assuming the limiting concrete compression strain of 0.003 is reached at the extreme compression fibre at the ultimate capacity
3a. For bars within the compression block depth 'a' determine the effective steel stress (i.e. actual bar stress minus (α1 x f'c)). This enables the compression force and more importantly the centroid of the compression area to be determined in the concrete ignoring the bars
3b. For bars between the compression block depth and the neutral axis the bars are still in compression but the actual steel stress is determined based on the strain.
3c. For bars below the neutral axis the bars are in tension, the bar strain is determined, if the strain is greater than the steel yield strain, the bars are assumed to be stressed to their yield stress, if not the actual stress is determined (actual stress = E x strain)
3d. Based on knowing the compression depth 'a' work out the area and depth to the centroid of a rectangular stress block in the shape of a circular segment.
4. Determine the forces in the concrete, and each bar
5. Sum the bar forces, concrete force and axial load (axial load assumed to act at the plastic centroid, i.e. center of column for circular column) which must equal zero for internal equilibrium.
6. Solve for the compression block depth 'a' using excels goalseek function based on the result of step 5 equaling zero
7. determine moment capacity by summing the component bar forces/concrete forces/axial load x their respective lever arms from any arbitrary location
8. multiply this moment capacity x the appropriate strength reduction factor.
Anyone see anything wrong in this approach that might account for the differences when the axial load is considered?
Its worth pointing out I've also downloaded a number of other free or demo spreadsheets from the net, all of which have a slight variations on the answer from SPColumn as well!
FYI, the differences between ACI318 & NZS3101 are that there is just a single strength reduction factor of 0.85 vs the different strength reduction factors in ACI depending on the type of failure. The α1 and β1 factors on the rectangular Whitney stress block are also ever so slightly different (probably just a metric/imperial thing), but for the purposes of the exercise can be taken as 0.85 for both.
So the crux of the question is that after writing the spreadsheet and going on and verifying it using SPColumn (Link), I get the exact same answer to a few decimal places with zero axial load for any cross sections, so all good there.
However when I factor in the axial load the answer starts to differ by a small margin (not much only a few percent). Depending on the arrangement SPColumn is generally giving me a few percent higher capacity, so the question is I think I am probably going slightly wrong somewhere, but for the life of me I cannot figure out where!
SPColumn allows a user to manually adjust the strength reduction factors, and the α1 and β1 factors. So I should be getting the exact same answer (on the assumption that the folks who write commercial software get it right 99% of the time (right?). But because SPColumn is written for ACI318 maybe there is something also going on in the background relating to the way ACI318 works out the capacity that I'm not factoring in. Perhaps SPColumn approximates the circular shape as a series of trapezoids, but then I would expect a lower capacity.
General procedure I'm following is outlined as follows:-
1. determine bar center line set out based on cover, column diameter, spiral diameter & longitudinal bar diameter
2. from bar set out determine depth of bar from the extreme compression fiber (i.e. say the top of the section)
3. Assemble a series of equations to determine the stress in relation to the distance from the neutral axis depth 'c' assuming the limiting concrete compression strain of 0.003 is reached at the extreme compression fibre at the ultimate capacity
3a. For bars within the compression block depth 'a' determine the effective steel stress (i.e. actual bar stress minus (α1 x f'c)). This enables the compression force and more importantly the centroid of the compression area to be determined in the concrete ignoring the bars
3b. For bars between the compression block depth and the neutral axis the bars are still in compression but the actual steel stress is determined based on the strain.
3c. For bars below the neutral axis the bars are in tension, the bar strain is determined, if the strain is greater than the steel yield strain, the bars are assumed to be stressed to their yield stress, if not the actual stress is determined (actual stress = E x strain)
3d. Based on knowing the compression depth 'a' work out the area and depth to the centroid of a rectangular stress block in the shape of a circular segment.
4. Determine the forces in the concrete, and each bar
5. Sum the bar forces, concrete force and axial load (axial load assumed to act at the plastic centroid, i.e. center of column for circular column) which must equal zero for internal equilibrium.
6. Solve for the compression block depth 'a' using excels goalseek function based on the result of step 5 equaling zero
7. determine moment capacity by summing the component bar forces/concrete forces/axial load x their respective lever arms from any arbitrary location
8. multiply this moment capacity x the appropriate strength reduction factor.
Anyone see anything wrong in this approach that might account for the differences when the axial load is considered?
Its worth pointing out I've also downloaded a number of other free or demo spreadsheets from the net, all of which have a slight variations on the answer from SPColumn as well!






RE: ULS capacity of circular RC column
For several reasons, I'd be curious to know if having more verts reduces the discrepancy.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: ULS capacity of circular RC column
I'll check out the other suggestion. So far I've been sticking to a couple of practical arrangements that I know the capacity of, but I'll increase the number of bars and see if I get the same deviation.
I'd have to recheck but I think I was also getting a lower capacity than SPColumn at lower axial loads, but slightly higher at higher levels approaching the upper limit. (if that helps to diagnose)
I'm also assuming neutral axis depth = a/beta1. On the assumption that irrespective of the shape the strain is still linear.
RE: ULS capacity of circular RC column
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: ULS capacity of circular RC column
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
You only need the modular ratio if you are calculating the 'cracked' second moment of area, and hence the effective section modulus at the serviceability limit state..... i.e. transform to one material, work out equivalent second moment of area, etc.
RE: ULS capacity of circular RC column
Because you assume that the section is made up of a number of trapezoid 'slices' it doesn't perfectly model the circular segment compression block shape and centroid (depending on how refined you get with defining the section and bar layout though). I have not put a really highly refined section in your spreadsheet yet to check if it agrees one way or the other.
A copy of the relevant isolated calculations are attached for what its worth. Entry cells highlighted grey.
For the arrangement in the spreadsheet:-
In SPColumn for N* = 0kN / ΦMn = 556.71kNm
In my checks for N* = 0kN / ΦMn = 556.60kNm
In SPColumn for N* = 1000kN / ΦMn = 630.70kNm
In my checks for N* = 1000kN / ΦMn = 621.34kNm
In SPColumn for N* = 2500kN / ΦMn = 633.19kNm
In my checks for N* = 2500kN / ΦMn = 639.10kNm
These numbers are a bit closer than some of the arrangements I have been looking at but you get the idea.
Its close enough for my liking and are happy to accept on that basis, but I'm wondering why the difference exists and I like to have the confidence I have the exact answer for something that you are able to calculate an exact answer for based on the assumed theory (which obviously includes some assumptions due to rectangular stress block approach)!
With less bars the difference in numbers is larger.
RE: ULS capacity of circular RC column
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: ULS capacity of circular RC column
The main difference between my spreadsheet and yours (other than you are using the exact circular segment area) is that I treat the axial load to be balanced as being phi.N*, rather than N*. In other words I am treating the axial load as being constant eccentricity and varying magnitude, whereas your calculation treats it as being constant magnitude and varying eccentricity.
The other difference is that my spreadsheet adjusts the stress block depth factor using:
If fcu > 28 Then
G = G - 0.05 * (fcu - 28) / 7
If G < 0.65 Then G = 0.65
End If
The difference in the axial load factor may account for your differences from SP Column, since it makes no difference at zero axial load.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
For N* = 1000/0.85 PhiMu = 630.0 kNm
For N* = 2500/0.85 PhiMu = 633.1 kNm
So that's the same order of agreement you got for the 0 axial load case.
My guess for the differences in the fourth significant figure is that maybe they make some deduction from the force in the bars that are below the base of the stress block but above the neutral axis.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
Especially in cases where the concrete section reduces in width as the concrete stress increases, the rectangular stress block can be very un-conservative. The biggest error occurs in the region central region of the column interaction diagram from decompression to balanced load points but it does affect the whole interaction diagram.
It is amazing how far the position of the decompression point is out compared to using a more accurate concrete stress/strain curve.
Some codes allow for this by adjusting the curve constants, eg Eurocode adjusts them for sections where the section width reduces towards the compression face.
RE: ULS capacity of circular RC column
I totally agree, and if the stress factor is not adjusted down for high strength concretes (which have a near triangular stress distribution, rather than parabolic-rectangular) it is even worse.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
Rapt, i'll check out the Eurocode provisions which might apply to the circular case.
RE: ULS capacity of circular RC column
I am not suggesting that the Eurocode solution gives the correct result either, just that it recognises the problem and tries to make an allowance for it.
The best solution for that problem and the one that IDS mentioned for high strength concrete is to use a good curvilinear stress/strain curve that allows for both effects. The Eurocode curvilinear stress/strain curve for concrete will do this for you. There are probably others that will allow for it.
RE: ULS capacity of circular RC column
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: ULS capacity of circular RC column
Even more so.
How it works out when combined with the standard simplified method of designing for biaxial moments I am not sure though.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
The simplified methods for biaxial have other problems from my calculations. The ones that look at combining the 2 orthogonal directions about the 2 major axes face real problems even with square columns. They were based on steel design logic from what I can tell, where the strength reduction factors are the same no matter what the element orientation.
However, with a concrete column, the strength reduction factors are dependant on section ductility. For ease of discussion, if you look at it in terms of an overall section strength reduction factor as ACI/AS codes do, the factor can vary from .8 to .6 depending on ductility. If you look at the strength of a square column about a major axis, its strength reduction factor may be .8 as it is still ductile. But if you then look at it at 45 degrees rotation, with the peak of the triangle in the highest compression strain region, the section then becomes much less ductile (according to code definitions of ductility) and the reduction factor then reduces and could easily be as low as .6.
However, if you use the method combining the orthogonal capacities it uses .8 in both directions, so the combined case is based on a strength reduction factor of .8!
So an accurate biaxial check at 45 degree rotation could easily show a 25% or more reduction in capacity compared to the combination method.
RE: ULS capacity of circular RC column
It seems they get around this in NZ by using the same reduction factor for tension and compression controlled sections.
But having just come back from looking at loads of photographs of failed columns in Christchurch at the Earthquake Engineering Conference, I am not sure that's such a great idea.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
From 1982 to 1995 there was the introduction of significantly lower levels of confinement in columns in NZ that weren't part of the lateral load resisting system. This lead to a lot of non ductile columns being constructed. It's been corrected since.
I assume by the curvilinear stress/strain relationship you are referring to the first method for stress/strain relationship from eurocode (screenshot attached). The 10% reduction due to decreasing width which is a page or two on in eurocode, only seems to apply to using the rectangular stress block unless I'm reading it wrong?
RE: ULS capacity of circular RC column
RE: ULS capacity of circular RC column
Regarding the Eurocode provisions, yes the reduction in the calculated capacity only applies with the rectangular stress block. The parabolic-rectangular block is a much closer approximation to the actual stress at each section, so no reduction is required.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
Our code is based on ACI318, but this is one area where we have an increased design requirement over ACI318, our code has a strong capacity design focus. Improved confinement generally leads to more ductility, which is sometimes required when columns yield under seismic loads (generally at the base but can be elsewhere in a frame as there is freedom to choose the ductile mechanism if we choose to do so).
RE: ULS capacity of circular RC column
Yes, that curve or the plastic one if you want to go to that effort.
Using those curves you are actually considering a different stress/strain relationship at every point over the compression depth, so where there is a varying width of section, the stress/strain relationship at that depth is applied to that width. So you do not need to try to fudge anything.
Also, those curves change shape for increasing concrete strength and the associated reduction in ductility. You will see from the constants that there is no horizontal part of the curve (the plastic bit) for concrete strengths above about 80MPa, so the curve is nearly triangular!
The .9 factor is a very inaccurate attempt to allow for the fact that the concrete width at the highest compression stress zone is very small while the concrete width at the lowest compression stress zone is much larger. The rectangular stress block was developed assuming they are the same width .
Actually, the best rectangular stress block for rectangular concrete sections is probably the Canadian one. The ACI and AS ones fall down badly in estimating the centroid of the compression force around the Decompression point, so they put the Decompression point in the wrong position on the Interaction curve. But the Canadian one still has the problems with non rectangular sections.
RE: ULS capacity of circular RC column
Do you have an example of that? I have always found that the AS rectangular block matches the EC2 parabolic-rectangular curve pretty well for all concrete grades (using the up-dated version of course!).
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: ULS capacity of circular RC column
The main problem is in the decompression point. If you look at the B1 term controlling the depth, by the time the concrete strength reaches 8000psi (about 55MPa), the B1 factor is .65, so the depth to the centroid of the compression is .325 times the depth of the neutral axis. So it is actually slightly above where the centroid of a triangular compression block would be. But at 55MPa the compression block is still parabolic and the depth of the centroid should be significantly lower, closer to .425 than .325 of kd.
As this is at decompression and there is no tension in any of the reinforcement, this has a very large effect on the moment at the decompression point, especially when you take moments about the plastic centroid. Around 40% difference in the cases I looked at.
There is also some effect at the balanced point, especially for lightly reinforced columns. The direct axial and pure bending points are not as significantly affected.