Rational Buckling Analyses to AS4100 or NZS3404 (Part 2)

In part 1 of this series, we briefly explored the requirements related to calculating the capacity of a column via the use of a buckling analysis. Introducing the general methodology to follow and showing agreement with the normal hand methods for a known effective length factor.

In this follow-up post we look at the power and beauty of this method in being able to assess any complicated design scenario, typically ones that don’t fit in the mould of the typical idealised restraints.

The example scenario

Consider we are working on a project where we have a column that is spanning several storeys in height to support some roof above. But the good old architect wants something as slender as possible, because it reminds him of a blade of grass in a meadow flapping in the autumn wind or some such nonsense (basically they were probably on drugs when they come up with the architectural vision for the structure, we’ve all been there (not the drugs, but being put in this position)….).

The column is situated at the end of an internal wing wall, and the wall supports some glass cladding or similar, so we are able to at least tie it back in one direction to an adjacent column. So, it’s all very ‘architectural’ to put a favourable description to it, architect thinks they’ve basically building the most iconic thing in the world as always. They are real keen to paint it a nice green colour, but that’s not so important to you. But you make a mental note to push them silently in the direction of a nice shade of brown instead as it reminds me of something else tied into my thoughts on this arrangement.

The column is tied into the roof and is essentially a gravity only column and they want to make it as slender as possible in one direction, there’s an adjacent column that’s hidden in a solid bit of wall that doesn’t carry any significant vertical load. So you have a bit more freedom to make it bit larger if it helps at all to support the slender column via a strut located at 3/4 point above the base.

Of course, with your engineering hat on you’d love to throw some bracing up to the roof to support the column. That type of logic is apparently against some architectural law nobody informed you of…. Because it will interrupt the internal visual flow and/or some other architectural bollocks. The architect has taken it as a personal insult that you even asked to put some bracing there.

You as the poor structural engineer need to make their fantasy come true, its what we do, performing your basic entry level miracles. Basically, engineering is like that some days, just coming up with something to hold up all the architectural rubbish without killing anyone in the process. Least that’s what the good ones are tending to strive for.

It looks something like this all said and done: –

Sorry, what am I thinking, like this: –

missed the roof label

A column supported by another column supported via a strut, the right column is tied into the wall and has support about the minor axis at mid-height. In this case will the two columns simply buckle together, or does the stiffer left column manage to provide some support to the right column buckling? All very good questions that make it hard to guess what the true effective length factor (k_e) might be!

If I really had to guess, I’d say that the right column is obviously going to be governed by the bottom 6750mm part of the span buckling about the minor axis, but given the shorter end span at the top, then surely they will give some additional support to the middle part of the span. Maybe the strut will buckle, maybe the left and right columns simply buckle together? STOP GUESSING DAMMIT!!!!!!

I’ve initially selected a member size for the right column that might work for the major axis direction with an effective length of the entire 9000mm. A quick back of the envelope check shows it would not work however if you took k_e=1.0 for the bottom 6750mm segment (see below). Simply because that might be the simplifying assumption you end up making if you were designing it by hand. The whole purpose of the buckling analysis is that we can (hopefully) do better than that simplified capacity calculation and demonstrate it quite easily!

So, in our model we’ve applied the actual loads we have, and we’ll factor the buckling loads from there.

Basically, I have no idea how this is going to work out as I’m kind of winging it and making it up as I go, but this analysis will demonstrate something: –

  • Hopefully that is that it’s bad to guess
  • …and that using a buckling analysis was a good way to work out the true axial capacity accounting for all the beneficial things on offer

Once we know the capacity, we should be able to back calculate out the effective length of the critical member and compare it to how we thought it was going to fail and theorise why it was a whole lot better (or worse) than we thought.

Let’s just randomly say the loading on the column of interest is 460kN and it’s a 200x100x9SHS because I have it on good authority, they actually look quite good in brown and we can more or less support the load required about the major axis eyeballing some capacity tables.

The load on the left column is essentially zero, and its initial size has been selected as a 150UC37. The strut is a 100x50x3RHS. All connections are envisaged to be nominally pinned.

Checking the minor axis capacity of the 200x100x9SHS using an assumption of L_e=1.0\times6750=6750mm:-

N_s = A_g f_y = 4800 \times 350 = 1680kN

\lambda_{ny}= \displaystyle{ \frac{L_e}{r_y}\sqrt{k_f}\sqrt{\frac{f_y}{250}}

\lambda_{ny}=  \displaystyle{ \frac{1.0\times6750}{39.9} \sqrt{1.0}\sqrt{\frac{350}{250}}



\phi N_{cy}=\phi\alpha_{cy}N_s

\phi N_{cy}=0.9\times0.184\times1680 = 278.2kN

So, the 460kN load we’ve selected is 1.65 times the capacity if we assumed k_e=1.0 for the longest span of the RHS. Hopefully were not being too ambitious here…. maybe I should have selected a smaller load, but no turning back now.

Our structural model below has been created, showing our interpretation of the above scenario (that right column looks stunning in brown by the way, we must send to the architect ASAP for consideration): –

Pink and brown, the architect better appreciate this effort, had to screw up my default colour settings to get that lovely shade of brown

So, hitting go we get the following governing buckled mode shapes, suggesting the columns are sort of doing their best to do some buckling together about the minor axis.

1st buckling mode about minor axis of RHS
2nd buckling mode about major axis of RHS

SpaceGass reports the following buckling load factors for the 1st and 2nd modes of buckling: –

 Load              Load                         Node at   Node at
 Case  Mode      Factor  Tolerance Iterations Max Trans  Max Rotn
    1     1       0.988   0.007812          8    28 (X)     1 (Z)
          2       1.244   0.007751          8     9 (Z)     2 (X)

So, it doesn’t bode well for the capacity that the load factor here is less than 1.0, however let’s do our capacity calculation just to be sure, following the same process as before: –


N_s = A_g f_y = 4800 \times 350 = 1680kN

N_{omx}= 460kN\times1.244 = 572.2kN

Therefore, the modified member slenderness for buckling about the major axis in accordance with CL 6.3.4 is: –


\lambda_{nx}= \displaystyle{90\sqrt{\frac{1680}{572.2}}}

\lambda_{nx}= 154.2

Therefore, we can now determine \alpha_{cx} using CL 6.3.3 noting that \alpha_b=-0.5 (note intermediate working not shown for brevity): –

\alpha_{cx} = 0.301

Therefore, the design member capacity for buckling about the major axis direction is: –

\phi N_{cx}=\phi\alpha_{cx}N_s

\phi N_{cx}= 0.9\times0.301\times1680

\phi N_{cx}= 454.4kN

Design capacity ratio: –


Therefore within 5%, say ok.


N_s = A_g f_y = 4800 \times 350 = 1680kN

N_{omy}= 460kN\times0.988 = 454.5kN

Therefore, the modified member slenderness for buckling about the minor axis in accordance with CL 6.3.4 is: –


\lambda_{ny}= \displaystyle{90\sqrt{\frac{1680}{454.5}}}

\lambda_{ny}= 173.0

Therefore, we can now determine \alpha_{cy} using CL 6.3.3 noting that \alpha_b=-0.5 (note intermediate working not shown for brevity): –

\alpha_{cy} = 0.243

Therefore, the design member capacity for buckling about the minor axis direction is: –

\phi N_c=\phi\alpha_{cy}N_s

\phi N_{cy}= 0.9\times0.243\times1680

\phi N_{cy}= 367.4kN

Design capacity ratio: –


Therefore, it’s no good, but the result clearly demonstrates that k_e  \leqslant  1.0.

So basically, the member(s) size is insufficient as it doesn’t work about the minor axis but let’s not forget that we got a significant increase in capacity by just strutting over to the adjacent column. I will add a note of caution here, that if the UC column were carrying some axial load that it would reduce the buckling capacity of the RHS.

Working backwards from Eulers equation N_{om}=\displaystyle{\frac{\pi^2EI}{(k_eL)^2}} to work out our effective k_e factor: –




Therefore, we get a nice little benefit from having the column adjacent propping the RHS so they have to buckle to some degree in unison. We no doubt also gain some benefit from the shorter top column span also acting to reduce the effective length of the bottom section.

So, going for gold in terms of maxing out the capacity we can achieve, let’s see what occurs if we beef up the Universal Column to the max. Changing it to the maximum available hot rolled size in these parts of the world, the mighty 310UC158. This lean-on column should in theory ‘prop’ the more slender RHS a bit more about the minor axis: –

1st buckling mode, buckling about the minor axis again..

Now a few things of note regarding the modes shape above: –

  • The first thing you might pick up on here is that the minor axis buckling mode is still the 1st buckling mode. This tells us the capacity about the minor axis is still probably lower than the capacity about the major axis straight off the bat.
  • The second thing you might notice is the mode shape is more “restrained”, with less deflection clearly observed in the mode shape at the strut level. It’s almost as if we put a support at the strut location, we’d get the same answer. Suggesting that we’d reached some theoretical limit if we did this, sounds a bit like increasing the stiffness of the restraining system has reached a limit where the capacity no longer increases…. sounds very familiar to the exercise regarding brace stiffness that was undertaken in this post, **hint** this is a very important concept to understand in stability theory **hint** .. 😉
  • If that hint isn’t enough, I hope you can appreciate that you could replace the left-hand UC column and strut with an equivalent spring and achieve the same thing. We can increase the spring stiffness until we get a higher buckling mode as outlined in the post linked to above.

But let’s follow it up with a calculation and see by how much the capacity may have increased: –

 Load              Load                         Node at   Node at
 Case  Mode      Factor  Tolerance Iterations Max Trans  Max Rotn
    1     1       1.215   0.007812          9    27 (X)     1 (Z)
          2       1.244   0.009521          8     9 (Z)     2 (X)


N_s = A_g f_y = 4800 \times 350 = 1680kN

N_{omy}= 460kN\times1.215 = 558.9kN


\lambda_{ny}= \displaystyle{90\sqrt{\frac{1680}{558.9}}}

\lambda_{ny}= 156.0

\alpha_{cy} = 0.294

\phi N_{cy}=\phi\alpha_{cy}N_s

\phi N_{cy}= 0.9\times0.294\times1680

\phi N_{cy}= 444.5kN

Design capacity ratio: –


Therefore within 5%, say ok.

If we back calculate the effective length factor again from this final analysis: –



Now if we think back to where we started where we might have typically assumed k_e=1.0 for the 6750mm span in the RHS column, we calculated a capacity of \phi N_{cy}=278.2kN.

After doing a buckling analysis to derive the maximum capacity we can achieve after increasing the UC size, we bumped this up to a capacity of \phi N_{cy}=444.5kN.

We only increased the weight of the UC column by a factor of 4.27 in getting this improvement, but maybe that’s the price of ..ahem.. good architecture…


So basically, we barely have a solution in terms of axial capacity vs demand, being a few percent over in each direction. Probably should have chosen a lower axial load…. fail.

In reality depending on how the axial load was delivered, there would also be some bending moment to consider in the RHS column. So overall it’s a failure if you were to go on and check combined actions.

I was hoping what we’ve done would show some important lessons in stability analysis: –

We showed the importance of stiffness, you reach a point where the restraining system isn’t going to increase the capacity any further. But even adding a small amount of additional stiffness, can really get some good gains in terms of reducing the effective length for compression buckling.

You could go back and check additional column sizes between the 150UC37 & 310UC158 and show that working down from the max size used here other larger sizes would still have the same capacity for the RHS buckling about the minor axis.

We also showed the use of a buckling analysis can account for the beneficial effects of other members and shorter adjacent spans offering considerable reductions in effective lengths for compression buckling. At least over and above any ‘conservative’ guess. No need to ever guess-timate again!

As noted in the last post, even though we are using AS4100/NZS3404 here, most standards allow you to use a buckling analysis. So, go have a read and stop with the guess-timation.

Oh, and lastly, brown columns please architects, at least give us that….. Part 3 here


  1. Hi,
    If my software requires that I divide the bars in several pieces, I should follow the same process, however my Ke would be higher? (since my L would be 1m if a 9m column divided in 9 pieces of 1m).
    When I increased the UC size, my RHS first mode of buckling was in the major axis. Not sure if I\’m doing this right. Do you have any knowledge in Robot Autodesk?
    Thanks..loving your blog.

    • Hi gabriel,
      Usually you will need to divide the members to some degree, but this is much more important for flexural torsional buckling analyses. But it could also be dependant on the analysis programme you use. Simply experiment and see if it makes a difference for axial buckling to the results. But yes, you would follow the same procedure whether you divided a member or not.

      Approach it from the point of view of thinking about the load factor, not Ke. As it is far more logical as Ke is a factor on the segment size in your model which is sometimes non-sensical to think about it in these terms as two members with different sized divided member as Ke factors are not directly comparable, but load factors are.

      For example a 1m segment as part of a 9m member could have a Ke of 9.0, and a single 9m segment could have a Ke of 1.0. But the buckling behaviour is obviously the same over the full member length between supports (say if simply supported). Back calculate the actual Ke based on the entire member length if you must know Ke for the member between supports. However, the load factor should be the same irrespective of the number of segments if that makes sense?

      Yes RHS major axis might become the critical buckling mode, if the UC is stiffer it is effectively making the RHS more stable about the minor axis (reducing the effective buckling length) and it is entirely possible that the major axis then becomes critical. I cannot remember but I may have excluded the effect of this to report only the effects on the minor axis buckling behaviour as that was the point I was trying to get across regarding the effects of the stiffness of the supporting UC member.

      I have access to Robot at work, but have not had much experience in using it. Glad you are enjoying the blog!

  2. Gotcha.
    I’ve also realised that if the critical load is dependant on the load case, I’ll have to input different K values in my steel design for different load cases. Or figure out if the software does that automatically.
    Thanks for the reply.. appreciate your time and effort.

    • Hi Gabriel, the critical load is not dependant on the actual load carried (not load case based).

      It is a function of the restraining system to a particular member, independent of the load applied.

  3. Great 2-part tutorial.

    What happens when your supporting member is also loaded? Presumably, you run an analysis with the additional loads modelled – however software often annoyingly likes to factor all loads whilst running a buckling analysis which will result in the wrong buckling factor and modes being provided.

    • Hi Dan. If you mean that both columns had axial load, then yes you need to model both axial loads. One concept in buckling is that the sum of both the axial loads contribute to the buckling of the system. In my example I only had axial load on one column, but if I was to split the load 50%/50%, interestingly the buckling load of the system (the sum of both loads) would be similar.
      Check out the sum of P concept in this video

      • Dan, if you are getting a lower buckling load factor and some other mode being the most critical. This just means that part of the structure is buckling and failing first. Basically, the lowest buckling capacity is found first via the first mode. So, in a way it helps you determine which member will fail first, and it may be another member you were not expecting provided you’ve modelled the restraints, constraints and supports accurately!

        You can imply if your member of interest is not the first mode that it at least hasa capacity higher than the current load factor.

  4. Do you think it’s reasonable to keep working up through the modes to find the first one that causes buckling of the member currently of interest to determine it’s Nom? Practically, I would probably be lazy and go conservative but for argument’s sake, let’s assume we’re trying to figure out if we can squeeze an extra row of books into the library which is miraculously balanced just on that column.

    • Yes you can do this, however you need to be mindful of the fact that if another member has lesser capacity that provided restraint to the higher mode failure member you won’t achieve that higher mode.

      For this reason I’d usually try separate out the particular member and its bracing system in another analysis just to confirm.

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