#### Web under local loads – Hand calculations

It’s very hard to calculate the capacity of small sections of any structure under concentrated loads. Luckily EN 1993-1-5 gives us a decent solution for webs under concentrated loads!

11 January 202120 minutes read

We all know (or so I hope!) the basics of buckling. You know, the classical buckling lengths in a non-sway structure, etc. However, life is “rich”, and oftentimes we have to solve more complex problems than those classically described in the textbooks. One of the great examples is this problem: How to calculate the buckling length of a column in which cross-section changes along with the height.

**A varying cross-section is a nightmare in buckling calculations. Mostly, since literally by definition buckling length can be calculated only for elements with constant cross-sections. This means, that the designer has to either make some better or worse assumptions… or use other methods of establishing buckling capacity.**

When I started teaching at university, this was one of the topics I was pretty scared about… luckily, I understand it more now, than back then! The complex column buckling is precisely, what I want to discuss today! So, let’s dive in!

I guess, that it’s only reasonable to start from the beginning! There is a lot we should discuss, and I’m first to admit, that I will have to discuss some “overall” things as well to get through this example. But the example is definitely a key part of our today’s journey.

So, the question we are asking is: **What is the buckling length of this column:**

I’m first to admit, that this may not be the most realistic of cases (especially with the horizontal support at the top), but I wanted to do something simple that we can follow with relative ease. The cool thing is, that the approach I will describe will work, regardless of the situation, so you can adapt it, as you need!

Also, the question we are asking isn’t “random”. If you would design such a column, your FEA software (with some code design rules implemented) would ask you to provide a buckling length for each element you wish to design… so without a doubt this is a useful topic, although the question itself isn’t “great”…

I admit, that I used a certain shortcut in the title… if your column has a changing cross-section, it won’t have a buckling length! At least not in a classical sense of the term!

This doesn’t mean, of course, that such a thing is undesignable. It’s completely doable, but you simply need to know how to do that! The best start is simple. You need to be super clear on what you need to design stuff due to stability and why. And this is where we will start!

You don’t need the buckling length!… yup, you just don’t! All you need is to know a slenderness of your element!

Currently, the most common way of designing stuff is divided in two steps. Firstly, you calculate the plastic capacity of the element (you know, when stuff breaks… literally). Then you apply a reduction factor for that capacity… due to stability (buckling etc.). You calculate this reduction factor based on the slendreness of your element – and this is what you need!

So why all the fuss with the buckling length?Well… using buckling length is one of the simplest ways to calculate critical force, thanks to the Eulers equation. When you have that, it’s easy to calculate the slenderness (the thing that you really need!). This is why your design software asks you to fill in the buckling lenght!

In other words, we want to know the slenderness of our column to design it. And the software we are using will want us to “fill in” the buckling length. I think this is a good start to this rather complex topic. Let’s take a look at this in more detail, just so you know how it works!

I do believe, that in order to answer the buckling length question, a wider context is needed. This is why I will discuss how the design is made. If you are working in the field for a few years, you may not even recognize some of the equations… since the software does that for you “in the background”. But I do believe that this is actually an important thing, to understand what is going on there!

This is why I will discuss the typical approach to the design of a compressed element. If you are interested in the final outcomes, I guess you may skip this part… but I would still try to convince you that it’s good to read this!

The typical approach (based on EN 1993-1-1) to designing a “simple” compressed column would be something like that:

**Calculate the plastic capacity:**You know, how much load you can apply before the thing “fails” because you simply squashed it. This is brutally simple, as you can see below. Such a “type” of capacity is usually referred to as “cross-section” capacity. Simply put, it’s the amount of load you would have to apply to a “slice” of the column to squash it… so obviously buckling is not the part of the problem (the slice of the cross-section in a hydraulic press won’t buckle!). Well… if our cross-section would be in class 4 there would be some buckling involved even here… but let’s say we are using “normal” hot-rolled cross-section, so class 4 is not a problem!

**The above is not enough!**It would be super cool if we could simply end here, but we need to take stability into account. Simply put we won’t even calculate the buckling capacity (!), we will just reduce the cross-section capacity above by a “buckling coefficient” as you can see below! Notice, that the safety factor changed from “type” M0 to M1… this only means we are dealing with stability problem, both are actually equal to 1.0!

**Let’s deal with the buckling coefficient!**This coefficient takes buckling into account and reduces the cross-section capacity as you could see above. This is how you will get it:

**It’s not so scary!**Ok, I admit that greek letters in equations may cause concern, but no worries – this is actually pretty simple. Funny enough, the above (and below) can be derived from differential equations of a compressed column deformation directly. I even did it once for training materials, but I never used it (obviously…). You can believe me that this is true… or derive this yourself to be sure – your call! Anyway, the above requires 2 things: Slenderness we will discuss in a second, and a “mysterious” parameter, that simply makes the above equation shorter:

**Don’t worry about imperfections!**Normally in FEA, this is a lot of work, but in the above, it’s extremely simple. Cross-sections are divided into groups, and you simply read the value of this parameter from the table (depending on a group your cross-section is in). Of course, in our case, it’s not so easy, since we have a column with 2 cross-sections – so you can either treat it as the “worse” out of both or simply the “worst” there is… which seems like a good idea 🙂 So finally we get to calculate slenderness!

**That’s all… for now!**As you can see, slenderness depends only on cross-section capacity, and critical force! So basically we know “everything” we need to know to calculate the column… apart of course from the freaking critical force. But it’s a longer topic, that simply requires its own section!

I don’t want to bore you with a lot of theory here. You can read some of it in this post about buckling!

If we omit how this all works, Eulers Critical Force can be reduced to such a sentence:

Critical load of an element is such a load that when you apply it, the element will fail due to elastic buckling.

This is always connected with bifurcation, you can see above. The Eulers critical force of the element is one of the basic engineering information, and the equation allowing you to calculate this force is relatively simple:

Firstly, now you see why you need the buckling length! It is simply needed to calculate the critical load (that is needed to calculate slenderness!). But with the same equation (and the schematic above) comes… why you won’t get the buckling length for our column! And that is because:

**Euler’s Critical Force Assumptions:**

- Member is perfectly straight
- There is a constant cross-section (so a constant moment of inertia along the length)
- Member is “only” axially compressed
- Material of the column in isotropic and homogeneous

You see this already don’t you? In our case, the column as a whole doesn’t have an Euler’s Critical Force… since it doesn’t “fit” into the assumptions!

Simply put, only the perfectly straight columns with constant cross-section have Eulers Critical Force… basically by definition! So if buckling length is used to calculate the Eulers Critical Buckling Force, and our column doesn’t have Eulers Critical Force… that it doesn’t have the buckling length either! Cool huh?

The fact that our column doesn’t have the Euler’s Critical Force… doesn’t mean it won’t buckle! It just means that it doesn’t “fit” into the solution Euler proposed.

There is still a Force, that when applied will cause an elastic buckling to our column… but we cannot use Euler’s equation to calculate this force in our case! Such a force will be Critical Force (by this I understand that it will cause elastic buckling), but will not be Eulers Critical Force (understood as a force that will cause elastic buckling, that can be calculated according to the Eulers formula)!

Interestingly enough, code EN 1993-1-1 doesn’t say that you should use “Eulers Critical Force” to calculate slenderness – it just says that you have to use the “Critical Force”. However, there are assumptions describing that the entire procedure works for straight members with constant cross-section, etc. But the reality is, that you need Critical Force to calculate slenderness… and that is that.

If we can’t use the Eulers equation… what else can we do? We can to FEA of course! And to be more precise LBA (Linear Bifurcation Analysis). Thanks to this awesome tool, we will be able to obtain the Critical Force for any case we may have! Let’s give it a shot!

Every time I start something new (or I explain something) I like to show that the tool I’m using actually works in a way I can verify with other means. You see, way too often I saw folks doing crazy complex analysis… that had no “verification” to them. It’s better to start simple!

In our case, the “simple” is an 8m long HEB 360 column. Why that? Of course, because such a problem fits perfectly into Euler’s equation. So we can use both the equation and LBA to compute the Critical force. The assumption is of course, that we will get the same answer from both!

The task is pretty simple if you want to do the hand calculations. While the column is easier to buckle on its weak axis, I will simply assume that it is sufficiently braced not to buckle in that direction. This changes nothing in the procedure itself – I just think that the “strong axis” buckling looks better (!).

The same can be achieved with LBA.

Since this is a “beam structure” I will use RFEM that is my default program for this kind of problem.

I simply modeled an 8m HEB 360 column and loaded it with 1kN od compression at the top. Since I used a buckling length coef. of 1.0, this translates to “simply supported” in a sense. There is a pinned support at the bottom (that also blocks rotation around the column axis to make it stable). On the top, only the horizontal translations are blocked.

The outcome from the analysis looks like this (linear static on the left, and LBA on the right):

But what is more important (in the case of LBA) is, that the critical load multiplier I’ve received was 13986. This is the value for the 2nd buckling form (since the first was for the “weak axis”, and I’ve already decided to ignore that). Since I applied 1kN of load, this means that the critical force is 1kN x 13986 = 13986kN… close enough!

So, as you can see, there is a possibility to calculate critical force with LBA. As I just demonstrated, it “fits” with the Eulers equation, so the method “works”. But there is also an additional cool thing! LBA works for any model you define… even if it doesn’t fit with Euler’s assumptions!

This leads us to our actual case!

First off, I will admit that I will dumb it down a bit. I don’t want to model the connection in detail, and I will even omit the eccentricity. This isn’t a perfect solution, so if you like you may play around and model it more accurately. But I do admit that I feel this will be “good enough” for many engineers working on such problems. I’m a modeling freak, and I love playing around with models and make them “perfect”. But I fully understand that often, there is simply not enough time to deal with details. Especially when estimates are needed NOW!

As you can see, I did a bit more here. On the left is the “pure” 8m HEB360 case, and then I’ve added HEA160 1m at a time, ending with “pure” 8m of HEA160 on the right. Above, you can see the critical multiplier in each case (that is basically a critical force in kN, since the load was always 1kN).

As you can see, obtaining the critical force for our case isn’t anything “impossible”, even though we could not use Eulers Formula.

This is what we set out to do here, but we are not done yet!

Since I had a simple model, I figured I could do a small case study. Instead of just showing you how to calculate a critical force, I actually analyzed different columns, as you could see above. Let’s take a look at how the critical force changes, as we add more and more HEA160 from the top to our 8m HEA360 column:

If this doesn’t look fascinating to you… then there is something seriously wrong with me!

I added a few points (that are not shown above), just to make the curve look smoother. And I do admit, that I find it fascinating! It looks like a very short piece of HEA160 (up to something like 500mm) doesn’t change a lot. Simply put, it’s too short to cause an impact, and since the support is “pinned” at the top, not much is “happening there” anyway.

But then, after around 750mm of HEA160 from the top, the “weaker piece” starts to wreak havoc! You can even see on the screens above, that the maximal deformations are where the “weaker” cross-section is… even if it is only 25% of total length! And of course, it’s completely downhill from there! After we reach 4m (half) of the column height with HEA 160… it’s basically almost as if the whole column was from a weaker cross-section! Sure, there is an impact (since the “half” HEB160/HEB360 column is twice as strong as a “full” HEB 160 column)… but the reality is we lost 10x of the capacity by moving with a weaker cross-section till half of the column!

I’m really glad that I did the above chart, as I never tested how things “change” if you do those sorts of things… I guess that doing such studies from time to time can teach me something (and I hope you as well!). It looks to me, that if the “top”, the weaker part can be “short”… it’s good to make it as short as possible. The return on the effort there is the best!

You may be surprised that this is the next step. After all… we need a buckling length to calculate the Critical Force… not the other way around!

That is a very good observation, but it doesn’t solve one big problem! And that is the **software!**

Sure, if you would do the design by hand, you could simply take the critical force, and be happy about it. But who does designs by hand? I like using RFEM over Femap in beam models because RFEM has the entire Eurocode design procedure already implemented! So instead of seeing cross-section forces etc. I can display the percentage capacity ratio according to EN 1993-1-1 as an outcome plot on my beams/columns*.

* If you read this, and you wonder why I mention such an obvious thing, this mostly means that you studied civil engineering in one form or another. I did training in a few companies with a mechanical background. They were super thrilled about such a fact!

But if I want to get my “capacity ratio”, the code procedure must be done! And I have to “tell” the software what is the critical force for each element. WAIT! That is not the case actually. The software you will use will assume, that you are not interested in calculating the critical force yourself! It will just ask you about a buckling length, and then calculate the “rest” for you!

But since you have the critical force, and you wish to input data to your software… you need to “change it” into the buckling length first!

A brilliant observation… and a true one too boot! And this is why I’m writing about it here!

It’s true, that our column has a critical force (although not an Eulers Critical Force, since it doesn’t fit into the assumptions as we discussed). But, the software doesn’t “care” about such nuances! It’s best to verify this in a manual of your program, but I’m pretty sure that your soft want’s to calculate the buckling length of each beam, using Euler’s equation. After all, this is why it asks you about the buckling length in the first place!

And if that is the case, stuff is pretty simple. All we have to do is to “remodel” the Eulers equation to get:

Awesome, we know the critical force thanks to the LBA! Now we can put all the numbers in, and calculate what we need!

Sure, the buckling length won’t have any “physical” sense, since our column doesn’t fit the Eulers Column assumptions. But that is a small problem! The software will take our “stupid” buckling length, and put it into the Eulers equation that doesn’t fit our case anyway. But the outcome it will get (the critical force!) will be correct in our case! And this is what counts!

Of course, before you start using this method, please do a single case study and verify if your program works this way! Better to be safe than sorry!

Did you notice this already? This nasty “teachers trick”? Think about it for a second… something is still missing!

Yeah, I purposefully omitted one single detail (such a common theme in free materials, isn’t it 😛 ). No worries, I’m a serious dude, so here it comes!

In the equation for the buckling length, you need to use Moment of Inertia. This depends on the cross-section… so should you use the small one from the top HEB 160, or the big one from HEB 360? Such a great question isn’t it?

**Of course… you need to use both** (and I beg you, don’t do an average!).

In your FEA model, you definitely have both parts of the column modeled separately (the only way to change a cross-section this way, right?). This means, that the algorithm, that will check for capacity will check each of those elements separately. This is actually a good thing! Think about it this way:

Each part of the column has the same critical force, but is verified independently!By this I mean: the column will buckle as a whole! There is no way only one part will buckle… the second part will take part as well! But… they have a different cross-section and all, so both pieces have to be analyzed separately!

How to deal with this? It’s actually pretty simple!

Let’s think about the top part for a second!

If we would make an 8m high column only from HEB160… it would buckle easier than the above, right? Simply put, HEB 360 (the lower part) is more rigid, and it “helps” the top part to stay stable. This means, that if we would do a column, that is ONLY made of HEB 160 that has the same buckling capacity (critical force) it would be shorter than the one above!

Strictly speaking… it would be 5.15m! I know this because I did:

Of course, you can do the same for the bottom part. This time, however, it will be “worse” than 8m! Simply put, if the entire column was HEB 360, the buckling length would be 8m. But since the top part is “weaker” it actually makes the whole thing less stable. So the column made purely from HEB 360 will have a buckling length longer than those 8m to have the buckling capacity the same.

Precisely speaking, the buckling length of the “fictional” HEB 360 column would be: 21.43m! Again, the calculations are simple:

This is the final stop of this post! Now you know how to calculate the buckling lengths you will need for design.

I know this may seem counterintuitive at first (and it’s definitely a bit of work!), but this is actually a pretty complex case!

To wrap it up, this is what we did in this example step by step:

**Model the column and do the LBA!**This way, we can have a critical load multiplier, that will give us a critical force for our column.**Shared critical force!**This is the fun part. All of the parts of the column will buckle “at once”. Regardless of the cross-sections all of the column components will always have a shared critical force. This is because the column fails as a team, so the weakest link decides. Simply put, you can’t buckle a “part” of the column. The column will always fail due to buckling as a “whole” from support to support (or from bracing node to bracing node if you wish).**Calculate the buckling length of each piece of the column!**When we realize that all of the column components will buckle “at once” thing becomes so much easier. For both parts, we can back-calculate the length a “hypothetical” column of a constant cross-section would have to have the same buckling capacity as our analyzed column. Thanks to this, we can design our column as a “collection” of “hypothetical” columns using the code algorithm implemented in our FEA software!

The sweet taste of victory!

I really hope that you liked this example. If that is the case, definitely let me know in the comments below!

And if you want to learn other useful things about FEA and design, definitely check my free online course! You can signup below!

##### Categories:

- Structural Design

10 Lessons I’ve Learned in 10 Years!

10 Lessons I’ve learned in 10 Years!

## Share

## Join the discussion