This is a new idea I have been working on, to show what is cool about design by hand, linear FEA, and nonlinear FEA, and I will demonstrate with several examples. I will solve each of them using all 3 approaches, then I will compare the outcomes, so we can discuss which method is useful for a given case and why! The first problem I will address is a web in a steel beam under local loads. Today we will solve the problem by hand!

Local failures are always a nightmare. 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! Let’s see what we get!

There will be some hand calculations ahead, but no worries, no higher-level mathematics will be required. I will also explain what we are doing step by step!

I also think that it’s actually good to be able to design stuff by hand! Surprisingly, you can adopt many of the sums we will discuss today to solve other problems as well! In other words, I think this may be useful for you. Let’s go!

The problem

Local loads of webs can have many faces. The classical example, that we will solve here, contains two perpendicular beams with one “simply standing” on another. The problem will look like this:

Please note one thing:

We are concerned with both beams of course. After all, the web of the top one is loaded just as the web of the bottom one. But I don’t want to spend too much time on equations! So I will assume that the top beam has a somewhat “stronger” web! This way we will focus only on the bottom beam here. It’s easy to make that assumption while watching the animation above, of course. Note, however, that “in reality” this is a very big assumption, and it shouldn’t be made without a good reason!

First thing’s first!

I fully admit that this may seem intimidating. You will soon see that I jump between chapters and codes while producing relevant formulas seemingly from thin air. While I strongly recommend simply reading all the codes, God knows that I haven’t done that myself. It just so happens, that I was teaching at uni for a decade, and this is one of the topics we were solving! Even after a few years I still simply remember everything (benefits of teaching others I guess!).

Right now, we are in chapter 6 of EN 1993-1-5. We need to start with assessing which of the possibilities fits us best. You can check all of them in fig. 6.1 in that chapter, but I will show them below for convenience:

It is quite obvious that from the selection we have here, we are in a “type (a)” situation. This gives us a chance to calculate the parameter kF. But since we have no idea why we need it, let’s not get ahead of ourselves! Also, note that there is a distance between stiffeners we need to know. The model you saw has some stiffeners, and we will use their span here. Interestingly, if we didn’t have stiffeners at all, it’s not quite clear what to do! Presumably, we could assume a=infinity, but the code doesn’t define that at all! Note that the difference between a few meters and infinity is small as far as the kF value is concerned!

In the procedure we trust!

I must say that this is a rather tangled procedure. But we are brave, so we will get through it anyway!

You can calculate the capacity of the web using equation 6.1:

I’m in love with parameters that are not so easy to “describe” in terms of what they mean. While yield strength has a weird symbol (as far as Eurocodes are concerned) this is explainable at least. You know it’s like “yield stress” fy… “for web” and hence “w” at the end). I’m not sure if there is a praxis of welding different steel grades together to form a cross-section… but if there is, the procedure is clearly ready for that!

The comment on the safety factor is that “M1” already tells us that stability will be involved. This is actually something I like about Eurocode! Usually, the “M0” and “M1” safety factors are the same (not so much in shells though!. I learned to “translate” the “M0” safety factor as “normal stress capacity” while “M1” is “capacity including stability”. This is a small nuance perhaps, but it can help you out a bit in understanding what you are reading if you notice such things!

Of course, the load on the top compresses our web. So technically we should expect capacity to be something like:

The above is super straightforward! We have a cross-section area times yield stress (so how much load our cross-section can safely transfer), then we further reduce this by a safety factor. Of course, if the buckling is possible, the amount of stress we can safely carry has to be reduced. This is done with the factor “X” to take buckling into account.

Knowing all that, equation 6.1 that we started with… “misses” the buckling factor. At the same time, the “effective length” Leff has to be the allowable “width” of the web we can take into account. I know this, because that width multiplied by thickness will give me the cross-sectional area I expected to see in the equation.

Also, if we move further in the code, we get to equation 6.2. It shows us that for whatever reason, the author of this part decided to hide the “buckling coefficient”! I expected it to be in the “main” equation 6.1. Instead, we will use it for the calculation of effective length:

OK, so at least we are getting somewhere!

Usually, it’s the buckling factor that is complicated to calculate. But here, we will have to think about effective width as well. Simply put, you usually know how big the cross-section of the element is that you are trying to calculate. Here, things are a bit more difficult! We will have to estimate how much of the web width will actually transfer the load. So, this is also something we will have to calculate/estimate as well.

Let’s take a quick look at that buckling factor now:

Of course, the “ease” with which our web will buckle will depend on its slenderness, as can be seen above. And it’s the “slenderness” that is the “tricky part” when you calculate the buckling reduction factor… let’s take a look:

As always slenderness is just a square root of plastic capacity divided by critical capacity. The plastic one is super simple… usually! Here of course, it will depend on the width of the web we can “assign” to transfer the local load. And we haven’t really established yet what that length will be (please be patient, we are getting there!). Usually, the trick is with the critical force, as you have to think about buckling lengths, etc. Surprisingly, it’s not so bad in our case:

Remember the “mysterious” kF parameter defined at the beginning (under the picture). Well, now at least we know where we to use it! Also, it seems that we know everything we need to know to calculate FCR, which is a good thing. But there is one thing I don’t like about this equation! Normally, I would expect the height to be “squared” (just like in the Euler equation). Checking such doubts by hand is tricky, but in the next post, I will do FEA! This means that we will be able to see how good this “fits” into the “reality” of various cases (yea… FEA can even help you with these types of things!).

The last thing we have to deal with is this freaking width of the web we will be allowed to use in our design. Let’s go through the procedure step by step!

Initially, this looks at least plausible. We have the contact width we can apply the load to. Also, the stress will “travel” at a 45deg angle, meaning that it has to go through the flange first. But then it will “reach” the web at a wider length (plus twice the flange thickness to be precise). And in the end we can further increase this “additional” width, based on two parameters:

The first parameter increases the “allowable width” taking into account the width of the beam. The second one clearly depends on the slenderness of the web. This is actually a funny part since the slenderness depends on the allowable web width too! So this is like a “double dependency”. But we will simply assume one case over another, and check if things work out in the end! Of course, if we “miss” the assumption we will make corrections if needed.

And that is it! The entire procedure of calculating capacity of the web in a real beam! It’s time to use what we’ve learned in our case to see what we will get!

The problem at hand

Before we will get to the actual calculation I owe you some geometrical details of the problem! It’s all fun and games to show you equations and describe what is what. But we need specific numbers to do the math we need!

The dimensions I give you below aren’t “special” in any way I can think of. In truth I simply started with the FEA model, and guessed all the dimensions as I moved along!

Also, there is a “trick” I’m using here. Normally, when I design a beam I would instantly go with the “height” from the center of one flange to the center of the other flange. This means, that I would model the bottom beam as 530mm high. This is 500mm for the web and an “additional” 15mm per flange. This would lead to the most accurate Section Modulus and is generally a good practice you should follow.

However, this also means, that my FEA model would “see” the web as 30mm “higher”. And since we are ONLY focusing on the web here, I will model the beam as 500mm high. This won’t have a “dramatic” impact on the outcomes, but I figured I may as well be “on point” here.

Just please note:

I do this ONLY because it’s a partial model that focuses on calculating ONLY the web and its stability. You want to use centerlines in all true design cases. Unless a very good argument can be made to do it differently!

Share this post with your Friends!

I hope it’s clear that the geometrical arrangement doesn’t matter that much. We are only interested in checking the local capacity of the web. So the length of the beams etc. isn’t as important. As you saw at the beginning of the post, I went with something like this:

The only thing needed here is, that the distance between the inside stiffeners (“a”) is 800mm in our case. Funny enough, the rest is irrelevant in the design that we are about to make!

Design for the win!

Technically the procedure I described above already explains how things should be done. But I don’t like leaving a single stone unturned. Plus if we want to compare FEA outcomes with the normal calculations… we kind of need an outcome! So, let’s do the actual math for our example!!

As you most likely noticed, I like to “present” the procedures “as we go”. This means starting from the capacity check and expanding into all the complexities of the problem. I think it’s easier to understand things this way. Of course, when we actually do the calculations, we will just “start from the end” and move “upward”. This means, that we will be starting from things we can actually calculate first. And we will move “towards” things we can calculate based on the sums we already did.

It’s only natural that we first set the values of M1 andM2 parameters. After all, this is the only thing we can do at this point. We don’t know what slenderness our web has, but I will assume that the slenderness is high. I do this because we have a thin web that is quite long! Simply put, I will assume that M2 actually has a non-zero value.

Armed with the above, we can calculate the width of the web. But we need to establish something I haven’t mentioned so far first. This is the contact width, and we will establish it according to figure 6.2:

As you can see we can consider various cases. Since we have an I-beam on the top as well I will go with the second case from the left. I will aim for minimal weld thickness. Mostly because I won’t have welds in the FEA models that I will do later! Let’s just say that I want to stay as consistent as possible. I will mark the horizontal size of the weld as “r1” (I think “z” is used in the UK for this). The value of 6mm means that the weld is a=4mm. Small for a 30mm flange, but I hope you will forgive me for this!

So now, we know everything to calculate the web width we can take into account in design:

As you can see above, I’m taking M2 into account assuming that the slenderness will be higher than 0.5. Let’s find out! But first we need to calculate a critical force:

As you can see I’ve used the equation from figure 6.1 I mentioned at the beginning to calculate kF parameter. With this, calculating the critical force is a breeze. So now, we finally get to see if the slenderness is actually higher than 0.5:

YES! We nailed it! Since the initial assumption is correct, we can swiftly proceed to verification of the capacity, as you can follow below:

Splendid! Unless I’ve made some mistakes above, it seems that we just obtained an answer according to Eurocode! It looks like the maximum reaction force the top beam can transfer to the beam we were analyzing is 315.8kN. This is generally the procedure that allows you to check local loads on the webs of beams in hand calculations.

Wrapping it up!

I hope that now you know how to do such calculations by hand! I must say this can have quite a few uses! But the best is yet to come! In the next post, I will do a linear FEA analysis of the problem! Later on, we will do the nonlinear analysis too! And finally, we will compare all of the outcomes and discuss what makes and doesn’t make sense! Let me know what you think about this series idea, and of course, stay tuned!

You can keep in touch with me, and also learn some awesome things at the same time! Just sign up to my FREE FEA Online Course below!

10 FEA lessons I wish I knew a decade ago!

Join the course, don’t learn the hard way!