Hello. In this video, we will see how we can actually include a truss in a beam, we will see what are the limitations to respect for it to work, we will see that construction materials such as reinforced concrete or steel are particularly adapted to the analogy of a truss in a beam. We first have the beam which we have seen in the previous video with 2 times 10 Newtons and we want to study the internal forces which act inside this beam. We already know that the lower part will be in tension, so I already place here, a little bit inside because I need a certain thickness, the tensile internal force. And then, I am going to study a free-body which is approximately located here, we are going to see why I cannot totally be accurate. This free-body is subjected to the load of 10 Newtons and then comes an inclined diagonal from the support. This diagonal cannot go till the edge. Why? Because it will be necessary to reserve a part of the matter to carry the internal force. The complete solution is not yet a truss in the beam, that is just an arch-cable, and you will see that it is enough for our reasoning. And we have here a certain amount of material which is reserved to be able to carry the internal force which acts in the beam. This internal force here, I call it... we are going to write it here, I call it N2 while the internal force in this part, which is larger, so I give it more space, I call it N1. So here, this internal force N1 spreads over a certain width within the beam. You remember that the truss elements which we had also had, eventually, widths when we were finished with our calculation. The width of this zone here, I am going to call it b1, and of this zone here, I am going to call it b2, remembering that our beam has a total depth of h, and a width t. So, on this basis, I can write that the internal force N1 is equal to the width of the diagonal times the thickness t times the compressive strength of the material. If I know the thickness and the material, I can thus determine that b1 is equal to N1 - in terms of absolute value, I do not write it but that is obvious that the thickness is going to be positive - divided by t times fc. Likewise, b2 will be equal to N2 divided by t times fc. What we can notice... I evidence the central part of the arch-cable. This arch-cable does not have the same depth than the beam, for an obvious reason: the whole arch-cable which supports the internal forces must be inside the beam, so actually, the real effective depth is located here and we will call it Z. It is very important to remember, obviously, that the effective depth z is smaller than the total depth of our beam. Let's look at how to obtain the internal forces N1, N2 and N3 in the lower part of the beam. Our dark blue free-body is subjected to the 10 Newtons. Then, turning counterclockwise, we first meet the internal force N1, here, then the internal force N2 which is horizontal. If we want to obtain the internal force N3, we have to look at this orange free-body which is subjected to the internal force N1 in the other direction and then to the support reaction here: we can easily guess that it is equal to 10 Newtons. I draw it in a staggered way. R on the left is equal to 10 Newtons, then finally, the internal force N3 which pulls rightwards and then we can see that the amplitude of N2 is equal to the one of N3. That is not possible to endlessly proceed in this way. If we have, like when we turned the beam before, a beam which still has the same loads but which has a smaller depth, what is going to happen when I try to draw the same arch-cable than before, still respecting a certain zone which will enable to distribute the stresses in our beam, I observe that suddenly, this angle that I call alpha becomes too small. So, this is a solution which is unacceptable. Why is it unacceptable? Not because we cannot calculate it, but because it has been observed in lab tests that the the structure is not going to behave this way. What is going to happen then? We are going to see another type of solution which is another truss which is going to enable us to solve this difficulty. Instead of directly going to the support, the internal force is actually going to first go down then to go up again, thanks to a tensile element to finally go down again towards the support. Then, we are going to have a compressive internal force here, and then the same thing on the other side. Then a vertical element in tension and a second diagonal in compression. Of course, all these elements are going to have a thickness, here the internal force is going to be maximum, it is going to decrease a bit here, and in the diagonals, however, it is going to be constant since actually the load of 10 Newtons must go through towards the support. Here, we have an angle which is acceptable if it is between 25 and 45 degrees. The building codes prescribe these limit angles but these values there are generally acceptable, for example, for reinforced concrete. So let's look at a reinforced concrete beam. Here, this is a beam that we have in our corridors, which enables us to show... a quite old beam, which enables us to show the types of reinforcement which act in a reinforced concrete beam. We first have the reinforcement, it is on both sides, the lower longitudinal reinforcement. That is the one which can carry tension as we have just seen it before. Here, we have another particular element, that is this kind of pipe, which actually contains, here we can see it, a lot of small wires, this is a prestressing reinforcement. We have already seen the word prestressing not that long ago when we were talking about the airplane of the Wright brothers which had prestressed their X-shaped diagonals. Prestressing is a system by means of which we introduce, in the present instance, compression in concrete by strongly tensioning these cables before loading the structure; that is why we call this prestressing. Concretely, the most significant consequence of prestressing here, is that we can introduce, thanks to these cables, a very large concentrated load which would otherwise need a very large number of reinforcement bars. We also have other types of reinforcement here which are vertical reinforcements which surround, and that is important, all the other reinforcement. In theory, these bars should be closed, here unfortunately they are not, but actually, they should be closed and these bars, we call them stirrups. They work as transverse reinforcement since, as we have seen it, it was necessary before. And then, we have a fourth kind of reinforcement here, that is the bars which remain here and also here. This is a constructive reinforcement. It is not, actually, strictly necessary for strength reasons but however, it is here to hold the other bars, in such a way that the reinforcement cage which it is going to constitute, remains stable when we are going to pour the concrete... because the concrete is very heavy material, you remember, with a density twice and half larger than that of water. Here, we have a model which is maybe a bit simpler to understand the behavior of a reinforced concrete beam. We have here, a beam with two loads, we have a lot of stirrups, here that is a beam with a rectangular cross-section and here the stirrups are closed, and then, actually, you can see, the stirrup is bent and it is closed in the other direction. There are several ways to do that, but this usually works quite well. Here it is, I am not going to draw them all, however, I am going to copy them here on the top. We have here stirrups, regularly distributed over the whole length. Here it is, and so forth... I do not correctly draw them, I need to correct them here. They must obviously extend over the whole depth and surround the load-bearing reinforcement, that is to say the lower reinforcement. This reinforcement should actually even raise upwards in a correct way, anchored. What behavior are we going to have? Well, what we have said before, that is to say a behavior with diagonals in compression. And then actually, we can see that since we have more reinforcements, we are going to have a behavior with multiple diagonals in compression and then, obviously, here on the top, a compression which is going to increase, it is going to be maximal in the central part, constant by the way in the central part and it is going to vary towards the supports and then still these multiple diagonals. Of course, you can ask yourself the question: "Does reinforced concrete really behaves concrete like that?" Well, what we can see here, on the bottom, is a similar test, also with two loads which act on this beam. We can notice the presence of a lot of cracks which are prettily parallel to each other and which let us understand the behavior of concrete. If we take a small element of concrete here, well, this concrete is cracked on either side, so it means that it is exposed, in this direction, to tensile stresses which exceeded its tensile strength and it cracked. And then, in the other direction, to compression stresses. It enables us to understand that indeed, the simplification that we have made, to imagine a truss inside the reinforced concrete beam, is actually not a real simplification. Here, on the right, you have the shape of the reinforcement, for example, we can also end the stirrup like this, thereafter, we are going to place a certain number of longitudinal load-bearing bars, and then, on the top, since I have drawn them in yellow, I keep going, there will be a certain number of non-load-bearing bars which are constructive bars. You can also see them in the central part of the illustration. A reinforced concrete beam thus behaves like a truss with N-shaped diagonals in compression. Here we have a nice example of a reinforced concrete beam just before its rupture. What happened is that here, the stirrups broke, I draw them in a discontinuous way and since the stirrups broke, they enabled the light which is behind to pass through but what is interesting is that despite this, we can see that there still were some concrete elements in a relatively good state, at the end, they broke, but these elements enabled to transmit compressive internal forces. That is a test by Michael Rupf, in the laboratory IBETON, at EPFL. We now want to look at how a steel beam behaves when it is subjected, let's say, to two loads like the ones we have seen before. Here, on the right, you have the cross-section of a steel beam which is typical, with two flanges: the upper flange and the lower flange. The upper flange is here to support compression, the lower frange to support tension and then the element between both, we call it the web. All the elements of this cross-section are quite thin because steel is quite expensive, so we try to save it and since it is very strong, that is absolutely possible. And then, also in the left part, that is a railway bridge in Switzerland and we can notice that indeed, we have here the lower flange, here we can see the upper flange, then we can see the whole web and also an additional element that I am going to draw here in light blue, which is a very one-time element, like this, there is one approximately every 2 meters. A vertical element, likewise, on the supports. If I have to draw this element, that is an element which is going to be here on the cross-section; it is typically welded to the cross-section and it closes it. Usually, this cross-section is open here, in this case, it is closed. What is the function of this element? We can see it here in a test, a test which has been carried out, there was probably a load or several loads which acted around there, we have a support reaction here, on the left. And then, the natural trend would be to have here diagonals in compression. But what happens when we introduce compression in a very thin element? Well, we can see it with this shape that we can guess here: this element, instead of remaining rectilinear, is going to bend like this. That is the same thing here. The result is going to be that this element cannot support any internal forces in this direction. So actually, I am going to erase my diagonal in compression because it cannot work because of this phenomenon. This phenomenon, we call it web buckling. That is a quite advanced phenomenon, we are not going to talk more about it here, but it thus means that for a steel beam, it is simply not possible to have N-shaped diagonals in compression. What can then happen? We at least know that we are going to have tension in the lower flange, and that we are going to have compression in the upper flange. To carry the loads, it is possible to have tension here, in these diagonals. In this case, to reach equilibrium, it is necessary that the posts here be in compression. The function of this element which we have here, which is called a stiffener, is to enable this part of the beam... Normally, it would also like to bend if it is subjected to compression, but thanks to this stiffener, it manages to resist and then the result is that this element, this post, can resist compression. The behavior of a steel beam with two loads is then as follows: we have tension in the lower flange, we are going to have compression in the upper flange... Because of the very particular shape of these elements, we can see that the effective depth is slightly smaller than h, but just a bit. That is a big advantage of this construction method. And then we are going to have stiffeners in compression here and then also in the left part, and finally diagonals in tension. So, the behavior of a steel beam can be assimilated that of a truss with N-shaped diagonals in tension. We have seen in this lecture how to insert a truss in a beam, with a clear process which generally leads us to trusses with N-shaped diagonals. For reinforced concrete, the diagonals are in compression. For the steel construction, they are in tension. In both cases, instead of taking into account the total depth of the beam, we have to consider the effective depth.