We can also obviously create variations changing the way we ditribute the depth for example, with a triangular variation of the depth. You can notice here that, since we make a division, we almost divide by zero. What is the consequence of this ? We are going to have internal forces which are going to be larger near the supports. They are not going to be zero in the middle, but they are going to be smaller in the middle than they are near the supports. And then it will be the same for the tensile internal forces. So we will have something which is a bit particular: till now, we tended to say: if we have uniformly distributed loads, the maximum internal forces are in the middle, but that is not the case when the depth is varying in this way. We can also decide to have a trapezoid variation of the depth, as I have represented it on the right, that is a solution which is often used, for example, for industrial halls. So we make the division, and we get the result, and here, actually, it must be seen case-by-case. Since we precisely have on one side, this trend to be close to a triangular solution which would involve larger internal forces in the chords, near the supports, and then to a purely rectangular solution which would entail larger internal forces in the middle, so these can offset each other, getting a little bit closer to the lenticular solutions, in which the internal forces are almost constant. Here, I have the solution for two of these configurations. For a triangular truss, we can notice that the internal forces in the chords are maximum near the supports. Now, let's look at the sign of the internal forces in the diagonals: the inclination of the arch should usually give us compressive internal forces, and here, we can witness a reversal of the sign of the internal forces in the diagonals. So, you should really be careful: it is going to depend on the inclination that we give to this truss and on the type of loads. Here, we have uniformly distributed loads. If we have loads which are a little bit more complex, we cannot really say where the diagonals will be in compression, and where they will be in tension. You need to be careful here, and we can notice that the internal force in the diagonals is larger in the middle that on the edges. And then here, we can say that the internal forces are more or less constant in the chords, but we can nevertheless notice that they decrease near the supports and then, the internal forces in the diagonals, we cannot say anything. Here you have two examples of variable depth truss structures and we can clearly notice that there are actually, at least on the left structure, two parts: a part which is linked to the external shape which we decided to give to the building, and then this, this is a shape which is induced, for example, by depth limitations, but also by the architectural appearance which we wish to give it, and then, we have another limit, here, which is linked to the internal shape. Here, this is a gymnastics hall, so it is probably necessary to have a certain depth on the edges, but we can be a little bit lower in the middle, given the fact that we are probably going to hang up equipments after that, and we thus have a shape which results from two famillies of constaints with a structure in the middle which adapts. On the right, this is a kind of structure which we can often see: this is a greenhouse, for example a greenhouse to grow vegetables, but it could also be a greenhouse to do light industrial work, and we have the external shape: a roof shape which is favorable to the snow evacuation, etc. then an internal shape where we want to have the largest possible depth. Everything with a very high transparency, that is obviously important if it is a hall to grow vegatable . The last type of truss which we are going to see together in this video is call the Polonceau truss. This truss has been invented to have a structure which is as light and as transparent as possible, so using as many as possible elements in tension. That is a structure which is going to be used for a roof, so the part of the roof on the top can easily carry compression, since we are going to have a relatively thick element, however, the lower part is often built with cables. For this resoltuion, I have decided not to place the loads under, as it was the case in the video, but over. That is also often the case for these trusses which will have to resist for example, a wind or a snow load. Since we have symmetrical loads, we will have reactions on the left and on the right which are equal and which have each time a value of 15 Newtons. I am going to start the resolution of this truss by this node on the right, which is subjected to the right reaction, R7, which is again staggered, for us to be able to see it well, but it should be superimposed on the same vertical axis than the applied loads, and afterwards, we turn counterclockwise around this node and we first meet bar 6-7 which pushes on the truss, therefore this bar is in compression, then bar 5-7, which pulls on the truss, which is thus in tension. Here is the contribution of node 7 to the Cremona diagram. We now move to node 6. Node 6 is subjected to the internal force in 6-7 in the other direction, to the applied load of 10N, to the internal force in 4-6 which is in the same direction than 6-7, which is also a compressive internal force and to the internal force in 5-6 which also goes in the direction of this node, therefore it is also a compressive internal force. Here is the contribution of the node 6 to the Cremona diagram. We now move to node 5, an unloaded node, but not without any internal forces, which is subjected to the internal force in 5-7, in the other direction, to the internal force in 5-6, in the other direction, then to the internal force in 4-5, which pulls on the node, thus a tensile internal force, and to the internal force in 3-5, I draw it in a slightly staggered way, for us not to lose our way, which is also a tensile internal force. The contribution of node 5 to the Cremona diagram is indicated here. We now move to node 4. Node 4 is subjected to the internal force in 4-5, in the other direction, to the internal force in 4-6, in the other direction, to an applied load of 10N, then to the internal force in 2-4 which goes in the direction of the node, which is therefore a compressive internal force, and finally to the internal force in 3-4, which is a tensile internal force. Here is the contribution of the node 4 to the Cremona diagram. We now move to the node 3. Node 3 is subjected to the internal force in 3-5, in the other direction, to the internal force in 3-4, in the other direction, to the internal force in 2-3, a compressive internal force, and finally to the internal force in 1-3 which I draw in a staggered way in relation to 5-7, which is a tensile internal force. And the contribution of this node 3 to the Cremona diagram is indicated in yellow here. We now solve node 2. Node 2 is subjected to the internal force in 2-3, in the other direction, to the internal force in 2-4, in the other direction, to an applied load of 10N, and to the internal force in 1-2 to close the polygon of forces, which is a compressive internal force and the contribution of this node to the equilibrium is indicated here in grey. Finally, we check the equilibrium of node 1. This Node 1 is subjected to the internal force in 5-7, in the other direction... the internal force in 1-3, in the other direction, to the internal force in 1-2, and finally to the support reaction R1 which is equal to 15N. Here is the contribution of this node to the global equilibrium. On this figure, we have represented a Polonceau truss which is a bit particular: you can see that, instead of being rectilinear, actually, this Polonceau truss goes a bit up. We have taken this intermediate bar and we have shortened it, inducing a slight curvature upwards. Reversely, if we had decided to lengthen it, we could have obtained a Polonceau truss which would be valid as well, but which we would have curvature downwards, which is absolutely possible, depending on the uses. On the left, we can see a Polonceau truss such as it has been built for the East train station in Paris. We can notice that the structure is opaque in its upper part, it is a good thing, we can put a lot of matter, timber, metal, which enables to carry compression, we have these elements here which are in compression, and all the others are very thin elements, cables, which enable to have a great transparency. It was particularly important at the time of these stations, since they used steam locomotives, it was necessary to ensure the ventilation and then to have the least elements which could block the air flow. If you look at the first image of this video, you will see the cover of another station which uses the same system. In this video, we have looked at trusses with a variable depth, we have seen the influence of the depth, both on the internal forces in the chords and in the diagonals, which can been quite different of what they are for trusses with a constant depth. We have studied in particular three typologies: lenticular trusses, triangular or trapezoid trusses, and then finally Polonceau trusses. But there are many more typologies: parabolic trusses, or with much more free shapes, depending on the uses which must be done of the structure.
