Hello. In this video, we will see a few variations around the theme of the arches-cables. We have seen in the previous video that an arch-cable is a combination of an arch and a cable, we will see in this video that it can also be a combination of a cable and a strut, and that it can take the shape of a structure which we call lenticular. We have here an example of a bridge made by an arch-cable, and first, it is important to notice why this bridge must be an arch-cable. Well, that is a bridge which has the particularity, you can see it going up, to be able to go up and to go down above the river, in such a way that it can let pass boats below. So, it would be impossible to have an arch which strongly pushes at a height of a few tens of meters. Thus, it is necessary that it should be an arch-cable, we have an arch in the upper part, and a cable in the lower part. And then, obviously, supports on these towers, which enable a rising and descending movement. However, I would like to draw your attention to the fact that there are other elements in this structure. There are vertical cables, about which we did not talk until now. What we are going to see is that the cables are part of the structure, that they work but that they do not change the functioning of the arch-cable. We want to look at a certain place where a load Q acts. This load acts at the level of the deck, where the vehicles pass on this bridge. Obviously, there is no vehicle now, since the bridge is going up. Let's isolate a free-body around this load, what we have horizontally is the deck, which is cut, vertically, we have the load Q, on the bottom, and a piece of cable upwards. We have a tension, T, in the deck, which is due to the effect of the arch-cable, and this tension does not change when it meets a couple of forces which are perpendicular, we are going to see it again later, and obviously, we have an internal force N in the cable. Let's make the corresponding polygon of forces. So, we have here the load Q, then the force T which pulls on the left, then the internal force N upwards, and the internal force T, which pulls on the right, thus my free-body is in equilibrium and both forces T must necessarily be equal : since both forces N and Q are equal and opposite, both forces T must also be equal and opposite. So, we can see the fact that there is a hanger does not have any effect on the cable of the arch-cable. Now, let's look at a free-body on the arch part. We have the arch, and I exaggerate a little bit, but it changes its direction at this place here, we can see that it is curved, and the force N which is equal to Q obviously act on the hanger, and we have a compression in the arch on the left, and another compression in the arch on the right. If we draw the polygon of forces, I start by the compression in the arch on the right, afterwards I draw the compression in the arch on the left, and finally I have the internal force N equal to Q, and my polygon is closed, we can clearly see that it is the orientation change and a little bit the magnitude of the compression in the arch, which support the internal force. So, it was really allowed to say that there is an arch which carries the loads, and simply a cable which only holds both ends of the arch together. We have here the structures of the general stores of Chiasso, which represents the symbol of the i-structures course, and we are going to see together why this structure is symbolic and important. To solve it, we are going to use the applet i-Cremona, specifically, we are going to use two applets i-Cremona, because we cannot solve this system in one go. We first want to start, in the applet on the bottom, to solve the problem of the cable. We have these vertical elements which introduce loads on the cable. I introduce each time a load of 100 which is the value by default, and I position the supports, for the cable on the left and on the right. We are going to stop the analyse now, the rest of the structure is a type of structure which we will see later, that is a frame, but for the central part, the analysis which we are doing is absolutely correct. I trigger the funicular polygon and I give it the proper shape. We can see that on the left part and on the right part, I am going to correct a little bit the position of my supports, we can see that, on the left part and on the right part, the shape of the cable is approximately right but that obviously in the middle, it is as though there was something which pulled the cable upwards. So, we are going to introduce a force to pull the cable upwards. I first introduce a force of 100, but it is not enough, pressing shift, I am going to change its magnitude, so we move it a little bit on the left, and there you go. When I introduce a force which is equal to roughly 200, 220, 230 here, then I get the proper shape of the cable. What is the element which can carry this load ? Well, it is precisely the arch, since for the moment, all the loads that we talked about are supported by the cable, since we have these loads on the roof, which go down and which create deformations, and thus internal forces in the cable, but for the moment there is no force which acts on the arch. Now, for the system to be in equilibrium, we are going to put a load of 228 downwards, which acts on this arch, and now, we have the internal force in the arch, and we have the internal force in the cable and both sets of forces give us all the internal forces in the whole structure. Note that these elements here are in compression, however, this element here, with a force of 100 which goes down and a force of more than 200 which goes up, this element here is in tension. Thus, we can now draw in this structure the internal forces which act, so there is compression in the upper chord, in the arch, there is compression in these posts, and then there is tension in the lower cable and in the central post. Why did the engineer Maillart decide to make this structure in this way ? Well, it is common to have inclined roofs and it is for evacuating rainwater. But it is not the only reason. The reason is that he was probably asked to provide a certain height, for his structure, and that if he had kept the cable as it was at the beginning, then he should have built all his structure higher to respect this. Introducing this small subtlety of change of the cable shape, he managed to decrease the total height necessary for the structure, and thus made it more competitive. Among the variations of arches-cables which we can make, we can also transform them in cable-and-strut. In this configuration, we still have a cable, but it is not rectilinear anymore, and we still have an arch, but now the arch has become rectilinear, so we still can call it arch, or we can call it a strut. A strut is an element which is used for the buttress, which is an element which only takes compression along its own axis. So, it is also an arch, we can clearly say it. In this applet, we can see one of the spans of the bridge of Shiosai on which act loads induced by the self-weight and by the traffic, I introduce a fixed support on the left, a mobile support on the right and I activate the solving and we can see that the shape which has been given to this bridge corresponds to the shape of an arch-cable. These elements, which carry the internal forces of the deck until the cable are all in compression. A more recent example has been designed by the professor Muttoni. It is also essentially an arch-cable with compression in the deck. We can notice that instead of having vertical elements, we have now inclined diagonals. These diagonals complicate a little bit for us the understanding of the phenomenon, but have the advantage to be able to slightly change the internal force in the cable along its length. Fundamentally, this bridge is still an arch-cable. There are lots of possible variations for the arch-cables, from the pure arch-cable which we have seen at the beginning to the pure cable with strut. Here we have diverse shapes, here a shape with the three quarters of the internal force which pass in the arch and 25% remaining in the cable. Here an interesting structure with half of the internal forces in the cable and the other half in the arch. And then here, only a quarter of the internal forces in the arch and the three other quarters in the cable. The purpose of this pictures is to convince you that all the variations are possible. You have to decide according to the structures you want to design, if you prefer a structure which has more the shape of an arch-cable or of a cable with strut or if you want something between both. Actually, the family between both with 50% on each side is very interesting, we can see here a bridge of the 19th century with a lenticular shape, we can see here that there are locomotives; we have a desk with very heavy loads which is hung up first to the cable, well, we can see that actually there are two cables which are linked by diagonals but this set constitutes a big cable and these cables continue to arrive until the level of the upper arch. So, since we really have a lenticular system, half of the loads is taken by the lower cable and the other half is supported by the upper arch. And, it is an arch-cable thus it only transmits vertical forces to the supports. In this lecture, we have seen that the arch-cable can take varied shapes from an arch-cable to a cable with strut with transitional possibilities, including one, the lenticular structure, which is particularly interesting. Hello. In this video, I am going to talk to you about the cable-stayed systems, the principle of their functioning, the way in which we can solve them by means of the graphical statics and about the diverse types of configuration which exist for the positioning of the cables. In this video, you can see a timber bar which is hung up to two inclined elements of chain. When I add two symmetric loads, this system is in equilibrium and we do not have any problem. Let's see how to solve this system. So, we have on the left and on the right a load of 10 Newtons which acts on our structure. First, look at the right part that I isolate with a free-body. In our free-body, we have an horizontal element of deck, an inclined element of cable and a load of 10 Newtons. We are going to solve the equilibrium of this configuration, so we are going to have here a load of 10 Newtons. Turning around the free-body, we first meet the chain, as it is a chain, I know that it is in tension thus I directly draw it in tension. And we have an element, a tensile internal force and we meet the horizontal element, which is in compression. We have here a compression, here a tension and we can color our cable-stayed system, well, I think we had already intuitively recognized it but now, we have demonstrated it, it is a system in which the chains are in tension, and the element which is used as deck is in compression. If we deal with the left part of our structure, we have the same force here of 10 Newtons, the compressive internal force in the other direction and the tension which is equal to the tension in the other cable. Here we have a structure which is more complex with 4 times 10 Newtons hung up. We can wonder if we can solve it. At least, if we identify this free-body here, we can see that it is perfectly identical to the one we had in the previous system. So, we can quietly solve it. 10 Newtons. I am going to number these tensions, I will call it T1, this one T2, this one T3 and this one T4 and I am going to take advantage of the opportunity to number the compression 1, 2 and 3 and I come back to the tension. So I first have this element here which is in tension. And I obtain here the internal force T1, then the next element is in compression. I obtain the compression C1. Now, I am going to take an interest in this free-body here which I draw nearby because it is interesting. So, we have the deck, we have an inclined cable and we have 10 Newtons. What we also know, is the compression C1 which comes from the right. Then, turning in the counterclockwise direction around this free-body, we are going to meet our force of 10 Newtons, our load of 10 Newtons, then the compression C1 in the other direction, then the cable T2 and finally the compression C2. Obviously, we could continue towards the left doing exactly the same construction. I spare you this construction, but I do it by myself to have the complete drawing. So, here, it will be C3 with both loads, the two last loads of 10 Newtons and then the tensions T3 and T4. Obviously, the compressions are used in both direction for each of these constructions. We can make a few interesting observations on the basis of this solving. What we can see is that the compression is not constant in the deck. It is lower at the end and becomes greater in the middle. Likewise, the internal force in the cables T1 is greater than the internal force in the cables T2. So, the cables which are more inclined support more loads than the cables which are less inclined. On the other hand, we can obviously see that the cables T2 are distinctly shorter than the cables T1. In this picture of the transporter bridge in Nantes, designed by the french engineer Arnodin, we can see that the stay cables have the same configuration than in our model, that is to say a configuration with a fan shape where all the stay cables are hung up on the top of the mast. This configuration has the advantage, we have seen it, that the internal forces are lower in the stay cables which are less inclined however, the stay cables are quite long and it is especially quite difficult to all hang them on the top of the mast, thus it is a configuration which has a few problems. Independently of this configuration, because the central span is much longer than the adjacent spans, it has been necessary to use vertical cables to anchor the structure in the ground. On the bridge of this illustration, we have another configuration of the stay cables which we call the harp shape, that is to say that the stay cables are parallel to each other. What is striking in this configuration is that the length of the stay cables varies very quickly. The external stay cables are very long, while the internal stay cables are very short. Let's look at what is about the internal forces. I just draw the right part of a cable-stayed bridge such as we have calculated it before, except that this time, the stay cables have a harp configuration and to get something, I also load each stay cable with 10 Newtons. Then, I can draw here this load of 10 Newtons for the free-body on the right. I trace a parallel to the stay cable, it is the tension in the external stay cable and this, it is the compression, so, we are going to call this T1 C1, T2 and C2. Here I have T1 and C1. If I look at the second stay cable, there is also 10 Newtons and its orientation is still the same. What is interesting is that T2 is equal to T1, so the internal force in the stay cables is the same. However, the compression C2 is quite a bit greater, in this case double, but it is really greater than C1. So, we can imagine that when we reach the end, when we have the shortest stay cables, we have a huge compression. However, we have the advantage to have internal forces in the stay cables which are constant. Note that before the internal forces in the stay cables rather tended to decrease, which was favorable. An advantage of the harp configuration is that we have more space to place the anchors of the stay cables which were not easy to hang up on the top of the mast in the fan configuration. As you have probably guessed it, it is possible to make transitional configurations, as we can see it here. Here, a half-harp configuration. The stay cables are not absolutely parallel but they are anyway well distributed on the height of the mast, which enables to have a little bit the advantages of both solutions. It is also possible to make cable-stayed bridges with multiple spans. However, if we remember of the necessity to anchor in the ground, as we had seen it for the bridge of Arnodin, it is actually necessary that the pillars should be stiff enough to be able to help support the vertical loads which are very significant in the exceptional cases. But the bridge in Millau which is in service since 2004 have shown that it was really possible to make this kind of structure in an efficient and elegant way. In this video about the cable-stayed systems, we have seen how to solve them by means of graphical statics, we have seen that there are several possible configurations for the stay cables, particularly the fan shape or the harp shape and we have seen that there is a large compression in the deck of this type of structure.
