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 does 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, we 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 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 more complex structure with four times 10 Newtons hung up. We can wonder if we can solve it. At the 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, thus 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 the advantage of the opportunity to number the compression 1, 2 and 3. 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 our free-body, we are going to meet out 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 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 directions for each of these construction. 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 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 quit 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. Which is striking in this configuration, it is that the length of the stay cables varies very, very quickly. The external stay cables are very long, while the internal stay cables are very sort. 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. This is the tension in the external stay cable and this, it is the compression, so we are going to call them, 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. Well, 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 are not easy to hang up on the top of the mast in the harp 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 help support the vertical loads which are very significant in the exceptional cases. But the bridge of 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.
