Hello, in this lecture I will show you how to use an arch-cable to determine the forces at the supports. It can be an auxiliary arch-cable, which will be used for determining the forces at the supports of other structures than the arches with ties. The method which we will use is simple and systematic. Let's start with an arch-cable with one force, it is an arch-cable a little bit more complex than the ones we have seen until now since it has an inclined force and the cable is not horizontal. Obviously, the method also remains valid for vertical forces and horizontal cables but I prefer this method which will enable us to be more general. So, first, we draw the force which acts on our system F1 in the Cremona diagram. Afterwards, we are going to take an interest in the free-body which surrounds this force and which is going to cut the internal forces N1 and N2 in the arch with tie. Turning around the arch-cable in the counterclockwise direction, we first meet the internal force N1. So, I trace a parallel to N1, I do not exactly know until where it is going to go. Afterwards, I trace a parallel to N2, there you go. So, here, I have the internal force N2 and here, the internal force N1. Let's now take an interest in the support B. What do we have just next to the support B ? We have the internal force N2 which pushes, then the tension T in the cable of the arch-cable, and finally, we will have a vertical force at the support. So, I first trace a parallel to T. And as I know that the last force is vertical, I go until the vertical projection of our force F1. Here, this is T and finally, here, I have the vertical force VB If I now take an interest in the free-body near the support A. The force T acts on this system in the other direction, the internal force N1 in the other direction and finally the force at the support We can see that RA is greatly inclined, almost horizontal. It is like that for our configuration. We now want to look at the case of an arch-cable subjected to two forces. We are going to take an interest in the free-body which includes both forces F1 and F2. We are then going to copy them in the Cremona diagram. So, two forces F1 and F2, they do not need to have the same magnitude at all. In this way, we can immediately obtain, in the Cremona diagram, the magnitude and the direction of the resultant which we copy in the sketch of the structure. Well, here, it is deliberately a simple case, that is to say in which both forces are converging, thus we do not have any complications to determine the line of action but you are able to do it if it is more complicated or if there are more forces, we can obtain a resultant in every instance for the set of loads which act on our structure. We are now going to work with an auxiliary arch-cable, which is the arch-cable of the resultant, and I am going to draw this arch-cable here, deliberately staggered in relation to the one which was outlined. And then, obviously, we will have tension in the cable. We will have here the internal force N1 and the internal force N2. We are going to draw in the Cremona diagram, the figure which corresponds to the equilibrium of this auxiliary arch-cable which does not really exist but which is going to be useful for us to determine the forces at the supports. So I draw here, in parallel with N1, and here, in parallel with N2. In this way, we obtain the internal force here, N1 and the internal force N2. Taking an interest in the the free-body near the right support, we have, acting on this free-body, the internal force N2 in the other direction, and the internal force in the cable T, which we extend until the vertical projection of the resultant. So this, it is the internal force T in the cable. And finally, we obtain the vertical force at the support V on the right, I would say, at the point B, VB. We will call the left bearing A. And afterwards, we are going to take an interest in the equilibrium of the right support, we will have the tension T and the internal force N1 in the other direction to finally obtain the force at the support in A, which I denote RA, which I can copy here, on the real structure. So RA, as before, is almost horizontal but it only depends on the loads and on the configuration which we had used. What is about the other types of structures ? Well, it is important to consider that if we look at a free-body which just cuts over the supports, well, what do we have inside this system ? We have have the resultant of the forces F1 and F2, maybe there will be more forces another time, they are going to make a resultant. And then, we are going to have forces at the supports. If we placed inside this system, instead of having a truss on the top or a beam on the bottom, if we placed an arch with tie, we would exactly obtain the same forces at the supports. So, we can define the following process to determine the forces at the supports. This process is called the process of the auxiliary arch-cable. First : Consider the free-body which passes just over the supports. Second : solve an auxiliary arch-cable and to do this, it is necessary to start by tracing the resultant of all the loads. Third, we will be able to determine the forces at the supports. You maybe asked yourself the question : yes but what would happen if I took another arch-cable, would I obtain the same internal forces ? Well, I encourage you, for the previous example or maybe for the one which has only one force, to take another arch-cable and to make the construction and you will notice that, unless you make a drawing mistake of course, you will obtain the same internal forces. So, any auxiliary arch-cable will enable you to obtain the forces at the supports. That is why I said : solve "an arch-cable", not "the" arch-cable. In this video, we have seen how to determine the forces at the supports for arches-cables and we have also seen that the method of the auxiliary arch-cable enables to determine the forces at the supports for all sorts of structures, as long as they have, as arches-cables, a fixed support and a mobile support. It will be useful for us thereafter, particularly for the forces at the supports of the trusses and of the beams.
