Hello. In this video, we look at towers and cantilevers, which are types of structures that we have not studied so far. We will look at the loads which act on them, we will look at how to determine the internal forces, we will see that the resolution method which we have previously seen remains applicable. Then, we will make some considerations related to the shape of these structures. We have, on the left, a type of structure which clearly relates to a tower. It is the mast of a boat, which is subjected to transversal wind. On the right, a structure classically considered as a cantilever, that is to say, an element which is subjected to a load at the end. In French we have an expression with means "carried wrongly". The word in French is "porte-Ã -faux". I will come back to the meaning of this expression later. In both cases, loads are transversal to the structure, that is to say roughly perpendicular to the axis of the structure. In this video, we can see a cantilever. There is already a small tension in the cable before I add the load of 20 Newtons, because obviously, the self-weight of the structure is acting, especially the self-weight of the lower timber element. This element already has a small internal force. In this video, we have taken the cantilever and we have made it rotate 90Â° to put it vertically and create a tower. Obviously, if I took off my finger, the tower would collapse. But applying a horizontal load, perpendicular, we can see that the structure is in equilibrium. We are going to solve here these two types of structures, using free-bodies, as usual. I first look at the solution of the tower, on the left. This tower is subjected to a horizontal load of 10 Newtons. I draw it here. Turning counterclockwise from my load, I meet this element which is going to pull on the free-body. So I am directly going to draw it in red, not to lose time. Afterwards, I will meet the right element which is in compression since it pushes on the free-body. Here, this element is in compression. And the other element is in tension. Here is the state of equilibrium for this free-body. We now move to the right, to the resolution of the cantilever for which we are going to look at a very similar free-body, even if we can see well that it has been rotated of 90Â°. This free-body is subjected to a load of 10 Newtons. Turning counterclockwise, around this free-body, I first meet this element which goes up leftwards. I know that there are tensions, so I am going to draw it in red. Then, I am going to meet the element which goes up rightwards, which is going to be in compression. We have here compression, and here tension. The equilibrium of this orange free-body is represented here, by this orange triangle. We can notice that both structures are very similar. If we take the left structures and rotate clockqise by 90Â°, we obtain this structure. Likewise, if we take this graphical construction, and that we rotate it by 90Â°, we obtain this one. That is why both these types of structures are structures with which we are going to deal simultaneously. I come back to the expression "porte-Ã -faux", a very interesting expression in French. I draw here the internal forces which we have. Till now, when we had a vertical load acting on the structure, then we had compression in blue on the top and tension in red on the bottom. However, what can we see here? Quite the opposite, we have tension on the top, and compression on the bottom. So we can understand well the meaning of the "porte-Ã -faux" (= carries wrongly). Obviously, this structures does not carry wrongly, but it can lead to make wrong reasonings, since, precisely, tension and compression are swapped. We can thus understand the expression "porte-Ã -faux" in this context. If we now have a tower which is subjected to two loads, we are going to do the same thing. On the left, I start with a free-body which is going to include the top of the tower, and the first load of 10 Newtons. I draw it here, smaller than before because there will be two loads of 10 Newtons. Then, turning counterclockwise, I am first going to meet the part which is inclined leftwards. I do not know exactly where it is, so I am simply going to draw it like this. Then, in a systematic way, I am going to draw the part which is going to be in compression. I did not know the shape of my tower, but now I know it. So I start from a point which is obviously at the level of this load of 10 Newtons, and then in the middle, between the supports, from a point here. I am now going to trace the parallel to the red part of my construction to obtain the shape of my structure. And till the level of the second load, it is going to take this shape for the left part. And the right part is going to take this shape in a more or less systematic way, if my drawing is correct. I now have a second load of 10 Newtons which is going to act here. So I add it. Second load of 10 Newtons here. So we are dealing with this free-body. I have an internal force which acts in the upper part, I have a load of 10 Newtons. Then I finally obtain this internal force, which gives the inclination of the lower part of the leg of my tower, which is going to be like this, till the level of the ground, of the support that I have on the left. What about the right part? It has not met any other load, so it is simply going to be rectilinear, till the bottom, where we will have another support. You can notice that I have not exactly reached the position of the supports drawn in grey. It is not a big deal, but it is a bit embarassing... - Remember, I had shown you the Eiffel Tower at the beginning of this video - And now, we have a structure which is not symmetrical, strongly inclined on one side, and rectilinear on the other side. You can note that this structure is in equilibrium. We have two free-bodies, maybe that is not useless to draw their contribution. Here, a first free-body, with this equilibrium, and a second free-body with the equilibrium of the lower part. Let's see how to solve this problem in a more satisfying way. Again, I am going to identify the mid-point as starting point. I am going to distribute this load of 10 Newtons differently. For the upper part, that will be the same thing. I am going to take the same inclinations. I have again here a free-body which includes the load of 10 Newtons on the top, and the top of the tower. Turning counterclockwise, I am first going to meet the upper part in tension, then the lower part in compression. I can already draw the part of the top of the tower till the level of the first floor. So here, there is compression. And here, there is tension. Then the load of 10 Newtons is going to be drawn differently: I am going to put 5 Newtons on the left, and 5 Newtons on the right. The first free-body is going to include the left part, with a first load of 5 Newtons which I add here, on the top. The equilibrium will be the upper part, the load of 5 Newtons, the lower part. I directly copy the lower part on my structure. So we arrive here again, at the level of the ground, again not on the support, but here. I look at another free-body in the right part which includes the second load of 5 Newtons. The internal force in the upper part, then the internal force in the lower part. Then the shape change is transferred to here. Everything in compression. We have here a quite symmetrical shape of the structure. And since it is necessary to make pass the 5 Newtons from the left to the right, this element is going to be in compression with 5 Newtons. To summarize, here are the three contributions to the Cremona diagram of the upper part, the left part and the right part. With this, we have a tower which does not really have the shape of the Eiffel tower, but which gets closer to it. We are going to use the i-Cremona applet to solve this structure. The supports are already positioned Let's now do the first resolution such as we did it. So we have placed two loads of 10 Newtons, with an orientation equal to 0, that is to say it is orientated rightwards. So a first load at this level. A second load at this level. When we activate the resolution and control the polygon to arrive to the top, on the symmetrical part, we can exactly see what we have seen before, so a curved part on the left, a rectilinear part on the right. That is not what we want So we start again with another resolution, where we are going to add a load of 10 on the top, and two loads of 5 on the bottom : one on the left and one on the right. If we now activate the resolution, we have to bring back the structure, and we can see that we have now a structure which is symmetrical. In the way we have placed the supports, we have directly obtained the correct shape. For the case of the manual resolution, we should have iterated till we find the proper shape. You can now notice how the applet i-Cremona works: if I take this force, and if I put it on the left side of the pink line, that is to say on the left side of the line which is perpendicular between the supports, so this force acts on the left. If I put it on the right, it acts on the right. And if I take both forces, and if I put them on the other side, I have one of these asymmetrical shapes with the internal force on the other side. Here, you can see a first sketch of the 300 m high tower which was designed by the engineer Koechlin of the Eiffel office. Mr Koechlin was a Swiss, hired by Gustave Eiffel. He came with this first idea to create a 300 m high tower with one, two, three, four, five and six storeys in total. While, as you know it, the final tower only has three. But it was a first idea, relatively early in the process. I am not going to show you how to solve this structure, because it is part of the exercise which goes with this video series. I wish you a lot of pleasure thinking about how the internal forces vary and how they are transmitted from the left to the right. In this lecture about towers and cantilevers, we have seen what kind of loads act on these types of structures, as well as how to obtain the internal forces. We have seen that the shape of the structure can result from the loads which are applied to it.