Hello. In this lecture, after looked at for quite some time at trusses with a the constant depth, we are going to lokk at trusses with a variable depth. We will see what is the influence of the depth, which is no longer constant between the supports, on the internal forces in the chords and in the diagonals And then, we will investigate in detail some of the typologies of trusses with avariable depth. What we have previously seen, is that starting from the shape of the arch-cable and dividing it by its depth which was constant, of our cross-section, we could obtain the internal forces in the chords, so compression which had a parabolic shape. So, for the loads I have indicated here, that is to say uniformly distributed loads, we had an increase with the largest internal forces at mid-span and smaller internal forces near the supports. These are internal forces in the chords. You also remember that we had developped rules for the internal forces in the diagonals. The idea we have now is to try with a variable depth so here I make it vary in the form of a parabola which looks like the shape of the arch, except that I made this shape symmetrical. You can note that we could have done it with the same shape than the arch-cable. And we can guess that it will more or less give us constant internal forces, maybe a little bit larger that the ones I have just drawn here, approximately constant in both chords. Here, we have the model of a lenticular truss on which I add three loads of 10 Newtons. And we want to solve the equilibrium of this structure. First, because it is symmetrical, it is obvious that the support reactions on the left and on the right are both equal to 15 Newtons. Before going further, I number the nodes of this structure from one to seven. To gain time, here, on the right, I have already drawn in grey the directions of the bars which will interest us for the resolution. And then, I am going to start the resolution by the right node, node 7, which is subjected to the right support reaction of 15 Newtons, it is represented here. It is staggered in relation to the applied loads of 10 Newtons, but obviously, it should be superimposed. This free-body, turning counterclockwise, makes me first meet 6-7, so I turn here, then I meet bar 6-7, in which the internal force pushes on the free-body, I already knew it, so this internal force is in blue in bar 6-7, then I meet bar 5-7 to close the polygon of forces for this node. And here, this internal force is an internal force which pulls on the free-body, it is therefore a tensile internal force. The contribution of this free-body to the Cremona diagram is then this one. I now move to the resolution of node 6, which is subjected to the internal force in 6-7 in the other direction, to the load of 10 Newtons, then to the internal force in 4-6 which pushes on the node, it is therefore a compressive internal force, then to the internal force in 5-6 to close the polygon of forces, which is also a compressive internal force. The contribution of this free-body to the Cremona diagram is indicated here in orange. We now move to node 5, which is an unloaded node but not without internal forces, since it 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 pushes on the node, it is thus a compressive internal force; and finally, to the internal force in 3-5 which pulls on node, it is therefore a tensile internal force. The contribution of this node to the equilibrium is then 5-7, 5-6, 4-5 and 3-5, as indicated here. We now move to the resolution of node 4, which 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 the applied load of 10 Newtons, then to the internal force in 2-4 which pushes on the node therefore it is a compressive internal force; and finally to the internal force in 3-4 which also pushes on the node and which is thus, also, a compressive internal force. I indicate in green the contribution of this node to the Cremona diagram. Node 3 is also an unloaded node. Well, I guess you have already understood that the resolution is going to be symmetrical since the loads are symmetrical, but we are going to proceed with this resolution anyway. So, 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, then to the internal force in 2-3 which is a compressive internal force, and to the internal force in 1-3 which is a tensile internal force. I indicate in yellow the contribution of node 3 to the Cremona diagram. We can see that it is symmetrical to the node 5 which we have seen before. We now look at the equilibrium of node 2, which is subjected to the internal force in 2-3 in the other direction, to the internal force in 2-4 in the other direction, then to the load of 10 Newtons, and finally, to close the polygon of forces, to the internal force in 1-2. And then, we have this contribution here to the Cremona diagram. Finally, we look at node 1, on the far-left. This node is subjected to the internal force in 1-3 in the other direction, to the internal force in 1-2 in the other direction, and then to the support reaction of 15 Newtons on the left, which I draw again in a staggered way, but it should obviously be superimposed. Here is the contribution of this node to the global equilibrium expressed by the Cremona diagram. I had forgotten to precise that bar 1-2 is in compression but I think that everyone had seen it. We can see here an example of lenticular truss of the 19th century, with clearly, here, an upper chord in compression, a lower chord in tension, and then numerous diagonals, which are going to be in compression, at least, when the train load passes. Here, below, I have also sketched it. So we have a truss with a nearly constant compression in the upper chord, a nearly constant tension in the lower reinforcement. I am going to make some additional comments at the end of this resolution. Now that we have the internal forces in the various bars, what do we notice ? We can notice that the internal force in the chord in compression, here, is approximately always the same. These length are roughly constant. And if we look at the chords in tension, that is a bit less the case because the internal force in 3-5 is nevertheless smaller than the internal force in 5-7 or in 1-3 but we also have rather constant internal forces. So, we have roughly constant internal forces in the chords. Furthermore, we can see that the internal forces in the diagonals which are all here, are relatively smaller than the internal forces in the chords. They are quite small in the diagonals. We can also see that they are relatively constant so we could choose diagonals which have the same shape. On this model, you can see a lenticular truss with this time, loads which are applied to the upper part. So I propose you to do it with two loads of 15 Newtons to have the same sum than we had before, so with forces at the supports R1 = 15 Newtons and then R7 = 15 Newtons. Here I have set up the calculation, but I have not given to you the grey bars, I am not going to do the resolution this time but, you have the opportunity to do it for yourself to understand well how it works. You will see, it is quite similar to what we have done but there are nevertheless some small differences, differences which make it worth to do it as an independant exercise. And here, you have the example of a bridge in the United States, with an upper chord which is always in compression, a lower chord which is always in tension, and we cannot see them well, but diagonals here, they are actually very thin, that is why we cannot see them well, very thin diagonals which are in tension. And likewise, here I have the resolution of the truss with a constant internal force in the upper chord, or almost, and also in the lower chord, and then internal forces which are nearly constant and which are small, in the diagonals. So, one of the advantages of these parabolic-shaped trusses is to have constant internal forces. We can particularly see it here for this bridge on which we can see that the upper chord which is very thin, has a constant width over all its length. Here, you have a more recent example of lenticular truss, this is the tribune of the Formula 1 circuit of Shanghai in China. I do not have a lot of pictures, but we can clearly notice the lenticular shape of both these parts of the building, and from this picture of the inside, we can also clearly see that there are diagonals, then, we can also see posts, it is thus a truss which we could also calculate with the methods we have studied so far.
