Hello. In this video, we continue to explore the solutions of typology for trusses with a constant depth. We are going to dedicate ourselves to trusses with X-shaped diagonals. You can see in this video how we pass from a truss with N-shaped diagonals in tension to a truss with X-shaped diagonals adding, in this case, four diagonals. On the final structure, the bars which were in tension are still in tension and the timber bars which have been added are in compression. This figure shows the same principle so we have a truss with N-shaped diagonals in compression on the left, on which we superimpose a truss with N-shaped diagonals in tension so if we put half of the loads onto each of both trusses, we obtain a complete truss on which all these loads act. We can already observe, because the results have been made for you, that the internal forces in the diagonals are smaller than before because now we have a diagonal in tension and a diagonal in compression; each time the internal forces are very small in the middle and even near the supports, they are significantly smaller. We are going to determine if this structure is statically determinate, se we have here two forces at the support, one here, so three in total to which we are going to add the number of bars, so let's count the bars: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40 and 41, so 3+41. And now, let's count the nodes: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 and 18. 2 x 18, so clearly, this is greater, that is equal to 44, which means 36+8, therefore our structure is eight times statically indeterminate. If you remember the previous video, that is logical, we had a truss with N-shaped diagonals, let's say in tension, and then we have added 1, 2, 3, 4, 5, 6, 7, 8 additional bars, diagonals in compression, to our solution which was initially statically determinate, it is now eight times statically indeterminate, which means that we will not be able to calculate this structure within the framework of this course but it does not mean that we cannot understand how it works and you can notice that the diagonals which have the same inclination than the arch-cable are still in compression, the ones which are in the other direction are still in tension and then the upper and lower chords have the largest internal force when the arch is the furthest from the cable. So the rules which we have established together essentially remain valid. Here, we have the example of a large building from the 19th century, and we can notice, particularly in the upper part, that they used very simple elements. They are probably flatirons, I am drawing their cross-section, that is an element which is just a flat steel bar. Why is it possible in this configuration ? Well, because the internal forces are small, there is a small compression in this direction, a small tension in this direction that is why the internal forces can be supported by such simple elements. We can notice here, where there are larger loads, that the elements are a bit more complicated Here, I have another example, this is not exactly a constant depth structure though we can clearly recognize the elements which are here in tension, which are, again, constituted by flatirons, and then the elements in compression which are constituted by two angles, that is to say two L-shaped elements. Obviously in this direction but the internal forces being quite small, we can notice that this structure is very transparent, that is an advantage of the structures with X-shaped diagonals. Since the structure is statically indeterminate, we can imagine taking off here all the posts. We have taken off 1, 2, 3, 4, 5, 6, 7 posts. We can imagine that the structure is thus one time statically indeterminate. We are going to check it anyway. So we have here two forces at the support, here one, so three support reactions in total. Plus 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34 bars and then we still have 2 x 18 nodes so it is larger but we have 37 which is equal to 36 + 1 so this structure is only one time statically indeterminate. We still cannot calculate it but we are closer and we can understand well how it works. You can notice that, one more time, the diagonals which are in compression are the ones which have the shape of the arch of the arch-cable. Here, we have a type of construction for which this type of truss is often used. That is for a temporary installation on a construction site. These elements will be used as support for the setting up of the concrete before it hardens. Here, we have a particular case with which the Wright brothers, who are the pioneers of aviation as you may know, have been confronted. It was necessary to create a wing which was sufficiently stiff to resist to the weight of the machine and of its pilot since it is the wings which carry the machine and also, on the other hand, which must remain sufficiently aerodynamic. So in this case, we can clearly identify posts, there are two rows, there are the posts of the front and of the back, which are in compression, I do not draw them all but I draw some and it was necessary to be able to have a system in which... so I am going to draw the air pressure, the air pressure, it acts in this way upwards, not only on this wing, also obviously on the wing of the top, so we have loads acting upwards, and then here, we have the weight of the pilot plus the one of the machine, that is essentially the engine which is heavy in this structure. And we would quite naturally have, tensile internal forces in certain diagonals and compressive internal forces in others. What has been done here, that is the diagonals, if we come back here, these diagonals should normally be in compression, these diagonals, here, should be in compression. So, what has been done here is that they used cables which were stretched before. Since they were stretched, they were then in tension and then when the compression arrived actually, they were just less tensioned since a tension plus a compression is equal to a smaller tension if the tension is large enough. And thanks to this, the internal force still remains a tension, which is a big advantage. So this is what we call prestressing, which is the preliminary tensioning, that is to say before putting the loads, of a cable element for it to remain in tension even if it is subjected to a compressive internal force. So we will have the internal force in the bar which will be equal to the internal force of prestressing which is positive, plus the compressive internal force and if the internal force of prestressing which we introduced before is larger than the compressive internal force which has just been added thus the internal force in the bar will remain in tension. This what the Wright brothers did, which enabled them to have, here, very thin diagonals since they are only constituted by cables and thus the structure is extremely light and aerodynamic since the engine was not really powerful either. In the next videos, we are going to continue the exploration of the typology of trusses with a constant depth.