Hello. The purpose of this video is to give you a short introduction about trusses. As we will see together, trusses are essentially structures derived from the arch-cable. If you remember well, the arch-cable is the last type of structures which we have seen in the course The Art of Structures I. It is the combination of an arch and of a cable which produces a type of structure with interesting particularities. The disadvantage of this structure, like for arches, is that it has problems of shape stability, that is to say that, depending on the loads, it must changes its shape to have a correct mode of resistance. First, since it is an extension of the arch-cable, I want to solve again a simple arch-cable so we have here, on the left, a fixed support, a mobile support on the right and two loads, one of 20N and one of 10N and I would like to have an arch-cable so the height will just be equal to the distance between the support and the lower end of the load of 20N. So this is the first segment of my arch-cable which has this orientation and since I am here, I can find the position of the resultant given the fact that this one is located on the extension of the line of action of this first segment. I am now coming to the Cremona diagram where I replace the orientation of the internal force on the left part of the arch-cable since I am going to need it. On the left part of the arch-cable, the internal force acts with this inclination and then it is added to the force of 20N. I first build here an auxiliary cable and I choose that the second segment of this auxiliary cable is horizontal from the upper end of the force of 20N. In the real system, it means that the segment, here, is horizontal. I have the right to do it because it is an auxiliary arch which is going to be used to determine the resultant. The last part of the auxiliary arch is determined by this shape on the Cremona diagram. This orientation, I draw it again, here, on the real structure. This last segment gives me the orientation of a second part of the auxiliary arch and the intersection of these two lines gives me a point on the line of action of the resultant. In the Cremona diagram, on the right, I draw the resultant in a staggered way, so we can see it well, its value is 30N. And in the real system, here, I have a resultant force, R, of 30N. I now end the auxiliary arch with a support for it to be in equilibrium. I now wish to determine the geometry of the arch-cable which passes by the left support, but also by the right support. We can see that we did not pass by the right support with this auxiliary arch. I know that the right segment of the arch must pass by the right support and by the intersection of the line of action of the resultant and of the one of the first segment. So I line up these three elements and then, it gives me the right part of my arch-cable. I take advantage of this situation to give names to the characteristic points The point A The point B The point that I just determined, C and, here, the point D. I now redraw the right segment of the arch-cable on the Cremona diagram which gives me the complete definition of the internal forces in this element. Here, I have the internal force A-B. We can erase it, we are going to use a dotted line. Then, the internal force here is the internal force in C-D while here, this is the internal force in A-B. The last segment of the arch, the segment BC, has this orientation - it also gives us these internal forces. And then this orientation here, in the real system, is necessarily parallel, at least if we have correctly made the graphical construction. This is the correct shape that the arch of the arch-cable must have for this combination of loads. It is an arch-cable so I now add the cable which acts between both supports : the left support A and the right support D. I am now going to check the equilibrium of my arch-cable node by node. I start by the node C. The force of 10N, and then, turning counterclockwise, the internal force in BC and the internal force in CD are acting on node C. The contribution of this node to the Cremona diagram thus corresponds to this triangle, here, which I indicate in blue. Afterwards, I want to look at node D where the right support is. On this node D, I know that the force C-D acts the other way. I had already drawn it before. Then, the internal force in A-D which is horizontal and leftwards. It a tensile internal force. Here. And then, finally, the vertical reaction RD is also acting on this node. I draw it again in a staggered way to see better what is happening. Here, RD which has a value of 12.5N. The contribution of the orange node to the Cremona diagram partially superimposes on the one of the node C. We are now going to look at the equilibrium of node B, which is loaded by a force of 20N. The force of 20N which is here, acts on this node. And then, the compressive internal force in A-B. The contribution of the node B to the Cremona diagram goes from there to there. And thus... I am now looking at node A which is also subjected to a support reaction. So, this node A is subjected to, counterclockwise, first the internal force in A-D in the other direction, then the internal force in AB and the reaction at the support which necessarily needs to be vertical for the system to be in equilibrium. Here, I have RA which is equal to 17.5N. So, RA is here, vertical upwards. Here, we thus found the internal forces in each of the elements of the arch-cable, as well as the reactions at the supports. This is a reminder. You have already done it within the framework of the course The Art of Structures I. If you have not done it before, study well this construction because this is a basic construction which we will use a lot and that we will develop in the following videos. In this video, you can see an arch-cable with two loads. When the loads are symmetrical, you can see that I can add a force, it remains perfectly stable. However, if I add a force on the left part only, the left node wants to go down and I cannot find an equilibrium position, only if I make it go up. This phenomenon of stability, which we have already well studied, is a problem with arch-cables. However, if I had a diagonal element, a chain in this case, I can see that I come back to the initial shape and that I can add a second load, or several loads while the structure remains stable. This is the interest of truss structures. So, what have we seen in the video ? We had an arch-cable here with three bars in compression and one in tension. And we have completed it. We have kept the three bars. We are going to see later, actually very soon, that they actually remained in compression. Then, the chain remained in tension and we have added a second chain. So on the left, we have an arch-cable. And on the right, we have a truss. The arch-cable has a variable shape since it must vary according to the loads for the arch to be stable. To the contrary, the truss has a fixed shape. So we will see in this case, depending on the loads, we can have a problem because this chain that I added can only carry tension, but in principle, the truss nevertheless keeps a fixed shape. There is a bit of vocabulary to add about trusses. When we talk about the upper part of a truss, we talk about the upper chord. In the lower part, quite naturally, is the lower chord. The inclined elements are called diagonals. And the vertical elements, here, are called posts. There are always diagonals in trusses. In some trusses, there are not posts. If, for example, you look at the illustration which is on the cover of this video, it is a truss with V-shaped diagonals without any posts. You have for example, here, a truss with V-shaped diagonals without any posts, this is the Palais de Beaubourg in Paris. We can recognize several trusses at each level, at each storey. Actually, this structure is supported by big trusses. We will see how to determine the internal forces in such a structure. And then we will also see how to determine the sign of the internal forces without proceeding to big calculations. In this video, we have seen that trusses are structures which are derived from the arch-cable. In addition to the properties of the arch-cable, they have the advantage not to have to change their shape to be able to resist various combinations of loads. The constituting elements of a truss are : the lower and upper chords, the diagonals and, if there are some, the posts.
