Hello. In this video, we are going to proceed to the analysis of trusses with seven and nine nodes, and by extension, we will see how to proceed to the complete analysis of a truss. In this video, you can recognize the truss that we have seen till here, which has been increased to become a truss with seven nodes; it only had five nodes a bit before. There is a difference in the loading, which will be effective for the further structures which we are going to see, that is to say that the upper left node now carries a load of 30 N, and not anymore of 20 N as previously. The second node still receives a load of 10 N. In this video, we will check that the trusses which we are studying are statically determinate, we will present a systematic approach to proceed to their solving in all the configurations, we will see how to obtain the internal forces, and we will make some observations about the amplitude of these internal forces, comparing them for example to the arch-cable. First, let's check if this truss with seven nodes is statically determinate. We have a fixed support on the left, and then we are going to have two support reactions, we have a mobile support on the right, and thus a total of three support reactions. Let's now count the bars. We have one, two (counts till 11) 11 bars. Let's now count the nodes. One, two... (counts till seven) seven nodes. 2 x 7 + 14, which is also equal to 3 + 11, our structure is thus statically determinate. Fortunately, because we want to calculate it with the methods of graphical statics. Let's now proceed to its solving ; here, I have already obtained the support reactions RA of 21.7N and RG of 8.3N, using an auxiliary arch-cable. We are now going to proceed to the solving, starting by the node G, on the left. I identify this node... and three forces act on this node : the reaction in G, which is equal to 8.3N and which is drawn on the far-left ; -- it has been drawn in a staggered way -- I am going to take this axis here to make the rest of the drawing -- and afterwards, turning counterclockwise, the internal force in F-G first acts on this node, and then the internal force in E-G which brings me back to the beginning... So, this, it is the internal force in F-G, and this in E-G. If we first copy FG, here, we can see that it is an internal force which is going to push on the free-body, so a compressive internal force. The internal force in E-G pulls on the free-body, it is therefore a tensile internal force. The contribution of the blue node to the Cremona diagram is indicated by this triangle. We now pass to the solving of the node F, which is a non-loaded node but we will see that it has however internal forces -- The internal force in F-G first acts on this node in the other direction, as well as the internal force in D-F, horizontally, and finally the internal force in E-F. The internal force in D-F pushes on the free-body, rightwards, so it is a compressive internal force. The internal force in E-F, however, pulls on the free-body; thus it is a tensile internal force. Here is the contribution of the node F to the Cremona diagram. It is a non-loaded node, but there are internal forces in all the bars. Let's now move to the node E, which is again a non-loaded node ; The internal force EG first acts on this node, in the other direction, then the internal force in E-F, also in the other direction, then the internal force in D-E, and finally the internal force in C-E -- we are going to go back to the beginning, I just stagger it a bit for it to remain visible. The internal force in D-E pushes on the free-body, it is therefore a compressive internal force. The internal force in C-E pulls on the free-body, it is thus a tensile internal force. Here is the contribution of the node E to the Cremona diagram, so this triangle, but, with, in addition, this small passing by the left part of the diagram. The node D now, a loaded node which is subjected to the internal force in D-E, in the other direction, then to the internal force in D-F, also in the other direction, to the load, here, of 10N, afterwards, to the internal force in B-D, which is horizontal, and finally to the internal force in C-D which is parallel to E-F. The internal force in B-D pushes on the free-body, it is therefore a compressive internal force, as well as the internal force in C-D, here. This free-body contributes in this way to the Cremona diagram. The node C, again a non-loaded node, which is subjected to the internal force CE in the other direction, then to the internal force CD, also in the other direction, then to the internal force BC, parallel to FG, and finally to the internal force in AC. Here again, I stagger it for us to be able to see it. The internal force in BC pulls on the free-body, it is thus a tensile internal force; as well as the internal force in A-C. And there is the contribution of this node to the system which is very small, but with these two very long elements, the internal forces in A-C and C-E. We now move to the node B, the last loaded node, which is subjected to the internal force in B-C, in the other direction, followed by the internal force in B-D, in the other direction, followed by the force of 30N, itself followed by the internal force in A-B, to come back to the beginning of BC. This internal force pushes on the free-body, so it is in compression. The contribution of this node to the Cremona diagram is composed of these two triangles Finally, we look at the node with the support A, which is subjected -- we can pass in this way -- to the support reaction RA which is equal to 21.7N, followed by the internal force in A-C in the other direction, followed by the internal force in A-B, in the other direction. To verify that the construction is correct, it is necessary to check that the segment A-B, which we had previously determined, is really parallel -- it is just a small error -- to the segment in the construction. The contribution of this node to the equilibrium of the whole system, is this grey triangle here. In this figure, you have the entire solution of this truss with 7 nodes, -- actually, entirely correct. Graphically, you can see, on the right, the Cremona diagram, and then, on the left part, we have, on the structure, indicated by colors, the type of internal force, but also the numerical values. Let's now move to a truss with nine nodes. Here, we have extended our model to add four additional bars and two additional nodes. Let's quickly check that this structure is statically determinate. We have a fixed support, so two potential components; one on the right, which give us three support conditions, to which I add one, two, three, (counts till 15) 15 bars, and I count the number of nodes : one, two, three... (counts till 9) 2 x 9 = 18, which is also equal to 3 + 15, therefore the truss is statically determinate. Let's quickly notice what happened when we moved from the truss with five nodes, which we knew was statically determinate, to the truss with seven nodes at first. Well, we added one, two, three, four bars ; so four, here, on the left part of the equation ; and then we added two nodes ; so, 2 x 2 = 4, it is logical that the structure remained statically determinate, and that when we moved from a truss with seven nodes to nine nodes, we have added again one, two, three, four bars and two nodes, so again, it is logical that our structure remained statically determinate. We could keep going, making a truss with 11 nodes, with 13 nodes and so forth, and the solving method would remain the same. I am not going to solve the equilibrium of this truss here but I gave it to you in an illustrative way, it is an another way to represent it, you can see : we have started again by the node which is on the far-right, I have called it A in this case, and I have drawn the Cremona diagram -- here you have, still in grey, the final diagram. I have drawn the contribution of this orange node to this first step. Afterwards, we moved to the node B, where we have discovered two new bars, which I drew in green since they have been discovered studying the node B, which is itself drawn in green ; here are their contributions, the node C has this contribution, with this return that we have already seen in the truss with 7 nodes ; and so forth for the whole structure. Here, on the bottom, the internal forces in each of the bars are represented ; so, again, under a load of 30 N, plus a load of 10 N. And here, we have this figure which is a bit bigger, on which we have represented, for the arch-cable, as well as for the truss, the amplitude of the internal forces using the width of the bars. We can notice that in the arch-cable, all the elements are essentially subjected to high amplitudes, they are quite thick. However, in the truss, we can notice that a certain number of bars, like these ones, are subjected to low internal forces compared to other bars which are subjected to much larger internal forces. What we can also notice is that there is a value which is absolutely identical in all the structures : the maximum tensile force at the bottom, which is constant in the arch-cable and has here a value of 31.3N. Other values, such as this one, 27.2 or even 18.9 are quite similar. The same phenomenon occurs for the compression. Here, we do not find exactly the same value : we have 45.1 and 42.3 but it is something which is quite similar anyway. Likewise, the compression in the upper part, has a value of 31.4N in the arch-cable, while it is 29.2N for the truss. What we can deduce from this is that the larges internal forces in the truss are similar to the ones in the arch-cable. Some of you have maybe been surprised by the fact that I consider this truss as a truss with seven nodes. This is true : we can see one node, (counts the nodes) seven nodes, but we can also see, here, an eighth node and a ninth node. Indeed, they are really nodes, the structure has really been built in this way. However, on this node 8, for example, except if there is a bird or someone who sits on it, the loads are quite unsignificant. So it is legitimate to simply ignore this bar and the node that belongs to it for the calculation. So, maybe you wonder why they built this bridge in this way. No, that was not unuseful, this vertical bar has two main functions. On the one hand, it is going to support this horizontal bar, which, otherwise, would be quite long and would risk, for example, to vibrate under the wind, or when there is some traffic. And on the other hand, it is going to transversally hold this bar, to prevent it to change its shape under the effect of compression. This phenomenon is called instability, we will talk about it later. And then, the purpose of this bar and of this node 8, is not to directly carry loads, but it has a function in the global system. Likewise, here, we have a truss which I called a truss with nine nodes, with (counts till nine) nine nodes, that is right, but actually, if we look at these vertical bars, we can see 10, 11, 12, 13 nodes. It is correct; and we could obviously calculate this truss as a truss with 13 nodes, it would simply be a little bit more complicated, but we have all the tools to do it. Why did they place this bar here ? Here, the purpose is not really to stabilize ; but it enables to reduce the internal forces in the deck of the bridge. Indeed, if I have a big truck, here, which is parked, or which simply passes on this part of the bridge, it corresponds to a very large load, and given the fact that in a truss, the loads can only be introduced into the nodes, well, this load will have to transit till these two nodes, 7 and 9, that is to say over a quite long distance, and this will induce higher internal forces inside the concrete deck. It is not necessary in the configuration which has been chosen, here, with this additional bar : when there is a truck here, the internal force goes up, by tension, till the node 8, and at the level of the node 8, the force of the truck is introduced. So the purpose of this bar is to carry a load which acts in this region here, and to bring it onto a node, which is closer, the node 8, especially because we can only reach it by tension. Therefore these nodes are not useless, we can make the calculation, or simply, as I did till now, we can introduce the load on the upper nodes of the truss. In this video, we have seen a systematic approach to solve all kind of statically determinate trusses, starting by a node which is a support for which we know the support reaction. Afterwards, solving the equilibrium node by node, we can obtain the internal forces in each of the bars of the truss. The internal forces, we have seen it, are similar to the ones of the arch-cable ; at least, the maximum internal forces. There are however certain bars in the truss which, sometimes, only carry small loads. We have observed that it is possible, in a truss, that a bar does not carry any loads ; it can have another function, for example the function of stabilizing the structure or of introducing a load onto a node.