Hello. In this lecture, we are going to solve our first truss, a truss with four nodes. As we have seen it in the previous lecture, it is an arch-cable to which we have added an additional cable. In the first course, we have seen that this is a method to stabilize the arch. It is indeed efficient but we did not calculate the internal force which acts in this cable. Now, we are going to see how to solve this structure proceeding node by node. We will ask the question : "By which node should we start ?" And then, we will obtain the internal forces in each of the bars of our truss. We have here a truss with four nodes and we can wonder : "By which node can we start ?" If I look at the node at the bottom on the left, I can see that it is not possible to solve it because I have here one, two, three and four unknowns, because the support force has two components, we have here five unknowns. However, the methods that we have seen do not enable us to solve a system with more than two unknowns. If I look at the right support, the situation is a bit better since I have one, two, and the vertical force, three unknowns. Unfortunately, I cannot solve a system with three unknowns, only a system with two. If I look at this node, I have a known force but one, two, three unknowns which are the internal forces in the bars. It is not going to be possible. However, if I look at this node here, we can see that we have one known force and two unknowns. It is thus by this node that we are going to start the analysis of the structure. We have here a structural scheme for this structure and we are going to start by the subsystem which I have just indicated to you, around node B. This node is subjected to a vertical force of 20N. Around this subsystem, we are going to turn counterclockwise, starting by what we know, that is to say the force of 20N. After this force, we are going to meet the internal force in the bar A-B. On the Cremona diagram, here on the right, we know that we are going to have an internal force which goes in this direction. To close the polygon of forces, we know that the internal force in B-C is going to be a horizontal internal force which is coming back to the beginning of the force of 20N. So, here. We thus have here the internal force in A-B and here the internal force in B-C. What is the sign of these internal forces ? We are going to copy these internal forces to the free-body. Here, the internal force in A-B comes and it pushes on the free-body. It is a compressive internal force, I am then going to draw this vector in blue. The whole bar A-B is in compression. The vector BC also pushes on the free-body. We have this component. Here, this is the internal force B-C. and here the internal force in A-B. We will be able to read these values to scale in the Cremona diagram. I am not going to do it now but you can do it with an accurate drawing at home. You will obtain the actual internal force in Newtons in each bar. The contribution of node B to the Cremona diagram is symbolized by this triangle. We can now move to node C because we had three unknowns but we just got one of those, the internal force in bar B-C. This node is subjected to a force of 10N, the internal force in B-C, the internal force in A-C and the internal force in C-D. I introduce the force of 10N. The internal force in B-C goes in the other direction and pushes towards the node. The internal force in A-C which I do not know, I am going to temporarily represent it with a black line. That will be enough. We are going to meet the internal force in C-D to come back to the top of the force of 10N. This is the internal force in A-C. And this, it is the internal force in C-D. In A-C, we can see that the internal force pulls on the free-body therefore this internal force is a tensile internal force. For C-D, the internal force pushes on the free-body, it is then a compressive internal force. The contribution of node B to the Cremona diagram is indicated by the orange zones. I am coloring bar A-C in red since it is in tension and bar C-D in blue because it is in compression. I am coloring bar B-C. We can now move to node D which is subjected to the vertical support reaction RD, because it is a mobile support. Turning counterclockwise, we are going to meet the internal force in C-D which we already know in the other direction, and the internal force in A-D to finish our polygon of forces. I have the internal force C-D in the other direction. We can see that the internal force A-D is going to be in tension because it pulls on the free-body going leftwards, here. And the vertical reaction at the support RD - I draw it slightly staggered to better see it but it should be superimposed - RD which is equal to 12.5N. You would see it if you draw it to scale. The contribution of the green node to the total equilibrium is indicated by this triangle. I am going to draw in red the bar A-D which is in tension. We can now move to node A which is subjected to the support reaction RA. I am not going to draw it because I do not know in which direction it is going to act but we are going to find it out soon. What are the internal forces acting on node A ? We are going to start by the internal force in A-D which is horizontal rightwards in the other direction, the internal force in A-C in the other direction, the internal force in A-B. To close the polygon of forces of the node A, we have the reaction in A which proves to be vertical and which has a value of 17.5N while RD was equal to 12.5N. The contribution of the node A to the equilibrium is indicated by the gray-colored zone, here. My drawing is quite complicated, that is why I made available for you the complete solution of the structure on the next slide, starting by node B with its contribution indicated by this orange triangle, then moving to node C with these two orange triangles, then to node D, this upper triangle and finally node A, the lower quadrilateral that we just made. You can also find the accurate internal forces in each of the elements of the structure. It is also possible to create a truss with four nodes by adding a timber diagonal, for example, instead of a chain, so an element which can at once resist to tension and compression. Let's see how we are going to solve this element. It is very similar to what we have done the last time. I am then going to do it quite quickly. We are going to start by node C because it is a node where there are only two unknowns. We know the force of 10N and this node is also subjected to the horizontal internal force in B-C which we are going to meet first, and to the inclined internal force in C-D. Coming back to the free-body, we can see that the internal force in B-C pushes on the free-body, as well as the internal force in C-D. So both internal forces, here, are in compression : the internal force in BC and the internal force in CD. I can draw these bars in blue to indicate that they are in compression. Here is the contribution of node C to the Cremona diagram. We can now move to node B since we know the internal force in B-C. We are going to use the internal force in B-C in the other direction. Then, we are going to add... Here is the force of 10N, I should have written it before. We are going to add the force of 20N. We are going to meet the internal force in bar A-B which acts in this direction. Then, we are going to meet the internal force in bar C-D which will bring us back to the beginning of the internal force in B-C. So here. Here, this is the internal force in A-B. Here, the internal force in B-D. If we copy these internal forces onto the free-body, we can see that A-B is in compression, as well as B-D. We can draw these vectors in blue AB and BD Likewise, here, on the actual system. And the contribution of node B to the Cremona diagram is this zone that I represent in blue. I can now solve node D on which a vertical support reaction RD acts but we do not know it for the moment, as well as two internal forces, C-D and B-D which we already know and the internal force in the lower chain, in all likelihood in tension, AD. We are going to have C-D in the other direction, BD in the other direction, AD in all likelihood horizontal in tension. We indeed see that if I copy this onto the free-body, AD is in tension. We can directly color this bar. The support reaction RD here, vertical, which is still equal to 12.5N. I remind you - we have seen that on the lecture about the obtaining of the forces at the supports - that whatever the arch-cable, the support reactions will be identical for a given combination of loads. It is not surprising that we get the same result for these two configurations. The contribution of node D to the Cremona diagram, we can see it here, is a quadrilateral which is a bit particular. We can now move to node A. We do not know yet RA. In all likelihood, it is going to be vertical. We are going to start by the internal force AD in the other direction, then the internal force AB in the other direction and then, to close the polygon of forces, we will need to have RA vertical. So, we are going to have RA equal to 17.5N like before. The contribution of node A to the Cremona diagram, we can see it here. On this graph, I have copied the results of the calculations of the three configurations which we have seen in these past two lectures. First, here, you have the arch-cable. Here, you have the truss with a diagonal made by a cable, a diagonal in tension. And then here, the truss with a diagonal in compression. Let's have a look at how to compare these various configurations. In the three configurations, we can first recognize an arch. In the case of the arch-cable, it takes a funicular shape. In the case of both trusses, it is a fixed shape, the one we have given to our truss. Note that the three configurations that we are comparing have the same height since the left bar always has the same inclination. If we get a closer look, the internal forces in the left bar 20N, 23N, 20N in compression are very similar. The internal force in the upper bar, 11.5N in the middle, 10.4N, is also similar. Here, it is only 5.8N but we can see that there is another bar in compression and both together reaches roughly 10N. If we look at the tensile internal forces, we first always have tension on the bottom of the structure. The internal forces, 10.1N, are identical in the configurations of the arc-cable and of the truss with a diagonal in compression. Here, it is 7.2, it is lower but we also have a diagonal with 5N. What we can thus say is that the order of magnitude and the signs of the internal forces are the same for the arch-cable and for both trusses. This is an important observation, it means that the system does not change in a dramatic way if we change its setup a bit. It also means that if you want to quickly estimate the internal forces in a future truss, a quick calculation simply using an arch-cable with the correct loads will already give you a very good approximation. I here have models of five configurations of trusses, always with the same distance between the supports and the same total height. We can recognize in all the cases an upper shape of arch with its timber elements. In the configurations presented on the top, we always have a cable at the bottom and then a diagonal in tension, or a câble at the bottom and two diagonals in tension, or else no cable at the bottom but two diagonals in tension. In the configurations presented at the bottom, we have a chain in tension and then a diagonal element which can be in compression or in tension. If we compare this configuration with this one, we can see that the diagonal, here, has been given to us as being in tension so this diagonal is going to be in compression. This diagonal, we have already calculated it before so we know that it is in tension. We can thus note a big variety in the possibilities of design of trusses. We can also wonder : "Can we calculate them all ?" If we look at this truss, here on the top, I cannot calculate it. I have two, three, four, five unknowns here. Here, I have one, two, three unknowns in addition to the force at the support. Even if I independently determined the force at the support, what I can do, I would not succeed to solve the structure because I have three bars. Here, likewise, I know one force and I have three bars. I know one force and I have three bars. Actually, we are not going to be able to calculate this type of trusses with the methods we have studied in this course. But how could we know if we can calculate or not a truss ? This will be the topic of the next video. In this lecture about our first truss with four nodes, we have seen that starting from a chain which was stabilizing the arch-cable, we have obtained a new structure which is called a truss. We have seen that the method of solving it to obtain the internal forces is similar to what we have seen till now, we have not introduced any new methods of calculation. Simply, we must start by a node where there are only two unknowns, for example two bars for which we do not know the internal force. We have seen that the diagonals in a truss can be in tension or in compression and that the order of magnitude of the internal forces is similar to the arch-cable. Finally, as we just saw, certain structures are not directly computable using the methods of this course. And the next video will enable you to understand the ones that we are able to calculate.
