Hello, in this video, we are going to see how to pass from an arch to a frame, a structure which has a shape which is different from the correct funicular structure corresponding to the loads. We will see this with the example of rectangular frames which will enable us to look at what is happening in a quite systematic and simpler way. We will see the importance of the position of hinges in an arch -- --or rather, in a frame with hinges. We will see how to solve a frame using the i-Cremona applet and I will set general principles for the solving of frames, especially to obtain the internal forces which act inside frames. We thus have here our reinforced concrete frame bridge whose pressure line moves away from the shape of the intrados. What we remember, at least for those who followed the course "The Art of Structures I", is that if we push on this side here and on this side there onto an arch, it makes the pressure line of the arch rise, so we will be able to make the shape of the arch rises into the intrados, in the part where there is some material, where there is some concrete. What we have also seen in the course "The Art of Structures I", is that when we have such a configuration, where the resultant of the pressure line gets out of the material, it is possible to obtain this resultant if we add at the same time tension in the cross-section. We will see thereafter that indeed, there is compression and tension which act at the same time in the cross-sections of frames. Not everywhere, but in lots of places. Let's take a simpler example, a rectangular frame with only one load in the middle. The pressure line corresponding to this load directly goes towards the supports. That is a frame with three hinges or an arch with three hinges, whatever how you want to call it, for the moment that still does not matter. The pressure line is clearly defined. As an aside, that is never bad to call it back to mind, the support reactions are inclined and correspond to this direct line which goes from the load towards the supports. We have a reaction on the left and on the right which goes in the direction of this pressure line. We can obviously notice that this pressure line cannot pass through the air, it must pass through the material. To that end, we are going to make act compression in this diagonal here and in this diagonal here. This compression is going to push on this arch. When I am going to push on this arch, it is going to go up towards the arrow in such a way that once I will have pushed enough on this arch, it is going to be inside the material, and at this moment, we will have a frame; we have not seen everything yet concerning the frame, but we will already have a frame, the arch part of the frame. There you go! We are really glad, the arch, which was outside the material, moved inside the material. Now, the problem is that we have here this compression. I am going to draw an arrow at the end of this compression because it is necessary to know, here, this compression pushes on the arch, no problem, it is at equilibrium. However, on the other side, on what does it lean? It is going to lean on a cable. We are going to fix a cable to the left support, which is going to run up to the corner of the frame, it is going to be deviated and then it is going to pass through the hinge since all internal forces must pass through the hinge, there are no possible exceptions. And then, that is the same thing in the right part of the frame. So, there you go, we have here the corresponding internal forces in a frame which are very different from the initial internal forces since we have pushed on the arch to make it get inside the material but to insure equilibrium of this diagonal element which has been added into the structure, we had to add these elements in tension. That is qualitative, but not only, it is really like this that the internal forces are distributed inside a frame. It is possible to model this using the applet using actually two applets. I first take the left applet, in which I make act a force, let's say of 10 Newtons, as usual, and if I ask for the resolution by means of the polygon of forces, we can see that it is just there. I am going to use the mode with the red ball to force the internal forces to pass just by this intersection. Now, I am going to add on the left and on the right, the force which corresponds to the internal force in the diagonal, so I need to make it rotate. I hold CTRL, in this way it keeps the same size, it is going to be important that these two forces have the same size. Likewise here, on the right, and then, I make them pass exactly by the diagonal and now we can see that we start having an arch which gets closer to the material. So, I select these two forces, on the left and on the right, I increase their amplitude, and we can see that at the same time, we obtain the passage of the arch just inside the frame. You should note that the force here is very large, I had put a force of 10, and now, I have a force in the diagonal of almost 6 times more than this force: 57. Now, we want to move to the right part. Here, I am not going to put the load of 10 Newtons but it is already carried by this system of arch. What are the loads which act in the right part? There is this force of 57, with an angle of 45 degrees, here, and of 135 on the other side, I carefully place them in the angles, then, I ask for the resolution, unavoidably passing by here. We can see that it is already solved. For this force of 57, we are going to have internal forces of around 40 in the columns and in the cross beam. Here, in the frame, the compressive internal force was around 45 in the column and 47 in the cross beam. The internal forces are much larger with this framed solution than with, let's say, a classic arch-shaped solution. But we managed to get the right shape. We do the same resolution for the case with a hinge which is not placed anymore in the middle of the cross-section, but on the top. That is a bit particular, because I have a very thick cross-section, if it was not very thick, the difference would be small, but here, we will have a significant difference. Of course, the initial resolution is very similar to the one we had before. And we add forces in the diagonals, so -45 here and -135 on the other side. We carefully position these forces, we select both. We can see that we can make the arch pass with a slightly smaller force, 28.14, actually quite smaller, I remind you that we had 57 before. Then, we are going to introduce in the second applet, forces which are equal to 28, with an angle of 45, and then, of 135 here. We are going to ask for the resolution and we can make the internal forces pass just through this hinge. Here, the tensile internal force is smaller of about 20 in the cross beam and on the right post. And the internal forces were 25 and 23 in the right post, respectively in the cross beam. We can see that the position of the hinge potentially leads to significant differences in the internal forces inside the arch. Here, we have a construction of a frame with a horizontal cross beam with the elements we are used to. Again, I solve it with both applets. Here, I have not placed the supports yet, so I place the supports. In the left part, I have a load of 10 Newtons which act just here. Then, I have a pressure line which is obviously going to pass by this intersection. Then, I add forces in the diagonal. So here, I make it rotate, not exactly 45 degrees, I roughly estimate the inclination of this diagonal, I select both diagonals and I carefully come to pull on the end until the internal force is centered. The internal force in the post is approximately equal to 23 Newtons, in the cross beam too. And the internal force in this diagonal is equal to 22 Newtons. I am going to introduce these 22 Newtons here in the right applet making the force here rotate, 22 Newtons, here 22 Newtons, I need to make it rotate more. And I activate the resolution -- I have forgotten to place the supports, sorry-- So, I place both supports. And now, I activate the resolution and I can see that I again have the internal forces which correspond to about 16 in the post and 16 and half in the cross beam, that is the same. We can also solve such a case, because we can see that it is a triangulated structure, like a truss. I am going to do it very quickly, because it may be useful for you, it can be useful to know how to quickly calculate a truss. I am going to use the fact that it behaves like an arch with 3 hinges to draw the pressure line which would correspond to the one of such an arch. And I remind you what I said before: the support reaction goes in this direction. If now I come into the Cremona diagram, we can see that we have here this load of 10 Newtons and then, turning, if I take a global free-body, turning counterclockwise around this global free-body, I will first meet the left reaction then the right reaction. Now that I have these reactions, we are maybe going to start the resolution of the truss by the left node, here. So, we have this node 1 on which act the reaction RG. Then, when we keep going counterclockwise, we are going to have the internal force in 1-3 which pushes on the node. So, this element here is in compression. Then the internal force in 1-2 which pulls on the node. So, the element 1-2 here is in tension. The contribution of the node 1 to the global equilibrium, respectively to the Cremona diagram, we can find it here. We can now move to the node 3, for example, we would actually have the choice between nodes 2 and 3. If we take the node 3, then we are going to have the internal force in 1-3 in the other direction. Then, the internal force in 3-4 which pushes on the node, thus it is in compression, here, 3-4. Then, the internal force in 2-3 which also pushes on the node, so which is also in compression. But we knew that there was compression here, in this diagonal. The contribution of node 3 to the global equilibrium can be read here in the Cremona diagram. We now move to node 2. This node is subjected to the internal force in 1-2 in the other direction, then to the internal force in 2-3 in the other direction, then to the internal force in 2-4 which is horizontal. So the element 2-4 is in tension. I let you solve the other part of the diagram. I am just going to indicate to you the colors since it is symmetrical, that is not very difficult. Here, we have compression. They are elements of the arch. And here, we have tension. They are elements of the cable. So, likewise, here, we have both these elements in tension and these 3 other elements in compression. So you can see that it is absolutely possible to solve such a frame by the classical method which we know for trusses-- Ha, I did not indicate yet, I need to do it, the contribution of node 2 to the global equilibrium which is obviously here. We can notice that there are really large internal forces in frames. Here, you have the applied load of 10 Newtons and you can see that the support reaction on the left and on the right are slightly smaller than 10 Newtons, but all the other internal forces inside the frame, whether it be tensile or compressive internal forces, are quite larger, that is also what we had seen with the applet. We had an amplification coefficient of almost 6. That is something we need to take into account. If we choose to build a frame, then the internal forces will be larger. This means that more material will be needed, more reinforcement in the concrete, more steel. Here, in this video, I have a frame with a horizontal cross beam, but that is the lower part of the cross beam which is horizontal. That is a frame which I could have used for the calculation, but since it turns out that the internal forces are actually very large in this frame, I thought that it was more useful to solve the one I have just shown you before. I would like to state four broad principles pertaining to frames. The first principle is that we are going to add internal forces-- internal forces to have the desired shape. For the moment, I took very simple shapes like that of a rectangle, but we can have more complex shapes and any kinds of shapes of this type can be obtained, simply, you need to know that we are going to add internal forces and we have seen that they were quite significant. Then, we distribute the loads between the arch system and the cable system. In the system I was showing until now, the arch system was carrying the whole load and you remember, the cable system was not carrying anything, it was just taking the internal forces which we had added, but we could have a system where, on the opposite, the internal forces are carried by the cable system and the arch system must carry the induced internal forces where the loads distribute between both contributions. Third, the internal forces are quite significant. I have already said it, but you really need to remember. The choice of a solution with frames is not necessarily very economical. To make a truss structure, for example, is more efficient. And finally, we have seen it but we did not talk that much about it, we have tension and compression in the entire structure. There will be places where there will only be tension or only compression, but generally, in a cross-section, we will have at once tension and compression. That is very similar to what we have for beams. Finally, I am going to deal here with this large example which was built by Pier Luigi Nervi. So, that is a large bridge which should ideally be an arch, at least if we consider the applied loads. This is the arch which would correspond to a uniformly distributed load. The shape which has been given to it is much more turned up. You can see here the outer side of the bridge, and here a cross-section of the bridge. The bridge is hollow, there is obviously material in the upper part, that is where the vehicles can run, and in the lower part, the intrados. And we can see here on the left and on the right, very inclined blades made of concrete, which really have a very important function for this structure. Indeed, these blades are going to introduce the compressive internal force here onto the arch on the left and on the right. And that is going to drastically change the way this structure works. Let's look at this using two applets as we have already did before. Here, on the left, we have an arch system. One more time, the arch carries all the loads, so I place here the hinge, that is the shape of the arch. All the loads are taken here. In the right part, we are only going to take the internal forces. We are going to introduce loads, here, which are inclined. I already make them a bit larger because I know they are going to be. Then, I am going to give it an inclination. I make it rotate. Once it will have rotated, I can put it again... there you go. I am going to place it a bit in the middle because, you are going to see, these loads are going to become quite large. I place a second force which is equal to it, but I make it rotate in the other direction for it to be lined up with the concrete blade of the right part. Here, that is not too bad. I am going to bring back this load here, in the middle. You remember that only the line of action matters. I can move this load on its line of action. By the way, you can see that it does not change anything for the applet. Now, I select these two forces and I make them grow larger. You can see that the arch very nicely changes its shape to snugly fit the shape of the intrados. Now, here we had ten loads of 1600. So, it was 16 000, and you can see that the force in the diagonal is approximately equal to the double, that is to say 31 500. And the internal forces in the cross-beam are in the range of 38 000. So we have internal forces larger than the applied load. In the right part, we are going to introduce this internal force of 31 500. So we are going to insert this internal force here on the left and on the right. We are going to increase a bit the scale for it to look a bit bigger. I have a force on the left, a force on the right. Now, I am going to incline them without varying their length. So, I push on the CTRL button of the keyboard to make them line up in the direction of the small concrete elements. Afterwards, I can make the cable pass, the structure in tension, through the hinge and we can see that we have elements in tension with also internal forces in the range of 13 000 here and of 24 000 in the central part, that is a bit smaller than in the diagonal elements but still very large. Here, I am going to copy the internal forces. We have seen that this element here is in compression at the level of the intrados. The diagonal element is also in compression. And the cross-beam is in tension in the upper part. We are going to materialize this tension using prestressing cables over the whole length of the bridge. There will be a significant amount of these cables, but that is not particularly a problem. And then, finally, we are going to have the equilibrium of the tension in the post, in the back part. In this video, we have seen principles for the resolution of frames which consist in adding internal forces to change the shape of the initial arch or cable and to make it enter into the shape which we wish to have. The consequence of this is that there are additional internal forces in frames. These internal forces are sometimes very significant. We have also seen how a frame can be solved using the applet i-Cremona.
