Hello! In this video, we will talk about frames and hinges So far, the frames which we have seen had three hinges, in this case, the pressure line is uniquely determined and it passes through the three hinges. We will see what happens when we have an arch with two hinges, or only one or even without any hinges. We will see that in these cases too, we can determine the type of internal forces which act within the frame. Finally, we will talk about a special type of cantilever frame which we will discover together. We have here a frame with three hinges. So in this configuration, the shape of the pressure line is totally determined since it must pass through the three hinges and change direction at the level of the line of action of the force. If we had several forces, it should change its shape several times but it would have to necessarily pass by the three red dots. We have not exactly determined our structure yet since we are going to have to push on the structure here and to pull on the structure there to make the pressure line enter inside the material. So placing a compressive internal force here, we manage to make the internal force come back inside material then here, on the right, we are going to make the internal force come back inside the material by applying a tensile internal force here. Then obviously, to carry the compressive internal force here, it will be necessary to have a tensile internal force here, likewise to support the tensile internal force in the right part. We are going to have this kind of internal forces which act in a frame with three hinges and an asymmetrical load. That is what we have represented in this picture. So I have put a very large load to be sure that everything is tensioned and what we can see is that we have here a diagonal element in compression, here a diagonal element in tension and then the effect of this diagonal element is to make the arch rise up in the left part and to make it go down in the right part, compared to the logical position of the pressure line. And we thus obtain a structure which has the shape which we have seen just before. Here we can see, because it is a structure which cannot lie, we can really see where we have tensile and compressive internal forces. Of course, it is a truss, we could calculate it with the calculation methods which we know since we have three hinges, that is a statically determinate structure. Thereafter, we are going to have structures with variable numbers of hinges. Here, we have three hinges. Here, we have two hinges at the base. Here we have one hinge and finally on the bottom, we have not any hinges. So three hinges, two hinges, one hinge and without any hinges. On all these structures, we are going to make act only one vertical load : Q in the middle of the cross beam, the cross beam is the horizontal part of the frame, while the vertical part is called the post. So simply, a vertical load. Under the effect of this load, I am going to have a pressure line in this arch with three hinges which is well determined since it must pass through the three hinges. Then, as we have previously seen, I am going to support this arch by means of these diagonals in compression in such a way that the arch is going to rise up and get inside the material, in the post as well as in the cross beam. And then, to equilibrate these compressive internal forces in the diagonals, I am going to introduce two tensile internal forces, one in the post and the other one in the cross beam, which are going to ensure that our structure is in equilibrium. In the case of the arch with two hinges on the right, the structure is statically indeterminate, so we cannot exactly calculate the internal forces within the framework of this course. But we however know that the pressure line must pass through these two hinges, and then it is going to pass somewhere, on a point, let's say here, in the part above the cross beam. How is the structure going to behave? Well, in a very similar way than what we have seen before. That is to say that we have a pressure line which is inside, so we must push on this pressure line by means of a diagonal in compression to make the pressure line enter into the posts. Then in the cross-beam, the pressure line is going to start inside, on the intrados. And then it is going to go up until the outer part to end up on the same side than the initial pressure line. For the structure to be in equilibrium, again, we insert a tensile internal force in the posts, and then in the cross beam, we are going to have tension which goes towards the extrados, in the corners, and which goes towards the intrados, inside, at mid-span. Finally, to reach the equilibrium here, we have a small element in compression. For the arch with one hinge, there, we simply know that the internal force is going to pass through this hinge then afterwards, it is going to get out. So, I am going to introduce a point at a certain distance, I am going to make it symmetrical, in this way, my drawing will be more correct. A point at a certain distance, but we cannot determine it. Those of you who will attend more advanced structures courses will definitely see methods to accurately calculate the internal forces in frames. We can understand anyway what happens without proceeding to an accurate calculation, so here, that is not the result of a precise calculation. Here, I need anyway a diagonal in compression to make the internal force rise up inside the material and then, we can see that the internal force in the post will be inside, on the intrados, in the upper part, and then it is gong to go outwards, towards the extrados, in the part near the supports. And then, in the cross beam however, it is the same thing than what we had here, we simply go up towards the hinge. Concerning tension, we need tension to equilibrate the diagonal in the corners. So we are going to have tension which is going to cross here the compression in the posts. And then, a tension exactly as we have seen above for the cross-beam. Finally, we have here a frame without any hinges. So, we are going to determine some points of passage. We are going to have a pressure line which is going to be outside, at the level of the supports, which is going to pass above the cross-beam and which is going to be outside, at the level of the right support. For the internal forces to be inside this structure, we must again insert a diagonal in compression. The internal force in the posts will be similar to what we have seen for the frame with one hinge, on the left, so, with compression which changes side. And then, at the level of the cross-beam, it is the same configuration than the one we have seen above. That is to say that the internal forces change side with an element in compression in the middle and tension which is opposite to compression. And likewise in the posts. So we can see that through a real understanding of what is the pressure line, even if we cannot entirely determine it, since there is only the case with three hinges, on the left here, which we can completely calculate on the basis of the methods of this course but understanding how the pressure line works, we can understand the sign of the internal forces in the various parts of this frame structure. What happens when we are confronted with asymmetrical loads? Asymmetrical loads could be placed, for example, on the left part of this structure, we only have its self-weight acting. And then, on the right part, a more significant load because it has been snowing and the snow has been blown. It only acts on the right part so we have the self-weight plus the snow on the right part. We still have, as we have seen it, three hinges here, how is this structure going to behave ? We are going to use this applet to see how this structure works. The supports have already been inserted, and the uniformly distributed load as well. We can activate the resolution and using the mode "point of passage", we force the pressure line to pass by this point here. What we can see is that obviously, we have a parabolic shape for the pressure line, and thus there will be a compressive internal force at the level of the intrados. The compressive internal force is maximum where the distance to the pressure line is maximum. And then we are going to have a maximum internal force here, and then a minimum internal force near the hinges. Now, what happens when we have a series of forces here? I select all these forces, and then I make them vary, increasing them. You can see that, when I increase the force in the right part, we are going to stop here, then in the right part, the pressure line rised up and the internal forces decreased, while in the left part, the pressure line went down, and thus the internal forces increased. That is a bit paradoxical but if we add forces on one side, the internal force increase on the other side of the structure. In this video, you can see what I called a cantilever frame, that is to say a structure which essentially looks like a crane, with a cantilever on the top, and then a kind of vertical element with tension and compression. You should note that the first thing I have done when I introduced this video, is to add a counterweight on the bottom of the crane to prevent it to overturn because of the load of 10 Newtons here. Very clearly, without needing a long explanation, that is clear that these elements are in compression, and then that the chain element is in tension. Actually, that is half a frame, which I represented here. You should note that because of the presence of this quadrilateral, this structure is not totally stable, it could deform in the way I made it move here. It was actually standing in a vertical position because I had tightened well this screw at the bottom. Let's concretely look at such a structure. It is very similar to what I have just shown you, but that is a real structure, that is a crane for a small construction site. If we first apply a load Q at the tip of the structure, what are we going to get as a type of internal force? Here, I am going to take advantage of this situation to repeat the things we have seen in different parts of this course until now. In the external part, that is a cantilever so we are going to have compression on the bottom of the cross-section, and tension on top of the cross-section. And then we are going to have diagonals in tension in this direction. This first part works like this, and obviously, the other diagonals are in compression. It works like a small truss. And then, we can see that we reach a larger structure, we are going to essentially have tension here. We are going to keep having a bit of bending in this part here but I am going to choose to ignore it because we are essentially going to have compression here, in this part of the structure. Here, tension is slightly deviated. So, to support this small deviation, we will need to have a bit of compression which is also going to be carried through the following part of the structure by a bit of bending. I keep looking at this compression here, this cable carries the vertical component of the internal force which is transmitted through here. We are going to also have an element in compression as well as two elements in compression, here, which are going to be used to deviate the horizontal internal force in this cable to make it go down towards the bottom of the structure. And then compression, will be deviated to go down into the mast of the structure. It should be noted that in this zone here, we have a large weigth W which is the counterweight of the crane. Sometimes, it is placed on top of the mast, here, it is placed atthe bottom. That is exactly what I personally placed to avoid the structure rotating, and becoming unstable. What happens when we have a load which is still called Q, but which is placed somewhere else on the crane? There are many crane elements which have not any internal forces. So, that is not that they exactly have not any internal forces, since, obviously, they keep having a weight. But, within the framework of this example, we only consider the part which is linked to the load. We are going to have tension in the diagonals and then in this cable, and then, compression in the diagonals in the lower chord. And then tension in the upper chord. And then, we are going to essentially have compression in this part of the crane. There is again compression here and there, as well as in the column. And then tension, of course, in the cable, from the ground, deviated by these elements in compression. But the remainder of the structure is not used. We can see that we have a structure which works in a quite different way depending on the position of the load on the crane. It seems to be quite natural by the way. In this video, we have seen what is the influence of the number of hinges on the position of the pressure line. We have seen that we can estimate the position of this pressure line when we cannot exactly calculate it, and that it enables us to determine the position of the internal forces and of the maximum internal forces, that is to say where the eccentricity between the pressure line and the structure is the largest. Finally, in the example of a crane, we have seen how a cantilever frame works.