Hello, in this video, we are going to dedicate ourselves to the specific analysis. The specific analysis will enable us to directly obtain the internal force in a certain number of bars in which we are particularly interested in. These bars, we have identified them in the previous video because they are the ones in which the internal forces are maximum. It will also enable us to do so whithout being obliged to calculate all the internal forces as we have done it till now. Here, we have a structure which we know well, which is our truss with 9 nodes, we already know by the analysis of the maximum internal forces that the largest internal force will be located in this chord because its depth with respect to the cable is the deepest compared to this one. How are we going to proceed ? We are first going to identify a node which is the node by which the largest internal force passes, corresponding to a value of 31.4N in the arch-cable. We are going to make the arch-cable pass by this node. It is very important: if the arch-cable did not pass by this node we could not proceed in this way. Then, we are going to isolate the internal forces in the bars which interest us by means of a free-body. There are three bars which we cut: There is the upper chord where the internal force will be maximal, the lower chord where the internal force will also be maximal and then this diagonal which does not interest us a priori but we are going to obtain it anyway in the process. When I cut this free-body, I obtain the following internal forces: I have the compressive internal force in the arch of 31.4 Newtons. And I have the tensile internal force in the lower chord which has a value of 31.3 Newtons. This internal force, I have already directly obtained it here: it is this internal force of 31.3N which is the maximum internal force in this bar. I have directly obtained it, simply thanks to this free-body. Unfortunately in the upper part, I cannot directly obtain this internal force because in fact, it is carried by two elements, the upper chord on the one hand and the diagonal on the other hand. But the good news is that these three elements coincide in this node, here I indicate them again in green. What does it mean ? It means that the internal force of 31.4N, we are going to be able to decompose it into a horizontal component in the upper chord and an inclined component along the diagonal. We are going to report these internal forces in the right part. 31.4 Newtons, it corresponds to 15.7cm on my scale. You certainly have another scale on your sheet. Whatever, we can do it with the scale we want. So, I copy this internal force of 31.4N It is a negative internal force, I draw it in blue so we can see it well and it is going to be decomposed in an horizontal component -- I have already drawn this horizontal line -- and a component parallel to the diagonal. These diagonals are all parallel so I can take any of them and I copy here and I have here a component in this direction. Now, I am going to see that the internal force which I am going to obtain here goes up rightwards. If I copy it here into the free-body, I have a compressive internal force in this diagonal. So, I can draw this internal force here in blue. And then an internal force in the upper chord which is horizontal. We can read the components of these internal forces in the upper chord: 29.2 Newtons and in the diagonal: 3.3N. So we have now obtained the internal force in the upper chord which has a value of 29.2N in compression, in the diagonal which has a value of 3.3N in compression and in the upper chord which has a value of 31.3N. So we have a method here which is very efficient to directly obtain the internal force in a truss element. Obviously, we could have proceeded to another decomposition, still using the same node in this case and then another free-body. With this free-body, we actually have the same internal forces than before: 31.4N in compression and then 31.3N in tension and we can decompose them again, the result will exactly be the same except that now, we draw with vectors oriented leftwards. So here, we have 31.4N A decomposition according to the horizontal, parallel to the upper chord and according to the diagonal. Again, we will exactly get the same construction, simply in the other direction, so here, with 3.3 Newtons and here with 29.2 Newtons. You can check these values, they are the ones we obtained in the lecture in which we solved this truss with 9 nodes. Let's now look at the example of another structure. This one, we have not calculated so far, it is slightly different, it is a truss which interests us. We can first see that the largest depth between the arch and the cable in the middle of the bar is going to be here for the lower chord and here for the upper chord where the internal force will be the largest. So, we will have to make the arch-cable pass by this point. We use the applet i-Cremona. Here, I have inserted this image, I have inserted two loads and I have already made pass the arch-cable by this place here and I can read here that the internal force has a value of 296 with an angle of 5.2 degrees. The internal force at the bottom has a value of 295, horizontal of course. So here, the internal force is equal to 296 with 5.2° and here the internal force is equal to 295 Newtons. Again in this free-body, we directly get the internal force in the lower chord since we cut it, that is the only one. Here, however, we cut at the same time the upper chord and this diagonal with this inclined internal force of 296N. I am directly going to draw it to the scale 1:1. Here I have 296 in compression. And then, there will be a component along the diagonal here which I copy here. And this component is going to be in this direction. If we copy it here on the free-body, this component pulls on the free-body, which means that this internal force is a tensile internal force and thus, this bar here, this diagonal, is in tension. Afterwards here, in this direction, we have the compressive internal force in the upper chord. This internal force here has a value of 315 Newtons in compression and this internal force here is equal to 35 Newtons in tension. You can notice that the operation which we are doing here is not an operation like the ones we did before. The sum of these two vectors, the sum of these vectors of 35N and of 315N is the 296N, so we do not have a vectorial sum which is equal to zero. Here, we have a decomposition of the internal forces in the bars. You can note that compared to the work we have done in the Cremona diagrams, here, there is no special convention of "rotation". I use quotation marks because we can turn, of course, in the way we want but we could have decomposed this internal force of 296N -- I am maybe going to quickly do it here -- into a component oriented in this direction and a component in this direction. So we would have obtained exactly the same result and both these solutions are equivalent. We have then already made this construction here. I propose you to make the constructions in these three other configurations as exercise, but for you to be able to do it in a simpler way, I am going to give you the internal forces in the various elements by means of the i-Cremona applet. For the solution number 2, we have an internal force here of 312N and an internal force here of 311N. In the configuration number 3, the maximum internal force is in this bar and it has a value of 286N with an angle of -5.2° and here 285N on the horizontal. For the configuration number 4, the internal force in the upper part is 323N in compression with an angle of -4.4° and here of 323N. I copy these internal forces. Here, we had an internal force of -311N while here we had 312N and an angle of 4.9°. Here, we had -286N with an angle of -5.2° and a tension of 285N. And finally, in the configuration number 4, we had an internal force of -323N with an angle of -4.4° and the tension has a value of 323N. So you can redraw it, I maybe just give you a small example here using the configuration which is here and especially, copying the angle which you have on the drawing, you can copy, for example here, this internal force of 312N. So, afterwards, the point for which we are going to do these operations is still the point which is here. But it is now necessary to identify the proper free-body to obtain the internal force where we want it and then afterwards, to proceed to the decomposition of these internal forces to obtain the internal force in both chords and in a diagonal into the bargain. In this video about the specific analysis, we have seen a simple method to directly obtain the internal forces which act around a node. So, we choose a node by which we make pass an arch-cable, we use an appropriate free-body which cuts the elements, the bars for which we wish to obtain the internal force. Then, once we have got the internal forces, if there are internal forces which are taken by several bars, we decompose these internal forces. So, we find both the internal forces in the considered bars, whose sum will be equal to the internal force, either in the arch, or in the cable. Generally, this internal force has an inclination. Obviously, this method will enable us to quickly proceed to the pre-dimensioning, since we will have directly obtained, without too much work, the maximum internal forces in a truss element.