Hello, in this video, I am going to introduce the superposition principle. This principle, which I am going to formulate, will enable us to quite simply obtain the effect of several causes by a simple addition. In this video, you can recognize the truss configuration with a cantilever, with which we have just dealt with, on which I add 2 loads, which was a configuration that we have often seen before in the trusses, with a load of 10 Newtons on the left and a load of 20 N in the middle, I mean... on the right but it has become the middle. Since we have already previously calculated the effet of both these load cases, either a load of 10 N on the cantilever, either a load of 10 and a load of 20 on the truss, we would wish to directly be able to obtain the combined effect of these two load groups. The superposition principle tells us that in linear elastic systems, the effect of the sum of two actions is equal to the sum of the effects of both actions taken separately. To be more clear, in our case, the internal forces induced by the loads "1", I use quotation marks, that is a group of loads, I can have one load or several loads, plus the internal forces due to the loads of a second group of loads which I call "2", are equal to the internal forces induced by the loads "1" + "2". That is to say that if I know the internal forces, in my case, induced in the structure by a load at the end of the cantilever, and on the other hand, the internal forces due to the loads I have put of 10 and of 20 N, on the structure itself, therefore, the internal forces in the structure due to the combination of the loads of the group "1" and of the group "2" will simply be obtained by addition. I have put in the statement that this principle is valid in linear elastic systems, which is perfectly correct. In our case, we will be able to apply it in all the cases which interest us. We have here these two load cases which we have previously calculated, we are simply going to apply this principle, that is to say that we are going to add the effect here, I start by bar A-B. We have, for the first load case, a compression of 14.4, which is negative, in the second case, a tension of 2.9. The sum is going to remain negative and will be equal to -11.5 N. I keep going with bar B-D. This bar is compression in the first case, in tension in the second one, but the tension is smaller so the result remains negative, a compression of -5.8 N. And then I keep going, for example, on the bottom. We had 7.2 minus 1.4, that remains in tension, so that is equal to 5.8 N. Here, 10.1 minus 4.3, that is also equal to 5.8 N. Here, that is interesting for the diagonals. Here we have 2.9, positive, minus 2.9, negative, so the sum is 0. Likewise, -2.9 plus 2.9, that is also equal to 0. The big diagonal here is the only one to be in compression in both cases, so it is obviously going to be in compression, with a slightly larger compression of -23.1 N. Concerning the cantilever, nothing changes since it was not existing in the previous structure, or at least it did not have any internal forces, so we have 5.8 N and then a compression in the diagonal which is equal to -11.5 N. What is interesting is that with this method, we can see that we have been able to obtain, by a simple arithmetical addition, nothing complicated, the effect to add a load of 10 N. What is interesting is that we could also very easily decide to add, for example, a load of 20N, and what we would do, is that we would multiply all this by two, then we would add it to the first case and we would get results which would be different. Maybe there would be some changings in the signs of the internal forces, at least their amplitude would vary but the calculation would be done in a very simple way. What is interesting to notice when we look at this configuration, is that in almost all the bars, the absolute value of the internal forces decreases when we add the effect of the cantilever. Here, the internal force decreases, here, it decreases, here, it decreases, obviously, both zeros, it also decreases here. There are only these three bars... these two bars here, that is exactly the same thing than in the cantilever, which is logical, and there is only this bar here near support E in which the internal force increases,only slightly in the present instance, since it increases by about 3 N, but that is interesting to note that in almost all the structure, the internal forces are smaller. This property will be used in the next video when we will talk about the Gerber beams. We want to look at the application of this principle to something more complex, and for this, we first want to come back to the level of arch-cables. For the first truss, on the top, there are 2 loads, one of 30, one of 10, which act on this truss. We know this truss, and then we have an arch-cable which looks like this. In the second configuration, we have a cable in the upper part and then a lower arch, since there is a cantilever, a load of 20 N at the end of the cantilever, and the sum of both, we can see here both components drawn to the same scale, it gives us an arch-cable which has a particular configuration since the arch is sometimes above and sometimes below the cable, this does not cause any problems. We then obtain this kind of configuration. What is interesting is that the principles which we have previously seen for trusses with a variable depth remain valid, that is to say that when the cable is under the arch, that is the lower chord which is in tension, while the upper chord is in compression, we are going to have a reversal of these internal forces here, in the right part of the truss. Likewise here, of course, these diagonals will be in compression while this diagonal, for example, will be in tension. And then here, we will also have a diagonal in compression. We are going to see this again in details. We can simply superpose our arch-cables and obtain a combined arch-cable from which we can get the internal forces. Here we have the detailed results for these 3 structures with the superposition of the internal forces which is induced and we can see that, for example, for this element here, we have a tension of 27.1 in the first case and a compression of 7.3, which gives us a tension of 19.8, while the largest tension on a support can be found in the case of the cantilever where there is nothing on the left, so, 58.3 N of tension just over the right support. In this lecture, I have introduced to you the superposition principle which says that the effect of the sum of the loads is equal to the sum of the effects of these same loads. We can apply this principle to all kinds of structures, we have applied it here to trusses but we will be able to apply it to all the types of structures which we have seen until now in this course.
