Hello. In this video, we will see how to consider beams and trusses with cantilevers. In fact, the phenomena which occur are the same for beam or truss structures, that is why I decided to gather these two topics together and to deal with them after the subjects, respectively, of trusses and of beams, unlike what is done in the book. We will see how the fact to add a cantilever to a beam changes the effects inside the beam, we will see that this is particularly important for internal forces, but also for support reactions. In this video, you can see how I turn a truss which we have often calculated into a truss with a cantilever, simply applying a load of 10 N at the end of the cantilever. When we apply this load, or at least when we apply a load a little bit larger, because there are friction forces and the self-weight of the structure is at stake, when we apply a load in this way, the lower part of the structure, which is composed of chains, turns into compression, so that the structure becomes unstable. For it to be stable again, I add timber elements in the lower part, which can support a compressive internal force. In the case of a truss, to which I add a cantilever at the end of which acts a load of 10 N, the result of our addition is a truss beam, with a cantilever and a load of 10 N at the end of the cantilever. So we have a truss, plus a cantilever -- we have already seen how to deal with these two scenarios -- and here we obtain a truss with a cantilever. Let's see how to obtain the support reactions of this structure, we are going to use the technique of the auxiliary cable. That is just going to be an arch-cable a little bit particular, since we are going to have an element in compression here, and an element in tension, here. Since it is a cantilever, to have equilibrium, it is still necessary to have a compressive internal force which acts here. We want to solve, now, this auxiliary arch-cable starting here, with a free-body around node F. This free-body is subjected, on the one hand, to the load of 10 N, then, turning counterclockwise, to the inclined tensile force -- I directly draw it in tension, since I have already indicated it as being in tension, that is clear that it pulls on the free-body -- and then, we are going to finish with a compressive internal force in bar E-F. So here is the contribution of node F to the Cremona diagram of this auxiliary arch-cable; now, what interests us is really to obtain the support reactions. We are going to start by support E, for which I look at this free-body; this free-body is subjected to the internal force in E-F, in the other direction; then to a horizontal compressive internal force, that is the internal force in C-E, or in A-E, actually, since we have simpler system, and then finally, to the support reaction in E which must close the funicular polygon I draw it slightly staggered, here, that is RE -- vertical since it is a mobile support. And then, it has larger value than the 10 N, actually, it is equal to 12.5 N. Ha, I am also going to indicate the contribution of node E to the whole Cremona diagram: so the equilibrium of the node E is here; now, we want to look at node A, here, which is subjected, still turning counterclockwise, to the compressive internal force in A-E, then to the tensile internal force in A-F, then finally, to the support reaction in A, so we can see that it is going to be downwards. That is RA. If I indicate the contribution of node A to the Cremona diagram, that is this part here. The support reaction in A is equal to 2.5 N, with a force which acts downwards. This is the effect of the cantilever; as you can see it in the next video, where I have increased the internal force -- so, you can see that I also stabilize the structure because it tends to go sideways, but you can see that pushing at the end of the cantilever, I can indeed lift up the left part, the left support; that is normal, there is a support upwards reaction. So this support can lift up; obviously, in our case, it was not lifting up, because there is a certain weight in the support and in the structure. We are now going to quickly solve this entire truss. We have already partially done it, so I am going to do it quite quickly: here, that is node F; this node is subjected to the force of 10 N, then to an horizontal internal force in D-F which is going to be in tension -- so I already draw it here, horizontaly, and then finally, -- so this, it is D-F -- and then the Cremona diagram must close, with the inclination of E-F, so necessarily, that is on this place here. So here, this is in E-F, so I can indicate the end of the arrow, then I quickly indicate the contribution of node F to the Cremona diagram. Now, I am going to consider node E. This node is subjected to the internal force in E-F, in the other direction; to the internal force in D-F -- in D-E, sorry. That is also a compressive internal force; then to the internal force in C-E, that is still a compressive internal force. So here, this is D-E and here, this is C-E. Then we finish by the support reaction on the right, whose we have determined the value before, that is 12.5 N. RE is equal to 12.5 N. We are maybe going to indicate the signs of the internal forces in the bars: so here, this bar is in compression, this bar is in compression, this bar is in compression, while this one is in tension. The contribution of node E to the entire Cremona diagram is here. Here you go. We are now going to move to node D, which is an unloaded node. This node is subjected to the internal force in D-E, in the other direction to the internal force in D-F, in the other direction, then to the internal force in B-D which is again going to go leftwards, and then we are going to finish by the internal force in C-D, which is going to be parallel to EF. So this, it is going to be the internal force in CD... and then here, we have the internal force in B-D. Here is the contribution of this node to the global equilibrium. We now move to node C, which is also an unloaded node; so, yes, I should have first indicated the signs of the internal forces -- so we have here tension in B-D and tension in C-D. And now, we move to the equilibrium of node C: so we are going to have the internal force in C-E, in the other direction, then the internal force in CD, in the other direction, then the internal force in B-C, which is going to be parallel to D-E. This is the internal force in B-C. And then we are going to end by the internal force in A-C, which I am going to stagger again for us to be able to see it... ... which is also a compressive internal force. So, in A-C, we have compression and in B-C too. The contribution of this node the Cremona diagram can be seen here. We now going to look at the equilibrium of node B: node B is subjected to the internal force in B-C, in the other direction; to the internal force in B-D, in the other direction; then to the internal force in A-B. And here, if I had drawn it well, everything should gather on the same point. There you go. This is the internal force in A-B, so we can see that A-B is in tension; the contribution of the grey node to the Cremona diagram is this part here. We finish by a control at the level of node A: this node is subjected to the internal force in A-C, in the other direction, to the internal force in A-B, in the other direction, and, finally, to the support reaction RA of 2.5 N downwards. And the contribution of this node to the Cremona diagram can be read here. So we have obtained the internal forces in this truss in a way which is absolutely similar to what we have done before, starting by the determining of the support reactions, and then attacking the structure from this node and progressing step by step, until we have solved everything. In this lecture, we have seen what is the effect of the introduction of a cantilever on a truss, concerning the support reactions and the internal forces. Regarding the support reactions, the introduction of a cantilever can lead to a lifting of certain supports, under vertical downwards loads. The internal forces generally have opposite signs, that is to say that we got compression in the lower part and tension in the upper part.
