Hello. In this video, we will look at the maximum internal forces which act in a beam or in a cantilver. To do so, we are going to establish together a table which links the main parameters of each of these structures to the maximum internal forces which act in. We will look at concentrated loads, and also at uniformly distributed loads. We are going to work on simple beams on the one hand, with a fixed support on one side and a mobile support on the other side. What we are interested in is the span, which is the distance between both supports, which we will call L, as usual. And then on this beam, in the middle, we are going to place a concentrated load Q, expressed in kiloNewtons; or else a uniformly distributed load over all the length of the span, and this load, we will call it q, in kiloNewtons per meter. We will also look at cantilevers. Cantilevers are supported by a clamping on one side. Their span is the distance between the clamping and the end of the cantilever. And here again, we will have a load Q in kiloNewtons, or else a uniformly distributed load q in kiloNewtons per meter. We are now going to establish a table which is going to link the loads Q or q to the maximum tensile internal force of our structure. And we have already seen that the maximum compressive internal force is going to be similar to the maximum tensile internal force. Our parameters are then first the static system with a simple beam, here; or else a cantilever. We will gradually fill the different parts of this table. We will have the type of load as a parameter; either a concentrated load, either a uniformly distributed load. The span is obviously an important parameter, such as the effective depth of the beam, named z. An important note to be able to do comparisons: we are going to admit, each time, that Q is equal to the uniformly distributed load times the span, both in kiloNewtons, so we will be able to compare the effect of a given load, whether it be concentrated or distributed, on the internal forces in each configuration. Let's start with the first case, the simplest one. That is a simple beam with a concentrated load. We can solve this inscribing an arch-cable, that is the simplest solution of the truss, an arch-cable with compression in these inclined elements, and tension in the lower cable. We can solve the internal forces in this structure looking at, for example here, the right support. I draw your attention on the fact that the depth which we have given to our arch-cable, is not the depth of the beam, but the depth z which is the effective depth so there is enough space for all the stresses to remain inside the cross-section of the beam. We come here on the right: our free-body is subjected to the right support reaction, which is simply the half of Q; and then turning counterclockwise, we are first going to meet the inclined compression, and then the tension in the cable. We are going to call it N, like in this figure on the left, because we are going to do some interesting operations with this element, now. If we compare this triangle here and this triangle in the Cremona diagram, this polygon of forces, we have constructed it by drawing parallels, obviously the support reaction is parallel to the vertical applied force, this blue part is parallel to the inclination of the arch, and then this, it is obviously parallel. So both these triangles are similar triangles. We are going to use the properties of similar triangles to derive formulae which are going to link what we have in the diagram of the real structure to what we have in the Cremona diagram. If we look at these two catheti of the triangle in the Cremona diagram we know that Q over 2N must be equal to z divided by L over 2. From this, we can directly obtain N which is equal to Q times L over 2 divided by 2 z or, to write the complete formula, we are going to write it here Q times L over 4 z, that is the maximum tensile internal force. I am going to surround these formulae each time for us to remember them well, and then I am directly going to copy these formulae on the previous page. So we are going to get the first solution here: N equal to Q times L over 4 z. We now move to the solution for a simple beam with an uniformly distributed load. What we have seen a long time before already in "The Art of Structures I", is that this curve here is a second order parabola, and therefore the tangent, either at the beginning, either, in our case, at the end of this parabola, is equal to a straight line which is twice steeper than the maximum depth of our parabola which is equal to z. We then solve one more time the free-body around the right support. This free-body is subjected to the right support reaction, which is equal to q times L, that is the total load, divided by two. And then, I have forgotten before to draw the compression and the tension. I do it quickly now. Here we have the compression in the arch, and the tension in the lower cable. So in our Cremona diagram, we have the support reaction q L over 2, then we have the compression in the bar on the far right of the arch, then we have the tensile internal force N to close the polygon of forces. We have here again similar triangles. This triangle in the real structure, and this triangle in the Cremona diagram. This enables us to write the relation: again, q L over 2N, this is in the Cremona diagram, is equal to 2 times z divided by L over 2. That enables us to directly obtain here that N is equal to q times L squared over 8 z. Again, an important formula, which is a result for us. So with a uniformly distributed load we have, for a simple beam, N equal to q times L squared divided by 8 z. We now want to look at the cantilever with a concentrated load. In this cantilever, we are going to have compression here on the bottom and then tension on the top. We are going to consider a free-body like this, which includes the whole structure with the load at its end. So this free-body is subjected to the load Q, which acts downwards. Then, turning counterclockwise, we manage to obtain the internal force N in the upper chord. Then, the compression in the diagonal. Again, we can recognize similar triangles, well they are not placed in the same direction, but we built them with parallels, so they remain similar triangles. So we have here this big triangle in the Cremona diagram, and then a smaller triangle in the structure. And we directly manage to obtain that Q over N is equal to z over L. That means that N = Q·L/z. I am going to copy this result on our table, for a cantilever with a point load, N = Q·L/z. We are going to look at the last case for this video. That is a cantilever with a uniformly distributed load, so we have a compression on the bottom which has a parabolic shape, a tension on the top, and then we are again going to look at a free-body which includes all the loads, so all the length of the beam, cutting near the supports. And near the supports, we will have again the same property: we have a tangent which is steeper than the shape of the curve. That is to say that this tangent intercepts the horizontal at a distance L / 2 from the support, just here where the load is applied. This, that is the geometry that says it to us. Now, let's come back to the graphical construction. This free-body is subjected to the load q times L which acts on the whole cantilever. Then a horizontal tensile internal force which I call N, then an inclined compression which I call C. This triangle here is similar, by construction, to the this triangle here. So we have q times L over N is equal to z over L divided by 2. So we obtain that N is equal to q L squared over 2 z. We insert this last result in our table. So for a cantilever with an uniformly distributed load, N is equal to q L squared over 2 z. We now want to look at what happens if there is a changing. Let's first look at the first changing, which is maybe the most obvious. What happens if, instead of having a uniformly distributed load, I have a point load? Instead of taking a load which I distribute over the entire length of my beam, I take the whole load and I put it in the middle. because we can see this very clearly through this formula, for a simple beam, the internal forces are multipled by two, since we have Q·l over 4 z and not Q·l over 8 z. I remind you that Q is equal to q times L, so we have q times L squared on both sides, if you prefer. If we look at a cantilever, that is also the case. So if instead of having unformly distributed loads we concentrate them in the middle of the span, the internal forces are doubled. Conversely, if we moved from concentrated loads to uniformly distributed loads, we would diminish the internal forces by a factor 2, that is not really negligible. And then now, well, that is a comparison which is a bit less usual: what happens if we decide to pass from a simple beam system to a cantilever system? Well, with a concentrated load, we move from Q·L over 4z to Q·L over z. That is very significant, the internal forces are four times larger! What happens if I do this for a uniformly distributed load? Well, that is the same thing, I have q·L squared over 2z, so that is also four times larger. That is very significant. This picture is an illustration of what we have just seen today. If we look at these two trees and that we imagine when they are full of leaves, that is going to give us here a relatively small crown but at a large elevation, and then here, that is going to give us a very large crown but a smaller elevation. The wind is going to act on our structure in this way, obviously much less low. Why? Because the wind is weaker on the bottom, and in addition to this, the exposed surface is larger here. We have a structure, a tree, which is extremely exposed, and a little bit exposed on the bottom. What is interesting is to see how nature equipped these two trees. This tree works like a cantilever with a concentrated load acting approximately on the top. And then, we have here tension on the external face and compression on the internal face. And then, for the right tree, that is going to be exactly the same thing, we are going to have tension on the external part, that is to say on the side from which the wind comes, and compression on the other side. What is interesting is to note that for these trees, the effective depth z is adapted according to the internal forces. When we have larger internal forces, for the right tree, because of the larger crown, even if this tree is less high, then the trunk naturally has a wider shape, which enables it to have a larger effective depth z. While the left tree, which probably has smaller internal forces, has a smaller effetive depth. In this video, we have established together a table which gives us the maximal tensile internal forces, but that would be equal for the compression, in the case of a simple beam or of a cantilever. We have subjected these elements to either a concentrated load, or a uniformly distributed load, and then we have made a comparison shows that it is more favorable to have uniformly distributed loads than concentrated loads, and very clearly more favorable to have a simple beam than a cantilever if the other parameters are obviously equal.
