Hello. In this video, we will see how to determine the internal forces in all the bars of a truss and to identify where the maximum internal forces in the chords and in the diagonals are located. On this picture, we have gathered together different representations, first, on the top, the arch-cable which we already know well, for our beam, with a load of 30N on the left, and of 10N, near the middle ; with larger internal forces near the left support, but pretty constant in the arch, and obviously constant in the cable. If now, I look at, underneath, the result of the complete resolution of the truss, well we have an internal force of 29.2N in the first bar, of 25N in the second one, and of 12.5 N in the third bar of the upper chord. So here, I can make a diagram of these internal forces, which are then the internal forces in the upper chord. Likewise, we can make this diagram for the four bars of the lower chord so the first one on the left, with 27.1N, the second one with 31.3N, the third one with 18.8N, and the one the most on the right with 6.3N. This is one more time a diagram with internal forces tensile internal forces this time, which represent the internal forces in the lower chord. What is interesting to notice -- I had superimposed the shape at the scale of the arch-cable -- is that indeed, this shape superimposes very well at once on the diagram of the compressive internal forces in the upper chord, and on the one of the tensile internal forces in the lower chord. This observation is important because it will enable us to use, very systematically, results from the arch-cable to apply them to the truss. We thus notice a similarity between the internal forces in the chords and the shape of the arch-cable. Here, we have copied again these same diagrams, with the internal force in the upper chord and in the lower chord, we have added another diagram, which is the profile of our truss beam, actually its depth. So here, I simply fill it to signal that we are taking an interest in the depth, and not in the particular elements inside this truss. We are going to introduce a quantity which corresponds to the internal force times the depth, which is the bending moment. On the top, this internal force is in compression ; on the bottom, it is in tension. We multiply it by the depth, which I have drawn in yellow, and we will get a quantity, which I am going to draw in green and which is the bending moment. So let's start : here, we first start by the part which is the most on the left, with the internal force in the lower chord, times the depth which is equal to 1.2, 27.1 x 1.2 is equal to 32.5 NÂ·m. Note that the unit of this bending moment that I just introduced, is "Newton", for an internal force, times "meter", for a depth. The next point, we will take it in this diagram, on the middle of the bar, times this depth, 29.2 x 1.2 equal to 35 Nm. Then, we will take the internal forces ; 31.3 x 1.2 equals to 37.5 Nm ; then here 25 x 1.2, it will give us 30 Nm. Here, 1.2 x 18.8 = 22.5 Nm, and so forth for the last quantities. We can notice that these quantities are distributed on this diagram which is continuous, and which we call "Bending moment diagram". That is not something with which we will work a lot, but we will refer to it a bit in this course, and certainly in other courses which will follow, that is why I introduced this notion here. Some important properties: in statically determinate systems, the bending moments, so the internal force times the depth, are independant of the depth. We have already seen this a bit when we played with the applet, with which we obtained a variety of arch-cables by increasing or decreasing the depth of the arch-cable, and if we look at the internal forces, indeed, they increased if we decreased the depth, and decreased it we increased it; and then now, we write this down in a formal way. And then, there is an affine transformation between the shape of the arch-cable and the bending moment diagram. Indeed, if we observe these two diagrams here, they have the same shape, but not exactly the same depth. That is it, an affine transformation: it means that if we multiply the lengths and widths by appropriate coefficients, we will get an identical diagram. We now want to obtain some information about the internal forces in the chords, that is to say the horizontal parts of our trusses. We have information, first, which is valid for all the trusses. For all the trusses, you have already observed it before, the chord which is on the side of the arch is in compression. We had already observed it, but we can now confirm it: the chord which is on the side of the arch is in compression. In a similar way, the chord which is on the side of the cable is in tension. So in this case, the lower chord is on the side of the cable, it is therefore in tension. This is not only valid for the truss that I have drawn, but also for all the trusses, so even for trusses with variable depth, the chord on the side of the arch is in compression. Now, particularly for trusses with constant depth: the chord with the largest internal force and since we have positive and negative values, we are going to talk about the absolute value; it can be in tension or in compression; is the the one for which the distance between the arch and the cable is the largest. So, what can we notice ? In this case of arch-cable, we obviously have, here, a maximum depth and we are going to wonder to what chord it corresponds, does it correspond to the upper chord or to the lower chord? Actually, I am going to add one more point, here: to know which is the largest, we consider the middle point of each bar. So if we look at the middle point, it is clear that it is this bar, here, whose the middle is exactly here, which has the maximum internal force. So, I am going to say that N(max)+, is the largest internal force in terms of absolute value, it is also the largest tensile internal force. Now, we can take look at two bars in the upper chord: the one which is just on the left of the load of 100kN, and the one which is just on the right. And here, we can clearly see that both depths are not the same. Why ? Because the left part of the arch goes down more slowly than the right part. Therefore the maximum compressive internal force will be in this bar, whose the depth, here, is maximum. So here, we have N(max) in compression. Obviously, this diagonal -- this chord, here, has a quite significant internal force, but it is a bit smaller than the one of the left bar. We can make similar observations, still valid in the case of trusses with a constant depth, but now, in addition, with a constant inclination of the diagonals. The diagonal with the largest internal force, still in terms of absolute value -- I am not going to write it but we always implicitly talk about absolute value; I use some parentheses -- is the one for which the inclination of the arch is maximum. So, you can measure it on your notes: indeed, the left angle is a bit larger than the right angle; and what do we notice? We notice that it applies to several bars. If we consider the middle point, well, we have this diagonal, this diagonal, and this diagonal, for which we have the same angle alpha(max). And thus we have three internal forces which are equal, two, in these diagonals, are in compression, one is in tension. I cannot yet tell you why, but we are going to see this very soon. Here, we have the maximum internal force in the diagonals for these three diagonals. In a similar way, the diagonal with the smallest internal force, still in terms of absolute value, is the one one for which the inclination of the arch is minimal. So we are going to clearly identify that here, we have alpha(min); it could even be horizontal. And in this case here, it applies to these two diagonals so here, we are going to have a minimal internal force here, in compression, and here, in tension. So here, we have the minimal internal force in the diagonals. Here, we go on with our reasonings, which are valid for the trusses with a constant depth, and with a constant inclination of the diagonals. What we want to see now is how we can know the sign of the internal forces in all the diagonals. So, we can figure out that this diagram, with the arch-cable, is characterized by three different parts, which I am going to call A, B and C; and we are going, in a corresponding way, to make three small diagrams, for these zones A, B and C. As you can see here, I have drawn an orthonormal coordinate system, if you remember to have studied, maybe, trigonometry, we talk about quadrants, -- here we have the quadrant 1, 2, 3 and 4; there are always four quadrants in the trigonometrical circle; you do not need to be a master at trigonometry, you just need to accept the notion of quadrant. And we have this assertion: if the diagonal is in the same quadrant than the arch, it is in compression. So, we are going to draw, approximately, the shape of the arch in this quadrant, and then we will draw, approximately, the shape -- ah, I need to number my diagonals! I am going to number them 1, 2, 3, 4, 5, 6 and 7. In this quadrant here, I have the diagonal 1, then the diagonal 2 in the other quadrant, and the diagonal 3 is the same than the diagonal 1, here, in this quadrant. If the diagonal is in the same quadrant than the arch, it is in compression, so the diagonals 1 and 3 are in compression. So here, I have the diagonal 1, and there, the diagonal 3, which are in compression. If the diagonal is in another quadrant, it is in tension. So, I have the diagonal 2, which is not in the same quadrant, and which is thus in tension. We now pass to zone B; in zone B, the arch has this shape, approximately, it goes down rightwards; and we have, in zone B, the diagonal 4, the diagonal 5, and we can see that the diagonal 4 is in the same quadrant, so it is in compression; the diagonal 5 is in another quadrant, so it is in tension. Same thing for the last quadrant, where we have the arch, which goes up again, a bit less steeply than before, and then we have the diagonal 6, which is in this quadrant; and the diagonal 7, in this one. So the diagonal 6 is in tension, and the diagonal 7 is in compression If, instead of having an arch-cable, we have a cable-arch, it is the same thing, by analogy. What do we have in our quadrants ? So here, we have quadrant A -- hum sorry, the zone A, the zone B and the zone C; so again, here, A, B and C. I draw for these three zones the position of the cable, so here, here, and here. And we can immediately see that the diagonals which are in the same quadrant than the cable are in tension. Here, that is this diagonal, here, that is this diagonal, and the others are in compression. We could have made this reasoning using an arch-cable, and we would have obtained the same results; I let you do this check by yourself. Se we can now determine the internal forces in all the chords, simply on the basis of our structural scheme, and of the arch-cable. What does the arch-cable tell us ? Well, the chords... The upper chord, which is on the side of the arch, is in compression; the lower chord, which is on the side of the cable, is in tension. The diagonals which are in the same quadrant than the arch are in compression. So here, it is this; here, it is this again. Here, it is this. Here it is this, here it is like this, and so forth till the end. And then the diagonals which are in opposite quadrants, if I drew here, I would cut, as well as here, if I drew, I would cut quite violently the arch... Well, these diagonals are in tension. Moreover, studying where is the largest depth, we can determine the maximum internal force, here in tension, we have N(max)+ here, in compression, N(max)- And then here, we have N(max) diagonal, here N(max) diagonal, and likewise, here, N(max). So you can see that, now, on the basis of what have learnt, you can very quickly obtain the internal forces in all the chords of a truss, even if it is complex, you just need to first obtain, simply, the arch-cable. In this lecture, we have seen that, one more time, the arch-cable is a central element, since it enables us to determine the sign of the internal forces in the chords, as well as that of the internal forces in all the diagonals. We can also, on the basis of the arch-cable, determine the locations where the maximum values of the internal forces occur in the chords and in the diagonals. These elements will be very useful for the pre-dimensioning of trusses which we will see thereafter.