Hello, in this lecture, we will see how a cable can carry a transverse load. Until now, the loads that we have applied on the cables acted in the axis of the cable. Now, we have, as you can see on the slide of the title, a rope which is held by two men, on which acts a load in the middle. In this lecture, we are going to study the free-body of the load to obtain the internal forces in both segments of cable on the left and on the right. We will take an interest in the forces at the supports. Then, we will make some variation studies about the influence of the load which is applied and of the geometry of the cable. In the case which interests us here, we have a load, which acts in the middle of the cable. And we are going to identify a free-body which cuts the cable in two places and which includes the applied load. I draw again this free-body nearby. So, we have both segments of cable, the acting load Q and the free-body cuts the segments of cable in two places. This implies that, at this place here, I have to replace the cut segment of cable by the effect which is inside this segment of cable, or more exactly, the internal force. Then, not knowing yet this internal force, I can already tell you, that is not a secret for a cable, that it is a tensile internal force which acts as until now, along the axis of the cable. We have here two internal forces, which I am going to call N1 on the left and N2 on the right. Let's now look at this little video of a cable first without load on which we add a load of 10 Newtons. To be able to work on this structure, it is convenient to model it replacing the real structure by what is called a structural sketch. The structural sketch represents, to scale, the structure with its good proportions, the loads which are applies and the supports. So, I copy both segments of cable in the part on the right, with the correct inclination. And at their intersection, acts the load of 10 Newtons. On the left and on the right, we have elements which are fixed supports. They are supports which does not enable any movements. Here, likewise, in the structural sketch, we have on the left and on the right a fixed support. Sometimes the fixed support is also symbolized in this way, like a small triangle, the support being located at the top of the triangle. We draw again the structural sketch here. That is quite easy because within the framework of this construction, the distance between the supports, which we call the span, which is designated by the letter l, is equal to 1.2 meters. And the vertical distance between the supports and the lower point of the cable, which we call the rise, f, is equal to 0.6 meters. Thus, the angle is 45 degrees, which facilitated the drawing. We have here, acting, a load of 10 Newtons. We first take an interest in the free-body of the load and we draw it again below. So, we have a segment of cable parallel to the piece of cable on the left, a segment parallel to the one of the cable on the right, a load of 10 Newtons, and then both internal forces on the left and on the right which are in the extension of the cut segments of cable and as I indicated before, we are going to have N1 and N2. Let's now continue with the Cremona diagram in which we want to draw the force of 10 Newtons which acts downwards. I draw it with a length of 10 centimeters to make a drawing to scale. So, I have 10 Newtons, equal to, we are going to say 100 millimeters. For the free-body of the load to be in equilibrium, it is necessary that the three forces which act on this free-body then, the load of 10 Newtons, the internal force N1 and the internal force N2, should be converging on only one point. Well, that is good, we can see well that indeed, they are converging on the point of hanging of the load. Then, on the other hand, that their vectorial sum should be zero, so we are going to construct the vectorial sum of the load of 10 Newtons and of both internal forces N1 and N2. So, I trace straight lines at 45 degrees in my graph which correspond to the inclination of the internal forces N1 and N2, thus this segment here is parallel to this and also to this. This segment here is parallel to the segment on the left in the free-body or in the overall system. I notice, indeed, that the internal force N2 acts in the direction I had supposed, that is to say that it pulls on the free-body. An internal force which pulls on the free-body is a tensile internal force. And the tension, for us, is red, then I am going to draw this arrow again, here, in red. And, since I have drawn to scale, I can see that the internal force, here, is equal to 7.1 Newtons. So, as I have drawn to scale, I directly obtain the numerical value of the internal forces. Likewise, for the internal force N1 that I also measure at 7.1 Newtons. Seventy-one millimeters give 7.1 Newtons. We can now copy these internal forces in the real system. That is to say that in the real system, we are going to replace the basic color of each bar by the conventional color for the tension, red. And we will write, we will sketch the internal force drawing a kind of small piece of cable with an element of tension and then, here, we will write 7.1 Newtons. Likewise, for the segment on the left, also 7.1 Newtons. We now want to take an interest in what happens near the supports. We are going to study a free-body which cuts the cable and which stops just before the support on the left. I draw again this free-body. Once again, there will be a segment of cable which is parallel to the segment on the right. We are going to draw the beginning of the support. What is the internal force in this cable ? Well, we know it, it is the internal force N1 which is equal to 7.1 Newtons. Of course, we could say that the internal force just near the support is 7.1 Newtons which goes in the other direction. Actually, in the field of construction, it is often useful to know the vertical and horizontal components of the forces at the supports and that is what we are going to do here. So, what would interest us, it would be to know the vertical component here and the horizontal one of these internal forces. And this, we are going to be able to do it directly in the Cremona diagram. In the Cremona diagram, we were taking an interest in the internal force N1. Now, in the orange free-body, we use the internal force N1 in the other direction, thus we place a head of arrow at the other end and we take an interest in both internal forces V and H which are in equilibrium with this internal force. And it thus corresponds to a vertical component V and an horizontal component H of force at the support. We can directly read their value in millimeters and see that V is equal to 5 Newtons and H is also equal to 5 Newtons. We are going to do the same with the free-body of the support on the right. I am going to be very quick here. The internal force is, of course, equal to 7.1 Newtons. And we have a vertical component and a horizontal component. We reuse the vector N2 but in the other direction which corresponds to our pink free-body. Then, we will have a horizontal internal force and a vertical internal force. We read once again their value. 5 Newtons for the vertical internal force and 5 Newtons for the horizontal internal force. To understand well, particularly if we want to come back to our solving later, I am going to indicate in the Cremona diagram, at which free-body corresponds which part of the construction. So for the first part of the construction with the force of 10 Newtons and both internal forces N1 and N2, it was the free-body of the load. Then, we have had the orange free-body with the internal force N1 and the forces at the supports on the left and finally the pink free-body with the internal force N2 and the forces at the supports, H and V on the right. You can note that I used the same symbol for H and V. I knew beforehand that these two quantities had the same value, otherwise I would have used, and we are going to do it later in other lectures, I would have taken H left and H right, V left and V right. We can note that the sum of both vertical forces, 2 times 5 is equal to 10 Newtons, thus equal to the applied load. That is logical, this applied load must be transmitted to the supports. What is interesting is to note that the sum of both horizontal forces, we have an horizontal force on the top of the Cremona diagram, which points towards the left, and an horizontal force, of the same value 5 Newtons, which points towards the right on the bottom. The sum of the horizontal forces is thus zero, in this case. Last point of glossary, the shape of the cable which we have thus obtained is called the funicular polygon. This is an important expression which will often come back in the rest of this course. Well, we have made this construction for a load of 10 Newtons. Now, what happens, if as in the video on the left, we add a second force of 10 Newtons ? So, I propose you to think about it and why not to calculate the results. What are the internal forces in the cable under the effect of a load, not anymore of 10 Newtons, but of 20 Newtons ? I meet you later to discuss about this solution. Indeed, we have noted, when I added a second load, on the system, that the shape of the system did not change. Thus, the construction that we have made before with 10 Newtons, we are going to be able to make it again. I am going to make it again quickly. Now, with a force of 20 Newtons. Then, we trace a straight line parallel to the first segment of cable. A straight line parallel to the second segment of cable. And, we can directly read the internal forces, which not surprisingly, are simply doubled. Of course we could have spared ourselves this extra work, simply saying, well, the drawing which I made before for 10 Newtons, I am going to use it for 20 Newtons. If now I say that here, it is equal to 20 Newtons, it just changes the scale of the drawing and I can read seventy-one millimeters which would correspond to 14.2 Newtons. It would be, of course, much more reasonable. If some of you said to themselves that we could proceed in this way, well, congratulations, you have well understood how this works. We now want to look at another parameter which has a large influence. So, I introduced to you already the concept of rise which is the height between the supports and the lower point of the cable. What happens if the rise changes ? We have already solved the basic configuration with a cable at 45 degrees which I draw in red, and whose the Cremona construction is already available on the right. Now, we want to look at a variant with half the rise. How am I going to proceed for the solving ? Well, exactly in the same way than before. So, I am going to make a free-body around the load of 10 Newtons. I am going to cut a segment of cable on the left, a segment of cable on the right and in the Cremona diagram on the right, I am going to copy parallels to both segments of cable to close the polygon of forces, which expresses the equilibrium. So, here I make the polygon of forces which corresponds to the pink configuration. And we can note that the internal forces increase. Here, it is equal to 11.2 Newtons. For the record, in the red configuration, we had 7.1 Newtons. We did not change the load, we just decreased the rise by half. What happens if we are even more drastic and that we divide the initial rise by four ? Well, we obtain this red configuration where the cable is very horizontal. Once again, we are going to make the same construction, we trace the parallel to the orange segment. And, likewise, the left part. And this time, the internal force reaches 20.6 Newtons. I am going to introduce now the concept of slenderness, which is equal to the ratio between the span and the rise, it is something which is important, which we will see again later. And what we can see is that if the ratio between the span and the rise increases, that is to say either if the span increases and the rise does not change, or, like in our case, if the span does not change but the rise decreases, so, the internal forces increase. Conversely, if the slenderness decreases, the internal forces decrease. We will see that it is not a linear increase but that it is an increase which can sometimes be very sensitive depending on the type of structure we consider. In this lecture, we have looked at the structural system which applies for a cable subjected to a single load. We have studied the free-body of the load, we have obtained the internal forces in the various segments of cable. We have seen that they were tensile internal forces. We have also studied the internal forces which act at the supports and the influence of the slenderness and we have seen the relationships which exist between the load, if the load increases, the internal forces increase, and the slenderness, if slenderness increases, the internal forces also increase.
