Hello. In this video, we will see what are the internal forces in a cable which is subjected to a load, which acts transversally, but not in the middle, but which acts in an eccentric way. In a very similar way that what we have done in the previous video, we will study the free-body of the load, we will obtain the internal forces in the segments of cable, then we will study the free-bodies of the supports, to obtain the forces at the supports, and we will finish by a case a little bit particular of a force which is not only eccentric, but also inclined. In this video, you can see what happens, when on a cable which did not have any loads, we apply an eccentric load. Once again, as before, the cable takes a rectilinear shape, between the supports and the applied load. As before, we construct the structural sketch tracing parallels to the different segments of cable in an independent sketch. So, we have a load of 10 Newtons which acts. And then, for the record, this segment is parallel to this one. This segment is parallel to this one. We start by identifying the free-body including the load and by drawing it again in an independent way. We have a piece of cable on the left, another piece of cable on the right, the load of 10 Newtons, and the internal force in the left part of the cable, N1, and in its right part, N2. We want, in the Cremona diagram, to draw with a length of 10 centimeters, the load of 10 Newtons, equal to 100 millimeters, there. Then, we trace the parallels to both segments of cable, N2 first, and N1 afterwards. So, here, we have an internal force N2 which is equal to 8.5 Newtons. Then, here, we have the internal force N1 which is equal to 4.5 Newtons. Then, we can take an interest in the free-body of the support on the left which includes a segment of cable with an internal force of 4.5 Newtons. That is this internal force N1 along which we travel in the other direction, and for which we are going to obtain a vertical force at the support V1 and an horizontal one H1. V1 is equal to 2.4 Newtons, while H1 is equal to 3.7 Newtons. We can do like before, that is to say to identify the part of the Cremona diagram which corresponds to the orange free-body. Likewise, we are going to take a pink free-body for the support on the right. We have a segment of cable on which acts an internal force of 8.5 Newtons. So, we will have the forces at the support V2 and H2. I should have drawn, before, V1 and H1 for the support on the left. In the Cremona diagram, using the internal force N2 in the other direction, it becomes an arrow with two heads, we obtain the force H2, whose we can see that it is equal to H1, thus 3.7 Newtons, and then, the vertical force on the right which is equal to 7.6 Newtons, the sum being equal once again to 10 Newtons, which is the applied load. We are going to copy again, I first indicate the free-bodies used in the Cremona diagram. So, the sky blue free-body, is the free-body of the load to obtain N1 and N2. Then, both free-bodies which have been used to get the internal forces at the supports. We copy the results to the structural sketch, so both segments of cable are in tension. Thus, we can draw them in red. The internal force N2 is equal to 8.5 Newtons in tension. And the internal force N1 is equal to 4.5 Newtons in tension. We can see that the method which we used to solve this case is very similar to the one we have seen in the previous video. Indeed, we do not have anymore values of the internal forces in the segments of cable which are equal, as well as certain components of the forces at the supports are different. Let's now look at, the case a little bit more complicated of a cable with a load which is neither centered, neither vertical. So, we have an eccentric load which is a little bit inclined. We introduce it in our relaxed cable, by means of a dynamometer. The force which acts is exactly of 10 Newtons. So, we exactly have the same applied load than before. However, it does not act vertically. Here, we have copied the structural sketch of our configuration. So, we have, here, acting on an inclined axis, a load of 10 Newtons. And as before, we take an interest in the free-body which includes the load. This free-body cuts the cable on the right and the cable on the left, and includes the load of 10 Newtons. We are looking for the internal forces N1 and N2 which act in the cable. We copy the load of 10 Newtons to scale, that is to say 100 millimeters. Then, the parallels to the segment N2, then N1. Which thus gives us, the internal forces N2 and N1. N2 is equal to 8.3 Newtons and N1 is equal to 4.8 Newtons. We copy these values in the structural sketch, also indicating the type of internal force in the structure. In this case, both elements are in tension. We now take an interest in the support on the left. I draw again the corresponding free-body, with the internal force of 4.8 Newtons. And this free-body will be in equilibrium thanks to the forces at the support V1 and H1. In the Cremona diagram, we had used N1 in this direction. We now use it in the opposite direction, and we close the polygon of forces thanks to its component V1 and to the force at the support H1. H1 is equal to 2.2 Newtons. V1 is equal to 4.2 Newtons. Likewise, we are going to solve the free-body corresponding to the support on the right; with the internal force of 8.3 Newtons in the cable and the forces at the support V2 and H2. We reuse the vector N2. Then, we introduce H2, and we finish by the force at the support V2. H2 is equal to 7 Newtons and V2 is equal to 4.5 Newtons. We can note that in this configuration, well, the sum of the horizontal forces at the support is not zero, since precisely, there is a horizontal component in the load which is applied. You maybe wondered, whether it be in this video, or in the previous video, why, after having drawn the load of 10 Newtons, I have systematically always taken the internal force N2 which was on the right of this load, and afterwards the internal force N1. And you maybe wondered, why did not I do the reverse ? Actually, I already applied, without officially saying it to you, a convention which we are going to use for the rest of this course, which is that, when we deal with the forces which act on a free-body, we take them in a systematic order which is the counterclockwise order. So, we turn around the free-body, in the counterclockwise order, starting by where the forces are known. So, here, we know the load of 10 Newtons. Afterwards, we will have the element two which is N2. And finally, the element three which is N1. And then, in the Cremona diagram, this is the first element that we have traced. Here, the second element. Here, the third element. We can, obviously, proceed with the opposite convention. That is to say, to turn in the clockwise order. Will the equilibrium be modified ? No, since regarding the treatment of each free-body, we insure the equilibrium, and we can of course ensure the equilibrium by adding first N1, and afterwards N2. Actually, that is what has been done in this figure, from the book The art of structures, in which we have dealt with this free-body. And if we closely look at this, indeed we have firstly dealt with the force, then the internal force N1 and finally the internal force N2. So, we have used the force which is indicated by Q, then N1 and finally N2. You can see that the diagram is similar to the one we did before. The results are obviously right, and are the same. You can try to do this job by yourself. To simplify the correction and the communication within this course, we ask you, for this course, for the exercises, for your reasoning, also with the assistants or with your questions, to systematically use the counterclockwise order, as I have shown you in this lecture. Within the framework of this video, we have generalized the approach method to solve the equilibrium of a cable subjected to a transversal load. We have first dealt with the case of an eccentric load, then of an eccentric and inclined load. We have seen that the order in which we consider the internal forces and the loads which act on a free-body, has an influence on the shape and on the construction of the Cremona diagram. For the rest of this course, we will always use the counterclockwise order for the construction of the Cremona diagrams. We have seen how to systematically obtain the internal forces in the segments of cable, and how to copy them in the structural sketch to communicate these internal forces. Finally, we have seen how to obtain the forces at the supports and their horizontal and vertical components.
