Hello, this video deals with the case of a cable subjected to two symmetric loads which act on it. We are going to solve this case identifying and solving the equilibrium of each load in an individual way using an appropriate free-body, then we will combine these results together. We will get the internal forces in each segment of cable and we will also obtain the forces at the supports at the ends of the cable. Here, you have a cable without any loads on which we simultaneously hang up two loads at predetermined positions. As you can see, the cable take a trapezoidal shape with the supports on the upper part of the trapezium and both loads on the lower part. We have, on the left, a structural sketch representing our structure and on the right, we are going to construct a Cremona diagram combining the various internal forces in a single graphical construction. We firstly identify a free-body around the force on the left. We can draw again this free-body just below, with an horizontal segment, an inclined segment, the load of 10 Newtons and, acting on the ends, on the edge of the free-body, the internal forces in the cables which we wish to determine. In the Cremona diagram, we can draw this force of 10 Newtons. Then, on this line of action, we know that the horizontal load will act, and finally, in parallel with the segment of cable on the left, we can copy the force which goes up towards the left. well, we can now finish the equilibrium of the blue free-body on the left, directly reading the value of the internal forces, here 7.4 Newtons in the horizontal segment of cable. We have tension here. 7.4 Newtons. Then, 12.4 Newtons in the inclined segment on the left. I identify the free-body which has given us this construction, with sky blue. We now want to study the force, the load on the right with this pink free-body which I also draw again below, with a horizontal segment, an inclined segment and the load of 10 Newtons. Here, we have seen in the blue free-body that this internal force had a magnitude of 7.4 Newtons. Here, it will be the same in the other direction. We already know it, here, we have 12.4 Newtons. And here, we will have a tensile internal force whose we do not know the magnitude yet. But we already know its inclination. Let's now consider the Cremona diagram, we draw the second force of 10 Newtons. In this Cremona diagram, we are going to actually start with the internal force of 7.4 Newtons acting horizontally. Then, we will have a load of 10 Newtons and we are going to close the polygon of forces corresponding to the pink free-body, here, by this internal force. Then, if we measure it, we are going to see that it has the same magnitude, it also has a value of 12.4 Newtons. That is quite logical since our structure is symmetric. I identify the contribution to the Cremona diagram of the pink free-body. We now want to take an interest in the free-body concerning the supports. Well, I draw again this free-body independently. We have a piece of cable, the end of the cable acting on this free-body, there is of course the internal force of 12.4 Newtons which pulls this support downwards. In the Cremona diagram, we now reuse the internal force of 12.4 Newtons pulling downwards then, we finish the construction with the vertical component on the left and the horizontal component on the left. We can see that this horizontal component is also equal to 7.4 Newtons and the vertical component is equal to 10 Newtons. We identify the contribution of the free-body to the global construction, and we now consider the support on the right with a yellow free-body with a piece of cable and the end. The internal force is equal to 12.4 Newtons. I did not copy it here, also 12.4 Newtons and I did not copy it in the free-body either. We have a tensile internal force of 12.4 Newtons. On the yellow free-body acts the internal force of 12.4 Newtons, going down towards the left. Then, the horizontal internal force on the right which is equal to 7.4 Newtons. Then, the vertical internal force on the right which is equal to 10 Newtons. I also identify the contribution to the Cremona diagram, of the yellow free-body. If we apply a second load on our system, actually it does not change its shape, but now, as it was the case for a system with only one load, the internal forces are doubled. Thus, we would just have to take again the Cremona diagram which we have previously made and to read it with a doubled scale to directly get the internal forces. Or, of course, we could integrally reconstruct it. Let's now deal with a practical exercise of dimensioning of the cable. It is now a cable on which acts not anymore 10 Newtons, but a permanent load of 12 kiloNewtons, 12000 Newtons so it is a quite significant load on which is added a variable load of 8 kiloNewtons. Let's see how to proceed to obtain the diameter of the cable knowing that the strength of the cable fsd is 320 Newtons per square millimeter. We first have to get the design load, Qd, which acts on this cable. Qd is equal to 1.35 times G plus 1.5 times Q or for us, 1.35 times 12 plus 1.5 times 8, that is to say 28.2 kiloNewtons. We are now going to have a design load, I draw it here in pink, a little bit staggered, Qd. By the way, I should not have moved it, we are going to draw it over, here, Qd equal to 28.2 kiloNewtons. Naturally, we could made a new Cremona diagram for the load of 28.2 kiloNewtons but it is not necessary since we have already the solution for 10 Newtons, and therefore we now want to multiply everything by the ratio between 28200 Newtons and 10 Newtons to directly obtain our internal forces. The larges internal force, we have obtained it before in both inclined parts of the cable, it was 12.4 Newtons for an applied load of twice 10 Newtons. Now, we want to see what is the result for 28.2 kiloNewtons and actually, because we do not want to have thousands of Newtons, we would be interested to have this answer in kiloNewtons, at least for the moment. This answer is simply going to be 12.4 divided by 10, this is a proportion, times 28.2. And this, it gives us 35 kiloNewtons. Thus the internal force in these two segments of cable, for the dimensioning, we have Nd is equal to 35 kiloNewtons. We now want to do the dimensioning of the cable. The dimensioning is the choice of the dimensions, actually we want to know the diameter of the cable. We have then Nd is equal to 35 kiloNewtons. That is to say 35000 Newtons and we have a material which resists to 320 Newtons per square millimeter thus the minimal required area is 35000 divided by 320, that is to say 109 square millimeter, so the diameter is still the square root of 4 times A divided by pi : it is equal to square root of 4 times 109 divided by pi, that is to say 11.8 millimeters, thus we choose a cable of 12 millimeters diameter which will be adequate to resist to the design internal forces. We also could obviously do the verification of the transitional part of the cable, you know, the one which had 7.4 Newtons for an applied load of twice 10 Newtons, but actually, this cable would have a smaller diameter, that is why we only did the dimensioning for the part with the largest internal force. In this video, we have seen how to solve the equilibrium of a cable subjected to two symmetric loads, we have proceeded expressing in the Cremona diagram, the equilibrium of each free-body around the load on the left and around the load on the right. We have seen how to obtain the internal forces in each segment of cable, then we have proceeded to the dimensioning performing a load combination to get the design load. Thereafter, a cross-multiplication, a proportion to obtain the design internal force in the cable and finally we have been able to determine the diameter of the cable. During the construction, we have also determined the forces at the supports in a very similar way to what we did before for the case of the cable with only one force. A question which maybe came to your mind is, but, what happens if I do not know the geometry of the cable ? Because that is true that until now, we have had photos of the geometry of the cable, well, we are going to take an interest in this question in the following lectures.
