Hello, in this lecture, we are going to see how springs behave, we will introduce the concept of stiffness and the modulus of elasticity. When we have a body of initial length l, which is subjected to a tensile internal force N, that is to say that we pull on the bar, we intuitively know that the bar is going to lengthen. This bar is in tension, let's characterize the tension by the color red, and the internal force inside the bar, also, by a continuous red line. The effect of this tension on our structure causes it to lengthen. We have shown it on only one side, here. It extends by a quantity delta l. Delta l is positive according to our convention. Let's also remember that the internal force is also positive. If, however, we introduce a compressive internal force in our structure, a negative internal force, which pushes towards our structure. Within the structure, we represent this internal force by a blue line which is often discontinuous. Why is it discontinuous ? Well, simply because we could build the element, which is in compression, with, let's say, bricks, piled on top of one another. They can resist to compression, while obviously, this same material, if it is not continuous, cannot resist to tension. One more time, a blue line means compression, or a discontinuous line. The effect of compression on the body, is to shorten it, and thus to get a delta l, smaller than 0. It is quite intuitive, since the final length of our body will be l minus delta l, and thus will be smaller. So, it is logical to say that delta l is negative. In this video, we are going to see how a spring behaves, on which we are going to hang up a weight. First, the spring, is initially 200 millimeters long, but the lower end of the spring, on which we are going to hang up the others weights, we are going to consider that it is at the position 0. If we introduce a weight of 10 Newtons, we get a measured lengthening of 40 millimeters. If we introduce a second weight of 10 Newtons, then for a total of 20 Newtons, the lengthening becomes 80 millimeters. If we add a third weight, for a total of 30 Newtons, the lengthening is 120 millimeters. If we introduced a weight of 15 Newtons, that is to say between 10 and 20, well, it is quite clear that we would get a lengthening which would be between the two, 60 millimeters. What does it mean ? It means that we could also put all kinds, a series of loads with intermediate values and which would always be aligned on one single line. So, we can trace a line which connects all these points, which enables us to predict what would be the behavior under any weight, at the least, between 0 and 30 Newtons. The slope of this straight line is 1 to R. The slope of this straight line is called the stiffness. And the stiffness R, it is equal to the difference of force, over the difference of length. Then, we have a difference of force of, for example, 10 Newtons over 40 millimeters, that is to say 0.25 Newtons per millimeter. If we unload the spring, we had then, this straight line, that we had established. But, if we run alongside this curve, and if we come back to zero, that is what is shown in the video, on the left, then, we come back to the position zero. That is maybe obvious, but it deserves to be mentioned anyway. In this video, we can see what happens, when we add a second spring to the first one, and when we follow, as we did before, with the lengthening. So, at the beginning, we are a position 0. When we add 10 Newtons, the lengthening is 80 millimeters, 20 Newtons 160. So, 30 Newtons, 240. With again, a linear behavior. If we have a look on the slope, hum, this straight line, the stiffness has a value of 10 Newtons over 80 millimeters, that is to say of 0.125 Newtons per millimeter. For the record, we had, here, a slope of 0.25 Newtons per millimeter. So, we can see that if we increase the length of the system, doubling the length of the spring, we increase this length, the stiffness of the system is divided by 2. In this video, we can also see a configuration with two springs, but instead of having put them one below the other, we have put them one next to the other, with a little device which enables to hang up the load directly to both springs in the same time. So, this time, we have a displacement, of only 20 millimeters for 10 Newtons, of 40 millimeters for 20 Newtons, and of 60 for 30 Newtons. Then, we have a behavior with a much steeper slope, than before. And, the stiffness, here, is equal, we are going to see, 10 Newtons for 20 millimeters, that is to say 0.5 Newtons per millimeter. So, what we can see here, it is that if we increase the section of our spring, or, if we double the number of springs, we also double the stiffness. If we write the stiffness, we have thus seen that it was inversely proportional to the length of the system, and then that it is proportional to the number of springs, to be more general, to the area of spring available, that is to say A. And then, we are going to introduce, here, a constant E, of which we are going to see the meaning very soon, but for the moment, we are simply going to say that R is equal to E times A over l. That is to say proportional to the section and inversely proportional to the length. What we have seen, now, regarding the stiffness of a structure, it is a behavior which on the one hand, followed straight lines. Thus, a behavior which was linear, where the lengthening increased proportionally to the loads. On the other hand, what we have seen is that we could increase the load, or decrease it, and that we are going to find again the initial position of our springs, at the end. This behavior, we call it an elastic behavior. An elastic behavior, is when a structure gets its shape back, its initial shape, when it is unloaded. And actually, here, we have got a behavior which was both linear and elastic. Then, we call it a linear-elastic behavior. On the left, we can see exactly the same graph, with this slope, which is equal to E times A over l, the stiffness of the structure. And, on the right part, we have a graph which does not show anymore the internal force, but the stress, and not anymore the lengthening, but the strain. What is the stress ? Sigma, it is the internal force N divided by the section A. As the section A, of a body is constant, we are dividing by a constant. We get a curve of the same shape. In the same way, epsilon, the strain or specific deformation, epsilon is equal to delta l, divided by l, where l is the initial length. So, actually, in relation to the graph on the left, we have divided the abscissa and the ordinate, each time, by a constant. Thus, we get, another straight line, with a constant. What is interesting, is that, after having done these operations, the slope with which we are left, here, since we have divided it by A and by l, we are left with E, this constant which is thus the stiffness of the material. On the left, we had graphs for the structure. And here, we have a graph for the material, and the feature of the stiffness of the material, we call it the modulus of elasticity. And it is measured, generally, in Newtons per square millimeter. In this video, we have seen that the lengthening of a structure is proportional to its length, and inversely proportional to its section. The stiffness R of the structure is equal to the ratio between the applied load N, and the lengthening that this N induces, that is to say, delta l. We can write this value, by the formula E times A over l, where A is the section of the element and l its length. About the materials, we have also seen that the stresses are proportional to the deformations. The modulus of elasticity is the stiffness of the material. And then, we have seen an additional formula. I summarize here the three simple formulas that we have seen so far. Sigma equals N over A. Epsilon equals to delta l over l And then, E, modulus of elasticity, equals to sigma over epsilon. These three formulas are independent, but combining them together, we will solve quite a few problems. That is what we will see in the exercises and the examples which are going to follow.
