Hello, in this video, we will see the concepts of internal force and of stress which will enable us to understand what happens inside the components of a structure. On the basis of the example of a person, standing on a stool, we will see how the weight of this person is in equilibrium with the gravitational force that this person exerts on Earth, through the stool. In this video, I stand up on a stool. My weight of 800 Newtons is not directly borne by the ground anymore. It is obvious that I still exert a gravitational force on Earth of 800 Newtons too. Then, these two forces must be in equilibrium. But now, between us, there is the stool. In order to sketch rather than to look at myself on the stool, we can simply, sketch. I am standing, on the stool which is directly standing, on Earth. Acts on me, my weight, a gravitational force of 800 Newtons. Acts on Earth, the gravitational force that I exert on Earth, which is equal and opposite, of 800 Newtons. And now, what interests us in this lecture, is what happens, is what happens inside the stool ? To answer this, we are going to isolate three free-bodies. One for the man, one for Earth. Well, it is not complete, but we include all Earth below. And then, finally, one for the stool. The forces that we already know, are the weight of the person, 800 Newtons, and the gravitational force that the man exerts on Earth, of 800 Newtons as well. For that free-body "man" to be in equilibrium, it is necessary that the stool exerts on him, a force equal and opposite to his weight of 800 Newtons, and on the same line of action. Thus, we have, here, 800 Newtons which are the stool's effect on the man. Likewise, below the stool, we must have a force of 800 Newtons which is the effect of the stool on the Earth. Our two free-bodies of the man and of the Earth are now in equilibrium, and now we have to deal with the free-body of the stool. The stool is subjected to a force of 800 Newtons which is the effect of the man on the stool. And then, below, also to a force of 800 Newtons which is the effect of the Earth on the stool. If we disregard, what is a reasonable hypothesis, the weight of the stool, we can see that the stool, which has no weight, is nevertheless subjected to forces which pass through it. These forces, inside a structure, we call them internal forces. This is the way I give you the following definition : an internal force is a force inside a structure. And, to be able to see this force, it is necessary to use a free-body. Then, we say that it is... this force is revealed by the cut of a free-body. Looking at an example, hum, of a man who is now on the top of a column. His weight is still 800 Newtons. And if we draw a free-body which passes just under his feet, it is obvious that, there must be an internal force in the column, of 800 Newtons, upwards, for the the man to be in equilibrium. Thereafter, once again, we are going to... it is a little bit less easy to admit, but we are going to do it anyway. We are going to disregard the weight of the column. If I take a free-body, in the middle of the column, and there is no force inside the free-body since we disregard the weight of the column. However, acts on the top of the column, an internal force of 800 Newtons which is the effect of the man on the column. And then, below our piece of column, we also have an internal force of 800 Newtons which is the effect of the ground on the column further to the person's weight. The internal force, inside this component, which we draw in blue, is a compressive internal force. And conventionally, now, and for the rest of this course, the compression will always be drawn in blue, and will always be negative. It is a convention, but it is important, and we will stick to it. How to know, if an internal force which acts in a free-body is compressive ? Well, we observe that, a compressive internal force pushes on the free-body. Let's look at the element of column once again, on which acts a compressive internal force of 800 Newtons, which we denote by the letter N. We have here, N equal to minus 800 Newtons. When I draw the force which acts on the subsystem, I do not need to indicate a minus sign, since if the force pushes, it is compressive. It is then, a minus sign, automatically. This internal force acts on the surface of the column, which we call its section. The section, it is the surface of the column, when we cut it perpendicularly to its axis. Here, we denote this section by the letter A. And it is expressed in units of area, typically, in square millimeters. Of course, it could be square meters, or something else. But, it is very often square millimeters. We want, now, to introduce the definition of the stress, as being the internal force per unit of area. If we take a little element of column, we recognize it, and if we say that it has an area equal to 1, a unit surface. Then, we can define the stress, as the internal force, per unit of area. We denote the stress by the greek letter sigma, and we write, then, sigma equal to N, the internal force, divided by the area. On this small element, we have a stress which pushes on the element. This is a compressive stress. And it is also negative per convention. You probably wonder if it happens to get positive internal forces and stresses. And the answer is yes. It happens when an element is in tension. Here, I have a free-body of a window cleaner hung up to a cable. We disregard the weight of the cable which extends below the window cleaner, which is, probably, very low. And, we notice that for this man... let's say that his weight is still 800 Newtons. For him to be in equilibrium, it is necessary that acts on the free-body, an internal force N, equal and opposite, thus of 800 Newtons upwards. If I take, now, a free-body of the cable in the part, above the window cleaner. We have, acting on this element, we have, acting on this element of cable, an internal force N of 800 Newtons, upwards, and downwards on his lower end. We draw, in this section, the tension in red. And, the tension is positive. How do we know if an internal force is tensile ? Well, in a very similar way than what we have seen about the compression, an internal force of tension pulls on the free-body. Then, the tensile stress is also positive. If we consider a little element of cable : sigma equal to N over A. Since N is positive, sigma will be positive. In this lecture, we have seen how to pass from the loads to the internal forces, and from the internal forces to the stresses. Internal forces are forces which act inside elements of structure. And stresses are internal forces per unit of area. Compressive internal forces occur when a force pushes on the free-body. Tensile internal forces occur when a force pulls on the free-body.