 Area moments of inertia, they are defined mathematically as the integral of y squared dA and the integral of x squared dA. And as we perform these integrals over the domain of a beam cross section, they are in fact geometric properties of that beam cross section. But what do they really represent and why do we need them? In this video we will try and break down the meaning of area moments of inertia. So far in this course, when we have analyzed beams, we have been presented with some form of structure like a bridge subjected to external loads and held in place by some form of supports, such as a pin connection or roller support. In order to analyze such a structure, we first need to draw a free body diagram. This particular free body diagram has all of the necessary elements of our bread acronym, a body, reaction forces, external forces, axes, and dimensions. But you can see that the thickness and other cross sectional dimensions of the beam do not come into play. This is because they were not required for our analysis of the reaction and internal loads. In order to perform equilibrium analysis and construct the internal force diagrams of the structure, our model of the structure, the free body diagram, only needed to capture the correct orientations of forces and positions where they were acting, which is accomplished by considering dimensions along the length of the beam. The cross section does not come into play. So why would we be interested in the properties of the cross section? The answer has to do with how the beam will deform under these loads. Imagine you wanted to cross a river with a makeshift bridge made out of a single plank of wood. With the plank oriented so that it was supported on its wider edge, we intuitively understand that the bending stiffness would be low and we may become unsettled by the large amount of deformation that occurs when we try to cross it. If instead we rotated the plank onto its shorter edge and carefully supported the plank with rocks so it wouldn't tip, the deformation would be much less when we tried to cross it, although it would require a little bit more skill and balance to actually walk across it. In both instances, the beam has the same cross sectional area, but the bending stiffness is dramatically different. So the area alone is not sufficient to capture this property of the beam. This is where the area moment of inertia comes into play. But in order to understand and derive this quantity, we have to briefly look at how the beam deforms under bending. Here I have a foam beam with a grid of lines drawn running along the length and through the thickness of the beam when it is undeformed. When I bend the beam back and forth, take a look at what happens to the grid lines, particularly the ones running through the thickness of the beam. On the screen you can see a snapshot of these grid lines in the undeformed state of the beam. What is important to note is that these lines remain straight, but rotate as a result of the bending. This results in the material on one side of the beam being in compression, while on the other side of the beam being in tension. We can model the material in this beam as a group of springs. For this particular beam, the cross section is replaced by four layers of springs through the thickness with four springs in each layer. In the deformed beam, the red springs are in compression, while the blue springs are in tension. If you recall from high school physics, the force in a spring is related to its change in length through the spring constant denoted by the letter K. We can use this result to examine how the beam is deforming. We will start by drawing a free body diagram of the spring model of the beam. The left hand side of the body was cut in the material of the beam, and thus will have an internal moment M acting on this cut face. On the right hand side of the body, we have cut through the springs, which are two force members, exposing the internal tensile spring forces on the lower side of the beam and the compressive spring forces on the upper half of the beam. As we observe the through thickness grid lines only rotate by an angle theta due to bending, the amount each of the springs elongates or contracts due to the bending is a simple geometric relationship. Each side of the spring translates a distance of y times sine theta. As both sides of the spring translate, the total change in length of each spring is twice this value, or 2y times sine theta. We can check that this result makes sense by noting that when y is negative, then delta L will be negative, indicating that the springs will be in compression. And indeed, in our model, we see that we are expecting the springs to be in compression in the upper half of the beam where y is negative. Next, we can look at the moments caused by each spring force about the x-axis. In this case, the moment arm for each spring force is the y-distance from the coordinate frame to the spring of interest. Thus, each spring will cause a moment about the x-axis equal to its spring force F multiplied by the moment arm y. We already established that F is equal to k times delta L, and that delta L is equal to 2y times sine theta. Rearranging these terms, we obtain that each spring causes a moment that is equal to 2 sine theta times k times y squared. Evaluating equilibrium for the cross section, we see that the moment on the left hand side of the free body diagram must be equal to the sum of the moments caused by the spring force on the right hand side of the free body diagram. This results in the following expression where i refers to the ith spring and 16 is a total number of springs in the cross section. Four layers of springs with four springs per layer. Let's take a closer look at this expression. We can see that the quantity to sine theta does not change for each spring, so it can be removed from the summation. It represents the global deformation of the segment of the beam we are analyzing. The remaining quantity in the summation is a sort of bending stiffness term. In fact, the entire expression is actually quite similar to our spring equation where the force in a spring is equal to the stiffness of the spring multiplied by the deformation of the spring. The problem with our simplified model is that the beam is not actually made of springs that have a stiffness k, but of smaller blocks of material with a cross sectional area ai. As a result, rather than the spring stiffness k, we should use the stiffness of an axial bar, which is given by the material stiffness e multiplied by the cross sectional area of the bar. Replacing k with ea, we obtain the following expression. We can further simplify this by recognizing that the beam is made out of a single type of material. Thus the material stiffness e can be taken out of the summation. This effectively divides the bending stiffness into a term dependent on the material properties and a term dependent on the geometric properties. The final step is to realize that the real cross section is not made up of 16 springs or 16 rectangular bars. It is a continuum, so the summation will only work in the limit as the area ai becomes infinitely small, turning the discrete summation into an infinite summation, otherwise known as an integral. And there we have it, the geometric bending stiffness or contribution of the shape of the beam cross section for bending about the x-axis is given by the integral of y squared da. This quantity is known as the area moment of inertia about the x-axis or i sub x. If we return to our example of a wooden plank bridge, we now have a quantity that can quantify the difference in bending stiffness. For the first orientation of the plank, bending will occur about the x-axis and we can quantify this stiffness as the integral of y squared da over the entire cross section of the plank. When we rotated the plank, bending would then occur around the y-axis of the beam instead of the x-axis. In this case, the bending stiffness would be equal to the integral of x squared da. We have not explicitly calculated these two integrals yet, but you can intuitively understand that the one on the right would be larger than the one on the left. Think back to the spring analogy and why this would result in the plank being more stiff in bending about the y-axis. In a later video, we will examine how to mathematically evaluate area moments of inertia. For now, it is important to focus on the meaning of the area moment of inertia. How it effectively captures the geometric stiffness of a beam cross section.