 Conceptually, we understand that the area moment of inertia represents the bending stiffness of a given cross-sectional geometry about a given axis. But how can we go about calculating it? It is just a mathematical expression, and you could argue that you simply need to evaluate it, and you would be right. However, with a bit of understanding of what the quantity represents and a bit of clever thinking, we can greatly simplify the calculation, as we will see in this video. First, let's briefly recap the two area moments of inertia we have encountered so far and what they mean. We encountered the first moment of area, or the integrals of x times dA, and the integral of y times dA, when locating the centroid of a geometric shape. We often denote these moments of inertia as qy and qx, where the subscript refers to the axis which the moment of area is calculated about. We can use these moments of area to calculate the location of the centroid for a shape with respect to a reference xy coordinate frame by dividing the first moments of area, calculated with respect to this reference frame, by the area of the shape. Conversely, the area moments of inertia, which is often referred to as the second moment of area, quantifies the geometric bending stiffness of a given shape about an axis. It is defined as the integral of the perpendicular distance from an axis squared multiplied by dA. So for calculating the area moment of inertia about the x-axis, this would be the integral of y squared dA. If we zoom into our element, we see that the area element has an infinitesimal width of dx and infinitesimal height of dy. The area of the element is thus simply width times height, or dx times dy. This means that our integration is in fact a double integral. We have to consider the summation of all columns and rows of differential elements. In most practical cases, we can reduce this back to a single integral by making some smart choices with our differential element, as we will see in a moment. Enough with the theory already. Let's just get to the business of calculating. Okay, well, I'll hand it off to you then. It's about time. So let's take a look at the following problem. We have a rectangular cross-section of width B and height H with two defined axes systems. We are asked to calculate four area moments of inertia associated with the two different axes systems. Ix and Iy are to be calculated for the x-y axes located at the centroid of the rectangle, while Ix-prime and Iy-prime are to be calculated for a parallel set of axes, the x-prime and y-prime axes system, located at the lower left corner of the rectangle. Let's start with looking at calculating the area moment of inertia about the centroidal x-axis. By definition, Ix is equal to the integral of y-square da. As we saw earlier, this is technically a double integral, as a standard area element will have an infinitesimal width dx and an infinitesimal height dy. However, we can be a bit smart in how we define our area element to simplify this integral. Our integration is effectively the summation of the function y-squared times area over the entire cross-section. So if we take a contour plot of the function y-squared and overlay it onto our geometry for the given axis system, we can visualize what we are effectively summing over the domain of the shape. As our integral is an infinite summation of area elements where the function being integrated does not vary, the area element needs to be infinitely small so that the value of our function, y-squared in this case, is constant over the area element. But we can see from the contour plot that y-squared is actually constant across the entire width of the section. Can we exploit this to simplify our integral? Indeed, we can take advantage of the fact that y-squared does not vary across the width of the rectangle and expand our differential element across the entire width of the section. DA thus becomes B times dy, which reduces our integration to a single integral as we do not need to worry about translating the differential element in the x direction anymore. In order to encompass the entire domain of the rectangle, we need to integrate between the limits of y equals negative h divided by 2 and y equals positive h divided by 2. This leaves us with the following definite integral. Now we are ready to integrate the expression. Integrating B times y-squared, we get B times y cubed over 3, which we need to evaluate between the limits of negative h over 2 and h over 2. Substituting in these limits, we get that ix is equal to B times open bracket h cubed over 24 minus negative h cubed over 24, close bracket. Simplifying this, we obtain the expression B times h cubed over 12. This is our final expression for ix about the horizontal axis passing through the centroid of a rectangle. Before we look at the area moment of inertia about the y-axis, let's first consider the area moment of inertia about the axis parallel to the x-axis, the x-prime axis. Since the x-prime axis is parallel to the x-axis, our formulation of the area moment of inertia is nearly the same as before, just with x-prime and y-prime as the coordinates rather than x and y. This results in the ix-prime being equal to the integral of y-prime squared da. We can use the same differential area element as before, only in the new coordinate frame da is now equal to B times y-prime. To evaluate this integral over the entire domain of the area, we can observe that the limits of our integral change with the new coordinate frame. Rather than varying between y equals negative h over 2 and positive h over 2, y-prime will vary between 0 and h. So despite the result of the integration being the same, the overall result is dramatically different due to the different limits. ix-prime is equal to B times h cubed over 3, which is 4 times larger than ix. I see you notice that ix-prime is 4 times larger than ix, but you determine that by performing the integral again. Don't you recognize that there was an easier way to calculate ix-prime if you already know ix? But I calculated the integral. Isn't ix-prime just the integral of y-prime squared da? How could that be easier? Well, the x-y coordinate frame is a centroidal coordinate frame. So ix is simply the bending stiffness about the axis we have to consider when we reduce a section to a single point. The rectangular cross-section is equivalent to the single point at the centroid with the same area as the rectangle and the bending stiffness ix we calculated for the rectangle. So rather than performing the integral on our rectangular area, we can do so on our simplified point model. Here we simply have to sum the various contributions to the area distribution about the axis in order to get ix-prime. So ix-prime is equal to ix, the area moment of inertia we have to consider in reducing the rectangular section to a single point, plus the distance from the x-prime axis squared times the area of the point. Here we have to recall that although we reduce the section to a single point, we still consider it to have the area of the original section. Although we did not rigorously derive this expression, you can indeed mathematically prove that it does hold, and it is known as the parallel axis theorem. Here we will use the bar symbol above ix to explicitly state that the x-y axis must be the centroidal axis of the area in order for this theorem to work. If you are interested in the mathematical proof, it is available in your textbook, or you can easily find it online. I'm not so interested in you knowing the proof, but understanding what the theorem means and enables you to do. If you think about how we calculated the location of a centroid for composite shapes earlier in the course, you might already recognize that this theorem allows you to effectively do the same for area moments of inertia. For completeness, this theorem also applies to other axes. The area moment of inertia about the y-prime axis can be calculated by adding iy about the centroidal axis plus the area of the cross-section multiplied by the distance between the y-axis and the y-prime axis squared. I guess that would have been easier. I'll make sure to take that into account in the rest of the problem when I calculate iy and iy. I will only need to integrate once to calculate iy. Since iy is equal to the integral of x squared da, I will have to define a new differential area element of constant x. This can be achieved by spanning the element along the entire height of the rectangle, resulting in da being equal to h times dx. After performing that integral, I can then use the parallel axis theorem to calculate iy-prime. I guess you could do that, but is it really necessary? Take a look at your shape and the new differential area element for a minute. What would happen if you rotated the entire system counterclockwise 90 degrees and then flipped them around the x-axis? Okay, so rotate the system 90 degrees and then flip around the x-axis. I get something that looks like this. Since I moved the axis and the shape together, it shouldn't change the solution. Correct. Now compare that to your previous solution for ix and ix-prime. Comparing to my previous solution and... Oh, well, both shapes are rectangular. There's just a swap in the dimensions and axes. So I can infer that iy and iy-prime for the image on the right would be equivalent to ix and ix-prime for the image on the left, with the width and the height dimensions swapped. iy would then be equal to h times b cubed over 12 and iy-prime would be equal to h times b cubed over 3. With that, I have then solved for everything the question asked for. We solved for ix and ix-prime directly using integration, although I could have saved some time using the parallel axis theorem, and we were able to infer what iy and iy-prime are due to the similar distribution of area about the x and y axes, just with the dimensions swapped.