The last topic I want to cover in Statics and Centroids and moments of inertia is moments of inertia and related polar moments of inertia. So, firstly, moments of inertia are important because we often need them in mechanics problems. For example, bending of beams, and other topics that will come up later on. In mechanics of materials. And this shows how they can arise and we'll use this later when we're computing bending moments in beams and torsion in shafts. But we don't use them very often in Statics so mostly in this section. The problems will just be related to calculating how you calculate moments of inertia. The specific topics in the reference handbook include moments of inertia, parallel axis theorem, radius of gyration, and product of inertia. And we'll just look at some of these. The most important one is calculation of moment of inertia, sometimes called the second moment of an area, and the definition of moment of inertia is given in this, extract from the reference handbook over on the right hand side here, and the definition of the moment of inertia about the x-axis. Is the integral over the area of y squared dA where dA is the element, is the area of an elemental area. And similarly, the moment of inertia about the y-axis is the integral over the area of x squared dA. In addition, we will need the parallel axis theorem, sometimes called the transfer axis theorem, which allows us to compute the moment of inertia about any arbitrary axis when we know the moment of inertia about the centroid. And the parallel axis theorem is illustrated in this diagram here. Let's suppose that the centroid here is c. And if we know the moment of inertia about the centroid c, we would like to get, for example, the moment of inertia about the x-axis and about the y-axis. So here's the statement of the parallel axis theorem from the reference handbook. And it simply states that the moment of inertia about say the x-axis, ix prime is equal to the moment of inertia about the, the horizontal axis to the centroid Ixc plus the distance squared between the two axis dx squared multiplied by the area. Similarly, the moment of inertia about the y axis, Iy prime, is equal to the moment of inertia about vertical axis through the centroid, plus the distance, dy squared, multiplied by the area. And this is the parallel axis theorem. But note that in that equation Ixc and Iyc are the moments of inertia computed about the centroid. So, therefore, with this theorem, we can compute the moment of inertia about any axis, about any parallel axis. The polar moment of inertia and radius of gyration are also mentioned. Although I don't think we make very much use of them, so I'll just mention the definitions. So here we have Iz, moment of inertia about the z axis, is equal to the integral of r squared dA. Which is equal to, we, we defined it is equal to p squared times A where rp is the so called radius of gyration. We also have another quantity called the product of inertia Ixy which is equal to the integral of xydA. But again, we won't be making much use of those here. Again, a extract from the handbook gives all of these properties for various shapes. But mostly, we'll be looking here at the triangle, rectangle, and the circle. And the moment of inertia of a rectangle about its centroid is the width times the height cubed over 12 and the moment of inertia of a circle about its centroid is pi times the radius to the 4th power divided by 4. And of course by symmetry, the centroids of both of those shapes are exactly in the middle. So those are probably the shapes that we'll be mostly concerned with. Let me show that by an example. We have a rectangle which is four meters tall by one meter wide, and the first question, the moment of inertia about an axis parallel to the axis through the centroid is most nearly which of these alternatives. So, here's a centroid. And, we want to get the moment of inertia about a horizontal axis through that centroid. So, from the previous slide, the handbook. We have the moment of inertia about the x-axis through the centroid is the width times the height cubed over 12. So, in this case, that is equal to 1 times 4 cubed over 12, which is 5.33 meters to the 4th. So, the answer is C. Second part, we want to compute the axis about the centroid, but now about a vertical axis through here. So, again the same, equation applies except that b and A just switched, so the moment then, Iyc is now h times b cubed over 12. Which is four times 1 cubed over 12, or 0.33 meters to the 4th. So the best answer is B. And, the final [COUGH] part. The moment of inertia about the x-axis, now. Is most, nearly, which of these? So, now, we want to compute the the moment of inertia about the horizontal x-axis here. So to do that we use the parallel axis theorem. Ix prime is Ixc plus dx squared times A. So in this case we've already computed ix prime here is 5.33 meters to the 4th. The distance dx is this distance. The distance of the centroid from the x-axis and this rectangle is 4 meters tall. So by symmetry, that is 2 meters above the x-axis. So it's 2 squared times the area which is 4 times 1 and the answer is 21.33 meters to the 4th. So the best answer is A. Now again, similar to what we did for centroids, we can compute composite areas by adding them together. So the moment of inertia of a composite area about an axis is just the sum of the moments of of its parts about that same axis. And emphasize here that it must be all referred to the same axis. And as we did when we're computing centroids, we can subtract those due to holes. So for example this T-shaped structure here and the. Moment of inertia about say the, the x-axis, Ixo is equal to the moment of inertia of area one about that axis, plus the moment of inertia of the area two about that same axis. Being able to subtract a juda holes often makes it for a simple kind of calculation. So let's suppose we have this shape shown here of height h width b with a hole cut out of it of width's b1, it should be b1 here, width b1 and height h1. So in this case, the centroid of this composite section about its centroid is equal to the moment of inertia of the outer area. Which is bh cubed over 12 minus the moment of inertia of the hole here. Which is width b1, height h1 which is b1h1 cubed over 12. So, let me do an example on that. We have a rectangle 3 meters by 3 meters, or a square, and it has a hole in the middle, which is 2 meters in diameter. The moment of inertia about an axis parallel to the x-axis through the centroid is most nearly which of these? So we want to calculate the moment of inertia about the horizontal x-axis through the centroid in this case. So here's our formula. The moment of inertia is equal to the moment of inertia of the rectangle minus the moment of inertia of the hole which is a circle. So if the moment of inertia of the rectangle is, about its centroid, is bh cubed over 12, and the moment of inertia of the hole, the circle, from the previous tables is pi r to the 4th, over 4. So substituting in the numbers we have the width and the height is 3 meters so it's 3 times 3 cubed over twelve minus pi and the radius of the hole is 1 meter. Diameter is 2 meters. So it's pi times 1 to the 4th over 4 is 6.75 minus 0.79. Is equal to 5.96 meters to the 4th, so the closest answer is B. And this concludes our discussion of Statics.