 Welcome back to the next session. The next session would be you know this one would be on area moment and mass moment of inertia. So, to start with why you know it is first of all why it is inertia. See that is one of the important concept because we have looked at the Newton's second law also where we said mass times acceleration and there you see the inertia term is typically used for resistance. So, that means the resistance to motion that can be defined as inertia. So, now if you look at statics and dynamics there are two different aspects. We only talk area moment of inertia where is in dynamics we only talk about mass moment of inertia. So, we are going to you know talk about this you know differently actually. So, we will first look at area moment of inertia and then after that we are going to study mass moment of inertia. Remember that in engineering mechanics especially you will not find the term I that is area moment of inertia. You see that mostly on strength of materials types of courses where for static problem we are actually solving for let us say bend you know deflection bending buckling and so on and so forth. So, you can clearly see that the resistance or inertia comes into play when the body is actually trying to deform in statics. So, that is the whole issue about it and that is why you know in engineering mechanics as such it is not showing anywhere in any text book. The second one in rotational motion when we consider that is in dynamics we also see the inertia. So, there we also refer to I as the mass moment of inertia, but it is taken in a different aspect all together because their body is under already motion. Therefore, we should have the resistance in the form of you know mass moment of inertia, but in statics body is not at all motion it is under static. So, therefore unless otherwise we go get into the subject of deflection or bending or buckling we are not going to really see the I terms that is the area moment of inertia coming into play. So, with that let us still try to look at what is area moment of inertia. How do I look at this? Remember there is a well connection in centroid typically what we have done. In centroid I kept saying that we are taking a very elemental force delta F let us say because if you think of a thin plate once I am talking about a thin plate I am interested in finding out of that thin plate where is the centroid. So, in that one really if you think of in terms of a small weight of an area right. So, the force or the weight is proportional to the area. Now if it is a thin plate that force is not varying with the distance that is the whole issue. So, in other word that force is only proportional to delta A. However, remember there are problems where this force that small force delta F will vary with the distance from the axis. So, we have seen this already in the hydrostatic pressure problem. However, we have not really thought about those problems in terms of moment of inertia. In case of a hydrostatic problem what is happening if you look at let us say A B C D is the dam or valve right that we already studied. Then remember we calculated a pressure. Pressure is equals to rho G Y. Therefore, the force is equals to rho G Y multiplied by D A. So, therefore you can clearly see that the force right on this elemental area is actually varying with the depth. So, therefore force is a function of Y distance from the axis A B. Therefore, if we really take the moment of this force what we are getting is integral of Y square D A. So, the moment about the axis A B in this case is Y square D A. Similarly, when you look at the strength of materials in the bending of a beam let us say a beam is bending and I am interested to find out the deflection or just to let us look at how the stress is distributed. You will see that stress will be distributed in a linear fashion from an axis which we call the neutral axis. So, from this axis O O we are going to see the stress which is nothing more than you know kind of equivalent to the pressure. Again it is varying in a linearly fashion. Therefore, what is happening again you see that the moment produced by this force is going to be Y square D A. So, moment is always proportional to Y square and that is why the resistance that the body sees it is always in terms of the Y square. Similarly, if you think of a torsion problem let us say I have a torsion problem. So, what it means that I have a shaft and I am simply trying to give it a twist. Again it can be shown that the stress distribution that we have or that is going to vary linearly with respect to the radius. So, you have the stress as if in terms of k r and if you take the moment of that then it becomes k r square. So, everywhere you see that I have a resistance from the body in terms of Y square. So, that resistance is a moment resistance given by the body in order to you know because it is trying to be under motion. So, it is either trying to deflect or buckle or twist therefore, it is given the resistance. So, with that basic concept now what we see therefore, that in case of the centroid studies I have only considered integral of Y D A or integral of X D A. But in case of moment of inertia we are really going to look at integral Y square D A. So, the Y square D A is nothing more than second moment of the area. So, in the centroid as I kept telling you that it is going to be only first moment of the area the reason being force is simply proportional to the D A. It is not varying from linearly from the axis of the and the concerned axis. So, therefore, we just concentrated on first moment. In this case however, you see when we take the moment of Y D A it becomes Y square D A therefore, we get the second moment of the area. So, in case of statics moment of inertia can be simply said to be second moment of the area. So, we call it area moment of inertia or second moment of the area. So, ultimately therefore, what it boils down to that if I want to define let us say this is my shape. Now, in this shape I want to find out the second moment of the area with respect to X axis with respect to Y axis. So, what does that mean? That means again we are going back to the integration approach as we have adopted for the calculation of centroid. So, there if we take all of this into account then we will see that I X is really equals to what? Y square D A where D A is the small area D X D Y and I Y is equals to integral X square D A. So, therefore, the second moment of the area can be simply found by these two formulas. So, one is about the X axis another is about the Y axis again problem is like this. Remember if I take D A as D X multiplied by D Y we have to do a double integration which is not easy and we have said therefore, during the study of the centroid typically we will always try to get into rectangular strip either it could be vertical horizontal or we will try to look at the problem in polar coordinates. So, therefore, now we are going to do the same thing here just understand this part correctly. So, what we have in hand I have again let us say I am trying to solve this problem and I am interested in finding out what is the moment of inertia about the X axis. Then I am interested to find out the moment of inertia about the X axis then my chosen strip is actually horizontal you see that the strip that I have chosen is horizontal. This has an area how much A minus X multiplied by D Y. So, if I want to calculate the moment of inertia of this small strip then I just take the second moment of this that means Y square D A where D A is clearly defined. Now, to get the I X all I had to do I have to integrate it out from 0 to some height. Now, problem is this how about if I choose this element and try to calculate the I Y. If I choose this element horizontal element and try to calculate the I Y it is going to be complex because first I have to know what is the moment of inertia of this element about its own centroid or rather about an axis that is passing through its centroid and parallel to the Y axis. So, I have to first know what is the moment of inertia of this parallel to Y axis that is passing through its own centroid. So, therefore, to avoid this problem what we do if we want to calculate the I Y we will prefer to choose a vertical strip. If I do choose a vertical strip remember that now I Y is simply equals to X square D A right where D A is given by this area. Now, this area is already defined this is Y multiplied by DX. So, we will go to simply we will take X square D A and our problem can be easily solved by carrying out the integration along X. So, now let us look at a quick example. Therefore, suppose I am interested in finding out the moment of inertia of this rectangle. So, it is a rectangular area I want to find out the moment of inertia about the X axis. So, as I said once I want to find out about the X axis it will be always Y is to get the strip horizontally that is parallel to the X axis. So, that the moment of inertia of this small element itself about its own axis X axis will be second order which I can ignore. So, only thing that we see is actually Y square D A. So, therefore, if I look at I X I X is simply Y square D A. So, Y square D A now what is my D A? D A is nothing but B multiplied by D Y. So, B D Y if I substitute and if I carry out my integration from 0 to H I am going to get B H cube by 3 which is a very well known result. So, at least what we have now understood if I have a rectangular area. So, for the rectangular area I know what is the second moment of the area that means moment of inertia about the X axis which is B H cube by 3. So, what would be the moment of inertia then about Y axis? The moment of inertia about the Y axis will simply be equals to just reverse this. Now, there B becomes H H becomes B. So, it will be simply H B cube by 3. However, to get that we will change the direction of the rectangular strip. We will just make it simply vertical strip. So, in some problems you would not be able to avoid maybe this vertical and horizontal strip issues. So, for example, now what is being demonstrated in this here right here. Suppose can I get the both I X and I Y based on one strip concept. That means either I choose a vertical strip or either I choose a horizontal strip. Can I get both the moment of inertia? So, what happens remember when I choose the vertical strip I should have no problem getting the moment of inertia about the Y axis because that is going to be simply X square dA where dA is given by Y dx right. So, for D therefore, what is happening the point here is D I Y the small moment of inertia that is developed about the Y axis due to this rectangular strip vertical strip that is D I Y that is simply X square dA that is X square Y dx. Now, you can easily integrate this out to get the I Y. However, if I want to get the I X what I have to know a priority what is the second moment of area of this vertical strip that small strip about its own base that is about the X axis and we have already studied that. We have already said that of a rectangular area it is always going to be BHQ over 3. Therefore, when you are looking at the X axis you can see the D I X of this rectangular area about its own base right that is equals to simply now B becomes DX. So, B is DX HQ is YQ that divide by 3. So, we are really taking these one from our previous you know calculations we are just taking BHQ over 3 that is really D I X for this small elemental strip. So, with this concept now you can clearly see everything kind of boils down to the integration. So, ultimately all I had to do an X square DA or Y square DA. So, ultimately the rectangular strip and vertical strip will again depend on the type of problem I have in hand for simple shapes. One more thing that comes into play when we have to calculate the moment of inertia about the Z axis remember here I have only talked about X and Y axis in this problem everything in the X and Y axis. So, when we are trying to calculate the moment of inertia about the Z axis. So, Z axis is perpendicular to the plane that is typically called polar moment of inertia. So, in polar moment of inertia we will be calculating the moment of inertia of this body as if the body is trying to rotate about the Z axis right. So, if the body is trying to rotate about the Z axis then I have basically I Z which is referred to as J 0 in this case. So, that is the polar moment of inertia and by definition you can clearly see that if I have a little force as I have explained the torsion problem if I do have a force which is equals to Kr. So, force is varying linearly from the let us say this center. So, therefore we have some Kr here. So, Kr multiplied by R that will create a moment which is Kr square right. So, therefore we are basically having R square dA. Now, what is R square? R square is nothing but X square plus Y square. So, therefore what we have essentially for the polar moment of inertia that is equals to X square plus Y square dA. So, if we break it then X square dA plus Y square dA that is equals to I Y plus I X ok. So, polar moment of inertia is nothing but some of the other two moment of inertia about X and Y axis. But remember in all of this problem we talked about the body is trying to rotate about an axis that is the whole idea. So, when we are talking about I X when do I need I X when the body is trying to rotate about X axis. When do I need I Y body is trying to rotate about Y axis and why do I need I Z which is referred to as J 0 polar moment of inertia body is trying to rotate about Z axis. So, when the body gives this resistance to the motion then we have to calculate this I X I Y for variety of problem such as bending, buckling, deflection and so on so forth ok in case of statics. So, for planar areas as we have discussed now there is another important concept the concept is as follows. If you now think of suppose I have found the I X now can I say like this what I can say now that I will consider a very long strip which has the same area as that of A. So, which has the same area as that of A and I am simply going to lump that area at some distance k X. So, there is a large rectangular strip which is parallel to the X axis such that I X of that strip about its own centroid can be ignored because I have a very large B and H is very very small for that particular strip ok. But at that area is equals to the original area then what we can say then what is the moment of inertia of that strip which is having an area A that is simply given by k X square multiplied by A then we will get a very important concept which we call the radius of gyration. See what it is in physical term it is basically the distribution of the strip. So, what it means as if I look at let us say rectangular body we can decide which way it has it is giving me the more resistance. Very quickly I can say if I have the radius of gyration which way it is giving me the more resistance ok. So, I can get the value of k X by simple this operation I X by A and I will be able to say wherever you know k X if k X is more than k Y then I can simply say ok you know my body you know having more resistance about that axis. So, radius of gyration will be very fundamental and you will see that only static problem involving buckling. So, buckling means let us say let us say I have a column. So, in case of a buckling let us say I have a column and I apply you know this vertical force. Now depending on its shape the column can buckle try to buckle in any direction either it could be about X axis either it could be about Y axis. So, what we can look at that which direction my k X that means radius of gyration is more is it along you know about the X axis or is it about the Y axis. So, wherever it is more it will have it will not try to deflect on that direction it will not try to buckle in that direction, but rather try to buckle in the direction where I have laid radius of gyration. So, conceptually it will be very easy to thought out if we know the radius of gyration. Similarly for the polar moment of inertia ok again it is a distribution of the shape all around the body. So, what is being done suppose you are taking a annular area of A. So, that area is equals to the actual area A. So, this is just an annular region then we can say what is the polar moment of inertia of this annular strip that is simply given by k0 square A r square dA. So, r square in this case is k0 square multiplied by A. So, I get k0 equals to square root of polar moment of inertia divided by A. So, therefore I should be able to get also the radius of gyration about its Z axis. So, let us try to study a few problem in stat to find out the moment of inertia that is second area moment in case of planar bodies ok. So, we are not going to go to the three dimensional body it is a 2D body as we have done in the case of centroids we are going to do also for the moment of inertia. So, let us say determine the moment of inertia of a triangle with respect to its base ok. So, with respect to its base if I you know study this then as I said it will be always preferred in this case to have the rectangular strip which is parallel to the x axis. So, we do that therefore as we can see the of this strip what is the moment of inertia of this strip about the x axis that is dIx is simply given by y square dA ok. Remember the moment of inertia of this strip about its own centroid will be ignored the reason being this dY is very small. Because you are going to get a dY cube when you are trying to think of the moment of inertia about this own you know about this rectangular strip of this rectangular strip about its own centroid. So, there will be a dY cube term that can be ignored. So, only thing that is coming into play is y square dA ok. So, therefore now what is my dA? dA is ldy now l is actually varying with the height ok. So, now you have to find l in terms of y. So, we get l in terms of y in this case alright and then we substitute into the dA and then dA will be substituted back to dIx. So, ultimately we are coming back to an integration process in that integration process we just have to integrate it from 0 to h. So, I will get the moment of inertia about the x axis which is BHQ over 2L. So, the moment of inertia of the triangle about its own base will be BHQ over 2L. So, now suppose I choose a vertical strip. So, as I have commented here could a vertical strip have been chosen for the calculation? What is the disadvantage to that choice? Now remember in this problem if I choose a vertical strip although I can get the you know moment of inertia of this vertical strip itself about its base that would be simply one third dx y cube, but remember there is a discontinuity in y as you touch this point. So, in the integration limit you have to break you have to first go up to here then again your integration limit on the dx should be from this you know height right here onwards. So, that is going to create a major disadvantage to solve this problem if we start solving from the vertical strip concept. So, now what we will do? First we will try to get the polar moment of inertia of a circular area by direct integration. So, I am interested to get the polar moment of inertia of this circle by direct integration and then from this result I am trying to determine the moment of inertia with respect to its own diameter that means moment of inertia with respect to let us say x y. However, since it has you know axis symmetry you can choose any diameter it will be same moment of inertia all around will be same. So, how do I attack this problem? How do I start solving this problem? Now you see I am changing the strip. So, now I have a annular strip. So, the annular strip was chosen at a radius u with thickness du. So, ultimately what is varying is u and therefore I have to integrate from 0 to r where r is the radius of this circle. So, therefore this is much easier because polar moment of inertia is already defined. So, if we look at it polar moment of inertia of this annular strip that is already u square dA because it is coming from the concept of r square dA. Now what is my dA? dA equals to simply the area of this annular strip which is 2 pi u du. Therefore, the polar moment of inertia is equals to can be just obtained by solving this integration. So, 0 to r that is my integration limit and therefore I can solve it pi r to the power 4 by 2. Now when the polar moment of inertia is solved I already know that J0 must be equals to Ix plus Iy. So, polar moment of inertia about the z axis which is perpendicular to the slide that should be equals to Ix plus Iy. We have already proved that and now in this problem Ix is equals to Iy because it has bidirectional symmetry. It has symmetry in both direction not only that even in this problem you have symmetry all around. So, for this problem we can clearly see that Ix equals to Iy and therefore we can simply say the either J0 equals to 2 Ix or 2 Iy and I should be able to get the either Ix and Iy. So, I diameter rather will be pi r to the power 4 by 4 which is simply half of the polar moment of inertia. In other words polar moment of inertia is twice that of the second moment of the area about any axis. Now I have also commented why was an annular differential area was chosen? Would a rectangular area have worked? Of course it would have worked, but problem would have been more complicated. We could have just chosen a rectangular strip also no problem, but mathematically it would have been more difficult to solve it that is all because then you have to have r square equals to x square plus y square. So, from that logic you had to either express x in terms of y or y in terms of x, but it would have worked, but mathematically we had to do more brainstorming. Now remember I can solve the same problem although the simplest possible way is what we have discussed in the previous slide. I could have solved it in this way also. Now what I have done here? I have taken a very small elemental area. So, here the concept of dx dy is coming into play. So, ultimately suppose I want to solve the Ix what is my Ix? Ix equals to y square dA. Now remember dA has dx and dy. So, therefore I would have performed also double integration. So, the problem is chosen because just to show that we could have done also double integration to solve some problems. Now when we do the double integration in this case, now how that is coming into play? Because since I have taken a small elemental area, remember that elemental area is taken at a distance r0 from the center. So, therefore this way it is dr0 and here I have rd theta. So, r0 d theta multiplied by dr0 is my area. So, the area is coming into play here nicely. So now remember since r0 and theta both are varying. So, I have to put the integration limit. So, the integration limit is taken from for r it should be from 0 to r. For r0 I have to take the integration limit 0 to r and for theta I have taken the integration limit from 0 to 2 pi. Now remember there is another important point here to consider what is now my y square. Since I am calculating Ix, I need the y square. So, what is the distance of this from this x axis? It is simply given by r0 sin theta. So, that has to be substituted here and once you do the full integration I am going to get pi r to the power 4 by 4 which is same as what I have got in the previous slide Ix is pi r to the power 4 by 4. So, again we can just leave it for the homework that can we find Iy now. So, Iy equals to x square dA. So, that can be easily done because now you have to say what is x? x is nothing but r cosine theta r0 cosine theta. So, you have to just substitute that carry through this and we are going to get the same result. So, result is not going to change. Now let us look at determine the moment of inertia of the area under the parabola about the x axis. So, you want to get the moment of inertia Ix of this planar area which is defined by x equals to ky square. So, it is now x equals to ky square. Now remember I could choose both strips in this case one could be vertical and one could be horizontal. So, we are going to explore in both ways. Let us first try to study in order to get the Ix I have always said that it is always advisable to go with a strip that is parallel to the x axis. So, if I take a strip parallel to the x axis for Ix then what I ultimately get? I ultimately have Ix equals to y square dA. So, what is my dA now? So, dA will be 4 minus x dy as we can see here. So, therefore we know the dA in terms of y and therefore we should know what is Ix. So, only thing again that is coming into play what is the area of this small strip and what is the second moment of this strip about the x axis that means what is y square dA. So, once we know that this is the simplest way to solve the Ix now we can solve the Ix also can we solve the Ix based on a vertical strip. Now if you want to do that remember now I have to know what is the moment of inertia of this vertical strip itself about this base that is the x axis. So, that is nothing but dx y cube by 3 right because the moment of inertia of a rectangular area about its own base is already learnt that is bh cube over 3. So, therefore now you can see dIx as I said is going to be dx y cube divided by 3 and now we simply have to integrate this now from 0 to 4 along the x axis. Remember since my variable is x in this case therefore y must be replaced in terms of x which I can easily do and that is what it is done we have replaced y in terms of x and therefore I can get the solution. So, from both approaches using horizontal strip or vertical strip I am going to get the same answer. So, what we will do now quickly one more important aspect that comes into play suppose I know about one axis the moment of inertia can I transfer to the other axis. So, there we have to get the concept of parallel axis theorem. So, for example what is being asked here suppose I have a general area I want to find out the moment of inertia of this rectangular area sorry this area rather arbitrary area about the axis A A prime. So, I know from definition that I A A prime. So, I A A prime should be equals to y square dA where y is the distance of that small area dA from the A A prime axis. Now, y can be broken into two parts y can be taken as d plus y prime. Now what is d this is very important now d is the distance from the A A prime axis to the centroidal axis B B prime. So, if we do that then we can see that I should be equals to integral y square dA where y is equals to y prime plus d. Now if you break it into you know just expand this one then we have first term as y prime square dA plus 2 d integral y prime dA plus d square dA. Now what is coming back here what is this term in the middle what is my integral y prime dA what does it mean. Remember y prime dA because this is a centroidal axis of this area we have already learnt y prime dA is nothing, but the first moment right. And the first moment of the area about its own centroidal axis should be equals to 0. So, what it boils down to ultimately that this term is equals to 0. And therefore what we have I A A prime equals to I cap. So, I bar is given because now this integral y bar y bar you know y prime square dA this is about the B B prime axis which is a centroidal axis. So ultimately we have I equals to I centroidal axis plus A D square. So, what it essentially mean that if I know the moment of inertia of simple shapes about its own centroid then I can calculate using the parallel axis theorem about any axis which is parallel to that centroidal axis. So, therefore we can take quick examples for example in this problem let us say I am interested in finding out the moment of inertia of this circular area about this line about this axis. So, how do I do that? So, remember we already know what is the moment of inertia about its own centroidal axis this one that is pi r to the power 4 by 4 right. And then all we have to do we have to apply the parallel axis theorem. So, now I have A times D square where A is pi r square and D is this distance. So, that is simply r square. So, I ultimately have the answer as pi r to the power 4 multiplied by 5 over 4. Similarly let us look at the triangular area in different way. Now what has been done we have already studied in this problem what is the moment of inertia of this triangle about its own base A A prime that is already known about the base axis we have already said it is going to be BHQ over 12. Now can I get it about the centroid? So, can I get it about the centroid axis BB prime. So, again we go to the parallel axis theorem we know that IAA prime should always be equals to IBB prime that is the centroidal axis plus AD square right. And therefore we just change this. So, we are interested in now to find out what is IBB prime rather I you know with respect to the centroidal axis. So, I just do IAA prime negative AD square. Now this was already known that is BHQ by 12 negative we have the area which is half BH multiplied by the D square. Now what is D? D is the distance between the centroid of this right to the AA prime axis. So, distance between these two parallel axis which is H over 3 that we already know. So, ultimately the answer is BHQ over 36. So, what we have learnt now that the moment of inertia about the centroidal axis BB prime is BHQ by 36. So, can we do this now? So, how to target this problem? See let us go step by step. So, what we have done from the earlier studies? So, I am interested in finding out the moment of inertia of this half circle about the x axis. Now I have to make use of some known results ok. What are the known results I have in the hand? First of all if I just choose the base of the half circle right I know what is the centroid location that is r bar that was already been done. So, that r bar is 4 r over 3 pi ok. Second thing I know is that what is my Ix prime? Ix prime since it is a half circle Ix prime should be just half of the moment of inertia of a circle with respect to its diameter. So, therefore I already know that it is going to be now this is for the circle. Remember pi r to the power 4 by 4 is the moment of inertia about its own diameter of a circle. So, for the half circle it is just going to be divided by 2 nothing is changed ok. So, I know this I know that now I have to utilize the parallel axis theorem very very carefully with whatever information I have in the hand. So, what I will do suddenly I will now try to go from this axis to that axis because I is known in this axis. For me Ix prime is known right from Ix prime I will try to go to Ix 0 which is the centroidal axis. So, that can be easily done. So, what I have here Ix 0 that is I bar. So, bar will always refer to centroid. So, centroid here that moment of inertia about centroid should be equals to Ix prime negative a times d square. So, a times r bar square because now I am changing from here to here ok. So, I will get this Ix 0 first. Now my job is simple and I have the Ix 0 I know this total distance. So, what is Ix? Ix will be Ix equals to Ix 0 plus area of this multiplied by the distance between this right. So, from the centroid to the distance of the x axis. Now therefore, what I have Ix as you can see here is simply Ix 0 that is my Ix 0 I bar plus area right multiplied by the distance the total distance is 15 plus r bar. So, I get this ok. Now what we will do quickly? Since we know now for the small small shapes ok now you tell me which one I do not know here all of these things are actually done in bits and pieces. I know the moment of inertia of all of these shapes or planar areas rather about the x axis and y axis. So, all are done ok. So, once I know the moment of inertia of all these sections or small areas now we can go to the composite areas. So, how do I solve problem involving arbitrary shape or composite areas? Therefore, depending on the type of shape given I simply have to discretize that area into the known shapes ok. So, that will be the main concepts. So, let us look at this ok. So, I am interested in finding out what is the moment of inertia of this composite area with respect to the x axis. Now why it is composite area? Because you can see that I can easily solve this problem. First I take the moment of inertia of this rectangle about its base right and then subtract the moment of inertia accordingly of this, but it is going to be challenging. It is not that simple at all because remember parallel axis theorem we have to adopt in these cases and parallel axis theorem only allows from you know that allows from changing the axis from the centroidal axis to some other axis. As long as I know the distance from the centroidal axis to the other axis then I should be able to you know get the moment of inertia about that axis. So, how do I go about solving this problem now? So, first compute the moments of inertia of the bounding rectangle and half circle with respect to the x axis. See what it is telling? We divide this in rectangle as well as let us say there is already of this shape what is the moment of inertia about the x axis. Then what we can do the moment of inertia of the shaded area? The shaded area can be obtained by subtracting the moment of inertia of the half circle from the moment of inertia of the rectangle. So, let us look at the process here. So, for the rectangle part it is very simple. For the rectangle if we look at this that is simply b h q over 3. So, b is 240 h is 120 that divide by 3. So, I will get the moment of inertia with respect to the x axis for the rectangle. Now, let us try to look at what happens to the half circle. What is the moment of inertia about the x axis? So, for the half circle now remember what is known to me is moment of inertia with respect to a a prime. So, moment of inertia with respect to this one is known if you go back to the previous shapes that we have already studied you can clearly see the moment of inertia is pi r 4 to the power 8 of this shape moment of inertia about the base that is x axis is pi r 4 to the power 8. So, we are going to take this help which is done here. So, ultimately I know the moment of inertia about this. Now, what I have to do? I have to get the moment of inertia about the centroidal axis of this area. So, again I apply the parallel axis theorem which will state immediately that I x prime because this is a centroidal area should be I a a prime negative a a square. Remember in other words I a a prime should be equals to I x prime plus area of this multiplied by a square. So, we are just taking it other way around because I know a a prime I want to solve for the centroidal moment of inertia about the you know this axis centroidal axis. So, now we are going to do this right now I a a prime is already known which is pi r 4 by 8 negative a area is also known we have already done in our previous study when we are talking about the area and centroid calculations. So, that can be done then this is my moment of inertia of this with respect to x prime. Now, what is my objective? My objective is simply to subtract this moment of inertia from the rectangle. Now, we have to go one more step ahead because I know about the x prime axis. So, what is the moment of inertia about the x axis? Let us find that so I x is simply equals to I x prime which is about the centroidal axis right plus a the area multiplied by b square. So, once we do that we get this one. Now, what happened in the process? I know two different result one is that of a rectangular area with respect to the x axis one is that of a half circle with respect to the x axis. Now, our original shape has simply a cut out of the half circle. So, therefore, I need to simply subtract the moment of inertia of that cut out which was found already with respect to the x axis from the rectangular area with respect to the x axis right. So, I am simply going to take I x minus this I x and that will be my net I x of the shaded area. So, therefore, what we have now this is kind of the ultimate you know picture that we are looking at. We have found the this area has been found rectangular area about the x axis. We have also found the moment of inertia of this shape with respect to the x axis. I just have to subtract this one from this. So, I am going to get this result is that clear now can I have any question on this let us say moment of inertia of planar area. How to find a irregular body moment of inertia with some arbitrary axis? See, we are not discussing for the arbitrary axis because it will take you know additional steps forward because in e make course you know we do not get into this principle axis and all that that will be required. So, we basically try to transform from parallel axis to the other parallel axis as much as we can and if you look at it know where are the real applications real applications of these will be some bending problem of beam in statics let us say or may be deflection problem in the you know in some kind of columns or beams. So, when we are trying to study those kind of problems just look at the problem we always try to look at the you know bending of beam about two perpendicular axis. So, there we are getting only about two perpendicular axis which are parallel to their own edges. We really do not take this thing forward towards you know arbitrary axis unless otherwise your loading pattern is very complicated such that bending may happen about that arbitrary axis. But, if required we solve this we again do this in the strength of materials. While dealing with the equivalent force systems you stated a system of forces is been reduced to single result in force and equivalent couple is it couple or moment. See it is always the reason we said couple you remember see it is the same thing because if you think of it it is a couple moment rather it is called you know couple moment it is in some text they will call couple some text it will be called moment. Say ultimately moment is a very general term you can use it. But, why we are saying couple can you give me the video please I will just explain why did we say it a couple. So, your question is why we are saying it a couple think of just look at this pen right. Now, how we are transmitting this force let us say I have a force here right how I am translating this force to here I am translating this force by taking this force here as well as another force which is acting from this. See why the couple is coming now this force and that force they are creating a couple. So, that is why we have introduced the concept of couple. So, it is a moment it is a couple moment it is a moment about the axis that we see which is perpendicular to the plane. But, it is a couple moment. So, there is no harm to use either moment or couple moment or simply a couple it is a equivalent force couple system. The reason being just because the transfer mechanism that we are adopting in order to get the effect of the force on to the other point clear is that clear. I have another fundamental verification here what is the difference between couple and torque. Again very. So, your question is what is the difference between a couple and torque. See couple again you explain in terms of this what is a torque first let us understand when we say it is a torque torque will be something. See it is basically applied in the problem involving torsion of a shaft. See if you think again of this pen right when I will say it is a torque when it will try to rotate this pen about its own axis is that clear it is better to use this word torque. Now, if I have a force on this system that is really tries to rotate the you know this pen about this axis which is this axis then I cannot say that is a torque. Now, couple is always the force that is the equivalent opposite force when it is on the body then we say it is a couple that means I have one force applied here I have other force applied here I will say it is a couple. Action of that couple will produce a couple moment, but torque is completely different issue or torque is something that is rotating about its own axis. Now, I can induce the torque let us say I can induce the torque if I go like this I have this pen and I have you know another let us say you know kind of bar that is like this and I will apply one load here other load here is that clear. So, in this bar that I have I will apply one load here another load here that will simply try to twist the entire body about its own axis then I will call that a torque. Remember moment is the most general concept, but torque is a specified nature of moment which we only study when we are talking about torsion problem it is a moment about its own axis okay it is a moment okay let us move on to any other question. In generally what is meant by inertia? I said when I started see inertia is the resistance okay resistance to motion it is as simple as that why it is resistance see it will be more clear when we talk about dynamics you know when I will be talking about dynamics it will be much more clear, but think like this again I can explain in a very simple way see let us say this is the axis you think in dynamic form okay and I have a mass here which is trying to rotate about this axis okay then what will happen it will try to you know if I produce if I give it a couple which is trying to rotate the mass about its own axis about this axis then there will be a resistance to that motion the resistance why it is giving the resistance because different mass will induce different types of resistance depending on the fact that how big is the mass and what is the distance from this axis of rotation right that will determine how fast I can move the body how slow I can move the body isn't that so in other words depending on the r and the mass that we have basically you are now talking about r square dm because you have now angular acceleration into the system so that will try to give a inertia force back onto this axis is that clear in a dynamic way you can think of in terms even particle if a particle is trying to move you have a inertia frame and you have a non-inertia frame inertia frame would be something that I am sitting here this particle is trying to move in the inertia frame it will always see some resistance but if I am as a person going to this particle and moving with the particle then I will make it a non-inertial frame by saying there is a inertia force onto the system it is a resistance okay so this study you know it will be more clear if it is in the dynamics if it is taught in the dynamic but inertia is just a resistance to the motion you can think of a bending problem if I have a bending think of I am trying to apply this you know movement equal and opposite the beam will try to deflect beam will try to bend right therefore immediately in the formulas you will start experiencing that I am getting this I term whenever I want to calculate deflection whenever I want to calculate stress I am going to get the item why because I am trying to get a resistance of this deflection that is given by the moment is that clear so inertia is nothing but the resistance to the motion in dynamics yes it is a complete motion of the rigid body in statics remember since we have considered the rigid body it is not actually deflecting so we need to look at strength of material to understand what is happening okay.