 We will come back again. So, in this session we are going to mainly discuss about the centroids or how to find centroids of planar area. So, our attention would be you know mostly on the planar area and that is why we use the term centroid. Remember the center of gravity or also known as center of mass that can be applied to three dimensional bodies. So, as we all know that that art exerts a gravitational force on each of the particles forming a body and consider how your weight is distributed throughout your body. These forces can be replaced by a single equivalent force that is equal to the weight of the body and applied at the center of gravity for the body. Now, so therefore, in a three dimensional system we use the term center of gravity. However, for a planar area it is typically known as centroid or rather center of area. Remember to find out the centroid the concept that comes into play is the first moment of an area should be used to find out the location of centroid. Now, why we are doing this analysis as we have said in the previous session that most of the engineering analysis will involve distributed loads. You can think of an example of a let us say dam. So, where you have you know the water pressure is being distributed in a triangular fashion because as you go down the depth then your pressure will also vary in proportional to the depth ok. So, in this way we can say that in order to replace those kind of distributed force system by a single force we should know the centroid of the distributed load. So, ultimately what is important is that area under the distributed load that should be the resultant force of the single load that we are looking for and its location should be at the centroid of this distributed load. So, we will try to you know expedite this session assuming that most of us are familiar with this subject. In fact, students have done you know already some kind of they should have some familiarity with the topic. So, ultimately remember the concept comes from the equivalent system. If we are talking about a center of gravity of a plate let us say. So, plate means it is a thin body and let us assume that thickness is uniform everywhere ok. So, now I can discretize this plate in small small area and therefore for each area I have this delta W ok. So, finally what is happening if we can see from the equivalent system perspective. So, this system which is considered to be distributed load and load is simply proportional to the area. Remember that the load is not varying with the distance it is proportional to the area, but it is not proportional to either x or y that may happen in some situations as well. So, if we consider this. So, therefore we can say that if you discretize it into small small area then the moment of this system right about the x and y axis should be equals to the moment produced by the resultant system. So, this is the total weight of the body that is passing through the centroid G let us say center of gravity G. So, then we can simply use the equivalent system concept to say that I the location of the center of gravity along the x direction. So, that is represented by x bar. So, x bar multiplied by W should be equals to sum of moment due to the individual weights of this small small area. So, in an integration fashion we can write x d W similarly if we you know equate the moment about x axis we will get y bar W equals to integral y d W. So, for the line of you know if we want to find out the center of gravity of an wire the same process follows except for the fact in this case I have a line element here I have a area element ok. So, ultimately using these two formulas we can find the centroid sorry center of gravity in this case. So, centroid is nothing, but we just take as I said that you can buy that concept of center of gravity basically then what we have if we think of it is a thin plate then W can be replaced by the specific weight multiplied by area multiplied by thickness ok. So, in this way what we can get rid of basically the specific weight gamma multiplied by T. So, ultimately what we have x bar A should be equals to integral x d A. So, this is my first moment as we can see the first moment with respect to y. So, that will give me x bar similarly the first moment with respect to x axis will give the y bar. Using the same analogy instead of the area A we can think of you know line element let us say delta L ok. So, in this way again we can say what is the weight of this delta L that is simply equals to gamma that is the specific weight multiplied by the area cross sectional area of this element multiplied by d L. So, volume multiplied by specific weight so, that will give me the small weight right. So, finally, from the both sides again gamma A will be cancelling out. So, we get x bar L equals to integral x d L and y bar L equals to integral y d L. So, we can find out the position of the centroid from our predefined axis x and y. So, what it eventually means that if we locate the centroid and if you have the centroidal axis that means, first moment about the centroidal axis has to be equals to 0. So, with respect to the centroidal axis the first moment must be equals to 0 ok. So, if we look at this concept particularly that if we think of the first moment of areas. So, what is being displayed in this slide is that can you find out the cross section which are symmetric how do I find out cross section are symmetric. So, look at the first example the first example is here. So, an area is symmetric with respect to an axis B B prime. So, this area can be called to be symmetric with respect to an axis B B prime if for every point P there exists a point P prime such that P P prime is perpendicular to B B prime and is divided into two equal parts B B prime. That means, this cross section will now be symmetric about B B prime and therefore, the centroid of this cross section should be at this line. We can clearly see the first moment has to be equals to 0 about this axis. The reason being if we consider from the line B B prime if I have an area located at some distance let us say D then the other area is located at some distance negative D and when you take the first moment then ultimately A times D multiplied by A times minus D that will cancel out. So, we get the 0 first moment about the B B prime which is the axis of symmetry. Similarly, in this case we can clearly see the line of symmetry is this and therefore, the centroid has to be at the in this axis O O I. Now, look at this problem if an area possesses two lines of symmetry. So, in this problem we have two lines of symmetry its centroid lies at their intersection and at the third remember if we just look at this particular cross section it looks like asymmetrical is not that, but ultimately we can again say that with respect to centre O if we just observe with respect to centre O if there is an element D A at x y then there is an element D A prime at the negative x minus y. Therefore, if you try to take the first moment about the x axis or first moment about the y axis both will be equals to 0. Therefore, the centroid will be at the intersection of the x and y axis which is the O. So, how do I determine the centroid by integration? So, this is a direct method to find out the centroid of different types of shapes and as we have seen that our basic formula is x bar A should be equals to integral x D A. Now, remember if I want to break the D A into D x and D y that is my small element D x and D y then I have to all the time perform a double integration. So, to avoid this what is usually done the D A is defined as a thin rectangular rectangle or strip. So, that means now what is happening if we have this you know if we want to locate the centroid of this shape then we will take the D A as a thin strip which could be like this. So, it has its own centroid right. So, when we take the moment of this particular strip then we have to basically look at what is the centroid of this strip from the x axis as well as from the y axis. So, we should be able to understand rather for this element which is a infinitesimally small element what is the centroid of that from the y axis that is the x bar element and what is the centroid of this that is the y bar element. So, we can clearly see if I have to now if the you know the line is known here. So, it is defined by some p x y let us say the shape is defined properly then if I use a vertical strip like this then we can say that x bar times A that is the x bar I am trying to determine. So, I have to take the moment about the y axis right. So, if I do that then I have x bar element D A. So, now what is my x bar element x bar element in this case equals to x and the area is y dx. Similarly, if I take the moment of this strip about the x axis in order to find out the y bar that is the centroid of the area then I should do y bar element is nothing, but y over 2 right because this strip remember is of length y and width dx ok. So, because defined by a point p x y. So, now I have choice to do this. So, I can either have a vertical strip or either I can have a horizontal strip. So, if we use a horizontal strip then you see we can we will try to complicate it. So, it is clearly displayed here. Now, if I have to find out let us say what is the centroid x bar ok. So, x location of the centroid then I have to find out first that what is the x bar of the element. So, in this case x bar of the element should be equals to A right plus x divide by 2 ok and the strip area is already defined that is A minus x multiplied by dy where dy is the width ok. Similarly, we can see what is the y bar element y bar element will be simply y ok. So, finally, we have to use this integration remember to perform this integration now since we have taken the you know integration limit along the y direction. So, I need to convert or I have to you know use x as a function of y. So, once we once we you know substitute x in terms of y then I can get I can do the integration from let us say 0 to some height as it is given in the problem. Similarly, for this there are problems specially you know when you look at a circular area or maybe you know a sector circular sector in that way it will be more appropriate to have a triangular element rather than vertical or horizontal element. So, if I use a triangular element. So, certainly what would be important now we have changed it to the polar coordinate r theta. So, the my integration we will operate from 0 to r and 0 to theta ok or defined by the limits of r and theta. So, in the triangular element what we have is that what is the center of this triangular triangle. So, we can clearly see that in order to find out the x bar. So, I have to take the moment of this triangular strip about the y axis. So, I have to find what is q y right in order to get the x bar. So, to do the q y how do I do that ok. So, now you look at x bar element. So, x bar element we can clearly see that it should be 2 r by 3 distance if we go in the r direction and the component of that should be 2 r cosine theta 2 r by 3 cosine theta as it is written ok. So, that is my x bar element and what is the area? Area should be defined by the this area of this triangle which is nothing, but remember this is r d theta that is the base and height would be equals to r. So, ultimately half r square d theta ok. So, now you can easily you know use the integration limit from 0 to theta in order to find out the x bar ok. Similarly, for the to get the centroid y bar we have to see that what is the y bar of the element. So, y bar of the element is nothing, but 2 r by 3 sin theta and area remains the same that is half r square d theta. So, these are the basic procedure in order to find out the centroid of an area that we can choose either the vertical strip horizontal strip or vertical strip the triangular strip depending on the type of problems given. So, we will discuss that in detail as we go through. So, let us just work out you know work on different problems. So, let us first problem is that of a triangle. So, triangle has a base of b and altitude or height equals to h. So, we want to determine the centroid from the base by direct integration technique. So, again the procedure will be evaluate the total area first that we can again do using the strips of our choice. But what is most interesting in order to get the centroid from the base that means we are interested in y bar from the base what would be most approach here do I choose a horizontal strip element or do I choose a vertical strip. So, that we have to be extremely careful remember in this case if I am trying to choose a vertical strip then there is a discontinuity at this as we. So, vertical strip if we choose we have to perform the integration along x direction and there is a clear discontinuity here. So, one function you have to adopt one limit will be 0 to some value here the other limit will be you know from here to this end. However, if I use a horizontal element then it will be much easier to do the integration. So, in this case therefore, what we will do we will choose a horizontal element. So, let us look at this step by step. So, first we evaluate the area. So, area is nothing but integral of dA. So, at a location y we have chosen and strip that strip has an width of you know depth of d y and the width of w. Now, remember this is the infinitesimally small strip. So, now what we have to find out w as a function of y and that we can easily calculate. How do I calculate w as a function of y? Remember we can resort to the similar triangle analogy. So, therefore, I have one triangle is this the upper part of the rectangular horizontal element and the other one itself is the large triangle. If we look at this I can clearly see that w must be equals to b by h multiplied by h minus y. So, we do the d y on this. So, the limit goes from 0 to a h. So, I have the area b h over 2. So, half b times h which is a very well known solution. So, now in order to get the centroid all I have to do I have to first find the first moment about the x axis. So, the first moment of this strip about the x axis should be simply y bar of this element. So, I have to find out what is the centroid of this element from the x axis. So, that is nothing but y considering this is a very infinitesimally small strip. So, therefore, we have y here directly and the area will be w d y. So, if we do this then they were going to get b h square over 6 that is the first moment about the base. So, to evaluate the centroid we simply say y bar a equals to q x and therefore, we get the height equals to h by 3. Remember it does not matter how I solve this problem. In all cases I am always going to find if I consider this to be my base let us say then I am always going to get h over 3 as my answer of the centroid from the base. So, another problem let us say we have a and b. So, we define this area we have to find out the area of this parabolic span braille which is defined by this curve y equals to k x square. So, what is first important that we find out the value of k once we find out the value of k we find out the total area and then we can either choose in this case either a vertical strip or a horizontal strip and simply take the first moments of that. So, q x or q y and evaluate the centroid based on the formula derived. So, first how do I find out the value of k? So, determine the constant k. So, remember y equals to k x square. So, we can clearly see that at let us say x equals to a value is equals to you know y value is b. So, from that b equals to k square. So, k equals to b s b divided by a square. So, I now have y completely defined in terms of b and a or x can be also defined in terms of y. So, to evaluate the area we are simply going to take a vertical strip. So, the vertical strip remember the area of the vertical strip is y dx as we see here. So, we will substitute for y as a function of x which we have done and integrate from 0 to a. So, that will give me area to be equals to a b by 3. So, once I get the area now I have to find out the centroid. So, therefore I will try to take the first moment of this rectangular element with respect to the x axis as well as with respect to the y axis. So, as we can see if I want to take the first moment of this area with respect to the y axis so, what is the centroidal distance of this small element from the y axis it is simply equals to x considering dx is very small. So, we simply have x multiplied by the d a d a is y dx now we can substitute y in terms of x. So, we will find the value of q y. Similarly, to get the q x that is the first moment of this area about the x axis we see that y bar element that is simply going to be equals to y divided by 2. So, we integrate from again y has to be substituted in terms of x because we are now operating along the x axis our integration limit. So, a b square over 10. So, ultimately for this parabola ok. So, now another procedure is adopted here if you really want to do this in terms of a horizontal strip. So, we have done in terms of vertical strip. So, the here the question is if we really use the horizontal strip do I get the same answer for the first moments or not and we can try this out we will see now in this case only thing that is happening your x bar element is changed. Now, what is the x bar element? Now, we discuss that in in the beginning also x bar element will be simply a plus x divided by 2. So, this is my x bar element of this horizontal strip and the area of it d a should be equals to a minus x times dy right. So, now what we have to do since now we are operating on y right. So, 0 to b is the limit. So, we have to replace x in terms of y ok and once we do that we are going to get the same answer if you look at it the previous one using the vertical strip we have this a square b over 4 and this one we also have the same. So, first moment will remain same and therefore, it does not matter which way I solve this problem. So, ultimate answer now we are going to you know have the x bar a. So, that is the equivalent system. So, in the equivalent system I am taking the first moment right. So, first moment of that area total area x bar a I am trying to find out x bar that is equals to p y right that is the first moment of this you know distributed area. So, we get x bar equals to 3 by 4 similarly y bar equals to 3 by 10 b. So, another problem. So, this is more you know kind of interesting as well. Now here there are two curves one is the linear function as you see y equals to m x and that is a parabola y equals to k x square ok. So, we are interested in finding out what is the centroid of this area the shaded area. So, remember the as the these distances are given a and h we should be able to first find out what is m and what is k how we are going to find it out. So, let us say that h equals to k a square. So, that means right here right at x equals to a y equals to h we substitute that this so we get the value of k. Similarly, for this function right here also we can say at x equals to a y equals to h. So, we can get the m ok. So, m and k can be known simply by saying at x equals to a I know the function y. So, how do I now choose the strip now I can either decide on a horizontal element or a vertical element. So, in this case a vertical element is chosen now if you choose the vertical element the main point to note here this vertical element will have two different heights one height will be y 1 right. So, that y 1 will be varying with this function y equals to k x square and we will have another height y 2 that y 2 will be varying with this function as we move on x ok. So, ultimately if we look at the centroid of this particular element. So, we look at x bar element that is nothing, but equals to x and y bar of this element should be just equals to y 1 plus y 2 by 2 ok. Remember y 1 and y 2 we know in terms of x already because they are defined by two function one function is y equals to m x the other function is y equals to k x square ok. So, now the area d a is nothing, but y 2 minus y 1 multiplied by d x. So, now you substitute y 2 and y 1 in terms of x which you already know. So, this is m x and this is k times x square. So, this is my d a. So, I can get the area. So, area can be obtained by integrating between 0 and a. So, that we get one sixth of a h. So, now we have to take the first moment which is done here. So, to do the first moment again everything is described in the previous slides. So, as we can see here that we all we have to do x bar element multiplied by the d a. So, that will be the first moment about the y axis. And similarly y bar element multiplied by d a that will be the first moment about the x axis. So, we are doing that separately and expressing y in terms of x because we are operating on x integration limit. So, we get this is the first moment about the y axis. Similarly, the first moment about the x axis that we can get ok. Remember here you know we have to do lot of substitution here because of the y bar element which is y 1 plus y 2 by 2 and area is y 2 minus y 1 d x ok. So, ultimately we have the first moments this one and that one about the 2 axis and from that we can find out the x bar. So, x bar a should be equals to u of y first moment about the y axis and y bar a is the first moment about x axis. So, we should finally get x bar and y bar ok. Now, we are actually talking about a circular sector ok. So, what is sector? So, you can see here that sector is defined by this alpha. So, it varies from let us say if this is my centroidal axis that I for sure know because it has actually symmetry about this line. So, that will be described between negative alpha and positive alpha. So, how do I get this? Now, in this case it will be always wise or advisable to get a triangular strip to get a triangular strip because now I want to perform in the polar you know integration in the polar coordinate. So, it will be always wise to get the triangular strip that is varying from negative alpha to the positive alpha. So, this was explained also previously. Now, only concern is here that ok what is the centroid of this triangular element as we know that this distance from the you know vertex right of this triangular element. So, this distance will be two-third of r. So, r is the radius of the sector right. So, therefore, my x bar element x bar element is nothing, but two-third r cosine of theta ok. So, two-third r cosine theta alright and the area is also defined area is nothing, but you have r d theta here. So, that is the base height of the triangular element is nothing but r. So, half multiplied by r d theta multiplied by r. So, this is half r square d theta. So, we get the area as simply if we look at half r square d theta and our integration limit goes from negative alpha to positive alpha. So, alpha r square ok. So, once we get the area then we can talk about taking the first moment. So, the first moment should actually be x bar multiplied by the area is already found r square alpha that should be equals to now you take the first moment of the element. So, as we have seen that x bar element is two-third r cosine theta. So, x bar element multiplied by the dA and integration should be negative alpha to positive alpha. So, ultimately I am going to get two-third of r sine alpha by alpha ok. Now, you can actually see that what happens if it is a semicircle. So, for semicircle what is my alpha? Alpha is simply going to be equals to pi by 2. So, if I substitute alpha equals to pi by 2 then you can see the centroid from the base should be equals to 4 r by 3 pi ok. So, you can also consider the quarter circle, but you know using the symmetry the center you know from the base is always going to be equals to 4 r by 3 pi. The question is can we also solve this problem by taking and strip which is also arc. So, if we choose a arc as an strip can we solve it? Now, this will give us you know little bit allow us to study also if I really choose the arc as an element then what is the centroid of that arc from the vertex. So, just look at carefully what has been done here. So, first thing that we can see now I am really going to you know work on a arc. So, this arc is defined here and my integration will be from 0 to r ok. So, now what is varying is that d r 0 ok. So, r 0 is a varying quantity and this r 0 you know coming to 0 as well as you know going to the r 0. So, ultimately to study this we have to do little bit more exercise to find out the centroid of an arc ok. So, how to find out the centroid of an arc? Remember we are discussing the centroid of lines right. So, here we really have an arc as a line and we are interested to find out what is the centroid of it. So, now we look at this part of the you know slide we do not go here at all. So, let us try to find out first of all what is the length of this element total arc the length of the total element would be simply 2 r alpha right ok. That is one thing now you can clearly see that is my l. So, l times x bar I am trying to find out this x bar should be equals to integral x bar d l. So, d l what is my d l d l in this case is r d theta ok. So, r d theta is here and what is the centroid of this small little you know length l or d l rather. So, that is going to be equals to r cosine theta ok and integration is from negative alpha to alpha. So, in this case the first objective is to find out the centroid of this arc a line element as an arc. So, we are able to see that x bar equals to r sin alpha divided by alpha. Now, this is now going to be introduced in this problem of an area ok. Because here we have chosen the arc as an element or strip ok and you can see the centroid of this element centroidal distance of this element from the vertex from the y axis is nothing, but r 0 sin alpha divided by alpha which we have obtained from this. So, now it is very simple. So, what you got here is the area of this thing right. So, that is simply equals to alpha r square. So, forget about this. So, it is going to be alpha r square ok that is the total area right and then we have x bar of this element that is r 0 sin alpha by alpha right and the d a d a of the element d a of the element is nothing, but 2 r 0 alpha d r 0. So, this is 2 r 0 alpha 2 alpha r 0 right multiplied by d r 0 that is the area of this strip ok. So, once we do that ultimately we are going to get the x bar. So, that x bar will be 2 3rd r sin alpha by alpha. So, what we have essentially done in this case we have looked at 2 different ways to solve this you know circular sector and ultimately we have 2 choices one is the triangular strip that is going from you know that is operating on the theta integration limit will be on theta. So, variable is theta here and in this case my variable is r 0. So, both ways I can get the same answer as we see. So, we always have a choice to choose those strips properly. However, depending on the problem we have to be extremely careful. So, now I think what has happened in the process we have actually studied all of this you can see clearly all these shapes right and we have studied all of this no doubt. So, their area is properly defined and their centroids are also defined here ok. So, this needs to be handy in order to you know go to the composite sections we need to have this data with us when we are going to study the composite sections. Similarly, we have work on the parabolic spandrel y equals to k x square we have found these are the centroids x bar y bar we have not studied this, but we can do as a homework exercise that what would be the centroids for a general you know parabolic spandrel of the order x to the power n circular sector we have also done from alpha to alpha. And I would really suggest that you write down this because this is going to be there in the tutorial also. So, this number please note down 2 r sin alpha divided by 3 alpha or rather 2 third r sin alpha by alpha that is the centroid of this sector from the vertex. So, vertex it at o we just need this number later on ok. So, now for the composite plates and areas. So, concept remains more or less same, but what we have learnt or what we have so far studied that how to get the centroids of simple shapes. So, by knowing the centroids of simple shapes now we can look at the composite plates and areas whereby we are going to discretize we are going to discretize this shapes into known shapes. And therefore, we know the centroids of individual areas in which we are discretizing the entire area. So, let us say now what is the objective again you are looking at two equivalence systems here remember. So, I have the total area a I want to find out what is the centroid of this total area. So, therefore what I am typically doing in this case first break it into known shapes and they are individual centroids are known with respect to corresponding you know axis or let us say even x and y axis we will know as long as we know from their base let us say here from base here you know for the rectangle we know it is at the centroid. So, individual centroids are known. So, essentially we are just equating the first moment of this with the first moment of this. So, you equate the first moment there is a mist type here. So, there should not be you know here I think there is a thing is misplaced let us work from here. So, what happens in this case it should be x bar multiplied by w i and then you have x bar i and w i. So, there is a w i. So, this is the total w something is missing here we just try to fix it. So, ultimately you have x 1 a 1 plus x 2 a 2 plus x 3 a 3 that should be equals to x bar multiplied by sum of a area. So, this is correct this expression correct there is little bit of problem right here as we can see. So, just look at this we are going to use this. So, we are basically equating the first moment of this and that. So, let us take this composite area now. So, we are interested to find out the first moment with respect to the x and y axis that is the first part of it and then once we know the first moment can I get the location of the centroid. Now, to do that as we have said that we have to break it into many pieces. So, how we are going to break it into different parts. So, one way of doing it would be let us say I take this triangular area below then I can have a rectangular area and on top of that I have the semicircular area. So, three areas I can have a 1 a 2 and a 3 and at the end I can simply subtract the circular area from that. So, what is being said here find the total area and first moments of the triangle rectangle and semicircle then subtract the area and first moment of the circular cutout ok. So, compute the coordinates of the area centroid by dividing the first moments by the total area. So, ultimately we are going to have four different areas. So, now to do this as you can see here I have three different areas now here, but these are all added. So, you can see the plus signs here ok this plus this plus this. However, we have to take out this cutout. So, you have to subtract this out ok. So, it will be always advisable that we put together a small table in order to work on this problem. So, the information that should have in this table the table should have the information on the individual areas ok which you already know. For example, this one is nothing but 120 multiplied by 80 as you can see here this shape has its own area written this shape has its own area written and this one has its own area, but remember we have to subtract that. So, there is a negative sign ok. So, all areas are written then the total area of the body is given by this ok. Now, in order to calculate the first moment what we need to do we have to find out the centroidal distance of individual components from the x and y axis. So, what does x bar and y bar means? So, it is coming with you can see and I here right. So, x bar let us say 1 that is nothing but 60 from the y axis similarly y bar 1 of the rectangle is 40 from the x axis. So, then we get the first moment of this rectangle or area about the x axis. So, x bar a we get that y bar a it is there. Similarly, for the triangle again very simple what is the location of that centroid from the x axis we all know right should be this 60 divided by 3 and this is 40 divided by 3, but remember x bar and y bar has to be inserted based on their signs because here y bar is on the negative y ok. So, therefore, we should put negative 20 for the triangle in the y bar ok. Similarly, for this so, what was it? So, the centroid from the base was 4 r over 3 pi that we already know it is marked. So, 4 r over 3 pi that is from the base therefore, we can now calculate from the x axis. So, from x axis it will be 80 right plus we have this small distance which is 25.46 4 r over 3 pi you can clearly see 105.5 46 is the number. So, this is the distance of the individual centroid from the x axis. So, likewise we fill this with proper signs. So, do not forget the signs. So, once we do that remember now what we are going to do we are simply going to get the x bar a we are simply going to get the y bar a right sum of all this. So, these are the first moment sum of first moment of the individual areas. So, we have value of q x, q x is nothing but this one and q y is nothing but this one that is the first moment about the x axis this is the first moment about the y axis. So, once we get that then we go to the next slide. So, all we have to do now we are trying to find out the centroid. So, x bar of the entire area. So, x bar multiplied by the total area should be equals to this. Similarly, y bar multiplied by the area should be equals to this and that is how we will get the x bar and y bar. So, ultimately we can see the x bar will be equals to 54.8 mm and y bar will be equals to 36.6 mm. So, in this session we have taken up the simple exercises. However, as we can see that we have really looked at various aspects of direct integration because what is really important for us to deliver to students are the concepts of direct integration. How should they perform the integrations on simple elements, simple you know areas and how should they choose various types of strips such as you know horizontal strip, it could be vertical strip, it could be triangular strip or it could be even an arc. So, we have studied all of these problems and tried to solve the centroids of simple areas using the direct integration method and then we have studied the composite plate case whereby we have taken the known results of the individual areas. But remember in the composite plates one has to be careful with the signs. So, we should if we are subtracting area that should go as negative if you know x bar and y bar of the individual centroids are not on the positive quadrant. So, we have to take the negative and positive signs accordingly. But it is always advisable that we put a clear table with all the information that are required and thereby we can find out the centroids of composite area as well. So, the again the concept was really based on the equivalent system that how we get into the business of locating the centroids of you know as individual elements as well as the composite plates. So, any questions you have anyone in the you know yes 1 2 5 6. Vector cross product the determination finding the determination the explanation was missing there I did not understand that over to you. The cross product was what we have discussed in the previous session right what was missing the explanation about finding the determinant. Yes. So, you basically have i j you know you have 2 vectors right p and q we want to do the vector product of that right. So, now you have basically instead of doing i cross i i cross j j cross a k like that we are simply you looking at in a matrix from. So, on the first row I have the unit vectors i j k in the second row I should put the p x p y and p z in the third row I have q x q y and q z. Once we get the determinant of that see instead of doing component wise the idea is you list like that and therefore, the determinant would be the solution for the moment ok. So, r cross f right that is the main objective here ok. So, basically you put the r in one row the components of r and the f in one row. So, second row should have r and the last row should have f and at the beginning you have i j and k on the first row ok. So, you can also do it you know one at a time you see my point like r 1 cross p 1 then again you know the way we use the distributive property we can use the distributive property, but that will be lengthy process ok. Instead of that we are just putting into the matrix form and then we get the moment from there clear. Thank you. Reva institute of technology 1 2 2 4. Hello, hello my question is many times the centroid is outside the lamina just can you clarify the students will always like ask the doubt how can it be outside the lamina. Yes. So, your question is that when the centroids are outside the lamina see what we are talking here about the distributed mass ultimately it is a distributed mass over the body right. Now, depending on the shape what we are trying to say the first moment if see remember the main idea here is that the first moment should be 0 about the centroidal axis. Now remember to do that suppose I choose now this is you know kind of problem I will just go to the slide little bit then it will be much clearer. Remember I talked about a you know L shaped cross section there was a L shaped cross section it is a L shaped right in that one if we look at it carefully their centroid will be outside that angle you know the angle cross section we talked about that centroid will be outside question is why see when we are doing x d a see you have to have the moment from one side should balance the moment from the other side see that is the main issue here such that your first moment is equals to 0 about the centroidal axis. Now how the area is distributed it can always fall even outside the shaded area or outside the lamina as you are asking. So, basically you are trying to divide the area not just dividing the area, but dividing the area in such a way that first moment becomes 0 about the centroidal axis. So, it is always possible that the centroid could lie outside the lamina is that clear this is 1, 2, 9, 5 from Hyderabad, Githam University. Good evening sir very good evening it is very nice to listen lectures from you from an eminent faculty from IIT sir it is very useful for us sir I need a fundamental clarification on the very concept to find out CG is told to be a point about which the object is supported right. Now to find out that location of CG how do we justify using the very concept of area multiplied by a distance from a particular reference point. How do we justify using this particular principle to find out the CG? Could you please elaborate on it. Ok it is a see as I said see the whole concept. So, your question is what is the concept of finding the center of you know centroid rather ok. So, first what we have studied is the equivalent system. So, typically what we are trying to study here is that I have let us say I have a distributed weight right on a body. Now I want to find out where is the resultant or what is the net weight through which it passes. So, what I am trying to do here I am just proposing two systems that are equivalent in nature. One is my original body where I have the distributed weight right which is proportional to the area small small area. And I have another system where I have the net weight with the total area is that clear. So, these are my two equivalent system. Now all I am saying that their first moment should be equal. So, that is the concept because if this is a their equivalent system then they should have the same resultant moment. So, ultimately I am simply equating the moment of the two system to get the centroid location is that clear concept comes from really the equivalent system. Because ultimately remember no matter what kind of a force system I have their resultant system. That means the equivalent should system should be defined by a resultant force and resultant moment. Now in this case I do not have a resultant moment why because I am talking about a parallel force system. Now you see my point. So, you have a parallel force system because your weight and you have the planar area and you have the weight right. So, your resultant force and the resultant moment they are perpendicular to each other. Therefore, I should always be able to replace that by a single equivalent load which is the net weight of the body clear. So, I am simply equating the moment of these two about the two axis to find the centroid. Sir I really thank you very much to clear clearing me this particular concept, but another clarification I do need on a particular section is that is angle section. As one of the colleagues from other institutions has mentioned the CG for L section is falling out of the cross sectional area. If I have to design a section using only angle combinations, but how do I justify because I have no point on the cross section to support it. No, no, no, no, no, no see that is a different altogether that is a different issue. See if you want to apply the force right. See this is a typical problem even in civil engineering. Let us say you have two channel sections right. So, even in a channel section we try to find the shear center that will be always outside the body. Typically speaking now when you are trying to apply the load actual load on the structure then you have to add some small member to it that is going to hold the load. But remember the contribution from that small element that is supporting the load will not affect your ultimate solution. So, these concepts are also delivered in solid means. So, why your question is absolutely valid reason that ultimately see what we are trying to essentially do here. I do not see that weight you know actually applied on the body. So, it is somewhere there right, but in in in a in an actual system how I apply that load then we have to make an appropriate connection to add that load in that area. So, centroid can be you know outside that region, but to make the you know to put the load if the problem demands that then we have to make some additional connections. It is a very typical problem if you are looking at a you know a beam. Let us say I have a beam and I want to prevent that from the twisting. If I want to prevent the beam from the twisting then it is required to put the load on the shear center. So, for a channel cross section I have shear center also which is outside the you know channel cross section then you have to make a special arrangement to put the load ok. Sir, one more question from Geetham University only. Yes, finding the centroid of that sector. Yes. In sector that in one case we considered triangular strip. In triangular strip the base is not a straight line sir, but it is a arc. So, you consider it as a base of arc finding area of that actually it is a arc you should take the length as arc. Yes, yes, yes because it is a infinitesimally. So, your question was that when I am considering the circular sector I am choosing a strip which is of the shape of a arc right. Remember that strip has a width of d r 0 which is very small right. So, ultimately if I know the centroid of that arc right what is the location of the centroid then I can take the first moment. So, what I have done first? First find the centroid of that arc ok. So, we find the centroid of that arc and then we can simply take the first moment and do the integration on the r 0 that is what it was done clear. So, first you have a line element remember you have a line first we find the centroid of that line ok and then we take that information in order to find out the centroid of this area clear. 1 0 3 2. Hello sir my question is that when we take the moment of a force about an arbitrary axis we take moment of a moment of the force about the point and then multiply it with unit vector along that axis. Yes. But we have we can have another approach to it that is finding the length of common perpendicular between the force and the axis which is nothing but the lever arm and then we can multiply with the magnitude of the force. Yes. But both approaches are valid but they give a different answer why? No they would not so your question is so if you are see your question is when you take the moment of a force about an axis about an arbitrary axis right. Now, you are trying to say you have one approach is the vector lambda dot r cross f the other one is the f time the perpendicular distance is that is that your what your question is right. So, there are two approaches so remember in the first approach what we are essentially doing we are taking the moment of the force first about a point right. So, that gives me the moment about that point. Now, I am simply taking the projection of that moment onto the line lambda. So, I have lambda dot m naught 0 lambda dot m right. So, that will give me the scalar solution for the moment about that axis. So, your next question was do I get a different answer if I do based on f multiplied by d absolutely not why should you get a different answer as long as you calculate that d you know the calculation of that d is not a trivial. So, that is the whole problem because you need to do lot of you know that will involve lot of mathematics to find out the perpendicular distance from the axis to the force, but you should not get a different answer that is absolutely not desired. Yes, sir because when I tried this one I got a slightly different it is not completely different, but slightly different answer that is why I wanted to know you think of a simple problem may be in 2 d planar case you know you should not have a different answer it is not possible ok we go to move. Thank you very much.