 Okay, so let us get on with the discussion on the different forces that we encounter in fluid mechanics. So usually we will have a surface force and a body force that we will deal with and as far as surface forces are concerned they will be expressed in terms of both normal stresses on a surface and a shear stress on a surface. On other hand when it comes to body force, we express a body force in terms of an intensity which is essentially body force per unit mass. The reason is because the body force, the example, classic example of which is the force due to gravity is supposed to be acting on the material throughout the body or throughout the mass that we are considering and for convenience we choose to take it at the center of gravity or center of mass if both are identical. So with this background let us start talking about what is meant by the specification of a state of stress at a point in a moving fluid. So what I have shown here is that a fluid particle has been isolated from a moving fluid and for the purpose of discussion and derivation the particle that has been isolated and shown on the screen right now is a triangular element and we will see the reasons for it in a few minutes. Capital G or is the body force acting on this particle and I have shown that in the chosen coordinate system the body force will have an x component and a y component gx and gy because we choose to use an intensity as the basis for denoting body forces capital G in x direction is written as the density times the volume element which is essentially the mass of this fluid element multiplied by that intensity small g in the x direction. In this particular case since the fluid element is a triangular one the volume element is simply delta x which is let us say the base times delta y which is the height divided by 2 that is the volume. The reason is because we are dealing with a two dimensional planar particle so on a per unit depth basis is what we are talking about as always. Similarly for the y direction body force so that has been shown essentially acting at the center of mass if you want for the flow for the fluid particle and then the normal stress and the shear stress acting on the three bounding surfaces of the fluid particle are also shown. Those who are familiar with notation and sign convention in solid mechanics or strength of materials will realize immediately that there is no difference between that sign convention and notation and the one that we are employing here. So normal stresses are going to be denoted by sigmas, shear stresses are going to be denoted by tau's and this is again it is a matter of choice. Many people will actually denote all stresses either by tau or by sigma. So depending on whether it is a shear stress or a normal stress the subscripts will be such that you will realize whether it is a normal stress or a shear stress but usually in elementary solid mechanics and strength of material courses we use the sigmas for normal stresses and tau's for shear stresses and that is the reason I have decided to go ahead with that notation. So what has been shown here for the triangular element is the three surfaces bounding surface like I said one is horizontal here, one is sorry vertical here, horizontal here and an inclined one and on each of these I am showing a normal stress and a shear stress, a normal stress and a shear stress, normal stress and a shear stress and the statement at the bottom of the slide is that all stress components shown are positive. Again those who are familiar with strength of material, solid mechanics will immediately understand what is going on but for the purpose of benefit to everyone let us talk about the notation and the sign convention. So a stress component will be either sigma if it is a normal or tau if it is shear and it will have two subscripts. The first subscript will denote the direction of the normal to the surface over which this stress is acting. So if the first subscript is x then the normal to the surface over which this stress is acting is essentially in the x direction that is as simple as that. The second subscript will denote the direction of the stress component itself. So on a surface the stress can act in whatever direction if it turns out to be again x direction in which it is acting the second subscript will be x. So that is about the subscript notation what the first one denotes and what the second one denotes. Sign convention now we will treat a stress component to be positive if both the surface normal and the stress component together point in either a positive coordinate direction or a negative coordinate direction. So let us see what this means. I have shown two cases here one is a vertical surface that I am drawing right now and for this surface the positive normal is shown to be acting in the negative axis x axis and the stress component that is shown to be acting on this vertical surface is a normal stress component and it is acting in the positive x direction. So this is a given situation let us say. First of all because the surface normal is acting in the x direction the first subscript is x because the stress itself is acting in the x direction second subscript is x. However the surface normal is pointing in the minus x direction the stress is pointing in the positive x direction so they are not pointing in the same direction and therefore this situation will be considered as a negative normal stress component. Let us look at the right picture here the surface normal is pointing in the positive x direction it is a shear stress component and it is shown to be pointing in the positive y direction. So it is tau because it is shear stress it is first subscript x because the surface normal is pointing in the x direction its second subscript is y because the stress itself is pointing in the y direction finally both the surface normal and the stress component itself are respectively pointing in positive coordinate directions. Of course one is in positive x and the other one is positive y that is fine but both are pointing in positive coordinate directions and therefore this shear stress component will be treated as positive. So let us go back and see why each of these is going to be positive. If you look at the left phase the vertical phase the normal stress component is pointing in the negative x direction its own surface normal will also be pointing in the negative x direction. So both the surface normal and the stress component will be pointing in the same coordinate direction which is negative x both are in x direction therefore it is sigma surface xx with sigma positive. Let us look at this shear stress on the bottom surface. As far as the bottom surface is concerned the outward unit normal for the bottom surface will be pointing in the minus y direction. The shear stress is shown to be pointing in the minus x direction. So both the surface normal and the shear stress are pointing in minus coordinate directions that is why this is a positive stress component. As far as the subscripts are concerned surface normal is in y direction hence first subscript y the stress component itself is pointing in the x direction. So the second subscript is x. Let us look at the inclined surface. For the inclined surface the normal unit outward normal will be pointing in a normal direction which is essentially what is shown for this sigma nn. So that will be the positive outward normal. Sigma is shown acting in the same direction positive outward normal direction therefore it is positive and since both the sigma as well as the surface normal will be in the n direction we are calling this as a positive stress component. On the other hand the shear stress component which is shown to be acting let us say pointing down in this picture is also positive because we normally go with a right hand coordinate system right handed coordinate system. So a normal and a tangential coordinate system that will be formed at this inclined surface will be such that the tangential direction pointing downward will be considered to be positive for being a right handed coordinate system and that is the reason that this component which is the shear stress is also positive. So as I said earlier those who are familiar with this double subscript notation and the sign convention from solid mechanics will immediately realize that that is the same convention that we are following. Going back to our Newton's second law of motion written in a few minutes earlier we had written this rho multiplied by the volume element of the fluid particle multiplied by the acceleration of the fluid particle in the x direction equal to the net force acting on the particle in the x direction similarly for y and so on. So now remember what this picture was we said that we have isolated a triangular looking fluid particle from a moving fluid and because we have isolated it we have shown as a free body diagram let us say all the forces that are acting on it. So now we want to write this f equal to ma for this particular particle. So what is f? It is essentially the net force acting on the particle. What we need to do then now is we need to sum all these forces in x direction separately and y direction separately so that we will have the net x force acting and net y force acting on the fluid particle and then equate that with rho multiplied by the acceleration in x direction and y direction. So that is precisely what I have written here. Now note that since the entire picture is shown with this fellow pointing this way and this fellow pointing downwards and so on what I will really expect now is that you will be able to proceed with this summation of forces in the x and y directions and come up with two statements of the Newton's law of motion one in the x direction and one in the y direction. What you need to remember is that when we want to sum these forces in x and y direction this sigma nn and tau nt which are the normal stress and the shear stress acting on the inclined surface will need to be resolved in x direction and y direction and then you take the appropriate component of each of these into the force balance. Other thing that you need to remember is that these are all stress components. So in order to convert these to a force element or a force they need to be multiplied by areas over which they are acting. So for example if you look at this sigma xx it is acting on an area element delta y multiplied by 1 into the depth of the paper. If you look at this tau yx on the bottom surface it is acting on an area element which has delta x multiplied by 1. Similarly this sigma nn is acting on a area element delta s multiplied by 1. So these things you keep in mind sum the forces on the particle and what you do is you let the particle shrink to essentially a point which will mean that mathematically you will take the limit of these summations as delta x delta y and resultantly delta s which was the inclined length all tending to 0. What you will realize is that when you do it the body force and the term on the left hand side which involves the acceleration term will actually drop out of the limiting process. The reason being that the acceleration term and the body force term are both proportional to the volume of the element and as you shrink the volume to 0 the volume actually will go faster to 0 in the limiting process than the surface area. So the surface will remain essentially in that limiting process and finally what you obtain is a pair of equations which are written down at the bottom of the slide. So this is a little bit of an algebra and I did not want to do this algebra here to save time. It is extremely straight forward algebra is just that it has to be done on a step by step basis and therefore those who are interested I will request that you please go back and using this figure performing the force balance and then taking the limit as is expressed here you can obtain these two equations. So let us see what these equations mean though. If you go back what is getting conveyed by these two equations is that the normal stress and the shear stress on that inclined surface is expressed in terms of the normal stress and the shear stress acting on both the vertical surface and the horizontal surface. So what is going to happen is that as you shrink the particle to 0 size these two surfaces the horizontal and vertical surfaces in the limit will appear as if they are passing through the same point mutually perpendicular to each other as you shrink the element to 0 size. So the vertical surface and the horizontal surface as you shrink the element to 0 size will appear as if they are passing through the same point as you shrink it. Similarly that inclined surface is also will keep coming closer and closer to that same point through which these two surfaces are passing and therefore the way to interpret the results shown by these two equations is that for a given theta of that inclination of the plane the third surface which can be arbitrary the normal stress and the shear stress on that inclined surface are completely specified in terms of four stress components in a two dimensional situation as we are discussing and these four stress components are such that they are acting on two mutually orthogonal planes passing through that same point. So functionally or mathematically we write that sigma nn and tau nt is a function then of sigma xx, sigma yy, tau xy and tau yx. Going back to our picture sigma xx, sigma yy are the two normal stresses tau yx and tau xy are the two shear stresses acting on two mutually perpendicular planes and as the fluid element shrinks to a size they will all be essentially passing through one single point in the limiting process. So this is for two dimensions immediately we generalize without really performing the analysis in 3D that in case of a 3D flow where my highlighter is standing we will now have the specification in terms of nine stress components instead of four acting on three mutually orthogonal planes passing through the same point. So we had four stress components on two mutually orthogonal planes in 2D in 3D we have nine stress components in or rather acting on three mutually orthogonal planes. And that is what has been shown here at the bottom of the slide using these nine stress components we form what is called as a stress tensor let us not bother about the terminology tensor etc. What it is is a collection of nine stress components three of them are the normal stresses acting in the x y and z direction six of them are the shear stresses acting in each plane if you if you consider three these three mutually perpendicular plane each plane there will be two stress components acting on each plane. So two stress components on each plane and one normal component on each plane that is the reason we have nine. And the way these are arranged in this stress tensor is a three by three matrix and if you see the row any given row you will see that the first subscript is always the same which means that in a given row the stress components are essentially acting on the given plane because the first subscript is identical and that is the first what we call as the unit outward normal direction for that particular surface. So if it is the same then we are essentially saying that we are talking about the stress components acting on a given plane if you look at a column the second subscript is the same any column that you choose and therefore stress components in a given column are essentially the stress components acting in a given direction because the second subscript was supposed to be the direction of the stress component itself. So this is what we call the state of stress in a moving fluid and as we just argued the state of stress is completely specified in terms of four components in a 2D flow and nine stress components in a 3D situation. Acting on now what we are going back is to our standard fluid element delta x times delta y and again I am showing all body forces and all surface forces acting on it. Remember that each surface it is a 2D element each surface there will be a normal stress and a shear stress. So there are total of eight surface stress components the two components of the body force acting at the center of mass. What I have shown here is explicit expressions for four of the surface stresses the remaining four I have left out on purpose hopefully you can figure out what would be the expressions for the remaining four which I have not shown. Now we just completed the discussion of the state of stress this point O at the center is the reference point let us say for this situation and for this reference point we are saying that let the state of stress be given by sigma xx sigma yy tau xy and tau yx we need four components remember for 2D flow and with respect to the reference point we are expressing the normal stresses and the shear stresses acting on the surface four bounding surfaces using a standard Taylor series expansion. So let us see one or two of these. So we have normal stress at point O as the reference point sigma xx and sigma yy. So if you move in the y direction by a distance of delta y over 2 the normal stress in the y direction for the top surface which is the north surface is sigma yy plus ddy of sigma yy times the distance is delta y over 2. If you move by a distance of delta x over 2 in the negative x direction the normal stress would be sigma xx was the reference value minus because you are moving in the negative x direction ddx of sigma xx times delta x over 2 or you can say that sigma xx plus ddx of sigma xx times minus delta x over 2 when we are the other this minus will show up. When you go in the minus x direction and minus y direction you will generate these minus signs when you go in plus y and plus x you will generate this plus signs. As earlier the body force components are g suffix x and g suffix y in the x and y coordinates and they are expressed in terms of the intensities small g and small sorry small g x and small g y which are also again on a per unit volume basis. The volume element here for this particular fluid element is delta x times delta y now because it is a rectangular one. One property of the shear stresses you can obtain by performing a moments balance for this fluid element. So, if you see if you perform or if you take moments of all these forces about the center point the only moments that will remain will be because of the shear stresses. The normal stresses and the body forces are actually passing through the center point O. Therefore, they will not have any moment arm the moment arm will be 0. However, the shear stresses will essentially form an anticlockwise couple and a clockwise couple. So, that you can form a net anticlockwise moment or a couple which will be given finally, if you perform that moment analysis by tau x y minus tau y x the whole thing multiplied by delta x times delta y. Just like Newton's law second law of motion says f equal to m multiplied by a where f is the net force acting on the fluid particle or a particle in general. m is the mass of the particle and a is the acceleration of the particle. When it comes to moments the net moment acting on a particle is expressed in terms of its moment of inertia in this case we are talking about a two dimensional particle. So, the moment of inertia will be about the z axis multiplied by the angular acceleration of that particle about the z axis. So, this is an equivalent expression to f equal to m a of the Newton's second law of motion. So, since our net moment in the counter clockwise sense if you sum it on this particular particle will turn out to be tau x y minus tau y x the whole thing multiplied by delta x delta y. And from calculus in the engineering mathematics courses we may have seen this we can actually obtain the expression for the moment of inertia for a rectangular particle such as this. So, I have simply taken the expression from the calculus book for this and I have substituted the expression for moment of inertia here to obtain an expression for the angular acceleration that this particle may experience. So, now see what is happening the angular acceleration comes in the form of the expression that is written out here. Now, let the particle shrink in zero size ok. So, if that is happening what we are saying is that delta x and delta y will tend to zero. If that happens the denominator in this angular acceleration expression will tend to zero which means that we are saying that the angular acceleration will essentially tend to infinity. And there is no physical reason to expect this there is no special constraints that we have placed this is a general discussion and on a purely physical terms it is unacceptable to expect that just because you are shrinking a particle to a very very small size it may start experiencing an infinitely large angular acceleration. So, the only way that you can resolve this paradox is to obtain a finite value of this ratio where there is a numerator and there is a denominator. The ratio is such that the denominator is tending to zero because we are saying that we are shrinking the particle size to zero. So, the only way the ratio will give a finite value is if the numerator is also tending to zero. So, that a zero over zero situation can lead to a finite value of alpha which is the angular acceleration. And therefore, using this argument we say that the numerator must also be zero. In other words tau xy minus tau x must be equal to zero or tending to zero which means that tau xy must be equal to tau yx. This is a very important result actually which says that the shear stresses occur in pairs. So, you can repeat this exercise in the other two planes. This was done in xy plane. If you repeat in the other two planes what you will realize that tau yz will be equal to tau zy, tau xz will be equal to tau zx. So, in general tau ij will be equal to tau ji. And therefore, if you go back to our 3 by 3 matrix you will realize that the off diagonal elements here tau xy and tau yx, tau xz and tau zx, tau yz and tau zy they are essentially equal to each other. Therefore, in this 3 by 3 matrix only six independent entries are present. The three diagonal elements which are the normal stresses and the three off diagonal elements which are shear stresses. So, this is something important and we will utilize this fact later when we connect the shear stresses to the rates of strain. Now, going back then over Newton's second law of motion for the fluid particle we said that rho multiplied by delta v multiplied by the acceleration which is m times acceleration is equal to the net force. What is done is going back to this particle again sum all forces independently in the x direction and in the y direction. So, that you obtain the net force in the x direction and net force in the y direction. Remember again that these are stress components on the surfaces. So, you need to multiply those by area elements to convert them into forces. The directions of each of these forces is already shown. So, when you sum those you take the direction into account as positive or negative. If you perform that analysis you will realize that what you end up with is rho times substantial derivative of u forget the middle part for now and what we will generate is a bunch of two stress terms with their derivatives which will essentially come from your summation of forces in the x direction and summation of forces in the y direction. I have shown this here on the two dimensional fluid element. However, when I have written out here I have written it directly to a three dimensional situation. So, if you want to perform this analysis you will realize that you drop this z terms if you are looking at our 2D element. The middle part where is this coming from? The middle part is coming from our earlier discussion where we had connected the conservative form to a non conservative form. So, remember that rho multiplied by substantial derivative of u is exactly equal to the conservative form which is right at the top where my highlighter is standing right now. And so the middle part here is essentially the conservative part of the I mean if you choose to use a conservative expression for the momentum equation you will use the middle part. If you choose to use non conservative representation you will use the left part. The right hand side of the equation is essentially coming from summing the forces on our chosen fluid particle shown here is a two dimensional particle shown in the slide are equations written directly for three dimensional particle. And these are what are called as the Cauchy's remember what these are? These are nothing but Newton's second law of motion expressed for the fluid particle. So, they are nothing but F equal to m multiplied by a expressions. F is the net force acting on the fluid particle and the net force has been calculated by summing forces in x and y direction separately and that is what has been written out here. This is a very important part of this differential analysis and I would like each participant really to perform this algebra where they take the force component rather that they first take the stress component convert it into a force component and then carry out this summation of all forces to eventually realize that this is what we are getting. Most of the standard fluid mechanics text books will actually have this derivation in there as well but it is always a really good idea to do it yourself once to realize that this is indeed how it comes. So, at this point what we have is an intermediate result in the sense that the acceleration times the density of course the volume element term drops out from all these expressions is equal to the net force acting on the fluid particle. Now what is left to do is to connect these stresses to the rates of strain which we had discussed yesterday in kinematics and that will complete our discussion. What I will do now is I will take a break.