 I would like to welcome you to the lecture today where we are going to basically talk about how to find shear stresses and normal stresses on surfaces which are arbitrary okay and the basic motivation for the problem is as follows. If we have a cylindrical jet or a cylindrical surface and that is a circle okay of radius r0 and if I am looking at this curved surface the direction of the outward normal is going to be in the radial direction okay and the direction of the tangent on the surface is going to be in the axial direction which is z okay. So this is the direction of the normal and this is a tangential direction. Now normally what we have to do is we have to apply boundary conditions on surfaces of this kind and these boundary conditions usually have a physical significance. So the boundary conditions would invoke things like continuity of shear stresses that is tangential stresses or would involve normal stresses okay. So when you write the boundary conditions which we will be doing later on in the course you will be the boundary conditions deal with normal stresses and shear stresses. So in this case the normal stress component is going to be given by tau rr this is the normal stress component okay. And what about the tangential stress component or the shear stress component that is going to be given by tau rz for example for one of the tangential stress components is going to be given by tau rz. So when you are going to write the boundary conditions you are going to write the boundary conditions involving tau rr which you are going to relate to velocity gradients depending upon the constitutive relationship. Supposing it is a Newtonian fluid you will write it as constant viscosity multiplied by the gradient of the velocity okay. So that is what you would do and since these are the standard directions r and z it is easy for you to actually know what these quantities are easily compute them okay. But suppose now suppose the surface is of a tapering jet. So now what is happening is the liquid is accelerating and the jet is getting tapered and it becomes becoming smaller and smaller. The radius here is r0 and r0 keeps decreasing as a function of z okay this is the z direction. So now if you want to look at the surface and now when you are going to apply the boundary conditions these boundary conditions are also going to be in terms of the normal stress components and the tangential stress components. But what about the direction of the normal component is going to be in this direction this is you draw a tangent plane and then you draw the normal to that. So this is the direction of the outward normal okay and what is the stress component that you are interested in if this is the vector n the stress component you are interested in is tau nn that is your normal stress component at that surface at that point. What about here the tangential stress component is going to be tau nt okay the point I am trying to make here is that tau nn has both component in the radial direction as well as in the axial direction. The tau nt also is having a component in the radial direction as well as in the axial direction. So one of the things which you would be doing is you would be using tau nn and tau nt when you have arbitrary shapes of surfaces and so you want to be able to calculate what tau nn is and what tau nt is and what we want to do is we want to relate tau nn and tau nt in terms of the classical shear stresses around a fixed coordinate system around the x direction or the r direction or the z direction or the y direction okay. So the idea is the boundary conditions now involve tau nn and tau nt these are not along r z directions unlike the earlier case r and z directions okay. So what we want to do is we want to relate to nn and tau nt in terms of and tau rz these are the classical directions that we have. So r theta z for the cylindrical coordinates is the you know are the different axis and what we want to do is we want to be able to relate the normal stress and the tangential stress components okay I think on an arbitrary surface to this. The reason is the differential equations they contain the shear stress components in the classical directions tau rr and tau rz and I also want to have my boundary conditions also in the same tau rr tau rz. So what we want to do therefore is to relate tau nn and tau nt in terms of tau rr and tau rz okay. So that is the idea that is the motivation as to why we are doing this. So what we will do is we will do two approaches first we will use a physical approach and we will get this relationship but then you will see that the physical approach involves using a specific geometry and getting a specific relationship what you want to do is you want to have a mathematical framework by which you can generalize this so that you can actually apply it across any arbitrary surface okay. So that is the idea so first we will do the physical approach try to get this relationship between tau nn tau nt in terms of some classical known components tau xx tau yy whatever it is tau xy and then we will generalize it okay. So let us use now we are going to simplify things and one simplification we are going to make is we are going to use a rectangular geometry okay and we are going to make a second simplification which is we are going to look at things in two dimensions okay and look at two dimensions. So this makes it easy for me to draw pictures on the board so it is going to be a better picture than what you had yesterday okay. So what we are going to do is x and let us say that is the x direction and this is the y direction and I am going to take a triangle okay this is definitely in the direction of the y and this is in the direction of x right. So this is my arbitrary surface so this side of my triangle represents my arbitrary surface the normal to this is going to be given by this direction okay that is the perpendicular I have drawn and let us say that this is theta what is this height this is delta y and what is this length here on the x direction that is delta x okay. What are the stresses which are acting on the vertical surface here this stress is basically tau xx because the outward normal is in the x direction and the direction is also x and this is tau xy which is in this direction normal and the y are both positive so they are writing all these stresses they are all positive okay the y direction is positive upwards this is positive that way what about here I have stresses here which is tau yy which is in this direction I have this surface here which is tau yx okay. Now what I want to do is basically find out what are the what is the normal component of the stress here and the tangential component of the stress here and relate it to this. So the normal component of the stress here I am going to denote that by tau nn so this tells you it is on a surface whose normal is in the n direction n and in the direction of n and tau nt is let me just say it is upwards so tau nt is this way okay this is tau nt. So if I had a boundary of this kind and if I wanted to write my boundary conditions I would write the boundary conditions in terms of tau nn and tau nt but what I want to do is I want to be in a position where I can write this boundary condition in terms of tau xx tau yy tau yy tau yx okay that is what we are trying to do. So our job now is to relate this to these stresses in the classical direction okay and the way we are going to do this is by simply a force balance a force balance which tells you that in this triangular element which is infinitesimal we look at all the forces we are acting on the system and say that that should basically give rise to an acceleration of the fluid element inside this okay. So basically we look at the acceleration okay so the force balance in the x direction is so maybe I have to write something here before I proceed we want to find tau nn and tau nt in terms of tau xx tau xy etc that is the job okay. So let us do the force balance in the x direction what do we have sorry yeah in the x direction that is right basically the forces acting on the system must be equal to the rate of change of momentum which will mass times the acceleration okay. So I am going to write the easy part first the supposing this fluid element is actually accelerating with let us say an acceleration in the direction of x given by ax then the acceleration of the system so this because it is a triangle the total volume element is half times delta x times delta y okay and the acceleration of the system is ax what about the forces acting on the element here you have tau xx which is acting on delta y and you have tau yx this is in the y direction so we do not worry about this we have tau xx acting on this we have tau yx which is acting over delta x and what I want to do is I want to resolve these the forces arising because of these stress components in the x direction okay and clearly the angle between two lines is equal to the angle between the perpendicular so what does this mean if I want to draw a vertical line here this angle is going to be theta okay this angle is theta so the component of the tau nn in x direction this is the negative x direction is going to be minus tau nn sin theta cos theta is vertical and then I have tau nt has to go with the cosine and that is also associated with the negative sign cos theta okay you also have the body force which is acting on the system which is going to be given by half times rho g times delta x times delta y and what I have done is I have also forgotten something here which is the length element dl okay so if the length is delta l I need to mention that this is multiplied by delta l and this is multiplied by delta l okay because this is the stress and just look at it as if you are looking at the unit depth into the board so that the other dimension is just unity okay so this is the equivalent of the delta x so now you have everything consistent this is also gx yeah but I am not going to worry too much about it and you will see why the reason I am not going to worry about that this term and this term is these 2 terms the one on the left and the last term on the right they are given by a product of 2 infinitesimal quantities delta x delta y okay all the other terms are of only a single power. So since these are basically infinitesimal quantities these are all of order epsilon so the first term and the last term are actually of order epsilon squared all other terms are of order epsilon so basically what this justifies and allows me to do is to drop these 2 terms and I am just going to write this here this is of order epsilon squared and therefore this goes off to 0 and this goes off to 0 or because this is of order epsilon squared and all these other terms are of order epsilon okay I like to do a y component balance a balance of the forces again in the y direction and if okay before I do the y balance maybe I should simplify this and write the x direction balance 0 as tau xy sorry tau xx is that what it is tau xx delta y tau yx delta x-tau nt with the sin theta delta l-tau nn I call it the other way sorry cos theta delta l okay that is my balance the other thing I can do is I can relate delta l to delta x and delta y okay by using metric geometry so what is cos theta cos theta is delta x by delta l and therefore sin theta is going to be delta y by delta l and what I will do is I am going to write delta y as sin theta multiplied by delta l and delta x as cos theta multiplied by delta l and this gives 0 equals tau xx times sin theta plus tau yx cos theta-tau nn sin theta-tau nt cos theta and the delta l I can take it out because delta l is not equal to 0 it is only tending to 0 okay so this goes off and so that means this is my force balance in the x direction okay so delta l is not equal to 0 remember that delta l only tends to 0 good so now I need to basically do the same thing in the y direction because I have one equation remember I know tau xx tau xy and all these quantities here I know tau yx I know tau xx I have 2 unknowns tau nn and tau nt I have only one equation I need one more equation and the other equation comes from the force balance in the y direction okay so what is the force balance in the y direction going to give me again we are going to neglect the order of epsilon square terms so I am going to just write the only the contributions arising from the forces on the surfaces okay because they are of order epsilon so taking only the terms of order epsilon we have 0 equals tau xy multiplied by delta y plus tau yy multiplied by delta x okay and I need to do the resolution of the forces arising from the surface in the y direction and clearly these guys are the positive direction the way I have drawn this outward normal so this is going to be tau nn cos theta but it is pointing downwards so it is going to be minus tau nn cos theta and I will remember to write my delta l and I get minus tau nt the tau nt sin theta component is going to be in the upward direction so that is going to be associated with the sin theta times delta l okay again do the same thing eliminate delta x and delta y by using your trigonometry get everything in terms of delta l and what do we have 0 equals tau xy times sin theta plus tau yy times cos theta minus tau nn cos theta plus tau nt sin theta okay. So, this how would I get this this is by using trigonometry okay so now I have 2 equations in 2 unknowns and the 2 unknowns that are tau nn and tau nt what I want to do is I can therefore solve for this right the fact is you know tau xy tau yy from your constitutive relationship from if the Newtonian fluid you know tau xy is the viscosity times the derivative of the x component of velocity with y plus derivative of the y component of velocity with x so you know all these okay. So, I am going to do a little bit of rearrangement of these equations and I am going to write this as tau nn cos theta minus tau nt sin theta I am moving this to the left hand side equals tau xy sin theta plus tau yy cos theta okay and this is my y component balance let me write my x component balance the same way move these guys to the left I have tau nn sin theta plus tau nt cos theta equals tau xx sin theta plus tau yx cos theta okay. So, these are right if I have not made any mistakes so what we have got to do now is you know I have got 2 equations in 2 unknowns you can solve this using anything that you have learned in mathematics earlier but I am just going to do one more step and leave it. I like to write this in a vectorial form which is I am going to write the left hand side as cos theta minus sin theta and sin theta cos theta multiplying this vector tau nn tau nt okay equals tau xy sin theta plus tau yy cos theta tau xx sin theta plus tau yx cos theta okay. So, the idea is very simple I mean what I have done is I have just written those 2 equations in a vectorial form so this is a vector now okay this is of the form ax equals b and what you can do is you can find x as a inverse b. So, I am not going to do that there is something for you people to do so find so the unknown is this. So, clearly the left hand side has my unknowns tau nn and tau nt I want to find out tau nn and tau nt in terms of my known quantities tau xx tau xy tau yy etc that was what we started off with so that is what we are doing here okay. So, all these are this is known and this is the unknown okay and x is clearly a inverse b and so you should be able to find out what this tau nn and tau nt are in terms of my classical directions x and y okay you can of course do the same thing in 3 dimensions also then you will have a 3 by 3 matrix okay. Now what is the problem with this method this is of course a method which works and you have of course found the solution for a surface which has an arbitrary angle theta inclined okay which is flat has an angle theta to the x axis which is what we started off with but clearly what you want is I mean when you are solving a problem you cannot be you actually do not know what the surface is most of the time what is going to happen is you would not know what the surface is okay. So, this is the reason why we want to go away from this physical approach the physical approach gives me an expression which is valid for this particular specific case but what I like to do is I like to generalize this in the form of some kind of a mathematical equation or a mathematical formula which helps me to consider things for an arbitrary surface okay and that is what we want to do okay. So, can this is valid for this particular surface which is an angle theta okay. So, how do we generalize this to an arbitrary surface that is the question and what I am saying is we have to use a mathematical framework I just want to emphasize that this mathematical framework is basically to help you do the generalization okay. So, you know you cannot have a formula so you can just do your calculations very quickly I mean you cannot be in a position where you are actually sitting and drawing surfaces and doing this physical argument of what the forces are and then trying to do this relationship okay and that is what we are going to see now. So, before we do this generalization to this mathematical framework so let me just write a few things maybe a small recap of mathematical framework when you have a vector let us say a velocity vector what is the typical thing you are possibly used in a course in physics you would write this as Vx times Ex by Ex is the unit vector in the x direction Vx is the component in the in that direction plus Vy times Ey plus Vz times Ez okay. Now rather than and these are the Ex is the unit vector in the x direction and so on and Vx is the component in the x direction etc. So, rather than talk in terms of x, y and z what people like to do is they like to talk in terms of indices because that kind of helps you make things more compact okay. So, because you have r theta z in the cylindrical coordinate system you have r theta phi in the spherical coordinate system and there are several other coordinate systems where you have different variables. So, what we want to do is we are going to write rather than use x, y, z we will use 1, 2 and 3 the index okay we will use 1, 2 and 3. So, I am going to when I say Vi I mean it is basically the component in the ith direction. So, V1 would be the component in the first direction the first direction could be x for the correction coordinates it could be r for the cylindrical coordinates and so on and so forth okay. So, what we can do is we can write the vector supposing instead of using x, y, z I want to write in terms of indices 1, 2 and 3 okay. So, let me just say that let me just write one more step here. So, i equals 1, 2, 3 correspond to x, y, z directions okay. I would like to therefore this velocity vector now becomes V1E1 plus V2E2 plus V3E3 okay. And what you would like to do is you would like to write this in a slightly more compact form as summation of Vi EI okay. I just wanted to introduce this concept of Einstein's summation convention which basically tells you that what does this say? It just says that if in a particular term you have an index which is repeated that means you are actually summing over that term okay. So, if in a term an index is repeated then we are summing over that index okay over the index. So, rather than write the velocity vector as I have written there as sigma Vi EI I can just write this as Vi EI to 3. This is what you would have normally done okay. So, here in this particular term the index i is repeated. So, it automatically implies that I am actually summing over all the i's. So, I do not have to explicitly write the sigma sign. So, whenever you see a term with a repeated index that means you are summing okay even if the summation sign is missing. Now, supposing you have another vector u which is therefore going to be given by ui EI okay. Consider u. Now, how do you find out the dot product or the inner product of these two vectors okay. What I would do is since this is being summed over i rather than write it in terms of i and since I am now interested in writing this finding out this dot product what I am going to do is I am going to use a slightly different index uj Ej. So, it does not matter because it is just an index. So, this is also going from 1 2 3 that is also going from 1 2 3 okay. So, I am going to write this as uj Ej dotted with Vi EI. What I am trying to make here is that this dot product is going to contribute only when the two directions are identical okay. If j is not equal to i the dot product of Ej and Ej dot EI is going to be 0 only if j equals i you are going to have a contribution to the dot product okay. So, Ej because normally we are working with directions which are perpendicular. So, Ej dot EI is equal to 0 if i is not equal to j and is equal to 1 if i equals j and that is I can write this entire thing in a very compact way by saying that EI dot Ej equals delta ij which is the Kronecker delta okay. So, the dot product EI dot Ej gives me delta ij which is Kronecker delta and the Kronecker delta is defined as being equal to 0 if i is not equal to j equals 1 if i equals j. So, this dot product now is going to contribute this dot product u dot v can therefore be written as since it is going to contribute only if i equals j I can write this as ui vi. So, what this means is you are summing over the corresponding components of the two vectors okay. Now, you are all comfortable with the way the vector is represented in terms of the unit vectors because only one thing you need to specify the direction of the vector. When it comes to tensors you need to specify not only the direction in which the component is acting but also the direction of the surface on which it is acting. So, basically what it means is you need to specify two quantities okay. So, basically a tensor like the shear stress or like the stress sorry the stress components the stress needs as to specify two directions. I mentioned this earlier the direction of the normal to the surface and to the direction of the component okay. So, that is what it was that you have the two indices the first index told you the direction of the normal the second index told you the direction of the surface where is normal or tangential or whatever it is. So, now when it comes to writing the stress tensor what you are used to is writing this as maybe in the form of a matrix where you write this as tau xx tau xy tau xz this as tau yx tau yy tau yz this as tau zx tau zy and z is different in different places it depends on how much I have to bend and this is tau zz okay. So, basically this is the way you possibly seen the stress tensor earlier written in the form of matrix. So, rather than having a one dimensional vector component you have a two dimensional matrix okay in this case is 3 by 3. What I want to do now is rather than talk about it in terms of x and y I am going to go back to using my notation of indices okay. So, indices wise this would mean tau 11, tau 12, tau 13, tau 21, tau 22, tau 23, tau 31, tau 32, tau 33 okay. So, I have just said what I did for the vectors instead of talking about x, y, z I am just doing 1, 2, 3 the 1, 2, 3 could be depending on your coordinate system whatever. So, I now want to write this again because see when I am trying to do my mathematical framework I want to be in a position where I can actually represent the directions of different components. But tau 11 for example represents the component which is acting on a surface perpendicular to 1 also in the direction of 1 okay. So, basically what I am going to do is I am going to write the stress tensor t as tau ij e i e j. So, what does this mean? So, remember this particular term has both the i component being repeated as well as the j index being repeated. The i index is repeated the j index is repeated that means you are going to be summing over both i and j okay. You are going to first keep i constant sum over j and then vary i and then sum over j again. So, basically you will have a total of 9 terms okay. So, this basically has 9 terms okay. So, what I am doing is I am just trying to tell you that tau xx or tau 11 is the component in the direction 1 acting on a surface whose normal I have chosen is also unity okay. That is what it is. What I will just do is make one particular statement now illustrate it and then we will possibly stop. I like to define just like we had the dot product which was defined. I like to define this particular quantity. We will explain this. I just want to illustrate to you what is in store for you on Monday okay. So, we have n dot t dot n and I want to evaluate this. What does n represent? n represents normal. So, n is a vector okay. This is a vector and this is also a vector and this is a tensor. What we want to show here is that n dot t dot n represents the stress component on a surface whose normal is in the direction n okay and acting along the n okay. So, what we will do is we will talk about how to evaluate this in the next class and you will then find out if this formula actually works for what we did using the physical argument. Using the physical argument we got the expression for tau nn and tau nt earlier on okay. That was just by doing a force balance. So, in order for you to verify that what I have written here is actually correct. You would evaluate the term on the left and then verify by whatever we did earlier okay. That would be one way. Now, if you are interested in getting the tangential component on a surface whose outward normal is n, then what you would do is you would just do n dot t dot t. So, this is the direction of the tangential vector represents the component on a surface whose normal is n and the direction of the tangent is t. So, what this allows me to do is if I have an arbitrary surface, if the surface is given you can from calculus you can actually find the direction of the normal and you can find the direction of the tangent. So, if you have a surface y is equal to f of x or r is equal to f of z in the cylindrical coordinate system you can actually find the direction of the normal and the direction of the tangent. So, on that surface if you wanted to find the normal stress component I would use n dot t dot n and if I wanted to find the tangential stress component I would use n dot t dot t. This would basically be because this is in terms of my classical directions x, y, z. So, what I am doing is basically whatever I did earlier in the class today by using a physical argument but now I just use this formula to get my components in the normal direction and in the tangential direction. Thank you.