 So, we have been discussing about earth pressure analysis finding out the active and passive earth pressures which are acting on the walls and in most of the situations we have talked about the Rankine earth pressure where the wall was smooth vertical backfill was horizontal and in the previous module I introduce the concept of sloping backfill and this is where we had computed the Ka value. So, the situation was like this that you have a retaining wall and rather than being the horizontal backfill the backfill is inclined at an angle of I and then I had shown a conceptual diagram that how the element or the triaxial element which normally we consider as this where we have sigma 1, sigma 3 acting gets modified to a different form of the element which looks like this in case of the sloping ground or the sloping backfill. So, this is the ground surface all right very close to the hills you may find this type of a situation one of the practical examples of this situation would be you have upside a lock and then this is where the road is. So, in the Healy Terrain and what we do is we create a retention scheme over here. So, this becomes a retaining system now one thing which is important over here is the free body diagram of this element which I did not draw in the previous module. I hope you will realize that at this depth z we have sigma z which is acting and what will be the value of this sigma z. So, let us complete the free body diagram first we have sigma z we have the earth pressures PA or PP depending upon the direction all right. And then once you have lifted this element or the ground surface by angle I what is going to happen there will be a shear stress which gets introduced all right. So, this is how the element looks like. Now, from this point onwards we should be knowing how to compute the value of sigma z. So, normally the procedure is if I consider this as a backfill and I want to find out the value of stress at a depth z can you compute this. Normally, the sign convention is along the length of the slope we take the slice width as d this is one of the slices which is constituting the entire soil mass all right. So, this is one of the slices which I have taken the entire continuum has been divided in small small slices and this is how one of the slices looks like. So, we take the width of the slice along the length b this angle is i. So, this becomes b cos i this is a convention which I will be using henceforth in most of the inclined situations. Now, what is the stress component which is going to come over here this is perpendicular to the surface all right. So, sigma z is perpendicular to the surface. So, this is equal to what is the weight of the block unit weight multiplied by b into z and cos of i divided by b all right. So, this is a stress which is sigma z and this will be equal to gamma into z into cos i okay. So, what we did is we took the help of the Mohr circle and we implied the concept of axis rotation. So, if you remember what we did in the previous class is this is the Mohr circle for the material this being a pure frictional material we have assumed this as the friction line all right and then we use the concept of rotation of axis. So, this is the plane on which the new sigma z is acting all right. So, this angle is i and we computed this was o, a and b at this point the value of sigma z is equal to gamma into z. Now, what you are observing is the sigma z on plane inclined at an angle of i that is at point b the value of sigma z is gamma into z into cos i. So, what is the shear stress associated with this? This is the shear stress and this is the normal stress axis all right. So, if this is gamma z this is gamma z cos i the shear stress associated with this would be equal to gamma z cos i into sin i fine. Now, this concept will be utilizing in slope stability analysis also everywhere. So, we know the normal stress which is acting on the plane we know the shear stress which is acting on the plane and the best thing is if you substitute i equal to 0 you can cross check whether you are going to get the same state of stress or not. So, if i is equal to 0 what is the shear stress is 0 that means in this case where i is equal to 0 this element assumes the form of principal stresses shear stress becomes 0 and you have sigma 1 sigma 3 acting on the system in which the equilibration is being done is this part okay. So, please remember this analysis is what we have done for purely pi material frictional material C component cannot be included over here because it is a Rankine wall the only thing which we are defining is the inclination of the angle i okay. This is one of the cases now you must be realizing that it is not so easy to deal with non-ranking earth pressure situations alright. So, under these circumstances what we need to do is we need to introduce the concept of Coulomb's earth pressure theory. So, I repeat in most of the situations what you notice is neither the retaining wall is going to be vertical in most of the real-life situation. So, this is the retaining wall you must have noticed when you cross by the high base you know there is some batter inclination this is taken as theta and the backfill is also not horizontal. So, this is also inclined at an angle of i now this phase could be vertical depending upon my requirements. So, this becomes a simple gravity retaining structure and what we want to analyze is how much earth pressure is going to get mobilized on this system. So, these type of situations can be taken care of very easily by adopting to Coulomb's earth pressure theory alright. So, Coulomb's earth pressure theory was given in 1776 quite much before the Rankine's earth pressure theory and what it assumes is again the similar assumptions homogeneous soil mass, isotropic granular material only thing is that the wall is inclined and the backfill is inclined alright. So, there are few assumptions which are made in Coulomb's theory what we assume here is that the wall is rough. So, the first assumption is that we have a rough wall and the backfill is sloping what is the significance of this the roughness of the wall is defined by an angle of delta and this delta is approximately equal to phi by 2 to 2 phi by 3 it is limiting value could be 10 degree or 15 degree alright. And I hope you can realize that normally I will be less than 5 this condition is imposed everywhere even here also when I say I is going to be less than 5 angle it is very important to understand how the wall friction is imposed on the retaining wall like we did earlier there are 2 cases one is active and passive situation active earth pressure passive earth pressure. So, active earth pressure when we talk about is draw the perpendicular on the wall surface and then under active pressure what is going to happen this is going to be the slip surface and this angle we assume as theta alright. Now under active earth pressure the block is pushing the wall out. So, that means the direction of the pressure would be like this. So, the reaction which is going to come on the block or the wedge would be opposite to this this angle as we defined earlier is delta which denotes the wall roughness this is what is taken as the positive better. So, this is how the earth pressure gets mobilized what we assume is that the direction of Pp and Pa is known and the point of application of Pp and Pa is also known and which is assumed to be H by 3. So, if the height of the wall is H we assume that the pressure is acting at H by 3 which is reasonable. Now if you do not assume the direction of the pressure alright then the problem would become redundant. The another thing what we assume here is that this slip surface passes through the heel of the wall this is the heel of the wall alright. So, the slip surface passes through heel of the alright this is the third assumption the first and the fourth assumption would be it is a isotropic material cohesion less homogeneous and uniform material alright. So, isotropic homogeneous material and which is cohesion less the basic theory was on cohesion less material okay anything else which is very important and which we have missed out. One more assumption here is that the wedge which is sliding either down or up. So, when I say passive earth pressure in this case what is going to happen in this case if this is the wall this is the perpendicular to the wall at H by 3 H is the height of the wall. The tendency of the wall is to push the block up inside the backfill correct. So, how that is going to happen that is going to happen when the point of application of Pp is like this. So, this is the angle delta this is the sign convention which is normally used alright. So, under these circumstances what is going to happen remember we had talked about when we were discussing the active and active earth pressure theory this is the case where the scooping effect is coming in the picture. That means when this type of pressure gets mobilized in the real life the wall is moving inside the backfill you will be having a sort of a hump formation on the surface. In this case what is going to happen in this case there will be a sort of a depression because the material gets adjusted. So, this volume of the material gets adjusted in the space which is getting created because of the movement of the wall on the left hand side and on the right hand side. Now suppose if I ask you to draw the slip surfaces actual slip surfaces because of the mobilization of friction of the wall. So, imagine when you are having a passive earth pressure acting like this the slip surface is bound to get become slightly deeper it is difficult to draw alright. And this is the point where you have so what is the angle associated with the passive earth pressure 45 plus 45 minus 5 by 2 alright. So, this angle is going to be 45 minus 5 by 2 this angle is going to be 45 minus 5 by 2 provided the backfill was horizontal clear. So, in this case we have to compute these values. So, this is how the log spiral curve surface gets generated. What about the case in the case of active earth pressure? So, active earth pressure the material has a tendency to slip down it has to be a bit of a concave surface and then it would join somewhere here. So, this becomes the surface. If I define this as a convex this becomes a concave surface. So, this is the influence of the earth pressure on the wall. So, by all means we are assuming that these blocks are rigid bodies. When you are assuming this to be a rigid body we can use the concept of limit equilibrium here to solve the problem. And hence the concepts of limit equilibrium can be employed. So, now I think you understand why I had emphasized upon the fact that we assume the point of application of the earth pressure at h by 3 because if you do not assume that that becomes another unknown. So, the magnitude of the PA is principle unknown. Theta is unknown which can be obtained analytically beta and delta will be known. Best way to obtain delta would be you conduct a direct shear test make half of the block of the sample by using the same material and half of the block of the sample would be of the soil and shear it along that surface to get the value of delta. These are real life problems where you know people will approach you to help them.