 I have been discussing about the earth pressure analysis by using wedge or trial wedge. So we already studied the Rankine earth pressure theory and the second procedure or the methodology which is normally utilized to find out the earth pressures on the earth retaining systems is what is known as trial wedge method and that is what I am discussing since last lecture. And the last one I will be discussing would be the Coulomb's method and there are some graphical techniques also which have become outdated in the due course of time with the advent of all powerful computational devices. So in continuation with where we left in the last lecture we were talking about the earth pressure determination by using a trial wedge and I had drawn the free body diagrams for, now this is the trial wedge and in most of the cases the statement of the problem is as long as this wall is vertical and a smooth we have to find out P and this P could be active or passive. So A corresponds to active earth pressure and P corresponds to passive earth pressure with the weight of the block known and what we wish to do is we wish to optimize the inclination of the slip surface. So this is the slip surface about which the failure takes place we draw drew the free body diagram for this situation. So this is the shear force which is getting mobilized it is a normal on the slip surface. So this is the n component normal component and the reaction of the two is resultant of the two is the reaction R. Similarly I drew the earth pressure for passive case. So in passive case the direction of the shear force will be changing and then you have the normal which is acting and this is the resultant. So once you have done the free body diagram life is not so difficult as I mentioned the previous lecture the principle unknown is Pa or Pp and this is the function of weight of the block and weight is the function of height of the wall we bring in theta component over here and principally this is what the statement of the you know this is the objective. There are three forces which are acting on the block A, B, C and there are few knowns. This angle is friction angle coefficient of internal fraction friction the tau could be C plus sigma t tau phi so this could be equal to C plus sigma tan phi and sigma is nothing but the normal component alright and right now we are dealing with the pure frictional case so C tends to 0 we are not going to take into account this. So once you have this objective function we can optimize the things this is where I introduce the concept of friction also. So if the wall is not smooth in that case friction gets mobilized and the only difference here would be if this is the wall and this is the block which is in contact with the wall. So one is the friction which is getting mobilized between the wall and the block this part we have analyzed and if I draw a perpendicular on the surface for active earth pressure condition this is the direction of Pa we call this angle as delta the friction angle between the wall and the block in case of the passive earth pressure what will happen is this will be the direction of PP now you do not remember this is very easy to obtain this will be again delta value because of the friction which is getting mobilized between the wall and the block in the direction of PP and pH changes alright this part we discussed in the previous lecture. So if the wall is in the active state this block is moving down the friction is going to be upwards and hence you have a normal component you have a shear component the resultant is Pa in case you are having a passive earth pressure condition the block is moving up inside the wall clear inside the backfill the shear stress is going to be acting like this alright. So these are simple see in passive situation the wall is pushing the block inside alright movement within the backfill so what is going to happen if this is a surface the tendency of the block is to move up slide up so that means the reaction on this is going to be like this this is how the shear stress is going to be mobilized remember the hump formation is a peculiar characteristic of the passive earth pressure alright and in case you have a depression here so this would be let us say the hump formation which is associated with the passive earth pressure and in case you find any depression over here this is going to be because of the movement of the material outside alright available in the backfill this is going to be active earth pressure clear this part we have discussed quite in detail now what I will do is I will just try to analyze the simple situation and then slowly and slowly we will move on to the complicated cases which are more practical cases. Now this is a system where three forces are acting and hence you can do the equilibrium analysis alright so let us try to do one of these cases active earth pressure let us say so I hope you understand now this is A B C weight is acting like this there is a reaction coming like this this is the normal component this is the shear component this is okay now with all of you this is PA alright so I can use this concepts of mechanics equilibrium of the soil mass or the block alright so I can say H or V I have two equations this is how you do the most of the rigid body analysis now one thing which you have to understand is if the height of the wall is known let us say this height of the wall is as H Rankine earth pressure theory gives me the point of application of PA which is H by 3 so this is known so first of all you try to identify what are the things which are unknowns so as I said over here the principle unknowns are the pressures which are coming on the wall because of the backfill number one number two unknown is the theta value because this has to be the critical value at which the slip surface will form and the failure will occur fine I am sure you will realize W is a known factor clear another principle unknown is R I hope you agree with this because you do not know what is the magnitude of R why several issues so R is a big question mark because what we have assumed is that this is angle friction which might get mobilized or not number one clear and then later on we will talk about the factor of safety is associated with CN5 parameters so the chances are that the entire friction may get mobilized or not in C5 soils what is going to happen is the shear strength might get mobilized because of mobilization of cohesion and friction 50-50 70-30 30-70 whatever combination I do not know so overall R is unknown and its point of application is unknown got it so it is a beautiful situation where R is unknown as a vector and its point of application is unknown so most of the time these type of analysis are done to obtain these questions answer to these questions by using the concepts of the mechanics where is the third equation which I can use I can use the free moment concept and that also I can equilibrate to 0 now this method is known as limit equilibrium method that means you are just talking about the limiting equilibrium of the block and where is the limit coming from it is not the failure has occurred is at the verge of failure and mechanics you use the term incipient motion is it not at the verge of failure at the verge of movement what is going to happen to the system clear so same thing is happening here also this block under all circumstances remains in equilibrium but it is on the verge of failure because of the plastic state of equilibrium getting formed in the backfill is this part okay so I can use these concepts of mechanics quite easily to solve the problem I can take moment about any point I have two equations which are going to help me quite a lot and then the third is the moment equation to find out what is the application of R provided I know PA provided I know tau provided I know N and R in fact clear and of course your theta term so the simply simplified model is where I am dealing with let us say a frictional material no C is coming in the picture right now that part we will deal later on because the moment C comes in the picture in the backfill what happens tension cracks clear although the less permeability consolidation will take place because of compact ability is not possible differential settlements and so on over the pressure development so we are not going to talk about the cohesion right now we are simply assuming that at the time of failure the shear strength which gets mobilized is because of the friction and 100% mobilization so subsequently I will be using a term that Phi m Phi m is nothing but the Phi mobilized clear imagine there is attack on the country you need not to send all the troops on that particular place you reserve some of the troops and then you send maybe 10% 20% 30% depending upon the situation now this is what the material does material will not mobilize the entire shear strength it will only mobilize certain amount of shear strength depending upon several situations which you have already studied so this m corresponds to mobilization or mobilized mobilized means utilized remember this concept that when we drew the Mohr circle this is what we were discussing the failure takes place at this point alright so this is the tau of sigma f but if you draw a perpendicular from here you will realize that there is more strength which is available at this point now I think it looks better so at this point at this normal stress you still have this much of the shear stress which is available but the material has a tendency to fail much before that is this part clear we discussed this the ratio of the tool is going to be a factor of safety so that means Phi mobilized will be equal to Phi divided by some factor of safety and this is my choice as a designer what factor of safety I am going to use so today for the first time I have introduced the concept of factor of safety in applications until now we just talked about the material properties clear so the sample say fails or the material fails at this point but at this normal stress you have so much of shear strength still available which is higher than the failure strength fine so what we will do is I know the w value this is half into h into h into cot theta is this okay so if this is h this is going to be h cot theta Cb so I know the w value multiplied by unit weight now unit weight is a characteristics you know how to compute it there could be a total submergence partial submergence depending upon the material variable submergence and so on so you know the value of gamma you can compute it very easily we have done this in the past clear now what is another equation which I will be getting so suppose if I say sigma fh equal to 0 I will be getting Pa will be equal to something you know you can resolve this and you can compute this similarly you will be getting r equal to something you can do it very complicated method is it not so what we will do is we will go for a simplified way draw the force diagram and the force diagram says just see how I am interpreting the force diagram Pa is acting over here the first weightage is given to active earth pressure passive earth pressure it does not matter whether it is Pa or Pp is acting in the horizontal direction this is the w this is part clear where is the third force which is balancing these two are what is r the reaction which is getting mobilized on this lip surface balancing w and Pa good so this is your r this is part clear simple now what we were discussing the last lecture suppose if I overload the system and if I say this is the surcharge qs one is due to the analysis which we did last time clear second would be take this component of the loading which is going to come on the system in w itself so this will become w plus let us say qs into h into cot theta is this part clear all right the same thing I can do if I apply a horizontal force coming over here let us say F a or whatever so what will happen this Pa will get added up with F a so imagine a situation where you are doing a construction all right and either due to vibrations or due to earthquakes or due to loading it so happens that the soil mass gets loaded externally that is the situation here imagine if you are having a system which you want to construct where you are excavating over here and all the men machinery and your camps are over here you are loading the whole surface and many times it happens during the construction itself the failure occurs why you are sitting on the branch which you are cutting that means you have mobilized more excessive weight for the less shear strength which is going to cause the failure fine having done this let us come back to the simple case situations can you compute the angle between w and r in fact sorry this force diagram is still not complete let us complete this first how many other forces are missing from this diagram very good so normal stress and shear stress clear so tau and n are the components of r if I draw it like this this is your tangential force and if I join it like this this becomes your normal component happy this is 90 degree let us compute the included angles compute the angle between r and n quick r and n r and n 5 clear this ok what is the angle between w and n quick normal stress and the weight so the weight is acting like this and you have a normal stress component over here ok 90 minus theta and what will be the value of w and r that will be theta so what is the value between r and w theta minus very good clear now henceforth you did not to compute anything of this sort just remember one thing that when we are talking about the active earth pressure the angles are theta minus 5 all right you can prove this so I can write a relationship Pa will be equal to w into cot of no Pa will be w tan of theta minus 5 is this correct fine w I know I have got an expression for Pa this was a simple case I got it very easily what is the earth pressure under active earth pressure conditions of k gamma h square is it not can you conclude something from here what is this k value that is it so simple thing is all these terms are nothing but k term and what is k 1 minus sin 5 over 1 plus sin 5 so once we have got this objective function what I would like to do this is where the designers concepts of the subjects that you know exposed now my question to you would be whether you like to have a situation where Pa is maximum or minimum so what is Pa active earth pressure the backfill is pushing the wall out just because of gravity easy thing to happen natural process is it not wall moving inside the system is going to be a difficult situation so truly speaking the component which is low has to be maximized and the passive earth pressure which requires lot of efforts have to be minimized as a designer you will require these concepts when you deal with the problems mathematically does not matter because what I will do is I can optimize this function now I am using the word optimization so if I say this tends to 0 what I have done by solving this I can get theta equal to theta critical is this part ok what should be theta critical go back to the Mohr circle this is sigma 1 a smooth wall vertical wall no shear stress agreed sigma 1 sigma 3 where is the pole at sigma 3 if I join these two this becomes my failure plane theta f and what was the theta value which we computed here Tashanka you told last time in the class this is ok so what you have to prove is that this theta critical under passive earth pressure condition is equal to 45 plus 5 by 2 which is nothing but mathematics no complications very nice I mean under gravity conditions prevailing alright when the only the weight of the blocks are coming in the picture the chances are that active earth pressures can be achieved without any efforts most of the hills are falling why because the gravity is allowing them to slip down this surface is it not there is a gravity effect you know if I keep a block on a horizontal plane and if I pack this point pivot this point and if I lift it what is going to happen still in a equilibrium the moment I lift to theta critical what is going to happen this will attain a impending motion clear that is what is happening so under natural gravity condition the chances are the system is going to fail under active earth pressure but all this is being done by the nature clear it is not in your hands the type of stress conditions which develop delta sigma as greater than 0 or less than 0 do not get detached from there that is causing all active and passive earth pressures to come in picture for a constant sigma v value you remember delta sigma v is 0 delta sigma h could be greater than 0 what is the meaning of this wall is being pushed out active earth pressure condition clear and suppose I reverse the situation what is going to happen delta sigma h is more than sigma v or it is increment wall is going to come in so this is the answer technical answer to your question fine and that is why we are maximizing and minimizing mathematically it remains same.