 Now, coming back to this situation, so what we have done is we have got the theta critical, once theta critical is known, W critical is known, what is W critical? W critical is the weight of the critical block which is going to be under plastic equilibrium and on the verge of failure. Now if you want to compute R, it is not difficult because there is a relationship between P, A, W and R also, is it not? So suppose if I write an expression you can do by this also and you can also compute the value that P A upon R will be equal to sin of theta minus phi. So that means R should be equal to P A over sin of theta minus phi, that is it. I know the value of phi, I know the value of theta critical, I know the P A value I have only computed, I can get the value of R. So I have solved the problem, fine. Reaction acting on the slip surface is known, its inclination is known, only point of application of R is not known. Use the third equation, take moment being a free vector about any point and get the value of R, application point, you know application of P is this, W you can obtain on a graphical paper, the CG of the wedge problem is deterministic, alright. Any questions? We would like to find out because look at this, if you remember when I was doing this explanation of the model, what I did, this was the box if you remember I filled up the soil over here and then I said each one of them is a critical plane, it depends from where the failure is going to take place, this is the condition at criticality, there will be several failure planes where the critical conditions do not agree, what is the critical condition, do not exist, this is the critical condition, fine, okay, that is it. So what we are trying to do is we are trying to find out the criticality which is all the system can fail from here also. This is a hypothetical plane if you remember, so you are walking on a, on a hilly terrain, a small boulders may collapse, subsequently what will happen, a big chunk will come out, again rains come, next portion, next portion and if you see the slip surface, they are all parallel, why? Because this is the plane on which the entire story is happening, fine, because you are trying to find out under what circumstances the whole thing is going to be the critical, yes, because this is under active conditions. So state of stress at every point on this plane is going to be guided by this equilibrium, so if I have understood your question correctly or maybe another way of looking at your question would be you are trying to play with W let us say, the logic which I have given you, a small block W is going to be less, if W is going to be less, PA is going to be less, agreed, this is a relationship between PA and W, keeping theta critical constant, why? Because that is a state of active earth pressure, so there is an interrelationship between W and PA which we wrote like this, PA and W, is it not? I can also write a relationship between PA and W, so truly speaking PA comes out because of W, so this concept we are going to use later on, finite decimal, you understand what is the word finite decimal? As small as possible, finite decimal, clear? W tends to 0, very thin slice PA tends to 0, PA becomes, so W becomes finite, PA has to be finite, I think this is fairly simple, is it not? So all concepts I have given to you, so you have to start from the basics and just develop these edges and free world diagrams, force vectors and that is it. Now let us move on to the passive earth pressure case, passive earth pressure case is going to be slightly tricky, imagination is required, so what people do normally is they analyze everything for active earth pressure and they convert it to passive earth pressure which is not correct, so let me start with the passive earth pressure case, now side by side you draw it on your paper and see whether you are getting the same answers or not for the free world diagram, this block remains same, PP is this vertical smooth Rankine wall, W fine, point of application is known or we are assuming whatever, the most tricky thing is reaction, passive earth pressure case, wall is moving in, wall is pushing the block in, the tendency of the block is to go up, the reaction is going to be like this, is this okay, that is the crux of the problem. Then you have a normal component over here and then you have the reaction, this is what is tricky, you see the way I have shown reaction and this is what I showed, I was demonstrating in the last lecture that if I have to lift a piece of a block how would I lift it, you know your calipers have a horizontal section also, so you have to lift it like this and you have to detach it from the parent body, so the more emphasis is on the R, this body or this block is going to be under equilibrium of PP, W and R and what you are observing is, how do you analyze this further, R and N what is included angle, fine, what indication it gives you when I was talking about this 5M and other things and somebody was asking a question, somebody asked a question why to optimize thing, which one is going to be easy, 5M less than 90 degree or 5M going to be not 90 I would say becoming much more as compared to the 5, now what you are observing here is 5M is going to be higher than what you can visualize, correct, so if you draw the free body diagram, oh sorry, a force diagram, this is PP, this is W, this is R, fine, what next, compute the angle between R and W, now can you conceptualize this thing that R and W are going to be theta plus 5 over here, is this correct, if this correct how your component of the shear stress and normal stress are going to get mobilized, draw it, are you getting the triviality associated with this system, so that is what the issue is because the way you have plotted here theta plus 5 cannot be designed, but let us prove that what mistake we are doing, mistake means it is not mistake, what triviality we are dealing with, it is not a mistake, so first you draw now TNN components and then you will realize what is happening, what is the angle between R and N, this is theta plus 5 that means your, this is going to be N, is this correct, so that means this is going to be 5 by definition, the resultant and the normal R inclined at pi angle, where is T, T is going to be somewhere here, I hope now you have understood, in an acute angle you are trying to show that there are two parts of that, so the way we have drawn it, if you really do it on a graph paper, you will realize that this thing will pop out, your direction of R will be taken care of, somebody was saying clear and what is going to happen now, this is 90 degree, this is phi, this is theta plus 5, check with N and W, what is this angle, so you have to conceive this idea that how are really acts on the system, we will talk about this later, let us still go, so we are not violating anything, we have depicted something with some angles and we are going ahead with this, fine, so I think now you have understood how to solve this, you can also get, now only thing is that when we do Pp, Pp will be equal to W and tan of theta plus 5, so now today onwards you should always remember in passive earth pressure condition, the inclination of the slip surface and the friction angle they get added up, fine, and in active earth pressure case subtracted, now what it says is theta truly speaking has to be more than friction angle, normally for sand friction angle could be 32, 35 degree, so I do not know whether you have done triaxial test or not, somebody was asking this question long back, after the testing of the triaxial sample if you take it out and see the failure plane, can you show me where the 45 plus 5 by 2 will be, more circle is there for your help, correct, so this is the interlinking of these concepts, the failure is going to take place from sigma 1 plane at an angle of 45 plus 5 by 2, in your triaxial testing this is sigma 1, this is sigma 3, what is this angle, we have proven it just now, got it, for your quick review, inclination of the slip surface with respect to the plane on which sigma 1 is acting which happens to be horizontal plane, let me read it out over here, state of stress sigma 1 0 acting on a horizontal plane, the moment you draw this plane cutting the circle at sigma 3 becomes a pole, any line passing through the pole at failure is the plane at which the failure is taking place theta f, theta f is equal to, this theta f is theta f, this is your 90 plus 2 theta f, so 45 plus 5 by 2, clear, this question somebody had asked long, long back when I was discussing triaxial testing and I had asked you to wait for some time, so now we have proven, what is the meaning of this, beneath the foundation in the soil mass a state of stress exist which is active in state, active earth pressure, do not goof it up, sigma 1 is acting on horizontal plane, clear and this inclination is with respect to horizontal for active earth pressure condition, in case of the foundations what is going to happen is when we will be studying next year foundation engineering you will remember me for whatever reasons, now if this is the soil mass, what is going to happen is this is how the failure is going to develop, alright and the question is what is this angle, so the question to you would be whether this is the active edge or passive edge, this block is resisting the movement of the foundation, clear, this is the active earth pressure, this is the passive earth pressure, again we are checking the things, correct, all these concepts are interlinked, so what I am trying to demonstrate to you, simple simple concepts which you have studied in C330 until now can be utilized for designing the systems, there is nothing great, fine. Now just for your quick understanding let me take one more case before we disperse today, let us take the effect of wall friction on the diagrams in this case, so wall friction case let us say, draw the direction of PA in your book, you know the direction of PA, fine, inclined at an angle of delta, we call this as a negative batter, this is known as negative batter, this is the word which normally we use in sublinging practice, inclination which is in the negative direction battered negatively, all right, it defines the slope of that particular line, so this is the normal, we have a negative batter, delta PA is done, what about W, okay, good, even if I am having QSH and everything that will come over here, it gets added up to this, so you may add QSH, whatever all these things, so this is the effective load, this is the reaction, can you solve this problem now? Now tell me quickly what is the angle between R and W, active earth pressure, R and W, theta-5, let us check this, so as long as your delta is finite, this angle is going to be 90-of that clear, is this correct, for horizontal case what will happen, delta is 0, so PA will be horizontal, this will be 90 degree, perfectly all right, what about this angle and the second question would be what about the included angle, so if this is, I hope you understand this is theta-5, what about this angle then, so this will be 180-90-delta-theta-5, is this okay, so this will be 90-delta-theta-5, is this okay, a quick question, what valve friction is doing to the earth pressures, quick, what happens because of friction, now this realization you should have, so more the friction angle what is going to happen, can you make out something from here, between these two friction angle does not come to the picture, are you getting this point because these are the gravity dependent issues, this is the material the way I will read this, this is the material all right, this is the material R, this is gravity, this is human efforts, so do a simple exercise, let us go for the sin rule clear, so this is PA over sin of theta-5, is this correct, equal to W over sin of 90 plus whatever, delta-theta-5, so I can straight away get rid of this fine and then truly speaking this I can transform over here, I can get PA equal to W sin theta-5 over cos of theta-5 plus theta-delta, delta-theta-theta plus phi, clear, quick check if delta is 0 what is going to happen, sorry, I have done some mistake over here, check it, so there is something I have done wrong over here, so 180 minus plus delta minus plus phi all right, sorry, same it is okay, very good, so if wall friction is not there, this is going to be phi minus theta at cos of minus correct, so this is going to be the same thing, so you just get this value and again your function is going to be same, so PA will be equal to half because let us put it like this W multiplied by KA and this KA is going to be inclusive of delta, so the question still remains what wall friction does and from where the wall friction comes in natural process or in real life, whenever you are making a wall, there will be some friction which is coming between the wall which is made up of RCC or wood or steel or whatever composites clear and the backfill material, so this is a more realistic picture but what we have done is we have used the simple concepts of mechanics to solve a complicated situation, so what is the answer, if the wall friction comes in the picture what is going to happen to the earth pressures, look at this, I mean like we drew the free world diagram of the wall and the backfill, so under active earth pressure the friction is resisting the movement, is it not, it is lifting it up, the tendency of the block has come down and this guy is lifting it up, so truly speaking if you draw the free body diagrams properly, this is the component of wall friction clear and the tendency of this force is to scoop it up, this force is trying to scoop it up, if I take moments about this point how it is going to be, this is going to get nullified but this component is going to be more or this component is going to be more, this component is going to be more because this is going to be a cos of delta term, got it, that is the point you have to remember.