 If I do a very slow shearing of the sample, what is going to happen? During shearing also the pore water pressure will be 0, why? Drain condition, all my walls are open, correct? So that means during consolidation pore water pressures are 0, during slow shearing the pore water pressures are 0. In other words, the total u is 0, that is why this is known as effective stress analysis. Now if the pore water pressures are 0, what is going to happen? Sigma 3 will be equal to sigma 3 and sigma 1 will be equal to sigma 1, clear? Now things are simple, if I plot it on a Mohr circle, so if you have tau versus sigma, what is going to happen? This is the first sample, this is the second sample, this is the third sample, sigma 3 is increasing and if I plot on this scale sigma 1, sigma prime also effective stresses and if I join this curve and if I depict it as tau equal to C plus sigma tan phi, now this is your Mohr Coulomb envelope. In the effective form what is going to happen? Because the pore water pressures are 0, the effective Mohr Coulomb envelope will superimpose on this Mohr Coulomb envelope also, that means tau will remain as same as tau prime, normally we do not write tau prime because this is your sigma 1 minus sigma 3, clear? So this will remain same as sigma 1 prime minus sigma 3 prime. So tau prime will also be equal to C prime plus sigma prime tan of phi prime but in reality prime indicates effective, effective is nothing but sigma minus u, remember the effective stress analysis which we have done. So because the pore water pressures are 0, the parameters which we are going to get from this test are going to be effective stress parameters, under what circumstances we are going to do this test. Stability of the dam for after 100 years, dams are normally constructed not for 20 years or 10 years or 5 years, you want to see whether it is going to be stable after 100 years. That means in 100 years I am sure the boundary conditions will develop because of the material which you are selecting that the drainage is going to take place completely, correct? And consolidation is going to occur. So ideal candidate test for to be performed on stability, long-term stability of dams are then dams. Somebody had asked this question we were talking about the heterogeneous and homogeneous dam cross sections and the seepage analysis, you know how do you select the slopes of the downstream and upstream sides of the dam, clear? Which one is going to be more critical? Look at this, the moment I have filled the water over here, you have already studied the seepage theory. Now what I am doing? I am clubbing the effect of seepage on the stability of the dam, which is being governed by the shear strength parameters. So this becomes a natural complicated problem. So truly speaking if I say that this is the critical section about with the failure of the downstream dam is going to take place, all the seepage force is going to act on this slip surface, clear? And this process is going to be dependent upon the way the pore water pressure is developing and the way the pore water pressure is getting dissipated. So when you are talking about the stability of the downstream side, which is very critical as compared to upstream side, because upstream side what is happening? Water column gives a stability to the, this side of the dam upstream. So that means for analyzing the downstream slopes, we have to do this test, CD test, is this part clear? Now let us take the second case, the second type of testing is CU test. Now as the name suggests the first one is consolidated and U corresponds to undrained, clear? So you are doing a triaxial testing in a 50-50 margins. You are allowing the sample first to consolidate at a certain sigma c, but you are shearing it in a very, very fast rate of shearing earthquake, there is no time for the material to understand how it has to behave, clear? So the fast testing is allowed to study the response of the systems which are not going to consolidate during their lifetime. So when the buildings are being constructed, you should go and see there is a natural formation and what do we do? We lay the foundation here first and then slowly and slowly the first floor, second floor, third floor keep coming over here, clear? So normally the curing time is 27 days, footings are very, very heavy in size and dimensions and the weight. Because of the self-weight of the footing, this soil is going to consolidate a bit, but after that what happens? The rate of construction is very rapid and you are not allowing soil mass to get consolidated particularly when the boundary conditions are not very conducive, both side clays, clear? So if I want to simulate this type of situation, the best test would be CU test and then I can measure the pore-water pressure and I can make it CU prime, that is the beauty. So in short, the first component remains same, uc is 0 under sigma 3. Now one thing which I have not written over here and I would like to correct is when you say shearing, what we have to do is, if this is sigma 3, I have to apply sigma f and this sigma f is also sometimes depicted as sigma d which is equal to sigma 1 minus sigma 3. So this is shearing process, that means after consolidating it, I am shearing it by applying a deviator stress, so this sigma d is known as a deviator stress, clear? And deviation is from sigma 3, so that means the axial stress you are applying in such a manner that the deviation between the sigma 1 minus sigma 3 becomes sigma d, that is what I wanted to explain, you cannot apply sigma 1 here, exactly. So the best thing is that truly speaking, this is the sigma d which is getting transmitted from this which is equal to sigma 1 minus sigma 3. So now you can compute sigma 1 from here which will be equal to sigma d plus sigma 3, clear? This is how we read this. What in short this indicates is having done the consolidation test, if I shear it, sigma d is increasing and the failure occurs, so this becomes your sigma f, clear? So once you have done the consolidation, this part remains in correct, so that means the sigma d corresponding to sigma f at which the failure is taking place will be the deviation with respect to sigma 3 of sigma d which is sigma 1 minus sigma 3. So what you have to do is starting from sigma 3, you have to apply this deviator stress to compute sigma 1, fine? So this component remains in, a UC test can, CU test can also be done to obtain the CV value. The second part is different, from this point onwards we shear it in a fast manner and fastness is in the form of undrained test. The rate of strain is so high that it does not get time to get, you know, drained. So this becomes an undrained test, till strain is going to cause the failure. You remember this, what is the significance of this? At this point, zero strain, axial strain, zero shear stress. As you move along this axis, what happens? The shear stress picks up, clear? And then ultimately what happens? The more and more axial strain, the more and more shear strength is getting mobilized and then the failure comes at this point, that is the answer to your question. So at this stage, if I say that this is sigma d which is sigma 1 minus sigma 3 and sigma 3, what will happen to the pore-water pressures? Because this is undrained test, Ud will not be equal to zero, there will be some value. Now what should I do with this? This is where we define a parameter A. So the A parameter becomes a pore-water pressure parameter. So the first time I am introducing a term A parameter which is a sort of a efficiency parameter. What is the efficiency and of what? The efficiency is how much cause and effect has been created. So what is the cause? Shearing, clear? And what is the effect? The pore-water pressure. So that means how much pore-water pressure gets developed for a given sigma 3? These are all efficiency parameters. Where I am going to use this term A parameter, you see ultimately all classification system I have forgotten now, I do not need any of those fine grained, coarse grained, organic material, non-organic material, specific gravity, density, nothing I need. What I need is how the system is corresponding or is responding. If I change sigma 3, how much pore-water pressure will develop and this becomes an efficiency parameter, clear? And this is nothing but A. So I can write now the pore-water pressure is equal to A into sigma 3. Once you have got the pore-water pressure, your effective stresses can be obtained. So that means your sigma 3 will be equal to sigma 3-UD and normally we write this as failure because A is at failure. The pore-water pressure is going to be a function of sigma 3 unless you achieve the failure, clear? And then sigma 1 prime will be equal to sigma 1-UD at failure. Effective stress develop also from here also. The parameters which you are going to get from here like here we had C, C prime, pi, phi prime and they were same because the pore-water pressure was 0. In this case what we are going to get is we are going to get C, C-U and phi C-U is a special category of the parameters which we use C, Consolidant drain, phi, Consolidant drain. And I think you can understand because this was 100% torture to the sample, this is 50%. Third we have to create is a situation where it is 0% torture. So that becomes your UU test. I will talk about this slightly later. Let me complete this story first. Is this part okay? Tractile testing is more of a philosophy rather than numerical modeling of something. When you do with the materials is more of a philosophy, it is the mathematics is only a tool which we utilize to represent parameters. What all can be done with C-U test? It is a very interesting test. So I hope you can understand if I am doing a C-U test what is going to happen to this share envelope? All these more circles are going to shift on the left hand side. So that means the first circle will shift on the left hand side and this is UD okay. So once all the circles shift on the left hand side because of the pore-water pressure that is one of the situations. There are possibilities that UD might be negative and the entire set of more circles will shift on the right hand side. So we will talk about that. Anyway so under general circumstances if you have a situation like this what is going to happen your envelope is going to get shifted and this becomes your effective stress envelope. So by conducting a test we have got different types of parameters depends upon me here. So when you do your blood test you know there are several parameters which are listed on the blood test report which you cannot make out. You go to a doctor you see everything is everything is alright you start taking this dose and this dose and do not eat this and do not drink this finished why because you must have seen some parameters and based on this he is recommending a future course of remediation clear or treatment is this part clear right. Now let me introduce here a concept of pre-consolidation pressure.