 In a triaxial test, a sample was consolidated, sigma 3 equal to 700 kilo Newton per meter square. Now, one of you should have asked this question that why we are so particular about selecting a sigma 3 at 700, 200, 300. So let me repeat, suppose you are taking out a sample from this point which is at a depth of z and what we are trying to simulate is by doing a triaxial test. The state of stress which the sample is exhibiting at this location, clear. So if I know the gamma d value of the soil sediments, I know the depth at which I am working. So gamma d multiplied by z is going to give me some idea about the sigma 3 value and this sigma 3 I am using for confining the whole sample. So the best possible simulation of the field conditions can be done like this. Well, there are a lot of limitations, everybody knows by the time you bring the samples and all. Now, let me introduce here the concept of back pressures. So in a triaxial test, a sample was consolidated at sigma 3 equal to this, consolidation is over followed by when I was discussing the schematic diagram of the triaxial testing, I had shown you a port which if I open, I can drain out the water during drain conditions. If I close it, it becomes undrained condition. I need another port to pump in saline into the body of the sample the way doctors do. The question is why we are applying back pressure to the sample. So that means if this is a sample, now rather than taking out something, I am applying a back pressure, now this could be done either by passing a gas or a fluid depending upon what is that you are working on. Mostly we apply water and we pressurize the sample from inside. You are doing testing at sigma 3 to begin with, clear? If the sample is very fragile and if I shear it from this point onwards, what is going to happen? This sigma d might result in instability of the sample itself or sometimes sample may get crushed. So what we do is we apply a manipulated pressure from the inside of the sample and what it is doing? It is getting added up to the port of pressures which are developing in the sample, clear? So this goes and becomes a part of the port of pressure in the sample. So I am basically taking care of the health of the sample, nothing more than that. Another possibility is it might so happen that my samples are not saturated and being a third year student whatever you are studying is the class of problem which is known as saturated soil mechanics. Unfortunately all this triaxial testing is not valid for the unsaturated state of the material, clear? So if you wanted to do the direct shear strength parameter analysis for the soils which are unsaturated then you have to come to laboratory like ours where we have facilities of doing unsaturated soil testing. Otherwise on conventional test setups you cannot do unsaturated soil samples. So this is where you can do a manipulation and the manipulation is I will saturate the sample by applying back pressure so that it falls into the category of saturated soils and camouflaging the situation in such a manner that unsaturated sample has become saturated and I will keep on doing the test, that is it. So this is the purpose of applying back pressure to avoid collapse of the sample, clear? So truly speaking now what I have? I have sigma 3 minus back pressure which is positive, number 1, fine? Even in your gas hydrants also this is valid. So we just make some finite difference let us say 10 kPa or so to make the sample stand over there and rest of the things are same. So let me now continue with the system. So I have introduced this concept of back pressure into the sample and what it does. Similarly apart from the physical integrity of the sample, it goes into the pore water pressures. Let us start this story again. Now having done all these things thereafter with drainage not allowed that means it is a undrained test, typical undrained test. The cell pressure was raised to 800 kilonewton per meter square. Sometime back I had introduced this concept of stress paths if you remember. I had drawn some arrows which were touching the Mohr-Coulombe envelope, clear? So triaxial testing is done to torture the sample the way you want and to reveal its response the way you want, clear? Now where all these things are going to be number 1 valid? You are seeing in bombasticity a lot of basements are being done everywhere. So this is the ground surface, this is the soil mass and my client says that I want to construct the underground facility over here. What has happened to the state of stress at this point from initial to final? We call it as a remover, clear? So that means at this point your sigma z and if I write incremental form it becomes delta sigma z and this is becoming less than 0, excavations delta sigma z becomes let it be sigma v. This z is also fine as long as I consider this as z direction no problems, clear? Or make it v because we have been dealing with v. So delta sigma v is less than 0, clear? So it is decreasing. I hope you can realize having done all these things rather than increasing the sigma 3 I would have brought it down to value less than 700 also. If I really wanted to see the response of samples, I would have tested that sample at 500 kilo meter per meter square sigma 3, you got it? So truly speaking what we are doing is we are simulating exactly what is happening in the nature in our laboratories on the specimens, fine? Another possibility would be you are doing a big embankment over here apart from this excavation, do not mix it together otherwise it becomes quite complicated. So what has happened to the state of stress at this point? This point I hope you understand delta sigma v has become greater than 0. So I want to understand now what is going to happen to the stability of this foundation system whether this facility can be built over there or not and of course the third situation would be you have very good neighbors who always like you, so you are constructing something over here and the next plot there is a deep excavation going on. Beautiful case of litigation, fine? Because you know what is going to happen. Now this is a beautiful geotechnical engineering problem, 1, 2, 3 all put together, how would I simulate it here and see how the material is responding that is it. Now how I do all this stress and all is a different issue altogether, concepts are simple. So starting from this undrained situation we have raised sigma 3 to 800 kPa or kilo Newton per meter square and there is a pore-water pressure equal to 445 kilo Newton per meter square as a result of this torture. Now we define a term which is known as B parameter, we have been talking about the pore-water pressures all the time and in my last lecture I think I gave you an idea that pore-water pressure come constitutes of two parts, the first part is during consolidation and second part is during shearing, normally we call them as u1 and u2, B parameter is normally used to understand what is the state of saturation of the sample. So this is equal to delta u by delta sigma 3, if you read the problem carefully what has happened is the final pore-water pressure is 445 minus the back pressure you have applied is 350 divided by 800 minus 700, yeah this is 95 correct. Now this is the catch, first part of the triaxial testing is done to understand what is the state of the patient and what I should be doing with it, though we have already applied the back pressures, by applying back pressure I have got a B of 0.95 for your information if B drops less than 0.8 it becomes unsaturated state of the soil, so who knows, the guy must be aware of this and in order to bring the sample to the saturated state all this has been done. Now the second part of the problem is relatively simple, having done all these things and this corresponds to from 0.95 to 1.0 range is correspond to saturated sample, so state of saturation of the sample was achieved by back pressure. The axial load was then increased, remember the axial load can be increased just by shearing the sample by lifting the whole thing from the bottom at a constant rate of straining, give a deviator stress of 575 kilo Newton per meter square, while the sigma 3 remains constant even sigma 1, sigma 3 will also vanish and we will be using only vectors which would be termed as state of stress or stress paths to deal with the problem, clear. So all this goes in the background and what I will do is keeping sigma 3 constant means delta sigma 3 is 0 from this state onwards and hence I am going to shear the sample keeping sigma 3 constant. At this state the pore of pressure reading comes out to be 640 kilo Newton per meter square, remember we define the term parameter A and A was the response of the material to U D upon sigma 3, is this fine and the way we read this is, this is the efficiency factor how much sigma 3 you are changing and what type of pore of pressure develops in the system during undrained shearing, so that is what is being done, the axial load was increased to give a ds of this much and the pore of pressure is this much compute the parameter A. So if you do quick computations you will find the parameter A comes out to be 0.339, now this parameter A I will be defining as A prime and A prime will be equal to A into B, so from here I will be getting A as A prime upon B and this comes out to be 0.3570, these are the nomenclatures you have to just remember, A is defined as delta U divided by delta sigma 1-delta sigma 3 and later on we will derive an equation which is known as Schemton's pore of pressure equation where this total pore of pressure can be written as delta U equal to delta U1 plus delta U2, now this is equal to B delta sigma 3 plus A into B which is equal to A prime into delta of sigma 1-delta sigma 3. This can be simplified as B delta sigma 3 plus A delta sigma 1-delta sigma 3, this is what is known as Schemton's equation, for saturated samples B is 1, so if B is equal to 1 for saturated samples this is how you characterize the soils, A parameter will become delta U-delta sigma 3 divided by delta sigma 1-delta sigma 3, understand the philosophy, delta sigma 3 is appearing both in the denominator and the numerator correct, so truly speaking delta sigma 3 nothing is sort of a cushion of the sigma 3 which you have applied and A parameter is nothing but because of change in delta sigma 1 and sigma 1 is coming from the sigma D because sigma D is equal to sigma 1-sigma 3, incremental form would be delta sigma 1-delta sigma 3, the moment I say sigma 3 remains constant delta sigma 3 gets knocked out, so this becomes delta sigma 1 and A parameter is nothing but the pore of pressure which is getting developed because of shearing process, so we use two parameters A and B and A prime to characterize the soil, the results of on the same material a series of, so by doing all these analysis what we have established is the sample is saturated only by applying back pressure, we have make sure on the soil the results obtained are, I hope you understand the meaning of these notations F, check this, be careful and see what I am writing over here and what is the significance, yesterday also we discussed this thing, what type of soil response is this, these type of dubious materials are OC materials and that is the reason we do not depend too much on OC materials and the OC response, there is a switch over, you are right, so if you remember we plotted A versus percentage we explained NC and OC, clear, so the OC character is Delhi Dalai, it switches over and shows instability, now rest of the problems, problem is simple whatever we have discussed until now, so you determine the shear strength parameters in terms of effective stresses and plot the variation of A as a function of OCR, if the pre-consolidation pressure happens to be, so for your quick reference OCR is sigma C divided by sigma 3, the pre-consolidation pressure alright, a very close look on the system is going to tell you that if I would have decreased the value of sigma 3 the response would have been something like this, so this is sigma C prime, this is OC response, this is NC response, NC response gives you only 5 and this gives you only C prime alright, so everything is dependent upon sigma 3 and if I extrapolate this thing to the one consolidation, one dimensional consolidation, this goes further down, this becomes your NC behavior, if I consider 2 points 1A and another one B at a given sigma prime value, so this corresponds to point number A and this corresponds to point number B, what it indicates is that the shear strength is a function of inverse of void ratios, this is okay, so from all sides we are safe, we have prospect like this 5 prime is 25 degree, C prime is 30 kilo Newton per meter square, this is the first time we are plotting AF versus OCR until now we have been plotting A as a function of percentage strain, so what you will be getting is this graph is useful for the designers, so we have discussed 5 types of triaxial testing problems, I hope I have given you enough ideas about the how to interpret the results, keeping in mind the material which you are using, the stress range in which you are working and the objective of doing a project.