 In last lecture I have talked about one class of problem where you know just by looking at the pore-water pressures which develop in the soil sample, we have tried to speculate the nature of the soil or the type of the soil. So in continuation to this theme of discussion of the problem, let me take another class of problem, you will find relatively simple and representation of the test results from the triaxial testing and how to develop the morculum envelope. So in a triaxial test, the pore-water pressure was measured during undrained testing. Now this sample happens to be a compacted fill material. So the first question is that you should realize that why these type of tests are being done. So as I have been emphasizing since last few lectures triaxial testing is mostly done to catch what has went, what has gone wrong or to prove your point related to some forensic examination which you are doing. So I hope this statement itself is clear, the pore-water pressure was measured during undrained testing on a sample which is from compacted fill material. So suppose there is some compaction going on, you are trying to create a facility by compacting the soil mass, until now you have studied compaction of the soil. And what you could derive from this is you did in situ density measurements, is it not? And you follow different techniques, that is the balloon method, sand replacement method, core cutter method, nuclear density gauge method or whatever, alright. So these are the methods which are normally used to check whether the fill has been compacted at the right in situ density or not, those 4, 5, 6 methods. So we got gamma D only from here, gamma D in situ, alright. And then we obtain moisture content also from here. So truly speaking this what we are going to discuss today is an extended part of these statements of problem. Having done the compaction of the soil for creating any facility, it could be an embankment, it could be an earthen dam, it could be a foundation pad, it could be a reclamation which you are doing, that is a Navi Uma International Airport. So having done this part, now I want to link these properties with the shear strength. This is the statement of the problem, clear. Use these type of tests are done to investigate the type of material which has been used so that you can certify the bills of the contractor, that is where the catch is. So you allow him or her to continue the progress of the work, but then I would go and collect maybe several samples, 200, 300, 500 depending upon the precision which I want, these samples I will bring to the lab, I will test them and I know what type of construction work is going on. So this is the compacted film material from an earthen dam, something clicks to you. In the CE323 course, we have talked about where the compaction should be done. So if you remember, the best way to link this type of analysis would be if this is the compaction curve, alright. And somewhere here we have the OMC, this side is the dry of optimum and this is the wet of optimum. And the question was that which side of optimum I should be compacting a material to create an earthen dam, agreed or a strip or a pavement or a highway. And this is where I had introduced this concept if you remember that for a given gamma D you have 2 moisture contents and some of you had asked this question that what is the significance of having 2 moisture content for the same gamma D. Subsequent to this what we did is we talked about the deformation characteristics of the soil also for stresses if you remember. So we had shown here that how deformation of the soil is going to be a function of sigma prime on the dry of optimum and wet of optimum. So this thing shear strength theory is continuing with this story. So the art of consulting is in the field of geotechnical engineering, you have to start from the particle size distribution, go to the compaction curve, go to the compressibility of the system, take out the sample, test it in a tri-shell setup and interpret the results that makes a full story, fine. This is the forensic examination. So this class of problems normally deals with estimating or evaluating the worthiness of the material. So the results are as follows just to add to the complexity of the problem I hope you understand where the economics gets added to the whole system because when you are filling something you are chopping off the mountains, clear? You bring that soil, crush it, make it to a certain particle size compact it that every stage of the process requires huge amount of money, fine. So this becomes a game of money. Now so these are the results. You have sigma 3, 2 samples were tested and 450, sigma 1 is known, 400 and 1000. The pore-water pressures are given as 30 and 125. It indicates that the pore-water pressures are positive, otherwise pore-water pressures will be written as negative. This is a straightforward problem, I hope you can solve this, there is nothing much complicated about this. Only thing is once you have got these sigma 3 prime and sigma 1 prime you can obtain it easily. The pore-water pressure is known, so sigma 1 prime will be 370 and this will be 875, this will be 120 and 325, fine. So the beauty of the testing is we have got the shear stress parameter, shear strand parameters both in undrained conditions, what type of testing is this and the effective stress. So you will be getting phi Cu, C Cu and here you will be getting phi prime C prime. In the last lecture whatever story we were discussing about the pore-water pressure, pre-consolation pressure, if I add this information over here and if I correlate that information with the pore-water pressure, I hope you can characterize it, whether it is going to be NC material or OC material, what is the response of the system. So if you solve this problem, what you will be getting is buy some graph papers or otherwise do it on your excel sheet, this is interesting. So what you are realizing is that there is a reversal of the parameters, 20 has become 25 degree in effective stresses and cohesion has reduced to 24, why? That is the interpretation part. So remember when we are doing the undrained testing, what we could have got from here, we could have got a parameter, but that would become a different class of the problem which I am going to discuss subsequently. Conditions are important, the same material used under different circumstances for different types of stabilities, you have to have effective or undrained parameters, long term, short term, I have discussed this. So when you talk about the long term stability, it is all the drain tests. And when you are talking about the short term stability, these are undrained tests. Remember the words that you have to utilize the material either as a cohesive material or a fixed material. On its own, there is no material which would be like X or Y. You will find it slightly intriguing. The data obtained from a CU test, soft clay, what comes to your mind? No, before you enter into all that game of numerals, the first question is how soft it is? And second question is why triaxial is being done on this? And third thing is how it was done, the concept of back pressure which I will be talking about today. Because soft samples are, they cannot stand on their own, they are so soft that they may get distorted, soft sensitive clays. So specifically it is written that there is a soft clay, it is not sensitive it seems. Had it been sensitive, triaxial would have been tough to conduct on it. So I would have done a UU test by using a bench here, either in C2 or a laboratory. So in the casings, you bring the sample from the offshore environment by using a boat or a ship, bring it to the lab, mount it on your setup and do a Venshia test and this gives you, where you would answer UU test, Venshia is always UU, clear, unsolided undrained test. Now with this preface, you have to determine shear strength parameters. So you are asking this question and philosophical answer would be, so that means they are ruling out few things from the algorithm, correct, which they do not want to tell you. So all these things we will discuss subsequently. So shear strength parameters in terms of effective stress, a CU test if you are performing this on a soft clay, you are going to get C prime, pi prime, photo pressures are measured. Now from this point onwards, I am adding something extra to this. So this part is okay. Now the second part is a different specimen of this sample, a different specimen of this soil is tested in undrained conditions under a sigma 3 of 150 kilo Newton per meter square and it fails, the deviator stress is equal to 75 kilo Newton per meter square. Calculate the pore pressures in the specimen at failure. Interesting problem. I have understood the, there are 3 parts of the problem. The first one is the statement, defines the purpose of doing it on the material. Now what goes in your mind, the moment it says that the soft clay, I do not know what type of response this soft clay is going to exhibit. But in the last lecture we have to discuss if you remember, we talked about NCOC states and the pore pressure and whether these NCOC material will be exhibiting phi value or C value or OC material will be exhibiting C value or Y phi value, you remember. So those things will eliminate the doubts from your mind and you can achieve the exact situation. So this is our impression about this material. Now what I want you to understand is that whatever material we are working with, NCOC or OC, that you should be aware of knowing, at least that you should be aware of. So for a given sigma 3, if you see the value of pore pressure, the moment you keep on increasing the pore pressure, what happens? The pore pressure keeps on increasing, for what type of material is going to happen? That is to be understood. It is left to the technologists to use this material the way they wanted to. Now suppose, if I want to extrapolate something on both the sides and if I really try to see 455, 227, 110 and by the time I bring it to 100 kPa, who knows what was the pore pressure over here and from 100 if I take it to let us say 50, the chances are that either this is going to be 0 or it might be negative. So you have all the rights to see what has happened in the lower stress ranges and what is going to happen in the higher stress ranges. Some of you were confused. Now this at failure thing is, everything is at failure now, clear? So first of all you should understand that we have 3 more coulomb and 3 more circles for 3 specimens, is this part clear? And we are going to have a unique more coulomb envelope. Now go back to the statement problem, try to understand whether C is going to be 0 or 5 is going to be 0 or C prime is going to be 0, 5 prime is going to be 0. Soft clays, you are working in the range beyond pre-consolidation pressure. So I do not know whether you are guessing or not. So somewhere over here you will be getting sigma C prime, are you getting this point? So everything is being done in the NC range and for NC material what is going to be 0? So I hope you must have realized that you might be understanding problem totally in a different manner, though the problem could be totally different. Rest is mathematics and geometry, clear? So this part is very important, so this is a typical NC material, 2 ways to prove this, positive power pressures, correct? Soft clays, pre-consolidation pressure happens to be extremely small value, no issues. So I will say 5 prime is not equal to 0, C prime is equal to 0 and now I can solve this problem. So first part of the problem has been given to you to understand the material, second will follow very easily. So once you have established the failure envelope, what is the interpretation of this? You have sigma 1-sigma 3 is known, clear? Sigma 3 is 150, that means what is the value of shear stress associated to this? Sigma D is known because ds by 2 or this is nothing but sigma 1-sigma 3 by 2, okay? So we have decoded everything now, things have become simple. So based on this assumption, what this assumption helps me in fixing the Mohr-Coulomb envelope, read the statement of the problem, a different specimen of this soil is tested in undrained conditions where sigma 3 equal to this, where is sigma 350 following in our results nowhere, this part is missing over here, clear? So I am testing it somewhere very close to pre-consolidation pressure and it fails when ds is 75, so I got tau value as 37.5, clear? Now what I should do further, everything is known and what you are supposed to find out? Compute the pore of pressure at failure in the specimen. So if phi prime is known, state of stress is known or not, what is failure? The Mohr envelope is known, Mohr-Coulomb envelope is known. Now this is the point where the failure is taking place, can you construct the Mohr circle? So this becomes your radius and this is the Mohr circle, job done, finished, okay? Reverse problem, so what we did yesterday was a forward problem, clear? And today what we have done? We have reversed the whole situation, first we identified the material, pore is soft clay and see material, we know state of stress what exists at the failure and from there now we are computing the pore of pressures, compute it, that should be a unique number. So this comes out to be about 98.8, phi prime will be equal to, I repeat, pressure angles are never written in decimal places and second decimal places, they are always absolute numbers, fine?