 I have been talking about triaxial testing and in the previous lecture I introduced the concept of pore pressure parameters and a bit of the stress paths. So these two things I will be covering in details in another two lectures, particularly in today's lecture I will be talking about the pore pressure parameters, how to determine them by doing triaxial testing and how to interpret the results that is more important. Now this subject is of practical importance and most of you must be dealing it with it either in academics or in the consulting field. So what I will try to do is I will try to balance both the sides and give you an optimum way which you can follow for understanding these concepts in a better manner. So yesterday we introduced this concept of a parameter and b parameter and a b parameter which we have defined as a prime parameter. Now normally we talk about two types of loading which are the genesis of this type of pore pressure parameters. The first one is what we call it as a isotropic loading and the second one is axial loading. I hope you can understand isotropic loading is the case where the sigma 3 is normally changed typically the consolidation state of the material because sigma 3 corresponds to the confinement. And if you remember the Skemptons equation, this is how we defined it that delta u equal to delta u1 plus delta u2 where delta u1 is during the consolidation and this is because of the change in the deviator stress. Many times change in deviator stress can also be written as delta of sigma 1 minus sigma 3 but in most of the Skemptons answers you will come across later on you will realize that mathematically delta sigma 1 minus delta sigma 3 might not be corresponding to delta of sigma 1 minus sigma 3. So sigma 1 minus sigma 3 is the ds term deviator stress term and sometimes people mistake this as like this depends upon the initial condition from which we are doing the testing. So it is always better to follow the incremental change in the deviator stress as delta sigma 1 minus delta sigma 3. So this sigma 3 is responsible for giving delta u1 isotropic condition alright. Static condition means if I consider an element of soil and if I load it from all round directions sigma 1 sigma 2 sigma 3 suppose if it is general case. So sigma 1 sigma 2 sigma 3 are going to be same equal to sigma 3 which is going to cause increase in pore pressure. So to be precise what we should be writing is change in delta sigma 3 is causing delta u1. The ratio of the 2 we have defined as d parameter this is equal to delta u1 upon delta sigma 3. So a bit change in confining stress gets resulted into change in the pore pressure under isotropic loading condition isotropic means all around in all the 3 directions. Now if you go into the analysis of this function you know what happens is b is normally defined as I am skipping out the derivation of all these things because this is at ug level is not required but you have to understand the parameters which are involved in constitution of b parameter as such. So b is normally defined as 1 plus porosity multiplied by cv upon cs okay and these parameters are utilized mainly for characterizing the soil mass alright. Porosity is the disturb cv corresponds to the compressibility of the fluid under isotropic compression and cs corresponds to the compressibility of the soil skeleton. So the mineral portion of the soil which is giving a solid phase is the compressibility of the soil skeleton under isotropic compression. I hope now you can realize that I can create a situation where all the pores are filled with water and in that case the compressibility of the pore fluid would be tending to 0 otherwise also because water is incompressible this is correct. So if this is a situation what happens is b normally comes out to be unity or close to unity. So yesterday we were debating upon this fact normally the range of saturation is from 0.95 to 1. If this is the condition which is whole filled we say the soil sample is saturated if not apply the back pressure and saturate the soil sample so that you can use the concepts of Terzaghi's principles and so on is this part okay. Now the A parameter is something different now A parameter happens is comes in the picture when let us say I apply a incremental change in sigma 1 itself uniaxial clear. So it is understood that this is uniaxial and this condition occurs when you are shearing the sample clear because sigma 3 is kept constant remember changing sigma 3 is not a easy task fine once you have consolidated the sample at sigma 3 you try to shear it by changing sigma d value or sigma 1 value so this is what is going to cause the shearing of the sample clear delta sigma 1 which is equal to sigma d. So because of this whatever pore or pressure develops gets translated into delta u2 so that means delta u2 is a function of this is due to the shearing this will be equal to delta sigma 1-delta sigma 3 this is the cause this is the effect and the parameter which we get is a prime which is equal to a into b and this is defined as delta u2 divided by delta sigma 1-delta sigma 3 I repeat delta sigma 1-delta sigma 3 should not be mistaken as delta of sigma 1-sigma 3 in most of the circumstances because now you can realize once we do not do that mistake I am free to change sigma 1 independently and I am free to change sigma 3 also independently clear unless the initial conditions are fixed and that is it now it so happens that this a b parameter a b parameter or a prime parameter is truly speaking turns out to be b into 1 by 3 because what we are doing is this delta u2 which is getting generated is to be distributed in all the three directions this is assumption alright so one third one third one third we apply everywhere and hence most of the time a parameter can be taken as theoretically one third but if you are doing an experiment you should be measuring a parameter separately fine now if this part is clear let us try to solve problem number 6 delta u1 at the consolidation stage delta u2 at the shearing stage that is it nothing greater than this alright and put together we have written this equation as b delta sigma 3 plus a b delta sigma 1-delta sigma 3 and this can be written as b delta sigma 3 plus a delta sigma 1-delta sigma 3 so those of you who are clever can utilize this equation the way you would like to in simplest possible form I can get rid of b if it is saturated soil sample clear s equal to 1 and then I can get the value of a parameter as total order pressure minus delta sigma 3 upon this is a delta sigma 1-delta sigma 3 now choice is yours the way you want to play with a parameter so in a conventional triaxial test what we do is we fix sigma 3 during consumption clear so by all means delta sigma 3 is going to be 0 so truly speaking a parameter is nothing but delta u upon delta sigma 1 now how I achieve delta sigma 1 we had lot of discussion on this and I gave you a logic that the sample is confined in a rigid frame and you are lifting it up clear and the reaction is being recorded by the load cell or by the proving ring whatever it is so that gives you delta sigma value directly fine so let us do the problem number 6 a compacted soil was tested what comes to your mind compacted soil was tested what compaction does to the materials now see we are talking about a pore-to-pressure parameters positive and negative clear so the moment we somebody gives us a hint that is a compacted soil mass what compaction does to the pore-to-pressures go to the compaction curve as your gamma D increases what happens suction it starts building up C323 concepts 2 concepts now have been clubbed together capillary action in soils and compaction process so the connotation is the more and more compacted material you create the more and more suction you are creating into it fine alright is this clear so this is the first connotation that compacted soil mean I am dealing with suction properly so was tested in a triaxial setup in an undrained condition at sigma 3 equal to 400 kilo Newton per meter square is this fine consolidation stage now there is a rider on this the rider on this is before the application of sigma 3 the pore pressure was 0 I understood this statement initial condition the pore pressure is given and which is 0 the following results are obtained you have strain we have deviator stress we have pore-to-pressure what type of pore-to-pressure this would be u1 or u2 that you have to differentiate you are straining the sample clear so u1 gets filtered out that part has been taken care of the shearing process is on so this is nothing but u2 what we want is delta u2 clear at 0 strain 0 deviator stress initial condition of pore-to-pressure is 0 the moment you are straining it the pore-to-pressure shoots up to 250 could be because of the disturbance caused to the sample and you are mounting it alright so you have just taken a sample your balance this to be the pore-to-pressure is 0 and then the moment you started sharing the sample the pore-to-pressure is 250 now let us write this complete series of tests hopefully failure comes at this point I have missed out 7.5 here that is the problem alright the pore-to-pressure are 250, 285, 150, 105, 75, 60, 50 alright normally the convention is everything is taken with respect to 0 strain so this becomes my initial value and this is the final value so when I have to compute delta u2 this will be final-initial always clear no confusion about this from this point onwards you are sharing the material so this becomes your origin clear at this state total consolidation occurred this becomes my background pore-to-pressure which I am filtering out from further values to know what is delta u2 value so for that matter 285-250, 150-250, 105-250 and so on change in DS is simple starting from this point this minus this this minus this and so on so the way to understand this would be I think some of you are debating the previous lectures and my logic remains same when you are sharing this material what is happening is starting from 0, 0 I am crossing these intermediate states of the material and then somewhere here the failure would take place alright so each of this state corresponds to the intermediate state in terms of shear stress versus epsilon a relationship so now this is fairly simple you get at 0 this is 35-100-145-175 and so on having done this I will like to compute delta sigma 1-delta sigma 3 why we are doing this because I want to get the a parameter clear the moment delta u2 is known this is your denominator term so this will be 570, 950 and so on I can get from here a prime parameter a prime parameter is equal to a into b so I should be obtaining b parameter first can I obtain b parameter by any chance this is a tricky thing but you have to understand if your concepts are clear you can obtain it easily so read the problem statement carefully and see whether we can obtain b parameter or not a compacted soil was tested in a triaxial setup in an undrained condition at sigma 3 equal to 400 before the application of sigma 3 the pore pressure was 0 the incremental change in the pore pressure happens to be 250 so that means I can obtain b as clear what you how do you characterize the soil this is the first characterization of the soil clear unsaturated sample and that is correct because you are working on compacted soils so that is what our hunch was so the way you started analyzing the problem is that you are dealing with the unsaturated soil sample which has been having suction in it by virtue of being a compacted material so for a quick review I know all of you have forgotten but please remember these things okay so this is the OMC now if I ask you to plot pore pressure as a function of W what is going to happen so the logic is simple the more and more you compact the negative pore pressure is developed clear if I plot is a suction at these OMC where all the pores are almost getting filled up what will happen this part you should remember so rest is all simple you have the pb parameter you have the a prime parameter compute the a parameter from a prime and then if OCR is given what we did in the problem number 5 I can relate this a to OCR value that was problem number 5 so this is one step ahead of the problem number 5 there we were knowing OCR but we are not knowing how to compute these parameters clear interesting thing would be if I plot the variation of a with respect to strain what I will be getting is you remember we studied this so if this is the response Nc if this is the response Oc try to see in what category the material falls so this retrates that most of the time the traction testing is done to characterize the soil mass fine see you should always remember 0 strain is this situation the moment you started sharing what somebody had asked the question sometime back how pore pressure develop I do not know who had asked this question that was a very good question and I think I gave you an answer that because of the change in the microstructure of the soil mass so 0 to 2.5 of shearing induces so much of pore pressure 0 to 5.0 of shearing induces this much of pore pressure this is how the material is this is a history of formation or whatever you cannot control it so we make this as a benchmark clear so everything is happening with respect to this change in ds is with respect to this change in pore pressure is also with respect to this and gain in axial strain is also with respect to this state you will realize that this was inbuilt over here so just a minute I got your point so rather than giving this value over here what I would have done is I would have given this value over here itself so the moment you get the pore pressure because of this process and then I would have sheared it does not matter but then the concept of back pressures I am trying to explain to you you will find this relationship slightly inverted so if you plot this relationship what they do is they plot pore pressure like positive and negative alright and this happens to be my OMC so I hope now you can realize this is how the pore pressure variation would be so when you come close to the saturation line or the maximum complexion state your pore pressure tends to become positive of 0 alright those of you who might get a chance to work in unsaturated soil mechanics will link this u versus w relationship with the k unsaturated with respect to moisture content so those of you who might get a chance to work on thmc another thing thermo hydromechanical coupling of soils because of high temperatures soil sample is becoming unsaturated sections are becoming negative sections are getting developed in the system clear the pore or pressures are becoming negative because of loss of moisture and the material might be transforming from a nc to oc state desiccation how would you characterize the soil based on the parameters so the thumb rules are like this b is equal to 1 for saturated soils a value is 0.5 to 1.0 for nc materials a if it is greater than 1 highly sensitive place if a is 0 to 0.5 these are oc materials slightly so I would say these are the lower compressible don't try to mug it up this just for your understanding nothing more than that and if a happens to be negative then this is a heavily over consolidated soils so I repeat most of the time triaxial testing is done to understand the condition of the soil clear you want to diagnose this.