 I have discussed quite in details triaxial testing and interpretation of test results. I have tried to expose to you the types of problems which we normally encounter in geotechnical engineering while designing and analyzing the systems which are made up of soil or which are located in the soil or which are coming on the soil mass and that was a very comprehensive discussion regarding the testing procedures, what care should be taken to select the parameters and so on. Now, in today's discussion I will be covering the concept of stress path which you will find very useful as a designer or a consultant. So, there are three types of stress paths which normally we talk about and I have introduced this concept a bit maybe when you are talking about the Mohr-Kuhlm envelope and this is where I had demonstrated what is the application of stress paths and how analysis becomes simple when you are following the stress paths. So, the three types of stress paths are the total stress path we call it as TSP. The second one is the effective stress path that is the ESP and the third one this is what is known as TSSP and the full form of this is total stress path minus static pore pressure. So I will introduce this concept of the stress paths to you and then try to analyze the situation where these stress paths become very important. One thing to remember is when we plot TSP, you remember what we did is we transformed our results of the shear test from a tau sigma plane to a PQ plane alright, so this is the transformation which we did and the added advantage was that rather than plotting the Mohr circles which appear over here which is a cumbersome thing to do particularly when you have several number of samples which you have tested what we did is very conveniently we picked up the PQ points where Q was defined as sigma 1 minus sigma 3 by 2 and P was defined as sigma 1 plus sigma 3 by 2 and sigma 1 plane plus sigma 3 plane by 2 and of course the pore pressure. This will be minus u component this will be P and P prime value so what we have written is that P prime is equal to P minus u alright. So when you transform the tau sigma plane to PQ plane we get points and each point corresponds to a circle Mohr circle and then what we did is we plotted this line and we termed this as a k line, sometimes it is also known as kf line the failure line which is equivalent to the Mohr Coulomb envelope alright. On this plane we have this Mohr Coulomb envelope failure envelope where we have defined c and phi in general I hope you understand why I am using the word general and here we have transformed this to Q and P if you remember we have done something like A and beta now what I will do further is I will get rid of this type of plotting of the results and I will deal with only PQ plane now. Now on PQ plane the kf line has some slope and this slope is beta. So one of the ways of defining beta would be if I say delta Q by delta P and this is nothing but the function of tan alright because the slope of the line in general form and this can be written as delta sigma 1-delta sigma 3 divided by of course 2 and this will be delta sigma 1 plus delta sigma 3 by 2 if initial values are 0 I can write this as sigma 1-sigma 3 divided by sigma 1 plus sigma 3. So that means tan beta has been related with the state of stress existing in the sample. If I introduce this concept of k parameter which is the earth pressure coefficient I can write tan beta equal to 1-k over 1 plus k and this will also give me k equal to 1-tan beta over 1 plus tan beta is a bit of geometry which we are discussing over here. So it depends either way you can work if you know the value of k I know the value of beta I can use the transform function and if you remember we derived this cos phi equal to tan beta and a equal to c into cos phi I think this is what sin phi. So we have all these type of mathematical transformations associated with us. Now the total stress path is defined as particularly a Cu prime test if you conduct this is Q plane sorry Qp plane the characteristics of total stress path is that always inclined at 45 degree and we will prove it subsequently alright. So the basic characteristic of TSP is that it is always inclined at 45 degree clear. So suppose if this is the kf line the failure line this is the starting point of the test hydro study condition sigma 1 equal to sigma 3 shear stress is 0 what I have written is TSP and this is equal to total stress path minus static power pressure static power pressure is normally defined as U0. Now when you are doing a consolidation test say Cu prime test during consolidation U0 becomes 0 no issues there could be a situation where you are applying back pressure alright. So during application of back pressure the value of U0 will be equal to the back pressure which you are applying there could be a situation I am taking the sample from the ground and there is a water table. So because of standing water table there will be pore water pressure which we have computed so that will be the value of U0 the static pore pressure alright. So in other words this TSP would get shifted to TSSP and the difference between these two would be U0 value alright. Now if I start shearing the sample under undrained situation this is how the ESP would be nonlinear curve because soils are nonlinear in nature this becomes ESP effective stress path what it indicates is from this point onwards if I start shearing the sample one of the ways to fail the sample under effective stress condition would be like this the difference between these two points is known as pore water pressure I will write U only alright U at failure now this is a typical response for a NC material pore water pressure is always positive clear. So because UF is positive this corresponds to normally consolidated soil on the other hand if I am dealing with let us say OC material what would happen is the pore water pressures are going to be negative if this is the failure line this will be the total stress path this would be the TSSP and then when you plot the effective stress path you remember the relationship which we plotted for U as a function of percentage strain in triaxial testing in CU prime testing and what else we did is we plotted A versus partial strain and this is how the graphs were NC material OC material and this was also for NC material and OC material. So what is going to happen is when you plot ESP for this is going to be something like this because of the negative pore pressures so wherever this cuts the KF line this pore water pressure is going to be U and depending upon the stress level this is either going to be positive or it is going to be negative. So at sometimes you will find UF coming out to be negative as well now all this is done normally when we do monitoring of the pore water pressures now what we are going to do in further discussion is we are going to use this concepts to see how the stability of the system gets governed. So if I start from this equation the Skemptons equation of pore water pressures where we define U equal to B delta sigma 3 plus A delta sigma 1 minus delta sigma 3 when you are doing triaxial test what is going to happen delta sigma 3 is 0 clear because normally we do not change the sigma 3 value so what it indicates is that U will be equal to the this term will be 0 and this will be equal to delta U upon delta sigma 1 and remember this term we have defined as U2 during shearing so one of the application the stress path is when you do triaxial testing you can draw the variation of A alright as well or you can obtain A value from Cu prime test for an example suppose if I do a little bit of geometrical construction over here now this value is going to be equal to let us say x y and z so delta U2 term will be equal to U parameter or I can write as x y z and what about delta sigma term delta sigma 1 is on Q plane which truly is half of the value so this is going to be 2 times the value of x y what it indicates is if I plot the variation of A line this is how A equal to 1 upon 2 will look like this is the stress path for the triaxial test alright where the A is varying in this way on the left right hand side of this you will be having U as negative okay on the left hand side you will be having 1 and this would be greater than 1 so this is one of the applications of stress paths in obtaining A parameter and I hope you understand what is the importance of A parameter okay these are new concepts they will take some time but anyway I mean you can as follow the notes and follow some book and then I think things will become quite simple this was the application one and of course now from this point onwards let us move on to another concept I hope you can realize that if I am doing isotropic compression and if I ask you what the TSP and ESP will look like interesting question to ask alright so under isotropic condition what is going to happen there is no shearing that means one of the situations would be that your stress paths would get translated like this so this will become the TSP and if I know the port of pressure if this is the value of U this will become ESP so this is a peculiar situation of isotropic compression we have been talking about construction activities alright and we have been computing the type of stress or the state of stress which exists in the soil mass this we are doing since long suppose if I give you a situation at this point P which is at a depth of Z in the soil mass ground surface or sometimes they call it GL we have sigma V and sigma H acting fine all of you know sigma V we have computed as gamma into Z and sigma H we have computed as K times sigma V which is equal to K times gamma into Z fine now suppose if I start constructing something on the top of this what is going to happen sigma V is going to change correct so that means if this is the activity of the construction you know I would say loading so what loading results in first of all loading results in enhanced sigma V value clear so this indicates that delta sigma V is going to be positive this is part okay keeping sigma H constant perfectly alright that means delta sigma H is equal to 0 this is part clear when you are loading the soil mass from the top confining a stress remains same what is going to change only the vertical stresses I would like to depict this whole process of loading and unloading in the stress path so that I need not to compute all the time I just want to know if I start loading it what is the limit of loading and this soil deposit before the system fails under shear clear so two things are inbuilt I have said system fails that means the failure envelope is known you are dealing with the soil mass that means properties are known gamma Z wide ratio RD C phi C prime phi prime phi Cu C Cu everything is known alright so material is known now I want to utilize it the way I wanted to utilize it so the question is some politician will say why not a train of 100 bogies should run on the railway track somebody has to resist no that is not possible we can only have a train of 56 bogies or 75 bogies why there must be a reason clear so loading on the tracks traction on the track all these things are interrelated somebody will say he wants to make a 200 story building in Kamala hills what is going to happen here fine yes it can be done either I will change the ground conditions ground modification and I will satisfy you if you are not spending money there I cannot help you and then you have to live with whatever is possible so one of the situation which I have created is if I load this soil mass delta sigma is always positive delta sigma is remains 0 reverse process of this suppose if I say excavate it so if you are excavating this is the situation clear so this becomes an excavation let me put it up to this point only because of a point of interest is only up to this point so again you have excavated that what is the state of stress at the point p what has happened to delta sigma v excavation unloading clear so this becomes a case of unloading is this part okay removal of material causes stresses to decrease external loading causes vertical stresses to increase keeping sigma h constant