 Welcome to lecture series on advanced geotechnical engineering and we are discussing module 4 and which is on the shear strength of soils. In the previous lecture we introduced ourselves to different types of triaxial test, they are namely consolidated undrained triaxial test, consolidated drained triaxial test, unconsolid undrained triaxial test and unconfined compression test and we have also discussed that there is several other possibilities also with the triaxial test wherein we can also perform this extension test in order to measure the tensile strength extension characteristics of a soil or tensile strength of a soil. So we are discussing about the stress paths in triaxial and thereafter we will introduce ourselves to octahedral plane and more coulomb condition and thereafter we will discuss about how we can actually determine elastic modulus from the triaxial test. So before looking into the discussion let us take an example problem wherein the problem reads like this, a drain triaxial compression test is carried out on a sample of soil known to have the effective stress strength characteristics as effective stress parameters and there basically C dash is equal to 10 kPa and 5 dash is equal to 22 degrees. If the cell pressure is 100 kPa and draw the Mohr's stress circle at failure and evaluate the failure values of T s dash q and P dash and draw the stress paths on both T s dash and Q p dash diagrams. And what are the slopes of the stress paths evaluate k dash or a and alpha dash or psi. So we should actually refer here that there are two types of parameters composite stress parameters one is actually put forwarded by MIT group and Cambridge group and both the groups actually have used unfortunately same you know symbols they are q and p. So for convenience here q is also defined as t and p is also defined as s. So as far as the composite stress parameters definition such as the deviator stress and mean effective stress are widely they are basically widely used in soil mechanics and in the course of the lectures also here and there we have used both MIT group and Cambridge group you know q and p symbols but here we are distinguishing with MIT group and Cambridge group the symbols put forwarded by MIT group and Cambridge group where t is equal to t dash means that is nothing but you know sigma 1 minus sigma 3 by 2 which is equal to q to the original you know definition put forwarded by MIT group s is equal to sigma 1 plus sigma 3 by 2. So in case of s is equal to s dash is equal to sigma 1 plus sigma 3 by 2 is equal to sigma 1 plus sigma 1 dash plus sigma 3 dash by 2 but in case of q it is q is equal to sigma 1 minus sigma 3 by 2 is equal to t and whereas Cambridge group if you look into it it is p is equal to 1 by 3rd of sigma 1 plus 2 sigma 3 by 1 by 1 by 3rd of sigma 1 plus 2 sigma 3 where p dash is equal to 1 third of sigma 1 dash plus 2 sigma 3 dash. So q is equal to q dash is equal to sigma 1 minus sigma 3 according to Cambridge group definition of composite stress parameters. Now from the given data the cell pressure was actually given as 100 kilo Pascal's and the soil actually was found to have you know has been given as the Mohr Coulomb failure envelope was given in the problem. So what we have done is that we have drawn the Mohr Coulomb failure envelope and on the y axis that is the tau and the x axis it is sigma dash. So here this you know ordinate is 10 kilo Pascal's which is the you know the cohesion ordinate and the 22 degrees is the friction angle. Now for the we know that the sigma 3 is 100 kilo Pascal's that is actually at this point it is given. Now what we do is that the radius of the stress circle at failure can be obtained like this. So radius is nothing but the radius of the circle is nothing but which is you know if you take this triangle which is nothing but 10 cot 22 plus 100 plus r you know sine 22. So if you take this one the radius is actually obtained here and once the radius is obtained with this as a center and with this is one of the points the Mohr you know circle actually is drawn and sigma 1f can be obtained as 100 is the sigma 3. So 100 plus 2r that is you know you know the radius 2 times radius we get the sigma 1 dash f that is the major principle stress at failure and once we get the major principle stress at failure you know this s dash f can be obtained as that is nothing but sigma 3 plus sigma 1 minus sigma 3 by 2 which is nothing but 100 plus r which is about 174.7 kilo Pascal's and qf is nothing but qf is nothing but 2r is equal to 149.4 kilo Pascal's qf is nothing but you know that is sigma 1 minus sigma 3 by 2 into 2. So because q is equal to sigma 1 minus sigma 3. So r is equal to sigma 1 minus sigma 3 by 2. So what we have done is that qf is equal to 2r is equal to 149.4 kilo Pascal's and p dash f is equal to one third of we know that this sigma 1 dash which is 174.7 plus 2 into 100 you know with this what we get is that that is sigma 1 dash is nothing but 249.4. So sigma 1 dash is nothing but 249.4 plus 2 into 100 we will get the p dash. So p so according to the problem we have been asked to determine q dash and p dash that is according to the Cambridge group we have determined like this qf is equal to sigma 1 minus sigma 3 and p dash f is equal to one third of sigma 1 dash f sigma 1 dash f is nothing but 100 plus 2r. So this is this point plus 2 into 100 we got this one. Now what we actually next we have said is that we can actually get the stress parts you know we have been asked to draw the stress part for you know ps and you know qp dash. So for both the things here one we have been given that these are the this is p dash and this is q and this we know that this slope is actually will be 3 to 1 3 1. So the q is 149.8 kilo Pascal's that is you know 149.8 kilo Pascal's that is this one 149.4 that is here it is plotted here and this is 100 you know up to the you know that is the self pressure which is applied. So it traverses from here to here to failure. Similarly when we take t is equal to q is equal to sigma 1 minus sigma 3 by 2 s is equal to p is equal to sigma 1 plus sigma 3 by 2. So this traverses like you know 100 plus you know this is nothing but sigma 1 plus sigma 3 by 2. So this can be obtained like this this is nothing but 174.7 kilo Pascal's. So that is indicated here. So the slope of this you know is one. So the t failure is around 74.7 kilo Pascal's and this is nothing but 174.7 kilo Pascal's. So it can be seen that t s dash stress path has a slope of 1 is to 1 and q p dash stress path has a slope of 3 is to 1. So t s you know with the MIT group definition the stress path actually has a slope of 1 is to 1 with Cambridge group definition of q and p dash it is 3 is to 1. Now we have been asked to determine k dash and you know k dash and alpha dash. So we know that we have a relationship between k f line and more coulomb failure envelope when we have the q and p and when we have the tau and sigma when we compare the from the geometry of the two circles we can actually we have derived that sin pi dash is equal to tan psi is equal to tan alpha dash and where c is equal to you know this ordinate c is equal to a by a or k dash by cos phi dash. So by comparison from here from the two circles we can get k dash is equal to from this equation k dash is equal to 10 that is you know c is given as 10 kilo Pascal's 10 cos 22 that is around 9.3 kilo Pascal's. So that means that we are actually getting that a or k dash as 9.3 kilo Pascal's in q and p you know q and p space and tan alpha dash is equal to tan psi is equal to sin phi. So sin phi dash is given as sin 22 phi dash is 22 degrees. So tan alpha dash is equal to sin 22 degrees which is nothing but alpha dash is equal to 20.5 degrees. So this alpha dash is equal to psi is equal to the inclination of the you know k f line is 20.5 degrees. So like this you know we need to you know you know look into the same for getting the you know different parameters at failure from the basic data which is actually deduced from the you know the traction test. Now let us consider the consolidated drained test and stress paths and when we have the isotropic consolidation the constant isotropic stress that means that we have got equal stress in vertical and horizontal direction. Then in that case the Mohr's circle is nothing but a point where the sample of soil and the isotropic stress basically has Mohr's circle of zero radius. So it is nothing but when we have sample of soil under the isotropic stress conditions the Mohr's stress circle has zero radius and which is nothing but a point. So A is nothing but the total stress point and A dash is nothing but effective stress point and here for the sample consolidated at constant confining stress the stress path is a single point with u is equal to 0 and sigma f is equal to sigma r dash sigma r is equal to sigma r dash at the end of consolidation. So this is the initial state of stress at the application of the you know stress sigma 3 then you know the Mohr's circle will be a point that we have discussed. Now you know when the stress traverses to along the line with the tau is equal to 0 so this increment in pressure is carried by the pore water itself the increment in pressure is carried by the pore water itself. So it actually traverses along the sigma line or sigma dash line and further you know once at the end of the consolidation the pore water pressure you know dissipates. So if the increment of pressure is actually carried by the pore water pressure is here and once at the end of the consolidation that u delta u actually dissipates. So if the total stress increment delta sigma is now removed the stress paths will be reversed in the opposite direction. So it has actually traverses like this but if the total stress increment delta sigma is now removed that stress paths will be reversed. So triaxial test is restricted as the cell pressure must be equal to the minor or major principle stress and wide range of stress paths are possible because of axial and radial pressure can be varied independently and you know axial and radial pressures may be increased or decreased or held constant so with that you know what we are actually deliberating is that you know number of types of tests which are possible and can lead to different stress paths. So the CD test with the drained triaxial stress paths compression where constant isotropic total stress. So here delta sigma r is equal to 0 compression where constant axial stress is 0 that is delta sigma a is equal to 0 and you know then extension the constant isotropic total stress where delta sigma r is equal to 0 and extension where constant you know axial stress where delta sigma a is equal to 0. So possible combination of drain loading in triaxial sample are discussed here. Now let us consider the drained compression the constant isotropic stress that delta sigma r that means that the change in radial stress is 0 delta sigma r is actually is 0. So in a drain triaxial test sigma a is equal to sigma dash a and the sigma r is equal to sigma r dash so and then u is equal to 0 that is because the pore water pressure dissipation is 0. So delta sigma a dash is equal to delta sigma a, so delta sigma a dash is equal to delta sigma a a dash which is positive and delta sigma r dash is equal to 0. So if you look into this delta t is equal to delta tau is equal to delta tau is equal to delta sigma a minus delta sigma r that is nothing but delta sigma 1 minus delta sigma 3 by 2 so if you look into this here delta sigma r is equal to 0 with that what we have got is that delta sigma A dash by 2 and this is delta tau is equal to delta A dash by 2. And when you have delta s dash is equal to delta sigma dash then delta sigma dash is equal to delta sigma A dash that is nothing but delta sigma 1 dash plus delta sigma r dash which is nothing but delta sigma 3 dash divided by 2 so with delta sigma dash is equal to 0, what we have got is that this delta sigma a dash by 2. So the slope of this one delta tau by delta sigma dash is nothing but the positive that is nothing but if they get cancelled then delta tau by delta sigma dash is equal to plus 1. So only the stress which changes is sigma a dash by an amount delta sigma a dash that is nothing but the actual stress is actually increased and the rest everything delta sigma r which is sigma dash is equal to 0 and in the drained stress path compression constant isotropic stress where delta sigma r is equal to 0 and with that you know the pore water pressure dissipation both during the shearing is 0. So this is can be indicated whatever we have deliberated with drained compression with constant isotropic stress where delta sigma r is equal to 0 with compression on upper axis and extension below so and sigma and sigma dash axis here. So m is that point where initial stress point and then what we have said is that you know the stress path actually traverses to m to p mp and the slope of this line is delta sigma by delta sigma delta tau by delta sigma dash is equal to positive 1. So that is nothing but because of this deliberation what we have discussed here in the in this slide. The slope of this line is you know 1 the slope of this line is one vertical one horizontal one vertical one horizontal and this is the kf line this is the kf line. So if you look into this the stress path is a straight line with the slope 1 is to 1 that is actually 45 degrees failing at point p. So the slope is actually meeting the kf line at point p. So mp is the stress path is a straight line with slope 1 is to 1 failing at point p. Now drained compression constant axial stress that means that here delta sigma a dash is equal to 0 that means that there is no change in the axial stress. So here what we have got is that you know the only stress with changes is delta sigma r dash by an amount delta sigma r dash that means that we are actually increasing let us say delta sigma r dash. So delta sigma a dash is equal to 0 and delta sigma r dash is equal to minus delta sigma dash. So delta tau is nothing but delta sigma a dash minus of minus delta sigma r dash by 2 with that you know what we get is the delta sigma r dash by 2 and delta sigma dash is equal to delta sigma n dash a plus minus of delta sigma r dash by 2. So with that what we get is that delta sigma a dash is equal to 0. So with that what we get is that minus delta sigma r dash by 2. So when we have let us say the sample delta sigma a dash is equal to 0 and when you keep on increasing delta sigma delta you know delta sigma r dash the sample experiences you know the then delta sigma delta tau by delta sigma dash is equal to thus the slope will be you know the is you know minus 1. So that means that you know one vertical minus 1 horizontal you can see here. So in this case when they when you have the constant axial stress delta sigma a dash is equal to 0 that means that there is a constant axial stress and if the stresses are initial isotropic the resulting stress path is given as MR that is that if the MR is the resulting stress path which is actually you know which has actually got a delta tau by delta sigma dash is equal to minus 1 with k0 is equal to 1. Now drained extension the constant isotropic stress where delta sigma r dash is equal to 0 that means that here in the drain triaxial test sigma a is equal to sigma a dash and sigma r is equal to sigma r dash and u is equal to 0. So here the only stress which changes is sigma r dash by an amount delta sigma r dash. So you know the only stress which actually changes is delta sigma r dash by an amount delta sigma r dash. So in this case delta sigma r dash is equal to 0 that is and then delta sigma a delta sigma a dash is equal to minus delta sigma a dash and delta tau is equal to minus delta sigma a dash minus delta sigma r dash by 2 and with delta sigma r dash is equal to 0 what we get is the delta tau is equal to minus delta sigma a by 2 and delta sigma dash is equal to with delta sigma r dash is equal to 0 what we get is that minus delta sigma a dash by 2. So delta tau by delta sigma dash is equal to plus 1. So if you look into this here what we have got is the delta tau is equal to minus and delta tau delta sigma dash is equal to also minus. So the stress path is here mq is a simple extension of the corresponding compression test mp into the extension stress zone. So here the delta tau by delta sigma dash is equal to positive but they have minus 1 vertical minus 1 horizontal and towards the extension side. So mq is the simple extension of the corresponding compression test mp into the extension stress zone. So for the drain extension the constant isotropic stress is delta sigma dash is equal to 0 is actually given as the stress path mq is a simple extension of the corresponding compression test mp into the extension stress zone that means that we actually have discussed this one which is in the compression test and this is actually resulting in the extension zone. Now we have another case that drain extension the constant axial stress delta sigma a dash is equal to 0 and where delta sigma a dash is equal to 0 where delta sigma r dash is equal to plus delta sigma r dash. So delta tau is equal to delta sigma a dash minus delta sigma r by 2 where with delta sigma a dash is equal to 0 delta sigma a dash is equal to 0. So what we have is that you know minus delta sigma r by 2 and delta tau dash is equal to delta sigma a dash plus delta sigma r dash by 2 with delta sigma a dash is equal to 0, what we get is that delta sigma r dash by 2 with the slopes delta tau by delta sigma dash which is nothing but minus 1. So the only stress which actually changes is sigma r dash by an amount delta sigma dash with what here delta sigma added that axial stress is actually not changed. So with that what actually we get is that the stress path MS is a simple extension of the corresponding compression stress path MR that is what we have indicated like this into the extension stretch zone. So this is nothing but the where delta sigma, delta tau by delta sigma dash is equal to delta tau by delta sigma dash is equal to minus 1, delta tau by delta sigma dash is equal to minus 1 for a k0 is equal to 1 condition where there is for isotropic case. So we actually have seen you know the various stress paths but in this particular slide a simple isotropic compression stress path is actually shown. We said that it actually traverses along T s dash that is s dash line and when you have got 1D consolidation 1 dimensional consolidation where you know there is only vertical strain and sigma epsilon r is equal to 0 axis symmetric case where you have the k0 line is like this, this is the k0 line. So this is you know on this the you know the 1D consolidation actually happens but you know when you have you know let us say for a direct shear test you know when the direct shear test when you draw initially you know this AB is actually very similar as the k0 line. So it traverses from A to B you know because you know in dry shear test what we do is that initially there is you know some vertical stress is actually applied and before coming shear there is elastic equilibrium or k0 condition is established. So AB is the you know along you know the k0 line so there is a point B where you know when the shear started you know what you can see is that BC is the path which actually takes the stress path which actually takes in direct shear test during shearing. So indicating that the you know the tau added to the sample horizontally does not change you know yes it does not change basically yes but just only T will change that what you see is that the T is changing but what is actually happening is that it only the T changes S will not change that means that the line traverses vertically upwards this is in case of a you know direct shear test. So in reality basically S may increase slightly due to the positive pressures which are actually developed at the sample ends in the upper portion of the sample. So that means indicating that the tau added to the sample you know horizontally does not change so this figure which is actually the stress path which is actually shown indicates that you know that tau which is actually subjected to the sample you know it does not change horizontally with S but only the T is actually changing. So in reality S may be subjected to slight increase due to you know passive pressure developed by the sample ends in the upper part. So this is actually one of the limitations of the direct shear test this we have discussed earlier. Now when you have got the triaxial compression and extension test what we have discussed is that either with one vertical one horizontal 45 degrees case or in this side this is you know in the towards the compression side this is towards the extension. So this is shearing in compression and this is actually shearing in the extension. So shearing in extension means how it comes. So AB isotropic consolidation isotropic consolidation is here and then you know when it actually BC it actually changes from here that is shearing in the stress path BC is the shearing in compression and stress path BD is shearing in extension. So consolidated undrained triaxial test so till now we have actually discussed about the stress path for the consolidated drain test. So in case of consolidated undrained triaxial test what we have discussed is that during consolidation the drainage is allowed but during shear the drainage is not allowed so that there is a possibility of the build up of the pore water pressure and then we also discussed that depending upon the stress history and the type of the soil deposit like a loose sand or a dense sand or normally consolidated clay or work consolidated clay. In case of loose sand and normally consolidated clay there is a possibility that the pore water pressure excess pore water pressure developed during shear is positive and when you have a dense sand and work consolidated clay you know the pore water pressure developed you know is negative in nature that is because of the phenomenon of the dilation. So the isotropic consolidation in increasing isotropic there is a stress path for isotropic consolidation remains identically same in both CD and CU triaxial test for a isotropic compression case initially they both CD and CU they are actually one and the same. So this we have discussed so in Mohr's circle actually initially a point you know with zero radius and stress path travels us along the tau is equal to zero this we have discussed during you know our discussion on the consolidated drain triaxial test. So this is at the end of dissipation of pore water pressure it actually you know the stress of the sample increases by delta sigma dash which is nothing but the amount of pore water pressure which is actually dissipated during consolidation and here we have two cases one compression the constant isotropic total stress delta sigma r dash is equal to zero and constant axial total stress where delta sigma a dash is equal to zero so possible combinations of undrained loading on the triaxial test are shown here where both are compression one is that constant isotropic total stress delta sigma r dash is equal to zero there is no change in the radial stress and in the other case is that there is no change in the axial stress. So first case is that you know the constant isotropic stress are delta sigma r dash is equal to zero when you consider then we have you know the slopes of the stress paths in conventional undrained compression are found to be by putting delta sigma r is equal to zero and then with that we get delta tau is equal to delta sigma a dash by 2 and delta tau delta sigma is equal to delta sigma a by 2 and then the ratio of delta sigma delta tau by delta sigma is equal to 1. So the effective stress paths will be separated from the from by these by pore water pressure value you at any time. So before you know discussing this stress paths of you know the you know in case of consolidated undrained triaxial test we should actually also introduce ourselves to two parameters which are actually put forward by Skempton they are called as you know the Skempton's pore water pressure parameters and like you know a simple way to estimate the pore water pressure change in undrained loading in terms of total stress changes is actually given by Skempton 1954 and which is given by an equation delta u is equal to this is delta u is equal to b into delta sigma 3 plus a into delta sigma 1 minus delta sigma 3 a and b the both are called as Skempton's a parameter a and b parameter. So b is a coefficient indicating the level of saturation. So if b is equal to 0 is basically a dry soil and b is equal to 1 indicating that the soil is completely saturated and a is nothing but the excess pore water pressure coefficient is a is nothing but the excess pore water pressure coefficient. So b is equal to 1 for saturated soils when you put b is equal to 1 when b is equal to 1 and then you know delta u by then we can actually write you know a is equal to and a f is equal to delta u by delta sigma 1 minus delta sigma 3. So at failure with b is equal to 1 we can write a is equal to a f is equal to delta u by delta sigma 1 minus delta sigma 3. So what we have done is that further this particular equation we have taken delta u is equal to b delta sigma 3 b into the brackets delta sigma 3 plus a into delta sigma 1 by delta sigma 3. Let us divide this entire Skempton equation by delta sigma 3. So what we get is like this delta u by delta sigma 3 is equal to b into 1 plus a by 1 plus a into delta sigma 1 by delta sigma 3 minus 1 delta sigma 1 by delta. So Skempton's equation is very useful in determining whether the soil is saturated in axis axis symmetric test. So during isotropic condition delta sigma 1 means delta sigma 3. So during isotropic condition delta sigma 1 is equal to delta sigma 3. So when you substitute in this what we get is that b is equal to delta u by delta sigma 3 when b is equal to delta u by delta sigma 3. So when during the isotropic consolidation we maintain the delta sigma 1 is equal to delta sigma 3. So with that you know when you substitute here delta u by delta sigma 3 is equal to b what we obtain. So for saturated soils b is equal to 1 and b is equal to 0 for dry soils. So a parameter at failure a f is nothing but for normally consolidated place a r is equal to 1. So normally consolidated soils the a r will be equal to 1 and for heavily over consolidated soils a f will be negative that the excess pore water pressure coefficient at failure will be negative. So the b parameter basically is a function of saturation and a parameter is basically function of over consolidation ratio. So over consolidation ratio if it is high then it actually changes into a negative for heavily over consolidated place that a f will be negative. So this particular table which is actually gives you know typical a f values for a heavily highly sensitive soils you know the 0.721 and normally consolidated soils place 0.521 and if you look into this compacted sandy clay that a f value is 0.252.75 and lightly over consolidated clay actually have got 0 to 0.5 and compacted clay gravel actually has got minus 0.252 plus 0.25 heavily over consolidated clay can actually go up to minus 0.520. So if you look into this here when you the plot which is actually gives variation of a f with OCR with increase in OCR you can see that here it is 1, 2 and 3. So OCR actually more than 3 that is from OCR is equal to greater than 2 is said as lightly more than likely over consolidated place. So when you actually have got OCR in the range of more than 3 the a f value will be negative. So this is actually you know that is the discussion we actually had earlier if you connect this is because of the negative pore water pressure development which actually takes place in case of you know during shear in case of you know the war consolidated or densely deposited soils. So the under end compression constant isotropic stress now what we with the background of the exkeptons pore water pressure parameters we actually write the pore water pressure equation again with delta u is equal to b into delta sigma 3 plus a into delta sigma 1 minus delta sigma 3 and at failure a is equal to 0.3321 for normally conserved clay. And so for b is equal to 1 for saturated clay when you substitute here what you get is that delta u is equal to a delta sigma 1 plus delta sigma 3 we have taken common into minus 1 minus a. So here what we have got is that delta sigma 3 into 1 minus a. So here delta u is equal to a delta sigma 1 plus delta sigma 3 into 1 minus a. So delta sigma r is equal to 0 and delta tau is equal to delta sigma a by 2. So this is you know delta sigma r is equal to 0 there is no change in the radial stress but delta tau is equal to delta sigma a by 2. So when you substitute this delta sigma a is nothing but delta sigma a is equal to 2 into you know delta sigma is equal to 2 into delta tau. So when you put here delta u is equal to a into delta sigma 1 with this when there is no change this becomes 0. So what we have is that delta u is equal to a into delta sigma 1. So delta sigma 1 is substituted by 2 into delta tau. So what you get is that delta u is equal to 2a delta tau 2a delta tau. So a is equal to 0 to 0.25 for heavily over conserved clay. Now further you know this delta u is equal to a into delta sigma 1 is equal to 2a into delta tau and where if a is constant during the test with effective stress path this is a straight line with slope as you know delta tau by delta sigma is equal to 1 by 1 minus 2a 1 by 1 minus 2a. So here if you look into it here you have got delta u by delta sigma 1 which is nothing but 2a by delta sigma. So what we get is that if a is constant during the test the effective stress path is a straight line with the slope is nothing but delta tau divided by delta sigma dash is equal to 1 by 1 minus 2a. So this is nothing but the slope of the you know the stress path which is indicated in this diagram and here in case of the this is the total stress path and this is the effective stress path you can see that and this difference is actually because of the you know delta u during the shear. So the slope of this line you know if a is constant the slope is nothing but 1 by 1 minus 2a the slope of this line. So the stress paths are assumed to be linear if a dash actually varies with the test once the test is in progress then ideally the stress paths are corroded in nature. So the effective stress paths will be corroded in nature that is what we have shown in the previous discussion the reason why the effective stress paths are corroded in nature because the value of the a will not be constant that is the stress effective stress the skimptons pore water pressure parameter a will not be constant that is the slope will be changing. So because of that what will happen if it is constant then it is linear and if a is constant the slope will be constant and then this is in effective stress path but if there is a you know the if there is a varies with the when the test is in progress then ideally the stress paths are corroded in nature. Now you know with the constant axial stress once you look into it so when you have a delta sigma then which is actually nothing but is indicated by a and which is nothing but when you have delta sigma and then no change in sigma dash effective isotropic condition and plus when actually c that is delta sigma is there that you know then a is equal to b plus c where delta sigma dash is equal to minus delta sigma r is equal to minus delta sigma 3. So delta tau by delta sigma sigma is equal to minus 1 and with this actually we can say that as b cause no change in effective stress and we can say that a is equal to c. So the effective stress path for under end compression are same as both for constant axial stress loading as well as the constant radial stress loading the effective stress path for the under end compression are the same for both constant axial stress loading as well as constant radial stress loading. So here with this slide what we are saying is that the effective stress paths for under end compression are same for both constant axial stress as well as the constant radial stress loading. So when this is the constant axial stress once you look into it and so total stress path you know sigma r then constant sigma a increase then it is like this and when you have the unique effective stress path is actually possible like this with variation of the AF and the total stress path you know total stress path sigma a constant and sigma r decrease then it is actually you know traverses like this. So influence of the stress paths on the laboratory measured under end strength which actually can be given by this particular tau sigma sigma dash plot here and this line is kf line this line is the kf line the failure line and this intercept is kf dash or a according to our notation alpha dash or psi and k dash cot alpha that is this ordinate. So we can actually get the delta tau by delta sigma dash as the slope of this line slope of this effective stress path is 1 by 1 minus 2af af is the you know the this slope of this slope of this line at af is that pore water pressure is kemptons pore pressure parameter at failure af. So delta tau by delta sigma dash is equal to 1 by 1 minus 2af that is the slope of this line. Now if you look into this you know the at failure delta tau by delta sigma dash is equal to delta tau by delta sigma dash is equal to 1 by 1 minus 2af. So at failure which is like this referring to the previous figure basically for the soil with initial stress tau 0 and sigma 0 dash. So we can write tau f is equal to k dash so k dash is nothing but this ordinate plus sigma 0 dash tan alpha sigma 0 dash tan alpha that is you know the vertical axis ordinate and minus 1 minus 2af you know tau 0 tan alpha dash. So what we have written is that this particular vertical ordinate we have written vertical ordinate we have written and that ordinate is actually working out to be k dash plus sigma dash sigma 0 dash tan alpha minus 1 minus 2af tau 0 tan alpha that is from the slope is actually known to you and with that we can actually calculate. So we are subtracting we are taking the net height that is you know tau f and so with this we can actually simplify saying that tau f is equal to c dash cot phi dash plus in terms of c dash and phi dash parameters. So we converting from the k f line to the tau sigma space completely where with that tau f is equal to c dash cot phi dash plus sigma 0 dash minus 1 minus 2af tau 0 divided by cosecant phi dash minus 1 minus 1 plus 2af. So undrained compressive strength can be obtained you know where phi dash is equal to 0 undrained compressive strength is equal to can be obtained as you know tau f is equal to c dash cot phi dash plus sigma 1 dash divided by cosecant phi minus 1 plus 2af. So here the considering the undistributed sample initially at negative effective stress where sigma i is equal to minus ue and initially tau 0 is equal to 0 initially tau 0 is equal to 0. So with that you know this term will get you know you know when this will be eliminated then what we have is that the sigma 0 dash is equal to sigma i sigma i dash. So what you have got that c dash cot phi dash plus sigma dash you know i divided by cosecant phi dash minus 1 plus 2af that actually remains here. So based on the deliberation let us look into the you know one problem here undistributed soil specimens are basically taken from a depth of say at z is equal to 5 meters in a soft lightly over consul clay for which k0 is equal to 0.7 undistributed soil specimens are taken from a depth z is equal to 5 meter in a soft lightly over consul clay for which k0 is equal to 0.7 and the unit weight of the soil is 16 kilo Newton per meter cube and c dash is equal to 0 phi dash is equal to 22 degrees and the water table is at depth gw is equal to 1 meter and assume that gamma w is equal to 10 kilo Newton per meter cube for specimens tested under a cell pressure of 40 kilo Pascal assume that you know sigma v0 is equal to gamma z and if af is equal to 0.8 find the undrained compression strength at cu and undrained compression strength cu. So here you know af that is the effective the stress parameter skimpton stress parameter at failure is given as 0.8. So the solution actually works out like this we can first get the vertical stress which is nothing but 16 kilo Newton per meter cube and 5 meter depth. So it is 80 kilo Pascal's the effective stress because water ground water table is given at 5 meter the ground water table you know water table is at a depth of 1 meter. So by taking these considerations what we get is the effective stress. So once you get once you know the effective stress and k0 is given as 0.7 and sigma dash v0 is actually 40 kilo Pascal's so with that what we get is that 28 kilo Pascal's. So ue that can be obtained by sigma dash v0 plus 2 sigma h you know sigma dash h0 divided by 3 that is you know the sigma v and plus 2 sigma h. So that is you know this one we are actually estimating and that is actually given as estimated as minus of 40 plus 2 into 28 by 3 which is minus 32 kilo Pascal's and tau f now is the c cot phi plus sigma dash that is nothing but you know which is obtained as which is given which is obtained as you know this one as 32 kilo Pascal's that is ue is equal to that sigma i dash that is 32 kilo Pascal's. So cosecant phi that is phi dash is given as 22 degrees cosecant 22 minus 1 plus 2 into 0.8. So with this the shear strength and failure works out to be 10 kilo Pascal's. So similarly another problem where undistributed samples are taken from a depth of 5 meters in a stiff heavily over conserved clay for which k0 is equal to 1.8 and because this is over conserved clay the k0 will be high and gamma 20 kilo per meter cube and c dash is equal to 5 kilo Pascal's and phi dash is equal to 20 degrees and water table is at a depth of 3 meter and assume that gamma w is equal to 10 kilo per meter cube and for specimens tested under a cell pressure of 100 kilo Pascal's and if af is equal to minus 0.15. So you can see that the af is negative here and find the undrained compressive strength cu. So the solution again the sigma v0 is actually 5 meter that is unit weight is 20 so 100 kilo Pascal's and the effective stress can be obtained as 70 kilo Pascal's and with 1.8 we will able to get sigma h dash as 126 kilo Pascal's. Similarly here now by minus of ue is actually calculated where the clay is over consolidated so sigma v dash plus 2h sigma 2 sigma dash h0 h dash divided by 3 so with that what we get is that minus 107.3 kilo Pascal's. So tau f is equal to c dash cot phi plus sigma dash divided by cos cosine phi dash minus 1 plus 2 af so by substituting this phi into cot 22 plus 107 because here sigma dash is equal to minus u this minus of minus you will get become press here and the cosecant 22 minus 1 plus 2 into minus 0.15 so with that the shear strength of the soil works out to be around 87 kilo Pascal's. So what we have done is that we actually have you know try to reduce the discuss about in the consolidated undrained triaxial test and then we actually have you know obtain this you know deliberation about. So then you know we also have this octahedral plane and where we have for if you are having the sigma 1 axis and sigma 2 axis and sigma 3 axis and then you know when you actually have this lines indicate that sigma 3 is equal to 0 and sigma 2 is equal to 0 along this line and this is sigma 1 dash is equal to 0 and hydrostatic axis actually passes from the median of this that is hydrostatic axis is equal to sigma 1 is equal to sigma 2 is equal to sigma 3 when it actually happens and sigma 3 dash actually passes through this one and sigma 1 dash actually passes through this vertex and sigma 2 dash is actually passes through this vertex but along this sigma 3 dash is equal to 0 along this sigma 1 dash is equal to 0, along this sigma 2 dash is equal to 0. So the octahedral plane, the octahedral plane is equal to sigma 1 plus sigma 2 plus sigma 3 is equal to constant, the octahedral plane is very near to the soil failure state so very useful basically to derive the failure theories of the soil. So this octahedral plane which is indicated as sigma 1 plus sigma 2 plus sigma 3 is equal to constant and which is a plane which is very near to the soil failure and so very useful to derive the failure theories of the soil. You know further sigma octahedral what we can actually get is nothing but sigma 1 plus sigma 2 plus sigma 3 by 3 and tau octahedral the shear along the octahedral plane that means that on this octahedral plane that normal to the plane that is sigma octahedral and along this that is tau octahedral. So we can get sigma octahedral as nothing but sigma 1 plus sigma 2 plus sigma 3 by 3 and tau octahedral is nothing but square root of sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square plus sigma 3 minus sigma 1 whole square by 3. So sigma dash octahedral is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 minus u and tau dash octahedral is equal to tau octahedral. So total and effective octahedral shear stresses are equal, total and effective octahedral shear stresses are will be equal. So from the Mohr Coulomb yield function in most generalized form for the Mohr Coulomb condition of failure can be expressed as which is nothing but sigma 1 dash minus sigma 2 dash whole square minus 2c cos phi plus sigma 1 dash plus sigma 2 dash sin phi dash whole square plus sigma 2 dash minus sigma 3 dash whole square minus 2c cos phi plus sigma 2 dash plus sigma 3 dash sin phi whole square plus sigma 3 dash minus sigma 1 dash whole square minus 2c cos phi plus sigma 3 dash plus sigma 1 sin phi whole square. So the Mohr Coulomb yield surface is indicated by you know this surface which is you know indicated as GFEDCB and A is the point of the intersection of the octahedral plane and you know the hydrostatic axis. This is the point through which through point A the hydrostatic axis passes where sigma 1 is equal to sigma 2 is equal to sigma 3 and this is the point of intersection of the octahedral plane. So you can see that these are the point of intersection of the octahedral plane. So this is the octahedral plane and the Mohr Coulomb failure surface is actually defined or limited by this GFEDCB surface. So this is called as the Mohr Coulomb yield surface Mohr Coulomb yield surface. So then you know different failure criterion are actually described. So what we do is that in the next class we will discuss more about the octahedral plane and relevance to the you know how this can be interpreted for you know the so called understanding about the failure criterion and then thereafter we will try to discuss about you know how we can actually determine E for the E from the triaxial test data and then we will try to do some example problems on the shear strength of the soils.