 So, welcome to lecture series on advanced geotechnical engineering and we are in module 4 and discussing about the shear strength of the soils. So, this is module 4 lecture 7 on stress strain relationship and shear strength of soils. So, in the previous lecture we introduced ourselves about the PQ space and discussed about the Mohr-Colem criterion in depth. Then in this particular lecture we will introduce ourselves to methods for determining shear strength especially in the laboratory by two predominant methods that is direct shear test and triaxial compression method and thereafter we will introduce ourselves to stress paths relevant to the triaxial compression test and the drainage conditions. So after having covered these contents we will be looking into the triaxial behavior and stress state and analysis of unconfined compression, unconsolidated undrained and consolidated undrained and consolidated drain and other special test and drainage conditions. So we have several methods for determining shear strength in the laboratory, predominantly they are like direct shear test and triaxial compression test and then there are some simple shear test and plane strain triaxial test. Like that we have several classes of tests which are actually can be done in the laboratory for determining the shear strength parameters of the soils under considerations both in disturbed or undisturbed conditions or for the remolded conditions and then we also have some methods for determining indirect methods for arriving at the strength parameters based on some parameters which are actually measured in the field. So the different determination of the shear strength parameters can be achieved through laboratory or through field test and the laboratory predominantly the direct shear test and triaxial test are performed traditionally and there are also some other classes of tests like direct simple shear test and torsional ring shear test and plane strain triaxial test and for very soft soils we wanted to determine where the sample making is difficult then in that case we have laboratory vane shear test and we also have a fall cone test where it also indicates the penetration of the cone indicates an interpretation of the shear strength measurement. In the field when we are having a soft soil when the sample pickup is very difficult then vane shear test at different depths is also popular and then we also have some pocket pentrometer test and pressure meter test will give the in-situ properties or in-situ elastic modulus of soil and as we discussed the static cone pentrometer test which actually gives the cone resistance and the frictional resistance or shaft resistance and with the by interpreting the shaft resistance and frictional resistance we will be able to get an idea about the shear strength parameters. Similarly, the standard penetration test which is the resistance of the soil is measured by the SPT n value so we have number of correlations which can actually can give for different n values by their undrained cohesion or the friction angle of a soil. So most common laboratory test used to determine the shear strength parameters C or phi in case of total stress parameters and C dash and phi dash in case of effective stress parameters. So total strength parameters are called as C and phi and effective strength parameters are called as C dash and phi dash and they are basically the direct shear test and traction shear test. So let us look into the first the direct shear test and it is also called as a shear box test and this box can be square or circular in shape and this box sizes are predominantly 60 mm by 60 mm or we have 100 mm by 100 mm or 300 mm by 300 mm. So depending upon the size of the particles which are actually involved and to prevent the scale effects the sizes of the direct shear boxes need to be adopted and it is also used to determine the soil stress strength and not the determination and deformations and different sizes of the shear box can be used depending upon the grain size of the coarse grain soil or the grain size of the soil under consideration or soil like material under consideration. And the sample is loaded first and with normal stress and then subjected to a shear force so that the shear can be applied on a predetermined shear plane which is actually induced the shear is actually induced along the predetermined failure plane. So this method of testing also makes possible for determining interface parameters like when you have got let us say that two different materials let us say that pile interacting with the soil or when you have got a retaining wall interacting with the soil and if you would like to have the interface friction angle then the particular material can be kept at the bottom and a dummy block can be placed and then the required interface surface actually can be put and then the required soil which is to be interacted with the particular surface can be placed. So with that we can actually determine the interface parameters and when we do when we try to determine the interface parameters then it is actually called as modified direct shear test and with that we get CA in case if you are having a cohesive soil then adhesion is actually measured and then interface friction angle. So this direct shear test actually has got the method of inducing failure along a particular plane is used to actually as a way for determining the interface parameters or if you wanted to determine interface friction angle between let us say a rock and surrounding soil outcrop then that can also be achieved through this particular method of testing. So this typical direct shear test configuration is actually shown here it actually has got a shear box and in order to some extent test can be performed with some drained conditions but they are actually regarded as you can actually do but not possible for us to measure any pore water pressure and other aspects and here the configuration is that you have got upper shear box and lower shear box and in order to induce to keep the failure plane at the predetermined surface the serrated plates are actually used at the base and then these serrated serrations actually have to be perpendicular to the direction of the shear so that the failure is actually induced along that predetermined failure plane and in order to allow the drainage you actually have the porous stones and then we have the loading plate and through this loading plate can go down or go up so it is which is started here and this actually with the lever arm arrangement we actually apply the normal stresses may be can the range of 25 kilo Pascal to 300 kilo Pascal to 400 kilo Pascal normal stresses can be applied and then what it is done is that the shear force is actually applied as shown here and with that the shear the resistance offered by the interacting soil or soil surface is actually measured so with that you will be able to get the shear stress versus shear displacement so what when we apply when we apply the shear force if this is the direction of force and then there is shear force is actually generated and the shear displacement is generated so when we do a different normal stresses we actually get set of curves for each normal stress which is actually performed and then depending upon the type of the soil whether it is a dense soil or whether it is a loose soil we have the variations in the types of the pattern of the stress curves changes similarly if you are having a undistributed sample let us say a over consolidated soil or a normally consolidated soil and the pattern of the stress distribution the shear stress versus shear strain variation is subjected to change. So here the mechanism is that at the start of the test what actually happens is that you have sigma n and sigma h so these are the major principle plane is actually here major principle plane is here and minor principle plane is perpendicular to this and at the beginning of the test when before the commencement of the test on the failure plane the major principle plane and the minor principle plane they do exist but when the test commences are during the test because of the mobilization of the shear and then those that particular failure plane can no longer be called as a predetermined plane over which the failure is being induced it can no longer be called as the principle planes. So in that principle plane in that case then we have to see that there is a chance that in the direct shear test the rotation of the principle planes can actually occur and here in this case the stress state at the end of the test is shown here. So you have got a normal stress and then there is a shear stress which is actually mobilized so this makes it actually not eligible to be called as the principle plane during the test and this is the normal stress remains constant for a particular sample of the test. So in this particular figure the typical shear stress versus shear displacement for loose sand and medium sand and dense sand actually are given so as can be seen that loose sand actually has got the continuously the hardening effect can be seen but in case of the medium sand and dense sand and particularly for dense sand there is a predominantly a peak is actually observed and then once the peak stress is actually observed and there is a decrease in the peak shear stress with an increase in the shear displacement or shear strain. So peak shear stresses are noted down at each normal stresses and then when we plot these peak shear stresses for each normal stress what has been done then what we get is that the tau sigma space we actually get the more coulomb failure envelope the slope of that more coulomb failure envelope joining all these tau 1 tau peak 1 you know sigma 1 and little sigma n1 sigma n2 sigma n3 and that actually gives the you know the slope of line actually gives the you know the internal friction angle and if it is actually intercepting at the tau axis then you know that intercept is actually taken as the cohesion of a soil. So when we plot in the direct shear test as we can see that here the vertical strain can be measured by putting a dial gauge and lateral movement can be measured by putting a you know the in case for dial gauge or a potential meter or LUDT so with that what will happen is that you will get shear displacement or shear strain versus you know vertical strain. So this vertical strain can be you know can lead to expansion or it can be lead to compression. So what actually happens is that in case if you are having a loose sand as can be seen here it continues to harden so in line with whatever we actually observed for tau sigma you can actually see that this undergoes a continuous compression so in this case the contraction actually takes place but in case of dense and medium sand for a given normal stress what will actually happen is that you know initially there will be a compression then thereafter what will actually happen is that has been discussed this was actually discussed in the previous lecture that the soil particles ride on each other and then induce or make the you know the loading plate to move upward because there is no other way you know the particles are actually can you know accommodate the disturbances which are actually created because of the shear. So in a way what will actually happen is that there is a dilation or in the sense that expansion in volume takes place and this actually get suppressed with decrease in particle size and similarly when we are actually having increase in the normal stress the tendency of the dilation actually keeps on changing. So volume keeps on decreasing for loose sand but in case of dense sand and medium sand what will actually happen is that initially there will be a compression and thereafter you know there is an increase in expansion of the soil takes place upon shear. So this actually you know phenomenon is actually attributed to the dilatancy effect or it is also the slope of that particular line is also called a dilatancy angle. So as we said and we can actually we can analyze or you know look into this as a you know saw blade you know you know analogy so it can be seen that when a shear is actually applied along the saw blade actually we have the movement actually can take place. So the apparent you know externally mobilized angle of friction on horizontal planes pi is actually larger than the angle of friction resisting the sliding that is pi u or in the pi u or pi i is equal to friction plus dilatancy. So the strength is actually nothing but friction plus dilatancy that actually what it actually turns out to be. Now as we have seen that it can be possible that the soil can be in very dense state and soil can also you know at particular situation that you know after you know a certain point actually where the shear can actually occur at you know constant volume then that is actually called as critical and when this actually attains at a constant volume the no change in void ratio takes place that is actually the void ratio at which you know no change in volume takes place then it is actually called as a critical void ratio. And then you know when you are having initially loose then you know as we have discussed that these all loose pockets which are actually they get filled with the soil particle so then that the compression actually takes place so the you can see this the orientation of the sub blade analogy is actually given in the downward direction and here is actually done in the upward direction this actually shows that you know the riding effect and this is also some depositing effect like you know one particle will get jumbled into other particle in a way what will happen is that there is you know a decrease in the volume takes place. So let us look into this interlocking and you know dilatancy phenomenon when soil is actually is initially denser than the critical state where critical state is the place where no volume change occurs which it must achieve then as particle slide past each other going to the imposed shear strain on an average they will actually tend to get separate and the particle movements will be spread about a mean angle of dilation that is called psi and in a dense sand there is a considerable degree of interlocking between the particles takes place and before shear failure can take place this interlocking must be must be overcombed in addition to the frictional resistance of the points of contact. In fact the peak actually which is attributed in case of a dense sand samples is actually attributed to the interlocking effect which actually you know results in the mobilization of the peak shear stress and thereafter you know the softening actually takes place. So in a dense sand there is a considerable degree of interlocking between the particles and before shear failure can take place the interlocking must be overcombed in addition to the you know frictional resistance at all points of the content. So in this particular slide you know we can see that the variation of the volume increase for a dense sand and loose sand is actually given here in this case of dense sand there is a contraction in this in the case of a loose sand for a given normal stress we can see that there is an increase in volume takes place. So this is you know the slope of this line that is d epsilon v that is the change in vertical strain or volumetric strain to the shear strain you know that is actually regarded as the angle of dilatancy. So in general the degree of interlocking is greatest in the case of very dense well graded sands and soil particles with angular which are angular in nature the texture of the particles if it is an angular and rough and there is a possibility that the degree of interlocking is the greatest and basically in the case of very dense sand or well graded soil mix and then the angular particles and the characteristics to strain curve for an initially dense sand shows a peak stress at a relatively low strain and thereafter as interlocking is progressively overcome the stress decreases with increase in strain. So the characteristic stress strain curve for an initially dense sand shows a peak stress that is what we have discussed at a relatively low strain and thereafter as the interlocking is progressively overcome the stress decreases with increase in the strain. So the The reason for having absorbed peak and then decrease in the stress value with an increase in strain is attributed to the breaking of that interlocking which actually was provided in that time and then upon shearing the once it is overcome that interlocking then what will happen is that there is a decrease in the stress with an increase in the strain. So the reduction in the degree of interlocking basically produce an increase in the volume of the specimen. So the reduction in the degree of interlocking so this particular reduction of or breaking of this interlocking results in increase in the volume of the soil specimen which is being subjected to shear. So as characterized by the relationship between the volumetric strain and the shear strain in the direct shear stress so the reduction in the degree of interlocking produces an increase in the volume of the specimen and during the shear as characterized by the relationship between the volumetric strain and the shear strain in the direct shear test. Now the term dilatancy is used to describe the increase in volume of a dense sand during shearing and the rate of dilation can be represented by the gradient which is nothing but what we have said is that d epsilon v by d nu which is the maximum rate of the corresponding to peak stress. The angle of the dilation psi is nothing but tan inverse d epsilon v by d nu and for a dense sand the maximum angle of shearing resistance pi max is determined from the peak stress is significantly greater than the true angle of friction. So we have now maximum peak friction angle that is actually determined at wherever we have the tau peak is there and when you pick up these values and plot the Mohr-Coulomb monologue with normal in the tau sigma space then you know what will actually happen is that there is slope of that line which is actually having a peak shear stresses which is actually called as the pi max and there is also true angle of internal friction that is actually nothing but you know the frictional resistance offered by the individual particles and for that actually it is pi u it is called and that is pi mu it is called pi mu, mu indicates for the frictional the sliding friction and the difference representing the you know there is a difference between these two you know between pi max and pi mu and that is actually because the difference representing the work required to overcome the interlocking and rearrangement of the particles. So pi mu is actually more than pi max is actually more than pi mu and this is basically the difference there is a difference between pi max minus pi mu this basically the difference is attributed to you know the work done which is actually required to overcome the interlocking and the rearrangement of the soil particles. Then we said that when you have got initially loose soil sample there is a no significant you know interlocking takes place and the particles always tend to you know arrange into the new position and you know in the way that the soil undergoes a contraction upon actually shearing and the shear stress increases gradually to an ultimate value without a prior peak and accompanied by a decrease in the volume. So the in case of initially loose sand there is no significant particle interlocking takes place. So that is the reason why there is no peak is actually observed and the shear stress versus shear strain variation is actually that and the shear stress continues to increase and after reaching the at large strains it actually tends to become constant. So in the case of initially loose sand there is no significant particle interlocking to be overcome and the shear stress increases gradually to an ultimate value without a prior peak and accompanied by a decrease in volume. But when soil is initially looser than the final critical state then the particles will tend to get closer as the soil is disturbed and the average angle of you know dilation is negative and indicating a contraction. So the angle of dilation is actually negative that is actually indicates a contraction actually takes place. So let us try to you know interpret between pi max and pi mu and pi cv. Pi cv is the friction angle at constant volume and cv indicates at you know at constant volume and so in this particular slide at tau versus you know shear strain is actually shown here and where for dense sand and loose sand the shear stress variations are shown here as can be seen that the shear stress is actually maximum here and then you know the peak is actually distinct peak is actually shown and then there is a decrease in shear stress with an increase in the shear strain and it tends to become you know in join with the curve wherein we have got the hardening actually takes place continuously for initially loose sand and becomes denser by virtue of shearing. So this thus at the ultimate or at a critical state the shearing takes place at a constant volume and the corresponding angle of shearing resistance being denoted as pi cv or pi critical it is called pi critical friction angle and the difference between pi mu and pi cv represents the work required so this is actually this gives the pi cv and this actually require this somewhere here you know somewhere here the this is due to the sliding friction the sliding that is because of the fractional resistance offered by the particles in arranging into the you know the into the denser configuration is actually is you know indicated with the pi mu and here the difference between pi mu and pi cv represent the work required to rearrange the particles in general the critical state is defined identified by the extrapolation of the stress strain curve to the point of constant stress and which should be also correspond to the point of zero rate of dilation on the voluminous strain stress curve. So this also corresponds to the void ratio at which this you know this particular phenomenon occurs this is actually called as the critical you know void ratio and this is the critical void ratio so this is the you know the critical state is identified by the extrapolation of the stress strain curve to the point of the constant stress and which should also correspond to the point of zero rate of dilation on the volumetric strain volumetric strain shear strain curve. Now you know we all done now we said that at constant volume when the friction angle is interpreted then it is also called as critical friction angle or pi cv and if the density of the soil does not have to change in order to reach the critical state there is a zero dilatancy as the soil shear at the shear of the constant volume but it is important to realize that a critical state is only reached when the particles have had a full opportunity to juggle around and then come into new configurations that means that a progressively you know this state is actually achieved and if the confining stress is increased while particles are being moved around then they will tend to finish up in a more compacted state so that arrangement will not actually happen. Now among a phi max and phi cv in practice the parameter phi max which is actually is called as a transient value and should only be used for situations in which it can be assumed that the strain will remain significantly less than the corresponding peak stress. So if for a given structure when it is actually being designed when the strain is actually when the strain is actually less than you know at which the peak stress is actually interpreted then you know the you know phi max is actually can be recommended for use. So in practice the parameter phi max which is a transient value and should only be used for situations in which it can be assumed that the strain will remain significantly less than the corresponding to the peak stress. If however the strain is likely to exceed the corresponding to peak stress if the strain is you know likely to exceed the corresponding peak stress a situation that may lead to progressive failure then it is actually a critical state parameter and phi cv should. So when we need to use phi cv when we need to use phi max is actually clarified in this slide if however the strain is likely to exceed the strain in the sense that induced for a particular geothermal structure is exceeded you know that the corresponding to peak stress a situation that may lead to progressive failure then the critical state primarily parameter phi cv should be used. So here when we look into this you know for the interpretation of you know the direct shear test and we actually said that when dense sands are over consolidated clays are sheared they dilate. So when dense sands actually are over consolidated clays you know when when they are subjected to shear they tend to dilate and larger the particle size greater is the dilation and the decrease of the particle size actually decreases the tendency of the part the soil to dilate and more coulomb idealization implies dilation at a constant rate when a soil is sheared and this is actually you know appear to be unrealistic. So when dense sands are over consolidated clays are sheared they tend to dilate and larger the particle size and greater is the dilation and decrease in the particle size suppresses you know the dilation tendency and also increase in the normal stress and increase in the over burden also suppresses the dilation tendency. Now you know let us look into the interpretation of the direct shear test results and then you know we will try to look into some example and here the peak shear stresses are noted from the shear stress versus shear displacement or shear strain curve and each normal stress the peak stress is actually when it is picked up and when it is plotted here and they line joining this you know this envelope is actually resulted as a more coulomb failure envelope and if it is a purely frictional soil and with the invisible amount of fines there is a possibility that you know the friction angle the envelope can actually pass through origin and important aspect we should remember is that while plotting these results the tau sigma curves have to be on the equal scale so that the proper you know interpretation can be done. So there will be you know like 1 or 2 or 3 basically minimum 3 are required and if you are actually trying to do like say 5 normal stresses and so you will actually get the 5 number of shear stress displacements and a plot of the peak shear stress versus normal stress do give the shear stress parameters phi and c from the particular soil. So we will able to get these c and phi or a particular soil from this curve and further you know there is a way where we can actually construct the more circles because here in this case what we are trying to do is that we can actually construct the more coulomb failure envelope and we know actually this particular point. So what we can do is that you know this is the this point you know more circle more coulomb hypothesis actually says that when more circle is in contact with the failure envelope then you know that is the point where you know the failure stress the tau shear failure shear stress actually achieved at a particular normal stress. So here for sigma n1 tau f1 is the failure shear stress sigma n2 tau f2 is the failure shear stress sigma n3 tau f3 is the so this is actually the dotted line which is actually shown is the you know failure envelope and now you know with particular you know dropping a perpendicular to the sigma axis from the failure envelope we can actually locate the center of the more circle and with this as actually a center we can actually draw a more circle then we will able to locate you know major principle stress and minor principle stress and from the known stress state on the more circle if you draw a line horizontal parallel to the sigma axis it actually intersects at a point p and that point is actually regarded as pole and from the pole when we join these points to major principle stress ordinate here and minor principle does ordinate here then actually we can get the inclination of the major principle plane and minor principle plane and so when we draw the more circle we will able to get the you know major principle plane and minor principle plane locations and magnitudes of the stresses and the direction of the principle planes and all. So as can be noted if you look into that initially when you start the you know the prepare the sample and then kept it for you know equilibration also at the time when the no normal stress is when the no shear stress applied when we have got along the predetermined horizontal plane you actually have you know the principle planes but upon shearing and during shear test actually in a case of direct shear test the principle planes are subjected to rotate and they rotate in the direction of shear and the degree of angle of rotation of the principle planes is approximately you know regarded as equivalent to the friction angle of a soil and so like this when we do you know different normal stresses and when you get register tau f2 and tau f3 the failure stresses you can actually draw these more circles and these more circles from here onwards so the line which is actually joining to this parallel to this one that is actually is nothing but the you know it indicates the failure plane and this indicate when you when you when it is actually meets at the point say p1 here and it is joining here then this is called the you know major principle plane inclination and when it is joining to this particular point that is called minor principle plane so we actually can determine the angle of the inclination of major principle planes and minor principle planes and then magnitudes when we know the stress state points at failure for sigma n1 and then tau f1 and sigma n2 and tau f2 and sigma n3 and tau f3 so this is the initial condition it has been as it been described here initially when we actually have you know sigma n0 that is the normal stress let us say a 50 kilo Pascal's or normal stress is applied and for a normally consolidated soil you know when you take a pressure at rest k0 as say 0.5 then in that case the horizontal stress which is actually there in the soil is you know sigma h0 is equal to k0 into sigma n0 so 50 kilo Pascal's into 0.5 let us say 25 kilo Pascal's will be there so when we draw that when we have the normal stress sigma n0 and you know the this is called initial Mohr circle and this much this Mohr circle is actually well below the well below the Mohr Coulomb failure and lobe then we actually have discussed that the this Mohr circle is regarded as the stable condition then further when with that same normal stress you know when we try to increase induce the shear then what will actually happen is that the Mohr circle actually you know the generates the you know the shear force and with that what will happen is that the Mohr circle you know becomes like this so with that what will actually happen is that you know you can actually get a point where you know it you know reaches to the failure so that is actually you know during the test and before the failure and so from here if you look into this here when you have this point at failure so what we have discussed is that you know when you draw in order to locate this and this one point is actually drawn so this is the center this is the center and this has the radius when you draw then this has the radius when you draw the Mohr circle then we can actually get a Mohr circle tangential to the so-called failure envelope and then passing you know this point and this point and then you know when you draw and locate a pole then when you join it to the major principle stress here then what you get is the major principle plane so the major principle plane is here and minor principle plane which is actually here so these are the you know directions or inclinations of the minor principle plane and major principle plane. So let us consider an example where in a direct shear test is actually run on a medium dense sand dense sandy silt with normal stress of 65 kilo Pascal's at failure the shear stress is observed as 41 kilo Pascal's and draw the Mohr circles at the initial and failure conditions and determine principle stresses at failure and the orientation of the failure plane and the orientation of the plane of maximum normal stress at failure orientation of the plane of maximum shear stress at failure so these you know issues can be addressed you know by the discussions whatever we have we had till now. So initially when normal stress is actually applied starting at the test then the 65 kilo Pascal's is actually located here and when we when we locate with being a silty sand and friction angle is actually given as 32 degrees so when we take using Jackie's formula K0 is equal to 1 minus sin phi and with that 1 minus sin 32 degrees is equal to then you know it comes out to be around 35 kilo Pascal's and with that you know 65 minus 35 that 30 kilo Pascal's as the diameter when you draw the Mohr circle you will get the initial Mohr circle for the initial condition. Then you know it has been given as you know the failure shear stress is actually located at 65 kilo Pascal's normal stress at a the normal stress is actually somewhere around 41 kilo Pascal's. So what actually happens is that you know when you locate that point and because it is given as silty sand and cohesion is actually assumed to be 0 and then it actually passes to the horizon so with that what we can actually get is that you actually have got tau sigma annulope this is the tau and sigma annulope actually here passing through the this is the point which is actually at 65 and 41. So when the so the angle of internal friction is equal to 30 degrees the slope of the line passing to the origin and point that is 65 and 41 and then you know in order to find the center of the circle you know with you know by dropping a you know perpendicular from here then you know we can actually locate point where it actually you know intersects the center then with this as the center and this as the radius we can actually draw a Mohr circle and that particular Mohr circle you know represents you know the Mohr circle actually becomes like this and then because we have taken this as the radius so it becomes tangentially here and this particular point where it actually intersects sigma and that is actually regarded as the minor principle stress so in this case about 40 odd kilo Pascal's is actually is the minor principle stress and here sigma 1 is actually here about 140 kilo Pascal's is the major principle stress. So we actually have located you know the Mohr circle and drawn the Mohr circle and the theta which is actually 45 plus 5 by 2 that is actually 45 plus 32 by 2 which is actually 61 degrees angle between horizontal and the line joining the center of the circle and point 65 41 is 180 minus 2 theta which is about 58 degrees so that is actually here is indicated here. Now you know horizontal line is extended through point 65 41 to the point to the other edge of the circle which actually gives pole and the line drawn through the intersection points between the circle and normal stress axis gives the principle plane inclinations to the horizontal so as we have discussed that when we draw this line for the point to this point and this point will actually get the inclination of the principle planes we you know with reference to the Mohr circle diagrams with that we will be able to get the we can locate we can see that you know initially when the sample is actually just prepared then we actually have got 65 degrees and about 30 30 odd kilo Pascal's as the minor principle stress but now the principle stresses magnitude has increased as well as the direction also changed in the direct shear test. So further in continuation of whatever we have been discussing you know we can actually look into you know the another way of plotting the shear stress versus you know the shear strain data as ratio of tau by sigma dash for a different sigma dash when we actually take with increase in sigma dash actually in this direction C actually has got lowest B and A then you know there is you know typical ABC plotted actually shown different sigma dash and same E naught initial void ratio is actually same for all the samples which are actually sheared at the three different normal stresses both maximum stress ratio and ultimate critical void ratio decreases with increasing effective normal stress then the difference between maximum and ultimate stress decreases with increasing effective normal stress. So the value of the pi max for each test can thus be represented by a secant parameter and the value decreasing with increasing effective normal stress until it becomes equal to pi Cv. So because of this particular phenomenon what actually happens is that at low effective stresses you know we actually have you know the difference between you know the critical friction angle as well as the you know pi max that is the peak friction angle but when it actually has got the normal stresses actually increase then the what actually happens is that the suppression of you know the you know the peaks actually appear and then you know predominantly the hardening actually takes place because of that what will actually happen is that you know they merge to the pi Cv axis. So you can see that you know at high normal stresses high effective stresses there is a beyond you know point B you can see that there is a you know merging of these angles takes place. So this particular phenomenon is actually attributed to the decrease in the ultimate void ratio. So the both maximum stress ratio and the ultimate or critical void ratio decrease with increasing effective normal stress and the difference between the maximum and ultimate stress decreases with increasing the effective normal stress and the value of the pi max for each test can thus be represented by a secant parameter and the value decreasing with increasing effective normal stress until it actually becomes equal to pi Cv. So after having discussed the direct shear test it actually has got you know first of all the direct shear test is a simple test and actually allows to determine this transparameters of soil very easily and the failure plane is actually want to be notice that the failure plane is actually predetermined and very difficult to you know control the drainage condition. But however to some extent what it can be done is that when you have got a shear box and if you wanted to get to some extent you know consolidated and drained parameters what it can be done is that the sample can be prepared particularly if you are having some silty sand or silt samples the samples can be prepared either under remolded conditions or undisturbed conditions and they allow to you know saturate under a particular normal stress over a period of time. Then you know the shearing can be done in a very very low strain rate so that the stress you know the there is no generation of excess pore water pressures during the shear. So with that you know with this exercise we can actually get drained parameters up to a maximum extent. Otherwise you know basically the total strength parameters are actually obtained and so in a way the pore water pressure measurement is actually difficult and only the total stresses can be determined in this particular case and shear stress on the failure plane are not uniform as failure occurs progressively from the edge to the center of the specimen. So one thing we need to notice is that the shear stresses on the failure plane they are actually not uniform as the failure actually occurs progressively from the edges to the center of the specimen and area under the shear and vertical loads does not remain constant throughout the test. So one more assumption where you know observation we have to make is that the area under the shear and vertical loads you know it does not remain constant throughout the test and the soil is actually you know forced to shear along the predetermined plane and we should not actually necessary the weakest plane. So if you look into that you know we actually have you know imposing failure along a predetermined plane which may not be the failure plane in the particular sample in such situation is actually you know we are actually regarding as you know the failure to take place along a predetermined failure plane and the another disadvantage is a demerit which actually comes as far as you know direct shear test is concerned is the rotation of the principal planes as we have seen that during shear the principal planes undergoes rotation in the direction of shear and the angle of rotation of this principal planes is approximately you know estimated as the you know the inclination of the failure envelope with the horizontal that is the friction angle and so these are the some you know disadvantages which are actually listed here and the only advantage with the direct shear test is simple and in case of sands is easy for the sample preparation. So another advantage is that you know the interface parameters and this is actually where you know when you have actually got multiple interface layers nowadays you know when we actually design a landfill lining system when we wanted to have you know the sliding friction values or in case of reinforced soil walls when we wanted to have a sliding friction between the reinforcement which is actually being which is interacting with the surrounding soil in order to get these parameters or interface between a geomembrane interface and soil under you know different conditions like you know without water or with you know with the presence of the water you know number of cases actually where you know the interface characteristics can be analyzed and this is actually becoming very handy as far as you know determining you know these the use of the direct shear test is actually concerned. Now as we said that the another way of determining these trans parameters is the triaxial compression test. In fact the triaxial compression test is the first physical model test which actually where in you know where in the in-situ conditions are actually attempted to be simulated. So here these triaxial sample actually simulates let us say that if you are having a soil sample collected from a certain you know certain death let us say at say 5 meters death or 10 meters death and then you know the soil sample will be actually having a vertical stress and the lateral stress which is actually equivalent to let us say an equilibrium conditions it is k0 times sigma v and with that what will happen is that it is subjected to these conditions. Now on these conditions let us say that there is an increase in the you know vertical stress which is actually happening and upon that increase in vertical stress if you wanted to see what is the angle of the resistance actually offered by the soil and that can be actually simulated by using this particular type of test which is actually called as a triaxial compression test. So this actually test actually has got you know a you know one of the widely used test where in you can actually get you know you know have the parameters which actually can have well controlled drainage conditions the soil strength parameters can be estimated very very accurately and mostly waged in for shear strength test is suitable for all types of soils and wherein you can actually do on very soft soil and very not very soft soil and where soil sample can be made to stand and then also loose sand dense sands yes different techniques need to be adopted and needs to be considered and one important point we need to actually highlight is the L by D ratio so in this case the L by D is actually maintained as 2 that means that length of the sample is always the 2 times the diameter and this actually you know one particular you know requirement which actually needs to be satisfied is that you know because of the L by D is equal to 2 what will happen is that the failure plane will remain within the within the sample length only the otherwise you know if you are noticing whatever we are actually applying on the horizontal surface and along the you know along the this vertical plane and these are actually regarded as the minor principle planes and they actually remain as minor principle planes during the test and even the end of the test so that means that you know in the case of traction test whatever the limitation which we actually have said the two major things one is that you know the non-uniformity of the shear along the failure plane and induced failure plane and the rotation of the principle planes those limitations are actually work very easily in a triaxial shear stress and the typical sample diameters for the you know the sample specimen diameters for triaxial stress are the 38 mm 100 mm and 300 mm and 38 mm is generally used for you know the small samples 38 mm diameter and 76 mm height or there is also one more intermediate size which is actually 75 mm diameter and 150 mm height or 100 mm diameter and 200 mm height or 300 mm diameter and 6 mm height in some special cases where we are actually having a large size particles then even you know 1000 mm diameter or 500 mm diameter and 1 meter height or 1000 mm diameter and 2 meter height very large triaxial test cells are also available nowadays so in this particularly what we do is that the sample is actually prepared under either from the disturbed or undisturbed conditions or remolded sample is actually prepared or undisturbed sample is actually placed and the sample sizes are such that you know the from the different sampler tubes we can actually can collect the samplers and then you know preserve the you know original soil fabric and original preserve the the degree of the disturbance and so that institutes shear strength parameters you know directly the samples which are actually corrected from the borehole investigations the same thing can be used here and with that when you can when you will perform and different types of drainage conditions we are able to maintain then there is a possibility that you will be able to determine the strength parameters suiting to the field conditions depending upon the the type of the you know you know the stress test which you are conducting can be adopted so this is the you know typical component of a triaxial test where you actually have got a sample and a chamber in which you know what you have is that you know the confinement is actually provided in the form of filling the water so initially when the sample is actually placed and to as an interface you know there is a membrane which is actually placed called rubber membrane and then you know the load cell which is actually placed here this is the pork loading ram and inside is actually filled with water and the water is actually pressurized to different pressures and you know when we are actually having you know sigma 3 1 sigma 3 2 and sigma 3 3 different cell pressures are actually applied and then for different cell pressures when it is actually applied initially the pressures are actually uniform and then you know to in order to simulate the increase in the load what actually happens is that you apply deviator stress so with sigma 1 is equal to sigma 3 plus you know sigma 3 plus p by a and you know that is actually you know what we get is that sigma 1 is equal to sigma 3 plus p by a and what we regard as sigma 1 minus sigma 3 as the deviator stress the difference between major principle stress sigma 1 minus sigma 3 is actually called as the you know the deviator stress so we get deviator stress versus axial strain and that is in case when you are actually doing a compression test so sometimes you know we can actually have controls on the you know you can actually arrest the movement of water so that means that no volume change is actually allowed then in that case we can simulate undrained conditions and we can also measure that what is the water pressure upon shear when the drainage is actually not limited in a saturated sample and when suppose if you are trying to you know drain the water and subject it to you know shearing condition then you know what will happen is that the you know consolidated drained conditions or drained parameters are obtained so in the next lecture we will try to look into further about the triaxial test and further we will try to look into the other aspects of the interpretation of the triaxial test results.