 Welcome to lecture series on advanced geotechnical engineering being offered by the department of engineering IIT Bombay Professor BVS Vishwanatham. So we are in module 4 and lecture 6 on stress strain relationship and shear strength of soils. So in the previous lecture we introduced about Mohr-Colem criterion and then introduced also the PQ space and in this lecture we will try to discuss more about the Mohr-Colem failure criterion and then interpretation and its correlation with the PQ space and further we will concentrate on definition of failure and interlocking concept and in its interpretations that is definition of failure particularly as far as the friction is concerned with interlocking concept and its interpretation. So as we discussed in the previous lecture we can actually express the principle stress relations shifts at failure as follows, sigma 1 is equal to sigma 3 tan square alpha plus 2 c tan alpha where alpha is equal to 45 plus 5 by 2 and which is also called as the failure plane inclination and in case if you are having sigma 3 is equal to another way as sigma 3 as the major and sigma 1 as the minor where sigma 3 is equal to sigma 1 tan square 45 minus 5 by 2 minus 2 c tan 45 minus 5 by 2. Now the Mohr-Colem idealization of geomaterials if you look into it we have a sample a cylindrical sample subjected to principal stress sigma 1 dash and sigma 2 dash and sigma 3 dash. So here as being sigma 2 dash is equal to sigma 3 dash in Mohr-Colem idealization what has been assumed is that sigma 2 is equal to sigma 3 has been assumed or sigma 2 dash is equal to sigma 3 dash and this is the stress strain variation d-way trick stress versus axial strain where axial strain in the direction of major principal stress. So this is the idealization and the slope of this is the elastic modulus and this is the curve which we may get experimentally and this is idealized in this fashion it is shown. So as we have discussed the Mohr diagram and Colem enolope if you look into this Colem has given during the investigations of retaining walls an equation which is actually called as tau is equal to c plus sigma tan phi and Mohr enolope actually has given as the tau is equal to function of sigma and this nonlinear enolope is actually for Mohr Colem Mohr enolope and this one is for the Colem equation. So this actually has been idealized to a straight line and Colem in his investigations of retaining walls proposed a relationship tau is equal to c plus sigma tan phi where c is the inherent shear strength also known as the cohesion c and phi is the angle of internal friction. So the criterion contains two material constants here c and phi and which are actually called as shear strength parameters and there are different methods to determine these shear strength parameters in the laboratory they are primarily called the direct shear test and triaxial test and the triaxial test actually has got you know the three different four different classes you can say and the first category is that where we do a triaxial test with no cell pressure then you know in that case it is actually called as unconfined compressive strength test is called UCS test and when we do this test with you know a cell pressure that is unconsolidated undrained triaxial test and when we do you know the triaxial test by allowing the consolidation you know during by allowing the drainage during consolidation and no drainage during shear then it is called consolidated undrained triaxial test. When we allow both you know the drainage during consolidation as well as the shear it is called consolidated drained triaxial compression test. So when we do this when we as we have discussed it that when we do a different cell pressures we get the different sets of Mohr circles if you are interpreting in terms of total stresses we will get total stress envelopes if you are actually interpreting in terms of effective stress effective stress parameters then you will get the effective stress envelopes and with that we will able to get the C dash and phi dash. So here in this particular diagram where the Mohr you know envelopes is actually shown along with the Coulomb envelopes and with the slope of inclination is phi and alpha f is that failure plane. So the criterion basically consists of Mohr Coulomb criterion basically consists of two parameters that is C and phi as opposed to one material constant for the Truska criterion as far as the Truska criterion is concerned and there is only one material constant. So yielding and failure takes place in the soil mass when mobilized axial stress at any plane tau m becomes equal to the shear strength tau f which is given by. So yielding and failure takes place in the soil mass when mobilized actual shear stress at any plane tau m becomes equal to when tau m becomes equal to the you know the shear strength tau f which is actually given by tau m is equal to C dash plus sigma dash n tan phi dash is equal to tau f where C dash and phi dash are the strength parameters then we can write this as in terms of function of sigma dash as tau minus sigma n dash tan phi dash minus C dash is equal to zero. So this we have written in terms of function of sigma dash we have written like tau minus sigma n dash sigma dash n tan phi dash minus C dash is equal to zero. So one point we need to note here in case of the Mohr Coulomb idealization we have discussed that because of the you know the non-consultation of sigma 2 dash as can be seen here when we have sigma 1 dash and then sigma 2 dash when with sigma 3 dash with the minor principle stress that this Mohr circle is actually you know is in contact with the failure envelope. So where sigma 1 dash is greater than sigma 2 dash greater than sigma 3 dash and here note that the value of the intermediate principle test does not actually influence the failure. So this particular you know the value of the sigma 2 dash actually does not actually influence the failure. So in the case of Mohr Coulomb idealization the intermediate you know principle stress sigma 2 dash is neglected. So further and to the Mohr Coulomb idealization of geomaterials and in this we consider a tau versus sigma and where you have got a Mohr circle and you have a sample which actually has got you know if the sample is homogeneous we actually have got the conjugate failure planes in both the directions and you can have a failure plane in this direction and you can have a failure plane in this direction and with that you know here this is the failure plane and the parallel to this is the failure plane inclination which is actually shown here. And so we can write from the geometry of the Mohr circle by considering a Mohr circle tangent to the line that is a stress state associated with the failure that is at this point and using the trigonometric relations the alternate form of tau is equal to c plus sigma f tan phi can be obtained. So in terms of principle stresses it can be written as tau f is can be seen as tau f is equal to sigma 1 minus sigma 3 by 2 and sigma f is equal to you know which this point is nothing but sigma 3 plus you know the sigma 1 minus sigma 3 by 2. So if you look into this this will become sigma 1 plus sigma 3 by 2. So this point is sigma 3 0 to it measures an ordinate of sigma 3 dash and this point actually measures you know this distance so that is actually obtained as sigma 1 plus sigma 3 dash by 2. So we can actually write this in terms of principle stresses sigma 1 minus sigma 3 dash by 2 is equal to c dash cos phi dash plus sigma 1 plus sigma 3 dash by 2 sin phi dash. So when we put this in one side and we can actually write f is equal to sigma 1 minus sigma 3 minus sigma 1 plus sigma 3 sin phi dash minus 2 c dash cos phi dash is equal to 0. So that is what actually to be stated here where f is equal to sigma 1 minus sigma 3 dash sigma 1 dash minus sigma 3 dash minus sigma 1 dash plus sigma 1 3 sigma 3 dash sin phi dash minus 2 c dash cos phi dash is equal to 0. So this you know we actually deduced from the trigonometric relation by satisfying the trigonometric relations and from the geometry of the Mohr circle. And this was actually earlier we have actually when we have done with relation between the kf line and Mohr Coulomb failure and all of and this also can be expressed in terms of you know principle stresses and when we plot pq plot with q is equal to sigma 1 minus sigma 3 dash by 2 and p is equal to sigma 1 plus sigma 3 dash by 2 when we plot so when q is equal to sigma 1 minus sigma 3 dash by 2 and p is equal to sigma 1 plus sigma 3 dash by 2 when you plot this is the kf line and which is actually inclined at an angle psi. So when we compare you know typical Mohr circle and this thing when we said that sin phi dash is equal to tan psi and c dash is equal to a by cos phi. So by using this also we can get the other form of tau is equal to c plus sigma tan phi as sigma 1 minus sigma 3 dash by 2 is equal to c cos phi plus sigma 1 plus sigma 3 by 2 sin phi dash. So with this also we will able to get the expression which we are talking here sigma 1 minus sigma 3 dash minus sigma 1 plus sigma 3 sigma 1 dash by sigma 3 dash sin phi dash minus 2 c dash cos phi dash is equal to 0. Now with no order implied by the principle stresses sigma 1 sigma 2 sigma 3 the Mohr Coulomb criterion can be written as in the form of this six equations and this represent to two equations, this represent two equations, this represent two equations wherein this is nothing but plus or minus sigma 1 minus sigma 2 by 2 is equal to a into sigma 1 plus sigma 2 by 2 plus b plus or minus sigma 2 minus sigma 3 by 2 is equal to a into sigma 2 plus sigma 3 by 2 plus b and plus or minus sigma 3 minus sigma 1 by 2 is equal to a into sigma 3 plus sigma 1 by 2 plus b where a is equal to M minus 1 by M plus 1 that is A is equal to M minus 1 by M plus 1 and M is equal to C0 by T0 which is nothing but 1 plus sin phi dash by 1 minus sin phi dash and B is equal to 1 by M plus 1 and C0 is equal to M by M plus 1 where T0 is equal to C0 by 2 into 1 minus sin phi dash where A is actually within 0 to 1 that is 0 less than or equal to A is within 0 less than or equal to A less than 1. So the values of this C0 and T0 are actually defected here, C0 and T0 are defected here and which is shown here the C0 by T0 which is nothing but the ratio of these two C0 by T0 and this is the typical more coulomb failure and all of with a Mohr circle C and Mohr circle D. So we have said that these six equations are there and wherein we actually have got A is equal to M minus 1 by M plus 1, M is equal to C0 by T0 and B is equal to 1 by M plus 1 where C0 is equal to M by M plus 1 and T0 is equal to C0 by 2 into 1 minus sin phi and where A is in between 0 to 1. So this T0 is the theoretical Mohr coulomb uniaxial tensile strength and that is on the negative side of the Mohr circle you can see that in tau sigma plot and that is not observed in experiments rather a much lower strength T is measured and the sigma 1 is equal to 0 and the sigma 2 is equal to that is a sigma 1 is equal to 0 and sigma 2 is equal to minus T that is actually shown here. So with the failure plane being normal to sigma 3 plane, so sigma 1 plane is here and the sigma 3 plane here so the failure plane being normal to the sigma 3 plane. So the C0 is the theoretical Mohr coulomb uniaxial compressive strength. So C0 is the theoretical Mohr coulomb uniaxial compressive strength, C0 is the theoretical Mohr coulomb uniaxial compressive strength. Now the shape of the failure surface in principal stress space is depend upon the form of failure criterion. So the shape of the failure surface in principal stress space is depend upon the form of the failure criterion, linear functions map as plants, planes and non-linear functions as curvilinear surfaces. So the shape of the failure surface in principle stress space is dependent on the form of failure criterion primarily it is linear functions map as planes and non-linear functions as curvilinear surfaces. So the following six equations the equations which we have said you know which we just discuss are represented by six planes that intersect one another along the six edges defining the hexagonal pyramid. So the following six equations which are actually discussed just now are represented by the six planes that intersect one another along the six edges of the defining a hexagonal pyramid. So that is these six edges which are actually defining this hexagonal pyramid these equations actually represent these equations represent the six planes that intersect one another along six edges defining the hexagonal pyramid. So this is the failure surface that means that this is the failure surface and this is the hydrostatic axis, this is the hydrostatic axis where sigma 1 is equal to sigma 2 is equal to sigma 3 and this is the failure surface on the equipressure plane sigma 1 plus sigma 2 plus sigma 3 is equal to constant or pi plane this is also called as the pi plane perpendicular to the hydrostatic axis. So perpendicular to the hydrostatic axis so if you take the projection of this one you will have the this is the pyramid surface which is you know hexagonal pyramid surface and which is also called as the pi plane which is actually perpendicular to the hydrostatic axis and where Mohr Coulomb criterion can be described as irregular hexagon with sides of equal length where Mohr Coulomb criterion can be described as irregular hexagon with sides of equal length. So here the pyramidal surface in principle stress space and cross section equipressure plane is actually given here and these the equations which you know are shown here they actually represent the all the six edges which are forming the hexagonal pyramid surface and the failure surface on the equipressure that is sigma 1 plus sigma 2 plus sigma 3 is equal to constant or a pi plane perpendicular to the hydrostatic axis where the Mohr Coulomb criterion Mohr Coulomb can be described as irregular hexagon with sides of equal length. So this is the irregular hexagon with is actually shown once again here where this is the hydrostatic axis and here is the point where the sigma 2 dash is there. So isotropy requires three fold symmetry because an interchange of sigma 1 sigma 2 sigma 3 should not influence the failure surface for an isotropic material. So if you have a isotropic material the requires three fold symmetry because of an interchange of sigma 1 sigma 2 3 should not influence the failure surface for an isotropic material. Note that the failure surface need only be given in any angle in any of the one of the 60 degrees regions that is one of the 60 degrees region that is what actually shown here. So note that the failure surface need only be given in any of one of the 60 degrees regions and Mohr Coulomb failure surface is a regular hexagon in principle stress space. Now consider the transformation from principle stress space sigma 1 sigma 3 to Mohr diagram sigma tau. So although the radial distance from the hydrostatic axis to the stress point is proportional to the deviatory stress a point in principle stress space does not directly indicate the value of the shear stress on a plane. However each point on the failure surface in principle stress space corresponds to Mohr circle tangent to the failure analog. So though the radial distance from the hydrostatic axis to the stress point is proportional to the deviatory stress a point in the principle stress space does not directly indicate the value of the shear stress on a plane. However each point on the failure surface in principle stress corresponds to Mohr circle tangent to the failure analog. So further extending the discussion for the particular case where sigma 2 is in the intermediate principle stress in the order sigma 1 dash greater than sigma 2 dash greater than sigma 3 dash or sigma 1 dash greater than sigma 1 dash greater than sigma 1 greater than sigma 2 greater than sigma 3. The failure surface is given by the side ACD of the axogonal perimeter. The failure surface is given by the side ACD of the perimeter that is the side ACD of the perimeter here which is actually shown in the surface ACD. D is this point sigma 2 dash. The principle stresses at point D represent the stress state for a traction compression test where sigma 1 and sigma 2 is equal to sigma 3 at point D and at point D is given by circle D in the Mohr circle diagram. The point D that is given by the Mohr circle D which is actually shown here. So this is the Mohr circle which is actually shown here point D. This is Mohr circle D. Similarly for point C with principle stresses sigma 3 and sigma 1 is equal to sigma 2 associated with the triaxial extension test. Mohr circle C depicts the stress state where point D and C can be viewed as the extremes of the intermediate stress variation. So where points D and C can be seen as the extremes of the intermediate stress variation and the normal and shear stresses correspond to failure are given by points Df and Cf. So this indicates the Cf and Df indicates the extremes of the intermediate stress variation. So it can be seen that the Cf and Df will give the failure points and so for this is for the triaxial test and where Mohr circle C depicts the stress state and points D and C can be viewed as the extremes of the intermediate stress variation and the normal and shear stresses correspond to failure are given by points Df and Cf. And points lying on the line CD on the pyramidal failure surface will be represented by circles between C and D. So points lying on the line CD that is this they are represented by the, represented by circles between C and D that is we are actually talking about circles between C and D, between circles between C and D. And as can be seen this hexagonal you know pyramidal surface has the corners that may sometime create problems in computations and however this particular difficulty is quite easily welcome by introducing a local rounding of the corners according to Griffith 1990. So the Mohr Coulomb in space that is Mohr Coulomb in principle stress space which is hexagonal pyramid surface has corners that may be sometimes create problems in computation. So this particular difficulty is quite easily can more come by the introducing local rounding of the corners according to Griffith 1990. So as we have discussed the other way of describing tau is equal to sigma, tau sigma plus tan phi in terms of principle stresses as f is equal to sigma 1 dash minus sigma 3 dash minus sigma 1 dash plus sigma 3 dash sin phi dash minus 2c dash cos phi dash is equal to 0. So this can be obtained further in order to make attempts to derive a consistent conditions of the Mohr Coulomb in space with p dash is equal to you know where specific volume is equal to 1 plus e and p dash is equal to 1 by 3 sigma 1 dash plus sigma 2 dash plus sigma 3 dash and q is equal to square root of sigma 1 dash minus sigma 2 dash whole square plus sigma 2 minus sigma 3 dash whole square plus sigma 1 dash minus sigma 3 dash whole square. So the whole divided by 2 and for triaxial conditions as well we have said trying when we have got a triaxial sample that is a cylindrical sample sigma 2 dash is equal to sigma 3 dash when we have this sigma 2 dash is equal to sigma 3 dash and what we get is that p dash is equal to 1 by 3 sigma 1 dash plus 2 sigma 3 dash and q is equal to sigma 1 dash minus sigma 3 dash. So when we put this into when we put this this we get simplified to q is equal to sigma 1 dash minus sigma 3 dash and p is equal to 1 by 3 sigma 1 dash plus 2 sigma 3. Now by substituting these here we can write sigma 1 dash is equal to sigma 1 dash is equal to 3p dash minus 2 sigma 3 dash. So sigma 1 dash can be written as 3p dash minus 2 sigma 3 dash. So if you put p dash is equal to 1 by 3 into sigma 1 by 3 into sigma 1 dash plus 2 sigma 3 we will get sigma 1 dash and sigma 3 dash is equal to sigma 1 minus q and if you substitute q is equal to sigma 1 dash minus sigma 3 dash we will get sigma 3. So by using these values with sigma 1 dash is equal to 3p dash minus 2 sigma 3 dash and sigma 3 dash is equal to sigma 1 dash minus q and substituting in this particular equation and what we get is that as the sigma 1 dash is equal to 3p dash minus when you substitute for sigma 3 dash here what we get is that sigma 1 dash is equal to 3p dash minus 2 sigma 1 dash plus 2q and where with this we can write sigma 1 dash as 3p dash plus 2q by 3 where p dash is equal to sigma 1 dash plus 2 sigma 3 dash by 3 and q is equal to sigma 1 dash minus sigma 3 dash and similarly sigma 3 dash is equal to sigma 1 minus q and by substituting these values we get 3p dash minus q by 3. So with this what we can write is that you know sigma 1 dash plus sigma 3 dash is equal to when you add these two sigma 1 dash plus sigma 3 dash is equal to 6p dash plus q by 3. So sigma 1 dash plus sigma 3 dash is equal to 6p dash plus q by 3 and which is you know when we add to this what we get is that when you substitute in this equation when you substitute in this equation sigma 1 minus sigma 3 that is nothing but q is equal to sin phi plus 6p dash plus q by 3 plus 2c cos phi. So by simplifying this we get 3q is equal to 6p dash sin phi plus q sin phi plus 6c cos phi. So what we get is that 3q when you do the cross multiplication is equal to 6p dash sin phi plus q sin phi plus 6c cos phi. So the equation actually has got pi dash and here also pi dash and here also pi dash and here also pi dash and pi dash. So q is equal to 6 sin phi dash divided by 3 minus sin phi dash into p dash plus 6c cos phi dash divided by 3 minus sin phi dash. So this is actually indicated as q is equal to n the new p dash plus c star where new is equal to 6 sin phi by 3 minus sin phi dash and c star is equal to 6c cos phi by 3 minus sin phi dash where this is actually called as formulation for the Mohr-Coulomb model in pq space. This is actually called as a formulation for the Mohr-Coulomb model in pq space and assuming that the flow rule and ideal plasticity condition and we can write f is equal to f function of p dash q is equal to q minus a new p dash minus c star is equal to 0. So by taking the differentiation with respect to p dash and dou f by dou p dash dp dash plus dou f by dou q into dq is equal to 0. So this is actually consistency condition which is called as far as the Mohr-Coulomb model in pq space is concerned. So what we have done is that from the alternative way of expressing tau is equal to c plus sigma tan phi which we have taken and from these considerations of by using these invariants p dash is equal to one third of sigma 1 dash plus sigma 2 dash plus sigma 3 dash by 3 and q is equal to root over sigma 1 minus sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 3 dash whole square plus sigma 1 dash minus sigma 3 dash whole square by 2 and then for traction conditions when you put sigma 2 dash is equal to sigma 3 dash we get p dash is equal to one third of sigma 1 plus 2 sigma 3 dash and q is equal to sigma 1 minus sigma 3 dash. So what we have written is that we have written this sigma 1 dash is equal to 3 p dash minus 2 sigma 3 dash. So this is actually expressed in terms of sigma 1 dash is equal to 3 p dash minus 2 sigma 3 dash and this is actually expressed as q is equal to that is sigma 1 dash minus sigma 3 dash. So sigma 3 dash is equal to sigma 1 minus q. So that is what actually we have written. Then what we have done is that sigma 1 dash is equal to 3 p dash minus 2 sigma 1 dash plus 2 q. So we actually substituted for sigma 3 dash sigma 1 minus q and then expressed in terms of p and q that is sigma 1 dash is equal to 3 p dash plus 2 q by 3 and similarly when we have taken sigma 3 dash is equal to 3 p dash minus 2 sigma 3 dash minus q and with that we have got 3 p dash minus q by 3 and then added sigma 1 dash plus sigma 3 dash then we have got sigma 1 plus sigma 3 is equal to 3 p 6 p dash plus q by 3. When you add sigma 1 dash plus sigma 3 dash we have got in terms of p dash and q where p dash is equal to sigma 1 dash plus 2 sigma 3 dash by 3 and q is equal to sigma 1 minus sigma 3 dash we get 6 p dash plus q by 3. Now this particular expression which is for sigma 1 minus sigma 3 this is actually replaced by q and then sigma 1 plus sigma 3 was actually replaced by 6 p dash plus q by 3 and then simplification yielded a consistency and more column model in the p q space which is q is equal to n that new p dash plus c star where mu is equal to this particular coefficient that is 6 sin phi dash divided by 3 minus sin phi dash and c star is equal to 6 c cos phi dash by 3 minus sin phi. So f is equal to the formulation for more column model in p q space can be expressed as q minus new p dash minus c star is equal to 0 and assuming the associated flow rule in and ideal flow ideal plasticity which we have started with by idealizing the geomaterial in case of more column criterion where a function of p dash q is equal to q minus new p dash minus c star is equal to 0 which is dou f by dou p dash dp dash plus dou f by dou q dq is equal to 0 which is actually called as the consistency condition. Now some limitations of more Coulomb theory this we have discussed but we are again elaborating. So one of the prime limitations we actually said is that the more envelope is a curl and the Coulomb equation is a straight line. So within the usual range of the experimental range in the laboratory what has been done is that it is actually assumed as a circular that is within the usual experimental range in the laboratory. So possible over estimation of the safety factor for the slope stability calculations can be interpreted and difficulties in calibration because of the linearization. So we actually have taken linearization so there are some difficulties in the calibration and then second is that the effect of the intermediate principle stress sigma 2 dash on the condition failure though we actually discussed that it is not influencing the failure but it is obvious that sigma 2 can have no influence on the conditions at failure for more failure criterion and no matter what magnitude it has. So though in the Coulomb criterion it is actually stated as a limitation but it is obvious that sigma 2 can have no influence on the conditions at failure and no matter what is the magnitude of the sigma 2 and the intermediate principle stress sigma 2 probably does have an influence on in real soil but the more Coulomb failure theory does not consider it. In real soil it may have influence but the more Coulomb failure theory does not consider it and the more Coulomb failure criterion is well proven for most of the geometricals but that of a place is still contradictory that is one of the limitation and the soils on shearing exhibit variable volume change characteristics depending upon the pre consolidation pressure and which cannot be accounted with more Coulomb theory and in soft soils volumetric plastic strains on shearing are compressive that is negative dilation takes place while the more Coulomb model will predict continuous dilation. So in soft soils or loose sands the volumetric plastic strains on shearing are compressive in nature that is the undergone negative dilation and whilst the more Coulomb model will predict the continuous dilation more Coulomb more Coulomb model predict the continuous dilation. So this is also stated as one of the limitations of the more Coulomb theory. Now the definition of failure when we further you know correct ourselves to friction and interlocking concept in order to correct that let us look into what is the definition of failure and failure along a plane in a material occurs by critical combination of normal and shear stress that is when we have tau is equal to function of sigma and tau is equal to C plus sigma tan phi the shear stress is the function of material cohesion or a soil inherent strength and an angle of internal friction. So this internal friction can be due to the sliding friction but the sliding which actually can happens between two particles or also due to interlocking that is one particle get locked into other particle and then there is or part depends upon the shape of the particle like an angularity and then angularity and size of the particles. So the failure along a plane in the material occurs by critical combination of normal and shear stress where tau is equal to C plus sigma tan phi where shear stress is the function of material cohesion C and angle of internal friction phi. So if you look into the definition of failure when we have got tau sigma annulope and when you have a point A stress point A and this is actually said that it is stable no failure occurs and but when actually no shear failure occurs for A and but when it is an actually on the failure annulope and when it actually on the failure annulope then it is you know the shear failure occurs and this is the tau is equal to function of sigma is actually represented here and when it is here that means that you know the stress state cannot exist above the failure annulope that means that the failure would have already taken place. The failure would have already taken place similarly when you draw the Mohr circle above the failure annulope that implies that the failure would have already taken place. So the point C is cannot exist and at point B which is you know at failure and point A is actually is stable the safe against the failure. Now we can also look into the flow rule for Mohr Coulomb for dilatancy angle and for Mohr Coulomb flow rule is defined through the dilatancy angle of the soil where G of sigma dash is equal to tau minus sigma dash n tan psi dash minus constant is equal to 0. So this is the plastic potential function so this is the you know for the dilatancy angle so G sigma dash is equal to tau minus sigma dash n tan psi dash minus constant is 0 where psi dash is the dilatancy angle and which is actually less than which is less than the angle of internal friction and which is less than the angle of internal friction. Now the interlocking concept and its interpretations when we look into it the frictional soil behavior is mainly influenced by the two factors one is you know frictional resistance or sliding resistance between the you know two particles upon shearing and to expand the soil against the confining pressure. So then the soil which actually can get undergo increase in volume upon the shear under confining pressure so and angular friction can be defined as pi is equal to pi u plus beta where angle of sliding friction between the mineral surfaces and beta is the effect of interlocking where that is also a combination which is can be said as because of the angularity of the particles and all. So where pi is equal to pi u plus beta where pi is the angle of sliding friction between the mineral surfaces and beta is the effect of internal interlocking. So the pi varies with the nature of the packing of the soil so denser the packing higher is the value of pi the denser the packing that is the in a given volume if there are more number of particles are there then it is called as denser packing higher is the value of pi if pi u for a given soil is constant and b must change with the denseness of the soil particle. So if pi u for a given soil is constant let us say for a particular angularity particular shape and if it is constant b must change with the denseness of the soil particle. So b increases with increasing denseness of the soil and because more work to be done to overcome the effect of interlocking. So because the particles are get locked in a dense configuration the soil actually has to work a lot in order to undergo this movement and so beta increases with increasing the denseness of the soil due to shear and because more work to be done to overcome the effect of interlocking which actually arises due to shear. So the effect of the angularity of the soil particles if you look into it soil possessing angular soil particles will show high friction angle effect of angularity of the soil particles if you look into it soil possessing angular soil particles will show high friction angles then that of rounded soil particles because angular soil particles will show a greater degree of interlocking and higher value of beta. So the so called this beta will be high for you know angular soil particles because of the greater degree of interlocking and higher value of beta. Now further the interlocking concept with interpretations with dilation so if you look into that here beta is the function of the you know the dilation of the function of the dilatancy of the soil. So if you take a loose arrangement of the grains if you take a spherical grain and you can see that these are the void spaces. So when they are subjected to a vertical you know in the vertical stress and a shear in the horizontal direction and this is the case the soil actually undergoes you know negative dilation in the sense that the soil undergoes a compression and which actually takes place like this. So this is you know called as a negative dilation so the volume decreases here but here when you have the soil particles which are actually you know dense then you can see that there is you know they increase in the volume there is an increase in the volume takes place. So you know this interlocking concept and interpretations can be interpreted by taking CISA concept or analogy wherein when we take this into consideration then we can see that you know the so called the dilatancy angle and other concepts can be discussed. So the interlocking concept and interpretation with dilation for loose sand the volume decreases with shearing for loose sand the volume decreases with the shearing that is called what we calling a negative dilation. And for dense sand the volume initially you know very marginally decreases and then thereafter it shows actually an increase in the you know the dilation so this is increase in the volume increase in the volume upon shearing. So this shows that this shows in this actually the tendency of this dilatancy decreases with increase in the confining pressure which increase in the confining pressure. So let us consider the interlocking concept and interpretations where dilation and direct shear response as we said that one of the tests if you do by using for determining the shear strength of the soil is the direct shear that is that you will be placing a soil mass in a box and then two boxes and even we allow the box to undergo movement then you can see that you know the shear is actually applied along the pre-determined failure plane which is horizontal. So here this is for the dense sand a typical variation for dense sand where Q by P versus X that deflection so Q is the shear force and P is the normal force and so this is for dense sand and this is for the loose sand. So loose sand actually has undergone you know there is a compression and then a dense sand so here where it can be seen that where it can be seen that the shear which actually undergoes a movement with this we can actually calculate what is the you know the friction angle with reference to you know application of this so called the dilation. So here the total work done is equal to DW which is nothing but P into delta Y that is in the vertical direction P into delta Y in the vertical direction and Q delta X that is in the horizontal direction so delta X is the horizontal movement in the X direction and delta Y is the vertical movement in the vertical direction so work done is net work done is equal to P into delta Y and Q into delta X. So we can write P delta Y plus Q delta X is equal to is counted by mu P delta X that is the friction which actually mobilized along this one. So by dividing delta Y by delta X is equal to mu minus Q by P when you simplify this what we get is that delta Y by delta X is equal to mu Q by P and which is nothing but which is nothing but tan psi and where tan psi is equal to tan phi m is minus tan phi c where phi c is equal to tan inverse mu. So this alternatively we can say that phi m is equal to phi c plus psi. So what we have done is that in the with reference to interlocking concept in interpretation the total work done is equal to DW is equal to P delta Y plus Q delta X and then we divided right through by delta X and with that we have got delta Y by delta X plus Q is equal to mu P and then when you simplify further and what we have got is that delta Y by delta X is equal to mu that is due to sliding friction by minus Q by P and is equal to tan inverse psi. Tan inverse psi is nothing but the so called inclination of this one that is called delta Y by delta X and which is nothing but tan psi is equal to tan psi is equal to tan phi m that is the due to friction minus tan phi c that is because of the interlocking and so with that phi c is equal to tan inverse mu and which we have written here and alternately we can write phi m is equal to phi c plus psi. So phi m is equal to phi c psi that is what actually we have say from the deliberation discussed just now. So how to understand this so called dilatancy so why do we get volume changes when apply the shear stresses and this can be explained with a simple analogy which is saw blade analogy can see that the orientation of these planes they are upward and this is the shear direction and this is the normal stress and similarly what will happen is that when the particles are under dense configuration when it is subjected to shear the particles will raid and another particle and be subjected to a similar analogical movement. So the analogy was actually we have been brought between the interlocking saw blades and dense packing of the grains so the simple analogy we can actually relate correlate is that interlocking saw blades with the dense packing of the grains when they are actually subjected to shear you know the inclination of that interlocking plane blades is the dilation angle and this angle and that angle is actually represented here this is psi and this is psi. So it is phi is equal to psi plus phi m that is the due to friction the apparent externally mobilized angle of friction on horizontal planes phi is larger than the angle of friction resisted resisting the sliding on the inclined planes so the strength is equal to friction plus dilatancy so it is actually nothing but the actual strength of the soil is nothing but the friction plus dilatancy the friction plus so the angle of internal friction is not only that you know it actually has got only friction but also the component of the dilatancy. Now this further you know can be the saw blade analogy can be extended to initially dense and a critical and initially loose conditions so you can see that in case of you know in case of initially dense the particles actually have no way to move except they have to raid on one another so the raiding of the particles takes place on the one particle raid on another particle and this particle raid on another particle so the orientation of the sea blade can be seen on the upward direction in case of critical condition where you know you can see that is horizontal but in case of say initially loose soil you can see that this is you know horizontal this is inclined down that means that it actually undergoes the compression. So when the soil is initially denser than the critical state soil is initially denser than the critical state it must achieve then as particles slide past each other going to the imposed shear strain they will average separate so when a soil is initially denser than the critical state it must achieve you know a denser configuration then as the particles slide each other going to the imposed shear strain then they actually have got a tendency to separate the particle movements will spread about an angle and that is actually called as dilation so see the orientation of the sub blade here so this is analogy and this is actually what happens in the soil in the real soil and when you have you know the orientation which is actually you know the initially loose soil when the soil is initially looser than the final critical state then the particles will tend to get closer together as the soil is disturbed and the average angle of dilation will be negative indicating a contraction. So that is why the sub blade you know inclination or plane of inclination is actually shown below so this is initially loose soil tend to become you know denser here and when we have you know have a place where it is constant you know in the sense that critical state if the density of the soil does not in the critical state is the where no volume change occurs so if the density of the soil does not have to change in order to reach a critical state then there is a zero dilatancy as the soil shear at constant volume. So the soil shear takes place at a constant volume so it is important to realize that the critical state is only reached when the particles they have the full opportunity to juggle around the and then come into the new configuration so if the confining stress is increased where the particles are being moved around then they will tend to finish up in a more compact state. So we have seen for initially dense state initially loose state and also in a critical state and in the critical state what we are saying is that if the density of the soil does not have to change in order to reach a critical state then there is a zero dilatancy of the soil shear at constant volume and it is important to realize that a critical state is only reached when the particles have had full opportunity to juggle around and coming to new configurations so if the confining stress is increased when the particles are being moved around then they will tend to finish up in a more compacted state. So in this particular lecture we try to understand about the Mohr Coulomb in the pyramidal hexagonal surface and also we discussed the consistency condition as far as the Mohr Coulomb condition is concerned and thereafter we actually introduced also to interlocking concept and then we also said that introduced to that when the soils particularly in the dense soil or in over consolid soil and when it undergoes shearing and we said that it is going to experience a phenomenon and which is called dilation and that angle is called dilation angle and that is actually the phenomenon is called dilatancy and the angle which is called the psi the dilatancy angle and this dilatancy angle decreases the suppressment of the dilatancy or dilation phenomenon can be observed once there is an increase in the normal stress so we will further discuss in the forthcoming lecture after having introduced about the different methods for determining the shear strength and further we will connect with these two spots.