 Welcome to lecture series on advanced geotechnical engineering and we are in module 4 stress strain relationship and shear strength of soils and we have discussed in length about the following topics like you know introduced ourselves with the stress state and Mohr circle analysis and then we have defined pole and then we introduce ourselves to principle stress space and the stress paths in pq space then we have discussed about Mohr column failure criteria and its limitations and then different stress strain behaviors and then you know under isotropic compression cases and definitions of the failure and interlocking concept and then we introduced ourselves to triaxial behavior and stress state and particularly in reference with you know unconfined compression test and consolidated and drain triaxial test consolidated, undrained triaxial test and consolidated drain triaxial test and other special test. We are not much covered on the special test and then we also discussed about the drainage conditions in this particular lecture we will be you know trying to concentrate on octahedral plane and octahedral you know stresses and interpretation of the elastic modulus from the triaxial test. So in the lecture in this lecture we are going to discuss about the octahedral plane and elastic modulus from triaxial test. We all know that you know the failure criteria actually which are used in soil mechanics were actually reduced from advanced materials mechanics of materials especially the comprehensive failure conditions are yield criteria are first developed for metals, rocks and concrete. So this comprehensive failure conditions are yield criteria basically they are developed for you know metals, rocks and concrete. Now let us consider the application of this yield criteria to soil and determine the yield surfaces on the principal stress space. One message 1913 proposed a simple yield function and which is given as F is equal to sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 3 dash whole square plus sigma 3 dash minus sigma 1 dash whole square minus 2 y square is equal to 0. So if you if we name this equation, remember this equation as 1 and you know this was actually proposed by one message in 1913 and you know proposed basically a simple yield function and that a yield function is F is given by sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 3 dash whole square plus sigma 3 dash minus sigma 1 dash whole square minus 2 y square is equal to 0 where y is nothing but the yield stress obtained in axial tension. However the octahedral shear stress can be given by the relationship which is actually given below which is tau octahedral is equal to 1 by 3 square root of sigma 1 minus sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 3 dash whole square plus sigma 3 dash minus sigma 1 dash whole square. So tau octahedral is equal to 1 by 3 square root of sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 2 dash whole square plus sigma 3 dash minus sigma 1 dash whole square. Now what we do is that if we square this one then you know we get tau octahedral square is equal to 1 by 3 square is equal to you know sigma 1 dash minus sigma 2 dash whole square plus sigma 2 dash minus sigma 3 dash whole square plus sigma 3 dash minus moment dash whole square. Now if you substitute this in equation 1 what we get is that 3 square into tau octahedral square is equal to 2 y square that means that this term will come outside this term will come out outside and this term will become octahedral square into 3 square then with this what will happen is that 3 square into tau octahedral square is equal to 2 y square and the tau octahedral can be given by square root of 2 by 3 into yield stress that is tau octahedral can be given by square root of 2 divided by 3 into yield stress. This means that what is the physical significance of this is that the failure will take place when the octahedral shear stress reaches a constant value which is equivalent to root over 2 by 3 into y so where y is the yield stress in tension. So what we have tried to do is that if you know when we equate when we substituted you know the tau octahedral in the yield function which was given by von Moshe's what we have got is that tau octahedral in terms of root 2 by 3 into y and this indicates that the failure will take place when the octahedral shear stress reaches a constant value which is equivalent to root over root 2 by 3 into y. Now what we do is let us plot this on the octahedral plane where octahedral plane is the plane on which the sigma 1 plus sigma 2 plus sigma 3 is equal to constant. So let us plot this on the octahedral plane where sigma 1 plus sigma 2 plus sigma 3 is equal to constant. So this is the octahedral plane which is actually shown you know the octahedral plane is obtained by passing a plane through the unit points on the principal axis. So in this particular slide what we see is that sigma 1 and sigma 2 and sigma 3 and this is the hydrostatic axis where sigma 1 is equal to sigma 2 is equal to sigma 3 so this is the hydrostatic axis and the line joining this points which are actually this particular plane is called as octahedral plane. So the octahedral plane is obtained by passing a plane through the unit points on the principal axis and the principal axis are nothing but sigma 1 sigma 2 sigma 3 and hydrostatic axis is nothing but sigma 1 is equal to sigma 2 is equal to sigma 3 which actually passes from the origin to the from the center point which actually erupted out which is shown in the octahedral plane here. So octahedral plane is very near to the soil failure state so very useful basically to derive failure theories of soil. So why octahedral plane has been adopted is that octahedral plane was found to be very you know very near to the soil failure states and so you know this is adopted to derive this failure theories of soil. So we have discussed that you know these theories actually were deduced for metals initially and then you know these are actually extended for the soils. So the octahedral plane is very near to the soil failure state and so you know the very useful to derive the failure theories of the soil. So octahedral plane has been adopted because it represents the you know close to the failure state in the soil and so this is actually used for failure theories derive the failure theories in the soil. Now here the yield surface in three dimensions is shown here. So here with sigma 1 sigma 2 sigma 3 and for example here the diagram shows the more coulomb failure surface as an you know hexagonal shape here and this is the bounded failure surface and here all failure stresses are assumed to be in the bounding surface. So all failure surface failure stresses are assumed to be on the, they are assumed to be on this boundary surface. So this is the principle stress shape showing the principle stresses at time of failure or yielding. So this is the principle stress space showing the principle stresses at the time of failure. So like this you know octahedral plane is normal to a space diagonal in principle stress space and there are eight such planes. So this octahedral plane is normal to the space diagonal. So you can see that when we have the sigma 1 versus root 2 sigma 3 this is nothing but deduced from the radulic plots and this is compression envelope failure envelope in compression and this is the failure envelope in extension envelope and this is the space diagonal through which sigma 1 is equal to sigma 2 is equal to sigma 3 and this plane which is right section perpendicular to this on this level the octahedral plane is represented and the normal to each of the octahedral plane has the direction of cos inverse root over 1 by root 3. So the normal to the that is the space diagonal is inclined to the each of the octahedral plane has the direction which is actually equal into cos inverse 1 by root 3. So what we have done is that in this we have actually presented the principle stress space showing the principle stresses at the time of failure or yielding. So on these boundary surfaces all failure stresses are assumed to be lie on this surface and this is the Mohr-Colomb failure surface was shown as actually as an example and further the any state of stress consisting of three principle stresses like sigma 1 sigma 2 sigma 3 may be resolved into two component states of stress and these are called octahedral stresses one is octahedral normal stress octahedral shear stress a component consisting of equal tensile stresses acting on in all directions and a component state of stress consisting of eight octahedral shear stresses. So we have this octahedral sigma octahedral this is the normal stress and this is the shear stress sigma octahedral and sigma this tau octahedral. So any state of stress can be represented by two component states of stress one is the component consisting of equal tensile stresses tensile or compressive stresses acting in all directions and the other one is that component of state of stress consist of consisting of eight octahedral shearing stresses. So the normal and shearing stresses on the octahedral plane are called as octahedral stresses the normal and shearing stresses on octahedral plane they are actually referred as octahedral stresses and the first invariant is indicated as j1 is equal to sigma 2 plus sigma 3 and sigma octahedral is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 is equal to j1 by 3 and tau octahedral is equal to 1 by 3 root square root of sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square plus sigma 3 minus sigma 1 whole square. So sigma dash octahedral is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 minus u and tau dash octahedral is equal to tau octahedral. So total and effective octahedral shear stresses will be equal like we have got q dash is equal to q. Similarly here tau dash octahedral is equal to tau octahedral and sigma dash octahedral is equal to sigma 1 plus sigma t plus sigma 3 by minus u that is sigma dash octahedral is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 minus u. Now using from the Mohr circle if you look into the earlier interpretations we can write radius is equal to sigma 1 minus sigma 3 by 2 is equal to c plus cos phi plus sigma 1 plus sigma 3 by 2 cos phi repeat r is equal to sigma 1 minus sigma 3 by 2 is equal to c cos phi plus sigma 1 plus sigma 3 by 2 sin phi. So by using this sigma 1 minus sigma 3 is equal to 2 c cos phi plus sigma 1 plus sigma 3 sin phi. Now this can be expressed in terms of more generalized condition more column condition of failure in more generalized form can be expressed as sigma 1 dash minus sigma 2 dash whole square minus 2 c cos phi plus sigma 1 dash plus sigma 2 dash sin phi whole square into sigma 2 minus sigma 3 whole square minus 2 c cos phi plus sigma 2 dash plus sigma 3 dash sin phi whole square into sigma 3 minus sigma 1 whole square minus 2 c cos phi plus sigma 3 dash plus sigma 1 dash sin phi whole square. Now you know this particular you know is represented as you know more column yield surface the point of intersection of the octahedral plane and hydrostatic axis is actually is indicated by A and B C D E F G is the more column yield surface this is the more column yield surface and this is the octahedral plane. So what we are seeing is you know when you see this in plan in the when we have the you know octahedral plane and this is the more column failure surface. So this when it is you know superimposed here more column failure surface is superimposed here we see like this where B C D E F G are the you know vertex at the point which are shown and A is the point of intersection of the octahedral and hydrostatic axis. So this is actually called as more column yield surface and this is octahedral plane. Now the failure surface defined by equation 2 that is basically this one is a pyramid basically is a pyramid with space diagonal which is sigma 1 is equal to sigma 2 is equal to sigma 3 as axis that is the isotropic line or hydrostatic axis which is actually called as sigma 1 is equal to sigma 2 is equal to sigma 3 as axis and a cross section which is an irregular hexagon with non-parallel sides of equal length. So this is actually the cross section is a basically the equation which is the more general form of you know more column condition which this equation 2 which is you know represents a pyramid with space diagonal with sigma 1 is equal to sigma 2 is equal to sigma 3 as axis and a cross section which is an irregular hexagon with non-parallel sides having equal length. So this is the you know this is the pi plane what is called this is pi plane and this is the sigma 3 is equal to constant plane and the stress points stress parts in the conventional triaxial tests are represented here stress parts in conventional triaxial tests are represented here and this is the failure locus this is the failure locus and this is the pi plane and this is the more column failure surface which is actually shown here. So this is basically the more general form of more column condition more column failure condition represents a pyramid with space diagonal sigma 1 is equal to sigma 2 is equal to sigma 3 as axis and a cross section which is an irregular hexagon with non-parallel sides of equal length. The projection of this irregular hexagon on the plane sigma 1 plus sigma 2 plus sigma 3 is equal to constant that is the plane right angles to the space diagonal or on the an act or an actahedral plane. The projection of this irregular hexagon on the plane sigma 1 plus sigma 2 plus sigma 3 is equal to constant that is on the plane at right angles to the space diagonal or an actahedral plane. When yield surface defined by equation 2 is plotted on the octahedral plane it will appear as an irregular hexagon in section with non parallel sides of equivalent that we have discussed. But point A is the point of intersection of the hydrostatic axis with the octahedral plane where point A is the point of intersection of the hydrostatic axis with the octahedral plane thus the yield surface will be hexagonal cylinder coaxial with the isotropic stress line. So because of this the yield surface will be hexagonal cylinder coaxial with the so it is like an hexagonal cylinder coaxial with the isotropic stress line. So the isotropic stress line passes through the center of the hexagon where you know that is it is coaxial with the isotropic stress. See octahedral plane which is the one which is actually according to one minus yield function if you look into it and this yield function when we define this one when this simple yield function which can be expressed as f is equal to sigma 1 dash minus sigma 2 whole square plus sigma 3 minus sigma 2 minus sigma 3 sigma 2 dash minus sigma 3 dash whole square plus sigma 3 dash minus sigma 1 dash whole square is minus 2 y square is equal to 0 as we have done here. So this circle which is actually indicates the one versus yield surface and the radius is nothing but the tau octahedral that is nothing but root over root 2 by 3 into y and y is the yield stress. So the one versus yield stress is something like a cylinder circular cylinder having you know diameter which is equivalent to 2 root 2 by 3 into sigma y or yield stress. So failure takes place when maximum shear stress on octahedral plane is equal to when maximum stress on the maximum shear stress on octahedral plane is equal to you know this root over root 2 by 3 into y. So the distance AB that is nothing but the distance AB that is radial distance AB. So according to now one versus failure surface is actually represented on the octahedral plane and from the earlier discussion whatever we have we actually have determined that tau octahedral is equal to root 2 by 3 y this is nothing but this radius of this one versus yield surface and this is represented further in depth here as this is the octahedral plane and this is the one versus yield surface and AB is nothing but root over root 2 by 3 into y and this is the hydrostatic axis sigma 1 is equal to sigma 2 is equal to sigma 3 and so this one versus failure surface is you know you can see like a circular cylinder having with coaxial with you know hydrostatic axis this cylinder is actually the diameter is the radius is equivalent to AB here where AB is equal to tau octahedral is equal to root over root 2 by 3 into y and note that the locus is unaffected by the value of the sigma octahedral. So that means that various values of sigma octahedral will generate circular cylinders coaxial with the hydrostatic axis which is a yield surface. So we can see that the locus is not getting affected by the values of sigma octahedral that means that the values of sigma octahedral will generate a circular cylinders coaxial with the hydrostatic axis which is a yield surface. So various values of sigma octahedral will generate a circular you know cylinder coaxial with the hydrostatic axis. The discussion which we continue further yield surface is a circle one versus yield surface is a circle and radius is equivalent to tau octahedral which is nothing but root 2 by 3 into sigma y and distance OA is the octahedral normal stress and locus is actually not affected by the values of sigma dash octahedral then you have you know the different values you actually have different cylinders coaxial with the hydrostatic axis. So various values of sigma dash octahedral will generate a circular yield surface which is coaxial with hydrostatic axis and sigma dash octahedral is equal to sigma 1 dash plus sigma 1 dash plus sigma 2 dash plus sigma 3 dash by 3 and Ob is equal to Ob is equal to sigma dash octahedral square plus tau dash octahedral square whole square. So the Ob is nothing but which is a distance which is shown here which is nothing but sigma dash octahedral square that is OA square plus AB square AB square is nothing but this radius which is nothing but root 2 by 3 into y. So this is Ob is given by this root over sigma dash octahedral square plus tau dash octahedral square. Now the octahedral plane is also given as represented by Tresca on the Tresca yield function. So which is nothing but the Tresca criterion or Tresca function is defined as sigma max minus sigma min is equal to 2k, sigma max minus sigma minimum is equal to 2k where factor k is defined by the case of a simple tension by Mohr's circle actually shown in the slide. So this indicates that failure takes place when the difference that is maximum shear stress reaches a constant critical value. When sigma x, sigma max and sigma minimum that is the maximum shear stress reaches a constant value which is nothing but sigma 1 minus sigma x minus sigma minimum that is nothing but sigma max minus sigma minimum by 2 reaches a constant critical value. So this constant critical value that factor k is defined by the case of a simple tension which is shown here for a simple tension for unconfined tension which is actually shown here where sigma minimum that is k where k is equal to this factor. So if the yield function is plotted on the octahedral plane sigma 1 plus sigma 2 plus sigma 3 is equal to constant the locus will be a regular hexagon, the locus will be a regular hexagon. This equation actually represents the regular hexagon equation. So the Tusca yield surface on the octahedral plane is actually represented where B, C, D, E, F, G is shown here which is very similar to Mohr Coulomb failure surface but A is the point through which the hydrostatic axis or where sigma 1 plus sigma 1 is equal to sigma 2 is equal to sigma 3 is ensured and where B, C, D, E, F and G so A represents the octahedral normal stress, the point passing through the octahedral normal stress through A the octahedral normal stress passes and B represents the failure condition in compression, B represents the failure condition in compression where sigma 1 value is greater than sigma 2 that is sigma 1 that is axial stress is more than sigma 2 is equal to sigma 3. So this is point B represents the failure condition in the compression and similarly point E represents failure conditions in extension where sigma 2 is equal to sigma 3 and sigma 3 greater than sigma 1, sigma 3 is greater than sigma 1 and point D represents failure condition for failure condition for sigma 3 greater than sigma 1 is equal to sigma 2 and point G represents failure condition where sigma 1 is equal to sigma 2 and sigma 2 is greater than sigma 3 and point F represents failure condition for sigma 2 greater than sigma 3 is equal to sigma 1 and point C represents failure condition for sigma 3 is equal to sigma 1 and where sigma 1 is actually greater than sigma 2. So we have on this Truska yield criterion which is also represented as a hexagon on the octahedral plane and since locus is unaffected by the sigma dash octahedral unit surface there will be for hexagonal cylinder and here also we have for different values of sigma octahedral and different coaxial to the hydrostatic axis we have got number of hexagonal cylinders are possible. So the yield surface and Truska criteria is actually shown here in this zone where sigma 1 – sigma 2 is equal to 2k where sigma 1 dash is greater than sigma 3 dash greater than or equal to sigma 2 dash in this zone sigma 3 – sigma 2 is equal to 2k where sigma 3 dash greater than or equal to sigma 1 dash greater than or equal to sigma 2 dash in this zone sigma 3 dash – sigma 1 is equal to 2k and sigma 3 dash is greater than or equal to or sigma 2 dash greater than or equal to sigma 1 dash and similarly here in this zone that is the zone between in this point and then this sigma 2 dash axis sigma 2 dash is equal to sigma 2 dash minus sigma 1 dash is equal to 2K where sigma 2 dash greater than or equal to sigma 3 dash greater than or equal to sigma 1 dash and similarly what we have is that here and here where sigma 1 dash minus sigma 3 dash is equal to 2K where sigma 1 dash greater than or equal to sigma 2 dash greater than or equal to sigma 3 dash. The Mohr-Coulomb-Van-Mises and Tresca criteria are actually seen to coincide for the compressive test. So the different failure surfaces are actually shown here and basically here we have the octahedral plane and the Tresca surface is actually shown here and Mohr-Coulomb failure surface is actually shown here and the round surface that is actually this is the Tresca and this is the Van-Mises failure surface which is actually a circular cylinder this is the circular cylinder and so if you look into this Mohr-Coulomb-Van-Mises Mohr-Coulomb failure surface yield surface and Van-Mises yield surface and Tresca yield surface from the Tresca criteria they are seen to coincide for the compressive test. However the strength in the tensile test is reported to be less for the Mohr-Coulomb failure theory. So the strength in the tensile test is reported to be less for the Mohr-Coulomb failure theory. Here in this particular slide the Scott 1963 plotted various Mohr-Coulomb envelopes plot in octahedral plane for values of pi values of 30 to 40 and 50. So here the Scott 1963 after Young and Workentyn 1975 you know the various values of Mohr-Coulomb envelopes have been plotted for different values of friction angles for 30, 40 and 50 and these were the based on the Tesson sand the results of the Tesson sand for varying normal varying stress conditions obtained by Kirkpatrick and Kesselmann are also plotted here. Kirkpatrick and Kesselmann are also plotted here and these points are actually were reported by Kirkpatrick and Kesselmann and as can be noted here the Mohr-Coulomb failure envelopes plotted by Scott and the measured values which are actually plotted by the are obtained by Kirkpatrick and Kesselmann are found to be in good agreement for the values which are actually shown like pi is equal to 37 and 39 degrees and this is for the pi is equal to 37 degrees and then this is for the pi is equal to 39 degrees. So after having discussed about the octahedral plane then we have actually have discussed about you know the result of the you know these values of the triaxial test or direct shear test. So in case of direct shear test we actually get shear stress versus shear strain variation for different normal stresses so there also we can actually get the initial modulus and tangent modulus then there can be a possibility that we can also get a secant modulus interpretation we can actually obtain from the test data. So in the initial modulus which is actually drawn for the initial portion of the curve where the soil stiffness is high and then the tangent modulus is actually drawn which is actually portion where which is actually shown here which actually represents you know this particular value here ET in the horizontal non-vetical. So at certain stress level we actually draw and then make this the initial tangent the tangent modulus the initial tangent modulus is nothing but for the initial portion where the when the tangent is actually drawn that is actually shown as initial tangent modulus and let us say that if you are actually drawing let us say at a point here and then a line which is actually joining this point and this and that it actually let us say that we are having a strain value of 2% at 2% strain or 50% of axial strain then we can actually get the E50 with the value the slope of that line joining you know the strain of at which is meeting at that particular deviated stress or particular normal stress for the shear stress we can actually get the you know the Young's modulus values and so with this you know you know in this particular slide which is actually shown the initial tangent modulus and tangent modulus at certain stress level computation and in addition to that there is also a secant modulus which can actually can be interpreted in case of triaxial test when we actually have got unconfined compression test or unconsolidated undrained test we actually get for you know the based on the in case of unconfined compression test with the sigma 3 is equal to 0 we get sigma 1 versus epsilon so from there we can actually interpret to some extent what is the initial tangent modulus and secant modulus you know up to at certain stress level. So the secant modulus which is actually defined as that modulus which is actually referred up to that particular stress level the slope of that line is actually valid. So it is very important for determining this you know this initial the soil stiffnesses correctly. Suppose if you are actually trying to determine these you know the soil stresses the stiffnesses in the initial portion then into the prevalent stresses in like as we have discussed in one of the modules the physical model test when the soil stresses are very low and then if you are actually having the you know we are dealing with the higher stresses higher stiffness values then resulting you know strains or the stiffnesses will be very very low. But in reality when we actually subjected to the real stress conditions the soil stiffness is low so the soil settlements will be very very high. So in this particular slide we are actually trying to discuss about the interpretation of the initial modulus and tangent modulus from the triaxial test data a typical triaxial test data for a given sigma 3 is actually shown here where sigma sigma minus sigma 3 is the dv stress and epsilon 1 is the axial strain. Then we also have some empirical equations where EI initial modulus which is also given as kPa into sigma 3 dash by Pa to the raise n where sigma 3 dash is equal to minor effective minor effective principal stress and Pa is the atmospheric pressure and k modulus number and the n is equal to exponent determining the rate of variation of you know EI with sigma 3 dash. So n basically indicates that the exponent which actually determines the rate of variation of EI with sigma 3 dash and then EI is equal to as we have given the values of k and n for a particular soil can be found can be found by number of triaxial testings and plotting EI versus sigma 3 on the logarithmic scale and the ranges k is equal to 300 to 2000 and n is equal to 0.3 to 0.6. So the value of the k ranges from 300 to 2000 and n is equal to 0.3 to 0.6. And according to Duncan and Chang 1970 where ET that is nothing but dou of sigma 1 minus sigma 3 by dou E. So Duncan actually has shown that the E value that is ET is equal to 1 minus RF into 1 minus sin phi by sigma 1 minus sigma 3 dash divided by 2 C cos phi plus 2 sigma 3 dash sin phi whole square into kPa into sigma 3 by Pa to the raise n. So if you look into this the Duncan and Chang actually modified the John Boo 1963 empirical equation wherein to this empirical equation the Duncan and Chang actually have added this particular term which is 1 minus RF into 1 minus sin phi by sigma 1 minus sigma 3 dash whole square 1 minus RF into 1 minus sin phi by sigma 1 minus sigma 3 dash by 2 C cos phi plus 2 sigma 3 sin phi whole square. When where RF is nothing but the failure ratio then generally the ratio is equal to 0.7521. So the Duncan and Chang is actually nothing but the modification of John Boo 1963 empirical equation. So further the Poisson's ratio can be actually obtained by mu is equal to epsilon axial that is minus epsilon v divided by 2 epsilon a and where epsilon delta epsilon axial is nothing but increase in the axial strain and delta epsilon n is nothing but increase in the volumetric strain which is nothing but delta epsilon a plus 2 delta epsilon r for the axisymmetric triaxial test and delta epsilon r is nothing but the lateral strain. So with this when you substitute this we get mu is equal to delta mu r this symbol which is actually shown here is equal to delta epsilon a minus delta epsilon a plus 2 epsilon r by divided by 2 delta epsilon a which is nothing but minus delta epsilon r divided by delta epsilon a. So this can be also determined by measuring or by pasting strain gauges when you are actually having unconfined compression test and with that we will be able to get the Poisson's ratio of a soil particularly with the ratio of epsilon r to epsilon a and so the mu is equal to delta epsilon a minus delta epsilon v divided by 2 delta epsilon a where the delta epsilon v is equal to delta epsilon a plus 2 delta epsilon r by substituting this what we have got is that you know this one minus delta epsilon r by delta epsilon a and there are the typical values of Young's modulus for granular material particularly here we have you know the unified soil classification system according to that if you have a granular materials which the Young's modulus values which are actually shown for mega Pascal's the values are shown in all these values are actually reported in mega Pascal's where GW and SW that is well graded gravel and well graded sand gravels and sand well graded when the loose state they actually have got 30 to 80 mega Pascal's and medium and dense state in dense state you can see that the EE values on the higher side that is 162, 320 mega Pascal's when you have you know the sand which is uniform that is called poorly graded sand where you actually have got all uniform size particles then the loose state we actually have got only 10 to 30 mega Pascal's and in the dense state we actually have got 50 to 80 mega Pascal's similarly we have you know silty soil gravel and silty gravel when we are actually have got GM and SM type of soils we can see that the Young's modulus values typically range from 7 to 12 mega Pascal's and the dense state or you know which is actually represented as 20 to 30 mega Pascal's so you can see that depending upon the groups the types in even in the case of granular materials you know the ranges of the you know in the different stress states the loose medium and dense configuration so is a function of density or the packing of the particles and with that we can also you can see that you know the how the values particularly the stiffness soil stiffness or Young's modulus value changes with the soil type particularly we have got well graded gravel or poorly graded sand or sandy sand and gravel in silty nature or sand gravel in silty nature. Now let us consider for the cohesion materials particularly vine grained soils where you have got the sills with low slight plasticity so we have got low plastic sills that is ML type of soils. In this the consistency is actually represented as very soft to medium and stiff to very stiff to very hard so in this case the E values range from 2.5 to 8 mega Pascal's to 40 to 80 mega Pascal's and similarly we have the sills with low plasticity they vary from 1.5 to 6 mega Pascal's to 30 to 60 mega Pascal's and CL that is clay with low medium clay with low medium plasticity they can actually have in a very very soft state the E value can be as low as 0.5 mega Pascal's and in hard state the CL type of soil can actually have 30 to 70 mega Pascal's and the CH the clay with high plasticity CH type of soils can have in very soft stage very low you know the Young's modulus values and the positions ratios for this type of soils under saturated conditions definitely can range from 0.45 to 0.5 and the clays with high plasticity CH will have the very soft state which actually has got 0.35 to 4 mega Pascal's to the hard state it can actually have as high as 20 to 32 mega Pascal's and organic sills O L which is actually in medium consistency have 0.5 to 5 very low organic clays also have actually got very low you know the very low E values even under the medium consistency. So the in this these two slides what actually have seen the distinct difference actually that what we have for the different soils where you have got you know the values which are actually for gravel soils and very high values are shown for depending upon the dense condition or loose conditions where in case of fine grained soils or cozier soils where actually have got low values when they are in the very soft to soft state and the values are on the higher state for higher order for the same soils particularly in the hard state. So in this particular lecture we try to understand about you know the octahedral plain and octahedral shear stresses and based on that three failure criteria namely one misses and Tusca and more coulomb failure surface. So the more coulomb failure surface we have seen as a hexagon and then on the you know this because it actually has got the capability of having different for different via angles we actually have got the different hexagonal cylinders are possible but in case of the Tusca it is also indicated as you know the hexagonal cylinder. So but we have seen for as far as the compressive in the soil in compression is concerned that both you know all the three Tusca and one misses and you know and more coulomb criterion were found to coincide but as far as intention is concerned the value which is actually predicted from the more coulomb criteria was found to be on the lower side. So in this particular module we have try to understand about the stress strain relationships for the soil and then we try to discuss about you know different stress parts particularly we have discussed about MIT based stress parts and as well as the Cambridge based stress parts and we have referenced these stress parts with reference to unconfined compression test and unconsolidated undrained triaxial test and consolidated undrained and consolidated drain triaxial test and then in the case of consolidated undrained triaxial test during this year we do not allow the pore water pressure to dissipate. So because of that there can be possibility that you will be able to measure the pore water pressure. So in that case when we have you know normally consolidated soil or a loose sand then there is a possibility that the entire pore water pressure is actually positive and the sample undergoes you know the volumetric compression when we have got very dense sand or a very stiff clay or highly over consolidated clay then there is a possibility that initially it undergoes compression and thereafter with increase in the axial strain there is a possibility that the soil undergoes a desiccation wherein you know the riding of the soil particles on each other actually happens and because of this increase in volume upon the strain there is you know a phenomenon which is actually called the negative in a dilation phenomenon which actually results in the negative pore water pressures. And the relevant stress pass actually were discussed then thereafter we connected ourselves to the you know the failure criteria particularly with the principal stress space with more coulomb failure, failure criterion and Truska and von Mises we have discussed and then finally in this lecture we discussed about how to interpret you know elastic modulus from the triaxial test data.