 Welcome to lecture series on advanced geotechnical engineering being produced by IIT Bombay. So this is module 4 lecture 5 on stress strain relationship and shear strength of soils. So in the previous lecture we try to understand about the different stress strain behaviors of mild steel and you know also we try to see and try to understand about the strain hardening and strain softening and elasto-plastic and rigid plastic or perfectly plastic behaviors. So after having discussed about the stress strain relationships now we will introduce into the more Coulomb criterion and we will discuss about the relevance to the you know how we can actually determine by knowing the state of the stress on the element how we can actually determine the stresses at failure particularly the shear stress and normal stress along the failure plane. So this is you know the contents which are actually given then the which we are going to concentrate in this lecture is the more Coulomb failure criteria and its limitations. So as we have seen you know in the previous lecture you know the stress strain relationship of a mild steel you know has a little hump you know after its proportional limit. So the little hump in the stress strain will occur for mild steel after yield is an example of work hardening and many soils also exhibit work hardening for example compacted clays and loose sands when they are subjected to shear and they you know exhibit this so called work hardening phenomenon. So sensitive clays soils and dense sands are examples of work softening so that means that after retaining certain peak value there is you know a sort of you know decrease in the stress with an increase in the strain. So at what point on the stress strain curve do we have failure that is you know the interesting we have to you know understand. In some situations what will happen is that if a material is stressed to its yield point the strains or deflections are so large that for all practical purposes the material has fail from the survivability point of view. So this means that the material cannot satisfactorily continue to carry the applied loads and the stress at failure is often very arbitrary especially for non-linear materials and particularly for you know those materials which are undergoing work hardening like compacted clays and loose sands we take arbitrarily at say some 10% strain or 15% strain. So when these materials like you know which are undergoing work hardening we usually define failure at some arbitrary percent strain 15% or 20% or it is strain or deformation at which the function of the structure might be impaired. So that you know the we define the strain it can be of say 5% or 10% or 15% or 20% or a strain or deformation at which the function of the structure might be you know effected. Now we can also define the strength of the material it is the maximum or the yield stress or the stress at some strain which we have called defined as a failure. So that is the strength of the material. So there are many ways of defining failure in materials or put another way there may be many failure criterias will be there but most of the criterias do not work for soils. So the most common you know failure criterion which can be applied to soils is the more coulomb failure criterion first the more hypothesis has been given and then the coulombs you know have coulomb has come out with you know the based on the experience with construction of the walls you know during his time he has come out with you know made attempts to determine the shear strength of the soil. So then it has turned out to be the more coulomb failure criterion which is actually being used widely even today. So the Charles Augustine D coulomb is well known for his you know studies on friction electrostatic attraction and repulsion. So he has put forwarded you know contributed to the more coulomb criterion similarly Otto Moore he hypothesized a criterion of failure of for the real materials in which he stated that materials fail when the shear stress on the failure plane at failure reaches some unique function of the normal stress on that plane. So Moore's hypothesis states that you know the failure for the real materials in which he stated that material fail when the shear stress on the failure plane reaches some unique function of the normal stress on that plane. So tau suffix ff is equal to function of sigma suffix ff. Now if you see this the two f's which are there two suffixes the tau is the shear stress sigma is the normal stress. So the first subscript f refers to the plane on which the stress acts in this case the failure plane the stress acting along the failure plane the second f is nothing but at failure. Suppose if you are actually trying to see tau along the tau f we write then that indicates that the shear stress along the failure plane may not be at failure. See similarly our prospective failure plane or sigma f we write the normal stress on the failure plane or the prospective failure plane. So when we write so the Moore's hypothesis states that the tau is a function of sigma ff where the shear stress at along the failure plane at failure that is the tau ff and sigma ff is nothing but the normal stress along failure plane at failure. So you know these are you know the earlier hypothesis which are actually put forwarded. And a failure theory basically is required to relate the available strength of a soil as a function of measurable properties and the imposed stress conditions. So the Moore Coulomb failure criterion is commonly used to describe the strength of the soils and its main hypothesis is based on the premise that a combination of normal and a shear stress creates a more critical limiting state than would be found if only the major principle stress or maximum shear stress were to be considered individually. So it is something like a combination of you know the normal stress and shear stress along the failure plane are considered rather than you know the major principle stress or maximum shear stress were to be considered individually. So the failure theory basically is required to relate the available strength of a soil as a function of measurable properties and imposed stress conditions and the Moore Coulomb failure criterion is commonly used to describe the strength of the soils. So the main hypothesis is based on the premise that a combination of normal and a shear stresses creates a more critical limiting state than would be found if only the major principle stress or you know the maximum shear stress were to be considered individually. So here in this particular slide an element which is actually subjected to you know major principle stress that is sigma 1f and you know minor principle stress that is sigma 3f is shown and a failure plane which is you know inclined at let us say here alpha f and this is the sigma ff is the failure plane you know failure normal stress along failure plane at failure tau ff is nothing but you know shear stress along failure plane at failure. So according to Moore's hypothesis it is a non-linear you know envelope where the tau sigma you know is ranging for the given ranges of the normal stresses in a non-linear way. So you can see that this envelope actually extends like this is a non-linear way so this is an equation for the tau ff is equal to function of sigma ff. So these stresses are actually acting here this is tau ff and this is sigma ff this is the normal stress along failure plane this is the failure plane at failure and this is the shear stress along failure plane at failure. So if you see this element at failure with the principle stresses that cause failure during the resulting normal stresses and shear stress on the failure plane. So we will assume that the failure plane exists which is not bad assumption for soils and rocks and many other materials. So these failure planes basically if the material is homogeneous and if there are no you know the if there is no there are no stresses along the major principle planes and minor principle planes then there is a possibility that you know we can actually have you know conjugate failure planes that means that we will have you know failure planes in both the directions. So if you know that the principle stresses at failure we can actually draw a Mohr circle to represent the state of stress for this particular element. So with that you know we will be able to get the you know the so called the strength of the materials at the strength of the material at failure. So in this particular slide what we have shown is that Mohr hypothesis according to Mohr hypothesis the envelope is non-linear and which actually indicates tau ff is equal to function of sigma ff and this is actually shown here and this is the failure plane and which is actually assumed for it exists you know for the soils and rocks and many other materials. Now if you look into this that as been told that if you have got a Mohr circle though generally we are actually indicating the upper half of the Mohr circle for convenience but if the material is actually having you know the homogeneity then you know this type of conjugate failure planes or failure planes can be possible that means that the failure plane one can be inclined like this and other the bottom portion of the circle it can have like this. So this is the Mohr failure envelope which is you know which is which appears like this. So the Mohr column failure criterion a failure point of tangency defines the angle of failure plane in the element of the test specimen. So the where you know this is the major principle stress and then when you draw this is the pole and from here when you draw a line where you know the Mohr envelope Mohr failure envelope is in contact with the you know the Mohr circle here and here and this inclination indicates that angle of failure plane in the element of the space plane. So this is you know we can see that along this these are the stresses which are tau ff and sigma ff and these are actually arisen because of you know the major principle stress sigma 1f and minor principle stress sigma 3f and when we have you know the homogeneity then you know the responsibility that you know we also have the failure plane in this direction so this is actually parallel to this one. So the Mohr failure hypothesis is actually illustrated for element at failure as actually shown in this particular figure and the Mohr failure hypothesis basically states that the point of tangency of the Mohr failure envelope with the Mohr circle at failure determines the inclination of the failure plane. So in another way Mohr failure hypothesis can be stated the Mohr failure hypothesis states that the point of tangency of the Mohr failure envelope with the Mohr circle at failure determines the angle of inclination of the failure plane. So in this you know the combination of now you know the Coulomb what he has said is that for the he said that there are two types of parameters are there one is stress dependent parameter and other one is stress independent parameter and the stress dependent parameter is nothing but some friction between the grains and stress independent parameter is you know the cohesion actually interaction develops between the soil grains. So this actually when it is combined with the Mohr and Coulomb then you know this indicated as a linear relationship within the stress range which so beyond at higher stresses the validity of this Mohr Coulomb you know this thing cannot be guaranteed but for the you know that means that the linearity whatever we assume is actually valid for the you know the assumed the stress range and in this the tau versus sigma plot here this indicates that you have to have a combination of sigma and tau f such a way that the stress at failure is tau f is equal to C plus sigma tan phi. So if this happens to be the failure shear stress then it will be tau f f C plus sigma f f tan phi. So the failure will thus occur at any point in the soil where a critical combination of shear stress and effective normal stress develops. So basically the failure will actually develop in any point in the soil where a critical combination of shear stress and effective normal stresses you know the develops. So tau f is the maximum shear stress the soil can take just before failure under normal stress of sigma. So the shear strength equation which is popular is actually is tau is equal to C plus sigma tan phi which is like y is equal to mx plus c where the intercept here c is this cohesion intercept and the angle of inclination of this Mohr Coulomb failure and allot is the angle of internal friction that is the friction angle it can be seen here. So when you have the combination of you know tau f and sigma the failure will thus occur at any point in the soil where a critical combination of shear stress and effective normal stress develops. So further it has been shown here the same with cohesion component which is the so the shear strength is actually contributed with two components one is the cohesion component and the other one is friction component. So we have tau f is equal to C plus sigma f tan phi higher the values of C and phi the higher the shear strength of soil. So if you have a material which actually can develop good cohesion and good friction then the material actually is prone to have higher shear strength. So let us consider some you know typical examples where Mohr circles and failure envelopes how we can actually construct. So let us see that if you look into this here we have two Mohr circles which are written one Mohr circle A and circle B and there is a line which is you know the Mohr coulomb failure envelope which is indicated as a straight line and this indicates that the circle B is stable though it actually has got the stresses more than you know the combinations is such that the circle A is actually at failure. So you can see that the possibility that the circle A can be located here that element here and the circle B is somewhere here so it indicates that the in a slope if you consider this can be at the tau so where you have the possibility of the failure here and the circle B is at the location away from the tau so you can see that this is can be under stable condition. So this is further discussed here where we have we show here the shear strength tau and normal stress on the x axis and with failure plane failure envelope which is the failure envelope with C dash and phi dash and these are the effective strength parameters which are actually called and we have a circle A and circle B and a circle A is well below the you know Mohr coulomb failure envelope and so it is said that the combinations of stresses which are major principle stress sigma 1A and sigma 3A is such that you know it is in the safe state of stress but when you look into the circle B you know it is tangential to the Mohr coulomb envelope and it is actually having a point of you know this point of coincidence here and when you draw the line this is the angle of inclination of the failure plane and you can see that this is the tau max this is the tau max and this is the tau f that this is the shear stress at failure. So if you look into this here even though the stress combination you know sigma n and tau max of circle A are obviously greater than the circle B it is actually circle B that is on the verge of circle B it is on the verge of failure. So if you look into that circle A and circle B even though the stress combinations of sigma n and tau max of the circle A is actually you know greater than the that of circle B it is the circle B that is on the verge of failure so circle A is in the safe state of stress. So the state of stress represented by the Mohr circles that exist beyond the you know Mohr coulomb envelope cannot exist. So that means that if you are having a Mohr coulomb you know the Mohr circle above the failure envelope that means that it indicates that the failure would have already take place. So if you are having a you know Mohr circle which is well below the failure envelope then it is said that it is in the safe state of stress the Mohr circle is stable and when it is in comes in contact with the Mohr coulomb failure envelope there is a possibility that the failure attains. So that is at failure that is the Mohr circle at failure and then there is no possibility of the existence of Mohr circle above the failure envelope because that by that time already failure would have taken place. So here we actually define the factor of safety which is nothing but tau ff that is available divided by the shear stress which is actually applied. So the shear stress can be due to the driving stresses either because of the self weight loading or because of seepage or due to you know certain perturbance or due to some certain external loading. So the factor of safety is tau ff available to tau f applied. So if the stresses increase so that the failure occurs then the Mohr circle becomes tangent to the Mohr failure envelope that is what we have seen with the circle B the combination is such that that is actually regarded as the failure circle. So according to Mohr failure hypothesis the failure occurs on the plane inclined at alpha and then when the shear stress on that plane is of alpha tau ff. So this is not the largest or the maximum shear stress in the element. So we can see that in the circle B here the maximum shear stress if you look into it that is actually more than the failure shear stress but the combination is such that you know where you have got a normal stress and tau ff is such that it actually you know comes in becomes tangential to the Mohr circle and then it is regarded as the so this is according to the hypothesis where if the stresses increase so that the failure occurs and then the Mohr circle becomes tangent to the Mohr failure envelope. So this is not the largest or maximum shear stress in the element so the tau f actually is less than the tau max but the maximum shear stress if you look into that acts on the plane inclined at 45 degrees is equal to tau max is equal to sigma 1 – sigma 3f by 2. So if you look into this here when we take you know at this point this is sigma 1 – sigma 3 by 2 and then it is acting on the plane which is inclined at 45 degrees here so it is not on the plane. So the maximum shear stress acts on the plane inclined at 45 degrees and is equal to sigma nf – sigma 3f by 2 which is actually greater than tau ff which we are actually have just now discussed. Now let us you know question we might come is that why you know does not failure occur on 45 degrees plane so when it is actually shear stress is actually maximum along that failure plane why not actually the failure is actually unique the failure plane why it is not common it cannot because on that plane the shear strength available is greater than the tau max because the shear strength available if you look into it here when you see here if you see the combination of this sigma plus you know the tau max the shear strength is actually available is actually more than you know though this on this 45 degrees plane the shear stress is actually maximum the shear strength available to avoid the failure is actually you know greater than you know tau max. So because of that you know what it implies is that the failure cannot actually occur on the 45 degrees plane so here we have shown a more coulomb envelope which is a linear version which is actually assumed in the so more failure envelope is you know you can see the nonlinear variation here now if you look into this here if you have got a circle which is has a major principle stress sigma 1 and sigma 3 and this is you know the tau f and this is circle is actually regarded as stable because the tau ff is actually you know is here this is vertical ordinate is tau ff the shear stress which is actually there is only tau f. Now let us consider you know if you are having a you know like Mohr circle where this is the envelope which is like this and then we are having the maximum shear stress which actually acts on that 45 degrees plane so this is the maximum shear stress and the but failure is actually occurring on that plane which is actually alpha f which is actually different from the 45 degrees maximum shear stress plane. Then we have discussed that the combination is such that the shear strength is actually available you know on the you know this plane is actually more than what actually you know the shear stress is. So this condition is actually represented by the distance from the maximum point on the Mohr circle up to the more failure envelope or more coulomb failure envelope. So this would be the shear strength available when normal stress available on the 45 degrees plane was sigma 1 plus sigma 3f by 2. So this is the shear strength available when sigma n is equal to sigma 1f plus sigma 3f by 2 that is at this point. So available is actually here what you can be seen is that this point is actually is the shear strength available and the maximum shear stress on the 45 degrees plane is tau max here. So because this ordinate vertical ordinate is actually if you look into this on the Mohr circle tau sigma plot is actually more than the tau max so it also explains you know why you know the failure plane will not be on the 45 degrees plane. But there are some exceptional cases when you are actually having a saturated and undrained conditions for a given soil there is a possibility that the failure plane occurs actually on the plane of maximum shear stress that is possible for when you are actually doing you know attempts to make determine the shear strength in the laboratory where you actually or you have a condition where you have saturated and undrained conditions then the particularly for either for pure clays or from these conditions provide and there is a possibility that you actually will have the you know occurrence of failure plane at 45 degrees. But with these two you know discussion what we have so this is you know a stable circle but when the circle actually you know comes in contact with the Mohr Coulomb failure envelope then that is the point this point from the sigma 3F to this is actually the angle of failure plane what we are calling. So this is the angle of failure plane so the question is that when we have the shear stress maximum shear stress on 45 degrees plane why not actually the failure plane you know occurs along the you know the plane of maximum shear stress. So for that what we are actually trying to explain for if you are having a given state of stress which are actually shown here from the graphically it is actually shown here you can see that this is you know the shear strength available when the normal stress that is at the center here so this is 0 to this is sigma 3F plus sigma 1 minus sigma 3 by 2 so that becomes sigma n is equal to sigma 1 plus sigma 3 by 2 so when this actually combination is actually there so you know what we have to attain failure is you know this much we should be there but what we have is only tau max. So because of that you know the failure cannot actually takes place on the so called plane of maximum shear stress but as it has been told that the only exception would be when shear strength is independent of normal stress that means that when the Mohr circle Mohr failure envelope is horizontal and phi is equal to 0 that means that when you have got tau is equal to sigma that means that when you have you know sometimes you have got an element such a way that the lateral you know that is the minor principal stress is 0 that is sigma 3 is equal to 0 and then you know this is the you know the stress at failure then in that case also you can see that when this is the point then this will be the Mohr Coulomb failure envelope which indicates that the angle of inclination of that Mohr Coulomb failure envelope is 0 and then this intercept is nothing but C so which is nothing but tau is equal to C and then this point which is you know is for failure plane which is actually you know alpha f is equal to 45 degrees here. So such materials are called purely cohesive or obvious reason or this may result in completely saturated or undrained conditions. So these materials are called as purely cohesive soils or pure clay soils when they exhibit this type of this thing then it is actually you know results in you know the failure plane occurs along the plane of maximum shear stress. So now consider an element which is actually subjected to the state of stress which is actually shown here sigma C is the confining pressure initial confining pressure let us assume that we have vertical and horizontal direction this can be you know achieved by applying a by pressurizing water and so that we will be able to get the identical pressures on the principal stresses and that is the principal planes both horizontal and vertical and so initially when we have applied sigma C so initially the Mohr circle is actually is a point then with that pressure when we try to increase delta sigma let us say that so this is actually possible when we have you know a certain element at a certain ground when what will happen is that when the that element is actually subjected to certain increase in pressure due to foundation loading or an embankment construction there is an increase in stress if that increase in stress is such a way that the state of the stress in the element is such a way that the Mohr circle actually becomes tangential to the Mohr column failure and load then there is an attenuation of failure and then that combination actually gives the shear strength at failure so if that situations actually happens in the field either from the stability of the slopes in the dams or in earth pressure problems the failure state is actually attained. So here a typical stress element is shown here and with the increase in the load which is sigma C plus delta sigma the Mohr circle is actually here and the state is such that it is still stable but when the sigma C plus sigma delta sigma is increased to such an extent that you know it becomes tangential to the Mohr column failure and load then the soil element attains to the failure so the soil element does not fail if the Mohr circle is actually contained within the envelope that means that it is well below the envelope then the failure is not attained in the sample. Now let us assume that the Mohr circle actually Mohr circle traverses because of the increase in sigma 0 plus delta sigma so sigma C plus delta sigma C is initial applied and then delta sigma is actually increased continuously then the circle migrates towards right with what we are doing is that we are actually maintaining same sigma C and then started increasing so with that what will happen is that you reach a certain state in the element in the failure plane where it actually attains tau ff that is this point and the wherever actually it becomes tangential so that is actually is alpha f that is inclination of the failure plane where the shear strength at failure is actually acting along the failure plane and this is the normal stress at failure along failure plane and it is actually on this thing. So failure occurs when Mohr circle touches the envelope so the circle traverses initially when we have when you prepare the sample and apply sigma C all throughout the Mohr circle condition is nothing but a point on the tau sigma envelope and then once we continue to increase the vertical stress that is delta sigma plus plus then there is a possibility that as the loading progresses the Mohr circle becomes larger and larger. Now so if you further look into it if you take the circle at failure let us say that sigma C plus delta sigma combination that is at the failure and the sigma C automatically it becomes at failure for the given state of stress. So this is the loading plane orientation it is actually acting on the vertical plane that is this parallel to this one parallel to this one there is a perpendicular so it is actually acting on this so this is the parallel to this plane and parallel to this plane is this one on which actually the stresses are acting. So if there are no shear stresses definitely they are eligible to be called as you know the principal planes major principal plane and minor principal plane and the failure plane is actually oriented at 45 plus 5 by 2 with the horizontal so that is the inclination from the pole here when you draw at the point of tangency that is actually 45 plus 5 by 2 and the plane of maximum shear stress is actually at 45 degrees. So in terms of effective stresses this we have discussed already but to illustrate further we have got let us say that vertical stress and horizontal stress on acting on a sample sigma H and sigma B and when we have the pore water pressure if it is then we have is subtracted from the total stresses then we have got effective stresses in vertical direction and horizontal direction and the pore water pressure acting in all directions. So for this you know when you have got you know the total stresses the Mohr circle actually shifts by a difference actually that is u so sigma H dash is equal to sigma H – u so by that you know the circle actually will become shift and so if you look into this here we actually have failure envelopes for effective stress conditions and you know in total stress conditions they are different so they can actually yield different combinations of so if you are actually having 100 total stress conditions if you are actually getting the strength comb parameters which is actually stress independent parameter C and stress dependent parameter friction then it is called C and phi and if you are actually doing with undrained condition then it is called as C u and phi u C suffix u, u indicates undrained, pi indicates the angle of internal friction or a friction angle and undrained conditions. So but when we have you know the effective stresses, effective stress conditions then it is actually said as C dash and phi dash where these are called effective cohesion and you know effective cohesion and effective friction angle or effective angle of internal friction. So these are actually shown here in order to get these parameters either in terms of total stress parameters or in terms of effective stress parameters we need to have different combinations of you know the stress conditions need to be applied to the soil because it is with one circle we will not able to you know ascertain you know for a given element. So in that case what we need to do is that we have to do it at different combinations here in order to build up this Mohr Coulomb failure envelope either in total stress or in effective stress conditions what is exactly we need to do is that sigma 3 1 sigma 1 1 and then we have to do it on sigma 3 2 and sigma 1 2 and then we can actually do it on sigma 3 3 and sigma 1 3 and the other one 4th combination is that sigma 3 4 and sigma 1 4. So when we put together you know when these circles actually attain these are the when these are actually obtained as a failure then you know when we draw the envelope you know tangential to the these all these Mohr circles and that envelope is actually is constructed where we can actually yield to the inclination of this envelope is actually lead to give friction angle and then this intercept actually gives you know the cohesion intercept but you know for some normally consolidated soils and you know sands when you are actually having particularly loose sands under drainage conditions there is a possibility that the C dash is equal to 0 then the envelope is actually you know is passing to the horizon. So when you have these in terms of sigma dash so the same when you measure the water pressure pore water pressure and when you subtract this from the total stresses then you will actually get effective stress circles so when you have the combination of these Mohr circles and that can lead to a construction of a failure envelope here which is shown here. So then you know we can actually further use this knowledge and try to get the principle stress relations at failure and these are actually worked out by using this you know Mohr circle which actually shown at failure where we have got tau versus sigma dash and this is the element which has been subjected to sigma 1 dash on the major principle plane and sigma 3 dash on the minor principle plane and the failure surface is actually assumed to be inclined at theta or another notation which you are following is alpha f where f is the failure plane and then this is the normal stress acting on the failure plane that is sigma dash f and tau f is the principle stress acting or the shear stress acting on the failure plane. So when we have this Mohr circle let us say that when we have got certain cohesion intercepts C dash and the angle of internal friction phi dash that is the effective stress Mohr circle and so the combination is that sigma 3 dash which is nothing but sigma 3 minus u when you take then we have got sigma 3 dash and here sigma 1 minus u that is sigma 1 dash. So here you know this point is the point of tangency and so here we can write from the by following the geometry tau f is equal to that is the major shear stress which is actually is sigma 1 minus sigma 3 by 2. So we can write tau f is equal to sigma 1 minus sigma 3 dash sigma 1 dash minus sigma 3 dash by 2 sin 2 theta and sigma dash f is equal to half sigma 1 dash plus sigma 3 dash plus half sigma 1 minus sigma 1 dash minus sigma 3 dash cos 2 theta. So we have the combinations of tau f and sigma dash can be you know at failure so this tau f is actually given by this ordinate vertical ordinate and sigma f is you know this is nothing but sigma 1 plus sigma 3 by 2 plus half sigma 1 minus sigma 3. So by using the geometry of the from here we can actually get the tau f and sigma dash f and if you are considering here C dash as this vertical intercept and with an inclination of phi dash this small horizontal intercept is actually called as C dash cot phi dash C dash cot phi dash. So theta is the theoretical angle between the major principle plane and the plane of failure. So from the you know the geometry we can actually extend from here that sin phi dash you know from the from this here the sin phi dash can be written as sin phi dash can be written as half sigma 1 minus sigma 3 dash by 2, half sigma 1 minus sigma 3 dash by 2 minus you know this ordinate divided by C dash plus cot phi dash plus half sigma 1 plus sigma 3 dash by 2. So that is C dash as cot phi plus you know we get half sigma 1 plus sigma 3 dash by 2 because this is sigma 3 dash plus sigma minus sigma 3 dash by 2 with that we will be able to get this horizontal ordinate. So now sin phi dash is equal to half sigma 1 minus sigma 3 dash by 2 divided by C dash cot phi dash plus half sigma 1 dash plus sigma 3 dash by 2. So sin phi dash is equal to half sigma 1 dash minus sigma 3 dash divided by C dash cot phi dash plus half sigma 1 dash plus sigma 3 dash by 2. So when we simplify further this is reduced to sigma 1 dash minus sigma 3 dash is equal to sigma 1 dash plus sigma 3 dash sin phi dash plus 2c dash cos phi dash that is simple by cross multiplication we have got that. The following equation you know is referred to as the Mohr Coulomb you know failure criterion and which actually gets simplified to sigma 1 dash is equal to sigma 3 dash tan square 45 plus phi dash by 2 plus 2c dash tan square tan 45 plus phi by 2 it is also called as sigma 1 is equal to sigma 3 tan square alpha plus 2c tan alpha and where alpha is the angle of inclination of the failure plane and this is also called as the Bell's equation sigma 1 dash is equal to sigma 3 tan square alpha plus 2c dash tan alpha or in terms of you know we also indicate by n suffix phi. So in that case you know if you write n phi is equal to tan square 45 plus phi dash by 2 we can write sigma 1 is equal to sigma 3 dash n phi plus 2c dash root n phi. So when we have let us say that c dash is equal to 0 then in that case sigma 1 dash is equal to sigma 3 dash tan square 45 plus phi by 2. So that indicates that the envelope runs like this when you have got a c dash is equal to 0 this component is actually equal to 0. So with that what we get is that you know sin phi dash is equal to sigma 1 dash minus sigma 3 dash by 2 divided by sigma 1 dash plus sigma 3 dash by 2. So this is actually also when c dash is equal to 0 and these identities also varied for you know also for granular soils. But in the special case when phi is equal to 0 when phi actually becomes 0 then what we say is that sigma 1 minus sigma 3 is equal to 2c. So when phi is equal to 0 you know then we indicate that this is actually is sigma 1 minus sigma 3 is equal to 2c. So what we have said is that for in case of granular soil we can say that sigma 1 is equal to sigma 1 dash is equal to sigma 3 dash tan square 45 plus phi by 2. But when we are having you know the general equation which is actually referred as the more coulomb failure criterion as sigma 1 dash is equal to sigma 3 dash tan square 45 plus phi by 2 and 2c dash tan 45 plus phi by 2. And this is also indicated as sigma 1 dash is equal to sigma 3 dash tan square alpha and 2c tan alpha where alpha is the angle of inclination of the failure plane. So here further this is actually indicated here wherein we actually have got you know the failure plane which is actually inclined with you know the conjugate failure planes also shown here this is the conjugate failure plane other where when the sample is actually homogenous as it is been told this is actually is possible. And so you actually have the failure planes which are you know which can actually have inclinations 45 plus phi by 2 and 45 minus phi by 2. So the important points which actually have cropped up from the discussion are that the coupling Mohr's circle with the coulomb's friction law allows us to define the shear failure based on the stress state of the soil. So that is what we are calling as the Mohr's coulomb failure criterion. So by coupling Mohr's circle with coulomb's friction law we are able to define the shear failure based on the stress state of the soil. So the stress state of the soil in a given level can be raised due to you know the how we actually apply the loading. So the Mohr's coulomb criterion which is you know for a let us say for granular soils when it comes out to that this is sin phi dash is equal to half sigma 1 minus sigma 1 dash minus sigma 3 dash by 2 divided by half sigma 1 dash plus sigma 3 dash by 2. So this when we simplify we get the relationships that is oblique relationships which are actually sigma 1 dash f, sigma 1 dash suffix f divided by sigma 3 dash suffix f is equal to 1 plus sin phi by 1 minus sin phi. So this from the trigonometry identities when you use this is nothing but tan square 45 plus phi by 2 and similarly when we have sigma 3 dash f by sigma 1 dash f then it is 1 minus sin phi by 1 plus sin phi wherein we have tan square 45 minus phi by 2. So failure occurs according to the Mohr's coulomb criterion when the soil attains the maximum effective stress of obliquity that is sigma 1 dash f by sigma 3 dash f when it becomes maximum then the failure actually occurs. So failure occurs according to the Mohr's coulomb criterion when the soil attains the maximum effective stress obliquity the ratio has to be maximum that is sigma 1 dash f by sigma 3 dash f has to be maximum. So what we understood from this discussion is that coupling Mohr's circle with coulomb frictional law allows us to define the shear failure based on the stress stress on the soils then we have also given come out with the oblique relationship which are actually valid for glenero soils or a frictional soils is sigma 1 dash f by sigma 3 dash f is equal to 1 plus sin phi dash by 1 minus sin phi dash is equal to n phi and sigma 3 dash f sometimes what we will have is that we have let us say that when we have got a earth pressure condition then the wall is actually moving towards the backfill with a constant vertical stress then there can be the horizontal stress that is lateral stress is actually more than that is more than the sigma 1 f. So in that case sigma 3 dash f by sigma 1 dash f so where that identity yields is that for minus sin phi dash by 1 plus sin phi dash and which is actually is tan square 45 minus phi by 2 this is what actually has been indicated here when you have a combination such that you know you actually have a failure plane which is actually the such a way that you know we get the so called you know sigma 3 dash f by sigma 1 dash f is equal to tan square 45 minus phi by 2. So the principal stress relations that failure is further given here that is for case where sigma 1 greater than sigma 3 that is sigma 1 is equal to sigma 3 tan square 45 plus phi by 2 plus 2c tan 45 plus phi by 2 and when you have a situation where sigma 3 is equal to sigma 1 tan square 45 minus phi by 2 minus 2c tan 45 minus phi by 2. So this is in case when you are having both c and friction angle so after having obtained this then we can also link this more column failure envelope with the with you know pq space and this we actually have discussed in our previous lectures and where p is actually indicated as sigma 1 plus sigma 3 by 2 where q is equal to sigma 1 minus sigma 3 by 2 when you have that you know when we have said that the kf is the failure line and which is actually indicated at psi. So here also and this indicates this combination indicates that you actually get you know the equation the point is actually tangential. So this intercept is a vertical intercept is a and this vertical intercept this inclination is psi. Similarly when you take the more column failure envelope and we have tau and tau sigma and where we actually have got a more circle with an inclination actually is phi and c as the vertical intercept then tau f is equal to the so called tau f is equal to combination of c plus sigma f tan phi. So sigma f is that normal stress at failure. So from the pq diagram the shear strength parameters phi and c may readily be computed. Suppose if you are having a pq diagram where p is equal to sigma 1 plus sigma 3 by 2 and q is equal to sigma 1 minus sigma 3 by 2. So from there also we can actually you know determine you know the parameters phi and c like this by equating c with a by cos phi this intercept and then you know the sin phi the sin phi is equal to tan psi. So sin phi is equal to tan psi so with that you know we can actually get phi is equal to you know from this relationship we can actually get. So this is you know once we know that the relationship between kf line and more column failure envelope can be obtained. So and the another issue which we need to discuss is that what is the effect of you know intermediate principle stress sigma 2 on the condition of failure. Though suppose if you are having a cylindrical sample what we say is that sigma 2 is equal to sigma 3 but if you look into it you know whether this intermediate principle stress actually has got an effect on the condition of failure because you know the more column failure envelope will not actually take this into consideration. So if you look into this here when you have got a sigma 1 and sigma 2 you actually have got a circle like this and when you have got a sigma 2 and sigma 3 the sigma 2 is nothing but the intermediate principle stress and the sigma 3 is the minor principle stress and sigma 1 is the major principle stress. So we have the sigma 1, sigma 2, sigma 3 they are the principle stresses. So by definition if you look into it sigma 2 dash actually lies sigma 2 lies between somewhere between the major principle stress and minor principle stress and if you are actually talking about the failure envelope it is the major principle stress the more circle which is actually resulted because of the major principle stress and minor principle stress that is the one which is actually gets tangential to the more failure envelope. So by definition sigma 2 lies somewhere between the major principle stress and the more circles for the three principle stresses look like those actually shown here and but the one which is actually becomes tangential to the more failure envelope at failure is resulting due to more circle which is resulting due to sigma 1 and sigma 3 combinations. So it is obvious that from this discussion that sigma 2 can have low influence on the conditions at failure for the more circle more failure criterion and no matter what magnitude it has because it falls within these two stresses and the intermediate principle stress sigma 2 probably does not have an influence on in real soil but the more column failure theory does not consider it. So the more column failure theory does not consider the intermediate principle stress and we also said that there is actually not having influence on the conditions at failure. So some limitations on the more column theory can be seen where we have got assumed that the linearization of the limit stress envelope. So the possible results in possible overestimation of the safety factor in slope stability calculations and difficulties in calibration because of the linearization and this is also valid for usual experimental range in the laboratory where in you can see that this linearization is actually valid up to certain range and then further we actually have more column failure criterion is well proven for most of the geomaterials but data for the clays the number of attempts are actually being made and the soils on shearing exhibit variable volume change characteristics depending upon pre consolidation pressure which cannot be accounted with more column theory and in soft soils volumetric plastic strains on shearing are compressive that is negative dilation whilst the more column model will predict continuous dilation. So these are the some of the limitations which are actually resulting due to the more column failure criterion or more column theory. We can actually look an example problem and this will be solving in the next lecture where we in the figure below shows a soil profile at a site or a proposed building. So determine the increase in vertical effective stress at which a soil element at a depth of 3 meters under the center of the building will failure will fail if the increase in lateral effective stress is 40% of the increase in vertical effective stress the coefficient of lateral pressure at rest is k0 is given as 0.5 and the friction angle is given as 30 degrees and the saturated unit weight is assumed to be 18 kilo Newton per meter cube both above ground water level and below ground water level then we need to calculate what is the delta sigma at which actually the failure occurs for a given condition. So here phi dash is equal to 30 degrees means the more column failure envelope is given to us so based on that we can actually calculate. So in this particular lecture we try to discuss about the in the in detail discussion we had on the more column failure criterion and we have thrown light on how the principle stress relationship can be deduced and then we connected with created initiated discussion on the limitations of more column theory.