 Welcome to lecture series on advanced geotechnical engineering and we are in module 4 and discussing about the stress strain relationships and shear strength of soils and in the previous lecture we introduced ourselves to how to draw the stress parts for different conditions. Now in this lecture we further discuss on the stress parts in PQ space and then afterwards we introduce ourselves to different stress strain behavior of different materials and thereafter we will try to introduce ourselves to more column failure criteria and thereafter we will discuss about its limitations and correlations with the PQ space. So in this particular lecture we will be discussing about the further discussion on the stress parts in PQ space then we will introduce ourselves to stress strain behavior of different materials and thereafter we introduce ourselves to more column failure criteria. So as we have discussed in the previous lecture we have considered an example where the hydrostatic stress conditions provide. The hydrostatic stress conditions means sigma v is equal to sigma h is equal to 0 but now let us consider we have a sample which is not having identical stresses in vertical and horizontal direction that means that sigma v is not equal to sigma h is equal to 0 or we can say that you know the sample which has been taken has been reconsolidated such a way that you know to represent the initial in situ stress conditions. So in this particular slide a sample which is actually shown with sigma v on the acting vertically and sigma h acting horizontally. So the initial condition here is that sigma v is not equal to sigma h not equal to 0 and this is called as the non hydrostatic compression. So we are actually taking that initially we have the non hydrostatic compression then q not is equal to sigma v minus sigma h by 2 that we have defined earlier and p not is equal to sigma v plus sigma h by 2. So during loading or unloading we have a tendency of increasing delta sigma v and by keeping delta sigma h constant or by you can actually increase the vertical stresses or horizontal stresses. Now we are actually required to develop a stress path for a axial compression where initially the sample is in non hydrostatic compression that means that sigma v is not equal to sigma h and which is also not equal to 0. So the final coordinates of the stress path A you know are you can say that qf is equal to q not sigma v plus delta sigma v minus sigma h by 2. So here we have actually increased sigma v and here also we increased in the pf nothing but sigma v plus sigma h by 2 plus delta sigma v by 2. So if you see that the delta q from the initial conditions they both are actually increased by about delta sigma v by 2 and delta sigma h by 2. So the stress path for this condition is something like this. When you have the condition where we do not have you know non hydrostatic compression previously when we have the hydrostatic compression we are actually on the p axis but now because of the prevalence of the non hydrostatic compression we have you know a certain ordinate here and from there for path A is that delta sigma h is equal to 0 and you know if you increase the delta sigma v then you know you actually have this condition here. So here what we have done is that delta sigma h is equal to 0 and delta x sigma h is equal to 0. So because of that you know we have increased delta sigma v in both the directions. So we have qf and pf they are the final coordinates of you know the stress path A and with that you know we actually have got the you know the inclination of the stress path is also because these both are same the delta q by delta p the slope of this line is actually comes to 1 that means that the stress path actually you know A is inclined at 45 degrees. So the stress path A is inclined at 45 degrees. Now for stress path D wherein the delta sigma v decreases while delta sigma h increases that means that you can see here stress path D. So stress path D where delta sigma v decreases and delta sigma h increases. Suppose if you say that for delta sigma v decreases and delta sigma h is equal to 0 then the stress path C is actually in this direction and this is also inclined at 45 degrees and you know this you know is can be reduced very easily and similarly by using the same concepts what we have discussed you know when we have got delta sigma v increases and delta sigma h decreases then we have a stress path in this direction. So let us try to derive you know what is this stress path for D that is in this condition we have delta sigma v decreases and delta sigma h increases. So for the stress path D the condition what we have is that delta sigma v decreases while delta sigma h increases. So the delta sigma h increases that is the condition which we are having here. So initially sigma v is not equal to sigma h is equal to 0. So initial conditions are that p0 is equal to sigma v, q0 is equal to sigma v minus sigma h by 2 and p0 is equal to sigma v plus sigma h by 2 and the final coordinates of the path D we can look into it here delta sigma v is decreasing. So we can write the final coordinates of the point D as qf is equal to one of the coordinates is equal to qf is equal to sigma v minus delta sigma v minus sigma h plus delta sigma h because this is positive because delta sigma h is increasing here divided by 2 and similarly pf is equal to sigma v minus delta sigma v because delta sigma v is decreasing and delta sigma h is increasing. So sigma h plus delta sigma h by 2 by simplifying this we get you know qf minus q0 if you take that is delta q which is nothing but minus delta sigma v by 2 and minus delta sigma h by 2 and delta p is equal to minus delta sigma v by 2 plus delta sigma h by 2. So the actual slope we are actually not defining the delta q by delta p is the slope of this stress path line D but the actual slope of the stress path D depends on the relative magnitudes of delta sigma v and delta sigma h but in general it trends downward and out. So the actual slope of the stress path D depends on the relative magnitudes of delta sigma v and delta sigma h but in general it is actually you know trending downward and out. So what we have done is that in this particular example in continuation of our previous example where hydrostatic stress conditions are there, there we have drawn the stress path a, b, c, d and similar to this what we have done is that when initially we have say initially non hydrostatic compression conditions then we have deduced and derived here stress path a and stress path b and we have said that the stress path b the inclination depends upon the relative magnitudes of delta sigma v and delta sigma h and so in the similar lines you know we can deduce the stress paths per b and c the procedure is same first is that getting the initial coordinates that is q0 and p0 and then depending upon the condition we have we have to get the final coordinates of the stress path then we have to get the delta q and delta p which is nothing but delta q is nothing but qf minus q0 and delta p is nothing but you know pf minus p0 and with that we will be able to get the delta q by delta p we will be able to get the slope of the stress path and we also can draw the stress paths at different you know on the qp space. Now one more you know information here is that the path a which is actually here also represents because it is initially you know non hydrostatic stress conditions so the initial in situ stress conditions in a for a sample when we have you know sigma v not equal to sigma h and this represents the initial in situ stress conditions. So path a also represents the reconsolidation of soil sample under k0 conditions so this path a this is also indicates the reconsolidation of a soil sample under k0 condition when we do because under k0 condition means we know we do not have the identical stresses and sigma v for example for a normally consolidated soil we have got you know sigma h is equal to k0 sigma v and where k0 is equal to it can be for 0.5. So in that case you know it will actually represent the path a also represent the reconsolidation of soil sample under k0 conditions. So we will try to you know look into for the different you know stress ratios it is convenient to express this in the form of stress ratios and sigma h by sigma v a sigma h is nothing but you know vertical stress sigma h is nothing but the horizontal stress sigma v is nothing but the vertical stress and you know depends upon like you know with the example of stress path starting from 0 where sigma h is equal to sigma v is equal to 0 then you know we have the you know we have written drawn different stress paths here and the one which is actually here which is indicates that k is equal to 1 that is what actually what we said is that the hydrostatic compression where sigma h is equal to sigma v where k is equal to 1 condition suppose if you are having a sample under water then you know that k is equal to 1 which is actually along this line then you know we have you know the line which is you know when k less than 1 that is per you know where we have got k greater than 1 which is actually indicates for the you know over consolidated soil samples and 0.5 to 1 it is actually normally consolidated place and up to maximum 0.8 for normally consolidated soils but for normally consolidated place this is k0 condition where k0 is equal to sigma dash h by sigma dash v and where in this zone above this p axis or pp dash axis sigma v is greater than sigma h and below this pp dash axis it is sigma v is less than sigma h. So in case of let us say that over consolidated soil there can be you know very high you know horizontal stress because of the stress locking which actually can take place because of the past stress history conditions or when we have got let us say the hard crust and where the soil has been subjected to a very high amount of work consolidation stresses because of the process of drying and there also it can be possibility that the some stresses can be locked and when and where the sigma h is actually can be more than sigma v. And we also have you know this particular line which is the uppermost line which is actually called as you know kf line and which is the failure line which is actually inclined at a psi and this line is actually inclined at an angle beta. So kf is equal to sigma dash hf and by sigma dash vf so this is nothing but horizontal stress at failure and vertical stress at failure. So the effective horizontal stress at failure effective vertical stress at failure. So and the samples below here this also indicates that you know the so when the sample actually has been subjected to extension at failure then we have the stress path actually extending below the below the pp dash axis and where with the angle of inclination is negative psi and this is also kf line under extension condition what we define and this is the kf condition kf line under the compression conditions. So the constant stress ratios appear as these straight lines on a pq diagram. So there could be you know these lines could also be stress paths for initial conditions of sigma v is equal to sigma h0 with loading of k equal to a constant sigma h by sigma v with loadings of k equal to a constant equal to sigma h by sigma v. So note that q by p is equal to tan beta so we can see that when you have got k you know here the q by p is equal to tan beta which is nothing but 1 minus k by 1 plus k in terms of k we can write that 1 minus tan beta by 1 plus tan beta in terms of k this expression can be written as 1 minus tan beta by 1 plus tan beta where beta is the slope of the line of constant k where k is less than kf at failure the kf line is indicated by the symbol psi at failure the kf line is indicated by symbol psi in compression or in an extension. But any point when we know at any point when we know p and q and sigma h and sigma v can readily be you know found out graphically that means that any point once we know the p and q let us say that at this point and by drawing 45 degrees line from here and here and then you know we can actually find out what is sigma h and sigma v. So with this particular conditions at any point when we know p and q the sigma h and sigma v can readily be found graphically for sigma h is greater than sigma v q is negative and k is actually greater than 1. So sigma h greater than sigma v that is sigma v less than sigma v less than sigma h q is negative and what it indicates that the k will be greater than 1, k will be greater than 1. So this is actually indicates the over consolidated state and this actually indicates the normally consolidated state. So this was the discussion about the stress parts for the in the pq space and stress parts during sedimentation sampling of normally consolidated clay where k0 is actually less than 1 if you consider. So when this normally consolidated clay the condition is that when soils are actually deposited in a sedimentary environment like a lake or at the sea and the sea in the marine environment the lake in the you know in a lacustrine the soils are called lacustrine soils and there is a gradual buildup of orbital stress as additional material is deposited above. So as a deposition actually happens continuously there is a gradual buildup of orbital stress as additional material is deposited above. So if the area of the deposition is relatively large compared to the thickness of the soil deposit then it seems that we can reasonably estimate that one dimensional compression takes place. So sigma h by sigma v is equal to k0 which is equal to 0.4 to 0.6 per granular soils and k0 is less than 0.5 and a maximum it can extend up to 0.8 or 0.9. So good average value what we can say is the 0.5. So the stress parts during sedimentation and sampling of normally consolidated clay we are actually intending to draw. So when the soils are actually deposited in a sedimentary environment like a lake or the sea there is a gradual buildup of orbital stress as additional material is being deposited continuously. So if the area of the deposition is actually large compared to the thickness of soil deposit then it seems reasonable to estimate that the compression is going to happen in one dimensional direction. And with that sigma h by sigma v is equal to k0 and where k0 is equal to 0.4 to 0.6 basically for granular soils and k0 can be less than 0.5 or maximum will vary up to 0.8 or 0.9 depending upon the type of the soil the average is actually about 0.5. Then the stress part is actually drawn like this with q versus p dash and where we have the because of the sedimentation consolidation the stress part actually follows the k0 condition because of the one dimensional. So it goes it is along the k0 line. So it is along the k0 line and then if your sampling is there and then there is a possibility that the stresses will get reduced. So you can see that the stress part actually takes this particular direction this is because of the sampling. So here this is actually due to the reduction of the overburden stress upon sampling and stress part actually basically this path actually takes place because this is actually reduction of the overburden stress because of the because we have taken the sample out the stress stress in the soil specimen ends up on the k is equal to 1 axis. So initially we are in the along the k0 condition then we have actually collected the sample and then the overburden stress got relieved. So because of that what will happen is that the path tends to traverse like this and it hits the p dash axis that is the k is equal to 1 axis. This is the k is equal to 1 axis what we have discussed. So instead of sampling sigma v0 decrease due to erosion let us assume that the overburden stress is getting relieved because of the continuous erosion process or some other geologic process then unloading the stress part is similar to BC would be followed. So the BC is also like you know unloading the stress part BC is also like an unloading the stress part and if the vertical stress continue to be removed if the vertical stress is actually continued removed then the path could extend to a point well below the p axis and soil would then be over consolidated. So when the vertical stress is continued to be removed then the soil at the stress part migrates towards the below the p axis and the soil would then be over consolidated and k0 will be greater than 1. So in this particular slide what we have actually drawn is these stress parts during the sedimentation and sampling of NC clay. So when you have a continuous sedimentation when it is happening and when say no sampling is actually taking place and no erosion is taking place or no other geological process is actually taking place then the stress part actually due to sedimentation process and consolidation process is in this direction. But in case at any point when the sample is actually is collected or in erosion or any some other geological process is being subjected then there is a reduction in the war burden stress upon sampling or due to the process what we discussed then the stress state in the soil specimen ends up on the k is equal to 1 axis. Now here in this condition where we have actually given you know the stress parts during drained loading on a normally conserved place and sand. So drained loading in the sense that you know during shear we are actually also assuming that the increase in excess pore water pressure is close to 0. That means that there is no changes in the excess pore water pressure and the sample is allowed to drain. So in this case we have different stress parts which are actually listed here and this is our k axis this is our k axis k is equal to 1 axis and you know this is you know k of line compression and this is our k0 line where we just know discussed about the sedimentation and consolidation and actually happens on k0 conditions and this is the k of line for extension case. This we have this and this we have actually already introduced. Now let us say that we have got a sample which has been subjected to axial compression and a geotechnical engineering example is that we have a foundation and the loading is actually increasing and the sigma h is constant and sigma v is actually increasing like you know we have two storage building or three storage building the column is subjected to say increase in loading. So in that case the element AC is subjected to axial compression and the example for this is actually the foundation loading increasing that is indicates for sigma v and with a sigma h constant. So then that case the stress path actually follows you know for this condition where you have k0 condition that is we have just now derived for you know non hydrostatic compression condition and AC follows at 45 degrees you know this particular you know path this is this is the stress path. So axial compression for foundation engineering increase the where foundation loading increase and sigma h constant and the next example is that le where there is a lateral extension is happening that means that you know when active at pressure conditions actually prevail then when the wall actually moves returning wall moves away from the backfill the soil element is actually subjected to you know extension lateral extension under a constant you know vertical stress. So this is actually indicated as active at pressure is decreasing sigma h is decreasing and with a sigma v constant. So because you know initially the wall is under elastic equilibrium elastic equilibrium with sigma h and sigma v related with k0 conditions that means that sigma h is equal to k0 sigma v but what happened is that is the wall when it is actually moving away from the backfill the pressure is actually decreasing in the sense that this condition is called active at pressure condition with a decrease in sigma h at a constant sigma v. So this you know is path is actually followed as le and this condition is actually indicated here. So this is actually a geological example in the geotechnical engineering where active at pressure condition then we have another condition called this is the stress path which is actually indicates for AE you can see that this is the stress path for AE where axial extension is taking place that means that here we do not have a wall but what is actually happening is that we are actually extending we are excavating the soil that means that we have got a certain ground surface and the soil you know the excavation is actually happening and because of that there is decrease in horizontal because of that there is you know the decrease in the unloading you know and because of that there is a decrease in the sigma v and the sigma h constant. So this condition actually will lead to you know the stress path actually like this AE the stress path is actually leading to this particular condition then you know when we have this particular example of you know LC which is the stress path for the geotechnical example is that passive at pressure in this case wall is actually moving towards the backfill wall is actually moving towards the backfill in this case the element undergoes horizontal element undergoes compression where sigma h increases with a sigma v constant. So in this particular case actually what is happening is that you know the lateral compression actually takes place so the stress path for that is actually indicated as the stress path line LC and it actually you know meets the failure line in the tension that is under extension ultimately and whereas active at pressure line actually meets ultimately the KF compression line failure line compression when it actually attains the failure and similarly the actual extension line meets also the KF extension line ultimately when it actually attains the failure under you know like example of unloading or excavation takes place. So where that LC basically with the lateral compression and you know LE with lateral extension you know the stress paths, stress paths are LE and LC and which is indicates that lateral extension and lateral compression and similarly axial extension and axial compression under initially under non hydrostatic stress conditions are actually plotted here and here one thing to be noted is that these stress paths are for the drain loadings and hence actually you know for drain loading as there is no change in the excess pore water pressures. So hence total stress paths for a given loading is equal to so in case of drain loading these whatever the stress paths we have drawn they are actually valid for both for you know total stress conditions as well as effective stress conditions. So for drain loading as there is no excess pore water pressure during the shear loading so the total stress paths for a given loading is identical to the effective stress paths. So the total stress paths for a given loading is identical to the effective stress paths. So PQ diagram allows you know one advantage with total stress paths and effective stress paths for NC and OC soils. The PQ diagram will allow you to show both total and effective stress paths on the same diagram because for normally consolidated soils particularly under drained condition as we do not have the variation of you know the excess pore water pressure so both are identical but in principle the PQ diagram will allow us to show both total and effective stress paths in the same diagram and for drain loading the total stress paths and effective stress paths are identical as the pore water increased by the loading was approximately equal to 0 at all times during shear. So why you know the total stress paths and effective stress paths are identical the point to be noted is that the pore water pressure induced by the loading was approximately equal to 0 at all times during shear. So because it actually is allowed to drain the pore water pressure induced is actually is you know approximately equal to 0 at all times during shear. However during unload undrained loading that means that when the sample is not allowed to drain when there is no volume change is actually permitted then the total stress path is not equal to effective stress path because excess pore water pressure develops because of the you know the undrained conditions. So because of the during undrained loading the total stress path is not equal to effective stress path because excess pore water pressure develops. Now if you look into the total stress path and effective stress path for normally consolidated clay basically for you know axial compression loading for the normally consolidated clay when k0 is actually less than 1. So this is the k0 less than 1 line and this is you know the kf line the failure line and you know in the normal consolidated clay when we do not allow the drainage to take place under undrained conditions the excess pore water pressure develops is positive in nature because you know there is the sample tends to compress. So for axial under compression so axial loading of a normal consolidated clay the k0 is actually less than 1 so the positive excess pore water pressure delta u develops and therefore the effective stress path actually lies to the left of the total stress path because sigma dash is equal to sigma minus delta u where delta u is the excess pore water pressure which is actually developed. So at any point during loading the pore water pressure delta u may be obtained graphically. So when we draw this is the total stress path what we have drawn but when we actually have the you know the effective stress path by considering delta uf at failure so we can see that this stress path is actually traversing to the left here the stress path is actually traversing to the left in the way it is actually shown here. So at any point during loading so we can actually determine the pore water pressure by the graphically we can say that by subtending these ordnance we can actually determine what is the pore water pressure depending upon the you know as the loading is actually happening. So this is a case for total stress path and effective stress path for normally consolidated clay where this is the total stress path AC is the total stress path and this is the effective stress path because for it this is actually true for normally consolidated clay under axial compression. Now let us say that in most practical situations you know in geotechnical engineering there exists static ground water table always. So that means that there is some initial pore water pressure u0 will always be there acting on the element. So in that case what will actually happen is that when you have got a k0 condition which is k0 less than 1 and the total stress path you know traverses like this this we have discussed by taking the initial pore water pressure into consideration then you know we have that you know this t-u sp is actually is this one and from there you know if it is subjected to the here loading without allowing the drainage then this stress path actually traverses like this. So then we are actually drawing here or showing here three stress paths one is the total stress path and other one is that t-u0 sp where u0 is actually is the initial pore water pressure because of the static ground water table and then that is actually called as t-u0 sp and the other one is the effective stress path. So if you see that the three stress paths need to be considered and they are actually effective stress path and t-u0 sp for axial compression loading but you know one point to be noted is that for static ground water table position u0 actually change does not affect significantly either on the effective stress path or on the conditions of failure. So hence you know this can be ignored and otherwise you can look into that you know the for static ground water table conditions the u0 actually does not affect you know either the effective stress path or the condition at the failure. Hence you know this is actually considering the practical situation we are actually have discussed that how that initial existing ground water table can be taken and then we have actually drawn three stress path the effective stress path and t-u0 sp and tsp but the static ground water table position u0 does not affect either at the either on either the sp or the you know condition at failure. So now after having seen for normally consolidated soil then let us look into for how it can be for over consolidated clay total stress path and the effective stress path for over consolidated clay where we have the k0 greater than one condition the k0 actually greater than one condition. So the sample is you know already where sigma h is greater than sigma v where actually is k0 is actually greater than one condition so if the clay is over consolidated basically what will happen is that the negative initially there will be some positive forward pressure and the clay the dense clay or a stiff clay tends to you know expand. So because of that what will happen the pore water pressure becomes negative the negative pore water pressure develops because the clay tends to expand during the shear but it cannot. So why it cannot because you know the no volume change is actually allowed during undrained loading so because of that what will happen the pore water pressure which is actually you know develops because of the you know undrained conditions and you know then the clay which is actually tending to expand so for that condition what will happen is that the pore water pressure access pore water pressure will be negative that is called negative pore water pressure actually develops. So in that case what will happen in the stress path is this is the total stress path now sigma dash is equal to sigma minus u and then because of you know minus sign the effective stress path is actually on the right hand side you can see that the effective stress path so this is the effective stress path for you know for over consolidated clay and this is you know the total stress path for the over consolidated so you can see that the origin of this stress path is from k0 greater than one line here. So now after having discussed about the stress paths in pq space now let us look into you know how this stress path in pq space can be linked with more coulomb failure criteria and then subsequently how we can actually define a failure criterion for soils. So we have already discussed about the Mohr circles and we said that the Mohr stress circle indicates the stress state in a given element and we can also determine at the stresses at failure. So these examples we have already looked into it in the previous lectures. Now if the load or stress in a foundation or earth soap is increased until the deformation becomes too large then we say that the soil under the foundation or the slope has failed. Let us say that we are having a foundation when we continue to load further then there can be possibility that the soil under the foundation or when you increase the instability to the slope the slope actually attains failure. So if the load or stress in a foundation or earth slope is increased until the deformation becomes too large we say that the soil under the foundation or slope has failed. So in this case we are actually referring to the strength of the soil and which is really the maximum or ultimate stress the material can support. You can say the strength of the soil is nothing but the definition if you look into it is the maximum or ultimate stress the material can support is defined as the strength. Then in the geotechnical engineering we are generally concerned with the shear strength of the soils because the most of our problems in foundations and at work engineering failure results from excessive applied shear stresses. When the shear stresses the driving shear stresses actually dominate the shear strength of the material then the failure inception of the failure takes place and the failure occurs. So in geotechnical engineering we are generally concerned with the shear strength of soils because in most of our problems in foundations and at work engineering the failure results from excessive applied shear stresses. So consider here in this particular slide we have a stiff footing and when it is actually subjected to excessive loading then there can be possibility that this footing undergoes a failure surface experiences a failure surface. So before attenuation of the failure surface there is a certain mobilized shear strength to counter this external loading and so this is what we call as a failure surface or slip line and this is the mobilized shear strength. But if you are having a slope which is because of the self weight when it is actually trying to counter this driving shear stress this tow m is nothing but the mobilized shear strength to counter the so this is also what we call a failure surface. So here the soil grains actually soil generally the fails in shear the moment actually the driving shear stresses due to this stiff footing loading overcomes the mobilized shear strength then the failure occurs. Similarly here also the driving shear stresses due to the self weight of the soil mass which is actually subjected to outward movement like this when it is actually the shear stresses actually more than the mobilized shear strength the failure occurs. And so then we say that the soils generally fail in shear and the soil grains actually slide over each other along the failure surface without any crushing of individual grains. So what will what is actually happening is that along the failure surface the soil grains actually you know slide you know over each other along the failure surface without any crushing of individual grains. So at failure the shear stresses along the failure surface that is mobilized shear strength reaches the shear strength of the material. So that means that tow m is equal to tow f. So at failure tow m is tow m actually mobilized shear strength actually equal to tow f that means that you know the failure inception takes place. So you know always for adequate stability either due to condition of example of stiff footing or due to condition of a slope or embankment see the tow f and tow m has to have certain factor of safety. Factor of safety you know is actually defined as you know tow f by tow m and if it is actually having adequate then you know it is you know found to be you know have stable conditions. So consider another example like retaining wall and when the retaining wall moves away from the backfill. So this is actually backfill and this is the typical example of a retaining wall and when it is subjected to say active condition active earth pressure conditions wall moves away from the backfill. So in that case what will happen is that the failure width actually is countered by the mobilized shear resistance along the failure surface which is actually shown here. Moment this is you know is dominated by this driving shear stresses in this case only self weight is there but if it is subjected to certain external loading and they also contribute to these driving shear stresses and then once these driving shear stresses dominate this mobilized shear stress then here also the failure actually occurs. So here also we say that the shear failure of you know soils. So now after having discussed about the importance of shear strength and shear stresses which are actually caused merely due to the external loadings. Now let us look into certain you know stress strain relationships and failure criteria and let us consider typical stress strain behavior of a mild steel here wherein we have sigma versus epsilon where we have got a stress strain variation of a mild steel is actually shown here and this is a point where the yield stress and what we call is the proportional limit. So you can see that there is a hump which is created and then the hardening takes place and then there is a some softening takes place here. So where the softening indicates that with increase in the strain there is a decrease in the stress and in the hardening in the sense but with increase in the strain there is a gradual increase in the strength. So from the stress strain curve of the mild steel you can see that the initial portion up to the proportional limit or yield point is linearly elastic. We can say that the initial portion of the plastic limit or initial portion of the proportional limit or yield point is linearly elastic and this means that the material will return to its original shape once the stress is actually removed. So as long as the applied stress is actually below the yield point we can say that the element will be under the elastic conditions. So the initial portion of the stress strain curve of the mild steel is actually in the linearly elastic conditions and this means that the material will return to original shape when the stress is actually released as long as the applied stress is actually is below the yield point. We need to note that even linearly elastic materials yield if sufficient stress is actually applied. So you can see that some yielding of these materials you know at higher order of stresses and at the proportional limit the material becomes plastic and yields plastically. So from at the proportional limit what will actually happen is that the material tends to become plastic and yields plastically. So the material actually experience a plastic failure. So this is actually yield point what we call and in this limit this point is called proportional limit and as proportional limit the material actually becomes plastic and yields plastically. Now let us also consider you know we have say we also have non-linear elastic variation the stress strain variation can be non-linear. So it is possible however for a material to have a non-linear stress strain curve and still it is being elastic. So note that both these stress strain by mild steel and non-linear elastic variations are independent of time and if the time is variable then the material is called viscoelastic. So some real materials such as soils and polymers are viscoelastic in nature. So some real materials like soils and polymers they exhibit the viscoelastic stress strain behavior. Viscoelastic stress strain behavior. So you know when the time is actually is a parameter but you know in case if you look into the stress strain behavior of a mild steel and the stress strain behavior which is actually shown here they are independent of time. So in people who can say that why cannot actually we use viscoelastic theory to describe the behavior of soils when we are actually saying that the soils actually real materials like soils exhibit the viscoelastic behavior. But the limitations is that the linear theory of viscoelasticity. So the limitation is that the limitations of the linear theory of viscoelasticity because of that we have some limitations in describing the behavior using viscoelastic theory. So here a typical two examples of perfectly plastic and elasto-plastic conditions shown here the stress strain behavior and here the perfectly plastic materials are indicated and sometimes it is actually called as rigid plastic so that means that yield stress is there on the y axis that is sigma axis and then material is actually reminds you know with increase in strain the strain actually the stress is actually is maintained at the yield stress. So behavior of the real materials can be idealized basically by two relationships stress strain relationships one what we call as the perfectly plastic other one is called elasto-plastic and this elasto-plastic actually is the more realistic stress strain relationship is elasto-plastic a linear region and then you know for example when you have got a mild steel beyond the proportion limit the particular portion can be also linearized by maintaining horizontal plateau here. So example is that elasto-plastic can be you know described elasto-plastic behavior can be idealized for a mild steel and linear elastic up to the yield point and then becomes perfectly plastic so beyond yield point it becomes actually perfectly plastic. So this is the more realistic stress strain relationship which is actually called as elasto-plastic and example is that for mild steel and which actually remains up to linear elastic up to yield point and then thereafter it actually has you know perfectly plastic state. Now we also have another typical stress strain behavior that is called brittle and work hardening and softening conditions different curves are actually shown here the curve which is on the left hand side which actually indicates for the brittle behavior variation so at some point the point of strain loading the material undergoes a catastrophic collapse and crushes so this is generally can happen for cast iron or concrete or some cement stabilized soils where they have very they tend to exhibit you know the brittle behavior so the brittle behavior is that the abruptly at particular after attaining particular strain the material undergoes a sharon collapse array you know it actually gets crushed and example is that the cast iron or concrete and lot of rocks are brittle in nature in that they exhibit very little strain as the stress actually increases so there will be an abrupt failure and this is actually a stress strain behavior which actually is shown for a brittle condition but here what we have for sigma versus epsilon and conditions like work hardening so we can see that we have the two curves here they actually has work hardening type of behavior and then we here we have the work softening type of curves here so you can see that this is you know typical work softening curve with increase in strain there is a decrease in the stress and in this case with increase in strain there is a decrease in the you know the stress you can see so here there is actually the hardening is happening and here is actually what we say is that softening is happening in work hardening basically material becomes so stiffer because they attain higher modulus as they are actually strained or as they are getting strained or worked out so work hardening materials they you know become stiffer as they are strained or worked out work softening materials show a decrease in the stress as they are strained beyond a peak stress as they strained beyond a peak stress so beyond a peak stress what will happen is that they experience the softening condition and with that what will happen is that the what we call the strain softening actually occurs when it is then they are strained beyond the peak stress so we have work hardening conditions and work softening condition in fact you know we in the mild steel we actually have first is actually has hardening and then it is actually followed by a softening actually takes place and so when you link this stress strain relationships and failure criteria so the little hump in this stress strain curve for mild steel after yield in example of work hardening what we are just discussing the many soils are also work hardening for example compacted clays and loose sands and loose sands when they are actually sheared they are actually exhibit also you know the work hardening you know when you have got some compacted clays they also show you know work hardening you know stress strain behavior and sensitive clays soils and dense sands they exhibit work softening that means that after the peak stresses what will happen is that there is a strain this work softening situation actually happens here. So at what point on the stress strain behavior do you do we have failure so that is actually you know what we say that in some situations if a material is tested to its yield point then the strains or deflections are so large that all the practical purpose of the material is actually fail. So if some situations if a material is actually stressed to its yield point then the we say that the strains or deflections are so large that all practical purposes the material actually has fail and 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. So the stress at failure is actually very arbitrary we take it for some non-linear materials we take a certain strain so means that material cannot satisfactorily continue to carry the applied loads and the stress at failure is often very arbitrary especially for non-linear materials. So in this you know we what we are trying to see that in some situations practical situation if a material is actually stressed to its yield point the strains or deflections are so large that all practical purposes the material actually attains the failure condition. So with these materials generally we usually define some the materials which are actually undergoing some hardening define some failure condition at some arbitrary percentage of strain like 10 percent or 15 percent or 20 percent strain or it is strain or deformation at which the function of the structure might be implied. So from the serviceability point of view and we can see and we can actually also define the strength of the material. So it is the maximum or yield stress or the stress at some strain which we have defined as the failure. So it depends upon that what arbitrarily the strain actually has been selected and there are many ways of defining the failure in materials or put another way there are many failure criteria and most of the criteria basically they do not work for soils. So most common failure criterion what we say is that the applied to the stresses in soils is the more column failure criterion where one of the traditional soil failure criterion and this is the only failure criterion in which actually the stresses at failure can be determined along the failure plane. So when the element or a soil mass is actually subjected to this shear stresses which are actually more than the strength of the material then the failure actually attains and what we are actually saying is that there are many ways of defining this failure in materials but that means that set of the criteria are there but most of the criteria basically they fail for soils and one of the common criteria which we are actually going to discuss is the more column failure criteria. This was actually introduced by first by the Colum 1736 to 1806 and well known from his studies in friction and electrostatic attraction and repulsion and because of the need for the design of retaining walls the Colum actually has postulated and defined the shear strength and then it was used successfully for after noticing the several failures of the walls and the Colum actually has formulated the shear strength and also recognized that there are two different types of inherent properties of the soil one is actually the stress dependent one other one is stress independent one and with that Colum actually has formulated and defined the shear strength and Christian Otto Moore during 1835 to 1918 hypothesized actually a criterion of failure for real materials in which he stated that material fail when the shear stress on the failure plane at failure reaches some unique function of normal stress on that plane. So Moore actually has opposed that a hypothesis actually given in which a set of criterion is that failure of the real material occurs in which he stated that material fail when the shear stress on the failure plane at failure reaches some unique function of the normal stress. So tau ff f tau is nothing but the shear stress and sigma is nothing but the normal stress. The first subscript f indicates that you know refers to the plane on which the stress acts the plane on which the stress acts and f is nothing but that is in this case failure plane and the second f indicates that at failure. So this is nothing but the shear stress at failure along failure plane normal stress at failure along failure plane. So Moore actually has given a relationship that tau ff is equal to function of sigma ff and where this is actually is you know given and then this was actually combined with the Coulomb criterion wherein the Coulomb actually said that this is what actually the tau versus sigma space where according to Moore hypothesis that tau ff is equal to function of sigma ff and here it is nothing but the tau ff is nothing but the shear stress along the failure plane at failure and sigma ff is nothing but the normal stress at failure and this can be achieved because at an element at failure when with the principle stresses that caused the failure the resulting normal and shear stresses on the failure plane. So we will assume that the failure plane exist which is not bad exemption for soils or rocks and many other materials basically we know that the principle stresses at failure we can draw a Moore circle and to represent the state of stress for this particular element we have also discussed that by knowing the principle stresses we can draw the Moore circles. So when we have got different sample stresses tested at different principle stresses we can actually draw the series of Moore circles. So in this particular lecture we introduced ourselves to discuss about the stress paths and we continued our discussion for OC place and normally consolidated place and thereafter we actually have tried to understand the stress strain relationship and different stress states conditions rigid plastic viscoplastic and perfectly plastic and elasto plastic stress strain behaviors and thereafter we introduced ourselves to you know the Moore Coulomb failure criterion. So we will continue further in this direction.