 Welcome to lecture series on advanced geotechnical engineering and we are in module 4 which is on stress strain relationship and shear strength of soils. Primarily we have actually discussed about the in the previous lecture about the Mohr circles but in this module 4 lecture 3 we will be concentrating and introducing ourselves to you know what is stress path and how these stress paths can be plotted and introduction to PQ space. The introduction to principal stress space and stress paths in PQ space so in this particular lecture of the module 4 we will be concentrating on principal stress space and stress paths in PQ space. So in the previous lecture we have discussed about the total stress circles and effective stress circles and we said that both in case of total stress circles total stress and effective stresses the Mohr circle diameter remains same. And when the pore water pressure is positive when u is positive the total stress circle effective stress circle is on the left hand side of the total stress circle but when the pore water pressure is say negative then the effective stress circle is on the right hand side of total circle. So it can be seen here tau versus sigma and sigma dash and where in the effective stress circle is actually obtained by using this half sigma 1 dash plus sigma 3 dash is equal to half sigma 1 sigma 3 minus u and half sigma 1 minus sigma 3 dash is equal to half sigma 1 minus sigma 3. So if you look into this here the total stress and effective stress more circles have the same radius or same diameter but are separated along the sigma axis by amount equal to the pore water pressure either it is positive or negative. In case of negative the effective stress circle on tau sigma space will be on the right hand side of the total stress circle and in case if the pore water pressure is positive then the effective stress circle is on the left hand side of the total stress circle. So the total stress and effective stress more circles have the same radius but are separated along the sigma axis by an amount equal to the pore water pressure. And inability of the pore water to resist shear stress so that the shear stresses are resisted entirely by the contact pore stress between the soil. So this explanation is given the shear stresses are not affected by the pore pressure. You can see that the shear stresses are not affected by the pore water pressure they are actually same both in case of total stress and effective stresses and this can be physically explained by the fact that the inability of the pore water pressure to resist shear stress so that the shear stresses are resisted entirely by the contact pores between the soil grains only. So you know we can see from this graph and then for the discussion we had in the previous lecture we can understand that the shear stresses are not affected by the pore water pressure. So this can be explained physically by the fact that inability of the pore water pressure to resist shear stress and so that the shear stresses are resisted entirely by the contact forces between soil grains. So now let us introduce ourselves to principal stress space. So the principal stresses sigma 1, sigma 2, sigma 3 experienced by a point in our soil continuum can be used as Cartesian coordinate to define D in a three dimensional space and which is actually called as principal stress space. The principal stresses sigma 1, sigma 2, sigma 3 experienced by a point in our soil continuum can be used as Cartesian coordinates to define a point D in a three dimensional space and which is actually called as principal stress space. So we said that sigma 1 which is the major principal stress and the sigma 3 is the minor principal stress and sigma 2 is intermediate principal stress. So this point in the principal stress space only displays the magnitude of the principal stresses and cannot fully represent the stress tensor because the three data establishing the directions of the principal stresses are not included. So the point in the principal stress space display the magnitudes of the principal stresses and cannot fully represent the stress transfer because the three data establishing the directions of the principal stresses are not included. Now let us consider for example a case of the point D where it has actually has coordinates like along sigma 12 and along sigma 2 axis 6 units and along sigma 3 axis let us say it has got 3 units. Then this forms the as can be seen from this figure here and D is the point in the principal stress space where sigma 1 is in this direction and sigma 2 and sigma 3. So this D is represented in the principal stress space with 12 units along sigma 1 axis and 6 units along sigma 2 axis, 3 units along the sigma 3 axis. Now this principal stress space the division of principal stress tensor into spherical and deviator parts can be done like this within the matrix 12 6 3 which is equal to matrix of sigma 1 sigma 2 sigma 3 and which is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 into matrix of 1 1 1 plus sigma 2 minus sigma 3 by 3 where matrix with 0 1 minus 1 plus sigma 3 minus sigma 1 by 3 with matrix of minus 1 0 1 plus sigma 1 minus sigma 2 by 3 1 minus 1 0. So by putting the respective units like for sigma 1 12 the units which we have defined here 12 6 3 in the example here then we actually get sigma 1 plus sigma 2 plus sigma 3 and once we get simplified by that this 7 will come then sigma 2 minus sigma 3 and sigma 3 minus sigma 1 and sigma 1 minus sigma 2 after simplification this represents the matrix like 7 7 7 plus matrix of 0 1 minus 1 plus 3 0 minus 3 plus 2 minus 2 0. So this represents in the principal stress space OA AB BC CD so which is nothing but OA is from here to here which is you know has 7 units here 7 units here 7 units here that point A and AB which is having I can say that towards the sigma 2 axis 1 unit here and towards the sigma 3 axis minus 1 unit here so this is B here and then when we have BC which is similarly 3 0 minus 3 so that is represented here similarly CD which is nothing but which is represented here then which is equivalent to OD so OD is nothing but what we have actually represented in the principal stress space. So the principal stress space is particularly favored for the representation of the theories of the yield strength particularly perfectly plastic materials so the principal stress space is particularly favored for representation of the theories of the yield strength of perfectly plastic materials and for a perfectly plastic material the principal stress axis can be converted into Cartesian axis as x is equal to 1 by root 3 sigma 1 plus sigma 2 plus sigma 3 is equal to root 3 p and similarly y is equal to 1 by root 2 sigma 2 minus sigma 1 and z is equal to 1 by root 6 2 sigma 3 minus sigma 2 minus sigma 1 so the principal stress space is particularly favored for representation of the theories of the yield strength of perfectly plastic materials. Now let us look the introduction to the stress paths in PQ space so before that let us consider two examples we have on the left hand side in the figure two marbles or which can be represented like a two spherical grains which are actually in figure on the left hand side here which is actually on the one grain is sitting on the other and assume that there is a vertical force is acting in the direction of the vertical direction and in this case the grain which is bottom grain is has the same position but upper grain is actually being pushed with a force applied at an angle theta the force applied at an angle theta. So in order to represent this one let us say that we keep on increasing this from 0 to certain F then the path of this load what we say that the direction the load path or force path traverses from 0 to A in this direction that means that in the vertical direction. In this case the force is actually applied at an angle which is theta so which actually traverses in this direction so this is for the path of the experienced by this marble where it can undergo tilting it can undergo riding and depending upon the frictional interaction between the two grains. So line O A is called the load path or force path and line O B represents a load path for the example which is actually shown here with a force applied at an angle so it is important to note that the response stability and failure of the system depends on the force path. So for the point B where which actually has got the y intercept of phi n phi f sin theta x intercept of f cos theta that is actually represented here. So in this particular example when the for the two grains which are either grains spherical grains or marbles when it is actually applied force in the vertical direction the load path is actually shown here and similarly the load path O B is actually shown with a force applied at an angle theta and this is for the force applied at an angle. Now we know that soils of course are not marbles but the underlying principle is same. So when you extend this principle the soil fabric can be thought of as a space frame with the soil particles representing the members of the frame. So soil particles representing the members of the frame and the particle contacts representing the joints. So here what we are having is that the soil fabric can be thought of as a space frame with the soil particles representing the members of the frame and the particle contacts representing joints. So the response stability and failure of the soil fabric or the space frame depend upon the stress path to which the soil specimen or soil mass is subjected. So what we have done is that from the example whatever we have taken for the spherical grains or marble grains we are connecting to from the our soil where the soil fabric is assumed as the space frame, the soil fabric means the particular arrangement is assumed as a space frame with the soil particles representing the members of the frame and particle contact points are represented as the joints. So the frame is actually connected with the joints and these joints are actually represented in the soil as the contact points at the grain to grain contacts and members are actually represented by soil particles. So the response and stability of a failure of the soil fabric or the space frame depend upon the stress path. So the stress parts are presented in a plot showing the relationship between the stress parameters and provide a convenient way to allow geotechnical engineer to study the changes in stresses in soil caused by loading conditions. So the stress parts are represented in a plot showing the relationship between stress parameters and provide a convenient way to allow geotechnical engineer to study the changes in stresses in soil caused by the different types of loading conditions and different combinations can be considered. So let us consider an isotropic compression loading condition in one, this loading condition is defined as one where in the isotropic loading condition where delta sigma 1 is equal to that means that vertical stress and horizontal stress sigma 1 delta sigma 1 and delta sigma 3 that is in the horizontal direction they both are same. So here in this particular chart where Q which is represented as P is equal to sigma 1 plus sigma 2 plus sigma 3 by 3 and Q is equal to sigma 1 minus sigma 3. So here consider the isotropic loading condition one, so by using incremental form of the stress variants so we can say that the stress invariant of, invariance of isotropic compression are so delta P1, delta P1 is nothing but the incremental increase in P1 that in the P space is equal to delta sigma 1 plus 2 delta sigma 3 because delta sigma 2 is equal to delta sigma 3 we have taken it as delta sigma 1 plus 2 delta sigma 3 by 3 is equal to where because delta sigma 3 is equal to delta sigma 1 then we are putting in terms of delta sigma 1. So because of that delta sigma 1 plus 2 delta sigma 1 by 3 which simplifies to delta sigma 1 and delta Q1 is equal to delta sigma 1 minus delta sigma 3 which is equal to delta sigma 1 and delta sigma 1 so it is 0 and the stress variants at the end of the loading one are so it increments with P0 so P0 plus delta P1 so where P0 is equal to 0 initial conditions are 0 so delta sigma 1 that delta P1 is equal to delta sigma 1 and Q1 is equal to Q0 plus delta Q1 both in initial and at the end of the loading they both are same so in that case what will happen is that Q1 is 0. So the stress path for this loading condition where isotropic compression is involved it traverses from O to A so it traverses from O to A where the slope of this line is equal to 0 because delta sigma 1 that is delta sigma 1 delta Q1 by delta P1 so delta Q1 is equal to 0 and delta P1 is delta sigma 1 so 0 by delta sigma 1 it is 0 so the slope of this line is 0 and this is OA is equal to stress path for the loading one that is loading one is isotropic compression which is indicated on the where we have got the stresses which are identical in all directions then that is indicated as path OA. So if you are having earth pressure at rest condition and with K0 is equal to 1 let us say and then in that case vertical stress is equal to horizontal stress in that case with isotropic kind of compression conditions are simulated then you know the stress path actually is along this P line along the P axis with delta Q is equal to 0 so this is for you know for this OA which is for isotropic compression. Let us considering the loading condition 2 where sigma 3 is constant sigma 3 is constant that means that there is a sigma 3 which is applied but there is no change in delta sigma 3 the sigma 3 is constant means rate of change of delta sigma 3 is constant that so delta sigma 3 is equal to 0 which is the loading condition is indicated within 2 here and delta sigma 1 is greater than 0 that means that once we held with certain sigma 3 with out at constant sigma 3 then you know the we increase the pressure. So in that case this loading condition is 2 where sigma 3 is equal to constant and with the delta sigma 3 is equal to 0 but continue to increase sigma 1 in that case the increase in the stress variance for loading 2 R delta P 2 is equal to delta sigma 1 plus 2 into delta sigma 3 that is said delta sigma 3 is equal to 0 so what we have is the delta sigma 1 by 3 and delta q 2 is equal to delta sigma 1 minus delta sigma 3 delta sigma 3 is equal to 0 so what we have delta sigma 2 is equal to delta sigma 1 and the stress variant at the end of the loading 2 R P 2 is equal to P 1 plus delta P 2 where P 1 is obtained as delta sigma 1 that is here in this case in the previous condition delta sigma 1 and delta sigma 1 by 3 so this simplifies to 1 plus 4 by 3 into delta sigma 1 so with simplifies to 4 by 3 into delta sigma 1 and q 2 is equal to q 1 plus delta q 2 which is nothing but q 1 initially we have seen in the previous case q 1 is equal to 0 so what we can say that q 1 is equal to 0 plus delta q 2 is equal to you know what we have done here increase in the stress variant for the type of loading condition which we have considered it is delta sigma 1 so q 2 is equal to delta sigma 1 so AB is the stress path for the loading 2 AB is the stress path for the loading 2 AB is the stress path for the loading 2 and the slope of the stress path AB is delta q 2 by delta P 2 delta q 2 is delta sigma 1 and delta P 2 is delta sigma 1 by 3 so you know by simplifying this what we get is the 3 so that means that the slope which is you know which actually has got 3 vertical one horizontal 3 vertical one horizontal is the slope for this line stress path from A to B for the condition where the loading where delta sigma 3 is equal to 0 and sigma 1 is continue to increase. So this is for a type of loading condition as can be seen this is initial condition isotropic compression and then there is an increase like this this is the stress path and for the next to this thing what we do loading condition 3 where delta sigma 1 is equal to 0 but we continue to increase let us say delta sigma 3 that means that we are date of change of delta sigma 1 is not there delta sigma 1 is maintained constant that is sigma 1 is maintained constant delta sigma 1 is equal to 0 in that case here what we are doing is that we are actually trying to see some sort of you know expansion where delta sigma 3 is actually some it also present something like a squeezing. So in this case for the loading 3 what will happen is that sigma 1 is constant that means that delta sigma 1 is equal to 0 but continue to increase delta sigma 3 and that means that delta sigma 3 is greater than 0 so increase in the stress variance for loading 3 r delta p 3 is equal to 0 plus that is delta sigma 1 is equal to 0 plus 2 delta 3 divided 2 sigma delta sigma 3 divided by 3 which is simplifies to 2 by 3rd of delta sigma 3 and delta q 3 is equal to 0 minus delta q sigma 3 which is nothing but 0 minus delta sigma 3. So there is an increase delta sigma 3 is being increased so you know delta q 3 is equal to minus delta sigma 3 and the stress variance at the end of loading 3 r p 3 is equal to p 2 plus delta p 3 is equal to 4 by 3 delta sigma 1 p 2 was obtained has here in this case p 2 is obtained has 4 by 3 delta sigma 1 4 by 3 delta sigma 1 plus 2 by 3 delta sigma 3 which is actually delta p 3 here. So q 3 is equal to q 2 plus delta q 3 delta q 3 q 2 was obtained as delta sigma 1 so by substituting here delta sigma 1 and delta sigma 3 you know with delta sigma 1 is equal to 0 what we will have is minus delta sigma 3. So this b c the stress path for the loading 3 which is shown here the stress path for the loading 3 is in this direction. So you can see that this path now takes a downward turn here because the loading condition here is that there is no change in sigma 1 that is delta sigma 1 change rate of change of then sigma 1 is 0 delta sigma 1 is 0 but delta sigma 3 is being increased that is greater than 0. So with that you can see that how the slope of the stress path b c is nothing but delta q 3 by delta p 3 which is nothing but minus delta sigma 3 by 2 third of delta sigma 3 with that what we get is the minus 3 by 2. So minus 3 by 2 the slope is actually indicated here 3 vertical 2 horizontal that is the slope which actually runs down here for stress path b c for the loading condition where delta sigma 1 is 0 sigma 1 is not sigma 1 is constant and delta sigma 1 is 0 and delta sigma 3 is continue to that is greater than 0. So that is for loading condition 3 but till now this is an example which we have taken and then we try to plot the stress paths but when we connect it to the soil when it is under drained condition or undrained condition it can be under during consolidation stage if the drainage is allowed then the consolidation takes place and during shearing if the drainage is not allowed then the excess pore water pressure develops and this is also depends upon the type of the soil if you are having normally consolidated soil then you have positive pore water pressure and when we have over consolidated soil there can be possibility that you know the native pore water pressures they do develop. So if the soil pore water is allowed to drain from the soil sample the increase in the stress carried by the pore water is called as excess pore water pressure delta u will continuously decrease to 0 and the soil solids will have to support all of the increase in the applied stresses. So if the soil pore water pressure is allowed to drain from the soil sample and the increase in the stress carried out by the pore water is called the excess pore water pressure and will continue this excess pore water pressure will continue to decrease g to 0 and the soil solids will have to support to the all the pressure which is increased by the increase in the applied stresses. Now let us assume that loading condition one represents the isotropic consolidation and since the excess pore water pressure delta u1 dissipates as the pore water pressure drains from the soil the mean effective stress at the end of each increment of loading one is equal to the mean total stress because delta p1 dash is equal to delta p1 minus delta u1 and at the end of consolidation because when the pore water pressure excess pore water pressure which is generated is dissipates to delta u1 when it is tending to 0 then we can say that delta p1 dash is equal to delta p1 which is nothing but which is both the effective stress path ESP and total stress path both are identical for the isotropic consolidation and which is represented by OA that is for during isotropic consolidation both OA represents for the effective stress path as well as for the total stress path. So in case of isotropic consolidation the slope of the stress path which is 0 and which is along the p dash axis and which actually the effective stress path and total stress path you know stress paths are identical and they follow in one line and this is for the total stress path and effective stress path for the isotropic consolidation. Now assume that during loadings 2 and 3 the sample is undrained now during loading during loadings in 2 and 3 the sample is undrained then we can see how things will change. Now let us assume that if the during loading so this is you know AB represents the stress path for total stress path and BC what we have deduced that is for the total stress path and then for effective stress path with sample remains undrained you know the effective stress path traverses from A to B dash and the effective stress path for this condition where delta sigma 1 is equal to 0 and delta sigma 3 greater than 0 the effective stress path traverses vertically down with slope almost 90 degrees you can see that BC dash and AB dash for the loading condition where delta sigma 3 is equal to 0 and where delta sigma 3 is greater than 0 you can see that how this path traverses downwards like this and this traverses upwards like this. But however this we will discuss while you know after having introduced the triaxial test for the stress paths in PQ space stress variable S and T are the two dimensional variables that do not capture the effect of sigma 2 and the whereas the mechanical response of soil is mainly expressed in terms of P and Q which are defined as the following which are defined as follows in terms of principal stresses. So this P and Q mechanical response of soil is mainly expressed by taking into sigma 1, sigma 2, sigma 3 and the stress variables S and T are basically they are two dimensional variables and basically they do not capture the effect of sigma 2 and the thus the P and Q are normal and shear stresses that are representative of the three dimensional state stress state at a point. So the P and Q are the normal and shear stresses and that are representative of the three dimensional stress state at a point where P is given by 1 by 3 sigma 1 plus sigma 2 plus sigma 3 Q is equal to 1 by root 2 root over sigma 1 minus sigma 3 whole square plus sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square. So in the two dimensional when you wanted to convert let us say that so this is in terms of Q which is 1 by root 2 into root over sigma 1 minus sigma 3 whole square plus sigma 1 minus sigma 2 whole square plus sigma 2 minus sigma 3 whole square. Now let us say that we have a Mohr circle with sigma 1 as the major principal stress and sigma 3 as the minor principal stress and sigma 1 minus sigma 3 is the diameter of the Mohr circle and the center of the Mohr circle on the sigma axis at a distance sigma 1 plus sigma 3 by 2 and let us assume that this is the stress point S T and the point here we can see that this point A which is nothing but when you join the pole here this is the major shear stress plane, this is called the major shear stress plane. And sigma this coordinate of this point is along the sigma axis it is sigma 1 plus sigma 3 by 2 along the tau axis it is sigma 1 minus sigma 3 by 2. So when you have let us say with the constant sigma 3 let us say that when sigma 1 is increasing then in that case what will happen is that with the constant sigma 3 you know the Mohr circle actually travels us like this. So you will actually get sigma 1 1 1 sigma 1 1 2 sigma 1 1 3 like this. So whenever we actually join these points which are actually of maximum shear stress then that indicates something like you know when you join this line and this is actually inclined at 45 degrees so a stress path is a plot in tau sigma or tau sigma space of the progression of S T points that means that when different states of soil stress states of the soil and the final stress state can be at failure, final stress state can be at failure and representing the loading process of a on a sample. So you know this is one stress point which is actually shown here but when we have got number of series of Mohr circles and they represent each and every time the Mohr circle is actually having new diameter and new stress points and but when you join these two and that actually represents the so called the stress path for this condition which is actually being defined. So a stress path basically is a plot which is drawn in the tau sigma space which is connecting a progression of S and T points where S is equal to S and T which we have defined from the two dimensional point of view representing the loading process on a sample. So where you have the S is equal to half sigma 1 plus sigma 3 by 2 and T is equal to half sigma 1 minus sigma 3 by 2. So this is again shown here with different notation which is P is nothing but S here, P is nothing but S here, S is equal to sigma 1 plus sigma 3 by 2, T is nothing but Q here sigma 1 minus sigma 3 by 2. So the representative of the successive states of the stress as sigma 1 increases with sigma 3 constant. So when you have the sigma 3 constant and this is for Mohr circle A, Mohr circle B, Mohr circle C, Mohr circle D, Mohr circle E. So series of Mohr circles, so the representation of the successive states of the stress as sigma 1 increases with sigma 3 constant points A, B, C, D, E they represent the same stress conditions in both the diagrams. So this points which are actually because it is not possible to indicate the number of circles when we do the conditions with different sigma 3 values. So when you will be having number of Mohr circles on the tau sigma space so that is difficult and can lead to confusion. So for that what it is actually simplified is that the stress path only simplifies a path where in point joining the points where the maximum shear stress on the Mohr circle is generated. So that points actually are picked up here and the A, B, C, D, E is actually known as a stress path for the condition where the sigma 3 constant and sigma 1 increases. So this is ST diagram or QP diagram where Q is equal to sigma 1, sigma 3 by 2 and Q is equal to sigma 1 minus sigma 3 by 2 where Q is equal to also T is equal to sigma 1 minus sigma 3 by 2. So the duplication of successive states of stress that exist in a specimen is represented by a series of Mohr circle that is what we have discussed and plot with the series of stress points and when we connect these points is called as a stress path. The locus of the stress points this is also called as the locus of the stress point. So the plot this stress path is nothing but a plot with a series of stress points and connecting these point when you connect these points and that is called as a stress path which is here indicated again as A, B, C, D, E which is actually inclined at 45 degrees because this is nothing but the plane of tau max and 45 degrees to the principal plane. So the principal plane, major principal plane is here and minor principal plane here. So it is actually 45 degrees to the principal plane. So in this case if you look into it and this is the major principal plane and this is the minor principal plane. So this stress path for this condition is inclined at 45 degrees to the principal plane. So the plane of tau max, so this is nothing but the plane of tau max and 45 degrees to the principal plane. So this stress path represents a state of stress and successive states of stress and the stress path need not be a straight line depending upon how we actually vary the incremental stresses. Sometimes we can actually vary non-linearly also. In that case the stress path actually is also formed as a curve. So stress path need not be a linear version, it can be also need not be a straight line, it can be non-linear also. So here when we have that ds which is nothing but small change in sigma 1 plus sigma 3 by 2 that is half d sigma 1 by ds is indicated as d sigma 1 plus d sigma 3 by 2 and dt is indicated as half d sigma 1 minus d sigma 3. When sigma 1 changes d sigma 3 is equal to 0, when sigma 1 changes that means that you are actually having sigma 3 but d sigma 3 is 0 then in that case ds is equal to half d sigma 1 and dt is equal to half d sigma 1. So you can see that ds and dt both are same, ds and dt are both are same, ds and dt are both are same. The change in ds and change in dt both are same. When sigma 3 changes and d sigma 1 is equal to 0 then ds is equal to d sigma 3 by 2 and dt is equal to minus 1 by 2 into d sigma 3. So dt is equal to, so if you look into here, in this direction sigma 3 is unchanged here and sigma 1 increase then the path goes like this, sigma 1 increase the path goes like this and sigma 3 unchanged, sigma 1 decreased then the path comes down like this. And when you have sigma 1 unchanged that is d sigma 1 is equal to 0 and sigma 3 decreases that means that you are actually releasing the confining pressure then this comes down like this. When you have sigma 1 unchanged there is certain sigma 1 and sigma 3 is continuous then the path actually goes down 0 here. So the stress path actually changes from this direction either this side or this side depending upon how the sigma 1 sigma 3 is the stress conditions which are actually changing on the sample. So in this particular slide what we have tried to explain is that ds is equal to of d sigma 1 plus d sigma 3 by 2 and dt is equal to of d sigma 1 minus d sigma 3 by 2. So here when sigma 3 changes and only d sigma 3 is equal to 0 then ds is equal to d sigma 1 by 2 and dt is equal to d sigma 1 by 2. When only sigma 3 changes d sigma 1 is equal to 0 then ds is equal to d sigma 3 by 2 only sigma 3 changes and d sigma 1 that no change sigma 1 remains unchanged then dt is equal to shear stress actually decreases so you can say that this is you can see that the path is actually coming down see here is coming down. So all possible stress paths when only one of the sigma 1 or sigma 3 changes are straight lines at 45 degrees with the horizontal with dt by ds is equal to 1 when sigma 1 alone changes when dt by ds is equal to minus 1 when sigma 3 alone changes. So here important point to be noted here is that all possible stress paths when only only one of the sigma 1 or sigma 3 changes are straight lines at 45 degrees and with horizontal and with dt by ds is equal to 1 the dt by ds is equal to 1 dt by ds is equal to 1 when sigma 1 alone changes and dt by ds is equal to 1 when sigma 3 alone changes. So this observation need to be noted. Now let us take an example problem where we have got initial condition which is hydrostatic condition sigma v is equal to sigma h you have got some cylindrical sample where you have applied vertical stress and horizontal stress that means that the sample is confined with in hydrostatic pressure which is actually having that means that the sample is actually with a pressurized system and then it will be subjected to sigma v in vertical direction sigma h in other direction but when it is with the hydrostatic condition the sigma v is equal to sigma h and during loading or unloading it can be like when you have the initial conditions and delta sigma v can increase or you can actually have delta sigma h can increase or decrease that means that here when you have the conditions are that during loading and unloading the sample will be experienced in something like increasing in delta sigma 1 decreasing delta sigma 1 and or when this is maintained constant there is increase in the delta sigma h or decrease in the delta sigma h. So the stress parts for this case with the different stress points parts for initially hydrostatic condition. So as we have said that initially hydrostatic condition means you know that path actually starts along the p axis or this is s is equal to sigma 1 plus sigma 3 by 2 q is equal to t is equal to sigma 1 minus sigma 3 by 2. So you can see that we have got different paths now a that is the if this is let us say a point t ta, tb, tc, td, te and tf are the different stress parts and out of this you can see that path a which is with condition delta sigma h is equal to delta sigma v and path b what we have done is that we have maintained delta sigma h is equal to half delta sigma v and in stress parts c what we have done is that delta sigma h is equal to 0 and delta sigma v is increasing that is what we have discussed in the previous example. And path d where delta sigma h is equal to minus delta sigma v here we are actually maintaining delta sigma h as minus delta sigma v and e where delta sigma h decreases and delta sigma v is equal to 0 that means that there is a certain sigma v on the sample and the delta sigma h is being decreased and in path f where delta sigma h increases and delta sigma v decreases. So this example you know this is the solution for this example problem when you actually have got you know variable loading conditions and this can be worked out like this the initial conditions of the for all the soil all the stress parts are p0 is equal to sigma v plus sigma h by 2 and with that sigma v is equal to sigma h. So this is equivalent to when you have sigma v is equal to sigma h which is nothing but p0 is equal to sigma v or sigma h and q0 is equal to 0 and the final conditions are which can be given as final condition which is indicated for q f is nothing but sigma v plus delta sigma v minus sigma h plus delta sigma h by 2 and pf is equal to sigma v plus delta sigma v plus sigma h plus delta sigma h by 2. So when we have you know this when we substitute for this conditions what we have for the stress path a in order to obtain the stress path a what we do we what we have been given the loading condition is that you know initially sigma v is equal to sigma h and delta sigma v is equal to delta sigma h that means that q f is equal to sigma v plus delta sigma v minus sigma v because sigma h is equal to delta sigma v minus delta sigma v by 2. So with that what we have got is that q f is equal to 0 what we have got q f is equal to 0 and pf which is sigma v plus delta sigma v plus sigma h I substitute as sigma v and delta sigma h is sigma v delta sigma v. So when I substitute this which is nothing but 2 into sigma v plus delta sigma v by 2 that gets simplified to sigma v plus delta sigma v. So we can see that the stress path the stress path a moves out of the p axis by an amount equal to delta sigma v is equal to delta sigma h that means that the stress path which is you know if you take q by delta q by delta pf it is 0 0 by delta sigma v so that means that the stress path actually travels us along this and it moves by a distance delta sigma h is equal to delta sigma v and this is nothing but the q f and pf are nothing but the coordinates where q f is equal to 0 that means that is on point is on the p axis or s axis which is nothing but s is equal to sigma 1 plus sigma 3 by 2. Similarly we have you know the stress path b the initial conditions are same but what we have done is here is that delta sigma h is equal to delta sigma v by 2 that means that delta sigma h is equal to delta sigma by 2 the stress path b is actually traversing at an inclination of we can see that 18.4 degrees. So in order to deduce this we follow the same conditions here q f is equal to sigma v plus delta sigma v minus sigma h plus delta sigma h by 2 and pf is equal to sigma v plus delta sigma v plus sigma h plus delta sigma h by 2. Now for substituting sigma h is equal to sigma v initially so sigma v plus delta sigma v minus sigma v minus delta sigma v by 2 because delta sigma h is equal to delta sigma by 2. So with that what we got is that 1 by 4 delta sigma v and similarly pf is equal to when you simplify this we get sigma v plus 3 by 4 delta sigma v. So these points are nothing but the coordinates of the at the end of the stress path q f and pf are the points coordinates of the at the end of the stress path b and q and p both increase by delta sigma v by 4 and 3 delta sigma v by 4. So both are actually increasing you can see that this is increased by delta sigma v by 4 and this is increased by delta 3 by 4 of delta sigma v 4 this implies that this stress path as a slope of you can see that this by this what q by p we get the slope as 1 by 3 or 18.4 degrees. So hence we can actually draw now this stress path b if t b as the you know with an incline and with the coordinates of what we discuss is 1 by 4 delta sigma v and then 3 by 4 delta sigma v it is increased by that much amount and then that is the stress path for the loading condition where delta sigma h is equal to delta sigma v by 2. The third loading condition here for stress path c is delta sigma h is equal to 0 that is no change in the sigma h but sigma v continue to increase. So this we have discussed but again we will actually solve this by using the method which you have discussed just now with delta sigma h is equal to 0 and delta sigma v greater than 0 that is increases by some amount then we can actually substitute here q f is equal to when you substitute sigma v plus delta sigma v minus sigma v by 2 because delta sigma h is equal to 0 it gets 0 so it becomes delta sigma v by 2 and p f is equal to sigma v plus delta sigma v plus sigma v by 2. So this also increases by amount you can see that both they are q f and p f they are the coordinates of stress path c and both increase by magnitudes which is equivalent to delta sigma v by 2 and delta sigma v by 2. So this implies that the slope is 1 so q by p which is 1 which implies that the slope actually the stress path actually has got a slope of 45 degrees. So now this Tc which is the condition where this delta sigma h no change in the cell pressure and that is the lateral horizontal pressure and delta sigma v increases and that is nothing but this stress path Tc where the coordinates of this which increases by the point which is what we have shown as delta sigma v by 2 and delta sigma v by 2. Now we have stress point D so in this stress path D the condition what has been maintained is that delta sigma h is nothing but minus delta sigma v, the stress path D where the condition here is maintained is that delta sigma h is minus delta sigma v. So by using the same principles substituting in this and writing here with sigma h is equal to sigma v and you know when we put this what is happened is that you actually have increased by delta sigma v so this actually has increased by delta sigma v and this point is nothing but sigma v plus delta sigma v plus we have sigma v minus delta sigma v because the delta sigma h is equal to minus delta sigma v. So with that what we get is that you know these two will get cancelled then what we have is that 2 sigma v by 2 which is nothing but sigma v. So the values of PQ are the coordinates of the end of the stress path D and Qp both increase by delta sigma v and sigma v and this implies that the stress path D has a slope of 90 degrees. So that is the reason why you can see that the stress path is actually traversing vertically up with a coordinate here what you can see that delta sigma h is equal to minus delta sigma v it is actually traversing up here with a 90 degrees slope. So for E let us consider for stress path E where delta sigma v is equal to 0 that is the change in vertical stress is 0 but in this self pressure delta sigma h is actually decreased by some amount. So by using the same principles where Qf is equal to sigma v plus delta sigma v minus sigma h plus delta sigma h by 2 by substituting here what we have what we will get is that sigma v plus 0 that is delta sigma v is equal to 0 minus sigma v minus delta sigma h by 2. So wherein we get Qf is equal to delta sigma h by 2 and Pf is equal to sigma v plus 0 plus sigma v minus delta sigma h by 2 wherein we have got sigma v plus delta sigma h by 2. So we can see that the values of Q and P are increased both by delta sigma h by 2. So this implies that the slope of E has a slope of minus 1 and 45 degrees so this traverses for this load condition which is what we have discussed is that like this. So when delta sigma h decreases and with delta sigma h decreases sigma h decreases with delta sigma v is equal to 0 you can see that this traverses in this direction towards the Q axis and which is with inclined at 45 degrees and increased by the same amount both in delta Q and delta P and when you have delta sigma h is equal to delta sigma v that is when you have delta sigma h is equal to 0 and delta sigma v increases it traverses in this direction when you have in this direction that means that here what is actually happening is that with a constant sigma 1 when you have and when delta sigma h is actually the when your sample is actually losing the confinement that means that there is sort of the sample is subjected to a tension. So you can see that is actually happening here whereas in this case where delta sigma h is actually 0 there is no change in the cell pressure or the horizontal pressure but when delta sigma is increasing this is actually towards the compression and this is towards the tension. So you can look into this the way what we have done is that we try to you know see you know these plot these test paths by actually following the fundamental principles wherein in this particular example wherein it is subjected to initially hydrostatic conditions are considered but suppose if you are not having hydrostatic conditions initially when say delta sigma v is not equal to delta sigma h or sigma v is not equal to sigma h then that means that you know the origination of this test path location itself somewhere here and from there again it actually follows the you know different paths can be drawn. So for the in this particular example what we have considered is that the initial condition is hydrostatic compression where we have actually sigma v is equal to sigma h and when we have this when we have got different variations of delta sigma v and delta sigma h then we have actually drawn and then we have actually also seen how this procedure can be adopted for plotting this stress paths. So you can see that here the stress paths are actually plotted and then they are nothing but the locus of the you know points on the Mohr circle and instead of this actually avoids the series of Mohr circles when you actually have different stress conditions. So the stress paths are the you know the straight lines joining the locus of the stress states on the Mohr circle and they are actually need not be straight line and depending upon how we exercise for example when you have delta sigma h is equal to half let us say delta sigma v square by half delta sigma v square then in that case it can actually undertake as a curve also. So you actually get a non-linear variation also so not necessarily the straight lines but when we have these conditions of these you know delta sigma h is equal to delta sigma by 2 and when we have delta sigma h 0 and delta sigma v increases and when we have delta sigma h decreases and delta sigma v 0 then it actually follows and it is actually nothing but the plane of maximum shear stress wherein it actually has got and it actually inclined at a 45 degrees to the principal plane. So in this particular lecture what we have done is that we discussed very briefly about the principal stress space and then we have tried to introduce ourselves to stress paths and seen some examples where how these stress paths can be plotted in pq space or st in the two dimensional space where s is equal to q is equal to sigma 1 minus sigma 3 by 2 and p is equal to s is equal to sigma 1 plus sigma 3 by 2. So further we will actually discuss on this and then when we introduce ourselves to the traction compression test then we will try to see when we have got a different drainage conditions undrained and drained how we can actually draw effective stress paths and total stress paths and with some examples and that makes very clear.