 In previous lectures, I have talked about state of stress and a bit of the Mohr-Coulomb envelope. I have discussed quite in detail the direct shear box test. We talked about the limitations of these tests and the strength of these tests. And I was trying to derive a relationship between sigma 1 sigma 3 by using the concept of Mohr-Coulomb envelope which also I defined as a failure envelope. So, to pick up from that point onwards, if I plot tau versus sigma and if this is the Mohr circle where I have defined this as sigma 1 and this as sigma 3 in the generalized form. So, if this is the Mohr circle and if you draw the failure envelope, so this is what we discussed in the previous lecture. This angle we have defined as internal friction angle. This thing we have defined as C. This is the perpendicular from the tangent. This angle is 2 theta. This point we have defined as tau f sigma f. f corresponds to the failure, alright and this is the Mohr-Coulomb envelope. Now this line is also defined as Kf line. Now the state of stress which we have been talking about is where sigma v is equal to gamma into z. You remember? We have taken out an element which is buried at a depth of z and this is the state of stress sigma v and sigma h and this is the depth z and corresponding to this we have defined this as K multiplied by sigma v. So, if you remember the expression which I derived, sigma 1 as a function of sigma 3, phi and C where we have defined this term Ka as 1-sin phi over 1 plus sin phi. Now this term is also written as tan square 45 plus phi by 2, 45 minus phi by 2, alright So, if I transpose it, this becomes Kp which is equal to 1 plus sin phi over 1 minus sin phi and this is written as tan square 45 plus phi by 2. Now what you are realizing is that this is the angle alpha. So, if I say that 2 theta is equal to this is 90 degree and this is phi. So, this is 90 plus phi that means theta equal to 45 plus phi by 2 and this is what I am denoting this as alpha. So, let me replace alpha with the theta term. So, basically this function can be written as now sigma 3 tan square 45 plus phi by 2 plus 2C Ka of no 2C tan 45 plus phi by 2. Now there are two ways of interpreting the whole thing. When you are doing direct shear box test, I asked you to plot the initial state of stress. What we do is we apply a normal stress first and when you are applying normal stress, there is no shear stress. So, we are dealing only in the x-axis. So, if this value happens to be sigma v which is equal to gamma into z, the sigma 3 component of this is going to be k multiplied by or k0 I had defined this as at rest condition multiplied by sigma v. So, indirect shear test what we have done is this is sigma 1, we are maintaining the normal stress and we are shearing the sample. That means keeping this sigma v constant if I start shearing the sample what is going to happen somewhere at this point the failure is going to occur. Is this okay? Now this point has to lie on the kf line. That means if this is the failure line if I say this is the kf line keeping sigma 1 constant which is equal to sigma v if I shear the sample. So, this is the pre-shearing stage and this is the shearing stage. So, the moment you shear the sample over here it fails over here and this becomes the kf line rest of the things remain same. Now what I am going to introduce today is the concept of switching over from a 2 dimensional to 3 dimensional situation that is from plane strength to the 3 dimensional situation where it will be very difficult for you to change sigma v sigma 1 and hence what is done normally is we change sigma 3. So, the moment you change sigma 3 this becomes a triaxial test. Sigma 3 is also termed as the confining stress alright. Sometimes we also write this as sigma c this is also known as cell pressure meaning thereby in the direct shear box stress we could not control the confining stresses from where the confinement is coming in the form of sigma h because what we have done is we have assumed this to be a state of stress as a state of stress at rest what I have discussed right now can be depicted in this form. If I start by keeping sigma 3 constant what I will have to do I will have to keep on changing sigma 1 to achieve the failure look at this is this correct. The second stage of stress would be something like this sigma 3 is constant. So, what I am doing is I am confining the sample to the same stresses and I am shearing it. So, truly speaking from 1 to 2 to 3 to 4 is nothing but depiction of how to achieve failure of the sample by shearing it is this part clear. So, the moment this circle touches the kf line the failure is going to occur all these points which are at the peak they become the part of the or the components of the stress path we use the word stress path. That means if I start shearing a sample from the state of hydrostatics when the shear stress is 0 that means these are hydrostatic condition at this point your sigma 1 is equal to or let me put it as sigma v equal to sigma h and hence the shear stress is 0. So, this is the typical hydrostatic condition alright starting from hydrostatic condition if I shear the sample the shear component is coming up and building up like this. So, what is going to happen this is how the failure is being achieved and somewhere here it goes and hits and the failure is achieved. So, what I have depicted is the stress path now I am sure you must be realizing this very difficult to draw the Mohr circles all the time it becomes very complicated. So, what we do is we transform from tau sigma plane to something which is known as a pq plane p becomes sigma 1 plus sigma 3 by 2 average of the stresses sometimes we call this also as the average stress if it is a three dimensional situation I will say sigma 1 plus sigma 2 plus sigma 3 by 3 clear and if I put a condition that sigma 2 is equal to sigma 3 a typical triaxial condition. So, what happens in that case this will become sigma 1 plus 2 sigma 3 by 3 what happens to the q axis the way q is defined is this is the deviation of sigma 1 with respect to sigma 3. So, we call this as deviator stress fine have you understood these two things. So, this is the deviation from the sigma 3 of sigma 1 which is causing failure of the sample. So, this is also written as sigma D now what is the beauty of projecting the Mohr circles on a pq plane if I point out a point over here like this each point is corresponding to what a Mohr circle. So, what I have done from a complicated graphical situation I have now converted the entire thing into a sample point wise depiction of each sample and if I connect this line this line becomes the kf line because at this point the failure is taking place for each sample remember here is starting from the hydrostatic condition we have shared the sample to achieve this point. So, this point is going to be unique each of this point is going to represent one sample. So, I have done identical 4 5 samples testing and then I got these results this should be divided by 2. Now, if I define this as let us say alpha and this intercept on the y axis as a can you prove that tan of alpha will be equal to sin of phi simple geometry and a will be equal to c cos of phi. Now, this is a transformation which we have done now one hint for you would be I can write this as sigma 1 – sigma 3 by 2 will be equal to sigma 1 plus sigma 3 by 2 into tan alpha plus a this is first equation this is okay y equal to mx plus c. The second equation comes when I have this relationship over here. So, I can also say that the way we have derived this function. So, we define this as sigma 1 – sigma 3 by 2 and sigma 1 – sigma 3 by 2 equal to 2c sorry this will be c cot phi plus this is sigma 1 plus sigma 3 by 2 into sin of phi this is okay please check my expression might be incorrect also hope is alright. So, sigma 1 – sigma 3 by 2 divided by c cot phi plus sigma 1 plus sigma 3 by 2 sin of phi is okay. If you solve this expression what you realize is we are basically transforming a tau sigma relationship to a pq plane now let me introduce the concept of the effective stresses sometime back we said that in direct shear box test the pore over pressures cannot be measured or even if you measure it is very difficult to introduce the concepts is it not or introduce the equipment. Now, suppose if I say I am interested in finding out the effective stresses this is the first time I am using this term in this class this happens to be a sort of a dry situation number 1 pore over pressure does not come in the picture or this could be a drain condition where I have allowed all the pore over pressures to dissipate but suppose if I ask a condition this material happens to be not so permeable which is not valid for direct shear test case the boundary conditions are so that there is no dissipation of pore over pressures taking place and pore over pressures develop. So, that means if I say sigma 1 prime will be equal to sigma 1 minus u remember effective stress concept sigma 3 prime equal to sigma 3 minus u that means q is equal to q prime is this correct what happens to p when we say p prime p prime will be equal to p minus what is going to happen u so today onwards now what we will do is we will talk about the effective sigma primes also and when you are dealing with effective sigma primes the best way is to transform everything on a p p prime plane and q happens to be q prime fine so this becomes simple so I have introduced 3 things today in the class one is the concept of k a and k p this is what is known as coefficient of earth pressure under active conditions or active earth pressure and this is what is known as coefficient of earth pressure under passive conditions or passive earth pressure later on we will discuss that k0 will sit somewhere in between so starting from rest I can achieve failure both under active and passive failure conditions only thing is this is the question which you are asking in the previous lecture I hope you are getting this point so this is the interplay between the k parameter which is going to define sigma 3 and sigma 1 the way you denote them sigma 1 is always the major stress sigma 3 is always the minor stress clear but what is going to happen is there is a switch over between sigma v and sigma h so remember when sigma h is greater than sigma v k p conditions prevail passive earth pressure when sigma h is less than sigma v the k a condition prevails so all this analysis which we are doing right now is valid only for the active earth when sigma v happens to be greater than sigma h clear when I start discussing about the state of stress in the soils and the plastic equilibrium I will introduce the concept of how this more circle will get transformed to a passive state where sigma v becomes lesser than sigma h the horizontal state of stress increases and vertical remains constant what I have done is I have switched over from a plane strain to a three-dimensional situation so I told two things you remember in a in a direct shear box test what we did is we applied normal stress first agreed and then we sheared it two stages are there okay so when you applied sigma this is sigma v which happens to be sigma 1 what is going to happen to the horizontal stress this is sigma h which will be equal to k0 times sigma 1 this is a state of stress at rest when you are not shearing the sample at all fine so this circle corresponds to this state failure has not occurred because you are not sheared the sample now slowly from this stage onwards if I apply the shear stress what is going to happen this point is going to jump over over here and this point is going to jump over here and until it meets the failure line so this is the path which is traversed by the state of stress to cause the failure in the sample clear now two things I introduce here as long as I am doing direct shear box test I keep sigma 1 constant maintaining sigma value I am shearing the sample you are traversing from here to here directly clear the second option is keep sigma 3 constant which can be done in the triaxial testing we will discuss this and then shear the sample so what is going to happen in that case sigma 3 remains constant which is the cell pressure triaxial testing and sigma 1 keeps on changing and ultimately you meet the failure these are the two ways of failing the sample but the conditions are different the first one happens to be a 2D plane strain third one second one happens to be a three dimensional suppose this is the small strip alright this is a small system let us say this is the sample I can fail it by tearing it very difficult to do soils by tearing them apart the reason is soils do not have tension clear so this is ruled out second possibility is compress it and see when the failure occurs this is normally we adopt compress the sample and see when the failure takes place along this plane this is what we have been discussing so unfortunately there is no other way to fail the sample this is correct statement so either you have to play with the sigma 1 or you have to play with sigma 3 this is the Mohr circle corresponding to what state please understand first thing this is no shear case and now you are slowly and slowly shearing it your assignment number 2 is on this concept only so what I have done is wait wait so what I have done is I am shearing the sample delta H is picking up delta V is picking up and I am trying to make you understand how the failure is being achieved so please remember another good way of depicting this was starting from this state when there is no shearing clear and now you are slowly shearing it but sample is not failing so what is happening you are picking up slowly keep on picking up picking up picking up picking up and ultimately the failure occurs at this point and what happens beyond this post failure dense ends other case you keep on shearing the sample there is no defined peak but the sample achieves the residual strength failure so all these states of circles which I have plotted over here might be depicting each incremental loading in the form of tau because delta tau is nothing but delta of this term so there are 2 ways of achieving this either keep sigma 3 constant sigma 1 very clear or keep sigma 1 constant and very sigma 3 this is a very interesting concept which we are discussing try to understand this the first situation which I talked about is when sigma 3 can be changed is only in the triaxial sample we will discuss this so you will follow it so you take a big chamber keep the sample inside and pressurize it clear so I am changing sigma 3 and then I am seeing when the failure is going to occur because sigma 1 is accordingly getting changed because the tau value is getting changed when you do direct shear test this is in your hands this is not in your hands so there are 2 ways of failing the sample in direct shear test going to be difficult that is what I told you because what you do normally is what is the limitation of direct shear test you must have realized you know your sigma 1 sigma 3 planes keep on changing when you are shearing the sample I think I discussed in the last class this starting from this state this is your initial condition of the sample this is sigma 1 and this is k0 sigma 1 clear and this is the point of the failure it is ok what did we do state of stress is acting on horizontal plane so this becomes the pole this is your sigma 1 sigma 3 plane unfortunately the plane themselves are getting rotated so we do not have much control on the state of stress which you wanted to have on the sample and that is the reason we are coming out of this and switching over to the triaxial state in short intermediate more circles cannot be drawn for direct shear box test this flexibility you get in triaxial testing and where you can observe how stresses are getting you know what how sample is responding to the state of stress which you are imposing on them and hence triaxial testing is supposed to be the best way to test soil samples that is right so that means failure is not occurred so what is going to happen this is a state of stress is it not this is also a state of stress this is also a state of stress but this happens to the critical one which we have defined as tau f sigma f this is some sigma 1 sigma n tau n sigma 2 tau n clear sigma 1 tau n but not the failure so they exist the state of stress exists but it is not critical it is not causing the failure so these are the two ways of representing one way of representing failure is this another way of representing failure is this and that is where the kf line becomes handy in short starting from this state I am achieving the failure by following this path path can be changed I can have another path like this I can have another path like this this we will discuss later on I might be having another path where I can touch this failure line by pulling it out look at this the easiest way to fail the sample would be tear it apart this becomes my stress path this becomes my stress path this becomes my stress path and so on that is the magic we have to do with the material