 In the previous lecture I started discussion on the plastic equilibrium in soils and I gave you basic concepts regarding what causes the plastic equilibrium in the soils. In continuation with where I stopped yesterday and I was discussing about the global state of equilibrium in the soil mass which is normally caused by the natural processes under the action of gravity and we define that as global you know state of plastic equilibrium in the soils. Now our interest is to understand how the manmade structures develop the state of plastic equilibrium and this is what is known as local plastic equilibrium in soils. So what we are going to discuss today is a subset of what we discussed yesterday that was for the natural phenomena and now I am going to focus on the mechanisms which cause plastic equilibrium to develop in the soils under local conditions. So the best way to understand the mechanisms is if I consider a box which is resting on a horizontal surface let us say this is made up of perspex, perspex is a plastic sheet which will not offer any friction. And this happens to be the top of the box and this is where I am filling up the soil mass. Now the way sample is created this is by sedimentation process or by very slow packing controlled packing which represents the elastic state and we have quantified this state as k naught at rest. So the first mechanism in which the plastic equilibrium state might occur in the soil mass would be if this wall which I define as let us say AB slides out alright this is what is known as sliding failure very common type of a failure where the wall AB would get slid to a finite distance and under this situation was going to happen this much of strain is caused in the system which I define as delta L. So this is the first mechanism now what has happened here is the wall has moved out of the backfill. So as we discussed in the previous lecture this is going to be a situation which is termed as active earth pressure alright. Now at this case or in this situation what is going to happen is because the soil mass is also moving out there will be a slip surface which is getting generated and this slip surface gets generated in such a manner that if this is the PA we define this as active earth pressure that is PA this angle is going to be 45 plus 5 by 2 we have proven this alright. Now if you take an element somewhere inside the soil mass of the soil and if I denote this as sigma v sigma h what is caused this type of a situation is that sigma v happens to be greater than sigma h alright. So this theory was given by a guy known as Rankine and we call it as Rankine earth pressure theory this was in 1847 I think long long back where what he assumed is that the shear force acting between the element and the hypothetical wall is 0 that means for a smooth surface. So there is no shear force which is acting between the contact alright there are few assumptions which this guy has made he says that there is a relationship between sigma h and sigma v number 2 this is a homogeneous isotropic soil mass number 3 this theory is valid for the frictional material that is pure frictional material that means C is standing to 0 which is dry so we call this as a non-cohesive material another condition which he has put over here is that the wall is smooth alright there is no friction getting mobilized on the wall itself because of the backfill and the backfill is horizontal there could be a situation where the backfill could be inclined at an angle of that is a beta. So this is what we will be studying later on this is the wall angle at 90 degree so when this type of state of stress develops where sigma v is greater than sigma h and sigma v delta sigma v remains 0 I hope you can realize that the equation which we derived was sigma 1 equal to sigma 3 1 plus sin phi over 1 minus sin phi plus 2c cos phi over 1 minus sin phi this is okay so in this case sigma 1 happens to be sigma v and sigma 3 happens to be sigma h and sigma v is equal to gamma into z so that means the equation which I am going to use for active earth pressure would be equal to gamma into z you have to do a bit of manipulation so sigma 3 goes on this side which is equal to Pa alright this becomes Ka minus 2c root of Ka so this you can derive very easily now in this case what is going to happen because of the movement of the wall away from the backfill what you will observe is this whole block is going to be under state of plastic equilibrium alright and we have yesterday talked about by using the concepts of Mohr's circle we have shown P1 and P2 and how these slip surfaces are going to be inclined so under active condition you know that this is going to be 45 plus 5 by 2 and this is the equation which I have derived now one of the features of this system would be that there will be always a depression over here because of the movement of the soil mass so once the wall moves out this soil mass being granular would have a tendency to flow out there will be a depression and what happens is that this material would like to get accumulated over here in this triangle meaning thereby active earth pressures are always associated with depressions on the top and bulging on the lateral side so this is the first mechanism of the failure now what I am assuming here is that there is no friction getting mobilized on surface AB and the base is also free of friction the smooth surface so there is no friction coming into the picture the principal stress condition remains maintained because the moment friction gets associated with this what you will observe in this case is if the backfill had been inclined at an angle of beta this element would have become like this and then we would have a sigma v over here sigma h over here and a shear stress developing over here so this is the basic difference between the two situations this situation we will be handling later on now there could be a second situation where I will say my efforts in drawing this picture again so there the movement of the wall is inside because of the state of stress and it may so happen that the wall moves in to let us say a double prime b double prime as we saw yesterday once the wall moves into the backfill this becomes the pp passive earth pressure now this is the situation where your sigma v is going to be alright so this angle is going to be 45-5 by 2 the slip surface this whole soil mass goes into the state of plastic lubrium rest of the body remains in the elastic state k0 condition what happens over here is sigma h is less than sigma h is greater than sigma v if this is the situation I can write this term as pp the passive earth pressure so sigma 1 remains equal to passive earth pressure sigma 3 becomes sigma v over here you are discussing this thing long back how the reversal is going to take place and this will become sigma z into kp plus 2c root of kp so what we have done is by using this simple model we have derived how the earth pressures developed in the soil mass now this was the discussion which was done by Rankine and there could be another methods of the mechanisms of the development of the earth pressure also what I may assume is that this point b acts as a hinge so what are the characteristics of the hinge moment about this point is 0 correct if the moment about this point is 0 what could be the mode of failure of the system so the mode of the failure of the system would be you have a b and this is the backfill if b is the hinge getting formed the chances are that there will be rotation about this point and the wall is going to deflect like this under active earth pressure clear so this is going to be delta l normally the wedge which is getting formed here is defined as ac equal to l and hence the percentage strains which get accumulated in the sample of the material would be delta l by l now in this case what is going to happen again the failure plane develops like this at an angle of 45 plus 5 by 2 the soil mass has a tendency to flow out so there will be a depression over here this much volume of the soil mass gets moved out and gets accumulated somewhere here typical active earth pressure condition alright and I can also assume rather than a sorry b as the pin formation or a you know a pin joint or the point of rotation I can also assume the mechanism number 3 so this is mechanism number 1 of development of earth pressures number 2 and number 3 would be you must have studied sluice gates in hydraulics so something of that sort if this is the wall so point A acts as a hinge these are assumptions so if point A acts as a hinge under active earth pressure condition was going to happen this is how the failure is going to take place wall is moving away from the backfill there will be a depression like this of the material so this volume gets accumulated over here this is the third mechanism of the failure where the hinge formation is taking place at point A alright so considering all these types of models these type of equations have been derived there is another theory which is proposed by Coulomb and what we call this as the Coulomb's rigid block wedge theory so basically this theory deals with the free body diagram of the wedge which is getting formed in the plastic equilibrium state and then by using the concepts of equilibrium you can solve it this part we will be taking up later on alright now this is of some special interest if I ask you to draw the pressure variation and suppose if this is the wall of height h and if I consider the z value from top and if the height of the wall is h so what you are observing is there are two components in the earth pressure alright so when z is equal to 0 what is the value of the earth pressure minus 2c root ka correct so that means this remains constant all throughout so this becomes the surcharge because of the cohesion of the material and this plus the first component which is a triangular variation you must have done in hydrostatics so this is equal to gamma into z and z will become now h multiplied by ka in short you can analyze this problems very easily if you have the pressure diagrams you know what is the CG of the application of these forces so this will be at h by 2 and this is what is going to be at h by 3 but normally we do not do it like this so what we are deciphering from this equation is that the total active earth pressure acting on the wall will have two components if I club them together this is how they will look like have you come across this type of a pressure distribution somewhere in civil engineering where beams so what is happening here in case of soils because of the cohesion which is getting mobilized in the soil mass there is a tendency for a tension crack to get developed so this is the z0 value which is defined as the depth of tension crack any cohesive material if it is compacted and left on its own in the due course of time the active earth pressure develops and during that what is going to happen this much portion of the wall is going to get exposed to the atmosphere because of the development of the tensile stresses now if I integrate this term to get the capital PA so this is equal to force per unit length of the cross section alright so this will be equal to 0 to h this tub so that means I am considering a point somewhere here of finite thickness dh and its length is per unit length let us say 1 meter so if you do this exercise what you will be getting is half gamma zh square minus 2ch root of ka alright so there is something known as a unsupported depth of the cut so this will be equal to 2 times z0 and I can prove this now so if this is the total force which is acting on the system I will try to find out when this becomes 0 so this will be gamma into z into h root of ka will be equal to 4 times c this is ok now this h is the total height of the wall and this z which I am considering is equal to the unsupported depth of the cut so basically z will be equal to 4 times c upon gamma into h into root ka and hope you can realize that this value if I substitute over here as z0 and z0 is the point where the point pressure is 0 so from this expression I can obtain gamma into z0 into ka equal to 2c root of ka and hence z0 will be equal to 2c root of ka upon gamma into ka is this alright so this becomes 2c upon gamma into this z actually will be what so this you have this will be delta z sort of a thing so this is alright not fine yeah so now what will happen is this will be h and gamma now it is ok now this z will not come so this h will be basically equal to 2 times z0 good that you pointed it out so now this term will be it is ok now root ka will be it cannot be because root ka I have cancelled it this is actually I wanted to write here ka term fine that gets cancelled so basically what we have proven from here is that the total depth of the unsupported cut is equal to 2 times the tension crack so this concept of tension crack I will be utilizing quite a lot so AO is defined as the tension crack what happens at this time when the tension crack develops in the soil mass this much portion of the soil will not be in touch with the wall and this is how the tension crack develops so this becomes a crack and this crack gets prone to accumulation of rain water so what is the message which we have learned from this simple null sis 2 messages 3 messages first thing is if you are dealing with a C5 material as a backfill material and particularly if cohesion component happens to be very high clay material it is prone to develop tension cracks which is not good for the health of the structure why number 1 you cannot compact the cohesive material as a backfill material number 2 it will consolidate number 3 there could be a water accumulation during the rains fine number 4 the entire system will settle because of self fit what we call it as a self fit consolidation clear it is not a freely draining system so that is the reason normally we avoid using a cohesive material as a backfill material.