 So, we have been talking a lot about the Rankine wall where we have assumed that the backfill happens to be pure granular material. In practice many times it happens that you have to retain cohesive soils because of the lack of availability of a pure granular material alright. So, under these conditions though it is not advisable to have a backfill material as a cohesive material, please make it very sure alright. So, we are just doing a hypothetical analysis to make ourselves understand that if cohesion is present in the backfill what would be its implications alright, otherwise by definition backfill material should always be a free draining material of very high permeability so that consolidation does not become a problem alright, long term consolidation should not become a problem water retention should not become a problem. The type of analysis which we have done in the previous modules where we have shown poised water table and the submerged situation where the water gets logged into the retaining wall these type of situations can be avoided by using highly permeable granular material alright. So, let us do analysis of a situation where the backfill happens to be a cohesive soil is a C5 soil let us say and it will be very interesting to see the case of you know a passive operation. So, if this is the trial wedge and we have all those conditions which we have considered earlier there is a Qs height of the wall is h inclination of the slip surfaces theta ok, we are trying to find out the passive earth pressure on the system this is the W weight and this material happens to be a C5 material that is the only difference draw the free body diagram of the forces which are going to act on the system so this part you are very well conversant with you have the normal force acting on this system then we have a component of the shear. Now this happens to be a passive earth pressure so shear force is acting downwards and then the component of C comes into the picture alright. So, the component of the C is getting mobilized in this form which will be equal to C prime multiplied by length of AB is this ok. So, this is the only difference which we have made and LAB is known this is h upon sin of theta now we can solve this problem by using the simple trial wedge analysis. So, try to work out the magnitude of PP what this should be equal to the first component is suppose if I take a worse situation here also if I allow water logics ok. So, the first one would be effect of water if I take it on the left hand side this becomes PP minus half gamma W h square PP prime effective earth pressure under passive conditions plus what is the component of what is the contribution from the soil go and force h square into k p term plus what is the contribution from the C. So, this will be equal to 2 times C prime h root of k p and what else is remaining the surcharge. So, the surcharge will be in the form of q s into h into k p is this fine try to solve this problem or this situation by using trial wedge analysis you know all the forces which are acting on the system and then compute this alright any questions now let me introduce the concept of tension crack in a C5 soil which you have studied already and if you remember we had talked about the soil mass of height h alright and if this happens to be a C5 type of a soil with unit weight of gamma ok what we did we drew the pressure diagram and this is how the pressure diagram looked like and this is the z naught the depth of tension crack. So, here the tension is prevailing here the compression is prevailing in the soil mass this is alright and the z naught value we computed as 2 C upon gamma root of k alright if you are working in pure cohesive soils k term disappears this becomes 2 C upon gamma if you are working in a C5 soil this is the contribution of C this is the contribution of phi in the form of k a. So, we get the value as 2 C upon gamma root of k a and if I multiply this by 2 so 2 z naught is equal to the critical height of unsupported cut alright. So, this concept we are now going to use for analyzing the stability of the walls for finding out active and passive earth pressures is this part ok. So, if I extend this logic to the stability block over here what you will realize is now this is the slip surface this tension crack I am depicting as a crack over here alright. So, this becomes C now C is this D and E. So, this is the tension crack which is developing because of the inherent property of the soil mass being cohesive what is going to happen during rains the water will seep through water gets logged over here and then the water pressure is going to act on the block and which is going to destabilize the entire system alright you remember all these things another property of E and D is this happens to be a free surface now what is the attribute of the free surface this surface has got detached from the parent block alright from the parent wedge that means this is a atmospheric surface there is no reaction which is going to come on the surface in the form of the normal stress that means if you look at the free body diagram of E B D alright here the normal stress is 0 and the shear stress is 0 it is a free surface clear. So, this is the case which is defying the law of mechanics the weight is acting downwards and suppose if this happens to be pure cohesive material this cohesion is going to balance. So, we have 2 forces and under which the body cannot remain in equilibrium. So, what is the way out we have to assume that this block is of infinite decimal weight where weight is tending to 0 then only it is possible this is part clear. So, because the weight is the component which was dealing with the induced cohesion and the way to justify this would be if E D B is infinite decimal that means the weight of the block is tending to 0 then only the equilibration can take place that means 0 weight no cohesion gets mobilized. I can extend this idealization in such a way that I can assume that the mobilization of the cohesion at point B is 0 when you reach the point D it becomes the full mobilization of the cohesion. So, this is the C prime value the maximum cohesion alright and beyond this point what is going to happen the C prime remains uniform. So, this is the variation of C prime starting from C equal to 0 attaining the maximum value at the tip of the crack which is known as tension crack and beyond which C prime remains constant. So, what we will do is we will try to utilize these properties in such a manner that they could be useful to us the total length of the wall is or height of the wall is h ok. This we have computed as Z critical alright is this ok. So, this is your Z critical which I had defined as Z naught or let it be Z naught and this is equal to 2 C prime by gamma root kA fine. Now, can I find out the average C which is going to act on the surface AB. So, C average which is acting on surface AB this will be equal to how would you compute the average C value total length of the surface is this ok I am assuming this as theta. So, h by sin theta is the total length what about this triangle this will be half C prime into this value. So, this will be half C prime into Z critical or Z naught upon let it be Z critical then because standard term is Z critical ok. So, Z critical divided by sin theta plus the rectangular portion. So, this will be equal to C prime h minus this portion goes out Z critical upon sin theta is this ok. So, if you solve this expression what is that you are going to get solve it quickly. So, this will be equal to C prime 1 minus ZCR open here divided by 2h this is right this is ok. So, this is the average value of the coefficient which is going to get mobilized on this surface why we have done all these analysis. We have done this analysis to make sure that we are using the right value of cohesion which is getting mobilized on the slip surface of a C phi retained soil mass for finding out the earth pressures. Suppose the soil happens to be pure frictional clear what is going to happen pure frictional soil no tension cracks C prime is equal to 0 ZCR equal to 0 if ZCR equal to 0 that means the average C is equal to the mobilized C prime this is part ok. And C prime can be neglected because what we are doing is we are assuming C prime to be tending to 0 for pure frictional material. So, value of C average will also be 0. So, this becomes a sort of a correction factor that is the critical depth of the tension crack divided by 2 times the height of the wall. Now using this concept we can find out the active earth pressures and passive earth pressures both. So, suppose if I give you the value of C average now can you write the expression for active earth pressure case because that is more interesting alright. So, this will be half into gamma w h square plus 0.8 half gamma b h square. Now what is going to happen to the C prime case that C prime gets replaced by the value which we have obtained. So, this becomes we are doing active earth pressure case. So, this will be 2 C prime or average value and average value is equal to 1 minus ZCR over 2 h into h into k a root of k fine. So, this term becomes what how would you define this term? This is basically average effective cohesion which is exhibited by the soil mass. Similarly, you try to find out the P P also. So, there is no difference I think you can just substitute the terms accordingly and you will see that most of the parameters remain same in this expression and we can obtain from P P also. So, this will become plus 2 C prime 1 minus ZCR upon 2 h into h into root of k p and then these terms will also we have done a mistake over here. So, this should have been k a this will also become k p.