 So, so far we have discussed few cases of finding out the earth pressure by using the trial wedge method. And remember that we have talked about only Rankine's wall that means the wall is smooth and vertical. There is another interesting application of the trial wedge methodology would be when we talk about analysis of completely submerged retaining walls and the backfill. So, I am going to discuss this about today. And this is the retaining wall and there is a surcharge on this. So, this is the surcharge and this happens to be completely submerged. So, the water table is up to this point. The height of the wall is h and we want to analyze it for let us say active earth pressure case. Now, this P could be Pa or this P could be Pp, we will complete the trial wedge. So, this is at angle of theta abc and because of the submergence what is happening is I can find out the weight. So, weight will be equal to half into gamma. Now, the question is which gamma I should be taking because it is a case of submergence. So, this gamma gets replaced by gamma T that is a total unit weight alright and this is multiplied by h square into cot theta. So, weight is known. Now, let us draw the free body diagram. The free body diagram would be there is a normal stress acting over here and why effective because it is a case of submergence. So, we have to superimpose here the pore or pressure in the force form not the pressure form and then there is a component of the force which is acting in the form of shear stress. So, this is the free body diagram of the system. Now, rest of the things remain as it is what we have been doing since long. The only question is how to obtain the u prime and remember n will be equal to n prime plus u prime. So, u prime is the pore or pressure which is getting developed in the system because of the submergence. So, a simple way to do this would be if I find out the pore or pressure at point A and pore or pressure at point B. So, remember the previous course if I want to find out the pore or pressure at point B what I have to do is I have to insert a piezometer and if you insert a piezometer over here what is the height of the water column 0. So, pore or pressure at B is 0 alright and what is the pore or pressure at point A? This will be equal to if I put a piezometer over here the height of the rise in the water tube or the piezometric tube would be H. So, this will be equal to gamma w into H alright. So, the variation of the pore or pressure from B to A is triangular starting from 0 attaining a value of gamma w H. So, if I want to define u as the force now this u as the force would be you need not to write it as a prime u will be equal to average of u A and u B. So, this is equal to half into gamma w into H alright this is in the pressure form. Now if I am trying to find out the u if I say capital U let it be small u force now this will be equal to this pressure is acting on this surface AB alright. So, this will be equal to half gamma w H and what is the length of the surface if this is H this is theta. So, this is equal to theta H by sin theta correct multiplied by 1 perpendicular to section. So, this is equal to H by sin theta into 1. So, this comes out to be half gamma w H square by sin theta this is the pressure pore or pressure which is acting on the slip surface AB because of the submergence is this part ok. Now rest of the things are simple mechanics put this condition sigma fx equal to 0 and sigma fy equal to 0 and solve this function. So, this is theta this will be T cos theta this will be T sin theta. Now we have here the two components like this ok and then we can compute the values of if this is theta this is 90 minus theta. So, this is also going to be theta. So, if you solve these two functions and one more thing which we require is we require a relationship between T and n prime. So, T and n prime can be written as T equal to n prime tan of theta this is part ok yes sorry I am sorry this is going to be phi. So, this is going to be phi prime ok the major unknown is n prime. So, what we have to do is from these two equations we have to obtain n prime and one of the ways to get rid of this would be I can substitute for the T value when I am taking the components equal to n prime tan phi prime. So, you will be getting two equations and you can solve those two equations and hence you can derive the relationship is between the p because the first equation would be fx equal to 0. So, this will be equal to p plus T cos theta equal to n prime plus u sin of theta this is fine and simply by using the resolve components of the weights I can take this also into account as qs into h into cot theta. So, both the components would be qs into h cot theta plus w w is known and this will be equal to this component cos theta now you solve this expression. So, ultimately what we should be getting if you if you solve these two equations what is that we are going to get we are going to get the values which are known as the earth pressure p which can be obtained alright you have to just do simple mathematics to obtain the pressure functions anybody in the class what will the final expression we have filtered out the effect of water. So, the p will be equal to half gamma w into h square plus when you are filtering out the effect of water what about the surcharge and the weight which is going to come. So, this will be half effect of buoyancy gamma bh plus qs is this fine h cot theta this has been so far the normal expression which we have been using in all these trial batch analysis and what is the component which is missing the 10 component hope you can realize what are the different components which have been presented over here in the form of mathematical terms. So, because of submergence we have the buoyant weight though we started with the total weight of the block the pore or pressures have been taken out from the component. So, when you do this type analysis what is going to happen this will be the pressure because of the water surcharge plus buoyant weight of the material. Now, suppose if you want to find out the maximum pressure which is acting on the system maximize this function that is it all right. So, when you maximize what will be what is that you will be getting the known terms that is half gamma w h square plus the effect of buoyant soil mass. So, this will be half gamma bh h square into K a plus the influence of surcharge which is acting on the system that is qs into h into K a is this okay without doing all this analysis also you would have been able to obtain this function just by using simple principle of superimposition 3 pressures which are being taken into account 1, 2 and 3 and that you are aware of fine. Now, this analysis we can easily extend to the cases when we have partial submergence in that case what is going to happen any guess the block or the trial which we have considered is now going to be a composite of 2 sections. Now, rather than having water table here the water table has moved down to this place. So, what will happen to the weight if I had written here a b c b and e the weight will be having now 2 components. So, this is w 1 and this is w 2 and we have discussed enough about the state of the soil mass in this block in the first course of soil mechanics. So, depending upon whether you have a sandy soil compacted clays, salty soil or whether I assume a variable saturation in this zone above the water table or completely dry mass long term short term stability if you remember you know we can attribute gamma values to w 1. So, this gamma could be dry this could be gamma partially saturated I can include the effect of capillary action also here. So, if I include the effect of capillary action what is going to happen there will be a going to be a zone of capillarity ok. So, we have now w 1 getting divided into w 1 1 and w 1 2 that can be taken care of there is no other difference that is all the things are almost similar fine. So, this is another situation which normally we come across in the real life. Now partial submergence could be because of several things this could be you are basically pumping out the water during construction and maintaining the water table over here it could be because of the drainage which is going to take place and hence the water table does not remain up to the top of the backfill under any circumstances the analysis can be done very easily. It is a good example of how would you extend sample analysis which is being used trial batch analysis for finding out the earth pressure on the system. If you remember the way we computed the stress acting on the soil mass I can also use the principle of super imposition over here how I know the pressure which is coming because of the entire system of height h and then I have water dropped up to this level. So, I know what is earth pressure because of this. So, in that case h gets divided into two parts we can have the h 1 as the height of the free water table alright and then we know the pressure distribution because of this and then from this point onwards what has happened the water table is only up to here. So, this is the earth pressure because of the soil mass the water table is coming only from this point onwards. So, this will be equal to gamma w into h 1. So, this is also you can do by all circumstances what is going to happen the p will be equal to half k a gamma h square fine subtract the effect of h 1 height. So, minus 1 k a into gamma h 1 square and then what is there the buoyant force plus half k a gamma b into h 1 square try to prove this function by using the sample mechanistic laws any questions any doubts fine. Now, there is another condition of purged water table in the system. So, when we are dealing with these type of situations the question is how would you construct the retention schemes or the retention system retaining walls which is going to be beneficial. And suppose if I take a case that this is the retaining wall we know that the critical failure surface is going to be like this idea is to minimize the earth pressure which are going to act on the wall alright. And suppose there is a rainfall which is taking place. So, because of this a situation like this or this might develop. Now, what you want to do is we want to minimize the earth pressure which are going to act on the system during rains submerges I have 2 options with me the one is I will create this u prime or the u value as 0. What is happening because of this if you take moment about this point of all the forces you will realize that the water which is present in the retaining wall in the form of the pore or pressure has a toppling effect correct. That means this u has a tendency to topple the wall the more this pressure p is going to be higher because then you require more reactions to balance this. So, in real life the best thing would be if you can create a situation so that the pore or pressure becomes 0. How will you do that in real life pore or pressure can be maintained 0 along AB which happens to be a slip surface when I provide a filter overly there. So, suppose if I create a layer of filter like this filter media sometimes you also call it as a drain you have done these things in your first course where we had talked about the non-homogeneous earthen damps where we had provided a filter layer to act as a tow filter alright. Now because of this what is going to happen if I ask you to draw the flow net for the situation which you have already done in your first course see page analysis what is going to happen here draw the flow net this is a flow line ok and this is how the flow net will look like these are the flow lines what about the equipotential lines equipotential lines are supposed to be cutting them perpendicular to each other I am sure now many of you will be unhappy to see this type of flow net why we are defining the conditions of flow line intersecting the equipotential lines at most of the places this is done look at this point at this point by virtue of being a filter media the pore order pressure is 0. So, this happens to be an equipotential line this is exposed to the atmosphere the surface. So, this is also an equipotential line alright. So, equipotential lines are cutting each other not correct flow lines you are seeing the ambiguity equipotential lines are also intersecting the filter layer line fine. So, it is a rough flow net which we have drawn just for the sake of understanding it is not the correct of flow net. Another issue is when we are creating a filter media layer like this it is extremely difficult to execute this type of a filter drain in real life during execution on black you can draw this very easy, but it is not very easy to execute this type of a filter drain in the field. But still what people do is they do for go for layer wise construction and then by maintaining the geometry they provide this type of filter drains, but I hope you can realize it is not going to be very easy situation. So, if I create this type of a filter drain and if I ask you to find out the pressure this is pressure 1 alright coming from the ball onto the block. What is another way of providing a filter drain in the retaining walls which is normally done the second situation would be which is rather easy to execute in the field the filter layers are provided in the form of a vertical chimney and this is how they are connected to the outside atmosphere even this can also go outside the atmosphere. So, this becomes a vertical filter drain and suppose if I ask you to find out the pore water pressure sorry the earth pressures acting on the system this is going to be P2. Each point along this drain is exhibiting the pore water pressure equal to 0 alright, but not along the surface of the slip. So, AB the pore water pressure is still acting as U what we have shown over here agreed. So, for the sake of construction feasibility this system is very good. Now if I ask you to draw the flow net over here this is an equipotential line because this is attached to the atmosphere. So, you have flow lines you know coming and intersecting like this and what about the equipotential lines now this becomes a difficult situation, but the equipotential lines are also going to intersect somewhere like this perpendicular to the flow lines. So, this is how the flow net has to be completed and then you can find out what is the pore water pressure acting at each and every point along the slip surface you can still use this function to average it get the value of U and do the analysis. The common sense is in second situation because the pore water pressure is acting the P2 is value is going to be higher than P1. So, you have to design the system accordingly this is fine.