 You must have noticed when you go to the offshore environment particularly let us say marine drive in Bombay, you must have gone and sat on the parapet wall and what you see on the seaside is tetrapods. Any idea why tetrapods are designed over there, one is of course they act as a breakwater. Number 2 of course they avoid erosion of the soil. Number 3 they will not allow the entire force of the wave to come on the wall. Number 4 which is more mechanistic and the answer is here and that is the beauty of you know practicing of geomechanics. So a careful understanding of this diagram will exhibit if I can nullify this by putting overburdle at this point. So why tetrapods are kept? You want to negotiate with this minus 2C root of K which is the culprit because this creates a easy path for environmental activities to happen inside the backfill. In cold countries what will happen? The water will enter, it will become ice clear and subsequently we will study that this block is going to be acted by the water pressure on the surface. So the shear stress is going to be 0 because of the cracking, the hydrostatic pressure is going to act on this block of the soil which was retained by the soil mass. So one of the ways to get rid of this would be you go for nullification of this by applying external loading. This could be a sort of a tetrapod. I hope now you can design the tetrapods easily. So the external pressure which you are going to apply on this soil mass will be equal to 2C root of Ka. So this whole term vanishes and you have compressive forces acting in the soil mass life becomes simple. Is this fine? Now I will take simple cases to extend this theory given by Rankine and try to analyze some situations which are very, very prevalent in geomechanics. Hope this part is clear. So what you should be careful about is the transformation of sigma 1 and sigma 3. How do they interchange? This we have been emphasizing since long. What you have to do is you have to simply overload the soil mass. So if this is your marine drive wall alright and this is the parapet on which you go and sit. So what you will observe is they put tetrapod over here. Why? Because of this reason. So this 2C root of K component I can nullify by applying this loading alright. Tension cracks will not develop. Imagine if this wall there is a tension crack which develops on this side the water will ingress and all sorts of problems will start. So let us try to attempt to solve now the application of Rankine earth pressure theory. The wall which I had drawn is known as Rankine's wall. Vertical wall having no friction backfill sloping 0 and this becomes a typical Rankine wall. So if this is the wall alright this is the pressure distribution because of the total wall and what we depict it as is PA and PP if the whole wall is H this will be acting at H by 3 and under this condition this excess of the pressure will be gamma into H into K. And because we are dealing with active earth pressure this becomes K and when we are dealing with passive earth pressure this becomes PP or rest of the things you can do easily until now you have been dealing with the fluids, water you have done lot of problems in mechanics where the water or the gas is contained in a container and then you are trying to find out the pressure coming on the vertical plane or horizontal plane depends upon the problem. So in case of water what is going to happen to K parameter this will be equal to 1 correct. So K parameter will be equal to 1 and hence this becomes your simple hydrostatic case alright and then you can solve the problem. Now the complications will start let us say now this point onwards do not write and just look at the board and try to understand once for all how the real life situations are dealt with. Now suppose if I say that there is a UDL fine uniformly distributed load what is going to happen? So you have the soil mass in which this type of pressure is developing active or passive depending upon the condition. Now this UDL acts as a surcharge and suppose if I define this as Q kilo Newton per meter or meter square depends upon the type of so because this is a line loading I will take it as a okay it could be aerial loading also so net effect gets added to this in the form of Q multiplied by K or KP that is the only difference a better way to understand this would be this whole situation can be transformed like I can always go for equivalent height of this Q provided if it is in kilo Newton per meter square what I can do I can replace this whole thing by a situation where this is a wall of height H and on the top of this there is another layer of the soil which is compacted and let us say this is the height Z. If the unit weight of this material is known what I have done I have done a equivalency between Q and Z fine so I can compute the height of the wall which is got extended because of the Q loading as Q upon gamma this type of transformations normally we do whenever we require whenever the situation calls for it alright. So no difficulty this much portion is going to create this pressure fine what is the axis here this axis will be equal to if gamma is known this will be gamma into Z into K depending upon active or passive situation if the unit weight of the material and C5 parameters remain constant right now let us I am dealing with the pure frictional material to make things easy if this is phi this is also phi there is no contrast so what is going to happen this overloading will remain constant and if you remember in the previous case what we have done we have taken this as a overloading because of cohesion component now this is the overloading in terms of the first layer plus the second layer so this is how it looks like so this remains as gamma into Z into gamma into H into K and this component gets added up to this so this is gamma Z into K and gamma Z is nothing but the surcharge which is acting on the system so I can say this is Q into K in total this will be Q plus gamma H into K which is equivalent to this function the rest of the things are simple mechanics you can find out the pressure you can find out the force and you can apply the equilibrium conditions that will come much later so if you venture into the real life situations I will create few situations and I will like to analyze that what we call it as a effect of water table in the backfill if there is a wall here of height H and this is the backfill and suppose there is a water table over here now what is going to happen it is a situation when it has rained overnight and water got logged in the backfill so is this situation good or bad that is what actually we are trying to analyze there is a complete water logging sometime back if I said had it been a C5 material with the predominance of C what would have happened the whole system might become saturated a saturated material under self weight is going to consolidate very difficult to compare cohesive materials cracking which we discussed the moment crack develops water seeps through and exerts pressure on the wall clear so those are the complications now when we analyze this type of a situation particularly for pure frictional materials because ranking theory is valid mostly for pure frictional materials what we did is we extended it to C5 soils in general so rule of the thumb is always filter out the effect of water remove the effect of water first so it is a case of submergence fine so the pressure diagram would be I use the word submergence this will be equal to gamma prime submerged multiplied by H multiplied by K and where the effect of water will go that can be added up to this so this gets added up in the form of the pressure from the water so this will be equal to gamma W into H so the total effect would be the submerged or the effective pressures which are going on the system and the water fine what we call it as a this is what is known as a purged water table now let us talk about a partial submergence water table drops down to this level so this is a partial submergence the moment partial submergence occurs the backfill behaves as if there are two layers of the system that is it that is for the principle remains same fine so what is going to happen this much soil mass is going to be a function of type of the soil and type of the soil is going to govern the capillarity yes or no both agreed so the more and more I compact the material what I am going to do I am going to create a capillary zone and this part we have discussed partially saturated system starting from this point this point below everything is submerged so drawdown of a water table creates a multi layered system in the soil mass what type of multi layered system this is creating even if the material property is C5 remain constant what is going to change is gamma so gamma could be a function of that depending upon the type of the material is this part clear so the simplest possible situation would be when you are dealing with sands coarse and not fine sands even fine sands will have some capillarity okay so coarse sands what is happening that is a special case when you have coarse sands as a backfill material and the water table drops in no time this whole soil mass will become dry and this remains submerged that is it so I have created a multi layered system where in you have the wall and this wall is having a layered backfill so let us say this is phi 1 this also remains phi 1 no issues this becomes gamma dry and this becomes gamma submerged because of water table so this is Z1 this is Z2 does not matter even if I have to superimpose the effect of cohesion I can take C1 and C2 so for the sake of simplicity we will not consider cohesion in the picture right now just to give you one concept and once that concept is clear you can solve any problem now suppose if I overload the system by Q what is going to happen another layer gets added up over here so this becomes equivalent surcharge a layer of thickness Z made up of that material or different material acting as a surcharge so this concept is normally utilized when we do 3 dimensional consolidation design band drills or PVDs because after inserting the PVDs you have to preload it by creating this pad alright so in continuation with this now let me introduce the last concept and after that you will be quite good enough to apply this anywhere in the life so I will consider a simple case of a layer deposit suppose this is phi 1 gamma 1 Z1 and this is Z2 phi 2 gamma 2 so if you have followed all this discussion which we discussed until now should not be very difficult for you to catch up with what I am going to discuss now if I ask you to draw the pressure diagram what is going to happen starting from this layer the first is this axis which will be equal to gamma 1 into Z1 into K1 it will be K1 for active earth pressure Kp1 for passive earth pressure so I am not writing all those things now suppose if I put a condition that phi 1 is greater than phi 2 what is going to happen so the moment phi 1 is greater than phi 2 what is going to change is K parameter correct so K is going to get changed so that means K1 is going to be more than K2 is this okay if friction is first one is layer is 1 more so this is going to be like this this okay yeah 1- sin phi over 1 plus sin phi so what it indicates is when you have a system of soil mass which is having a higher value of lesser value of friction angle how this pressure is going to get redistributed in the second layer so gamma Z1 X as a Q if this condition is valid K1 is going to be lesser than K2 that means you will be having a stepped out function like this and this will be equal to gamma 1 Z1 into K2 because K active has increased fine reverse situation if I can create a material where phi 1 is less than phi 2 what is going to happen this K1 is going to be greater than K2 and this kink which has gone out will be inside and this becomes the second situation this will be the first situation so we have got we have taken care of the first layer and what is going to happen because the first layer equivalent model of this would be if I can shift it over here this becomes a wall and this whole thing I can superimpose as Q1 but then the material is lost there because C phi cannot be taken into account that is only static pressure and then followed by this layer so this is going to be equal to gamma 2 Z2 into K2 if I want I want I can make it more complicated now what I will do is the whole thing is submerged no issues you can solve this problem filter out the water so the way we did it over here everything becomes effective clear submerged and superimpose the effect of water on this that is it so the moment this happens you deal with effective stressors effective weights and superimpose gamma w over there partial submergence no issues fine depends upon the type of material you have capillary fringe saturation find out the gamma Z as a function of depth of the wall superimpose it over here so this way you can now deal with multi layered systems the question is where I am going to use this concepts so from the next lecture onwards what I will do is I will start analyzing more complicated real life situations which are known as sheet piles so suppose if I say that this hypothetical wall becomes a sheet which is embedded in the ground fine this is what we will analyze so I have created a three layered system 1, 2, 3 and this side yes we have created other type of system so it becomes a multi layered system but the basic concepts remain same.