 I will start discussion on a new topic which is stresses in the soil mass due to external loadings. Until now we have discussed about the internal stresses in the soil mass and these internal stresses are let us say pore water pressures or because of the self weight. Now what we are going to do is we are going to talk about the situation where I would like to know how the soil is going to respond when external loading is applied to it. Now in engineering applications this concept has lot of significance and most of the problems which are being tackled by the consulting geotechnical engineers would require these type of concepts to be practiced. So let me first bring to your attention the types of situations where the stresses in the soil mass due to external loadings become very, very important. The first situation could be or we may say this is the real life. Say I have the ground surface somewhere here there is a footing which is being placed. So this is the column and this is the foundation and this happens to be the soil mass. It is understood that this column is going to take load P and hence the footing is going to transmit a pressure Q on the soil mass. I would like to understand what is the state of stress at this point P and this point P can be defined in such a manner that I can say that this is the radial distance r and the depth of this point is z. So basically it has the coordinates r, z I can include theta also here it could be r, z and theta. So this is situation number 1 when you construct foundations I would be eager to know what is the state of stress in the soil mass. There could be another situation like suppose if I have a multi-storey building, so this is a multi-storey building, situation like let us say copyright and most of the populated cities in the world where the cost of land is extremely high. So you cannot afford you know higher spacing between the buildings. So suppose if I ask you a question what should be the safe distance at which I should be designing or I should be constructing another building. Remember space is extremely important very expensive alright. So this is the building which is sitting on the foundation system. Now because of construction of building B if the building A was already existing what has happened? The state of stress at a given point P which I have taken over there has changed. So PA has changed to PB alright and I would like to understand whether this system is soil is capable of taking this type of loads or not alright. So this is again the soil mass. There could be a situation number 3 where I am sure when you travel all along the highways or the railways what you observe is there are the pipelines which are carrying water. So suppose this is the embankment of the railways, so this is the let us say embankment on which the train is moving is an embankment made up of soil. And this is a railway track it could be an airstrip also or this could be a highway, a pavement and somewhere you must have noticed what we do is we carry let us say some utilities. So this is the utility pipeline and I want to see whether do not go by the elevation this could be you know x this could be y most of the time x is going to be less than y and hence I will try to understand because of the movement of the traffic what I have done is I have simulated 3 types of traffic one is the railways another one is the aircraft the third one is roadways. So I would like to understand whether this pipeline is safe or not that means because of the movement of the vehicles here what is going to happen to the pipelines alright this could be buried pipelines also. So I would like to find out what is the state of stress somewhere here again the point p shifts now point p could be anywhere in the domain and I would like to do the analysis. There could be a third situation we have been talking about fourth situation this is a ground surface and as we discussed in the last lecture the application of sheet piles I would like to create a bulkhead over here so this becomes a bulkhead this is the soil mass or water this could be storage and I would like to know what is the state of stress at a point let us say somewhere here somewhere here somewhere here now suppose if I remove the piles this is a sheet pile and if I say that in a simple situation where I have the ground surface I would like to retain an embankment the same embankment which we have created over here so this is the embankment and this embankment can be utilized for creating let us say oil foundation, tank foundations, storage whatever so suppose if I want to find out the moment I have created this embankment what is the state of stress at different points in the soil mass so p might be varying here you know p, p1, p2, p3 and so on. What has changed over here is that the type of loading which we have has become a triangular loading at this point the loading intensity is 0 as you go this side the loading increases to q max and then it remains constant so this becomes a typical loading condition which we are imposing on the soil mass there could be another situation where we might be doing excavations so this is let us say the loading of the soil mass and this could be let us say excavation I would like to excavate the soil and I would like to create a facility like this so this is the initial ground surface and this becomes the final ground surface if I remove this much of the material this much material has been removed and I want to find out after removal of this material what is the state of stress at this point because ultimately this system has to stay stable is unsupported here this is supported with the help of a sheet pile alright so these are different examples of the situations which we might come across in today's you know civilization what I want to know is what is the state of stress at point P define the state of stress point P and why we want to do this because if I know the state of stress at point P I can link it with the yield strength of the soil you may think of any other situation which you think is practical I will create a situation like this also suppose if I do a let us say detonation of an explosive alright so what I will do is I will bury the charge over here and I can charge it so what is going to happen this is going to create shear waves compression waves and because of this the state of stress is going to change at point P so these are very critical situations which you may come across alright so what we will do is the general principle remains same except for this situation where I have talked about the blast loading which is a different scenario altogether rest 1 to 6 we can easily solve by considering the concepts which I am going to discuss in the class today alright and as an undergraduate student as a beginner in the subject you should be aware of situations 1 to 6 how to analyze them fine the best way to define the state of stress in the soil mass because of the external loading is by considering the Bosonist equation or Bosonist philosophy so as I said we are talking about the stresses in the soil mass due to external loadings this is where the Bosonist theory is used in 1858 this was proposed by Bosonist but this theory states is that for homogeneous soil mass which is isotropic in nature which follows the constitutive relationship that means elastic modulus is defined Poisson's ratio is defined and hence we are assuming the soil to be linearly elastic material okay one interesting thing is that this soil mass has been assumed to be weightless did you notice this that the soil mass has been assumed to be weightless why we are making the soil mass divide of its weight because weight causes the internal stresses in the forms of either cell plates or the quarter pressures so it has been assumed that as if we are computing the state of stress in ether sometimes these type of soils are also known as ponderable soils check it on net what is the meaning of word ponderable these soils are talking to themselves they are pondering they do not know the way they have been modeled their properties are different but somebody is modeling them in a different manner so if these relationships are valid homogeneous isotropic constitutive relationship is valid weightless material and the point load the external load which is acting on the soil mass is a point load where do you come across point loads as I said foundations of columns so the moment I say raft this is not a point load so foundations of the columns are point loads alright if I convert it into a raft this becomes a uniformly distributed load area wise so as per this if you have the soil mass which is exposed to an external loading of q and if I want to find out what is the state of stress at a point p in the soil mass the radial distance is r this distance is capital R another assumption here is that the soil mass happens to be semi infinite soil mass that means the extent of the soil masses semi infinite we call it as a semi infinite soil mass alright soil mass happens to be semi infinite semi infinite means if this is the direction of z this is the direction of x and if this is the direction y we are considering the domain so soil mass is nothing but the domain in which we are doing this stress analysis this happens to be all right semi infinite so z is tending to infinity x is tending to infinity y is tending to infinity so this is a domain in which we are doing the stress analysis what is the state of stress at point p the state of stress in point p would be normally let me change the nomenclature we take this as z and this as y this is z so basically this is sigma z the normal stress which is acting at point p sigma r I am sure you are doing a course in RCC also this semester so suppose if I give you a ring of finite thickness alright and suppose this is a pipe which is carrying some fluid and if I ask you to show the annular stresses we call them as hoop stresses so hoop stresses will be acting in the periphery circumference alright so these are hoop stresses so we have sigma z we have sigma r there could be a stress in the direction also but because I have assumed this to be isotropic in situation and hence sigma z is equal to sigma y clear if you draw the state of stress at this point you have done engine mechanics and you have done mechanics of solids this is sigma z sigma r we will be having the shear forces tau zr and if I want to come out of the this coordinate system I can always use the Cartesian coordinate system if I define this theta and if I say that this is r and theta this is okay so this I am assuming as z this distance and if you look in the three-dimensional at point p there is a component of r which is a function of x and y this is okay if you look in three dimension this state of stress has been defined as this will be x and x and y correct this will be an xy plane because that we have already taken alright so r will be under root x square plus y square so this is the plane in which this point is sitting so as per the bosonics the equation is for the point load sigma z equal to 3q upon 2 pi into z is cube upon r square plus z square to the power pi by 2 this can be simplified to 3q upon 2 pi z square and this becomes 1 upon 1 plus r by z square to the power pi by 2 I think this one is easy to remember alright q is the load z square is the sort of a area so this becomes a pressure term 3 upon 2 pi is the solution normally what we do is we replace this term as ib and this is what is defined as the influence factor alright influence factor ib I am sure you must be realizing that state of stress in the soil mass is totally devoid of its properties what are the properties normally we define the properties of the soils or the soil mass as density void ratio porosity saturation unit weight over the pressure okay what else Poisson's ratio engineering properties these are the fundamental properties bulk properties engineering properties are elastic modulus and Poisson's ratio so these are the soil property none of them appear in this expression because the way the bosonics theory has been derived this is for weight less material alright. Now sigma r is q upon 2 pi this is 3r square into z upon but yes if you can remember that is okay 1 minus 2 mu upon r square plus z square plus z square plus z square to the power sigma theta is defined as q upon 2 pi 1 minus 2 mu negative sign z upon r square plus z square to the power 5 by 2 minus 1 divided by this term if I use this term as let us say capital X this becomes capital X rather interesting thing if Poisson's ratio is equal to 0.5 which is for saturated soils saturated clays particularly sigma theta becomes 0 the fourth stress which we have not still considered is tau zr the shear stress the shear stress would be 3q upon 2 pi rz square over r square plus z square to the power 5 by 2 if you solve this equation you know the bosonics equation which has been used you will get this sigma theta as the negative term the state of compression and you must be realizing that all these stresses are compressive. So if you are let us say if you take a plate and then compress it from both the sides there could be a bulging so the bulging indicates the negative pressure so sigma theta is going to be tensile in nature. Now what we can do is so you can replace all these coefficients with influence factor alright IB hence I can write sigma z equal to 3q upon 2 pi into IB and z square also 2 upon 3 upon 2 pi also I can take into IB and I can say that this is a function of q upon z square now do you find any abnormality in this expression see we started with a point load because the bosonics equation is for the point loads so suppose if I say that z tends to 0 or for that matter if let us say I say r tends to 0 and z tends to infinity where this point would be r tends to 0 means on the z axis and z tends to infinity this is one of the situations in which I will be interested the second equation would be the second thing in which I will be interested is z if tending to 0 r tending to 0 so z tending to 0 r tending to 0 is this point clear so what is the stress at this point sigma z tends to infinity clear so this is the issue because what you are doing is when r and z tends to 0 this becomes a point load the stress at this point is going to be infinity one of the abnormalities of the solutions anyway no problem we as an engineer we will still go ahead with the solutions and we try to see how these things can be sorted out so if you look at this influence chart if I substitute the values of r and b and sorry r and z and if I get the values of ib in that expression for r by z equal to 0 the ib is 0.475 you can compute this for different values up to 6.15 and this will be tending to 0.001 0.001 so for all practical purposes I can write that at r equal to 0 now this is a very interesting solution which we have obtained there are two possibilities of getting this either z tends to infinity then only we can get r by z equal to 0 or a better possibility will be r equal to 0 itself so this is a solution when r equal to 0 we say that sigma z equal to 0.4752 upon z square now the point is this is the solution which normally is used to obtain the sigma z now what I want to do is I want to use the Bosnian theorem or Bosnian relationship for developing three things the first thing which I want to develop is the isobars for the point load we call it as a isobar diagram the second thing is we want to do the pressure distribution so when I say pressure distribution this is your sigma z sigma r sigma theta tau rz this is a state of stress acting at a point p because of the external loading q so pressure distribution on a horizontal plane and the third thing would be to get the pressure distribution on a vertical plane fine so these three things I want to do now let me introduce the concept of the isobars isobars are the contours of equal vertical stresses that means if sigma z remains constant and if I plot the pressure lines or the pressure contours they become isobars so these are the contours of equal sigma z vertical pressure how do you plot them now I am sure you can realize from here if I what is the form of this equation between sigma z and z q is a constant is it not so what is the form of this equation if I write like this z square into sigma that is constant suppose this is y square into x equal to constant what is this graph hyperbola clear so that means all the contours of the stresses are going to be hyperbolic functions that means if I take let us say q and this is the point and this is my z direction if I plot let us say 10% of q as sigma z not sorry 10% 90% let us say where this would be sitting it would be sitting like this please mind the discontinuity at this point which we discussed just now so at this point the function is discontinuous that would look like this it is not a circle it is a hyperbolic function so this is corresponding to 1 as the intensity of the pressure decreases what will happen somewhere here you will have this would be let us say sigma z equal to some 50% of q and this will go up to certain value where I may say that sigma z has become let us say 5 or 10% of the q value because this influence factor is doing all the tricks you are right but see if you look at this point again this is nothing but the point p1 p2 p3 and so on so r is changing z is changing which is getting absorbed in this term ib is a constant term which represents r by z so that means a better way of writing this would be I can say ib into q upon z square the stress term so this I can write as stress term sigma so ib is a function of rz so the basically the contours of these pressure bulbs we call them sometimes we call them as isobars are inclusive of r and z and they depict the state of stress alright now this is one thing which you wanted to do the second thing which you want to do is we want to do the pressure distribution so if this point is q and if I ask you to draw the vertical pressure distribution and the horizontal distribution of the pressures this is the ground surface at a depth of z if I create a plane this is the plane at which I want to see the variation of sigma z as if you are cutting all these pressure bulbs along this axis so if there is a plane which is passing through this this is the same plane as what I am showing over here so the pressure distribution would be like why the rutting takes place of the pavements this is the answer so excessive loading the tire pressure from the vehicles creates the maximum intensity of the pressure at a certain distance z just below the point of application and beyond which what happens it fades out so the moment r tends to infinity what happens to your sigma z value this tends to some constant value at this plane otherwise this is going to be the maximum value so this is the pressure distribution in the horizontal plane as a transportation engineer particularly if I am working on the materials I will be very much eager to know if I am designing a runway let us say I always give these examples when I travel particularly with pilots I do not know whether you are aware or not but most of the runways are defined I have designed based on the layered theory multiple layers of the material sitting over each other this is layer number 1, layer number 2, layer number 3, layer number 4 we play with the stiffness of the material so stiffness of the material is defined as epsilon and sorry Poisson ratio and elastic modulus and the thickness ok. So if I want to minimize the influence of the sigma z on a horizontal plane what I should be doing is I should be playing with this cross section so that I can nullify the effect of sigma z and I can make it a point that at this point sigma z p is going to be much less than sigma yield this is what you will be studying in your transportation engineering course this is how the pavements are designed fine. Now another interesting thing which I would like to do is I would like to plot the variation of stresses on the vertical plane so as if I am cutting this whole thing across a vertical plane so this is let us say a a this was b b so this is b b and the a a would be if this is q these are the contours of this is what I have written there these are the pressure bulbs contours of equal sigma z known as isobars so this whole yellow colour bulb is going to be representing sigma z equal to 90 95% of the stresses which are getting transmitted to the soil mass not the inside part these are the these are the contours of the points you know he was discussing each and every point on this surface is going to be of 90% q value so that is why we call them as contours of equal sigma z fine so as you come deeper the pressure intensity is decreasing this becomes only 5 to 10% of the q value and you can use this expression which I had written over here to compute what is the state of stress and you can plot it over here so this would be a function like this it cannot increase so I have to curtail it somewhere here okay so starting from a very high stress value the sigma z is decreasing as z increases so this becomes the variation on a horizontal plane this is the variation on a vertical plane now depends upon what type of problem you are dealing with suppose I am designing a pipeline barit pipeline there could be a situation that this is my barit pipeline this project I did recently for Santa Cruz airport so I hope you are aware that the airport used to be shut down for several hours during March and February so it is a active runway at runway cannot be stopped permanently but I am sure next time when you are travelling you should see there have been two pipelines which of the oil which have been taken across the active runway that means the landing and taking of operations are occurring and you would like to take out some utility across the runway simple analysis like this can help you I can define what is the zone of influence of the pressures and I can negotiate by locating the pipelines there every time I cannot be so lucky so then you have to do engineering based on the material property than whatever anyway so give you an example suppose this is how the load comes alright and I want to see whether this pipeline is going to get influence because of the external loading or not if I know the value of q I can find out what is the state of stress at the crown of the pipe pipe is made up of some element either structural element would be steel or mild steel or concrete or whatever and then I can compute what is the yield strength of the pipe and then we can design it accordingly fine is another good example of where the application of these type of computations would be in real life so it is always the vertical stresses which are going to be much more detrimental as compared to in plane or perpendicular to the plane stresses but suppose if I am designing a tank foundation which you must be seeing big big tank forms in your structure so this becomes a ring foundation normally the oil tanks are seated on the ring foundation I need not to spend so much money because rafts are going to be expensive what I will do is I will cut them into ring is this ok so as if a ring is bearing the entire load of the tank in this case your sigma theta is going to be very important r theta because this is going to be the hoop stress here so this design will be based on sigma r theta however coming back to your question why we are only interested in sigma z because sigma z happens with the most critical stress which is going to govern the stability of the structure as well as the soil mass in case yield occurs at any of these points your structure cannot remain stable we will discuss these things in the second course fine now let us discuss the concept of superimposition of the loadings how many types of superimposition principles you have studied until now so there you superimpose is it not suppose there is a cantilever and I can superimpose the load in such a manner to get a certain value of the deflection what is the name of this method instruction analysis suppose I started with the first situation there is q1 acting here and there is q2 acting here and find out what is the state of stress at any point let us say somewhere here so the best way of doing this would be you assume them independently find out the state of stress corresponding to q1 first at this point q2 at this point and sum them up so sigma z will be equal to summation of if I have series of loads over here this will become sigma z at point p because of q1 so this becomes i equal to 1 to n have you understood this so what I have done a point load has got converted into a line load practically there is no difference between the situation you know where several q's are acting all along the line okay it could be udl also why not suppose if you take a plane on which there are several q's are acting several q's now if I integrate on the surface this is the integration over the line now if I integrate over the surface this is going to give me the net effect of a loaded area of certain area of cross section so very simply I have converted the context of the situation from point load where we were defying the stresses at r equal to 0 z equal to 0 I have created a uniformly distributed load in the form of the line load and uniformly distributed load on the area it is okay the rest of all integration this is okay so if I integrate over these things I will get the same effect in other words as long as the situation is very simple I can keep on clubbing for q1 q2 q3 q4 q5 q6 and I will say that this is a total sigma that acting at this point so suppose if I ask you to find out the second situation from point load we will move on to the uniformly distributed circular area until now we have talked about a foundation system you know where this is a column so this is the point load the best possible approximation would be that the q is acting at a given point though the foundation also is of finite size but this is approximation as compared to the domain of semi infinite soil mass clear now when I come to circular area circular footings is a good example I might be having a situation like this this is the area which is carrying a uniformly distributed load of q where q is equal to capital Q upon area of cross section and suppose if I ask you to find out what is the state of stress at a given point p so this whole thing is resting on the ground and because of this you want to find out what is the state of stress at point p you have done a lot of integration in your 10 plus 2 or whatever I have in 10 plus 2 so suppose if I consider an element at a depth as a radial distance of r small r this is the radius of r entire area and if I consider a small element over here as dr okay r into dr clear area cross section of this can be known that will be 2 pi r into dr and if I know the loading intensity q it gets multiplied by small q this is a load nothing so great what I have done is the entire circular area which is uniformly distributed loading I have converted again it to a point load and I am assuming here that the circular loading is also an assembly of several point loads acting on the soil mass fine this is okay so your function sigma z equal to 3 q upon 2 pi z square Ib can still be used I will replace this by del of sigma z and now Ib is a tricky term so this is a function of r by b r by r by z clear so this q is equal to this one so I can substitute this value over here and I can integrate so I can get the expressions for the sigma z due to the circular loading is this part clear fortunately I hope you will realize the way we assumed Ib over here in case of point loading if you remember what we did we simply said sigma z equal to q upon z square into Ib okay so this term you can form a table and that table you can always refer to for different r by z and you get the values of sigma z the same thing you can do over here also conceptually things are same another interesting thing sometimes what we do is rather than having a circular footings we might be having let us say strip footings whenever you see your fencing in the hostel or in the buildings you know fencing or maybe compound boundary walls or compound walls they are all sitting on a foundation which goes in the one dimension why this is the foundation part which will be extending up to infinity when we say infinity when the width of the foundation b and thickness of the foundation is much smaller than the length of the foundation clear this becomes a strip footing one dimensional footings because their width and thicknesses are very small as compared to the length is you need not to worry now this again has become a line load situation clear so you can integrate over all intermediate q values and you can find out the stresses acting at a point what is going to change is I hope you will realize that this dr term which is going to come over here when you integrate it this is going to be dr so here r has to be from 0 to r here and what about this z term this will be from the given point to the point where you are interested in finding out the pressure so that means 0 to z same thing you can do over here so this becomes a case of the strip footings circular footings now whenever you make embankments as far as the foundation part is concerned this is nothing but the strip foundations embankments embankments are nothing but the dams which you are discussing in the previous lecture as far as the see page losses was concerned so you have certain specific width at the base slopes 1 is to n 1 is to m if you remember we have to define the height we have to define the top width of the embankment and this embankment also tends into infinite length railway embankment so railways are passing through kilometers of lengths of these type of embankments agreed so suppose if I ask you to find out what is the state of stress at the bottom point no problem you can still model it the entire loading can be now there is a trick the load is going to act over here the railways are going to run over here the roadways are here okay aircraft loading is here we have to delegate the stress at the base of the embankment clear I use the word delegation so for a better word for this is the dispersion of the loading so suppose there is a queue load which is acting and I want to find out what is the state of stress in the gross terms as a designer at a depth of let us say z and I do not have time to do these analysis I mean I am just doing first DPR to tell you what is the cost of the project I do not want to go into the entry cases I do not know how much time is going to take me to analyze all these things what I can do simply there are 2 3 ways of dispersion of the loads the first one is we normally use the concept of 45 degree dispersion angle that means if this is a queue I will disperse it on this plane at 45 degree these are thumb rules so I am sure you can realize now if this is z 45 degree this is also going to be z clear and this is also going to be that so basically q has been delegated to 2 z and then the third dimension is perpendicular to the board so this becomes your loaded area what I have done is I have delegated the stresses from this point to this point so when you design foundations normally we give a foundation pad so what this is going to do I get 2 3 advantages I have created a pad for the foundation system and I have reduced the stresses this is engineering another method of dispersion would be sometimes they have a tendency to disperse it at 30 degree angle I mean do not ask me from where it has come this is just based on people's whims and fancies so this 45 degree becomes a 30 degree angle alright this is also valid for let us say there is a finite foundation of width B thickness and if I want to disperse this load on this plane which is at a depth of z so if I use this angle of dispersion as alpha this is going to be what z tan alpha and this is also z tan alpha plus B so this basically this whole stress is getting transmitted now on a plane which is of area of cross section 2 times z tan alpha you must be wondering how these constructions are done on the marshy lands this is a trick hope you are getting the point mechanistic point is this clear if I expose the marshy lands to direct loading nothing can be done you cannot even stand there but if I create a good foundation pad I can disperse the load depending upon the z this area is going to be extremely higher and hence the bearing can be obtained now something of interest would be uniformly loaded rectangular areas to make our life easy there was a person known as fadam this person has already given charts so we call them as fadams chart the beauty of these charts is if you have a uniformly loaded area which is rectangle of length L with B and please understand that these methods have limitations or they are particular methods so you can find out the state of stress only at the edges at a depth of z point P alright now let me first interpret this graph is a very longish expression the influence factor would be 1 upon 4 pi 2mn m square plus n square plus 1 upon m square plus n square plus m square n square plus 1 multiplied by m square plus n square plus 2 over m square plus n square plus 1 plus tan inverse 2mn m square plus n square plus 1 to the power half upon m square plus n square minus m square n square plus 1 fine so please do not remember ever where m is equal to B by z and n is equal to L by z it does not matter if you interchange B or L so it does not matter whether m or n get replaced or interchanged conceptually these are nothing but the graphs which look like this the fadams chart this is the influence factor Ib this is m and then this is how the n varies this is a peculiar solution one of the solutions I would say if you have a uniformly loaded area of B and L and if you have to find out the state of stress exactly at the edge or the corner point of the loaded area alright manipulations can be done in several ways I hope you will realize and you may enjoy that if I have a uniformly loaded plate or let me not segmentize it first suppose this is a plate or this is an area where q is known so I hope you remember that sigma z will be equal to Ib into q by z square clear so the moment you get Ib from this charts of this equation just substitute over here you get the value of sigma z now suppose if I ask you to find out what is the state of stress just beneath the CG of this loaded area at a depth of let us say 10 meters can you use fadams chart no is also correct because look at this situation so it has to be at the just beneath the corner ok so what I can do is I can discretize this whole thing in such a manner that I will create 4 units out of it clear now I think you can understand I can apply this concept fine is this clear so 1 unit at the edge no issues I can find it out another unit at the edge I can find out another unit at the edge I can find out 4th and I will superimpose all of them as if there were 4 sectors q1 q2 q3 q4 done no issues very good suppose if I say find out the this is the loaded area this is a foundation and I want you to find out the state of at this point please remember when I say state of stress point this point means in the z direction add a certain depth clear how will you use this concept now do a bit of geometry is this ok so I can do like this I have made this section as a subset of a bigger square and I can solve it apply your mind a bit and try to understand in what way the context of the application of Adam's chart has changed the loaded area is only this much agreed so if I use method of superimposition what I have to do if this is a b c d and what I have done is I have projected on a bigger scale a prime b prime a double prime a triple prime and d prime clear so if I have to find out the state of stress at this point using these charts at depth z I will find out with the whole thing as the rectangle so a double prime a prime b prime b c d d prime a triple prime a double prime clear so this becomes my rectangle first compute the state of stress through this this place area does not exist clear now how would you use the superimposition principle you will have to subtract the effect of stress which is coming because of extended portions that is it so method of superimposition can be done in two ways either you club the stresses make them summed up or you can use them as a subtraction also so this portion this portion this portion is not contributing and hence I can subtract the stresses which are going to come because of this strip this strip and the process what happened this strip has come two times so add once clear and then you have the bigger one apply the correction you remember what we did this was the loaded area and you are trying to find out what is the state of stress over here so this is just a way to apply fadams charts for computing the state of stress is very simple as far as the mechanics is concerned but the application part is going to be very interesting so when you enter into the field where you are a consultant and when you are applying these concepts how do you apply these concepts very interesting situations look at the modern day failure I do not know how many of you are watching the you tubes where the major failures are occurring in the eastern part of the countries or the dams are you aware of this what the system is going to do over here suppose see where the litigation comes in the picture if you are getting this point where you learn money I do not know whether you have ever realized this or not this is your house you are living comfortably over here and then one day one builder comes and starts constructing something over here now what would happen is this situation not similar to the one which we have analyzed just now you are living over here and then some activity is going to happen over here so this q has become q plus delta q and because of this q plus delta q at this point the state of stress is changing and we have done lot of analysis you know we have computed sigma total we have computed sigma prime we have computed order pressures so that means all this is going to create a detrimental effect okay so this is one example problem second example problem I might be having my house here and tomorrow real ways decides to take a railway track close to my home are you allowed and do this computations in a clandestine manner and do not disclose that your house is going to be safe so I can find out because of this extra loading what is going to happen at this place vibrations in other issue as far as the loading is concerned I can find out what how the state of stress is going to change and whether this is going to be acceptable to me or not fine because from next lecture onwards we are going to talk about the application of all these concepts in the form of consolidation so if I am constructing something over here and if this system you know gets distressed and God forbid if this happens to be a heritage building but then somebody has to safeguard that let this operation go on and we take guarantee that nothing is going to happen to the heritage building so you cannot just say like this what you will be doing you will be doing simple stress analysis and show here we are talking about loading look at another situation your house is here and I am doing excavation inside so I have created three situations now a person like me cannot sleep over here because I am an expert in my subject things which appear to be very simple on the blackboard and the paper when they get delegated into the real life is the practice of the subject which is very very interesting