 So, welcome to CE330, we have been talking about slope instability and I gave you an overview of what is meant by instability of the slopes and how to stabilize the slopes alright as an engineer geotechnical engineer in most of the situations. I gave you several examples of what type of failures or the failure mechanisms are considered when we talk about slope instability alright, there was sliding, there was a block, there was a flow and the creep and then we discussed different situations where instability can be caused to the slope alright just by cutting at the toe or by loading it from the top and I also discussed about how to stabilize the slopes that means unloading of the slope alright and providing the drainage provisions and so on. I also gave you an idea about what are the internal and external forces that cause instability of the slopes. So, we discussed under two heads as internal forces and the external forces, external forces are mostly caused by the external agencies and internal forces are because of the shear stress which is mostly a stabilizing force rather than destabilizing force which is caused by the external agencies and of course the physical, chemical, mineralogical phenomena which go on in the soil mass or the rock mass which are the principle constituents of the slopes. Now having done all this, I wish to now discuss about the stability or instability of the slopes and when we talk about this, we normally consider two types of slopes. So, the first category of the slopes is finite slopes, it is a misnomer that this is of finite dimensions, most of the time people mistake this, they think finite slopes are finite in size and infinite slopes are infinite in size, it is not correct, it has something else to do with which I will be discussing today. The second part of the or the second type of slopes normally we deal with is the finite slopes alright, mostly I am sorry the infinite slopes let me say infinite slopes and the second one is the finite slopes, mostly finite slopes are man made mostly, you take the soil mass compacted, you remember we were talking about compaction process in CE323, the best possible application of that is to create an embankment. So, suppose this is the natural formation and I want to create a facility over here for roadways or railways, it could be a strip also yes, so rather than laying the foundations of the pavement on the ground surface what we do normally is we create a elevation, this is what is known as embankment and this part we have discussed quite in details ok, this could be made for water retention also and on the top of this you will be exposing the railways or highways or roadways and so on. Now, this becomes a very interesting problem, so this is the embankment which has been created by compacting the soil mass in layers, you know we talked about you know what type of compaction may lead to what type of consolidation and what type of drainage conditions when we are talking about the shear strength parameters, so depending upon the foundation material and depending upon the embankment material I might come across a situation where the slow drainage no drainage may occur, consolidation may occur, consolidation may not occur, we have talked about the long-term stability of embankment, short-term stability of embankment and so on. Anyway in context to the finite slopes which will be discussing in the second half of this course or this topic I will be taking it up later on, normally the properties of the material change as a function of depth, so material properties are a function of depth, more compaction more c, more phi, more density, so more c, phi, gamma with more compaction effect more density, these are typically finite slopes, to begin with our discussion on slope stability I will be talking taking up first the infinite slopes, how do you produce them? This could also be man-made, suppose you remember the concept of repose angle, yes we discussed long-long back you are right, so suppose if I take a granular material and if I pour it from the top with a funnel air pluviation is also a technique yes you are right, so air pluviation is also a technique very good, so if I am pouring a granular material from a certain height what happens, the grains fall they form this type of heap where the angle of inclination is going to be angle of repose correct, it cannot be more than this angle because each grain has a tendency to roll down instability, so whenever you pour materials like grains granular materials the ores after mining, so whatever heap is formed this is a classical example of a infinite slope alright, yes this is the angle of repose concept. Now what we assume here is that the material is going to fail parallel to the face of the slope always, so these are the possible slip lines or failure planes, so what is that you are observing here, the failure plane is always parallel to the face of the slope first condition yes always, now if I take an element here suppose this is the mass which has failed and if I magnify this, this is how it looks like, this is the mass which has failed, this is the length of the affected portion and normally we define this as D, the depth of maximum possible depth of the slip surface, yes you are right, so suppose if I draw a tangent here not you will be realizing that this is parallel to this line, so it so happens that if D by L is less than 10% this condition should be satisfied, thickness of the unstable mass D is much less than the height of the slope, the height of the slope could be this much, so thickness of the affected zone is much less than height of the slope and followed by this condition D by L is less than 10%, the next condition is failure plane is always parallel to the slope, natural formations are good example of this type of failure, yes, so if you check on internet you will get several situations where the infinite slopes have failed, the properties of the material along the depth and the state of stress along the depth remains constant, properties and the state of stress as a function of depth do not change, so material properties and stress conditions are not a function of, you may also add here that truly speaking infinite slopes are not a infinite length slope, soil properties remain constant, stress conditions remain constant as a function of depth, if more than one strata is present what is going to happen, if more than one strata is present then everything is going to be again of the uniform characteristics, each plane has to be of uniform characteristics alright, so boundary of each layer is going to be parallel of to the ground, so suppose if I may say that this is the strata number 1, strata number 2 and 3, within the strata the material properties remain same, overall what happens is the slip surface the failure plane is always parallel to the slope surface, yes, so in case of more than one strata with uniform characteristics, each strata is having uniform characteristics and the boundary of each layer is parallel to the ground surface alright, the next characteristic would be let us show the state of stress, so stress between any vertical element of the soil is identical, so suppose if I take an element over here or an element over here or an element over here, this becomes my one element alright, this is another element, this is another element, so stress between any vertical element of soil is identical, what is the significance of this, now suppose if I take it out and zoom it and try to draw the free body diagram, try to draw the free body diagram, yes, you have this surface okay and we have taken a slice we call it as a slice, this is also I, this is also I, draw the free body diagram, so the weight is acting over here, what else is acting over here, pressure from the base perpendicular, let us define this as the base pressure, what else is going to act, uniform resistance from the left hand side and right hand side, because what is this going to be, suppose if I say this is P left, pressure which is acting from the left to right, but we write normally this as pressure coming in the right direction and then PL, both of them are going to be parallel to the planes or the surface, I hope you can realize if AB is parallel to CD, what should be the condition in mechanics you have studied for equilibrium of this type of a situation, PR should be equal to PL and they must be coaxial, yes, so this becomes a typical two dimensional problem or situation, let us not, I mean the problem is not the real sense, problem is a situation, so this is a two dimensional situation, the third dimension is going to be perpendicular to the board, so the third dimension is going to be in this direction, let us say it could be like this, okay, it is better to show it over here, this is how the third dimension goes perpendicular to the plane of the black board, green board, now this is what is known as plane strain problem also, we were talking about this situation when we did direct shear test, alright, so this is a plane strain problem, what is the significance of this, what clicks to you, one layer is sliding against the second layer, detached surface or detached block and the parent block and the contact surface remains maintained, fine, so this becomes a simple mechanistic model, if I want to find out the shear strength parameters which test I should be doing, you have to think about it, direct shear box test, yes, so as long as the material is uniform it does not matter, you take the material, soil mass compact it in the direct shear box set up and shear it and get the friction angle and get the cohesion angle for the granular materials, fine, suppose this happens to be the layer 1, layer 2, layer 3 and I want to find out the shear strength parameters along the contacts, interfaces, so the moment interfaces come what we have to do, the moment interfaces come what we have to do, again you shear it along that plane, so I can take a sample of this soil mass from here, if it is a C5 material, cut it from this plane, size it, trim it and place it in the direct shear box test, make sure that this surface is aligned with the three-way surface of the direct shear box, do the direct shear box test and get the parameters, as simple as this, so in today's discussion I will be talking more about how to use these concepts to solve more critical problems, there is one more concept which I would like to introduce over here, when we are talking about this situation is this free body diagram complete, no, why, what is missing from here, so one shear stress has to come, this becomes a normal stress, very good, this is a free surface, no normal stress, no shear stress, what happens on the sides, this side is being resisted passively by the soil mass, so given a chance this is going to be the passive resistance, now passive resistance means the shear stress is going to be acting upwards or downwards, you can find it out, so generically I will show that the shear stress acting like this let us say, what is the conjugate shear stress for this, yes, these two are equilibrated most probably, what about this and this, so you are defining the law of mechanics, this is the state of stress which is acting on the slice and this slice if you remember is nothing but a portion or a slice which we have cut out from the infinite slopes, so this condition is correct, similarly if I define this as ER and this as EL, they are also supposed to be equal, fine, now if this shear stress is 0, the conjugate shear stress is going to be 0, how the equilibrium will be achieved provided this tau also tends to 0 and hence this also tends to be 0, the moment this shear stress tends to 0, what is going to happen, the slip is going to take place, the failure is going to occur, loss of shear strength causes failure of the slopes, I hope this clears all your fundamentals, think about it, okay, yes, so this is a situation where the normal stress is equilibrating with the weight of the block, so if you resolve this what is going to happen, we have W cos I W sin I, if I tends to 0, I hope you can understand or second thing is W itself tends to 0, thin sections, so from this point onwards we have introduced the concept of thin sections, infinitesimally thin sections of the slopes are considered for doing the analysis and all these infinite testimony thin sections create an infinite slope, this was an interesting paradox which we have discussed just now, is this okay, shall we move ahead? So suppose if I compute this W, what will be the value of W, I have to define something, so normally the convention is we take the horizontal distance between the 2 slice vertical surfaces as the lateral distance, alright, this is A and this is D because just now we discussed C D becomes a slip surface, shear stress is 0, so this much of the soil mass is vulnerable to the failure and parallel to the slip surface which is parallel to the slope surface, okay, what will be the value of W, A into D multiplied by unit weight, you can resolve them in components sigma F X equal to 0, sigma F Y equal to 0, fine, what else can be done, what else you have an equation, there is a relationship between T and N also, what is the shear strength of the material, the value of T in the form of shear force is equal to normal stress multiplied by tan of I, yes, this equation is going to be very very useful, now can I derive from these equations N equal to W cos I and T will be equal to W sin I, just try to see how would you get this, fine, so why we are computing N and T because I can compute from N sigma the normal stress, what will be the value of normal stress, this will be equal to W cos I divided by what is the length of the base, if this is A I what is the length of the base, A by cos I and what about the shear stress, shear force is known, so this is W sin I over, is this correct, check it, what is the magnitude of W, A D gamma, alright, so this will be equal to A D into gamma cos square I, A cancels out, what is this gamma D into cos square I and this will be equal to gamma D sin I cos I multiplied by gamma D, is this part okay, so what we are doing is we are assuming here dry sands, simple example, simple cases to deal with first, yes, so now can you apply this equation tau equal to C plus sigma tan phi, shear strength envelope, substitute the values and see what we are going to get, now one question remains, if somebody asks you to draw the force diagram can you do it, true and then you have T component and N component, what is the included angle between W and N, I, you get easily these equations, I have skipped those relationships, sorry those steps, so if you solve this for pure sandy material C tends to 0, so what is T, this is gamma D sin I cos I and this will be equal to gamma into D into cos square I into tan phi, now the way put we put it like this, the shear strength for stability has to be greater than sigma tan phi, the moment shear strength drops below this or tends to 0, what is going to happen, the slips is going to happen fine, so what this indicates is gamma D can be cancelled, so we have sin I cos I equal to cos square I tan phi, now let us use the term factor of safety, we discussed this long long back, what is factor of safety, if factor of safety is equal to 1, this condition is valid, otherwise what is the factor of safety, this will be equal to cos square I tan phi divided by sin I cos I which is equal to tan phi over tan I, very interesting relationship we have derived, what is the philosophy behind this, what do you understand from this relationship, factor of safety depends upon the coefficient or the friction angle, internal friction angle and the inclination of the slope I, so what is the limiting condition of I, I cannot be greater than phi, remember simple block sliding on a surface and you keep on changing the inclination of the surface, what is going to happen, after the critical angle of I what is going to happen to the block rather than sliding it has a tendency to topple, true, so this is the critical value, so factor of safety for a dry slope is the ratio of friction angle and the inclination angle, so what is that you realize here, gamma is not coming in the picture, what is the significance of this, the depth of the critical surface is also not coming in the picture, failure can take place along any plane where the shear strength becomes critical, this is part okay, material properties do not come in the picture, now from this situation, suppose if I create a situation where the entire slope gets submerged or there is a seepage which is induced in the slope, what is going to happen to the factor of safety, the factor of safety is going to drop because this is a limiting value for the sands, correct, C component is not coming here, there is no cohesion here, this is purely frictional material, so for a purely frictional material under dry condition the factor of safety is tan 5 over tan i, now you will realize soon that this function is very very important and critical because when I say factor of safety at most what is going to happen is some multiplier multiplied by or sorry multiplier multiplied by this term is going to emerge, multiplier cannot be more than one, so all sorts of seepage conditions, all sorts of submergence of the slopes will lead to determination of this multiplier and that is what we are going to do subsequently, fine, this also indicates that when full shear resistance is mobilizing and sliding is going to take place, your i value will be equal to friction angle.