 Now, let us complicate the thing, rate of shearing plus depositional history, the way the formation took place, how the sandy deposit was formed clear and this is where I will introduce a term which is known as k parameter and to be specific, I will say that this is k0, we write this as earth pressure at rest, alright. Truly speaking, this k is a function of sigma h upon sigma v that means sigma h is equal to k0 into sigma v. What is the meaning of at rest? I am not shearing the sample at all, at rest, no movement, no displacement clear. So, this is at rest condition, from here we will derive active and passive situations later on, kA and kP. So, kP we become passive earth pressure condition and this becomes active earth pressure. So, to begin with I will deal with only k0 condition, what is the interpretation of this? The interpretation would be the initial state of the stress in the sample is the moment I applied sigma and there is no shear, the value of sigma v is sigma and value of sigma h will be k0 sigma, before shearing process at rest condition, the sample is just static. Draw the Mohr circle for this, you can draw it easily, no issues and normally k0 is going to be a number which is less than 1. So, that means on a tau sigma plane the initial state of stress of the sample would be, this is let us say sigma value and this will become k0 times sigma, initial state of the sample, now comes the shearing part. So, when you are shearing the sample what is going to happen? Your tau is picking up, alright. So, starting from this point of sigma 3, you know if I draw a vertical line what is going to happen? I require a higher value of sigma v to fill the sample, is this okay? Is this correct? So, that means what I am going to do is if I have to increase the value of sigma, k0 sigma also changes, it is not going to be feasible. So, what should I do? That is a big question. What will be the answer now? Anybody could think of this? k cannot be changed because k is the ratio of the two states of the stress. Now please remember sigma h is going to be a sort of your shear strength or the shear stress. Now sigma v is equal to gamma times z of the material, unit rate multiplied by the depth of the point, correct? So, that gives you sigma v value and the all possibilities your sigma value is going to be constant, but now I want to shear it what is going to happen. Look at this point. As long as this situation is concerned, it is fine. Now the moment you shear it what is going to happen is this is sigma 1, this is sigma 3 and just now we have discussed that this is the Mohr-Coulomb envelope, this is the point of tendency. Point of tendency indicates that the failure has occurred, peak point. In this case it is going to be the residual value. So, this is the state of the stress where you have sigma f tau f, this is part okay, the failure. If this is the center, this becomes the radius, this is 90 degree, where is the pole? Now this state of stress you are forcing to be acting on the horizontal plane, remember and that is the biggest limitation of the whole test. That means this state of stress if it is acting on the horizontal plane what I have to do is I have to draw a horizontal line. What this is going to give me? Starting from the given state of stress you get the pole, agreed? What I have shown here? I have shown the rotation of the state of stress on the sample. Now if you draw a tangent, so not tangent sorry, if you join this point what this becomes? So passing from the pole if I connect sigma 3, this becomes the major sorry minor principal plane. If I draw sigma 1 passing through P what is going to happen? This becomes the major principal plane. This is what we call as rotation of planes. So this is where sigma 1 is acting and this is where sigma 3 is acting fine. If you have followed these concepts solving numericals is not going to be a big issue. So that part is left up to you. Please solve the problems. I think I have discussed all finer aspects over here and it should not be a difficult task. Any questions here? Hope you are enjoying the strength of the materials where the materials are soils, lot of ambiguities, lot of expert eyes are required to understand what type of interpretation we are going to do with the results. Because you have to understand the material fine. If this is clear, should I move ahead? If interlocking comes, so do one thing you please do like this. This is the interlocking effect and now try to give torque. No, this should not be slippage, interlock them. This is the interlocking and now rotate. What is happening? So it is a sort of apparent cohesion clear? Interlocking as I said would always give you apparent cohesion. Friction is because of the rubbing of the particles together. Here you have long T2 particles. They are not going to slip over each other clear? When slippage occurs then only friction develops. So we came only up to point number 1, point number 2 and k0, depositional history. Number 3, the most important thing on which the shear strength of the material would depend upon is the type of the soil and the fourth one would be moisture content. Very soon I will be introducing the concept of OCR. So depositional history is depicted mathematically as OCR. OCR is over consolidation ratio, over consolidation ratio. Another interesting way to characterize the soils which would include the type of the soil, the history of formation and history of deposition also mathematically. Very advanced way of characterizing the materials clear? So over consolidation ratio. Until now we have talked about only consolidation characteristics. Now this is the first time I am using the term how the soils are formed. So suppose if I take you to the geological journey of the soils, remember these are the Himalayas and then this is the Gangetic plane and somewhere here you have deltas alright? Bay of Bengal and these are the Himalayas. Take a point somewhere here. What is the state of stress? Something let us say sigma v and sigma h. What is happening here? Weathering process. So I will get defragmented clear? Water, very high speed of the water. So truly speaking most of the erosion takes place over here. When erosion occurs, the state of stress at this point which is today. So sigma state of stress I am writing today is always going to be less or more, less as compared to what it was in the past. The ratio of the two is OCR. So is this fine? So the state of stress acting on the sample today divided by what it was in the past is overconsolidated ratio. These materials are going to be under-consolidated. Oh sorry over-consolidated. By the time soil particle moves and comes over and gets deposited over here now what is happening? It is a reverse process. Now today's stress is going to be more than the stress in the past. Why? Because rivers are bringing continuous loads of sediments and these sediments keep on getting deposited one over the other. Consolidation process might not be over clear? So in this case the OCR value is going to be more than 1. So what we have done is, yeah this is more than 1. Today's is more than past. So OCR is greater than 1. OCR is less than 1 over here because today is less than past clear. So I have now used the state of stress to characterize the soil deposits. And I have a strategy now in my mind rather than doing all these sorts of tests which I have been talking about by using only the state of stress in the past and present I know how to characterize them and how to deal with them. So this itself is going to guide to the shear strength of the soils. Now one of the ways to depict all the results would be if I plot tau versus sigma and the way I have done over here I will be getting let us say for pure sands. When I say pure sands friction angle I am assuming to be 0. This is the coefficient of internal friction. Rubbing of particle with each other causes friction. Take a water bottle fill it up with water keep on adding drop some particles of sands clear. Close the bottle and shake it. Sound comes why? Because of friction clear. That is the easiest way to explain to you what friction is keep the particles of the soil on your hand and rub it. Now if you look at this there is enough sound coming out of it clear granular material this is the frictional mobilization friction mobilization of the soil. It may cut your pumps also why? Because all sudden done particles of the sands are quarts and the crushing strength of quarts is 2030 MPa. You have to be absolutely careful when you are rubbing this thing in your pump. A good geotechnical engineer would go to the site take a sample put beneath the pump rub it and you can tell you what is the C5 parameter but never do this. So I hope now you understand that this is what is going to be for the peak strength if this angle is 5 peak what I am doing is I am consuming this value tau peak sigma is fixed. So tau peak upon sigma is a very rough estimation this is what is going to give me the peak strength profile or envelope. However this is for the dense sands and this is for the loose sands. If I remove this effect of peak I will say 5 and then what I can do is I can subscribe this to RD relative density which is a function of white ratio. Now let us analyze a bit more on the friction angle there is something known as angle of repose alpha. What is angle of repose? If I take a funnel and if I pour sands into it and if I open the other end and let the sand fall out approximately let us say 30 centimeter distance the distance should be small whatever heap gets formed you must have seen in most of the production units this could be grains, it could be sugar, it could be rice, it could be metal slag, it could be soil or whatever. This is what is known as the hopper the slope which is formed naturally when you drop sands from a certain distance why distance is so important so that there should not be any impact of falling grains on the grains which are already sitting in the heap you are not imparting any energy. This slope is defined as alpha this is known as angle of repose. Angle of repose is approximately equal to internal friction angle of the soil coefficient of internal friction of the soils. So this concept is very useful in making you know different type of silos for storing grains or the granular material and then we can link all this information with the model which we are going to develop now on a tau sigma plane if this is the Mohr circle stop writing and please concentrate for one minute on the board your concepts will be clear forever. If this is the Mohr pull-up envelope alright this is the point of tendency which corresponds to state of material at failure fine. If I connected with the point sigma 3 which happens to be the pole of the material pole of the Mohr circle this is the failure plane this angle is theta f if I connect this point sorry if I draw a perpendicular to this Mohr pull-up envelope starting from the point of tendency it cuts the x axis at this point which is the center of the Mohr circle this is sigma 1 minus sigma 3 by 2 if you remember this is sigma 1 plus sigma 3 by 2 this we have just now defined as C the apparent cohesion and this angle is phi now if this is the theta f failure plane inclination this angle is going to be what 2 times theta f correct let us do some simple construction now and before I move on to that there is a peculiarity about this material if I draw a perpendicular from this center and extend it in such a manner that it cuts the Mohr Mohr pull-up envelope what you are observing this line cuts over at this point which is point number A the Mohr pull-up envelope says that this itself defines the state of stress of the material at failure. So theoretically state of stress at failure at point A occurs but practically before point A is reached sample has already failed why look at this before you achieve point number A the failure is already occurred at f point now this is your sigma f sorry tau f and this is sigma f so this is a fictitious point what it indicates is you have enough strength for the material even corresponding to this confining stress because the shear strength available theoretically is this much suppose if I say O so O A happens to be a theoretical strength which is available for the material but it so happens that the material fails before theoretical shear strength is achieved and hence this becomes the point of failure I can define the term factor of safety now first time I am introducing this concept so what factor of safety says the factor of safety says that you have so much of strength available but the failure is occurring at this strength only so if I say this is tau f and this is the total strength which is available theoretically the factor of safety is going to be the ratio of the two and this is the term which people utilize for designing most of the foundations and sublinging structures is this part okay there is another interesting way of defining the factor of safety there is something known as brittleness index of the soil this is normally depicted as IB if I consider this is strength as tau peak and this strength as tau residual the brittleness index is defined as tau p minus tau r divided by tau p and this becomes another way of classifying the soils when you use them as engineering material what it shows is the state of the material where the brittleness is too high then systems will always show higher brittleness as compared to loose systems fine so I have defined now factor of safety is two forms this itself is a factor of safety term how much safe I can be without you know failing the systems and another factor of safety now I have defined in this way so for all practical purposes this is your sigma 1 minus sigma 3 by 2 now if I say that this is point B and this is point P can I use simple relationship geometrical relationship sigma 1 minus sigma 3 by 2 equal to if this is the friction angle this is going to be this is C tan phi alright so sigma 1 minus sigma 3 by 2 will be equal to sigma 1 plus sigma 3 by 2 plus C tan phi okay I am sorry this will be C cot pi this is okay and then this is 90 degree so this multiplied by sin phi this is okay so solve this expression and what you will be getting is sigma 1 minus sigma 3 will be equal to sigma 1 plus sigma 3 sin phi plus 2 C cos phi I can reschedule the terms so this becomes sigma 1 minus sin phi equal to sigma 3 1 plus sin phi okay and plus 2 C cos phi this answers your question Sashankar what you are asking last time we have started from a state of stress where sigma v is equal to sigma 1 sigma v happens to be a major stress clear so this equation is valid for a condition when sigma v is sigma 1 and sigma h is sigma 3 the moment we reverse this situation the more circle will change we will discuss about this later on so now you do this analysis and what I can do is for pure frictional materials I can put C equal to 0 so phi only prevails pure frictional material so this becomes sigma 1 equal to sigma 3 1 plus sin phi over 1 minus sin phi this term is k term k parameters the way we have used k parameter now is we have to relate sigma 3 to sigma 1 clear so you have to transpose it and this will be written as sigma 3 equal to 1 minus sin phi over 1 plus sin phi into sigma 1 so this becomes k a coefficient of earth pressure at active stage fine this is part clear two more things before we disperse today if I am assuming a pure frictional material this line the Mohr Coulomb envelope is going to pass through the center and if I draw this line over here like this the point where the envelope touches the Mohr circle becomes the point of maximum obliquity this is ok so this becomes the maximum possible friction angle a system can mobilize in reality what is going to happen is some fraction of phi will be getting mobilized and the material is going to fail few limitations and strengths of direct shear box test so I am sure you must be realizing that lot of things can be obtained from a simple direct shear test the limitations are as we discussed you know drain conditions cannot be simulated not valid for fine-grained materials area of cross section changes and hence you cannot share it much the pressure keeps on changing at the interface it is not constant because area of cross section keeps on reducing alright and so on the strengths are simple test and whenever you are dealing with the field situations and suppose there is a stratified deposit alright what I can do is I can take out a sample from here capturing the stratification bring the sample put it in the box share it and obtain the shear strength along this plane that is the beauty it could be inclined plane also does not matter if the stratized like this I will still try to take out a sample from here these are the exploration techniques I will bring the sample in the lab put it in the direct shear box and I will get the shear strength at this plane so in short there are lot of beauty of doing this test though the disadvantage are many and what we will do is to overcome these disadvantages we will now switch over to triaxial testing that will be from next lecture onwards.