 We have discussed a lot about the soil characteristics and we have devised different types of strategies to understand how the origin of soil occurs, how soils get the redeposited and how they are classified based on their redeposition and the transportation agency. Later on we talked about some indices which characterize the soil mass and they were at a bug limits. That in short was the interaction of the soil with water and we wanted to study what really happens when soils come in contact with water. From there on we moved to discuss about the compaction characteristics of soils. And this was the first time when we talked about how the microstructure or the porous structure changes because of the external impact of the loads or the stresses. Then we talked about the seepage characteristics and then we try to understand based on the coefficient of permeability how the soils can be characterized. I am sure you must be realizing that all these discussions which we had done until now are valid for characterization of the soil mass. Subsequently, we talked about the stress which develops because of the external loading in the soil mass and this is where the compressibility comes in picture. That means when the soil mass is exposed to external loading how it responds and this could be a sort of an engineering behavior. So I am sure the common thread between all these discussions which we had until now is to understand the response of the soil mass to the external stimuli or to the fluids or to the pressures which are being exerted on them. So in today's lecture I will be discussing about the compressibility characteristics of the soil mass and what I have written over here is this is compressibility and consolidation because once we have understood the compressibility characteristics of the soils we will move on to discuss about the consolidation characteristics of the soils. As the name suggests the compressibility is the compression undergone by the soil mass and this is mostly because of the external loading. In the previous lecture we have talked about how these stresses can be computed in the soil mass because of the external loading. Now if I consider the compression of the soil mass this is normally we talk about or we are interested in the volumetric compression. Now volumetric compression of the soil mass would depend upon the state of the soil mass and the type of loading which acts on it. We have already studied the type of the loadings. This could be the point load, this could be the uniformly distributed load, this could be different types of loadings which we have talked about in the previous lecture. So as far as the state of the soil mass is concerned normally we deal with a dry state, it could be a partially saturated state or it could be a completely saturated state. So when we define the compression characteristics of the soil mass it becomes important for us to understand soil is in what state dry state partially saturated state or the saturated state. Now suppose if I consider a surface, this is the ground surface let us say and there is some loading which is coming on the top of this. So normally we say that this is delta sigma. Now delta sigma is the one because of the construction of this facility over here. Now if the soil mass happens to be dry what is going to happen? The compression is going to be due to the expulsion of air, why? So air is incompressible as compared to the soil grains which are incompressible and so is water also. If I am dealing with a dry soil and if I load the dry soil system the chances are that the air will be incompressible will go out but the soil grains which are incompressible however when I change the scenario I will be talking about the third case directly that is when you have been saturated state of the soil this is because of the expulsion of water volumetric deformation is going to occur because water is incompressible and it takes time to move out of the pores and there is a time lag that means this process of expulsion of water from the soil pores is going to be a function of time and this is mostly because of the fact that hydraulic conductivities of the geomaterials are extremely small particularly for the fine grain soils alright. So the situation is like this that the moment you load the system of soil mass from outside in the form of delta sigma the amount of some air may expel out and there could be a compression the major contribution of the compression is going to come because of this process and hence we say that volumetric deformation of the soil mass is going to take place because of either the expulsion of air expulsion of water or the third one could be the compression of or reorientation soil grains alright. So these are three responsible factors now when I was talking about the particulate nature of the soil I had this small model that suppose there is a loading P which delegates to sigma and I had talked about two situations suppose this is the soil mass and the lateral sides are unconfined alright. So this is the unconfined soil mass if I keep on increasing the intensity of the load say from P to Pn where Pn is greater than Pn minus 1 is greater than Pn minus 2 is greater than P what is going to happen if I do not have a confinement the tendency of the particles would be to first you know the whole soil mass will get compressed and then because there is no confinement the tendency of the soil mass is to flow out. So if I restrict Pn to such a limit that Pn itself is less than the yield strength of the soil now we do not write sigma y over here we will be writing the shear stress so suppose if I say the shear stress y alright in this case only the compression is going to take place why because the shear strength of the soil is higher as compared to the stressor which are getting delegated into it and hence there is no you know slippage there is no rolling over there is no reorientation of the particles and hence this is a pure case of the compression this is what actually we are going to study today a part however if reverse is true and if I say Pn creates a stress which is greater than the yield strength of the soil this becomes a shear strength analysis so shear strength dominates the mechanism of load transfer hope this part is clear to you. So part number b we are going to do in the second course this will be dealt with under shear strength theory and not today however the part 1a will carry on with the discussion on a that means we are restricting to a situation where the state of stress in the soil mass because of the external loading is not going to exceed the yield shear strength alright and hence the compression only is going to govern the process is this part okay oh this is the P that means the load and if you compute the stresses along with this this will be sigma n greater than sigma n-1 to sigma so these are the stressors these are the loads below this is the yield strength of the soil yield shear strength of the soil alright so coming back to the issue see what we have done until now is we have defined the compressibility in such a manner that this is the fundamental characteristics of the soils as engineering property and it basically tells you about the response of the soil mass to the external loading a follow up this would be the consolidation process which will be discussing in the subsequent lecture. So this is how actually the compression has to be defined now what I intend to do in today's discussion is I will be talking about how to quantify the compressibility characteristics of the soils so the question is how would you obtain the compressibility of the soils so that is the determination of compressibility characteristics of the soil mass there are two ways the first method is which is known as a odometer test we call it as odometer test find out what is the meaning of the word wedge so this is the hollow ring of steel stainless steel normally 75 mm in diameter and 25 mm in height and made up of you know about about 0.5 mm metal now this is a odometer ring where it is basically a sort of happiness so the name is a interesting name this is what is known as a odometer ring all right so this is a stiff stainless steel hollow ring of dimension 75 mm and 25 mm sometimes people use 50 mm diameter and the moment it becomes 50 this will become 20 mm so these are the standard things so one is by conducting the odometer test and filling the soil sample in this ring and then applying a normal stress alright so sigma is the stress which is applied on the soil sample which is contained in a hollow ring which is made up of steel now steel gives enough stiffness and hence the radial strains in this sample tend to 0 that means there is no deformation of this steel ring so when you apply the load on the soil mass there is no radial deformation taking place in the system so this becomes a opposite case where we had talked about the unconfined this becomes a confined boundary and when confined boundary is there you know this is what gets simulated with the odometer ring when I load it the only possibility is that the soil grains are going to get compressed there is not going to be a shear failure there is not going to be any rolling and there is slippage along the plates of the fine grain materials now unfortunately this is not a very realistic situation because most of the situations which prevail in the nature or in practice are three dimensional I hope you can realize that this is a one dimensional loading and one dimensional deformation of the sample so that means if I keep my sample like this so if this is a sample to begin with the moment I have loaded it and normally rather than having a point load what we do is we distribute this load on the surface so this is how the P load is acting which gives a sigma stress now this is not a very realistic situation so in order to create a realistic situation what we can do is we can obtain the compressibility characteristics by doing a triaxial testing so this is three dimensional in nature what we do is we to take a cylindrical sample so this is the cylindrical sample you must have noticed here what we call is the diameter d and this is the height of the sample h so we maintain d by h ratio as 3 this is what the typical situation is and I hope you can realize that why d by h is kept as 3 because in the previous lecture when we are talking about the stress which is getting induced into the soil mass you know when your dimension of the loaded area is three times the height of the sample all the stress balls which are going to get confined into the sample would be 99% you got this so geometrically we force the full all the pressure balls to remain confined in the system and hence I can assume that the loading is uniform on the sample from top to bottom clear that is the reason so when we do triaxial testing we apply let us say sigma 1 over here sigma 2 over here and then sigma 3 over here normally we maintain sigma 3 equal to sigma 2 so this also becomes a baxial testing but it is in three dimension then we will be talking about the triaxial test for obtaining the shear strength characteristics of the soil mass so let us again remain confined to odometer test which is one dimensional in nature okay and as I discussed when you apply this stress over here so the deformed sample would look like this this is the deformed sample and I can say that there is a deformation of let us say delta h and delta h is reduction in height is specimen specimen of the soil because we have taken some small portion of the soil mass we have put it in the ring and then we are trying to find out how the compression characteristics look like now delta h is the compression also rather than writing reduction in better scientific language this can be replaced with the compression what is the significance of this if you have a soil mass sorry if you have a soil mass and suppose if I lay a foundation on the top of this all right so this much of these compression of the soil we are going to observe due to external loading provided the soil is same so this soil is same as what you have in the nature you have taken out some sample over here filled in the ring either I can remold the sample or I can take rings which I can insert into the soil mass I can take them out and this is how we can collect a undisturbed sample and that sample can be inserted into the loading frame and I can load it and I can get the deformation fine so this becomes the odometer testing which is what we are going to discuss in today's class so let me discuss a bit about the one dimensional odometer test or the consolidation test we call it there are two methods of doing the analysis particularly when we are working with fine-grained materials like either clays or you might be having sills or you might be having clay sills or silty clays so on normally we do not talk about the compressibility characteristics of the coarse-grained materials because as I said the grains are incompressible and hence there is no point in talking about the compressibility of the dry soils particularly the granular material so when we talk about the fine-grained materials in order to simplify things what we do is we saturate them how saturation is done this ring which is containing the specimen of soil normally this is placed in a on a pad in such a manner that this pad has a groove in it alright so this is what is known as base plate and this base plate is connected with a drainage tube or the drainage we call this as a drainage port and this drainage port can be connected to a water column what we do is in order to distribute to spread the load which is coming on the sample we put it on porous stones so we place porous stones in this groove this is the porous stone and one similar porous stone is kept on the top of the specimen which is contained in the ring so if I show you the ring like this in which the specimen is contained so this is the odometer ring with a specimen of the soil mass we will put a porous stone at the top also so this is porous stone number 2 or you can say top and bottom alright on the top of this porous stone comes a loading pad this is made up of steel and there is a groove in which I can keep a ball so this is the loading this is the ball which is made up of steel stainless steel and this is connected to entire thing is connected to a load frame so this is the loading pattern or sorry loading assembly we can call it so this is the machine or the loading frame I can read how much load I have applied or I can device a system by which I can apply a certain known amount of the load on the plate I also attach a dial gauge over here and this dial gauge gives us the deformation undergone by the specimen when it is loaded so dial gauge is DG is used for obtaining the delta H alright so dial gives you the readings that after you applied the stress on the sample or the load how much the deformation the specimen is going on that is delta H and I hope you can realize that area of cross section remains constant of the specimen remains constant and hence delta H is also equivalent to delta V the volume now choice is mine if I want to do a study on a dry sample I can do it you can prepare the sample you can take dry soil you can pack it in the odometer ring and you can apply stress and you can obtain delta H I can vary the moisture content of the fine-grain materials and I can fix them in the odometer ring and I can study the compressibility characteristics normally when we deal with the fine-grain soils as I was saying to make life easy we saturated make sure that the sample remains saturated always so when we do this test we open the wall and we allow water to remain always filled up in a container which is a bath so this becomes the water bath and this remains flooded with in fact the water bath should be up to the higher limits of the specimen so the water can be maintained up to here so sample remains or the specimen remains submerged in water all the time and hence we are ensuring that we are obtaining the saturated state deformation characteristics compressibility characteristics of fine-grain soils in particular we should do this test to understand things better or maybe seen some videos where these type of demonstrations have been done so as far as the analysis of test results is concerned basically what I want to do is I want to obtain a relationship between the void ratio and sigma prime so sigma prime is in kg per centimeter square sometimes we also write it in kPa void ratio is going to be in percentage and whatever relationship I get this becomes the compressibility characteristics okay there are two methods of doing this test the first one is where at the end of the test so when I say at the end of the test it is understood that the objective or the target effective stress level has been exposed to the sample normally effective stress is exposed on the sample in steps so when we say at the end of the test that means number of loadings which are normally doubled so suppose if you start with sigma prime 2 times sigma prime 4 times sigma prime 8 times sigma prime 16 times sigma prime and so on and between the each loading incremental loading we wait for 24 hours for the dial gauge readings to stabilize in other words we give 24 hours of waiting time for specimen to undergo the maximum possible deformation all right or the compression so we can find out how much sigma prime is required that much load you have to apply because sigma is the area of cross section multiplied by sigma prime will be the loading so normally you take first load 2 times second loading and so on so at the end of the test that means once the number of loadings have been achieved you find out the moisture content how it is done suppose if I am a designer and I want to design less than embankment which is going to be coming on the ground surface so remember what we have done is we have obtained these stresses coming at this point because of the external loading so that is delta sigma so this becomes my design stress okay so this design stress I am going to expose onto the specimen to get this response and hence I know what is the total magnitude of delta sigma which I have to apply on the specimen I can break it small cycles number of cycles or number of steps so I can apply delta sigma by n and follow this pattern of 2 times of the loading which is the standard procedure and complete the test so once the test is over you remove the specimen containing sorry you remove the ring containing the specimen and wait and find out its moisture content so suppose this moisture content is w in the second procedure what we do is at the end of the test take out the ring containing the specimen and dry it up and find out the dry weight of the specimen which is ws so these are two methods fine so in the first method when we are dealing with moisture content if a specific gravity is known multiplied by moisture content this will give me s into e so I am making sure that the specimen remains saturated all the time by doing this arrangement of water bath and water bath remains filled up with water specimen remains saturated so s is 1 that means the void ratio r g into w if I know the initial void ratio e0 this is the final one after the test so if I say that delta e is e0-ef clear and making sure that I am doing one dimensional analysis if you remember here area of cross section is constant so I can say that delta v upon v the volumetric strains which the sample or the specimen is undergone are equal to delta h upon h where h is the initial height so suppose if I replace this by capital H and delta h is the change in height of the sample this will be equal to delta e over 1 plus e0 so this is a relationship which can be utilized so here what I have done I have got the final void ratios I know the initial void ratios delta h can be obtained at every change of stress so if I try to re-complete or complete this graph I know the initial void ratio which is known at a very small stress we call this as the seating load or the seating pressure this is just to maintain the contact between the sample and the top plate you know complete because if you do not allow any seating load there will be initial disruption in the readings so this is the point from where we are starting the compression process so suppose if you take the first increment of the load sigma 1 prime y prime because the sample is saturated and submerged clear so you have to reduce the pore pressure and compute the sigma 1 prime but I am sure you must be realizing when we are doing the odometer test and we have put porous stones at top and bottom we are ensuring that the pore pressure does not remain contained in the specimen and gets dissipated so basically u tends tau to be 0 and hence sigma prime is same as sigma so this is how the relationship would look like okay so this is the function which I have created from e value now how would you get e from this relationship so delta e will be equal to 1 plus e0 into delta h upon h e0 is known change in height of the specimen which can be obtained directly from the dial gauge reading and I know the initial height of the sample I can compute delta e and from there I can back compute starting from this point to this point I can fix the relationship so once I have e versus sigma prime relationship I will tell you how to do analysis further the second method is at the end of the test you find out the dry weight of the specimen which is ws so if I know the ws so if I know the ws this thing divided by area of cross section multiplied by g into gamma w what this value will be sorry very good so this will be the h final is this correct nice why because once the test is over you have taken out the whole thing you can subtract the weight of the ring from the total dried weight of the specimen you get ws of the specimen ws this is the area of cross section and g into gamma w is the density clear so weight upon density is multiplied by area of cross section will be the height so hf I have obtained I hope you can realize we have fixed this dial gauge to get the change in the h value and what I am getting from here is final hf so at each incremental loading I know the value of h also so this can also be utilized to compute the wide ratio by using this relationship choice is yours whatever you find easy can be utilized so nothing is changing except for ws the weight is corresponding to the final stage so hf is fixed so in most of this situation what we do is after getting the final point we back traverse to the initial value so hf once you have obtained delta hf will be equal to h0- hf clear now delta v upon v is the volumetric strain which is same as the vertical strain the sample or the strain the sample which is equal to change in wide ratios upon 1 plus e0 if you remember when we are talking about the three-phase system of the soil we had used 1 plus e0 term as a specific volume okay so it is a sort of a normalization of the change in the wide ratio undergone by the system because of the compression for a specific volume of 1 plus e0 so this is how you normalize the things so 1 plus e0 is a normalization parameter which is known as a specific volume so implicitly what we are not plotting on these scales is I hope you can understand the time because each step is a function of time so this time we assume to be 24 hours and we make sure that the dial gauge readings have subsided normally the practice is you keep on observing 3 to 4 dial gauge readings and once they are constant we can go for the incremental loading or unloading so if 3 dial gauge readings are constant we assume that the sample has equilibrated clear none any questions these tests look very simple but you will be surprised to know that till date the profession depends upon these tests to get the correlation between what is happening in the field once you have taken a sample out from down below and testing it in the laboratory there is no other way to do the analysis of what has caused the failure of the geotechnical structures foundations embankments any structure which we are making out of it now I would like to interpret this graph further because this is the crux of the things so what we have done is by conducting these experiments we have got a relationship between wide ratio and applied stress so let us do the analysis of these relationships and see what all can be deciphered from here so typically we get a relationship like this what I have shown there if I plot e versus sigma prime this is the initial compression curve all right at this point which is corresponding to a certain sigma prime value if I stop loading the sample and if I unload it reverse process to see what we have been doing here is we have been doubling up this stresses so this is the loading process and suppose if I start removing the load from 16 to 8 8 to 4 4 to 2 2 to sigma so this will be unloading here so at this point if I unload the sample this is how the response looks like so this is the loading curve this one is the unloading curve or unloading response this point is defined as pre-consolidation pressure and we define this as sigma p prime this has lot of interesting you know knowledge in built into it we will discuss about this this is also known as compression curve and this is known as swelling curve this is compression curve or compression characteristics you may say these are the swelling characteristics now one interesting thing to observe here is this is how we started the test with the e0 value what you observe is that point a and point b will not match a and b do not coincide what is the reason the first reason is because the soil sample is non-linear because truly speaking e upon sigma prime is nothing but a sort of a stress strain curve is it not you must have noticed that if I normalize it with the e0 value this becomes e upon e0 or I can do even delta e upon e0 which is the strain versus stress so this is first of all because of the non-linearity of the soil number 2 regain of yes correct so what you have done from a to c you have squeezed out water from the soil sample by applying the external load and water being incompressible as moved out of the specimen now c to b is a sort of a unloading behavior little bit of water can be sucked by the sample yes so that is also valid that is you know what we call it as a capillary intake and which is less than the water which has been expelled out because the uptake capacity of the soil in compacted form is less the most important thing is why a and b do not match because of the friction so this is a friction which is offered by the ring due to the friction associated with the odometer ring which is not allowing b to coincide with a is a in a in a in an elastic material fine so this is known as unloading curve or swelling characteristics also so now suppose if I reload the sample from point b onwards what is going to happen is this is how the reloading would be so b to c to d a cb is loading and unloading b cd is I would say reloading or sometimes people also call this as a recompression now one interesting thing here to see is if I reload the sample from b it passes through point c and goes up to d I am sure you will realize that bc has formed a hysteresis and hysteresis associated with the memory of the material so from this point onwards I respect soils for having memory so this point will get clarified very soon soil behaves behaves as if nothing has happened to it and if you look at a cd this becomes a continuous curve so a to c to d remains a virgin compression curve as if nothing has happened to it all right so for all particular purposes I can assume this to be something like this okay so this happens to be its past all sorts of torture which you have given to the sample goes into the memory and it maintains it now this concept is used quite a lot when we talk about the shear strength theory of the soils it covers the interpretation of c to b to c so unloading and reloading cycle why we have called this as virgin compression curve as if nothing has happened to this from this point onwards so pre-consolidation pressure is the point clear or the pressure above which the sample is going to behave as normally consolidated we call it as NC response from this point onwards you will realize that the material vanishes the state of stress in the material guides the entire mechanics of the material so we do not bother whether this is clay or this is another soil as far as the pre-consolidation pressure is concerned that becomes the basic characteristics of the soil mass which is going to help us in designing the systems from c point onwards when you did loading sorry unloading this creates the OC behavior of the soil which is known as this is the OC behavior we call it as over consolidated soil all right so by definition if I define a term OCR which is over consolidation ratio if the sigma p prime is greater than sigma prime you know at this point what has happened the material has already experienced a very high effective pressure as compared to unloading part clear so these type of soils are over consolidated they will show less settlements less deformation less compressibility however if your sigma prime happens to be greater than sigma p prime this is the case of the normally consolidated soils where the deformation the settlements and the consolidation is going to be much more fine so what I have done is by utilizing the concept of the pre-consolidation pressure I have characterized the soils and as an engineer technologist I will be more interested in utilizing the soil mass when it is exposed to the stresses rather than anything else so this is the best possible engineering characterization of the soils based on the history of stresses the system has already been exposed to there is a third category which is known as so this is the first one over consolidated response normally consolidated response and the third one is you know which is known as under consolidated we call it as a UC in this case the present pressure is higher than the previously exposed pressure to the system sigma p prime and hence this is NC material later on we link these three things with the pore water pressures so in OC materials normally the pore water pressure is always negative why because you rightly said they have a tendency to suck water so here the tendency of the material becomes to suck as much as water it can uptake or it can suck so this becomes a situation where the negative pore water pressures prevail in the soil sample in NC normally we have positive pore water pressures so suppose if I give you two situations the soils which are formed in the Himalayan region and the soils which are getting formed in the Sundarban delta which we discussed in the beginning of the lecture so imagine the case of the deltas everyday fresh load of sediment is coming and it is getting dumped into the ocean beds clear so that means the present pressure is always higher than the past pressure so this becomes a normally consolidated material however in upstream in the Himalayan region what is happening the soil is getting eroded day by day clear so the present pressure is always lesser than the past pressure clear so this is the situation of OC material so the soils which are in the Himalayan regions are over consolidated as compared to the ones which are in the Bay of Bengal clear so have you understood that from a simple test how much information we have deciphered about the material and that too in one dimensional compression situation under consolidated it is the one where you know the past pressure is equal to the present pressure is something in between so this is what is known as under consolidated response so these one two are normally used in defining the state of the material and how to utilize this as a construction material