 We have been discussing the compressibility characteristics of soils in the previous lecture and this is where I had introduced the concept of consolidation also. Specifically I talked about the difference between consolidation and compaction and I think I highlighted that these are two mechanisms or different mechanisms which occur in the different states of the material. After talking about the computation of settlements in the soil mass because of the external loading, I discussed about the one-dimensional consolidation theory which is proposed by Terzaghi and this is a spring analogy which I discussed in details followed by the assumptions which are made in the one-dimensional consolidation theory proposed by Terzaghi. In today's discussion I will be talking about the first of all derivation of the one-dimensional consolidation equation which is used to obtain the pore water pressures as a function of time and z and once the pore water pressures are known I can obtain the settlements undergone by the soil mass because of external loading. To begin with let us try to derive the one-dimensional consolidation equation if I have the soil mass and if I take an element of the soil mass of depth dz located at a depth of z up to the cg and as you remember what we did is we exposed this soil mass to external loading maintaining the two boundaries which are freely draining. So this is the boundary number 2 and number 1 and there is an external load which is applied on the system which is causing delta sigma to occur at this point at the middle of this element. So if this is dx and the third dimension if I consider to be dy because of the loading the pore water pressures are going to develop and because of the development of the pore water pressure and due to the vicinity of the draining boundary there will be a tendency of the soil mass to dissipate the pore water pressures. So suppose if I assume that in this element there is a influx of the seepage or the discharge and this is vx plus del vx upon del x into dx. So this is a continuity of the flow which is entering in the system in this direction I can say this is del vz upon del z into del dz. So I have to reverse the directions of the arrows. So this is vz and this is a discharge which comes out. I can consider this element as a three dimensional element the way I have shown earlier and this axis is dy. I can write that is dv by dt is the change of volume. So let me put this as capital V so this is the capital V which is the volume of the element of the soil. I can prove that this dv by dt equal to velocity into the area. So that means del vx upon del x into dx multiplied by dy dz plus del velocity these are all velocities del v by del y into dy del x del z plus del v by del z into dz into dx into dy. I can simplify this as del vx upon del x plus del vy over del y plus del vz over del z into dx into dy into dz alright. I hope you will realize that I have already subtracted the discharge which is coming out of the element with the discharge which is entering into the element. So this is the mass balance alright otherwise this will be vx plus this term multiplied by dy dz minus this term. Now if I consider that the soil element is incompressible under that condition what is going to happen this will be 0 alright but most of the time soils are compressible as we discussed. So del v by del t is not equal to 0. Now the question is if I assume that the one dimensional consolidation occurring in the system can I obtain a relationship where I can show what is the value of rate of change of volume with respect to time. Now this is where we can help take help of the relationship which we have developed delta H upon H equal to delta E upon 1 plus E naught equal to you know this was our basic premise is this okay. Now analysis of this leads me to a situation where I would say delta H is equal to mv into delta sigma prime into H you are agree this we derived in the last lecture. What is delta sigma prime? Delta sigma prime is the effective stress which is developing at the CG of this element which we have considered because of the external loading. So this delta sigma z is equal to delta sigma minus delta u and this delta u is nothing but delta uw fine that means now if I try to write this function del v by del t can I write this that del v by del t will be equal to if I differentiate this function by substituting it over the here this will be mv is practically constant we are assuming multiplied by delta sigma will be constant what is changing is delta uw. So this becomes minus delta u I will remove the term w now is understood that the power of pressure so this is going to be del of t multiplied by H is constant okay. So what I have done is I have got a relationship between the coefficient of volume compressibility the power of pressure which is developing in the system and its rate of change multiplied by thickness of the sample. Now is there a way to get rid of the velocity vectors I hope you will realize that velocity is nothing but minus k into i so I can write this as minus k into del H upon del x where H is the head and this can be written as minus k over gamma w into del u by del x this is part clear because H is equal to gamma w into H into gamma w is equal to u for the pressure. So this is the relationship I am using so if I substitute it over here I can and if I assume a one dimensional situation your other terms will disappear and this is the volume term of the element I can write that minus k upon gamma w into del u by del x into volume alright and what we are doing is we are taking derivative of this function. So this function has to be differentiated with respect to x so this will be equal to minus mv del u by del t into H this is okay so this is your rate of change of volume with respect to time we have obtained. Now this is also going to be a function of volume because in one dimensional we have written like H so what I can say is that del u by del x square into k upon gamma w this will be equal to mv into del u by del t now this expression is of some use to us what we can do is we can write this as del u by del t equal to k upon mv into gamma w into del u square by del x square now this equation is known as one dimensional consolidation equation which is proposed by Terzaghi this term which is appearing over here is also defined as Cv. So where Cv happens to be the coefficient of consolidation now this is what is known as one dimensional consolidation equation which was proposed by Terzaghi. Now interesting thing here would be this is a transpose in the one dimensional case so this volume in three dimensional is volume this has to be multiplied by area of cross section because this equation is valid for delta H upon H remember area of cross section is constant so I can say this is area and this is also area so this becomes del v by del v this is how your this volume term gets cancelled with this volume term okay this is one dimensional case we are talking about only height area of cross section of the sample remains constant so what we have done is we have obtained the one dimensional consolidation equation and now to understand that what we should be doing with this equation number one what is the form of this equation suppose if I say del square u by del x square multiplied by Cv equal to del u by del t truly speaking if I solve this expression or this equation I hope you understand what is second law of fix or the fixed second law you must have come across somewhere is it not so this is u the pore pressure I can substitute it with the concentration I can substitute the temperature or whatever you want and then this is your second order fix equation so truly speaking u is a function of x and t x happens to be incidentally the z so to make it more apt to our discussion I will say this is z, t in simple words the one dimensional consolidation equation is the value of pore-water pressure at a given time and at a given depth in the soil mass so that means if I consider a given point at a given time because of the external loading whatever pore-water pressure is developing in the system is depicted by one dimensional consolidation equation this is part clear so what we have to do is we have to solve this differential equation to get u so this is the first objective when we solve this u as a function of z and t I have to apply some boundary conditions what will be the boundary conditions the moment I apply load because of the hydraulic conductivity and this is saturated soil so the assumption is the moment you apply load delta sigma gets generated and this delta sigma is at t equal to 0 initial condition so delta sigma is equal to or it is generating delta u now if my initial value of u is 0 delta u is nothing but u so this becomes u initial and if I say minus of this so basically delta u is nothing but the ui term initial pore-water pressure is 0 clear this is the first equation which I am having now this is existing at z equal to 0 at this boundary as well as at this boundary because both the boundaries are draining so this is t equal to 0 delta sigma equal to this and delta u equal to ui in the entire domain so this will be less than equal to d normally we define the thickness of the layer as 2 times d for a specific reason I will tell you why so if the thickness of the layer is 2d so the boundary condition is going to be z less than equal to 0 in the entire domain the physical significance of this is the moment you compress the whole system by applying external loading the initial pore-water pressure develops which is equal to the delta sigma value and at t equal to 0 the entire system shows a constant pore-water pressure alright this is the initial condition so that means if I want to plot let us say this is z and suppose if I plot pore-water pressures so in the initial condition what is going to happen is the value of pore-water pressure which is developing here as well as here is going to be a constant I hope you agree with this so this is what is equal to ui the moment you apply the pressure the pore-water pressure is equal to ui at t equal to 0 now what is going to happen something intermediate is going to happen at t equal to t let us say very difficult to estimate or if I want to estimate what is the value of pore-water pressure at a time at this time what I will have to do is I will have to solve this expression so this is what fz t would be intermediate state the third state is when I say t is tending to infinity so the moment you say t tending to infinity what is going to happen because these two are drainage boundaries the pore-water pressures at this point in the soil mass and at this point in the soil mass is definitely going to be 0 and the pore-water pressure is going to be completely 0 because t is infinity that means the other bound of the plotting would be u equal to 0 which is t tending to infinity and this is the case when you have t equal to 0 initial case this is part clear in between what is going to happen let us try to discuss what is the form of this equation what type of equation is this it is a parabolic function alright so that means in between whatever is going to happen is a parabolic distribution of u as a function of that and t so suppose if I consider the line of symmetry of the sample this line of symmetry also defines the symmetry of the drainage boundaries so whatever I am doing right now is valid only for the drainage boundaries 1 and 2 put together if I change the boundary conditions of drainage this solution is going to change now it so happens that the moment t increases from 0 to some finite value this is how so this is how it is going to look like is this okay now this line is known as an isochrone isochrone what we have done is we have just represented the u as a function of depth for a given time so this boundary please remember corresponds to t equal to 0 and this boundary corresponds to t tending to infinity alright so these are isochrones as time progresses what will happen the isochrones will keep on shifting towards the left hand side so in this direction the time is increasing and a stage comes when the pore order pressure becomes 0 at t equal to infinity fine so this is the line which you are meeting this is the initial value so what is the interpretation of these isochrones if I consider a point over here on this graph or on this curve can you tell me what this point is going to correspond to this point is nothing but this equation where is the z this is the z and where is the time this isochrone corresponds to a certain time okay so at this point the pore water pressure is this value there is a symmetry of the isochrones along the axis of the sample provided the boundary conditions are same both sides training now if I draw a tangent over here what is this going to give me and if I draw a tangent over here what I am going to get del u by del z at a given t what is del u by del z velocity vector got it so this is the velocity of the water which is present in the pores at this point and what is the direction it is moving up this is moving down because there is a free draining condition here there is a free draining condition over here at this point what is happening if I draw a tangent this is a velocity vector and this is a velocity vector okay so this is the interpretation of the isochrones there is one thing more which we would like to discuss about the isochrones and how they can be utilized in day to day practice of geotechnical engineering I hope you are realizing one fact that isochrones are nothing but a form of a solution the graphical form of this equation and ultimately as t tends to infinity starting from t equal to 0 isochrones are also going to tell you how the consolidation process is occurring that means if I draw a lateral horizontal line suppose here now this is the value of out of so much of the pore water pressure look at this x axis is ui all right suppose certain this is the value of u now what is remaining is ui-u that means if I want to define the term degree of consolidation okay so degree of consolidation is defined as normally capital U and this is equal to the total amount of pore water pressure which is available to get dissipated to the one which is remaining so I will be defining this as ui-u over ui have you understood this now this can be written as 1-u upon ui imagine that I wanted to find out what is the settlement undergone by a soil mass because of the external loading of known properties like K, MV initial wide ratios gamma D and all these things are known moisture contents saturation is 1 because we have saturated the sample okay all these properties are known what I have to do is I have to get the values of pore water pressure as a function of z and t is there any other way to do this rather than going for all these mathematical analysis experimentally yes there is a way ultimately what I am measuring is pore water pressures which you have measured until now in several cases so what I have to do is I have to just insert several piezometric tubes all along this is okay and whatever amount of pore water pressure comes over here I can use this for drawing these contours is very difficult to conduct this type of experiment because the size of the piezometric tubes is quite big and the size of the sample if you remember is hardly 25 mm in the laboratory but suppose if you want to see what is happening in the field you can insert these type of in potential means sorry potential meters or sort of you know capillary tubes and you can obtain what is the pore water pressure which is developing because of the external loading so you are free to do whatever you want to do the most interesting thing which we have got from these simple analysis is first of all the equation it is nature the solution of this equation is going to give me the pore water pressure and using this pore water pressure I can find out the degree of consolidation now degree of consolidation is going to tell me how much the settlement of the system has occurred that means degree of consolidation can also be defined as suppose if I remember suppose if I extend this analysis to E sigma prime curve starting from E0 I attain EF and corresponding to E0 the sigma is sigma not prime and this is sigma F prime ok. So this is a sort of a situation where I can say EF minus E0 this will be in the denominator and something intermediate which I can define as ET will be coming over here so this is another equation which I can use for defining the degree of consolidation this is ok that means the void ratios at a given time minus initial void ratios upon EF minus E0 I hope you understand EF is going to be lesser than E0 ET is going to be lesser than E0 so negative negative will cancel out and this will become E0 minus ET upon E0 minus EF so there are several ways of doing this now let us complicate this function a bit because in real life these type of boundary conditions are rare so what we have discussed until now is the laboratory setup where if you remember we had taken a consolidometer ring we call it as a oedometer ring we kept the specimen there we put two porous stones there saturate the entire thing apply the external load clear. So this equation is specifically simulating what you did in the laboratory if the sample or the specimen in the water bath can be maintained at elevated temperature if I do the whole exercise by increasing the temperature of the sample now we have a coupled effect you agree no you are not asking this is what people are doing in the western world this is ok THMC of everything that means this whole test has been done under STP standard temperature pressure conditions I can raise the temperature of the sample to certain the value and I can see what is going to happen to the consolidation or the pore pressure are going to change. So those of you who might get a chance to work in thermo active structures would be still analyzing the whole thing only thing is that you would become a function of temperature also and that becomes a coupled phenomena which is complicated so at undergraduate level you are not supposed to learn all these things a bit more of analysis of this fact which we were discussing so at t tending to infinity what is going to happen the value of u tends to 0 and this is going to happen only for z equal to 0 and z equal to 2d because in between we are not sure about what is happening so we have to do it either analytically or experimentally now suppose if I say here we have well defined distribution of the pore water pressures now if I say that the u is a function of z I do not know what type of variation of the pore water pressures is so I can also write this as 1 minus if I say u z and from 0 to 2d it is varying in certain fashion I can average this out and I can integrate it from 0 to 2d into dz the ui remains same if I integrate ui also in the same form what is going to happen this 1 upon 2d will get removed and this function will also get changed to 2d into ui into dz that means if I say that the initial pore water pressure is also a function of z so this type of manipulations can be done the triviality comes suppose if I give you a situation where the clay sample is draining only from one side this is the impervious strata alright and this side is draining in nature how it is going to happen and when it is going to happen suppose there is a clay layer clay seam which is sandwiched in sands so this is sands and this is sands both sides draining condition what we have discussed over here the porous stone simulates both side draining I change the context of the problem the clay layer is underlain by let us say another silty layer no not silty let us say this is another clay layer or a rock and hence this is going to be impervious now if this is a situation this becomes one side draining so where I said intentionally I am using this 2d term this 2d is defined as drainage path so drainage path is going to be half of the 2d value why because there is a line of symmetry and we have already proved that all these points are they going to drain or not this point remains confused whether it should get drained here or whether it should get drained over here so in other words how many of you have baked a cake ever in your home so if you really want to learn consolidation process go and make a cake and I hope you will realize that despite good amount of heating which you have done in the microwave oven what happens to the central portion of the cake it remains uncooked why this is the answer. So look at the water molecule it does not know whether it should go on the on this side or it should come on this side so that means the central layers are always going to remain wet there is no draining taking place so it is very difficult to consolidate the sample fully alright so the drainage path is defined as half of the 2d let us say d so this is 2 times the drainage path and the d becomes the drainage path through which the drainage is taking place in this case the drainage path is going to be equal to h because this is one side draining only now suppose if I ask you to plot the variation of u with respect to time what is going to happen start from the basics at t equal to 0 the entire load comes on the system the pore water pressure develops this is the value of ui which is equal to delta sigma the external load as time passes by what is going to happen the top layer will quickly dissipate the pore water pressure but this being impervious surface this is not going to dissipate anything and hence what you have done is you have created a sort of a profile of pore water pressure like this is this okay so I will say that this is the profile f as a u as a function of z and t this case is known as a partial drainage case or half drainage case and as I said this is possible when the clays are sitting over impervious system or a clay system which is much more impervious than the clay in which you are trying to find out the settlements the one more thing which is still bothering us is CV is unknown you agree this is a bigger culprit so we do not know how to obtain CV because there is some coefficient now you can always say that if I conduct one dimensional consolidation test and if I use this expression which I had written for CV so CV equal to k over mv into gamma w all right if I know the hydraulic conductivity of the soil if I know it is mv value which I can obtain by conducting the one dimensional consolidation test gamma w is known CV can be obtained simple method normally this is not done we never obtain CV like this what we do is we obtain CV we knows mv we know gamma w we compute k so k is equal to CV into mv into gamma w hydraulic conductivity can be obtained by knowing mv value CV value and gamma w so we differed the problem the problem is not solved yet we still do not know CV clear one interesting thing you must have noticed if I ask you what is the dimension of CV so this is del square u by del z square del u by del t u is truly speaking a pore-water pressure parameter so the units of CV will be l square by t and hence this is known as a diffusion coefficient so those of you might get a chance to do higher research in THMC, THMCB and all those series of contemporary geomechanics you can replace this term by different types of coefficients if I replace u by c concentration gradient in the direction of z concentration gradient as with respect to time change of concentration diffusion coefficient d clear if I am interested in doing how heat migrates in the geomaterials del square theta theta is the temperature rate of change of temperature with respect to distance rate of change of temperature with respect to time and this coefficient is going to be thermal diffusivity clear that becomes alpha so this is how you can interpret the equation now let us come back to the consolidation theory and relook at what we need to do further is this part ok if I solve this equation in analytical form the solution of one dimensional equation can be written as the pore-water pressure equal to n equal to 1 to infinity 1 upon d half of the drainage path 0 to 2 d ui sin n pi z upon 2d into dz and this term is multiplied by sin n pi z upon 2d exponential minus n square pi square cv into t upon 4d square so this is the solution which is known as a Fourier series which you obtain when you solve one dimensional consolidation equation if I simplify it further if I assume that ui is constant all throughout this can be written as n equal to 1 to infinity 2 times ui upon n pi 1 minus cos of n pi sin n pi z upon 2d exponential this term if I assume that tv is equal to cv into t upon d square this is what is known as a time factor it is a dimensionless factor now what is the interpretation of this parameter t normally we have used v term here and this defines the drainage which is taking place in the vertical plane because of the vertical loading so one of the characteristics of tv is that this is material dependent that means if I conducted laboratory test and if I take out the sample of the soil from the field and the same sample has been tested in the laboratory I can say that this will be equal to tv in the field there are several possibilities the possibilities are this tv is cv into d upon d square in the lab and this will be equal to cv into t upon d square in the field as I said as long as the material is constant same material cv cancels out so you have a relationship between t and d in the laboratory in the field so basically you are doing a sort of a modeling of models which you normally do only thing you have to keep in account is that this should be the drainage path in the field condition I might be doing a laboratory test by maintaining the you know double drainage condition so this is a double drainage condition both sides the drainage is allowed so in this case as we discussed just now the drainage path is going to be this however there could be a situation in the field where the d will become let us say the total thickness of the deposit as we discussed sometime back this is equal to h the total thickness of the deposit and this happens to be a draining boundary the sand layer so that is the only difference clear so another interpretation could be this is regarding the drainage paths the second interpretation could be about the time physical time because this t is the physical time you remember the time at which a pore of pressure is being determined at a given depth now what this might tell you is to achieve a certain value of TV which is linked with you and you we have defined as the degree of consolidation in other words what we are going to get is I am going to get the time required under the laboratory condition and in the field condition which is going to tell me how much amount of degree of consolidation has occurred in other words I can simulate by doing a simple laboratory one-dimensional consolidation test the real-life situation so this is IQ balance which we use okay now if you solve this equation further you will be getting some more interesting results the first one is if the degree of consolidation is less than 60% we get TV equal to pi by 4 into u square for u greater than 0.6 the TV value is defined as 0.933 log 1-u-0.085 in literature you will get or in the books you will be getting relationships between u 0 to 1 100% I can put it in percentage also or it is up to you and suppose if I put the TV value here which is equal to CVT upon D square x axis this is how the graph looks like the first thing this graph tells you is as physical time increases the TV value increases TV value increases means degree of consolidation increases clear and this is what is getting depicted over here so please be careful as long as the observing the y axis is concerned this is from 0 to 1 0 to 1 and this is the TV value I may have different boundary conditions which you can substitute the interesting thing here is the relationship between u and T this again is a parabolic equation provided u is less than 60% the moment this becomes more than 60% there is a linear nature coming in the picture constant and 1-u of log this is going to be an exponential term or logarithmic curve so this culprit is still unknown is it not so we have to do something to get CV value because what we have done is very conveniently we have used this term over here also to define the non-dimensional time factor which is an integral part of the one-dimensional consolidation equation.