 Welcome to lecture series on advanced geotechnical engineering course. We are in module 3, this module is on the compressibility and consolidation and lecture 3 of module 3. So in the previous lectures we have discussed about how the loads which are actually applied on the surface can cause stresses and then so called displacements. Now in this lecture we are going to look into how these stresses can cause settlements particularly for fine grained soils like clay soils. So in this connection we will try to look into the concepts of consolidation and Terzaghi's one dimensional consolidation theory and application in different boundary conditions. We will discuss about the ramp loading and how we can actually determine the consolidation characteristics of a soil and then that is coefficient of consolidation there are two methods are there and then how we can distinguish between normally consolidated and over consolidated soils and then we look into the several aspects like compression, corrosion, secondary consolidation in the subsequent lectures. So in this lecture we are going to discuss about the Terzaghi's one dimensional consolidation theory. So consolidation basically when the materials are loaded or stressed they deform or strain so that we have actually seen in the previous lectures when the materials are loaded the stresses are actually subjected and transferred to the ground. And response of the under load is instantaneous for sandy soils for certain soils like granular soils or sandy soils where the permeability is very high the response under the load is very high. But fine grained soils like clay soils require a relatively long time for undergoing deformations. So we have to demarcate here the response under the load is instantaneous for sandy soils and clay soils require a relatively long time for undergoing deformations. So when the soil is loaded it will compress because of the following reasons one is the deformation of the soil grains and compression of air and water in the voids so that means that the air which is actually with the voids and water within the voids will get compressed and the squeezing out of water and air from the voids. So because of the deformation of soil grains and compression of air and water in the voids and squeezing out of water and air from the voids is one of the causes for the these are the causes for the you know the soil why the soil undergoes compression. So according to Terzaghi in 1943 he defined consolidation as a decrease of water content of a saturated soil without replacement of the water by air is called a process of consolidation. So the expulsion of water from a saturated soil upon loading and without replacement of the water by air is called the process of consolidation. So when a saturated soil or you know which have low coefficient of permeability are subjected to compressive stresses due to foundation loading or due to any structure loading the pore water pressures will immediately increase then it takes some time for the pore water pressure to dissipate. So because of the low permeability of the soil there will be a time lag between the application of the load and the expulsion of pore water and thus the settlement. So this phenomenon is also called as consolidation. So because of the low permeability of the soil particularly a fine grained soil like clay soil there will be a time lag between the application of the load and the expulsion of the pore water and thus the settlement. So this phenomenon is called consolidation. So the total vertical deformation at the surface resulting from the load is called settlement. So we have to understand in this case particularly the total vertical deformation at the surface resulting from the application of the load is called settlement. So that means that if you have got a footing then the footing of particular nature it undergoes settlement. So the movement may be downward with an increase in load or upward with a decrease in the load. So sometimes when the load is actually decreasing there can be a possibility that the swelling or heaving can take place. So the downward movement is actually called as settlement. The total vertical deformation of the surface resulting from the load is called settlement. So the temporary construction excavations and permanent excavations will cause a reduction in stress and the swelling may result. And lowering of the water table will lead to settlements due to increase in the effective stress within the soil. So lowering of the water table will also lead to settlements because of the increase in the effective stress within the soil. So here in this slide what we are trying to determine is the settlement the total vertical deformation at the surface resulting from the load is called settlement and the movement may be downward with an increase in load or upward with a decrease in the load. So if you have got temporary construction excavations or permanent excavations because of the removal of the material will cause a reduction in the stress and the swelling may result. The lowering of the water table will lead to settlements that is due to increase in the effective stress within the soil. So the design of foundations for engineering structures it is required to know how much settlement will occur and how fast it will occur. So for any soil deposit when we are actually trying to construct a structure for a designing foundation we have to calculate what are the likelihood of settlements and how fast they will occur. So excessive settlement may cause structural as well as other damages especially if such settlement occurs rapidly. So excessive settlements may cause structural as well as other damages especially if such settlements occur rapidly. So the total settlement ST, S suffix T of a loaded area has three components. So if you look into that ST is divided broadly into three this thing ST is nothing but SI plus SC plus SS and SI is nothing but the immediate or distortion settlement and we also call it as elastic settlement that is the readjustment of the particle or rattling of the particles with that the immediate or distortion settlement will be there. And SC is nothing but the consolidation settlement which is known also known as the time dependent settlement and SS is called the secondary consolidation or secondary compression settlement. And this is also time dependent but this occurs at the end of the process of consolidation under the constant effective stress. So that is nothing but the creeping of soil particles will occur and the void ratio will continue to change under the constant effective stress. So this is evident for some examples like municipal solid waste which is a man made of the solid waste material which will undergo very large secondary consolidation settlements because of its own characteristics because of the heterogeneous mixture of the matrix as well as the ongoing biodecomposition which actually takes place because of the biochemical changes which are actually happening in the municipal solid waste with time. Another example is the PT type of soils because of the fibrous nature they undergo this secondary consolidation very high. So that is the reason why the marshy lands you know particularly the construction marshy lands they are very highly prone for secondary consolidation settlements. So the total settlement is actually divided into SI plus SC plus SS. So the immediate or elastic or distortion settlement is SI then SC is nothing but the secondary consolidation settlement which is a time dependent and SS is nothing but the secondary consolidation this is also a time dependent settlement. Now let us look into if a volume of a soil changes the effective stress must change. Since the soil grains and water are assumed to be incompressible the volume of saturated soil can only change as the water is squeezed from or drawn into the pore space. So since the soil grains and water are assumed to be incompressible the volume of a saturated soil can only change as the water is squeezed from or drawn into the pore space. So as the water flows from the innermost pores towards its boundaries flow will be governed by Darcy's law but since the rate of flow must be finite the soil volume will change with time. So consequently the effective stress must change with time so will be the hydraulic gradient and rate of flow. So if you look into this the variation of soil volume is a function of effective and total stresses and pore water pressure and seepage and compressibility nature of the soil. So the seepage and compressibility of the soils, so variation in soil volume is a function of effective and total stresses the pore water pressure as seepage and compressibility. So if the volume of the soil changes the effective stress must change that is if suppose if the water is expelling out of soil then there will be a reduction volume then the effective stress you know will have to change. Since the soil grains and water assumed to be incompressible the volume of a soil saturated soil can only be changed as the water is squeezed from or drawn into the pore space. So as the water flows from the innermost pores towards its boundaries the flow will be governed by Darcy's law but since the rate of flow must be finite the soil volume will change with time. So the time dependent process of volume change in soil as the water is squeezed from the pores is known as consolidation. So this is also another definition which we look into it the time dependent process of volume change in soil as the water is squeezed from the pores is known as consolidation. During consolidation process the transient flow of water from and through the soil material occurs. So during the consolidation process this process is also called the transient because the pore water pressure keeps on changing with time. So the seepage forces experienced by the soil structure will vary with time and the soil structure itself may deform under the varying load it sustains. So during the consolidation process and this consolidation phenomenon is also called a transient flow phenomenon where the transient flow of water from and through the soil material occurs. The seepage forces experienced by the soil structure will vary with time and the soil structure itself may deform under the varying load it sustains. So the relationship between the volume of the soil and the effective stress and which is the relationship independent of time is known as compression. That means that the relationship between change in volume of soil and nothing but the change in volume of the soil is nothing but change in void ratio because if you assume that the area is A then in that case the effective that is change in void ratio and the effective stress if there is a which is the relationship independent of time and this is basically known as compression. So compression is nothing but a relationship between the volume of the soil and the effective stress and which is the relationship independent of time is known as compression. And the time dependent process of volume change in soil as water is squeezed from the force is known as consolidation and during the consolidation process the transient flow of water from and through the soil material occurs and because of this what will happen is that the seepage forces experienced by the soil structure will vary with time and the soil structure itself deforms under varying load it sustains. So if you compare clay versus sand we said that clay is one material where the fine grain nature and the permeability is very low and the sand where the example part of a coarse grain material where with 0 percent fines let us say and the compressibility if you look into that is medium to very low and it is less than few centimeters per sands. In case of clays is actually medium to very high up to 1 meter so the settlements actually can go the compressibility can actually go up to 1 meter to 1.5 meter in certain type of soft clays and permeability is high and the drainage happens during the construction in case of sands permeability is very high and the drainage actually happens during the construction. In case of clays very low and nearly undrained during construction that means that in case of clays the permeability is very low and nearly undrained during construction and if you look into the stress strain behavior for the sands and clays essentially the same basic principles are applied. So we have to note here that the sands actually have got high permeability and drainage actually happens during construction and clays actually have very low permeability and they behave like nearly undrained during construction and the compressibility can be very high up to 1 to 1.5 meters for certain type of clays and the it is actually rated as medium to very high for clays and for sands medium to very low and sometimes is less than few centimeters compressibility for sands. So in this particular slide if you look into this the time versus compression of a sand is given and you can see that when the within fraction of few seconds then you can see that the sand actually is under about 80% of compression. So that means that the settlement of the granular soil takes place very very instantaneously. So consider the case where if you consider the case where the granular materials are one dimensionally compressed the deformation takes place in a very short time due to the relatively high permeability of granular soils. The compression of sand occurs during the construction stage itself. So the compression of the sand occurs during construction stage itself. As settlements of granular layers occurs relatively fast they may be detrimental to a structure when sensitive to rapid settlements. Suppose if a structure is actually sensitive to rapid settlements then the settlement of a granular soil can be detrimental in affecting the structure behavior. So compressibility of the sands if you look into the deformation takes place in a very short time due to the relatively high permeability of granular soils and the resistance is more of frictional in nature and the compression of the sand occurs during the construction stage itself. Now let us look into this particular behavior of the soil compression and consolidation model. So here we have got two graphs which are actually plotted in this slide where the upper one is actually versus time versus the stress and the bottom one is that time versus volume and V0 is the initial volume of the soil and this is divided into four stages that is stage 1, stage 2, stage 3 and stage 4 and this particular stage where the general hydrostatic water table is there that is the total stress is equal to sigma naught dash effective stress plus hydrostatic water pressure that is U0. So in this particular stage at time you know the change in volume is 0 and the settlement is 0 the change in volume is 0 and the total stress is equal to sigma naught dash plus U0 where U0 is the hydrostatic pressure because of the prevalent ground water table in a given depth. Now when at time t is equal to 0 in stage 2 let us assume that this is the initial effective stress that is sigma naught dash and the soil surface is have been subjected to an increase in load which is actually called as delta U it is a delta sigma, the delta sigma is the incremental stress which actually has been applied and once this is applied the immediately at time t is equal to 0 the water in the water within the voids will be subjected to the pressure equivalent to that of the intensity of the pressure which has been applied on the surface of the soil. So that is nothing but becomes delta sigma becomes delta U. So that is the reason why at that particular moment at time t is equal to 0 the pore water pressure is U0 plus delta U and effective stress is sigma naught dash. So in this stage also no volume change occurs, no settlement occurs and the effective stress is sigma naught dash but the pore water pressure is U0 plus delta U. Now in the stage 3 where the commencement of the dissipation of pore water pressure takes place that is that once that the delta U starts decreasing that means that the soil grain starts supporting the or attracting the load. In such situation this actually can be at that particular point if you can see that this so called sigma naught dash plus delta sigma t is the stress already delta sigma t dash is actually the stress which is already transferred to the grains and this portion of the pore water pressure is at to be dissipated and wherein you have the delta U t is actually at to be dissipated. Then further when you go further then it reaches to the so called U0 plus delta U tends to become to U0 that means that this is the portion this is the time when t infinity where the process of the consolidation takes place. So in this circumstances what will happen is that the total the effective stress now changes to sigma naught dash to delta sigma that means that once the pore water pressure you know rises and dissipates over this particular in this pattern and during this particular time lag there is a volume change will occur that means that volume change is nothing but the settlement and you know that this process actually is told can be said as the consolidation the consolidation process. So at the end of the consolidation process sigma naught dash plus delta sigma is the new effective stress and the U0 is the original hydrostatic more or less the water table remain if the water table remain constant then it is the original hydrostatic pressure conditions are maintained. So in this way what we are saying is that if you have a soil deposit and if it is subjected to a certain time of loading and it takes for a period of time to you know the transfer the load to the effective stresses and you know the effective stresses tries to pick up that load and then you know gain to or reach to that particular sigma naught dash to delta sigma at the end of the process of consolidation. Now this also can be explained you know by considering a spring energy model let us assume that we are having a rigid container in which a spring which is actually place and the container is actually filled with water and we have a piston which is actually connected to the spring and the piston actually has a hole and which actually represents the rate at which water flows out of the soil that is nothing but the permeability of the soil and initially assume that the you know we have the piston and the spring and the container is within the portion within the container below the piston is filled with water and no stress is applied no delta sigma is applied in that case the total stress is equal to if due to the sigma naught dash is equal to if a total stress is actually equivalent to that is applied then sigma naught dash is equal to sigma naught dash plus u naught and where in here where in here the soil grains actually they are represented by the spring and water as the water in the cylinder as the pore water. Now moment to the delta sigma is actually applied then you know the pore water pressure the water is within the cylinder rises to delta u and moment once the you know the drainage conditions commences that is in stage 3 then you know the solely the spring starts undergoing compression that indicates that the raise in the effective stresses here you know nothing but the indication of this compression of the spring where in the sigma naught dash continues to increase and reaches to sigma naught dash plus delta sigma. So you know this particular moment you know the final hydrostatic conditions are actually reached then you know the original hydrostatic equilibrium conditions are maintained and then you know the spring is now compressive to such a extent that it is the stress is now sigma naught dash plus delta sigma and u naught is the hydrostatic pore water pressure. So this model explains the soil compression and consolidation behavior you know. So this is you know this what actually is the process of consolidation explains the process of consolidation. The process of consolidation is actually explained here if you are having a spring which is loaded the piston which is loaded with sigma naught dash plus delta sigma and the wall when it is actually close then the water pressure which is nothing but sigma naught dash plus delta u. That means that initially the delta u is 0 here and moment the water pressure the load is actually delta sigma is applied the delta u plus delta sigma is actually is born by the water. Now what will happen is that once this wall is open then this yellow portion starts you know covering the blue portion that means that that is nothing but sigma naught dash but delta sigma t at any time if this happens then what will happen is that this is the portion which is actually transferred to the soil grain and this portion is at to be dissipated. Moment this reaches to this level this point then it is called as the so called the completion of consolidation and that is called sigma naught dash plus delta sigma. So in this case if you look into this the variation of the pore water pressure with time and with depth is actually also called as the isochrone. So the moment at time t is equal to 0 when instantaneously when the pressure is actually applied and this is actually called as the first isochrone and moment as the dissipation of the pore water pressure that is as the water is actually leaking out of the cylinder then there is a possibility that the isochrone actually migrate towards this and this is the called as the final isochrone. So this is at a time t and this is the isochrone at this particular time and this is at time t infinity that is at the end of the consolidation and this is the final isochrone. So the isochrones are nothing but you know identical pore water pressures at a particular time. So identical pore water pressures at identical times. So if we are having you know that is for single a small clay layer where we have spring analogy but the same spring analogy can be extended if you are having a clay layer which is you know have got W open layers both the top and bottom and clay layer is actually saturated and clay which is actually having a thickness 2H and if this is this clay is actually completely saturated and where we have got you know the upper layer and bottom layer or the open layers or also called as you know the drainage layers. So this is actually is indicated by a spring analogy with multiple springs connected in a series where in 1, 2, 3, 4, 5 these springs are actually placed like this and then each and every compartment is connected and the upper and bottom let us say that both the layers are you know sand layers then both the walls are actually open simultaneously or if the one layer let us say that if you are having a rock layer at the impervious layer at the base then this wall will be kept as close then you know the water will have to escape through you know this particular upper wall only. So if you are actually having if these walls are actually you know let us assume that open simultaneously simulating the double open layer double drainage condition then in that case what will happen is that first the spring 1 and 5 and undergo compression then subsequently spring 2 and 4 will under compression and finally the last spring to undergo compression is 3 that means that the springs which are actually close to the you know the wall they actually attract the you know the effective stress you know very rapidly and followed by the springs which are actually further away from the drainage phase. So this is you know indicates say for example this particular is indicated by a pressure variation with the depth so this is the initial effective stress variation in case of spring analogy it can be indicated like a uniform pressure but for convenience for the clay layer this is indicated with certain surcharge and then this is the effective stress. If this is the effective stress then what will happen is that at time t is equal to 0 when it is subjected to increase in load due to some delta sigma then the water pressure that means that this is the it will happen that the first isochron will develop here and then if you notice here at time t greater than t is equal to 0 that is t1, t2, t3 when it actually happens then what will happen is that immediately the pressure at the drainage phases actually drops down to 0 and that means that the soil which is actually close to the clay soil which is actually close to the drainage phase is the first one to you know transfer the stresses attract the stresses and the soil the water transfers the stresses to the soil rapidly and you know the effective stress here is nothing but delta sigma and here also it is delta sigma but when it comes to a central point here and you can see that the being at the mid depth of the clay layer and also the hydraulic gradient which is actually there at the center is very, very low and also is delta u by delta h is actually 0 so because of that what will happen is that the flow is actually 0. But in the case when you look into this particular portion that delta u variation with the delta z the hydraulic gradient is actually infinity here. So because of the you know because of this you know sharp hydraulic gradient changes the rapid flow actually happens in the drainage phases and the transfer of the effective stresses becomes very, very you know rapid at the boundaries of the clay layer. You see in case if the clay layer is having let us say that half closed layer that is at the rock layer is at the base that means that you know this peak will be at this center that means that approximately this will become the you know the drainage path the water tries to flow and then you know the water pressure escapes only in this direction provided if you are having only one dimensional consolidation. So this is you know the first isochrone at time t is equal to 0 that is t is equal to 0 t is equal to t0 is the time at the moment the application of the load has been applied. But once t1 t2 t3 times are there this is the isochrone and t2 is greater than t1 t3 is greater than t4 then this is for t infinity. So what will happen is that this if you look into this at this particular depth this is the sigma dash z plus delta sigma d dash so this much portion of the effective stress is already transferred to the soil. But at this portion here you can see that the soil is still having the effective stress which is actually at sigma not dash whatever the initial effective stress is there and that is still prevails at the center. So here we will look into this here the migration the movement of the isochrone actually happens here ultimately and this process this is actually the time which is required for the dissipation of the pore water pressure also called as the time lag and which is you know depend upon the permeability of the soil. So at the mid depth the decrease in excess pore water pressure is small compared to the change at the top and bottom. So as a result what we are trying to explain is that it takes a long time from the center of a doubly drain layer or at the bottom of a single drain layer to dissipate the excess pore water pressure and the slope of isochrones at mid depth is almost like delta u by delta z is equal to 0 and no flow condition actually occurs at z is equal to 0 and z is equal to 2H that is at the both top and bottom the delta u by delta z is infinite and the flow is largest right at the drainage spaces. So because of the because the flow is actually largest at the right at the drainage spaces that means that because of that there is a possibility that you know the you know the soil undergoes so you can say that consolidation very rapidly at the boundaries. So the Terzaghi's one-dimensional consolidation theory the basically the assumptions which are actually involved the clay is homogeneous and 100% saturated and drainage is provided at both the top and bottom of the clay layer and Darcy's law is valid and this is what which has been assumed but afterwards the theory can be derived for different boundary conditions and Darcy's law is assumed to be valid and the soil grains and water are assumed to be incompressible and compression and flow are one-dimensional that means that here in this case one-dimensional is assumed but in the later part of this lectures we will also discussing the three-dimensional that is this one-dimensional as well as the two-dimensional flow if that actually happens there is a possibility that the consolidation can be accelerated and the deformation of the soil occurs in the direction of the load application. So here the deformation of the soil occurs in the direction of load application and the unique relationship between delta E and delta sigma so here during the process of consolidation that is the during the process of consolidation there is a there exists an unique relationship between delta E and delta sigma. So the deformation of the soil occurs in the direction of the load application and compression and flow are one-dimensional nature and the soil grains or water are incompressible and Darcy's law is valid. So here the relationship between delta E and delta sigma are given so where the value of the dE by d sigma dash that is the change in effective stress and change in void ratio. So initially if you are having a void ratio say E0 and at the sigma0 dash moment the delta sigma is applied the sigma1 dash is now nothing but sigma0 dash plus delta sigma so because of the increase in effective stress in soil there is a reduction in the void ratio from E0 to E1. So E0 minus E1 that is nothing but the dE or delta E. So the value of dE by d sigma dash is constant for relative range of effective stresses and the permeability k is also asymptotic constant for the relative range of effective stresses and the time lag in consolidation is entirely due to the low permeability of the soil. The time lag in the consolidation is entirely due to the low permeability of the soil. Terzaghi's one dimensional consolidation theory is essentially a small strain theory that is the applied load increment produces only small strains in the soil and therefore both the coefficient of compressibility Av and k remain essentially constant during the consolidation process. So the applied load increment produces only small strains in the soil and therefore both the coefficient of compressibility that is Av which is nothing but delta E by delta sigma dash which is actually shown in the previous slide where Av is equal to delta E by delta sigma. The slope of this line is defined as coefficient of compressibility which is nothing but delta E by delta sigma dash and k the coefficient of compressibility Av and k remain essentially constant during the consolidation process and the constant Av implies that there is no secondary compressor compression if the secondary compression occurs that the relationship between delta E and delta sigma dash would not be unique. So secondary compression as we defined earlier it is the change in void ratio that occurs at constant effective stress. So the Terzaghi's one dimensional consolidation theory essentially the small strain theory and this means that the applied load increment produces only small strains in the soil and therefore the coefficient of permeability and coefficient of compressibility remain constant during the consolidation process and constant of compressibility means that there is no secondary compression during the process of consolidation which we what we call is the primary consolidation. If the secondary consolidation or secondary compression occurs then the relationship between delta E and delta sigma dash would not be unique. So now we will try to derive the so called the Terzaghi's one dimensional consolidation equation. Now consider when the clay is subjected to an increase in vertical pressure delta sigma and the pore water pressure at any point A will increase by U. So consider a soil strata where here for example the delta sigma is equal to 10 kilo Pascal's has been shown here but assume that you have got a clay and the water table is at this point and it is doubly drained and clay layer is the portion which is in between sand which is between two sand layers and on the top in the ground surface there is a delta sigma which actually has been applied. Now consider a small element having volume dx dy dz and z is this direction so x and y directions are z is this direction and the plane perpendicular to this z direction is the x and y. So in this x and y direction the flow actually happens that means that the water enters the element and water leaves out the element. So the water enters into the element is actually nothing but q is equal to A v continuity equation where in q in is equal to dx dy dx dy is nothing but the element area into v z and q out is nothing but that v z plus dou v z by dou z into dz into dx dy that is because of the pore water pressure change. So in consider what we are discussing is that consider a small element having the volume of dx dy dz at A so v is equal to dx dy dz at that is at point A. So in case of one dimensional consolidation the flow of water into and out of the soil element is one dimension only that is in the z direction. So in that case q x is equal to q y is equal to change in flow in x and y direction also 0 because this one dimensional consolidation is valid if you are having a soil deposit where it has been subjected to large area loadings. The area loading is such that it is assumed that it is you know to the for the convenience assume that it is instantaneous in nature and it is extending large area extent. In that case what will happen is that the one dimensional consolidation only predominantly takes place. If you are having a finite area which is actually loaded there is a possibility that the three dimensional flow in y direction x direction as well as the you know so called z direction it can happen. So but in case with the large area loadings the one dimensional consolidation is actually possible. Let us assume that for this one example for the one dimensional consolidation is that if you are having a soft clay when it is actually loaded with this area fill which is about say 1 or 2 meters extending over the entire area then that case you know the area extent of the area is much larger than the thickness of In that case what will happen is that the one dimensional consolidation only predominantly happens. Now here in this particular slide the change in flow that is outflow to inflow that is difference qz plus 2qz minus qz is given as rate of change of volume of soil element which is nothing but dou v by dou t. So qz is defined as we have written in the next slide as vz into az which is nothing but kz iz into az and kz is the coefficient of the permeability in vertical direction that is z direction and iz is dou h by dou z and flow is actually happening in dx dy perpendicular to dx dy so az is equal to dx dy. Now qz plus dqz is equal to kz plus iz plus dz into dx dy so by writing kz into dou h by dou z plus dou square h by dou z square into dz into dx dy so by taking the subtraction with qz plus dou qz minus qz what we will get is that kz into dou square h by dou z square into dx dy dz. So dou v by dou t is now nothing but dou v by dou t is nothing but by substituting and simplifying dou v by dou t is nothing but k into k is indicated as now here kz is indicated k dou square h by dou z square into dx dy dz so using h is equal to u by gamma w so we can write differentiate this partially dou h by dou z is equal to dou u by dou z into 1 by gamma w where gamma w is the unit weight of water and differentiating once again dou square h by dou z square is equal to dou square u by dou z square into 1 by gamma w. So this is dou square by h by dou z square so the substituting for dou square h by dou z square here it becomes k by gamma w into dou square u by dou z square is equal to this if you bring it here it becomes 1 by dx dy dz into dv by dt. So here this particular equation now turned out to be k into k by gamma w into dou square u by dou z square is equal to 1 by dou x 1 by dx dy dz into dou v by dou t. Now here during the consolidation the rate of change of volume is equal to the rate of change of the void volume so the dou v by dou t will be equal to dou v v by dou t so where v v is the volume of the voids in the soil element so we can write v v is equal to e v s and now what we can do is that the particular equation we can write in terms of dou v by dou t we can write like dou v by dou t is equal to this by differentiating this we get dou u by dou t v s into dou u by dou t so for dou v by dou t if you substitute v s into dou u by dou t so further v s is rearranged as v s is equal to v by 1 plus e and with that where e is equal to v v by v s with that what we can write is that v by 1 plus e we can write so this is nothing but v s the volume of the solids is written in terms of v by 1 plus e into dou u by dou t and this is equal to dx dy dz which is nothing but v is nothing but dx dy dz by 1 plus e into dou u by dou t now equating this equation 2 with 1 that is k by gamma w dou square by dou z square with this one what we get is that dx dy dz will get cancelled and what we will have is that k by gamma w into dou square u by dou z square is equal to 1 by 1 plus e into dou u by dou t so here what we have is that the equating 1 and 2 we have got now k by gamma w into dou square u by dou z square is equal to 1 by 1 plus e into dou u by dou t now further what we do is that the change in void ratio dou e is due to the increase in the effective stress change in void ratio that is reduction in the void ratio when it is actually happening from e0 to say v e1 the due to the increase in the effective stress or increase in the effective stress and also we have to find note that the decrease in the excess forward pressure increase in the effective stress which is preceded by the you know the decrease in the excess forward pressure so assuming that there are linearly related so we can say that dou e is equal to minus a v into dou dou sigma dash by a minus a v into a v is nothing but coefficient of compressibility dou into dou delta sigma dash so where a v is the coefficient of compressibility again the increase in the effective stress is due to the decrease of the excess forward pressure so we can write dou e is equal to a v delta u so by you know substituting this dou e by dou t is equal to a v delta u by dou u by dou t so we can write the previous equation that k by gamma w dou square u by dou z square is equal to 1 by 1 plus even dou u by dou t so what we have done is that we you know substitute for dou u by dou t is equal to a v into dou u by dou t so k by gamma w dou square u by dou z square is equal to a v by 1 plus e into dou u by dou t so this a v by 1 plus e is nothing but defined as m suffix v and that is the coefficient of volume compressibility so coefficient of volume compressibility is a parameter and where the units of this are nothing but meter square per kilo Newton a v by 1 plus e naught because coefficient of compressibility units are meter square per kilo Newton so this is actually followed exactly the same and dou u by dou t so k by gamma w into dou square u by dou z square is equal to a v by 1 plus e into dou u by dou t or which is nothing but m v is equal to dou u by dou t so in this figure you know this pressure void ratio relationship is actually explained here suppose p1 which is nothing but sigma sigma naught dash in our case and if that happens delta sigma and this is sigma 1 that is sigma 1 is nothing but sigma naught dash plus delta sigma this is end of the process of consolidation so this portion if there is increase in the pore water pressure this is the pore water pressure increase this is the void ratio which is actually decrease at any time t and this is the pore water increase in the effective stress and this is the pore water pressure at to be dissipated so the using that relationship this is what actually this explains that you know the increase in effective stress is due to the decrease in the excess pore water pressure so because of that you know what we have written is that dou u is equal to a v delta u a v dou u and which by writing by dou u by dou t is equal to a v by a v into dou u by dou t so that's how we have got k by gamma w into dou square u by dou z square is equal to a v by 1 plus e into dou u by dou t. So now further continuing we can say that now the dou u by dou t is equal to k by gamma w m v dou square u by dou z square so this particular equation is called as dou u by dou t is equal to c v into dou square u by dou z square is the basic differential equation of the Terzaghi's one dimensional consolidation theory and we can be solved with proper boundary conditions that means that we have to see whether you know the pore water pressure whether it is having open layer at the bottom at top and then what is the variation of pore water pressure so for the different variations to the pore water pressure within the soil and this can be solved. So this dou u by dou t is equal to k by gamma w m v so if you look into this here k is equal to coefficient of permeability is related as c v gamma w into m v so coefficient of permeability k is equal to c v gamma w m v is the coefficient of volume compressibility and gamma w is the unit weight of water and the basic differential equation dou u by dou t c v is nothing but the coefficient of consolidation the units of this coefficient of consolidation are the meters per second. So if the c v value is high that means that you know if you look into this here k if the k is actually high the c v value is high and k value is say low very impervious soil the c v value will low so the larger the c v value the faster is the settlements can actually taking place and lower is the c v value the you know that much time that much delay will actually happen in happening the settlements. The c v actually indicates the you know the time rate of the rate of settlements basically in the parameter there is only parameter the soil parameter which is actually there in the Terzaghi's one dimensional consolidation equation. So the c v the coefficient of consolidation which we will discuss further how to determine this in the laboratory and how this is actually found to be dependent on the effective stress where for the range of the loadings which actually can be subjected in the laboratory where c v is equal to k by gamma w into m v. So this is the basic differential equation of Terzaghi's one dimensional consolidation theory and this can be solved with proper boundary conditions. So to solve this equation we assume that you basically the pore water pressure to be the product of two functions and the product of a function of a z and function t are given as u is equal to a 1 cos b z plus a 2 sin b z a 3 and to the exponential of e to power of minus b square c v t is equal to a 4 cos b z plus a 5 sin b z exponential minus b square c v t where a 4 is equal to a 1 a 3 and a 5 is equal to a 2 a 3. So the constants and equation 3 can be evaluated from the boundary conditions which are as follows. So by considering the proper boundary conditions can be solved. So at time t is equal to 0 u is equal to ei that is u is equal to ei means that at time t is equal to 0 the moment the load actually has been applied instantaneously the initial access pore water pressure at any depth is you know is taken and at z is equal to 0 that is u is equal to 0 and at z is equal to h t is equal to 2 h bottom of the clay layer the pore water pressure is again 0 the increase in pore water pressure is 0. So note that h is the length of the longest drainage path and in this case which is the two way drainage condition top and bottom of the clay layer h is equal to the half of the total thickness of the clay layer. So water actually half of the portion of the water in the voids actually flows upward and half of the portion of the water flows actually downward to the bottom drainage layer. So here from the above a general solutions can be obtained where u is equal to n integer is equal to 1 to infinity summation a n sin n pi z by 2 h exponential of minus n square pi square t v by 4 where t v is the time factor t v is the time factor to satisfy the first boundary condition we must have the coefficients a n such that u i is equal to n is equal to 1 to n is infinity where a n is sin n pi z by 2 h and this is regenerate as 5 and with that equation 4 is a Fourier series and it is a n can be obtained by a n is equal to 1 by h 0 to 2 h that is the total thickness of the clay layer u i sin n pi z by 2 h into d z and combining equation 4 and 6 we get u is equal to n to 1 to infinity 1 by h 0 to 2 h u i sin n pi n pi z by 2 h d z into sin n pi z by 2 h into exponential minus n square pi square t v by 4 where t v is the non dimensional time factor and which is equivalent to t c v by h square. So h is nothing but here the drainage path in case of a clay having a double drainage layer it is actually is h by 2 in case of a single layer the h is equal to h. So far no assumption have been made regarding the variation of u i with the depth of the clay layer but there can be a number of variations can be assumed like assuming that if you are having a uniform variation if it is u i is actually constant with the depth and if you are having a double drainage layer then h it is equal to 2 h in this case when we look into this and this converts into u is equal to n to 1 2 u 0 by n pi cos 1 minus cos n pi by sin n pi z by 2 h exponential minus n square pi square t v by 4. So here this n is equal to 2 a plus 1 where m is an integer when we put it to this and this is for the one way drainage this is for one way drainage and this is for you know two way drainage and this is actually is assumed to be uniform variation but there is also another case where upper layer is actually impervious bottom layer is previous but that situation may not actually arise in practical terms. If you are having a impervious soil surface or impervious rock here then it is actually one dimensional flow only happens here. So further this actually gets simplified to this one where the Terzaghi's one dimensional consolidation theory where we actually finally narrowed down to u is equal to m is equal to 0 to m is equal to infinity 2 u 0 by m by sin m z capital M z where capital M is nothing but 2 m 1 into 2 m 1 within brackets into pi by 2 exponential minus m square t v. So at a given time the degree of the consolidation at any depth can be determined. So u z is degree of consolidation at any depth z within the thickness of the clay layer is nothing but the excess pore water pressure dissipated by the initial excess pore water pressure. So excess pore water pressure dissipated that is u i minus u, u is actually the pore water pressure had to be dissipated. So u i minus u is you know is the excess pore water pressure dissipated to the initial excess pore water pressure. Now this is nothing but 1 minus u by u i and which is nothing but delta sigma dash by u i is delta sigma dash by u 0. So by using this u by knowing u z at any point of any at any level we can actually construct the isochrone that is for a given time if you are having u z then in u i into 1 minus u z we can actually find out u is equal to u i into 1 minus u z. So in this particular lecture we try to understand about the definition of the consolidation and the consolidation phenomenon we understood that is a transient condition, transient seepage condition and wherein we actually have discussed and introduced about the Terzaghi's one dimensional consolidation theory. Then further we will actually apply how this Terzaghi's one dimensional consolidation theory can be applied to different boundary conditions and also further we extend to ramp loading and other conditions.