Welcome to lecture series on advanced geotechnical engineering course offered by IIT Bombay. We are in module 3 compressibility and consolidation. So this is the lecture 5 on compressibility and consolidation and we have introduced ourselves in previous lectures stresses in soil from the surface loads and we also discussed about the Terzaghi's one-dimensional consolidation theory and application in different boundary conditions and what will happen if you are having a constant rate of loading or ramp loading. Then we proceed further how we can actually determine the consolidation characteristics in the laboratory particularly by using oidometer test. Then we will introduce ourselves to normally and over consolidated soils and then we try to discuss about a typical laboratory virgin compression curve. Then we will try to look into the determination of the coefficient of consolidation what are the different methods which are available and all. So in the previous lecture we have discussed about the instantaneous loading where in the at time t is equal to 0 itself the load will be assumed to be acting on the surface of the load. Surface of the soil but in reality or in practice this placement of this load takes or occurs over a period of time. So if that is done in single or constant rate it is called constant rate loading or single stage ramp loading or if it is done in two stages or bund is constructed in two stages then it is called two stage ramp loading. So in this particular slide the comparison of constant rate of loading and instantaneous loading which is actually discussed by Sew Kogan at all 2014 wherein they said that the average degree of consolidation at any time is significantly less for constant rate loading than the instantaneous loading. So the average degree of consolidation at any time is significantly less for the constant rate loading than for the instantaneous loading. So this is for the instantaneous loading if theoretically if you assume is what we get but constant rate loading we get the u versus t plot or u degree of consolidation versus t v plot for instantaneous and constant loading is shown here and this is the time factor and this is the degree of consolidation. So the average degree of consolidation at any time is significantly less for the constant rate loading than the instantaneous loading. Now we try to look into the response of soil for the stress controlled stresses which are applied like if you are having sandy soils we said that the sands undergo compression relatively faster. So in this particular slide e versus sigma dash and compression time plots for sand are shown here. So as can be seen here within a short span of time of 1 or 2 minutes the sand undergoes the compression. If you are having a loose sand and a dense sand you can see that the loose sand reduces its void ratio from 0.75 to somewhere around 0.52 or so. In case of dense sand there will be less reduction in the void ratio but ultimately at the larger effective stresses they tend to be toward the same void ratio. So the sand deposit compresses immediately on load application, sand deposit the response is very instantaneous and loose sand compresses more than dense sand and loose and dense sand deposits tend toward the same void ratio. So this particular slide illustrates the response of the sand particularly loose and dense sand for the applied loading whereas if you look into for a clay or a fine-grained soil where the soil is having very low permeability the compression varies with time and it takes place over a long period of time. So even after 10 minutes the degree of compression is actually very very less. So it can be seen that it takes long time for undergoing compression for a given load. Similarly here the void ratio reduces relatively less where at 50 kPa pressure and tries to reduce when we start increasing the pressure and even at 250 kPa the void ratio reduction is only 0.6. So these are actually shown just to as an example how the E sigma dash and compression time plots will be there for clay and time dependent compression takes longer time compared to sand and the magnitude of compression is also large for clays. So we need to understand that the magnitude of compression is large and it also takes over a long period of time. So if before looking into the idometer or consolidation test let us re-look into the definition once again if the soil is of low permeability the application of a surface load results in an increase in pore water pressure. So the resulted increase in pore water pressure gives rise to a hydraulic gradient and it responds to which the pore water flows out of the soil and the soil deforms and as the water flows out of the soil the pore water pressures gradually return to their equilibrium values and after which no further deformation takes place. This is what we are actually we have been discussing about the consolidation phenomenon. So the process of deformation of the soil over time due to the dissipation of non-equilibrium pore water pressures is termed as consolidation and again the compression is used to describe changes in volume due to changes in sigma dash without the difference to the time scale over which they occur. So compression and consolidation here in case of consolidation the process of deformation of the soil over time due to the dissipation of non-equilibrium pore water pressures is termed as consolidation and compression is used to describe the changes in volume due to changes in sigma dash that is effective stress without reference to the time scale for which they occur. So in order to determine the consolidation characteristics in the laboratory the test which is used is called idometer test or consultometer test and in this slide a typical idometer is shown and wherein this is the characteristics of a soil during one dimensional consolidation or swelling can be determined by means of the idometer test. The name idoma is actually derived from the Greek word that is swelling the idoma means swelling. So the characteristics of a soil during one dimensional consolidation or swelling can be determined by means of the idometer test. So hence the name idometer has come like this and the idometer test or consultometer test is actually used for getting the compressibility characteristics or consolidation characteristics of soil samples in remoulded state or in undisturbed state. So generally the samples are actually having the porous stone here and the porous stone the pore water pressure increase in porous stone is 0 which is actually similar to the open layers at top and bottom and this is the moving loading plate and one interesting thing we should note here is that this confining ring is actually rigid enough and will not actually allow the lateral deformation hence whatever the strain which actually undergoes is axial strain. So if you are not allowing the epsilon r that radial strain then it turns out to be epsilon r is equal to 0 so epsilon a will become epsilon v. And in order to you know inundate under water all these water is actually filled up till here and the sample is allowed to consolidate under a given volume under given load. So the load is also in order to operate large pressures up to let us say 800 kilopascals or 600 kilopascals in the labor of technique is actually used in the laboratory. So the idometer or consolidation test in this test the stress is applied to the soil specimen along the vertical axis and while strain in the horizontal direction is prevented. This is actually shown here the no horizontal movement and the sample to the thickness to the diameter that original thickness to the diameter is normally maintained as 3. The T by D is about the ratio is 1 by 3 or you can say diameter to thickness ratio is maintained as 3 and so the cylindrical sample of thickness T and diameter D that is T by D approximately equal to 1 by 3 is confined in a metallic ring and loaded with a vertical axial pressure. So due to the rigidity of the metallic ring the radial strain of the sample epsilon r is equal to 0. So epsilon r means that in this direction in the radial direction will be 0. Since the axial strain is epsilon a is not equal to 0 thus the epsilon a is exactly equal to the volumetric strain so epsilon a is equal to epsilon v and the confining ring imposes a condition of 0 lateral strain on the specimen and the ratio of lateral to vertical effective stress being k0. So sigma v is equal to sigma a is equal to k0 sigma r that is sigma r is the radial stress. So there are attempts by the several investigators to measure the lateral resistance offered by the ring by placing some strain gauges and load cells and all. In the idometer test we need to look into the two types of idometers are there one is called fixed ring idometer and floating ring idometer or idometer with fixed confining ring or idometer with floating confined ring. The major experimental difficulty with the idometer test is side friction that is the wall friction. So shear stresses develop along the cylindrical surface of the specimen as a vertical strain occurs. So as the vertical strain occurs you know the cylindrical you know surface of it is specimen the shear stresses actually develop along the cylindrical surface of the specimen as the vertical strains occur. So the presence of the side friction disrupts the one dimensional state of strain and prevent some of the axial force from reaching the bottom portion of the specimen. So the presence of the side friction disturbs the one dimensional state of the strain and prevent some of the axial force from reaching the bottom portion of the specimen. So to minimize the effect of the side friction forces, the thickness diameter ratio, so one of the reasons why the diameter to thickness ratio is maintained as 3 is that to minimize the side friction forces. So to minimize the side friction forces, the thickness to diameter of the specimen is kept as small as practicable, so that is maintained as 1 is to 3 and the use of the idometer with the floating ring container also helps to minimize the effect of the side friction so that the ring will not offer any resistance. So use of the idometer with the floating ring container also helps to minimize the effects of side friction and many attempts have been made to minimize the side friction through the use of lubricants and plastic liner sheets. So by using the plastic liner sheets or by using the lubricants the side friction can be reduced, this was actually attempted by several administrators. So basically the many codes they recommend these fixed ring idometer and with thickness to diameter ratio as 1 is to 3 basically this helps to limit the side friction forces. In addition to this the walls of the confining ring can be lubricated so that the side friction forces can be minimized. So here in this particular slide this is idometer with the fixed ring, so the ring is actually fixed here and in this case the idometer is actually provided with a floating ring, you can see that the idometer is provided with a floating ring here. So there is no connection between the base plate and the floating ring, in this case the ring is actually attached rigidly to the base plate. So in the fixed ring idometer the friction gradually decreases to 0 towards the bottom. So as the load is actually high here the friction will be very high here and reduce to 0 to the bottom of the specimen. In case of a floating ring idometer the plane of 0 friction that is the neutral plane is at the middle of the sample because the sample is compressed from the both sets. So as the sample is compressed from both the sides as the reaction is actually offered from here and here in the case of a floating ring the neutral plane will be at its mid plane. So a floating ring idometer has the plane of 0 friction at the middle of the sample because the sample is compressed from the both the sets. As the sample is compressed from both the sets so floating ring idometer has the plane of 0 friction at middle of the sample because the sample is compressed from both the sides. case of fixed ring hydrometer the friction gradually decreases to zero towards the bottom so maximum here and then reduced to zero. So the process of the procedure of the hydrometer test is described in this slide the whole assembly with confining ring and in other accessories which sits in an open cell of water to which the pore water pressure in the specimen has free access. And the initial pressure will depend upon the type of soil then a sequence of pressures is applied to the specimen and each being double the previous value. So why it is required to be double why not you know what will happen if you are actually having you know more than the double value but what will happen if it is less than the you know the designated if suppose if sigma 1 sigma a pressure of P1 is applied if it is not applied to P1 or if it is applied less than to P1 what will happen all those things we will discuss subsequently but here we can take it as the initial pressure will depend upon the type of soil. So for example if the soil is at its liquid limit and you know we cannot actually apply very high pressure. So the pressure has to be as low as possible then sequence of pressures is applied to the specimen each being double to the previous value and each pressure is normally maintained for a period of 24 hours. So here also we will discuss about what will happen if the same load is maintained for more than 24 hours or 7 days what will happen or if it is you know removed within 24 hours what will happen. So in exceptional cases a period of 48 hours may be required so for some certain type of soils exceptional cases a 48 hours is required otherwise 24 hours you know period is actually normally maintained and the compression readings being observed at suitable intervals during this period. So for each load what we need to is that we have to observe the compression time versus compression and moment the you know the end of the 24 hour before applying the new load we need to note down the compression value and the load need to be applied carefully. And the axial stress is varied in a stress control manner. So in the idometer test the consolidation test the axial stress is varied in a stress control manner. So at the end of the increment period when the excess pore water has completely dissipated the applied pressure equals the effective vertical stress in the specimen. So at the end of the increment period when the excess pore water pressure has completely dissipated the applied pressure equals the effective vertical stress in the specimen. So in this particular slide time dependent plot during the consolidation for a given load increment is shown for a given load increment it can be a pressure of say P1. So this is the deformation initially the sample actually has higher thickness and let us say higher void ratio and it undergoes compression and then you can see that there is a straight line portion is actually extending up to certain extent and then you know it tends to have another curvature and then tends to go towards this. So we can see that there are three stages are there in the process of consolidation. One is defined as a stage one and stage two is actually called as which is the major part which is actually called primary consolidation. So mostly when the consolidation phenomenon means but stage one and stage two is considered and stage three the secondary consolidation for certain type of soils is very prominent like examples like peat and which is non like soil means the municipal soil waste which actually undergoes very high amount of creep. So stage one, stage two and stage three these are the three stages of consolidation. So initially the occurs because of the readjustment of the particles and some elastic strains are raveling of the particles actually takes place. In stage two basically in the primary consolidation where the soil undergoes particles undergoes irreparable changes and in the secondary consolidation under the constant effective stress the load is not the you know void ratio continues to fall. So this is because of the certain nature of the soil so this actually is prevalent. So the stage one is actually basically mostly caused by preloading and in stage two which is primary consolidation the excess pore water pressure is gradually transferred to effective stress by the expulsion of pore water. So excess pore water pressure is gradually transferred into the effective stress by the expulsion of pore water. In stage three basically this commences at the end of you know primary consolidation. So this occur after the complete dissipation of excess pore water pressure for a given load and this is caused by the plastic adjustment of the soil fabric that means that soil fabric is nothing but the soil grain structures. So readjustment of the soil fabric actually happens here and because of that this secondary consolidation or a stage three takes place. So here in this particular slide the successive load increments of height high versus locked time are shown. So the pressures are typically like one can actually start with 5 kilo Pascal's and 10 kilo Pascal's or you can say that 12.52550 you can see that the load increments are actually always doubled here 50, 100, 200, 400, 800 and 1600 kilo Pascal's. So each time the sample undergoes consolidation and reaches to the new thickness and to the new effective stress like that you can see that the sample undergoes the reduction in the thickness. So this actually happens by the same way you know solids will remain same but only thing is that the water which is there in the three phase system of the soil will get expelled out. So that basically the results of this odometer are presented by plotting the thickness of the specimen or the percentage change in the thickness like we can say that percentage change in the thickness. If it is the thickness is you know let us say delta h is the change delta h by h or which is nothing but you know the strain actual strain the plotted in the thickness this is the results are plotted by plotting the results are presented by plotting the thickness of the specimen or the void ratio at the end of the each increment against the corresponding effective stress. So the effective stress may be plotted either natural or logarithmic scale. So preferably these curves which are when you plot with E and effective stress is generally plotted with E log sigma or it is also E log sigma dash or E log p curves. So if you decide the expansion of the specimen can be measured under the successive decreases in applied pressure. So once the sample has been subjected to loading and unloading the sample during the process of unloading the sample undergoes expansion. So if you decide the expansion of the specimen can be measured under successive decreases in applied pressure. However even if the swelling characteristics of soil are not required the expansion of the specimen due to the removal of the final pressure should be measured. So even if the swelling characteristics of soil are not required the expansion of the specimen due to the removal of the final pressure should be measured. So here in this particular slide how the analysis can be done you know so for example here a three phase system, two phase system where water and solids and with this the volume is 1 plus E0 and so initially height is H0 and this actually changes to because the volume this thickness reduces to H0 minus delta H. So delta Epsilon V the change in volume strain is nothing but delta H by H0 which is nothing but delta E by 1 plus E0 so which is nothing but delta E by 1 plus E0 since the specimen of the soil is only due to the change in void ratio the vertical strain delta Epsilon V can be expressed in terms of the void ratio of the soil specimen at different stages of the test and the void ratio at the end of each increment period can be calculated from the dial gauge readings and either the water content or the dry weight of the specimen at the end of the test. So since the settlement of the soil is only due to the change in void ratio the vertical strain delta Epsilon V can be expressed in terms of void ratio of the soil sample at different stages of the test. So we can actually say that delta H by H0 where H0 is the initial thickness is equal to delta E by 1 plus E0 E0 is the original or initial void ratio. So here the procedure is actually given water content measured at the end of the test W1 void ratio at the end of the test because it is under completely saturation so E1 is equal to W1GS moment once you know the water content at the end of the test by knowing the specific value of the solids we can actually calculate what is the void ratio at the end of the test and the thickness of the sample at the specimen at the start of the test is H0 and the change in thickness is delta H so void ratio at the start of the test is E0 and which actually changes to E1 plus delta E so delta E by 1 plus E0 is equal to delta H by H0 we can write like delta E by delta H is equal to 1 plus E0 by H0. So in the same way delta E can be calculated up to the end of any increment period. Then in the second step the dry weight measured at the end of the test that is mass of the solids and the thickness at the end of any increment period is say H1 so area of the specimen is A so equivalent thickness of the solids we can actually get as HS is equal to MS by AGS row W so with that what we get is that once by know the equivalent thickness of the solids E1 is equal to H1 which is the thickness at end of any increment period by HS so H1 by HS is equal to 1 H1 by HS minus 1 so E1 is nothing but H1 by HS minus 1. So in the oidometer test here the first E versus sigma dash curves are shown on the left hand side of the slide we see the isotropic compression curves where sigma 1 is equal to sigma 3 the sample is actually compressed in all directions identically so this state is called isotropic compression and this is confined compression vertically you can see that this is you know the sample is restrained and laterally and confined in the vertical direction. So then the deformation actually occurs in the vertical direction so you can see that the distinctly different you know the E sigma dash plots for the isotropic compression can be seen so here there is here also there is a you know compression takes place and the sample unloading is taking place and reloading is taking place and again the sample is undergoing compression here you can see that the sample is undergoing compression and unloading and then reloading so compression unloading and reloading and then it is actually again going into the compression mode. So this is this plot actually showing the initial compression followed by the expansion and recompression and the shape of the curves are related to the stress history of the the related to the stress history of the clay and here as a result of the oidometer test what we get is that we get E sigma dash or void ratio effective stress relationships on either an arithmetic scale or in the logarithmic scale. So you can see that initial part is flat and then there is a a steep portion which actually commences here and then once we unload here after reaching certain pressure and the sample undergoes expansion and then the sample is subjected to recompression and then goes into the compression mode again here. So this straight line portion this portion is called the virgin compression for a soil and the slope of this virgin compression curve is actually called as compression index the slope of this virgin compression curve is actually called as this virgin compression curve is called as the compression index and this is actually this portion is this is called recompression index or this is also called as the recompression index and this is actually called the swelling index that is called swelling index when the load is being relieved and you can see that this is the initial void ratio and this is the void ratio. So in the process of this application of this load as the soil particles or grains have been subjected to the continuous rearrangement of the particles and then irreparable changes have been subjected because of that what will happen is that the sample cannot actually meet this particular point in the sense that it will never be possible to achieve the same again. So this is the unique for a given soil for a given type of soil or a clay wherein you can see that the straight line portion and then there is a change in curvature here and then this portion is actually called this gives the virgin compression, this is virgin compression curve and the slope of that is actually called as compression index. So the E log sigma dash relationship for a normally consolidated clay is linear and the E log sigma dash relationship for a normally consolidated clay is linear or nearly so and is actually called as the virgin compression curve because that E log sigma dash relationship for a normally consolidated clay is linear and it is called as the virgin compression line. the clay is over consolidated its state will be represented by a point on the specimen on the point on the expansion or recompression parts of the E log sigma dash plot. The recompression curve ultimately joins the virgin compression line and further compression then occurs along the virgin line. So it joins banks back to the original virgin line and during the compression the changes in soil structure continuously take place and the clay does not revert to the original structure during the expression. So as during the compression the changes in soil structure continuously take place and the clay does not revert back to the original structure during the expansion and the plots show that the clay in over consolidated state will be much less compressible. So one can see that in the recompression state the sample will be less compressible than the normally consolidated state. So basically in the previous slide we have seen that plots show that the clay is over consolidated in the clay in the over consolidated state will be much less compressible than that in the normally consolidated state. Then we will try to look into the define the different parameters like we have introduced ourselves to while discussing the theory of one dimensional consolidation the coefficient of volume compressibility and this can be obtained by using eidometer test and this is nothing but MV is nothing but AV by 1 plus E naught where AV is nothing but coefficient of compressibility which is delta E by delta sigma dash. So it is defined as the volume change per unit volume per unit increase in effective stress. So this is coefficient of volume compressibility is defined as the volume change per unit volume per unit increase in the effective stress and the units for the MV are meter square per kilo Newton or meter square for mega Newtons and the volume change may be expressed in terms of either void ratio or specimen thickness. If for an increase in effective stress from sigma naught dash to sigma 1 dash the void ratio decreases from E naught to E 1. So we can actually say that here initial void sample thickness is say H naught and the after certain compression the sample thickness reduced to H 1. So H naught minus H 1 is delta H and H s is the height of the solids. So MV is nothing but E naught minus E 1 by sigma 1 dash minus sigma naught dash that is delta E by delta sigma dash by this 1 plus E naught that is this 1 plus E naught. So this is actually given as MV is equal to 1 by H naught into H naught minus H 1 by sigma 1 dash minus sigma naught dash. So the value of MV for a particular soil is not constant but depends upon the stress range over which is calculated. So the value of MV so this is actually used in settlement calculations also and once we know the MV value we can actually calculate the estimate the consolidation settlements. So the value of MV of a particular soil is not constant but depends upon the stress range over which this is calculated. And as we have actually discussed in the slope of this virgin compression curve is actually called as compression index and here different portions of the curve is shown here 1, 2, 3 and 4. So here the soil is described as normally consolidated when it is state exists on the steeper line. So 1 and 4 so this is actually called as normally consolidated and soil described as over consolidated when it occurs on the flatter portion that is 2 and 3 here. So it can be like after here when the unloading takes place again it will be less flatter than this one so this process actually continues it undergoes continuously. So here the compression index is defined as the slope of the linear portion of the E log sigma dash plot and is dimensionless and for any 2 points on the linear portion of the plot we can actually find out E naught and what is sigma naught dash that is the corresponding sigma naught dash E 1 or this corresponding sigma 1 dash. So Cc is nothing but E naught minus E 1 divided by logarithmic of sigma 1 dash minus sigma naught dash. So the rearrangement of the soil particles you know in this portion the rearrangement of the soil particles takes place a permanent or irreparable changes takes place and elastic strains in particles are partially recoverable and compression of bounded water layers is recoverable. So in this particular slide what we have seen is that the definition of the compression index and in how this can be determined and we also have seen that the normally consolidated and over consolidated terms we have been introduced so we will actually look into that how these terms can be defined. Before that we actually have to look into the another output from the consolidation test or iodometer test is pre consolidation pressure. So Kasekande proposed an empirical construction to obtain pre consolidation pressure from the E log sigma dash curve for an over consolidated clay the maximum effective vertical stress that is acted in the clay in the past referred to as the pre consolidation pressure. So the maximum effective vertical stress that is acted on the clay in the past referred to as the pre consolidation pressure. So here once we have got the data is plotted for the E void ratio and logarithmic of sigma dash. So we actually have got you know the portion AB and then portion BC and this is relatively a straight portion. So we are not taken unloading and reloading components. So here this actually portion is shown here. So in order to determine the pre consolidation pressure of a given soil suppose if you can actually get an undistributed sample from the side and then if the test is actually done without much sample disturbance then there is a possibility that we will be able to assess what was the stress the soil has been subjected in the past. So the procedure is like this first what we need to do is that we have to draw a tangent to the straight line portion of the E log sigma dash curve extend this backward. So extend this tangent backward and secondly what we have to do is that we have to locate a point D of the maximum curvature on the recompression part of the AB curve. So this involves some sort of you know judgment and by proper judgment the determination of the point D of the maximum curvature of the recompression part AB of the curve can be obtained. So you know then we have to draw the horizontal which is parallel to the logarithmic of sigma dash axis through point D and a draw a tangent passing through D. So and bisect an angle between this you know this tangent which is actually drawn through this point D which is located on the maximum curvature point of the E log sigma dash curve and this horizontal and one the point where this you know the extended back tangent of the linear portion of this compression curve wherever it meets this bisected line passing through point D and when you drop the vertical line below and that gives the sigma C dash or Pc is called pre-consolidation pressure. So the vertical through the point of the intersection of the bisector and the CB produces the approximate value of the pre-consolidation pressure. So with this procedure which is actually given by Casagrande we can actually obtain what is the you know the pre-consolidation pressure and what was the stress of a soil, what was the stress the soil would have been subjected in the past. So the pre-consolidation pressure as we just now seen how this can be determined in the laboratory. So it is the pressure it is the previous maximum effective stress to which the soil has been subjected in the past. So here we actually introduced ourselves to two terms and we have been actually discussing that the normally consolidated and over consolidated soils. A normally consolidated soil is nothing but a soil is called as normally consolidated. If the present effective over burden pressure is the maximum to which the soil has ever been subjected. So a soil is said to be normally consolidated if the effective present effective over burden pressure is the maximum pressure to which the soil is ever been subjected. That is sigma dash present effective stress is less than is greater than or equal to sigma dash past maximum. Sigma dash present greater than or equal to sigma dash past maximum and over consolidated soil is defined like this. A soil is called as over consolidated if the present effective over burden pressure is less than the maximum to which the soil was ever been subjected in the past. That is that sigma dash past maximum is much more than the sigma dash present. That is sigma dash present is less than the sigma dash past maximum. So in the natural condition in the field the soil may be normally consolidated or over consolidated. So mostly in India along the coastal our peninsula mostly the soils which are actually there are normally consolidated nature and in case of Europe and other countries because of the glaciation and other natures the soils can actually be in the over consolidated state. So along the coastal line the most of these soils they remain in normally consolidated state in Indian peninsula. So in the natural condition in the field a soil may be either normally consolidated or over consolidated. And normally consolidated we say that when the sigma dash present is greater than or equal to sigma dash past maximum and over consolidated we will say that sigma dash present is less than sigma dash past maximum. So a soil in the field may become over consolidated. If the soil with when the continuous deposit is actually taking place it can actually become over consolidated through several mechanisms. First is that you know a structure may might have been existing till today but may not be there. So the past structures and removal of the overburden pressure removal of the overburden pressure are glaciation and deep pumping that deep pumping of and the desiccation due to drying mostly in the water zone in the upper portion of the soil. The soil is actually said to be in the over consolidated state and that is because of the desiccation due to drying. And the desiccation due to plant lift the lift of water in the water zone and this can also cause you know the over consolidated state to the soil. And the change in soil structure due to secondary compression and change in the PHE value and salt concentration and change in temperatures and weathering ion exchange and precipitation of cementing agents. Suppose if the cementing agents are getting precipitated into the soil deposit and it also lead to the formation of over consolidated soils. So particularly you know what we can say that the glaciation and the past structures are removal of the overburden pressures are the major causes for the soil deposits to change into over consolidated state but the particularly the deep soil status. And the other things what we call what we come across normally is that in hard crust what this is basically because of the desiccation due to drying this will be in the over consolidated state naturally. So whenever possible the pre consolidation pressure for an over consolidated should not exceed in the construction. Whenever possible the pre consolidation pressure for an over consolidated should not exceed in construction. If that is the case then it will undergo settlements again it will go into you know the compression again and the soils will be subjected to irreparable changes again. The compression will not usually be great if the effective vertical stress remains below sigma dash C and only sigma dash C is exceeded compression will be large. If sigma dash C is exceeded then only compression will be large. So if the given for example if you are having a certain soil and if you are actually trying to construct say one or two floor building and the compression may not actually you know will be there and if the so if this does not cause you know serious problem within the design life of the structure then may not be issue. So compression will not usually be great if the effective vertical stress remains below the sigma C dash and if only if sigma C dash is exceeded the compression will be large. Then we are actually trying to determine one more term which is called as OCR. OCR is nothing but ratio of sigma sigma dash C the pre consolidation pressure to the present effective overburden pressure. So normally for normally consolidated soils OCR will be in the range of 1, 2 you can say 2 and the any value of OCR greater than 2 is called likely over consolidated soils and there can be some soils because of the past existence of the structures and some glaciation activity which might have taken place in that particular site because of the history. There is a possibility that the OCR can have value up to 9 to 15 or so. So highly over consolidated soils they actually have they would have undergone the consolidation and the water content in those soils is very less. In case of normally consolidated soils they are soft in nature and have very high water content and compressible in nature. So before looking into this example problem which we will discuss let us look into this particular slide wherein we have distributed we have got a bottom which is actually having impervious and assume that there is a soil which is actually has got a stage 1 deposition and then the soil is actually getting deposited here and the deposition process is actually happening and this is the water level in the sea lake or estuary where estuary is nothing but where wide portion of the river where it meets the sea. So the sigma versus z this is nothing but this is you know pore water pressure and then this is effective stress in stage 1. So when the deposition actually happens then the deposition is happening in stage 2 the height of the soil deposited increases with that what will happen is that this element undergoes increase in the effective stress and what will happen is that the effective stress increases from sigma 1 dash to sigma d dash so the element is subjected to increase in effective stress from sigma 1 dash to sigma 2 dash. So this is plotted for the element here for this element which is at a certain point which is actually selected in stage 1 and stage 2 and wherein it actually says that E and versus log sigma dash so this is the stage 1 point where up to that the stress actually has been subjected in the element A is sigma 1 dash this is because of the deposition of the soil and stage 2 because of the stress increase from sigma 1 to sigma 2 that is delta sigma is increase in incremental effective stress this is stage 2 and if this actually is point A is sigma C dash that is the pre consolidation pressure so this is said to be in the normally consolidated state. So this is the past stress history and this is for the E sigma dash which is actually shown here past stress history and the present state and then loading when it actually happens again it goes into the normally consolidated mode here. So this is a true NC clay would only be exist in current deposition environment one more thing is that this normally consolidated soils the continuous deposition actually take place mostly these are also called as Young deposits such as in river, estuary, lake or coastal region where the water level has always remind above the soil level and where it would not have been affected by any other process which might produce over consolidation. So a true normally consolidated clay would only exist in current deposition environment such as river, estuary, lake or coastal region where water level is always remind above the sea level, above the soil level where it would not have been affected by any other process which might produce over consolidation. And during the deposition what will happen is that the grain structure of the soil element will be adjusting to changes in void ratio mostly the structural rearrangement of the soil particles takes place. So that is why the along the compression line there is you know the constant you know compression undergoes and then as the because of the deposition the stress keeps on in the effective stress keeps on increasing. And this process where the particles move closer is an irreversible process so that the original particle arrangement could not be recovered even if the stresses are removed whereas in case of over consolidated clay what we said is that this is the present ground level but the past maximum ground level was here. This can be with the ice or with the soil where either it would have been eroded it means subjected to erosion or this ice which was actually there was subjected to say glaciation. So in that case now the war burden which was there as the soil where the soil strata was subjected will not be existing now but the past you know it was actually subjected. So in this case what will happen is that with the present war burden this is sigma naught dash and the after deposition you know this is the path. So what will happen is that we can see that the ACR actually decreases with as the sigma naught dash increases or with the Z or sigma naught dash the ACR variation will be there with increases which decreases with an increase in sigma naught dash. So here OCR is actually defined as sigma C dash by sigma naught dash which is nothing but we can write 1 plus sigma C dash minus sigma naught dash by sigma naught dash. So in this sigma C dash minus sigma naught dash which is actually constant and this is actually for the over consolidated portion. In the case of OC clay basically this is the clay that has been subjected to effective stress in the past stress history and sigma C dash larger than the effective stress existing at the point at the point sigma naught dash. So on removal of the stress the soil skeleton swells but only the reversible components like elastic strains in particular and compression of water molecules attracted around each particles they are recovered and so that the void ratio increases are smaller and the erosion line follows a flatter path. So what we have understood is that the normally consolidated and the work consolidated soils are distinctly different and normal consolidated soils undergo you know the large compression and they have very high water contents and by knowing liquid limits we can also determine what is the compression index from the order of the compression index values we can actually determine. In case of work consolidated soils the soil would have been already subjected to consolidation. So here in this context we can actually say that the soil particularly fine-grained soil clay is very sensitive to the stress history. So then in this particular lecture what we try to introduce ourselves to the you know the so called you know how we can actually do the idometer test for determining the consolidation characteristics of soils. So from the idometer test what we can actually get is that compression index and coefficient of volume compressibility and then also we can actually get by we have popular two methods are there for determining the coefficient of consolidation with those methods we can actually get the variation of the coefficient of consolidation with time and as we have discussed that coefficient of consolidation and coefficient of permeability are related. So by using that relation we can also obtain what is the change in coefficient of permeability when actually happens with increase in the effective stress.


Sir, please tell us about the canonical conjugate. Okay, so canonical conjugate is actually connected to Heisenberg-Eisenberg principle. So the way we have studied Heisenberg-Eisenberg principle yet, so we generally try to cover like if there is a uncertainty in the position mapping of a quantum mechanical particle and also its momentum. So if we try to find them, we'll find they actually follow this particular equation where h cross is h by 2 pi. So now what this is actually saying that if we can measure one of these parameter precisely, that means one of them will become 0, either delta x or delta p. If one of them becomes 0, the other one you can see will become undefined because you have to divide a finite number with 0, which will become an undefined number, that means it is not possible. So that is why mathematically it is saying that you cannot have either delta x or delta p close to 0. That means you cannot have a system where you can measure the momentum or the position precisely. However, these are not the only two parameters that are connected with this particular scenario. So what it is found that there can be multiple different combinations of parameters, different combinations of parameters and parameters are such that they are actually nothing but a variable that you can measure. So you actually can find that there are different combinations of parameters or variables that is possible to that they are following this particular equation. So one variable is a, one variable is b and they are following this particular equation and they are multiple vectors. Such combination of variables which actually follow this particular law or this particular rule that is actually known as the complementary canonical conjugate variables. And what is the other important term? So generally this a or b will be such parameters that they actually can define all the four different dimension for that particular system you are actually working on. So whatever the quantum mechanical system you are looking into, these two parameters will be such a way that they will be defining all the four dimensions that is required to define a system, x, y, z, three dimension and the time dimension, this four dimension. So one example over here the delta is defined x, y, z. The momentum has the velocity where there is a time hidden there. Similarly, there could be energy and lifetime, one of such parameter. So don't always look into Heisenberg-Answery principle such way that it is only valid for delta x and delta p. It is one of the examples of that, but it can be valid for any particular combinations of complementary or canonical conjugate variables. So does it clarify your doubt or no? Yes, sir. Thank you. Any more questions from anyone? Okay, so if not, we will start in a little bit where we actually left in last class about the line resolution. So now as we have just discussed the Heisenberg-Answery principle and we know delta e, that means the uncertainty in the energy value. The energy value, that means the energy difference between the two states excited and ground state and delta t is the lifetime at which the lifetime of the electron or particular nucleus in this case remaining in the excited state. So the excited state lifetime. So if this is the two parameters that we are talking about that is also following this Heisenberg-Answery principle. Now over here, what we also found that whenever we are seeing a signal because an excited state system is coming back to ground state and it is giving an energy and if I want to plot this energy with respect to intensity versus energy, that is how the graph looks like. And we can find at the half maxima, that means there is a 50% of the maxima. What is the line width? So this is known as the line width at half maxima and this is defined as the term as we said the capital gamma. And we found this capital gamma is connected with the uncertainty in energy with this following equation. So higher will be the uncertainty in the energy. Obviously this energy uncertainty increase, you will see a much more broader peak. And with respect to that, obviously this line width will also increase and because we are measuring at 50% so that is why it follows this gamma by 2 equation. So now what we are going to do, we have two equations equation number one, equation number two. We are just including this in the place of delta e. So we are going to find delta 2 into delta sorry gamma by 2 into delta t is delta nickel h plus by 2. And further that gives me this equation. So now what is happening? We are trying to find that there will be certain amount of line width always present. This line width is important because over here at the end we are actually emitting a gamma ray from a source and trying to getting it absorbed with a sample. And over there the minute difference we have on the electronic environment which is affecting my nucleus that is I am trying to measure. So that is why what I want to have that this line width should be very sharp so that if there is a certain particular change over here I can actually see that. However, just imagine if your line is already very broad and if there is similar amount of changes happening over here. So now this will be much more tough to follow because it is already very wide and the wide scale you are trying to find a very small difference. So it will be always better if your line is pretty sharp and that is why this term energy resolution coming to the picture. That means can you find the difference you are getting due to the interaction between your sample and source and can you figure it out on the background of the original absorbance. That is the point we are having. And this energy resolution is given by half with a line width at half maxima and the energy the original energy. This particular ratio gives you the energy resolution. Now what we are going to do with this thing in our hand we are trying to find a rough number where it surely comes one. So again energy resolution is gamma by energy. Now over here we already know how to find gamma because we know gamma into delta t greater than h cross. So delta t defines what is my lifetime t is my lifetime. So whatever the lifetime I can find because lifetime is a very small number delta t will be very similar to that. So now if I take iron 57 as an example that has an excited state lifetime around 141 nanosecond. So that means that delta t would be very similar to 141 nanosecond at least. And that is the largest value we can consider not the smallest. The largest value it can have is the same as the lifetime itself. And if I consider that I will find my gamma will be h cross by delta t which will come around 5 into 10 to the power minus 9. By putting the value of h cross the Planck's constant value. So this is in the region of this gamma. So don't worry about the numbers I am just trying to give you an idea like at which particular range of numbers will be given. And over here now if I look into delta by e that also should say that energy gap that you are talking about for iron 57 for a transition between i equal to 3 by 2 to i equal to half that is generally coming around 14.4 kilo electron volt. There is a kilo over here so that means it is 14.4 into 10 to the power 3 electron volt. So with respect to that now I have gamma value now I have e value I can find what is the resolution. So I am going to put this numbers over there and you can find the number is going to come around 3 into 3 into 10 to the power minus 13. It is unitless because it is canceling each other out. So it is actually you can see a very small number the resolution 3.3 into 10 to the power minus 13. So you can imagine how tough it will be to gain an idea how much difference you are going to see. Because that is the resolution value pretty low. So how this number matters. So to understand that let me compare between optical spectroscopy that we are doing all the time versus Mosbyr spectroscopy. And let compare how the energy differences and resolutions are going to come. So now if I look into an optical spectroscopy in this side an optical spectroscopy the gamma value is generally comes around the region of 10 to the power minus 8 unit. Whereas you can find that it comes around 10 to the power 9 unit for Mosbyr spectroscopy for i. I am just taking the numbers without any unit side. And then the energy that we are talking about it is around 0.121 electron volt that is optical spectroscopy energy. And this is having an energy around 10 to the power 3 or 10 to the power 4 electron volt. So all together you can see now if I take gamma this is going to be in the region of 10 to the power minus 7 to 10 to the power minus 9. Whereas this one is going to come around 10 to the power minus 13. So you can see whatever the optical spectra I am looking into this one is 10 to the power 6 to 10 to the power 4 times much weaker. That means that will be that much tough to get the resolution out. And now imagine you can measure say 1 nanometer difference in an optical spectra if you are seeing the difference. And over there this is saying the gamma ray energy resolution is at least 10 to the power 4 to 10 to the power 6 times lower. So you have to understand then you have to have a system that can even record 10 to the power minus 4 nanometer difference in a wavelength. Only then you can go to see the energy difference if it is happening or not. Or just imagine you have a ruler in your home with you where you can easily measure 1 millimeter. And then say that is the optical spectroscopy in the region of 10 to the power minus 7 to the minus 9. And in most spectroscopy it is 10 to the power minus 4 times weaker. So that means in your naked eye you have to see 100 micrometer or 1 micrometer system which will be very tricky to see. So that is the way we are talking about the energy resolution and that is why it is very much important to understand where do we stand with respect to energy resolution. So this is the reason that most spectroscopy we cannot do for every other element. Why? Because over here what it is saying that energy resolution you have should be such a number that this energy resolution is a very high. Higher number is better because 10 to the power minus 13 is actually lower number compared to 10 to the power minus 7 or 10 to the power minus 9. Because this is a minus sign is over there. So that is why what you want to have you want to have a higher value for this. And only then you can see a spectra very easily and some difference very easily. And to make sure that this is actually a high number what you need to have that there is two things. So this delta the gamma value the line width value should be high and this E value should be low. It is a normal denominator and numerator system. So you want to have your numerator should be high denominator should be low so that overall your number is actually high number. So how I can make sure that may gamma value is actually a high value. So you already know that we have the gamma into delta t is greater than equal to h cross. So that means gamma is equal to h cross by delta t. So in this way it is saying that if your delta t value is a very low value only then your gamma value will be high because h cross is a constant. So it is saying your excited state life time should be very low and only then your delta 2 will be very low gamma will be very high. So this is condition number one. Second is this E value should be low. That means that the energy change you are seeing between the excited state and ground state of the nuclear states. The nuclear state energy state that you are seeing and where it is leading the gamma energy. If this gamma energy has to be low that means they should be very close by. So that means you are looking for a low translation energy. So these are the two parameters you want. Your excited state life time should be very low and your energy of transition over here that should be also very low. If these two parameters are present only then you can have a practical chance to even look into a system where you have enough resolution to see a spectrum. And that is why in reality you don't have a lot of the systems a lot of elements that can actually be used for MOSBAR spectroscopy. Only a few of them can do that. So over here I am showing you an example of the elements that is there you can see the MOSBAR spectroscopy. So you can say it is actually a MOSBAR spectroscopy p-o-dictator. And over here the darker ones are actually inactive and white ones are actually active for MOSBAR. And over here as it is shown the number over here it is showing like how many transitions you see. That means it is 3 by 2 to half there is one transition or there is another 5 by 2 to 3 by 2 another transition. So that is showing how many transitions you have. And this one is showing how many nuclei can or isotope can actually show this particular MOSBAR active transitions. So because it is a nuclear phenomena so if you change the isotope you are changing the composition of the nucleus. So the properties will not be same for MOSBAR spectroscopy. So that is why in MOSBAR spectroscopy you have to define that particular isotope. So that is why I am saying it is iron 57 not iron 56 or just simple iron. And over there you can see what are the nuclei are actually practically possible to have it. And even among them not all of them are very easy to use. I am coming into that pipe. So there are the most common ones that is actually used worldwide. One of them is this iron 57, 119 tin, 121 antimony, 129 iodine, 197 gold. So these are one of the few ones which actually have quite a wide use for MOSBAR spectroscopy in the reality. And among them iron 57 is the most common one that has been used for the MOSBAR spectroscopy. So it has been used so regularly that nowadays if you are doing a particular iron complex chemistry where you are talking about oxidation state change or spin state change people will invariably going to look for that do you have the MOSBAR spectroscopy to support your hypothesis. So MOSBAR spectroscopy and iron chemistry right now became almost inseparable at this moment because they are wide usage. So from now on that is why we will mostly keep our discussion based on iron. So iron 57 MOSBAR spectroscopy. So that we are going to discuss. So now we are going to dissect the overall spectroscopy for iron 57 and find out how it is actually going to happen. So first of all you need an iron 57 isotope. So this is the isotope and as you know iron 56 is the most common one naturally abundant. So this is actually not really abundant a lot in nature. So the natural abundance for this particular isotope is pretty low actually. This is around only 2.14%. So if you find 100 of the iron atoms naturally only two of them will be iron 57. So that means if you want to do a chemistry with iron 57 you need to artificially enrich your sample. Otherwise you cannot even see the signals particularly. Because you need to have enough sample which actually can undergo MOSBAR spectroscopy. Because only the iron 57 work the iron 56 is actually a nuclei which will not participate in this MOSBAR spectroscopy. Now as we have discussed so far in the MOSBAR spectroscopy what happens that this figure will come again and again. You want to have an excited state already available which will come to ground state while it will leave the gamma ray. And this gamma ray will come and excite your sample. So this will be the sample, this will be the source. So this source and sample has to be identical. So both of them has to be iron 57 and only then it will work. So now the question is that if they are very same then what is the difference? There will be a minute change in this energy of the ground state and excited state depending on the electronic environment around the system. And that change we are going to capture over there. And that is the change we are seeing that should fall in the resolution range that we are actually talking about. Now first of all we need a source which actually going to leave a gamma ray. That means I have to have an iron 57. We start with a system where the nucleus is already in the excited state and how that is possible.