 Welcome to lecture series on advanced geotechnical engineering, being offered by the department of civil engineering IIT Bombay, we are in module 3 lecture 6 on compressibility consolidation and we have introduced ourselves to the normally and over consolidated soils and we have discussed about the methods for determining, method for determining pre-consolidation pressure. So in this particular lecture we will continue our discussion on normally and over consolidated soils and then we will try to discuss on over consolidation ratio under consolidated deposits and thereafter we will try to extend our discussion to the determination of the coefficient of consolidation, the only parameter in the differential equation put forwarded by the subject is one dimensional consolidation theory. So in the previous lecture we have defined pre-consolidation pressure and we said that it is the maximum previous maximum effective stress to which the soil has been subjected in the past. So the pre-consolidation pressure is the previous maximum effective stress to which the soil has been subjected in the past. So based on this we can classify predominantly two types normally consolidated and over consolidated and there are also some under consolidated deposits that means that very very very young deposits. The normally consolidated deposits you know soil or normally consolidated soil is actually defined as a soil and it is called as normally consolidated if the present effective over burden pressure is the maximum to which the soil is ever been subjected that is sigma dash present is greater than or equal to sigma dash past maximum. And over consolidated soil a soil is said to be over consolidated if the present effective over burden pressure is less than the maximum to which the soil is ever been subjected in the past that means that the sigma dash present is less than sigma dash past maximum that means that in the past the soil would have been subjected to very high you know the effective stress and you know whatever the current stress is very much less than that. So in this case normally consolidated soils are what we have discussed is that these are very prominent in along the Indian peninsula and all along the coastal line the most of the marine clay deposits are normally consolidated in nature. So this becomes a challenge for a geotechnical engineer for designing you know the structures on these place. So in the natural condition in the field a soil either may be normally consolidated or over consolidated in some cases under consolidated. So soil generally you know become over consolidated through several regions and we have several mechanisms or several reasons one is that removal of over burden pressure that means that if a past structure or if a over burden was existing and if that is removed then there is a possibility that the soil might have actually gone into over consolidated state. And if some of the past structures when they were removed the soil still experiences or remembers the stress which actually has been subjected in the past and because of the presence of the large chunks of ice on the soil and subsequently when it gets undergoes a process of glaciation then the soil also you know in the over consolidated state and either because of the deep pumping or desiccation due to driving, desiccation due to drying and desiccation due to plant lift of water and the change in soil structure due to secondary compression and importantly change in pH of the soil and salt concentration also you know can change these you know the convert the soil into over consolidated state and weathering some ion exchange and aging of soils and precipitation of cementing agents all these things you know lead to you know make the soil over consolidated. So in this context actually whenever possible the pre consolidation pressure for an over consolidated state should not be exceeded in construction and compression will not usually be great if the effective weight stress remains below the pre consolidation pressure and if the pre consolidation pressure is exceeded then the compression will be large. So if for any soil if the pre consolidation if the loading which is actually is less than that pre consolidation pressure the settlements will be small but if the load actually exceeds you know to more than the pre consolidation pressure then the soil can change into normally consolidated mode. So in the field the over consolidation ratio is actually defined as sigma dash C by sigma naught dash, sigma dash O. So if OCR is equal to 1 that indicates that normally consolidated soil and OCR greater than 1 is called over consolidated soils. In fact OCR is equal to 2 it indicates likely over consolidated soils and OCR also can be like you know 5 and 9 more than 9 and they are heavily over consolidated soils, heavily over consolidated soils that means that the clay is very very stiff in nature and the water content is very less and it has been subjected to very high amount of past very stress in the past and there are also some cases particularly it occurs in some lake beds and all where OCR less than 1. So this is called as under consolidated soil that is for example recently deposited soils either geologically or by man so the soil has not yet come to the equilibrium under the weight of the over burden load. So the status of these you know the pore water pressure is soil is that the soil is actually not yet attained the equilibrium under the weight of the over burden above that particular layer on particular point then pore water pressure would access the of the hydrostatic pressure the pore water pressure would be excess of the hydrostatic pressure. So these under consolidated deposits are you know something like very recently deposited soils either geologically or by man made activities. So in that case OCR less than 1 is called but predominantly OCR is equal to 1 it is actually called as normally consolidated soils up to OCR is equal to 2 like they are also treated as lightly over consolidated soils and the OCR value can also go up to 5, 9 and greater than 9 so where they are heavily over consolidated soils. So let us look into some example problems before discussing you know the methods for determining coefficient of consolidation. So in this example problem the laboratory consolidation data for an undistributed clay samples are given and which is actually given as E1 is equal to 1.1 at a sigma 1 dash 95 kilo Pascal's and E2 is equal to 0.9 at sigma 2 dash is equal to 475 kilo Pascal's. So that sigma 1 minus sigma 2 the sigma 2 2 minus sigma 1 is the increase in the stress that is 475 kilo Pascal's minus 95 kilo Pascal's. So what we need to do is that when there is an increase in effective stress then they you can see that the void ratio decrease from 1.1 to 0.9. So the calculate the coefficient of volume compressibility m suffix v and what will be the void ratio for a pressure of 600 kilo Pascal's and note that you know we have to see that sigma dash c is less than 95 kilo Pascal's that is the soils pre consolidation pressure is less than 95 kilo Pascal's. Now the solution runs like this we have been given the data which is plotted on void ratio on the bi-axis and pressure on the logarithmic scale that is pressure on the sigma dash and logarithmic scale in kilo Newton per meter square and by putting 1.1 for E1 and the point lies here and for 0.9 the point lies here that is at 475 kilo Pascal's. Now we can actually determine the slope of this line is delta A by delta sigma that is delta sigma if it is in arithmetic scale but coefficient of volume compressibility can be obtained like this which is nothing but 1 by 1 plus E0. So here E1 is taken as the initial void ratio that is 1 by 1 plus 1 plus 1.1 that is 2.1 then E0 minus E1 that is nothing but E1 is 1.1 E2 is 0.9 and the pressure is that 475 and 95. So with this expression which is nothing but AV by 1 plus E0 what we are doing is that AV by 1 plus E0 and AV is nothing but delta E by delta sigma dash into 1 by E0. So with this what we have got is that MV we have got. So by getting this MV we can actually after having so the soil's volume compressibility is known then you know in this particular pressure range then you can actually calculate what is the settlement and all. Similarly the slope of this E log p curve the straight line portion of this virgin compression curve is nothing but indicated as compression index. So this slope can be obtained like this Cc is equal to the slope of line E1 minus E2 divided by log sigma 2 dash minus log sigma 1 dash. So even minus E2 so even is 1.1 and E2 is 0.9 so 1.1 minus 0.9 divided by log 475 divided by 90. So with this what will happen is that the compression index is actually coming as 0.286 compression index is coming as 0.286. Now as the soil is in the same normally consolidated state and 600 kilo Pascal's is you know the right next 2, 475 kilo Pascal's. So what we wanted to know is that what is the void ratio of a soil at pressure of 600 kilo Pascal's. So further by taking the same slope that is Cc is known to us 0.286 and initial void ratio is known to us that is 1.1. So E3 is required to be found out so it is nothing but Cc is equal to E1 minus E3 divided by log of sigma 3 dash by sigma 1 dash. Now the sigma 3 dash which is nothing but you know that is new pressure that is 600 kilo Pascal's. So 0.286 that is nothing but E1 is 1.1 minus E3 divided by log of 600 by 95 and Cc is known to us which is 0.286. So with this what we can actually get is that when the pressure increases to 600 kilo Pascal's at the end of that you know the process of consolidation what you get is that E3 value is 0.87. Now it decreased further from 0.9 to 0.87. So what we have done in this problem is that the given data is plotted and then we have determined what is the coefficient of volume compressibility that we have been asked. Then afterwards we determined compression index from the given data and then we were asked good to find out what is the void ratio of a same soil with undergoing the consolidation and what is the void ratio at a pressure of 600 kilo Pascal's. So that works out to be 0.87. From the given data this portion belongs to lecture 5 of module 3. From the given data of pressure versus void ratio we have to plot E log p curve and logarithmic of pressure on the x axis and void ratio on the y axis and once we get these plots we have to identify a point where the maximum curvature provides and from there we have to draw an horizontal line and then a tangent need to be drawn from that point of where the maximum curvature exists and we need to bisect this line, bisect this angle let us say if that angle is say delta alpha that angle need to be bisected and then from this point D a line need to be drawn and this tangent which is the straight line portion of the virgin compression curve need to be extended and the point where it actually meets that bisected line and from there we can actually draw and that point is regarded as pre-consolidation pressure according to Kassagrande's method. So in this particular problem from the given data what we have done is that we took point D and then we have drawn horizontal line and from the point D we have drawn a tangent forward tangent and then we bisected that angle and then we have drawn a line through point D and then we extended this back tangent from the straight line portion of the virgin compression curve and the point where it meets that is actually regarded as pre-consolidation pressure in this problem we have got as 117.5 kilo Pascal's. Then what we did is that we determined the pre-consolidation pressure from the illogic plot. So here at P2 500 kilo Pascal's E2 is 0.8 and P1 300 kilo Newton per meter square even is 0.9. So the pre-consolidation pressure is works out to be 117.5 kilo Pascal's. Now from the compression index we can actually find out here even minus E2 by log of P2 by P1. So 0.9 minus 0.8 by logarithmic of 500 by 300 you will get the compression index of 0.451. So the compression index is actually obtained from the given data as 0.451. Now after having discussed about these problems now we said that you know we will try to look into the methods for determining coefficient of consolidation and we have been introduced ourselves over the coefficient of consolidation term for the first time in the consolidation and compressibility module in the consolidation equation wherein we said that dou u by dou t is equal to cv dou square u by dou z square. So the cv term is the coefficient of consolidation. So the coefficient of consolidation cv is the only term in the consolidation equation that takes into account the soil properties which govern the rate of consolidation. So this cv is the only term or you can say the soil mechanics term in the consolidation equation and which govern the rate of consolidation. So the coefficient of consolidation CV signifies the rate at which the saturated clay undergoes one dimensional consolidation when subjected to an increase in pressure. So one more physical significance of coefficient of consolidation is that it signifies the rate at which the saturated clay undergoes one dimensional consolidation when subjected to an increase in pressure. So a knowledge of CV is essential for predicting the rate of primary consolidation settlement. So you know the knowledge of from the consolidation test data there are methods which are actually there also in the laboratory as the methods are also there in the field also. Particularly when you have got a soils which are isotropic then it is assumed that horizontal consolidation as well as the vertical consolidation they are same. But if the soils are say anisotropic in nature then the horizontal coefficient of consolidation, vertical coefficient of consolidation they are different. So naturally for CH that horizontal coefficient of consolidation is actually more than CV that means that this is analogous to when we discussed the permeability the horizontal permeability permeability in horizontal direction is actually greater than permeability in the vertical direction. So at that time we have discussed that the ease with which the water can flow through the soils is actually less in the horizontal plane than the vertical plane. The reason what we discussed is that you know the as the sigma H is equal to K0 into sigma V at K0 is the coefficient of at pressure at rest because of the less horizontal stress. So the less you know you can say that less confinement so because of that the coefficient of permeability in horizontal direction is actually more than in the vertical direction. As permeability and coefficient of consolidation are related we are based on the same analogy we can also say that the coefficient of consolidation in the horizontal direction is actually you know will be more than the coefficient of consolidation in the vertical direction. So the coefficient of consolidation basically signifies at which the saturated clay undergoes one dimensional consolidation when subjected to in the rate at which a saturated clay undergoes a one dimensional consolidation when subjected in increased pressure. So for isotropic cases where identical properties are there then we can take CV is equal to CH. So knowledge of CV is essential for predicting the rate of primary consolidation settlements there are also you know the field methods are there for determining you know coefficient of consolidation by using piezo cones one can actually determine the you know the coefficient of consolidation in the field. So there are several methods available for obtaining coefficient of consolidation over a period of from 1940 onwards you know the number of methods have been evolved and these methods compare the characteristic futures of theoretical time factor T or TV and the degree of consolidation U relationship with time compression data obtained from the laboratory. So from the whatever the data we obtained from the laboratory so what people compare is that the U versus TV or U versus T curve data are compared. So these methods compare in principle characteristic features of the theoretical time factor TV and the degree of consolidation U relationship with time compression data obtained in the laboratory. So the square root of time fitting method which is also called as root T method proposed by Taylor 1948 and the logarithmic of time fitting method that is log T method and also called as Casagrande's method which is after Casagrande and Fadam 1940 are the most widely used methods in practice and are considered as standard methods. So we have two broadly classified methods and one is you know root T method which was put forward by Taylor in 1948 and log T method which is called as Casagrande's method and which was put forward by Casagrande and Fadam 1940 and these are the most widely used methods in practice and are considered as the standard methods. So we will discuss you know these two methods and then also some couple of methods we will discuss which are actually evolved in the recent literature. So for determining the coefficient of consolidation by using the root T method so first let us consider the root T method. In the root T method the CV values are larger than those obtained from the Casagrande's method so the reason we will actually discuss but what we need to understand is that we have to plot the dial gauge readings for the specimen deformation for a given load increment that means that you know we know that we have discussed that in the consolidation or hydrometer test each load will be kept for about 24 hours and let us say that if you are keeping 100 kilo Pascal's load and the less load increment is 200 kilo Pascal's. So each load increment is kept for about 24 hours or given different time and so for a given load increment whatever the time versus compression data which is the dial gauge readings the beginning of the placement of this load and till the end of this load and time has to be plotted on the semi logarithmic graph paper. Now what we need to do is that we have to plot two points A and B on the upper portion of the consolidation curve which corresponded to time T1 and T2 respectively note that T2 is equal to 41. So this is nothing but you know we have to put plot two you know points on the consolidation curve. So this is dial gauge readings on Y axis and logarithmic of the time in minutes on the X axis and this is a semi log curve and this is the U versus log TV that is U versus TV or U versus TV theoretical curve wherein you can see that this is the 0% consolidation and somewhere here when you extend this portion this portion is into the secondary consolidation after the end of primary consolidation. So this 90% consolidation or 0.9 is somewhere here and this portion is you know so in this so this is you know elaborated one with the time versus deformation is actually shown here. So you can see that the important part of this method is that we have to plot two points A and B on the upper portion of the consolidation curve. So the upper portion of the consolidation curve means somewhere here and where the curvature is actually changing from the beginning portion you can say that two points have to put such a way that the upper portion of the curve such a way that which correspond to time T1 and T2 so note that T2 is equal to 4 T1. So the difference in times the T1 and T2 so T2 is equal to 4 T1 with that we have to identify two points. So after having done that the difference in dial readings between A and B is equal to X and so we have to locate point R which is at a distance X above the point A. So what we need to do is that the difference in dial gauge readings you know is X here let us say that if these two dial gauge vertical compression readings between A and B is say some X and above some X again a locate point R that means that another this is one X and this is the another X. So here what we see some initial compression some primary consolidation region and this is the secondary compression and secondary consolidation region. So locate point R so we have located point R above A which is X units above the X units vertically above this one and that from through this point a draw in horizontal line and you indicate that as the D0 that is the dial gauge reading at 0% consolidation take that as the dial gauge reading at 0% consolidation D0 and this is regarded as D0 at 0% consolidation. After obtaining that draw the horizontal line RS and the dial reading corresponding to the this line is D0 which we have already said and which correspond to 0% consolidation. Then what we do is that the project the straight line portion of the primary consolidation curve and secondary consolidation curve and intersect intersect at T then the dial reading corresponding to T is D100 that is 100% consolidation. So what we need to do is that the secondary consolidation portion and primary consolidation portion this tangent we have to extend further like this and this tangent we have to extend backward and this point when you take it back and this is the point what we indicate is that for that given load increment this point is the dial gauge reading corresponding to 100% primary consolidation that is at the end of 100% primary consolidation. So beyond this it is actually entering into the secondary consolidation and secondary compression zone. So this D100 is nothing but which is at the end of primary consolidation D0 at the 0% primary consolidation the average of you know initial beginning of the primary consolidation at the end of the primary consolidation is called as D50 which is nothing but D0 plus D10 by T and the time corresponding to D50 is regarded as T50. So note here the D50 is equal to D0 plus D100 by 2 and the time corresponding to the D50 is regarded as T50. So this line indicates that which is extended towards for example in this case it is 10.2 minutes. So we need to determine the point V on the consolidation curve such that the reading correspond to a dial reading of D0 plus D100 by 2 which is equal to D50 the time corresponding to point V is called the T50 that is time for 50% consolidation curve. So in order to determine CV from the equation what we need to do is that for 50% consolidation TV is nothing but 0.197. So by using TCV by h square and h is nothing but h drainage square that is h is nothing but the depending upon the drainage for example in case of consolidation test we have got you know this filter stone both top and bottom. So it is HDR by 2 so we can write CV is equal to 0.197 that is that I factor for 50% consolidation into h square divided by T50. So from the graph what we do is that we actually simply measure what is T50. Once we get that one we can calculate by knowing the thickness of sample divided by 2 in case of doubly drainage we can get the CV value. So as for example here for D50 let us say for D50 that let us say the time here is shown as 10.2 minutes. So CV can be obtained as 0.197 into 2.06 by 2 that is nothing but this is you know thickness of the sample divided by 2 whole square divided by 10.2 into 60 then you know we will get the coefficient of consolidation in centimeter square per minute. So in the logarithmic time method what we have done is that for a given load increment dial gauge readings versus logarithmic of time we have to plot and then what we need to do is that we have to indicate two points mark two points on the you know the dial gauge reading log T curve and such that T2, T1 is at A or T2 is at B let us say and T2 is equal to 4T1 and that is the time lag difference and then you know that the vertical dial gauge reading difference between A and B points is let us say X then with another distance X above locate a point R and then draw line RS through that and the point along that particular line where we along that line where it meets to the dial gauge reading indicates that D0 and then when we extending the end of primary consolidation tangent and then the straight line portion of the primary consolidation portion and the secondary consolidation portion when they meet at point T and this is actually is indicated or regarded as D100 and D50 is equal to D0 plus D10 and with that we can actually locate a point you know D50 here on let that point is B when we drop a vertical that is the T50. So with that what we can actually get is the by knowing T50 we can actually calculate for 50% degree of consolidation TV is equal to 0.197 so with that we can actually get what is the you know the so called coefficient of consolidation. The next method is Taylor's root time method here what we need to do is that this is also for a given load increment plot the dial reading and the corresponding square root of time. So here dial reading on the y-axis and on the x-axis square root of time is actually required to be plotted then that we have to draw the tangent DQ to the early portion of the plot. So here in this particular this is the Taylor's root time method so you can see that the x-axis is actually is marked with a root T that is in minutes root T which is in minutes so whatever the time which we take that we have to plot it in the root T scale on the x-axis and the dial gauge readings whatever we record we can actually plot here. So in this one assumption is that the initial portion is actually assumed to be a straight line. So normally for you know certain soils where you know the secondary consolidation or clayey soils you know the getting you know this straight line portion is difficult but in the presentation with where if the organic pattern is say not prevalent if the soil is actually silty in nature there can be possibility that the straight line portion can be obtained. So in this case you know we can actually draw a line a tangent such a way that passing through the initial portion of the curve which is the straight line portion a draw a line which is actually is DQ is here. So this is what the DQ is mentioned draw the tangent DQ to the early portion of the plot and draw the line DR such that OR is equal to 1.15 OQ. So we will see how OR is equal to this is O and if this is O here OR is equal to 1.15 OQ that means that OR is equal to 1.15 OQ. So this is you know we draw the initial portion the DQ and another line which is actually originates from D again which is you know the OR here. So how this has been obtained is that if you look into this theoretical curve where U versus root Tv again we you know we said that you know this is analogous to U and Tv and U and root Tv and U and the settlement or compression versus root T. So these two are analogous so because of this what we do is that this U and you know the root Tv the nature of the curve is like this. So this is 0% consolidation and this is 90% consolidation so this vertical height difference is 0.9 and this is 1. So this triangle has this ordinate is 0.9 this ordinate is 1. So here the upper part of the U versus root Tv is almost straight line you can see that the upper part of the U versus Tv is almost straight line this is for you know for this is based on Tv is equal to pi by 4 U by 100 whole square. So by using that one we can actually say that you know this slope of this line you know is nothing but 2 by root pi the slope of this line is nothing but 2 by root pi. So with this what we can actually obtain now is like this the slope of this straight line portion of this U versus root Tv is 2 by root pi this is actually obtained from you know for primary consolidation equation for U less than 60% we said that U is equal to pi by 4 U by 100 whole square. So that one when you simplify we will actually get you know 2 by root pi as the slope. So now for the straight line portion so we can write that Ac, Ac is nothing but Ac is nothing but that is here for 90% consolidation, 90% consolidation Ac is nothing but root Tv is nothing but 80% for 90% consolidation time factor is 0.848 so which is you know root over 0.848 and AB, AB is nothing but vertical ordinate this triangle if you look into that OAB then OA is equal to 0.9 and the slope is 2 by root pi then AB becomes 0.9 into 2 by root pi. So with that what we will get is that Ac by Ac by AB, Ac by AB you know is obtained as Ac by AB is nothing but Ac is nothing but root over 0.848 divided by 0.9 into 2 by root pi which once we simplify we will get Ac by AB as 1.15. That means that the theoretical curve U versus root 2V is a straight line up to 60% consolidation and the axis of the curve of 90% consolidation is 1.15 times the extension of an, absence of an extension of the straight line. So the theoretical curve U versus root 2V is a straight line up to 60% consolidation and the absence of the curve at 90% consolidation is 1.15 times the absence of an extension of the straight line. So using this logic what we have said is that the OR is such that OR is equal to 1.15 times UQ is actually considering the analogy and then by using the characteristic of U versus root 2V we can say that that OR is equal to 1.15, 1.15Q can be set. Once we get that you know once we draw this particular you know these two lines the obsessive of the point E that is the intersection of DR and the consolidation curve will give you root T90 that is the square root of the time for the 90% consolidation that is the point where this DR will intersect the consolidation curve that is this point that is this point is actually regarded as E and the vertical when you drop we get is that time required for 90% consolidation. So this is 90% consolidation this is 100% consolidation and this is regarded as the 0% and this point D you know is regarded as the 0% consolidation that is there is reading at 0% consolidation. So the value of the T for UV is 90% is 0.848. So CV can be obtained as 0.848 into h square by T90. So here also in this thing h is depending upon the drainage path single drainage or double drainage and based on that we can actually calculate CV and which is nothing but 0.848 into h square by T90. So this is was discussed already. So with that by using that procedure which we have discussed in the Taylor's root T method we can actually find out what is the equation of consolidation. And if you look into the chief differences between these two methods and long time method makes use of early pre consolidation, makes use of early primary consolidation and the latter time responses that is secondary compression and in comparison the root time method only utilizes early time response which is expected to be a straight line. So log time method makes use of early as well as latter time responses but in comparison root time method only uses the initial portion of the consolidation and which is assumed to be a straight line and root time method should give good results except with nonlinearity problems arising from the secondary compression caused substantial deviations from the expected straight lines. So if there is a secondary consolidation you know situations when they are actually happening then you know the root time method you know will not be able to give the good results because the basic problem is that the nonlinearity which actually arises because of the secondary compression. So most pronounced for clay soils with organic so this nonlinearity is actually pronounced and you know of the dial gauge reading versus root T curve or U versus root T curve is actually tend to be nonlinear and the straight line portion may not be expected so that is for marine clays. So basically you know we can actually we can put like that as you know logarithmic time method says for primary consolidation and it takes into consideration latter time response that is secondary consolidation. We can say that the log time method is most suited for clays especially marine clays and root time method is for silt and where the organic matter is you know even for organic silt is a question but if there is an absence of organic matter then there can be considered. So log time method is actually most suited for clays and root time method can be said as the most suited for silt and in the recent past then there are other methods which are actually have come they are called early stage log T method and these methods you know this is one of these recent methods this was actually put forward by Robinson and Alam in 1996 and this is an extension of logarithmic time method and the early stage of the logarithmic time method. So here what they have done is that you know the early stages of so this is you can say that this you know the logarithmic time method you know was actually extended by this Robinson and Alam in 1996. This was done basically by looking into the field data where the quotient of consolidation equations quotient of consolidation values are very high compared to the laboratory data. So the early stage of log T method that is after Robinson and Alam 1996 is an extension of the logarithmic time method and is based on the specimen deformation against the logarithmic of time plot. So in the first step we follow the logarithmic of time method to determine D0. So here also to determine D0 we have to follow the procedure which we adopted. So what we have to do is that we have to you know we have to first plot the dial gauge reading versus the logarithmic of time then we have to plot and indicate 2 points such a way that A and B and such a way that the vertical compression difference is say X and the location horizontally on horizontal axis this points A and B are located such as that Tb – Ta, Tb is equal to 4 times Ta and then we can actually determine D0 which is corresponding to 0% primary consolidation. Then what we do is that draw the horizontal line DE through the you know this D0 point corresponding to the 0% consolidation and then they draw a tangent through a point of inflection that is through a point of inflection where the straight line portion is there and beyond which actually there is you know the secondary consolidation that is the 100% 90% consolidation and 100% consolidation will actually come here. So extend this tangent like this and the point where this meets you know this horizontal line DE is indicated as G. So this particular point is regarded as the tangent intersect the line DE at G. So this tangent intersects the line DE at G and then determine T time T corresponding to G which is the time at UAV is equal to 22.4% that is the which is this is the time which correspond to UAV is equal to 22.4% for so for 22.4% consolidation and the time factor is TV is equal to 0.0385 so by using this and we can actually calculate the CV which is nothing but 0.0385 HDR square divided by the time for 22.14% consolidation. So with this what we can actually get is that we will get the you know the by using the early stage of logarithmic either the consolidation test for a given loading element we can calculate the consolidation. We can calculate the coefficient of consolidation and we need not actually wait for the consolidation test to complete as for a given loading element as the data starts coming and we will be able to plot and then try to get you know the coefficient of consolidation from the early part of consolidation itself. So in most cases for a given soil and pressure range the magnitude of CV determined using logarithmic of time method provides the lowest value the highest value is obtained from the early stage of the logarithmic time method. So this is because for a given soil and pressure range the magnitude of CV determined using logarithmic of time method provides the lowest value and this is the highest value is obtained by the early stage of logarithmic. So if you compare with the conventional logarithmic time method the early stage logarithmic method value is on the higher side the reason is that this is because the early stage of logarithmic time method uses the earlier part of the consolidation whereas the logarithmic time method uses the initial as well as the lower portion that is the lower portion where you know the component of secondary consolidation also comes into picture. So because of this the you know the early stage logarithmic method log t method will actually has higher values predictive values are on the higher side and when the lower portion of the consolidation curve is taken into consideration the effect of secondary consolidation plays a role in the magnitude of CV. So several investigation also reported that CV value obtained from the field is substantially higher than that obtained from the laboratory test conducted using conventional testing methods that is logarithmic of time and square root of time method. So use of early stage logarithmic time method may provide a more realistic values of the field that means that what this author's contention is that by using this early stage logarithmic of time method you know it may give us a realistic values which may match with the field values. So when the lower portion of the consolidation curve is taken into consideration the effect of the secondary consolidation plays a role because of that the magnitude of CV decreases. So several investigators also reported that the CV value obtained from the field is substantially higher than that obtained from the laboratory test conducted using conventional testing methods like logarithmic of time and square root of time method and by adopting this early stage logarithmic time method the values may provide you know have a good match with the realistic values in the measured in the field. So after having discussed you know this you know methods for determining coefficient of consolidation you know let us look into what we have done is that in the determination of methods for consolidation we have discussed about the two standard methods and one early stage method which is given by Robinson and Alam in 1996. Now let us consider an example problem wherein the problem statement works out like this we have got 8 meter depth of sand or nice a 6 meter layer of clay below which is an impermeable stratum. The water table is 2 meters below the surface of sand and over a period of 1 year a 3 meter depth of the fill is to be dumped on the surface over an extensive area that means that the fill extends to large areas the saturated unit weight of the sand is 19 kilo Newton per meter cube and that of the clay is 20 kilo Newton per meter cube and above the water table the clay the unit weight of the sand is 17 kilo Newton per meter cube. So for the clay the relationship between the void ratio and the effective stress in kilo Newton kilo Newton per meter square can be represented by the equation which is given in the problem statement as is equal to 0.88 minus 0.32 logarithmic of sigma dash by 100. So here we have the 8 meter depth of the sand or nice a 6 meter layer of clay and below which there is the water table is 2 meter below the surface of the sand and here if you notice here the fill is actually not dumped over instantaneously it is dumped over a period of 1 year. So the construction period is takes about 1 year for the fill to reach about they maintained about 1 year to reach a 3 meter height on the soil strata which is actually given and the saturated unit weight of the sand is 19 kilo Newton per meter cube and that of the clay is 20 kilo Newton per meter cube and this unit weight of the soil above the water table of the sand is 17 kilo Newton per meter cube and then here if you look into this the relationship between the void ratio and the effective stress are given that relationship is actually worked out from which is obtained from E log sigma or E log p curve which is nothing but E is equal to 0.88 minus 0.32 into log sigma dash by 100 where the sigma dash is measured in kilo Newton per meter square and the is also given that coefficient of consolidation of this soil is 1.26 meter square per year we know that the coefficient of consolidation the unit weight is meter square per units are meters per second meter square per year or centimeter square per minute this is the units are the coefficient of consolidation of this soil in the of the clay is 1.26 meter square per year. So what we need to do is that calculate the final settlement of the area due to the consolidation of the clay and the settlement after period of 3 years from the start of dumping. So we need to calculate you know the final settlement of the area due to the consolidation of clay and the settlement after period of 3 years from the start of dumping then if you are actually having a very thin layer of sand freely draining type and existing it is existed 1.5 meter above the bottom of the clay layer then what will be the value of final and 3 year settlements. So first is that clay is homogeneous but naturally there are some lengths of the clay and where you know which actually has been located generally this when you have got some clay deposits they are stratified in nature and sometimes there is a possibility that large thin sand lenses can actually come. Suppose if these thin sand layers are not located from the soil investigation and if the settlements are actually estimated without considering this presence of this thin sand layer then the settlements will be catastrophic in nature that is what actually we are going to demonstrate to this through this problem. So here if a thin layer of sand and freely draining type and existed 1.5 meter above the bottom of the clay layer then what would be the values of the final and 3 year settlements. So this data which is actually given the problem is explained here bottom layer is impermeable and the upper layer is here is sand so 8 meter layer is sand here and water table is 2 meter below the ground surface and we can see that here the clay is 6 meter thick and the second portion what we have is that we have got impermeable layer but a thin layer of sand is actually shown here. So this is 4.5 meter this is 1.5 meter so this is 1.5 meter below the base layer and this is the homogeneous layer. So since the field covers the wide area the problem can be considered to be one-dimensional the consolidation settlement will be calculated in terms of Cc considering the clay layer as a whole layer and therefore the initial final values of effective vertical stress at the center of the clay layer are required. So we can actually calculate sigma naught dash that is at this center of this layer we know that because here the excess pore water pressure dissipated is minimal here so we calculated the center of the clay layer sigma naught dash so this is actually sigma naught dash is equal to 17 into 2 that is above water table 9.2 into 6 plus or 10.2 into 3 so this is sigma naught dash that is taking pore water pressure out of it. Then E naught which is you know 0.88 minus 0.32 into log 1.198 because here sigma naught dash is you know 119.8 divided by 100 we will get this much. Then you know what we have is that initial void ratio from the E log p data is 0.855 and sigma 1 dash because here we wanted to say that 119.8 plus 3 into 20 so this actually has been loaded over a period of 3 years so the sigma 1 dash is 179.8 equivalent per meter square so the logarithmic of 179.8 divided by 119.8 which is 0.196, 176 so with that what we get is that the settlement is you know about 182 mm. So in this particular problem we will actually stop here wherein we actually discussed about how we can actually determine the final consolidation settlement and this is the part of the problem only wherein we have calculated the you know the settlement which is coming out as if this clay undergoes consolidation the total settlement is a consolidation settlement will be 182 mm. And then further we actually have to account for you know if you notice that the time required for filling the you know the placing the fill on the soil is about one year. So then we have to do correction for the construction period correction. Once this construction period correction concept is introduced then we will try to solve the part 2 of this problem and then we will try to take into the effect of say thin naturally occurring thin layer of sand and how that affects in accelerating the consolidation. Generally if you know this is one of the classical example sometimes if it is not recognized during the site investigation if the structure is constructed thinking that the soil is you know homogeneous and free from this lenses of sand layers and if it is freely draining type the settlements can be you know the catastrophic in nature.