 Welcome to lecture series on advanced geotechnical engineering which is the course you know being produced by the Indian Institute of Technology Bombay department of civil engineering. And we are in module 3 lecture 7 on compressibility and consolidation and we have been actually discussing about the methods for determining coefficient of consolidation in this particular lecture also. We will try to discuss about another method which is the rectangular hyperbola method and then we will try to discuss about the secondary consolidation concepts and then how to determine settlements of a compressible soil deposit from the compression curves and that is settlement of the compressible soil layers. And we also look into you know when we have you know construction period when it actually happens for a certain period of time how the construction period can be accounted or corrected in the you know in computing time rate of settlements then we will continue the problem which we introduced ourselves in the previous lecture. So the correction for construction period as we can see in this particular slide in practice the loads which are actually applied to soil they are not instantaneous but over a period of time they are applied over a period of time that means that if you wanted to construct in a bund in an embankment of about 6 meter high it cannot actually happen all of sudden. So in that case you know it takes over a period of time and it can be you know a constant rate at which the height will increase or when we look into the stages like you know two stages or three stages depending upon that when you take the rate it can be also set as constant. So initially when we actually acquire the site there can be some excavation or you know so in this case there can be some you know the relieving of pressures that is removed of the load and then reloading again because of the some restoration profile and then you know the fill will start commencing up. So this is you know the load versus time let us say that you know TC that is the time required for construction let us say that 6 months may be in the field or 1 year in the field that is called time required for constructing let us say an embankment of 6 meters or h meters height. So in this case if you are having in a embankment of certain height and which causes a pressure of say P dash then TC is set as the time period of construction. Now what we if you look into this you know as per the Terzaghi's theory of one-dimensional consolidation and it assumes that you know the load is placed instantaneously that means that here you will see that the entire you know this is due to instantaneous loading but as we know that this particular portion is not actually subjected to load you know first 6 months let us say that the load will not exist at all. In that case you know we actually end up you know having the higher order of settlements but over a long period of you know after a long period of you know elapsing long period after construction then again they tend to be same that is what you can see that. So here what we see is a corrected curve so how to you know obtain this corrected curve that we look into it and then we apply this in the problem which we are going which we are discussing from the previous lecture onwards. So in practice what we are submitting is that the structural loads are applied to the soil not instantaneously but over a period of time. So Terzaghi proposed an empirical method for correcting empirical method of correcting the instantaneous time settlement curve to allow for the construction period. So Terzaghi has come out with an empirical method of correction for instantaneous time settlement curve to allow for the construction period. So what we do is that the net load P dash is the gross load less the weight of the soil excavated and the effective construction period Tc is measured from the time when P dash is 0 that is when P dash is 0 means the construction period is actually measured from the period this is not the excavation all those things from this stage onwards the construction period is measured from this stage to this stage that is Tc that is the construction period. Now it is assumed that the net load is applied uniformly over the time Tc and the degree of consolidation at time Tc is same as if the load P dash had been acting as a constant load for the period Tc by 2. So it is assumed that the net load is applied uniformly over the period of time Tc and the degree of consolidation at time Tc is same as if the load P dash has been acting as a constant load for the period Tc by 2 that is of the construction period that means that the settlement any time during the construction period is equal to that occurring for instantaneous loading at half the time that means that you know if settlement is there the settlement at any time during the construction period is equal to that occurring for instantaneous loading at that half the time. However, since the load then acting is not the total load the value of the settlement so tend must be reduced in proportion of the load to the total load. So in view of this let us define here that the TC is the time which is required for construction that is to raise to the pressure of P dash and if that settlement is say SC and what we are saying is that at TC by 2 also the consolidation so what we are saying is that if the settlement or degree of consolidation at TC by 2 and TC are assumed to be same. So what we try to look into this is that consider at any time T between 0 to TC a pressure T and which is having an ordinate P1 here the loading pressure and at this point let us say the time T1 and if you know if you are trying to get this T1 by 2 the T1 by 2 the settlement is say SC1 here SC1 here so the settlement at time T1 should be less than that. So in order to account for that we can write that the settlement at that particular point is nothing but S1 is equal to SC into P1 by P dash what we have done is that we actually apportion with we have apportion with P1 by P dash so P1 is actually less than P dash so P1 by P dash into the final consolidation settlement we said that that is S1. So this is nothing but what we have done is that we actually use this principle of triangles for this area for this zone and then this zone and then we try to compute the settlements here. So from the similar triangles we can write P1 by P dash P1 is actually pressure within the construction period P dash at time TC so and then time is that T1 by TC is equal to S1 by SC so for the period subsequent to the completion of construction the settlement curve will be instantaneous curve offset by the of the effective construction period that is what we are saying is that this particular one it is of the construction period that is TC by TC by 2 but a long period of elapsing long period after the construction they both remain to be same. So thus at any time after the end of construction the corrected time corresponding to any value of settlement is equal to the time from starting of the loading less of the effective construction period. So from here what we understood is that thus at any time after the end of the construction the corrected time corresponding to any value of the settlement is equal to the time from the start of the loading less of the effective construction period that means that what we have to do is that we have to take or deduct you know after the starting of the loading let us say that if it is 3 years is given and construction period say it takes you know 6 months then 3 years minus 6 months is around 2 and half years for example is the period that means that what we are doing is that we are deducting this effective construction period. So after a long period of time the magnitude of settlement is not of the effective construction period that means that if the construction period takes 1 year that is 3 years divided by 1 by 2 it is 2 and half years after a long period of time the magnitude of settlement is not appreciably affected by the construction time. So what we are doing is that 2 important things which we have understood is that the instantaneous curve and corrected curve are initially during the construction period they are not the same. So what we are doing is that as the loading intensity is less than what it is actually assumed in instantaneous curve computation so we correct the settlements such a way that you know the and then also any time after the end of the construction the corrected time corresponding to any value of the settlement is equal to the time from the start of the loading less of the effective construction period and after long period of time the magnitude of settlement is not appreciably affected by the construction time. Now let us look into this example, an 8 meter depth of sand over lies a 6 meter layer of clay below which an improbable status is there the water table is 2 meter below the surface of the sand and over a period of 1 year so here the construction duration is 1 year please note the construction duration is 1 year a 3 meter depth of the fill which is actually having an unit weight of 20 kilo Newton per meter cube is to be dumped on the surface over an extensive area that means that the area is the fill area actually spread over large areas. So 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 unit weight of the sand is 17 kilo Newton per meter cube. So for the clay the relationship between the void ratio and effective stress is given and this is nothing but E is equal to 0.88 minus 0.32 logarithmic of sigma dash by 100. So what has been asked is that calculate the final settlement of the area due to the consolidation of the clay and the settlement after a period of 3 years from the start of dumping and the coefficient of consolidation is given that is 1.26 meter square per year and if a very thin layer of sand freely draining type existed 1.5 meter above the bottom of the clay layer what would be the values of the final and 3 years settlements. So this is the figure based on the problem what we have discussed we have got a 6 meter thick clay layer and which is having a thickness of 6 meters clay layer having 6 meters thickness imperable stratum here and so this is one way drainage water flows in this direction water flows in this direction and this is the open layer for this clay layer and this layer is having a thickness of about 8 meters and 2 meters is the depth of water table from the ground surface and on this a fill of 3 meter height is placed over a period of 1 year and it is also possible that in some locations there can be chances of thin lenses of sand layers particularly in alluvial deposits they can exist and these deposits this type of thin lenses of sand layers can cause what is their effect we can actually see. So in this given problem we have a layer of this is 4.5 meter and this is 1.5 meter so this problem is actually after Craig 2004 very classical problem where it has been discussed what is the effect of this thin lenses of sand layers on the time rate of settlements. So this we have discussed but anyhow we will try to recapture what has been discussed as we have seen that the field covers a wide area the problem can be considered to be one dimension that is one we have discussed the consolidation settlement will be calculated in terms of Cc considering the clay layer as a whole and therefore the initial and final values of the effective vertical stress at the center of the clay layer are required. So with this once we calculate we get 1 I2 to mm as the settlement and which is you know which is the ultimate consolidation final consolidation settlement. Now here because of the increase in the construction of you know the fill having height of say 3 meters or a period of 1 year now what we need to do is that we need to calculate the settlement after 3 years that means we actually have to correct by using the Terzaghi's method what we discussed where T is equal to 3 minus Tc by 2 where Tc is nothing but the time required for constructing 3 meter embankment that is 1 year so 3 minus 1 by 2 2.5 years in the calculation of degree of consolidation 3 years after the start of dumping the corrected value of the time to allow for the 1 year dumping period is nothing but 3 minus 1 by 2 that is 2 and half years. So the layer is half closed that nothing but a single drainage therefore T is equal to 6 meters. Now from Tv time factor is equal to Tcv by h square or Tcv by d square wherein what we get is that 0.0875. So by using Tv is equal to pi by 4 u by 100 whole square where we can actually get for u is equal to average consolidation as 0.335. So as we know the final consolidation settlement which is 182 mm and settlement after 3 years will be 61 mm settlement after 3 years will only be 61 mm that means that 0.335 is the degree of consolidation we have computed based on the data which you are having and the settlement after 3 years is say S is equal to 0.335 into 182 that is 61 mm. Second problem what has second part of the problem has been asked that the final settlement will still be 182 mm for even for the second problem but only thing is that the rate of the settlement will be affected why because if you are having a thin layer of sand and where it actually has got the let us say that the layer receives the water and has got the drainage capability then it can actually work as from the bottom portion it actually can work as one way drainage and the portion above 1.5 meter it can work as a two way drainage. So in the sense what will happen is that here we actually have the positive effect of this clay layer the sand layer on the clay layer if it is directed in advance then you know there is a possibility that the settlements can be accelerated. So from the point of view of the drainage there is now an open area open layer of thickness 4.5 meter that is above a half closed layer of thickness d1.5 meters. So that is if we look into this here this portion is behaves like a one way drainage the water from here shunts down to this one and this portion which is actually having a drainage path of 2.25 meter half of the water goes this side half of the water goes this side and we assume that this layer actually has got a capability of pumping the water out from this layer this side this portion to this portion and in this case now let us see what we do is that this upper portion of the layer actually having 4.5 meters thickness and bottom portion is having 1.5 meters thickness and total thickness of the clay layer is 6 meters with this you know we can actually calculate the effect of this like this. So upper portion of the curve that is TV1 for this you know we can actually calculate you know what is the you know the based on the drainage T Cv by h square that is nothing but 2.25 by 2 that is T is nothing but 3 minus 1 by 2 so and Cv is 1.26 meter square per year divided by that is 4.5 by 2 you will get 0.622 for 0.62 by using you know the we can actually get the degree of consolidation as 0.825 and then the bottom layer which is you know which is nothing but half closed layer so which is nothing but TV2 can be obtained like 3 minus 1 by 2 into Cv is 1.26 divided by 1.5 square with that what we get is that 1.4 so with this what you get is the degree of consolidation as 0.97. Now if you look into this now for each layer SC is equal to U into SCF now we have to get the you know by using the weighted average method you have to get the U bar. Now upper layer is having a 4.5 meter thickness and bottom layer is actually having 1.5 meter thickness so 4.5 into U1 plus 1.5 into U2 is equal to 6 U bar so U1 is equal to 0.825 U2 is equal to 0.97 and so U bar is actually obtained as 0.86. Now for so settlement after 3 years can be obtained as final consolidation settlement we have actually obtained as you know 182 mm and when we have only 6 meter thick layer of consolidation 6 meter thick layer of one way half closed layer we said that the settlement of only 61 mm will occur but if we are actually having a thin sand layer then SC is equal to 0.86 into 182 that is 157 mm. So that means that if we are in from the soil investigation data if we fail to recognize this type of occurrence of thin layers of sand and the design of this building is done from the bearing capacity and settlement point of view and if the settlements of this order of 157 mm occurs within you know period of say 3 to 4 years of from the start of the construction of the building or a structure and the structure which is you know going to be you know the which is going to be place going to have experience the distress because the allowable settlements in the clay are not more than 50 mm. So in this case what we can see that you know the settlements are very excessive in nature and they can actually cause distress and failures to the structures. So this example clearly demonstrates the effect of the thin layers of the sands on the you know how they can actually have accelerating effects on the rate of consolidation but unfortunately if it actually occurs in your site and if it is recognized you know value in advance yes there is a possibility that the settlements can be accelerated but if you fail to recognize then ill effects are discussed but if these layers are there then there is a possibility that the settlements will be accelerated but it is hard to find this type of things you know occurring repeatedly so in view of that you know there are several other methods which are there for accelerating consolidation so those things will be discussing while addressing about the methods for accelerating consolidation settlements. Now after having discussed about that particular problem let us try to discuss about the another method which is rectangular hyperbola method and which is after Sridharan and Prakash 1985 and based on this equation which we have shown here uz is equal to 1 minus summation m is equal to 0 to infinity 2 by capital M sin mz by h exponential of minus m square Tv. So it can be shown that the plot of T by UAV versus Tv T suffix v will be of the type shown in figure a which will be showing and in the range of 60 percent to 90 percent of average degree of consolidation the relation is linear and can be expressed as Tv by UAV is equal to 8.208 into 10 to the power of minus 3 T suffix v plus 2.44 into 10 to the power of minus 3. So this is what actually shown here then rectangular hyperbola method T suffix v Tv versus Tv by UAV when it is plotted here you can say that this portion is non-linear and then this portion is linear and the UAV between 60 to 90 percent what we are actually saying is that this is in the range of the linear range the relation is linear and can be expressed as this particular expression. Now similarly by using this we can actually plot T by delta ht with the time required for consolidation then we actually get this part of variation like O, B and C. So this slope of this line is m to the vertical ordinate and 1 to the horizontal and this point B and C that is actually 60 percent and 90 percent and D and this ordinate is D. Now using the same analogy the consolidation test results can be plotted in graphical form as T by delta ht versus time where T is the time and ht is the specimen deformation which will be of the type shown in the figure B that is what actually we have shown this is the plot which is T by delta ht versus time t. Now the following the procedure can be used to estimate the CV so once we established once we plotted from the sample deformation data ht and delta ht is nothing but the sample deformation data which is recorded for the particular load increment versus time of consolidation. So identify the straight line portion BC and project it backward to D determine the intercept. So what we need is that we have to plot T by delta ht with the time and then identify the linear portion extend this backward and determine what is this intercept D. Now determine the slope m of the line BC that inclination of the slope m. Now calculate CV as CV is equal to 0.3 into m h square divided by D capital D. Now here we can check whether it is giving units of meter square per second or not. The authors while setting this one h is the length of the maximum drainage path that is if it is a single drainage or double drainage and note that the unit of m is L minus 1 that is unit of length is 1 by L and unit of D that intercept is T by L unit of D is T by L. So if you look into this 1 by L into that is L 1 by L into L square into T by L on simplification what we get is that CV as the L square by T that is nothing but meter square per second meter square per year or meter square per day. So that is how you will actually try to get the coefficient of consolidation by using rectangular hyperbola method. Now we actually have in the beginning while discussing the consolidation and we said that the consolidation component of a fine-drained soil is 3 components one is elastic and the second component is the major component which is the primary consolidation settlement and primary consolidation portion and thirdly what we said is that due to secondary consolidation or secondary compression. And this total summation of all these three put together is called total settlement which is nothing but due to elastic settlement, consolidation settlement and then secondary consolidation settlement. The secondary consolidation settlement they do occur on soils like Pt type of soils or certain type of soil like materials, municipal solid waste they undergo very high degree of secondary consolidation. So clays are certain type of soils or soil like materials continue to settle under sustained loading at the end of the primary consolidation and this is due to the continued readjustment of clay particles and this phenomenon is called secondary consolidation. Clays particularly this is also called as creeping of soil or say you know the secondary creep or secondary compression. So clays of certain type of soils or soil like materials continue to settle under sustained loading at the end of the primary consolidation and this is due to the continued readjustment of the clay particles and this phenomenon is called secondary consolidation. Now if you look into this here this particular slide where figure where time versus delta v, delta v is nothing but the volume change is plotted here and that is nothing but delta h and here a into delta h is nothing but delta v. Now up to e we can say that the process of pre consolidation, primary consolidation it actually happens. So initially at small times the hydraulic gradients are very high, the flow is very rapid and as the time elapses and it comes closer to the completion of consolidation the hydraulic gradients you know they drop down to almost to 0 and no flow conditions occur and the settlements almost the 0 excess pore water pressure conditions provide and it is also it will be that you know when there is no change in the loading the excess effective stress also changes will be minimal. Then we can actually say that you know the soil consolidation is complete and all but it then the curve also tends to become asymptotic to the horizontal and then it pretends as if that the consolidation is complete but in certain type of soils that is actually beyond point e even under the constant effective stress the soil undergoes you know the multiple structural changes and because of readjustment of the particles under because when the particles undergoing creep there is a possibility that this secondary compression or secondary consolidation or creep can result. So at point e excess pore water pressure is 0 and the constant sigma dash appears and no change in delta v occurs but however a small changes in delta v occur due to the soil creep. So this is a typical time dependent compression of the soil is actually shown here. So secondary consolidation settlement is more important than the primary consolidation basically if you are having organic and highly compressible in organic soils. So if you are having a organic particularly a marine clay or highly compressible in organic soils there is a possibility that the secondary consolidation settlement will be much more than the primary consolidation settlement. In work consolidated in organic clays the secondary compression C alpha is very small and of less significance for example if you are having a work consolidated in organic clays the secondary compression C alpha is very small and it is of less practical significance. So in this particular slide wide ratio versus time is actually shown here on the logarithmic scale. So this is the end of the primary consolidation but beyond this time you can say that time t1 is the time at the end of primary consolidation the soil undergoes a change in water ratio that is delta e and the time let us say is t2. So t2 minus t1 let us say is you know that is the slope of this line is indicated by C alpha where C alpha is nothing but delta e divided by logarithmic of t2 by t1. So C alpha is nothing but logarithmic of delta e by log t2 by t1. So the settlement actually is written as secondary consolidation settlement is nothing but C alpha into hc by 1 plus ep where ep is nothing but the void ratio at the end of the primary consolidation that is ep is nothing but the so initially we will start somewhere here and at the end of primary consolidation we reach here so from ep to delta e the change will be there. So that is nothing but here is indicated as sc is equal to, ss is equal to you know C alpha that is the secondary consolidation coefficient into hc by 1 plus ep into logarithmic of t2 by t1. So t1 is the time required for completion of primary consolidation and time 2 is the time at which the secondary consolidation is being determined. So it can be of let us say one logarithmic cycle that is let us say t2 is equal to 10 times t1 then it is C alpha is equal to you know that is C is equal to 1. Now for according to Messery 1973 for sedimented undisturbed soils for sedimented undisturbed soils delta e by delta log t decreases with increase in the final consolidation pressure. So for sedimented undisturbed soils delta e by delta log t decreases with increase in the final consolidation pressure and the remoulding of clay creates a more dispersive fabric. So this results a decrease in the effect of coefficient of secondary consolidation at lower consolidation pressures as compared to that of for the undisturbed samples. However it increases with consolidation pressure to a maximum value and then decreases finally merging with values of normally consolidated undisturbed samples. So remoulding of clays create a more dispersive fabric the clays you know soil fabric changes and this results in a decrease of the coefficient of secondary consolidation at lower consolidation pressures as compared to that for the undisturbed samples. However it increases with consolidation pressure to a maximum value and then decreases finally merging the values for normally consolidated undisturbed samples. So pre-compressed clays or preloaded clays show a smaller value of coefficient of secondary consolidation. So the degree of the reduction appears to be a function of the degree of pre-compression. The degree of the reduction appears to be a function of degree of pre-compression. So there are empirical correlations are actually available you know for secondary compression index C alpha for inorganic clays and sills which is C alpha is equal to 0.04 times C c where C c is the compression index for inorganic clays and sills. For organic clays and sills it is regarded as 0.05 times C c and for peats that is fibrous nature type of soil types which is actually peat which is the part of marshy lands C alpha is equal to can be as high as 0.075 into C c. For more consolidated clays with more consolidation greater than 2 to 3 the you know the secondary compression index will be only 0.001. Organic soils it is 0.0252 or more and normally consolidated soils will have 0.004 to 0.025. Then as I said that municipal solid waste is a material which is generated from the is a man made waste and where in actually has got a combination of 40 to 50% of biodegradable waste and then other inert materials like construction waste and so in this mixture of the heterogeneous mixture of materials causes you know very high normally for you know these ranges from 0.024 to 0.03 but even up to 0.163 to 0.35 are reported by Shokomar Babu at all. So the creep deformations are small and normally neglected it is required to be however you know we need to notice that you know the creep deformation if you look into that this compression index values are very very small when compared to compression index. So the creep deformations are small but normally neglected but it is required to be noted that the small time dependent deformation that are not due to exclusively to change in X mod dash do occur in soils and soil like materials. So in case of municipal solid waste this occurs due to you know ongoing biodecomposition where in this actually you know results in the changes in the you know these you know very large changes in the secondary compression index in fact in the case of municipal solid waste it actually is defined as C alpha 1 and C alpha 2, C alpha 2 which is the due to the you know biodecomposition and C alpha 1 is the initial part of secondary compression which is in the you know if you are having a C alpha of say 0.163 and you know out of that 10 to 20% is of C alpha 1 and the rest of the portion is C alpha 2. So after having discussed about the secondary compression and how this can be used in calculating then you know we can calculate you can see how the settlement of a compressible layers can be calculated in one dimensional settlement this is basically one dimensional settlements and so if the clay layer of thickness say HT is subjected in increase in pressure, pressure increase due to sigma naught dash at the mid depth of the clay layer to sigma 1 dash. So settlements are always computed at the mid depth of the clay layer the reason we have discussed it is that at the mid depth of the clay layer there is you know during let us say 50% degree of consolidation or 90% degree of consolidation there is a certain amount of pore water pressure had to be dissipated. So in view of that you know the settlements are actually computed at the mid depth of the clay layer. So if the clay layer of layer of total thickness HT is subjected to increase in average effective overburden pressure from sigma naught dash to sigma 1 dash it will undergo a consolidation settlement of H1 hence the strain can be given by the strain is nothing but a vertical strain which is nothing but delta HT by HT and so this is actually shown here schematically when you are having in the field a soil where it is subjected to pressure from sigma naught dash to sigma 1 dash delta HT is the settlement and HT is the original thickness so the strain is nothing but delta HT by HT. But if you are having you know volume of the voids and volume of the solids here which is actually indicated as 1 and total volume is 1 plus E0 or specific volume which is defined as 1 plus E0 is small v that is specific volume is nothing but 1 plus void ratio is the specific volume and if the change in void ratio is say delta E and we can actually say that change in volume to original volume which is nothing but delta E by 1 plus E0 so this is also you know equal to strain so again if an undistributed laboratory specimen is subjected to the same effective stress increase then the void ratio will decrease by delta E so the strain is actually given by delta E by 1 plus E0. So with this what we actually obtain is that by equating you know this strain due to change in thickness and change due to strain in void ratio both appears to be same now epsilon is equal to delta HT by HT and delta E by 1 plus E0 so we can actually calculate delta E is equal to delta HT into 1 plus E0 by HT. Now in this particular figure you know it is actually shown here like we are having a clay layer and which is having a distance of say H1 from the base of the foundation H2 from the you know from the bottom of the clay layer is H2 meters away and top of the clay layer is H1 meters away the thickness of the clay layer is H2 minus H1 so HE is the thickness. Now here what will happen is that whenever we load a certain structure and the increase in stress is nothing but you know whatever we have discussed in stresses due to loads on soils that we have actually discussed so based on the appropriate whether it is a circular shape or whether it is a strip loading whether it is a uniformly loading or is a raft loading accordingly what we need to do is that we have to calculate the increase in stress in different portions of so what exactly we need to do is that we have to divide this clay layer into suppose if it is 6 meters each layer is actually divided into say 6 1 meters of thickness at each 1 meter you know at 0.5 meter within that layer we calculate what is the increase in thickness and we know the effective stress at that particular point then the sigma 0 dash 1 plus this delta S sigma 1 that is actually is the increase in stress so we can calculate what is the small increase in consolidation settlement that is delta 1, delta 2, delta 3 like that up to delta 6 the total which actually is resulted as the you know the consolidation settlement the more we actually have like you know shorter thickness of layers then the more you know effective is the determination of settlements. So now we can actually have the two cases one is that the soil can be normally consolidated completely that means that you know the we can actually have a virgin compression curve like this then the slope of this curve is the compression index then we can actually calculate the consolidation settlement but it can be also like we can have a work consolidated portion and normally consolidated portion sometimes we are actually have if this is the sigma 0 c dash and if this is the pre consolidation pressure then you know if the load is only up till here then we actually have you know the settlement only due to over consolidated portion but if you are having settlement here then if you are having a loading here sigma 1 dash here then we have to calculate settlement from here to sigma 0 dash that is initial pressure to this one from here to here. So the total resulted due to over consolidated portion as well as the you know the as well as the normally consolidated portion sometimes if you are not able to capture the over consolidated nature of the clay from the log from the consolidation test data and then you know we actually calculate settlements by assuming that the soil is normally consolidated then we tend to predict very high order of settlements and sometimes it results in the conservative mode of estimation of settlements. So from this slide the deliberation what we can discuss is that if you are having sigma 0 dash is equal to sigma c dash then you know we can say that that is consolidation settlement is equal to delta e by 1 plus e 0 is equal to cc by hc by 1 plus e 0 where hc is the thickness of the clay layer so here whether it is half drainage or full drainage entire thickness is going to undergo the settlement so we have to consider either full thickness of the total thickness of the clay layer and cc is the compression index e 0 is the initial void ratio at an effective stress of sigma 0 dash. So log of sigma 0 dash plus delta sigma average this delta sigma average which is like you know it can be at the mid depth of clay layer and sigma 0 dash so for normally consolidated clay that is sigma 0 dash sigma 0 dash c and over consolidated clay suppose if you are having you know only up to sigma 0 dash c and then there is over consolidated portion then we actually have to use this Cs that is the slope of this line the slope of this line in the over consolidated portion is Cs so the settlements will be very low if you are actually having a over consolidated soil and if the loading is only up to sigma c up to the pre consolidation pressure then the settlements will be very minimal but if you are having like the loading is say somewhere here then we have to calculate this portion and we have to calculate this portion so that is actually here it is listed it is nothing but Cs into hc by 1 plus e 0 log sigma c by sigma 0 so here sigma 0 to sigma c that means that we have accounted from here to here then further sigma c to the next one that is increased here so that is nothing but Cc by here it is the normal consolidated portion that is the compression index is actually accounted here Cc by hc by 1 plus e 0 into log sigma 0 dash to the delta sigma av by sigma c dash so like this by using the compression curves one can actually determine the secondary consolidation settlements one can determine the consolidation settlements so this one need to this is actually based on the compression curves which are actually deduced from the testing data and once these data are available then it is possible that one can calculate what is the how long it will take to occur the settlement and what is the magnitude of settlement. So the empirical methods for to obtain Cc in the absence of laboratory data let us say that if you are actually having this scanty laboratory data or field data then one can actually you know estimate there are methods are the empirical methods are there to estimate compression index so then from there we can also estimate now secondary compression index also so here say according to Skempton 1944 Cc is equal to 0.009 LL minus 10 this is for undisturbed clays for remolded clays it is 0.007 into LL minus 10 so liquid limit has to be given in percentage here that means the liquid limit is say 50 percent 50 minus 10 into 0.0007 which if you use then you will get the compression index for a remolded clay. Similarly Wartt and Wartt and Wood 1978 they postulated Cc is equal to 0.5 times Gs into Pi percentage by 100. So Gs is the shear modulus and this is actually used for calculating for postulating the Cc value from the plasticity index as well as the Gs. And Rendon and Herrero 1983 they have given again based on the Gs here is not the shear modulus it is the specific gravity of the clay Gs is the specific gravity of the clay and E is nothing but the initial void ratio so Gs is nothing but the specific gravity of the clay and Cc is equal to 0.14 into 0.141 into Gs to the raise 1.2 into 1 plus E naught by Gs to the raise 2.38 and similarly the Park and Kowmoth 2004 they actually have given a empirical method to calculate the Cc from the in situ porosity of the clay which is N naught in situ porosity of the clay from the porosity value is that they have given and Kulhawai and Maynay 1990 they have given Cc is equal to Plasticity index by 370 where plasticity index is taken as you know in percentage. So you can see that the compression index actually having you know direct relationship with the plasticity index most of our soil deposits actually they do exist at plastic limit. So the plasticity index is you know if it is high then the compression index is also very high that is for the clays of high compressibility CH type of soils the plasticity index value will be very very high. So then the compression index of such type of clays also is expected to be on the higher side. Now let us consider an example where the results of an odometer test of a normally concerted clay are given below in the two way drainage and we have a sigma dash kilo meter square the void ratio E that is at 50 kilo Pascal's and 1.01 and at 100 it is 0.9. So the time for 50% consolidation for the load increment for 50 to 100 kilo meter square was 12 minutes and the average thickness of sample was 24 mm and we need to determine the coefficient of permeability and the compression index. So in this particular problem the time for 50% consolidation for the load increment from 50 to 100 kilo Newton per meter square was 12 minutes. So for time for 50% consolidation for the load increment from 50 to 100 kilo Newton per meter square is given as 12 minutes and the average thickness of sample was 24 mm and determine the coefficient of permeability and the compression index. So based on the data here we can actually calculate Tv is equal to Tcv by h square and the u average is equal to 50% so it has been asked actually the two way drainage and 50% consolidation. So for uav is equal to 50% Tv is equal to that is average degree of consolidation 50% Tv is equal to 5 by 4 u by 100 whole square so with that we can actually calculate time factor as 0.197. So with this by knowing 0.197 Cv this time is what we need to say 12 minutes so the 12 minutes divided by 2.4 by 2 with that what we get is that Cv is equal to 0.236 cm square by minute so we can actually get 0.0236 into 10 to the power of minus 4 meter square per minute. So with this what we can actually get is that by knowing the Cv value we can actually calculate now coefficient of permeability as we know that by knowing k is equal to Cv mv gamma w or we can write also Cv is equal to k by mv gamma w with k divided by mv is nothing but delta e by delta sigma dash into 1 plus av into gamma w. So with this from the given data we can actually get delta e as 1 minus 01 minus 0.9 so that is nothing but 0.11 and delta sigma is nothing but 100 minus 50 that is 50 and by using this we can actually also get the so coefficient of consolidation as well as the coefficient of permeability that is 0.2605 into 10 to the power of minus 7 meter per minute and the compression index is now can be calculated by using delta e that is 0.11 divided by logarithmic of sigma 2 dash by sigma 1 dash which is nothing but 1.01 minus 0.9 divided by logarithmic of 100 by 50 which works out to be 0.365 the compression index is actually works out to be 0.365. So it is also like this from the by knowing the consolidation properties of soil in the laboratory like by using let us say we have got from here further from the laboratory data and we can also calculate what is the time required for the same consolidation to occur in the field like for example then in that case what we do is that we actually project our analogy of the results from the laboratory to the field. So in that case how we get is that the time required for the field is equal to time required in the laboratory into h f divided by h laboratory whole square is nothing but if the time required in the laboratory once you know like here in this case say 12 minutes into suppose if this is the result which is actually obtained for a two way drainage layer having say 6 meters thick layer so 6 by 2 divided by 2.24 by 2 that is 24 mm divided by 1000 by 2 and by whole square which we were able to get what is the time which actually will take for the soil layer to undergo consolidation in the field can be forecasted. So in this particular lecture what we have discussed is that we discussed the coefficient of consolidation determination based on you know rectangular hyperbola method which is according to Sridharan and Prakash 1985 then we also have discussed that how the construction period correction can be done according to method which is actually postulated by Terzaghi and Frolic in 1936 and then we have also solved a problem and we also have discussed that effect of the thin layers of sand layers on the time rate of settlements not the magnitude of settlements the time rate of settlements they will be very fast in the case that what will happen is that in the given not the final consolidation settlement but you know the rate at which for example for a given period of 3 years if you do not recognize the thin layer of sand the settlement will be very less but the structure will be subjected to high distress because of the prevalence of the neglected or ignored thin layer of sand then thereafter we also have discussed about how we can actually compute the settlements and also we discussed about the secondary compression and secondary creep in this particular lecture.