 Welcome to lecture series on advanced geotechnical engineering, we are actually in module 3 lecture number 4. So we are actually discussing about the compressibility and consolidation in module 3 and we introduced ourselves to the Terzaghi's one dimensional consolidation and then in the previous lecture we have reduced the consolidation equation that is dou u by dou t is equal to C v dou square u by dou z square and this is for one dimensional consolidation and this one dimensional consolidation is assumed to prevalent if a clay layer is subjected to large area loading on its surface. And if you are having a restrained or a finite dimensions loaded area then there is a possibility that the three dimensional consolidation can come but however we assume it like one dimensional consolidation and then we try to solve the and calculate the settlements. This example of the possibility of two dimensional consolidation can be in a footing resting on a soft clay. So in the previous lecture we have deduced in equation dou u by dou t is equal to C v dou square u by dou z square and then the C v is the coefficient of consolidation and it is related if you look into this it is related with permeability and coefficient of volume compressibility. So the k the coefficient of permeability or Darcy's permeability as the Darcy's law is assumed to be valid so k is equal to C v m v gamma w. So if you look into this if k is high that means that the permeability is high then the C v will be high then the time rate of settlements will be very high. If k is low like say meren clay where 1 into 10 to power of minus 9 to 1 into 10 to power of minus 10 meter per second then the C v will be very low and then the consolidation will take long time. So this is the basic differential equation of the Terzaghi's one dimensional consolidation and it can be solved with proper boundary conditions. So we have solved this by assuming the proper boundary conditions and this was obtained and further we have assumed these boundary conditions that at time t is equal to 0 it is assumed that the load is instantaneously applied on to the surface of the soil that is at that time the initial excess pore water pressure within the soil is equal to u is equal to ui is equal to delta sigma delta sigma is the increasing load which is applied on the surface of the soil. Then u is equal to 0 at z is equal to 0 and u is equal to 0 at z is equal to hi that means that once the consolidation commences moment the consolidation commences the pore water pressure at the boundaries reduce it to 0. This reasons we have discussed in previous lecture. So h is the longest drainage path so if you are having a clay layer sand which between let us say sand layer at top and bottom then if the thickness of the clay is say 2 h then top portion of 2 h will flow toward the upper layer and the upper the bottom portion of h of the clay layer the water in the bottom portion of the h of the clay layer will flow downward. So this is called double way drainage that is double way that is 2 way drainage or you know 2 way drainage or it is also called as you know double open layer. In this case the consolidation will be relatively faster and h will be equivalent to h t if it is the thickness then h will be equivalent to drainage path length is equivalent to h by 2 or h t by 2 which is you know the effective drainage length. Let us assume that we are having rock base at the bottom and upper layer is you know the sand layer then all the way water flows for the entire thickness so it takes longer time then the h t is equal to h so that means that here the water flows for you know takes long time to reach to the boundary and then it will take longer time if you are having a CV which is actually very low value considering let us say we are having a low permeable soil. So from the above the general solution can be obtained and this general solution is given by u the excess pore water pressure is n is equal to 1 to infinity a n sin n pi z by 2 h exponential minus n square pi square T v by 4. So T v the term what we are seeing is called the time factor and which is nothing but T c v by h square T is the time required for certain degree of consolidation and C v is the coefficient of consolidation and h is equal to h dr square h dr is the drainage path length. So to satisfy the first boundary condition we must have the coefficient a n such that u i is equal to n is equal to 1 to infinity a n sin n pi z by 2 h this we have put as 5. Now equation 4 is a Fourier sin series and a n can be given by a n is equal to 1 by h 2 0 to 2 h u i sin n pi z by 2 h dz combining equations previous slide 4 and 6 then we get an expression which is u is equal to n is equal to 1 to infinity 1 by h integral of 0 to 2 h limits u i that is initial excess pore water pressure sin n pi z by 2 h dz bracket close into sin n pi z by 2 h exponential of minus n square pi square T v by 4. So the T v is the non-dimensional time factor and is equal to C v T by h square that is what we have discussed. So so far the no assumptions have been made regarding the variation of u i with the depth of the clay layer but there are you know the number of variations which are actually possible for u i with the depth it can be the simplest thing which has been assumed by Terzaghi is that the constant variation of u i with the depth that is mostly used in all the solutions that means that if u i if delta sigma load is applied then is assumed that the excess water pressure is assumed to occur uniformly throughout its depth but there are also some assumptions like u i assumed to increase with like a parabolic shape or it can be like a 0 at the top and you know at the center it will be it is approximated like a triangle so the different variations are there but however we are actually discussing on the uniform variation of u a with the depth. So this is actually shown in the slide where u i is equal to u naught which is actually for a doubly drainage layer which is actually shown here h t is equal to 2 h h t is equal to 2 h so this is for the doubly drainage layer so this is you know what the initial excess water pressure which has actually increased upon loading. So if u i is constant with depth if that is u i is equal to u is 0 then we can write 1 by h to the integral of 0 to 2 h is equal to into u i sin 2 pi z by 2 h into dz is equal to 2 u naught by n pi into 1 minus cos n phi is equal to u naught. So this will become like u is equal to n to 1 to infinity 2 u naught by n pi by 1 minus cos n phi and sin n pi z by 2 h in exponential of minus n square pi square t v by 4. Here note that the term 1 minus cos n pi is in the above equation is 0 for cases when n is even when n is equal to 2 4 6 it is 0. So therefore u is also 0 for the non-zero terms it is convenient to substitute n is equal to 2 m plus 1 where m is an integer m is equal to 0 1 2 3 it will take so n is equal to 2 m plus 1 which what we have assumed. So accordingly now what we write is that you know by putting in terms of m then u is equal to m is equal to 0 to infinity 2 u naught by 2 m plus 1 by pi into 1 minus cos 2 m plus 1 into pi sin 2 m 1 plus pi into pi z by 2 h into exponential of minus 2 m 1. So we have substituted n is equal to 2 m plus 1 so minus 2 m plus 1 whole square pi square t v by 4. So this we can write and simplify like this with m is equal to 2 m 1 into pi by 2 so we can write u is equal to m ranges from 0 to infinity 2 u naught by m sin m z by h exponential of minus m square t v where m is equal to 2 m 1 into pi by 2. So at a given time the degree of consolidation at any depth z in z within the clay layer that is at z is equal to 0 at the top and z is equal to 2 h at the bottom of the clay layer if h t is equal to 2 h then in that case degree of consolidation at any depth is defined as excess pore water pressure dissipated to the initial excess pore water pressure. So initial excess pore water pressure is nothing but u i that is the excess pore water pressure dissipated divided by initial excess pore water pressure so which is nothing but u i minus u by u i so we can actually write like 1 minus u by u i which is nothing but delta sigma dash by u i which is nothing but delta sigma dash by u naught. So this is delta sigma dash is nothing but the net you know the stress which is actually the increase in the incremental area which applied and u naught is the initial excess pore water pressure. So by simplifying this one u z is equal to you know when you write in terms of u then u is equal to u i into 1 minus z so from u z is equal to 1 minus u by u i so if you look into this here by knowing u i u i is nothing but the initial excess pore water pressure that is let us say if you are assuming delta sigma is increase in the effective state at a depth z due to consolidation then this u i is equal to that much particular stress. So u i is equal to let us say if a delta sigma of 100 glow Pascal is applied and if the constant variation is assumed then u i is equal to 100 glow Pascal then u z is the degree of consolidation at you know that particular in depth within the clique. So if you look into this with this now where if you take at any time t slightly greater than t is equal to 0 then you know you can see this u z is equal to 1 because you know the moment at the boundaries the moment consolidation commences and the boundaries are the one where respond to the where the soil actually transfers the you know the water transfers the you know soil grains at boundaries. So that the pore water pressure will get dissipated like 100 percent and the increase in effective stress will be equal to delta sigma dash. So at any given time the degree of consolidation is given by u z excess pore water pressure dissipated to the initial excess pore water pressure and u is equal to u i into 1 minus u z. Now in most cases if we however we need to obtain the average degree of consolidation for the entire layer and so generally though we have actually defined at u z but most cases what we do is that we calculate the average degree of consolidation for the entire clay layer thickness. So in most cases we need to obtain the average degree of consolidation for the entire layer and this is given by u av that is the average degree of consolidation is 1 by h t because h t is the thickness of the clay layer 0 to h t u i d z minus 1 by h t 0 to h i u d z divided by 1 by h t 0 to h t u i d z. So the solution of this is obtained as you know this system of curves which is shown here variation of u z with z by h and t v is actually shown here. The variation of u z with depth for various values of non dimensional time factors so this t is nothing but t v, t v is 0.05 and this is t v is equal to 1 here which is the degree of consolidation will be 1 that is t v is equal to here which is per degree of consolidation is equal to 100%. So the variation of u z with depth for various values of non dimensional factors is shown here and these curves are also called as isochrones and the diagrams which represents the successive stages of the process of consolidation by means of isochrones is also briefly called as piezographs. The diagrams which represent the successive stages of process of consolidation by means of isochrones will briefly be called as piezograph. That means that if we are actually having let us say you know the so called you know at t is equal to 0 that initial excess pore water pressure rises to from the hydrostatic pressure the excess pore water pressure increases to say u0 plus delta u and then it tries to fall in such a way that at the center of the clay layer the pore water pressure will remain to be dissipated and then at the top and bottom boundaries if it is open layers then you know it dissipates rapidly. So the system of the migration of these curves which is the system of successive stages of process of consolidation by means of isochrones is actually called as piezographs. So let us consider an example based on the chart we have discussed and in this consider the case of an initial excess hydrostatic pore water pressure that is constant with the depth that is what we have assumed that ui is equal to u0 and for tv is equal to 0.3 determine the degree of consolidation at a depth z by 3 measured from the top of the clay layer that is z is equal to h by 3. So we have derived you know the equation for the excess pore water pressure as in terms of u is equal to m0 to m infinity to u0 by m capital M sin mz by h exponential of minus m square tv where m is equal to 2m1 into pi by 2. So here z is equal to h by 3 or z by h is equal to 1 by 3 and m is equal to 2m plus 1 by 2. So m is equal to 2m plus 1 into 5 by 2 here 5 by 2. So here what we do is that we have given the solution in terms of the in terms of a table where z is equal to h by 3 and z by h is equal to 1 by 3 m is equal to 2m1 2m plus 1 into pi by 2. So here what we do is that for different values of m for example if you see m is equal to 0 then m is equal to 0 means it is like 3 pi by 2 that m is equal to 0 here that means that this will become m is capital M capital M will become pi by m is equal to 1 here that 3 plus 1 3 pi by 2 that is correct m is equal to 0 it is pi by 2 and m is equal to 3 it is m is equal to 2 it is 5 pi by 2. So the capital M value is pi by 2 3 pi by 2 and 5 pi by 2. So what we have given is that we have given 0, 1, 2, 3 like that this we have to see whether where it will be we can actually stop this iteration and z by h is 1 by 3 that is constant, T v what we wanted is 0.3 and m z by h after having got this one z by h is equal to 1 by 3 so m z by h is pi by 6 pi by 2 pi by 6 then 2 by m 1.273 1.424 and exponential of minus m square T v we have got this sin m z by h that is also obtained here then the equation which is gives that 0.3036 0.0005 triple 05 and almost equal to 0. So when you submit these things you do this and this summation it comes to 0.3041. So that means that the value of 0.3041 calculated in step 9 above is the degree of consolidation at depth z by 3. So that means that u h by 3 is equal to 1 minus 0.3041 that is 69.59% of consolidation is already occurred that means that 69.59% of consolidation is already occurred. So u h by 3 is equal to 1 minus 0.3041 which is nothing but 0.6951 which is nothing but 69.59%. So here what we are actually doing is that we actually have calculated now this we can actually also see from the chart here so this is z is equal to 0 and z by h is equal to 1. So z by h is equal to 1 by 3 so somewhere here we can actually pick up for T v is equal to 0.3 so 0.2 and 0.3 so we can see that the value of this u z value will be about 0.69 here between this is for 0.3 and this one it is about 0.69. So the u is equal to initial excess pore water pressure the excess pore water pressure which is at to be dissipated is equal to u i into 1 minus u z. So u z is the degree of consolidation so now if you look into this suppose let us say if 0.31 times u i is at to be dissipated that means that 60% of the pore water pressure already got dissipated excess pore water got dissipated and it is transferred to the soil grains and 31% is at to be dissipated at that particular depth z by h is equal to 1 by 3. So this how we use this charts for determining this the consolidation at any depth but for we actually also have you know the average degree of consolidation generally we take but this is an example with determining the consolidation you know this for time factor of T v that means that after certain time of application of certain initial u i then you know we are calculating that as time factor T v for that we have determined what is the degree of consolidation at a particular depth z is equal to h by 3 by using the equation with what we derived and also tells that if m greater than 3 is actually you know this almost like the degree of consolidation which is obtained is negligible hence you know we can actually stop at this particular you know iteration with m is equal to 2. Now the Terzaghi's one dimensional consolidation as we are talking about the average degree of consolidation we let us define that the average degree of consolidation is also the ratio of the consolidation settlement at any time to the maximum consolidation settlement. Note that in this case if h t is equal to 2 h u i is equal to u naught so combining the following equations u is equal to m to 0 summation 2 u naught by m sin m z exponential minus m square T v and u a v where we have just now defined that 1 by h t 0 to h t u i d z minus 1 by h t 0 to h t u d z by 1 by h t u 0 to h t u i d z. So combining these two equation and then completing the integration we will get the expression for average degree of consolidation as follows u a v is equal to 1 minus m 0 m ranges from 0 to infinity summation 2 by m square exponential of minus m square T v where m is equal to 2 m plus 1 small m into pi by 2 where T v is the time factor which is nothing but T c v by h square. So Terzaghi suggested the following equation for u a v to the approximate values from the u a v is equal to 1 minus m ranges summation of m ranges from m 0 to infinity 2 by m square exponential of minus m square T v. So for u a v is equal to 0 to 53 percent that is actually 52.6 percent it is rounded to 53 percent in some references also say that 60 percent but let us say according to Terzaghi it is u a v is equal to 0 to 53 percent T v is equal to 5 by 4 u a v by 100 whole square. And then for u a v greater than 53 or equal to 53 200 percent that T a v is given by 1 minus 1.781 minus 0.933 within square brackets log 100 minus u a v percent is the parenthesis closed the square brackets closed. So the u a v expression is given here which is T v is equal to 1.781 minus 0.933 so there is a reason for giving this. So if you plot the logarithmic of this T v the time factor with u a v then we will get a curve which is shown like this. The curve represents the curve goes like this initially the initial part of the curve looks like parabola and then you know it beyond 90 percent 95 percent it will try to you know take the shape of the it will get asymptote with the horizontal. So you can see that this is asymptote here which is shown here. So this indicates that you know the consolidation will never complete this infinitely long consolidation and up to 53 percent or so if this is the initial portion of the parabola and when you extend the tangent it goes and hits at the negative you know the T v and this is the this resembles the equation of a parabola that is a reason why u by 100 u percentage by 100 is equal to root over 4 C v by pi h square root T so this is the equation of parabola and this is the initial part of you can say that primary for the consolidation up to 60 percent this is actually valid and for this graph or this condition is actually for two you know conditions one is that open layer at top and bottom and initial excess pore water pressure is constant with the depth. So this particular graph will actually you know beyond that it actually it straightens and beyond 90 percent you can see that this gets asymptotic. So the explanation for this is given like this by for values of u greater than 52.6 which is rounded as 53 percent of the log T v versus u curve is almost identical with a curve with the equation that is for greater than that so that is the reason why this curve was suggested by Terzaghi this particular equation suggested where for values greater than 52.6 the curve the curve is actually almost identical with the curve with the equation T v is equal to 1.781 minus 0.933 log 100 minus u a v so it should also be noted that the radius of curvature of the curve u percentage versus log T v increases steadily until you becomes approximately equal to 50 then decreases once and assumes a second minimum at u is equal to 85. So the curve thus obtained has a point of inflection at about u is equal to 75 so the point of inflection is about at that particular point where you have got one curvature here and you have got another curvature here that is about 75 percent somewhere here. And in the vicinity of u is equal to 95 it flattens rapidly that is what we are saying and approaches a horizontal asymptote corresponding to u 100 percent. So now overall the curve represents an equation we can say that log of base 10 T v log T v is equal to function of u percentage log T v base 10 is function of u percentage. So the substituting T v is equal to T c v by head square so we can write log T v is equal to log of base 10 T c v by head square when you apply the log for both sides we can write log T to the base 10 plus log c v by head square. So we can say that you know as c v and the thickness are constant so that is constant so we have log T v to the base 10 is equal to log T that is the time required for consolidation is equal to function of u that is the degree of consolidation. So if the degree of consolidation of the two beds of the clay with the different values of c v by head square is plotted again is the logarithm of time that time consolidation curves thus obtained have the same shape but they are separated from each other by horizontal distance which is nothing but log to the base 10 c v by head square. So this equation implies whatever we have written like log to the log T v to the base 10 is equal to log T that is time required for consolidation to the base 10 plus constant is equal to function of u percentage it implies that if the degree of consolidation of the two clay beds with the different values of c v by head square is plotted again is the logarithm of time that time consolidation curves thus obtained have the same shape but they are separated for each other by horizontal distance which is nothing but log of the base 10 c v by head square for T v by T is equal to 1 the time that is the T v by T is equal to 1 the time consolidation curve becomes identical with the time factor consolidation curve. So that ends the because that is the reason why we use this because you can see that T v by T for T v by T is equal to 1 the time consolidation curve becomes identical with the time factor consolidation curve. So this important property of semi logarithmic time consolidation graph facilitates the comparison of empirical consolidation curves with theoretical standard curves for the purpose of detecting deviations of real process from the theoretical ones. So therefore in many cases the semi logarithmic plot is comparable to the arithmetic plot. So the important property of this semi logarithmic time consolidation curve facilitates the comparison of the empirical consolidation curves with a theoretical standard curve for the purpose of detecting the deviations of the real process from the theoretical point. So let us take an example again where we have got a soil strata which is gamma 19.5 kilo per meter cube that is the unit weight here the water table is here and this is the proposed fill and this is the soil strata this is the ground surface on this the soil is to be proposed to be reclaimed for a weight of 5 meters. Now we will consider before fill and short term after fill long term after fill. So we wanted to calculate what is the effective stress at point A. So here this is the silty sand layer and this is water table that is 1 meter below the ground surface and once the fill is placed then it will become like 6 meter but then this will become the new ground surface and below the fill that is the gamma which is given that is saturated 19 kilo meter cube and 16.5 kilo meter cube is the unit weight of the clay and the depth at below point A is located is 4.8 meter vertically the clay layer is 12 meter and so before placement of the fill the total stress can be calculated we know that that is 18.7 that is before placement of fill so this will not be there 18.7 into 1 plus 19 into 2 that is this thickness plus 16.5 into 4.8 so we are actually interested in at this point so there is 136 kilo Pascal and the pore water pressure at point A by taking U gamma is equal to 9.81 kilo meter cube 9.81 that is 2 plus 4.8 67 kilo Pascal. So the initial effective stress is 69 kilo Pascal shortly after placing the fill now sigma A becomes now 136 plus 19.5 into 5 that is 234 kilo Pascal. Now the you know the delta sigma is increased by about you know 100 kilo Pascal then UA is equal to UA naught plus delta U so the UA is which is nothing but that is 67 that is initial hydrostatic pressure plus that shortly after placing the fill entire the pore water pressure is the drainage actually is not started so 19.5 into 5 so pore water pressure is 165 kilo Pascal and as the drainage is not started the effective stress will remain constant so because of that sigma dash A is equal to sigma A minus U that is 69 kilo Pascal. Now long term after the fill assuming that the clay undergoes consolidation and the complete consolidation is occurred then what will happen is that now the total stress is 234 kilo Pascal and the pore water pressure drops down to 67 kilo Pascal this was provided there is no change in the ground water table levels this UA is equal to UA naught is equal to 67 kilo Pascal. Now sigma dash A is nothing but now sigma A minus UA now we can see that now this particular effective stress at this point after a certain after elapsing the time required for the completion of consolidation the soil grain the strength by you know the increasing the enhancement of the effective stress by the amount which is nothing but 136 to the 69 into 167 kilo Pascal. So this is about the effective stress increased by almost 2 times. So this is actually given this solution this is the solution but is actually given in a graphical form where the construction period though instantaneous loading is assumed but the fields are actually constructed you know over a period of time. So will be also you know in this lecture the ramp loading will be considered. So this is you know the so called the construction period and this is the delta sigma which is nothing but you know this you know is maintained constant. So when we place this one and there is an initial excess pore water pressure and then the consolidation occurs that means that the pore water pressure dissipates. Now what we have done is that we actually calculated the effective stress gradually as the consolidate as the excess pore water pressure dissipates the effective stress keeps on increasing. So this is actually whatever before and after and shortly after the fill that is at this point and then after a long term after the fill so this is actually at this points. So this indicates this actually shows this example in a pictorial form. Now take another example wherein we have a soil strata which is shown here and we have the sand layer at the top so this is the sand layer and this is the sand layer at the bottom then sand layer. So this is called double open layers that is the two layer system where the clay is sandwiched between two layers and water table is 1.5 meter below the ground level and clay is of having 12 meter thickness so 0 to 12 meters is the clay thickness this is the top of the clay this is bottom of the clay and 6 meters is the middle of the clay layer. And in this actually proposed fill is about delta sigma is about 100 glow Pascal's kilo meter square and the E naught and initial water content the properties are given but important we will what we use in this is that CV coefficient of consolidation is 8 into 10 to the power of minus 8 meter square per second. So here the pore water pressure in clay after 5 years is actually can be given like this the pore water pressure that is for clay after 5 years you know so if you look into this so this is the initial excess pore water pressure wherein what we can say is that you know this is the water table so at 0 so 3 plus 2 150 kilo Pascal's will be there so this is the initial excess pore water pressure this is provided if there is no you know the disturbance of the soil on this particular area then the hydrostatic pore water pressure conditions are prevalent but you know this clay layer is subjected to certain type of fill and randomly and then there can be you know the pore water pressure can be more than the hydrostatic pressures but assuming that if there no such activities then we actually have got this variation then at time t is equal to 0 once we apply at you know the pore water pressure increases to 130 kilo Pascal's here and here it is because ui is equal to delta sigma is equal to 100 kilo Pascal's so it increases to 250 kilo Pascal's or kilo Newton per meter square. So this is the first isochron at t is equal to 0 so what will happen is that the clay the undergoes consolidation over a period of time so we wanted to know what will be the pore water pressures in clay after 5 years after so moment the you know once the you know once the load is applied and if no drainage is actually taking place then the pore water pressure at the upper region is 0 to 30 and then increased to 130 and then 250 it actually this is the this is at you know immediately after placing the fill so this is immediately after placing the fill so now consider this particular portion only so this is will be the final isochron and this will be the first isochron this is the first isochron and this will be the final isochron which can happen once 100% consolidation which is not possible and if that happens then that is you know going to achieve. Now using tv is equal to tcv by hdr square and for t is equal to 5 years we can write cv in terms of meter square per year when you convert then so this is 3.1536 into 10 to power 7 meter square per year and hdr is equal to 2h by 2 that is 12 by 2 that is double drainage and tv can be obtained as 0.35 now what we can actually calculate is that at z is equal to 0 meters 3 meters 6 meters 9 meters 12 meters means at these depths that is 0, 3, 6, 9 and 12 what is the you know the degree of consolidation so by not assuming as you know the uz we will try to find out so this can be obtained from the jar which we have given but uz is equal to 100% uz is equal to 61 and uz is equal to 46% uz is equal to 61 so this is at 3 meters this is at 9 meters and these are the top and bottom so you can see that this 100% consolidation has already occurred. So these are obtained like this here for this is the top of the clay z is equal to 0 meters for the top of the clay and z is equal to ht is equal to 2h meters below the for the top for the bottom of the clay from the this is the bottom of the clay. So we can actually find out here at z by h is equal to 1 let us say z by h is equal to 1 for tv is equal to we have got as 0.35 so for that we can actually obtain uz is equal to 0.46 so for the z by h is equal to 1 which is nothing but here uz is equal to 0.46 this is how we are obtained. So by using u is equal to ui into 1 minus uz at z is equal to 0 meters u is equal to 0 kilo Pascal and 3 meters 6 meters 9 meters and 12 meters if you look into it so what we can actually do is that this 39 kilo Pascal's 54 kilo Pascal's and you know the 0 39 54 39 0 this is the pore water pressure which is at to be dissipated that means that what we have found out is that this excess pore water pressure which is at to be dissipated is actually is obtained and this is actually after 5 years so we actually calculate the time factor based on the tea after applying the surcharge of 100 kilo Pascal's what will be the pore water pressure in the soil. So this exercise actually has helped us to plot this isochrone after 5 years this is the excess pore water pressure at to be dissipated so you can see that this is at z by h is equal to 1 that is at this point. So this is at 3 meters and this is at 6 meters 9 meters and this is at 12 meters and this is at top. So this system of the migration of the isochrone actually happens here and this portion which is actually ash with light blue lines this actually shows that portion is already transferred to the soil grains that means that the water pressure the excess pore water pressure already transferred to soil grains and the portion within this zone is at to be dissipated. So this system of you know these isochrones is actually called as piezographs and so we can actually draw innumerable number of piezocrones as we actually traverse from 0 years to you know let us say that second year can be here third year can be here and maybe after 10 years it can be here. So in this example what we have done is that we have tried to calculate what is the excess pore water pressure after let us say 5 years of time period so by what we have done is that we actually have used this the solution of the differential equation of the one dimensional consolidation theory and then we tried to use that and then try to find you know this you know the pore water excess pore water pressure to be dissipated after certain period of time. Now consider now as has been told here that the loading which is you know we assume that suppose if you are having a clay layer of certain thickness ht is equal to 2h and if you are having a double layer at both top and bottom that is porous layer and then ground water devil as it is this point here and if you are actually trying to put the fill is not possible to instantaneously which actually can happen at time t is equal to 0. So there is you know in fact in this particular you know 0 to tc that is tc is the time required for construction so for filling up to height and particularly if the soil is very soft it is not possible to load you know the entire this fill in one go we have to do it in different stages. So in a way what will happen is that it requires certain time period of construction that means that what we can do is that what one need to do is that you have to fill certain height h1 and wait for certain time and again fill for h2 wait for certain time. So that waiting period is called waiting period of you know for you know dissipating the excess pore water pressure which actually has occurred for a given weight. So for a given fill height so once we do that incremental stage wise construction then we can actually achieve that. So the Wilson 1977 presented a mathematical solution for one dimensional consolidation due to a single ramp load but if you are actually assuming you know the instantaneous load then the you know we will be actually looking into that wherein it actually has got a you know the deviation from the you know the instantaneous load conditions. So Wilson 1977 have given a mathematical solution for one dimensional consolidation due to a single ramp load where the stage wise loading has not been considered wherein a single ramp has been constructed and with time TC but when you look into the this can be approximated when you have got multiple stages let us say that the Qc is reached in Qc1, Qc2 and Qc3 and where final Qc3 Qc1 plus Qc2 plus Qc3 is equal to Qc then you know if you take the same slope and all that can be approximated but this is for a single ramp load where TC is the construction time for placing the fill or attaining Qc. So the expression for the excess pore water pressure for the case where ui is equal to u0 is given by that is what we have derived that u is equal to m is equal to 0 to m infinity to u0 by m capital M sin mz by h exponential of minus m square Tv where m is equal to 2 m plus 1 into pi by 2 as stated above the applied load is the function of time and Q is equal to function of time where Ta T suffix a where Ta is the time of application of any load. So for differential load dQ applied at any time Ta the instantaneous pressure increase will be dui is equal to dQ whatever is been applied that will be you know mobilized that is dui is equal to dQ so at time t the remaining excess pore water pressure du at depth z can be given by the expression du is equal to m to the raise 0 to infinity substitute now for u0 to dui du by m sin mz by h exponential of minus m square Cv T minus Ta divided by h square. So Ta is the you know the time of application of any load so now you know by simplifying this we can get m is equal to 0 to infinity to substituting for dui is equal to dQ so 2 dQ by m sin mz by h exponential of minus m square Cv into T minus Ta by h square and this is termed as equation A then the equation B is obtained by calculating the average degree of consolidation which is settlement at time t is equal to settlement at time t is equal to infinity. So UAV is equal to alpha QC minus 1 by HT 0 to HT UDZ by QC where alpha QC is the total load per unit area applied at the time of the analysis and the settlement at time t is equal to infinity is of the of course the ultimate settlement. So t is equal to infinity means the ultimate settlement and settlement at any time t is the average degree of consolidation. So note that the term QC in the denominator is equal to instantaneous excess pore water pressure UA is equal to QC that might have been generated throughout the clay layer had the stress QC has been applied instantaneously. So that is the you know the QC is the alpha QC actually accounts that the partially you know generated you know the factor alpha indicates the partial generation of the applied low excess pore water pressure in the soil. So for the proper integration of equations A and B what we get is that for TV time factor less than or equal to TC then U is nothing but 0 to infinity to QC by M cube TC sin M Z 1 exponential of 1 minus exponential of minus M square TV and UAV can be obtained as TV by TC 1 minus 2 by TV M 0 to infinity to 1 by M to the power of 4 1 minus exponential of minus M square TV and similarly for TV greater than that is time factor greater than the time of construction and which is actually given here. So where TV is equal to TC that is the time required for the construction of a particular field of having intensity QC CV by H square. Now this is actually given in the form of a charts here for plot against the time factor for single ramp loading and this plot is for TV that is time factor and for different TC. So TC is equal to nothing but TC T suffix C CV by H square. So if you know the TV and then by knowing the construction period then you can actually calculate what is the average degree of consolidation. So here you know the construction time actually has been accounted here. So we look into the an example here wherein we can say that based on the one dimensional consolidation test results on a clay the coefficient of consolidation for the pressure range was obtained as 18 to 10 to power of minus 3 meters mm square per second. In the field there is a 2 meter thick layer of same clay with a 2-way drainage. So based on the assumption that a uniform surcharge of 70 kilo Newton per meter square was to be applied instantaneously that total consolidation settlement was calculated as 150 mm. However during the construction the loading was gradual and the resulting surcharge can be approximated as Q in kilo Newton per meter square as 70 by 60 with time t and time t measured in days. So estimate the settlement at time t is equal to 3 days 30 days and 120 days after beginning of construction that means that after beginning of construction. So in this if you see that Tc is given as the time of construction is actually given as 60 days. So based on the one dimensional consolidation test on a clay where here in this case the load is not actually placed instantaneously. If the load would have been placed instantaneously the settlement was actually calculated as 150 mm. But if the for the given properties of the soil which is actually considered but if the load is actually placed over a period of time and that is actually approximated as 70 by 60 into t where t measured in t in dimensions are in days then we can actually what is actually has been asked is that the settlement at 30 and 120 days after beginning of the construction. So here Tc is equal to Cv Tc by h square so where Tc is the time required for the time of construction of a ramp load and now by taking Tc is equal to 60 days which is nothing but 60 into 24 into 60 then it is nothing but you know we actually converted into this time of construction into seconds and h t the thickness the clay is very shallow in thickness h t is equal to 2 meters it is 2 h is a 2 way drainage and h is equal to 1 meter that is 1000 mm. So by substituting we can actually get Tc as 0.0414 and time t as 30 days. So for at time t is equal to 30 days you know what we get is that Tv is obtained as 0.021 0.0207. So time factor for which accounts the construction time of an ramp load is Tc which is for a time of construction of the ramp is 60 days so that has been calculated and time at time t of 30 degrees 30 days and by conventional time factor is actually obtained as 0.0207. Now what we have to do is that from the chart which we have obtained for Tv of 0.0207 and Tc of 0.0414 what is the average degree of consolidation. Once you know the average degree of consolidation then ultimate settlement is what we have defined in this case also is 150 mm that means that up to 30 days only 7.5 mm of settlement takes place as the construction is actually proceeding of way of the construction period only 7.5 mm of settlement takes place. So how it is being calculated for let us say that we will see for 60 days if you look into it for Tv is equal to 0.083 and Tc is equal to 0.014 we can see that we will get you know this UAV is equal to 27% that is actually here. So this chart is for Tv and Tc different curves are here so this is actually valid for this is after Wilson 1997 this is the plot against the time factor of the single ramp loading the ramp loading single ramp loading in the sense that the ramp loading actually has taken place in a single phase. So this is for 120 days here similarly what we have done is that for 120 days we calculated Tv and then for you know we have determined Tc that is for the we actually have got that is from here for 60 days of construction we have got 0.414 then for Tv of 0.083 and Tc of 0.414 we can say that this is the average degree of consolidation about 27% and once you know the average degree of consolidation so that means that after 120 days of construction or the fill it actually has got 40.5 mm of settlement only. So this shows the you know marked difference of you know assumption of instantaneous loading as well as you know the so called you know the ramp loading conditions. So once having looked into that the limitations of one dimensional consolidation can be seen in the derivation of in the derivation of one dimensional equation the permeability of key z and coefficient of volume compressibility are assumed to be constant but as the consolidation progresses void spaces decrease and this results in the decrease of permeability therefore the permeability is not constant. So coefficient of volume compressibility changes with the stress level therefore Cv is not constant. Similarly the flow is assumed to be one dimensional but in reality flow is three dimensional and the application of external load is assumed to be produced excess pore water pressure over the entire soil stratum but in some cases the excess pore water pressure does not develop over the entire clay stratum. So what we have done in this particular lecture is that we continued the theory of Dirge's theory of one dimensional consolidation and we have seen the ramp loading and we deduced the equations and then relevant equations and we also have solved the couple of example problems as an application for what we studied. And then finally we have looked into the limitations of one dimensional consolidation theory wherein we said that the some parameters which are actually assumed like permeability, coefficient of compressibility are assumed to be constant but as the consolidation progress is actually happening void spaces decrease and the result in decrease in permeability therefore the permeability is not constant and similarly the value of the Cv is actually not constant. So further we will actually look into the different aspects of consolidation in subsequent lecture.