 In the last class, I have discussed about that modeling of stone column supported embankment and then how the different components of that total system can be modeled by using this mechanical element like spring, dashpot, shear layer. Those things has been discussed and today I will discuss further how these things can be developed in equation form and then how those equation can be solved. Now, in the last class, in this embankment, so if this is the embankment, in this one is resting on a granular layer and then this granular layer is sand pad or sand cushion and then this is resting over the stone column improved soft soil. So, these are the stone columns. So, this embankment is modeled by Pasternak shear layer, this is by spring stone column which is non-linear, then soft soil by spring dashpot, then granular layer also by Pasternak shear layer. So, these are the different components which is modeled. So, then if this is center line and this is L c, this one is also L c, this is H e height of the embankment, H f height of the granular field or thickness of the granular field, H s is the depth of the soft soil. S is the spacing between stone column d or d c or equal to 2 b is the diameter or width of the stone column. So, this is a plane strain analysis. So, last class it was derived that for the shear layer d delta n x del x plus q minus q t that was 0. Now, q t means, q t is the diameter of the cylinder in the embankment. So, if at the top is q, then at the bottom reaction is q t. Similarly, for the granular layer on the top it is q t and in the bottom it is q s. Similarly, for the soft soil over stone column, this will act as a q s over this. Stone column reinforced soft soil. Now, these things I mean already discussed that the stress acting on the stone column and state acting on the soft soil are not same. So, now in this expression q t and this total expression we have to write in this form that you know that n x is the total shear force and that is the shear force. We can derive 0 to d to tau x z d x, because last class it is derived that n x that here n x if it is total shear force per unit length, then n x will be 0 to 1 tau x z d x and then d z here also d z. So, because it is in the unit length. So, you will get this is simply tau x z which is nothing but g p into del w del x. But here it is unit length, but here we consider this is in terms of d where d is the depth of the embankment. Now, this d is not constant throughout this length, because d is varying up to l c distance it is uniform h e then it is decreasing 1 is to n component. Now, this d value is 1 is to n. Now, next one so if I get this n x value this will be 0 to d tau z x into x z into d z. So, now here we consider a non-linear relationship of tau x z. So, in a similar to the k s relationship that k s non-linear relationship that was given that q is equal to k s is equal to x z. So, k s 0 w 1 plus k s 0 w into q u. So, that expression is already given for the non-linear form. So, similar expression was suggested by Ghosian models in 1994 to express this expression is originally proposed by Konder 1963. Similar expression was proposed by E 0 Ghosian model d by d x divided by 1 plus G 0 d w by d x divided by tau u E. So, this is proposed by Ghosian model 1994. So, now in this expression if we look at this expression. So, this is similar to the non-linear expression where G e 0 is initial shear modulus and tau u e is the ultimate shear strength of the embankment material and tau u e is the ultimate shear strength of the embankment material. So, now this is the non-linear expression that we are using. So, once we use this expression. So, final expression that we will get that n x that will be equal to 0 to d G e 0 then d w w del x divided by 1 plus G e 0 del w del x tau u e into d z. So, once we derive this expression. So, after the derivation and the integration putting this d value we will get this is equal to now this is here. So, this will be equal to G e 0 divided by del w del x 1 plus G e 0 del w del x divided by tau u e into d because just to integrate 0 to d you will get d z after integration d. Now, again del n x by del x that will give a G e 1 d w d x d dash plus G e 2 into d del 2 w del x square where just to derive this expression we will get in this form where G e 1 is G e 0 1 divided by 1 plus G e 2 into d del 2 w del x square where just to derive this expression we will get in this form where G e 1 is G e 0 1 plus G e 0 del w del x into tau u e and G e 2 is equal to G 0 e 0 1 plus G e 0 del w del x divided by 1 plus G e 0 divided by tau e into whole square. So, we will get in this form where d dash is nothing but d d by d x the rate of change of this d with respect to x. Now, this q at the top we can write that is q 0 plus gamma into d. So, where q 0 is the surcharge gamma is equal to unit weight of the soil and d is equal to d by d x depth of the or height of the embankment. So, once we write this expression then the final form we can write because the final expression is del n x del x plus that is the final form of the expression that plus q minus q t is equal to 0. So, once we put this value so q t value will be equal to del n x del x plus q that is equal to q g plus gamma e d unit weight of the soil means embankment plus G e 1 del w del x d dash plus G e 2 d plus del square w del x square. So, similarly this is the expression where this is the q t this q t is acting over the embankment layer. So, now for the embankment layer we have to derive similar type of expression. So, for the embankment layer if we consider a similar type of expression that for embankment if I consider this shear layer because both the granular layer also used granular layer we have also used this Pasternak shear layer. So, now for the granular layer if we consider a particular strip of thickness h because h f is the thickness of the granular layer from here also the shear force that will act tau f into tau f plus del tau f d x del x because this distance thickness is del x. Similarly, this one will be w and this one w plus del w and it is G f is the shear modulus and here on the top q t will act and in the bottom q s will act. If it is the reinforcement layer then here the reinforcement layer also there will be shear stress and there will be normal stress top and the bottom. So, then this will be q t and this will be q b and if we placed say reinforcement layer on the top q s will be normal stress. Or interface between the granular layer and the embankment then this will come fast then this expression will come later on, but here in place of q t you have to use q b this is q s. So, now in this expression for a small segment if we consider the take this expression then we can write q t del x minus q del x minus q del x minus q del x minus q del x s del x and from here tau f type is cancel out plus del tau f del x and this is del delta x now here this is h f. So, this will be h f. So, this one will be equal to 0. So, now the q t q s del x del x cancel out this q t is equal to q s minus q s minus q s minus del tau f del x to h f. Now, similar to for the delta f similar to for the tau f also if I consider the f 0 del w del x non-linear expression 1 plus g f 0 del w del x tau u del x f where g f 0 is the initial modulus of shear layer or granular layer and tau u f is the ultimate shear strength and this is for the granular layer. So, finally, once we write this expression then after the derivation we will get this expression. Thus del tau f tau x then we have to derive this expression. So, we will get this form that is g f 0 then 1 plus g f 0 del w del x divided by tau u f into square del square w del x square. So, once we derive this expression we will get this form that is del tau f del x we will get in this form. Now, if we put this expression here then we will get that expression that g f 1 is equal to g f 0 del x square. So, once we derive this expression 1 plus g f 0 del w del x tau u f whole square then this del tau f del x that is equal to g f 1 into del square w del x square. So, these are the expressions that we can use. So, finally, so these we can write this is q t equal to q s minus g f 1 h f into del square w del x square. If I put this a value in this expression then this expression is we will write in this form. So, finally, we can write q t value in this way also. So, q t q f and this value. So, final form of the expression is will be that in terms of q t and total force that we can write that gamma e d plus q 0 plus gamma e d plus gamma e d plus gamma e d plus is equal to c w minus g e 1 d dash del w del x plus g e 2 d plus g f 1 h f del square w del x square. So, where because c q s value is equal to k s into w. If it is equal to linear spring now if it is a non-linear spring and if you want to incorporate the consolidation effect then we can write that for consolidation effect we can write that q s will be equal to effective stress and the pore water pressure q s is the total stress that is equal to effective stress and pore water pressure. And this effective stress is taken by the spring itself and then pore water pressure. So, this increase of q s if effective stress is taken by the pore water pressure spring and the spring if I consider it is a non-linear then this q s sigma bar is dash equal to del q 1 plus k s 0 w q u. This q u is the ultimate bearing capacity soft soil and then finally, we can write here that q s that is equal to k s 0 w 1 plus k s 0 w by q u plus u e u which can be written in this form that is u 0 1 plus degree of consolidation. So, q e is the effective pore water pressure at any stage and u 0 is the initial pore water pressure. So, finally, once we put this expression in this form and then initially pore water pressure q 0 that is also equal to q s. So, finally, once once we get this expression the final expression of q s will be k 0 k s 0 w u 1 pi k s 0 w by q u where this u is the degree of consolidation. So, once we write in this expression then if I put this expression in the final form then this expression gamma e d plus q 0 that is equal to c w minus g e 1 d del w del x plus g e 2 d plus g f 1 h f into del square w del x square. So, that we can write that if embankment is this one and this total width is b then this expression is valid up to total is 2 b then if this is the b then this expression is valid in between 0 x and b is greater than 0. So, this is 0 and less than b and c where c is equal to k s 0 u 1 plus k s 0 w q u this is for the soft soil region and in the stone column regions k c 1 plus k c 0 w q u c 1 plus k that is for the stone column region. So, because here we have two separate regions one is within the stone column region and another in between the stone column this soft soil region. So, soft soil region the degree of consolidation will play a very major issue. So, this degree of consolidation we have to use here and this stone column region we will get this k c 0 is the initial spring constant or modulus of stone column material. So, and this is q u c is the ultimate load carrying capacity of the stone column. So, then we will get and the for the within the if it is x is greater than b and so on. So, sorry less than a. So, if x is greater than b and less than l l is the width of the total model from the centre then we will get expression that is because in this expression this total force is 0. So, we will get expression with the q s 0 w u 1 plus k s 0 k c 1 plus k s w by q u s then minus g f h f del square w del x square that is equal to 0 because here this part will not be present this part will also not be present because in this part sorry this one also d dash because in this part we will get because embankment is not present here beyond this point this is b to l this region embankment is not present. So, we will get this expression. So, this expression number one and expression number two these are very major to expression and how to solve these two expressions. Now, first we will express this expression we can express this expression in non-dimensional form and we can express non-dimensional this expression but in terms of k s and the width of the embankment. So, these things this total derivation and this expression how to non-dimensional this part can be explained is explained by Deb in 2010. So, this things can be explained which is can be find this is 2010 this the paper which is now once do we non-dimensional this part now we have to put the d value because d value is varying from place to place and d dash also this dash is the rate of change of the depth and d is the depth of the embankment at any condition. So, now if I put that part if it is l c. So, we can write d is equal to h e height of embankment for x less than l c less than equal to l c and greater than 0. So, this is 0 similarly d will be equal to h e minus x minus l c divided by n for l c less greater than x but x greater less than b less than equal to b. So, that means if x is within this g g n then d equal to h e and d dash is equal to 0 because the rate of change of this depth is 0 because the depth is not changing. But in this region this will be the d and d dash value will be minus 1. So, n by n. So, now we will get d and d dash value for different condition. So, now this expression this k s term is present k s 0 and k c 0 this both two are the initial sub grade modulus of the stone column material and the sub soil material. So, these two things can be we can convert into in terms of e and mu. So, that we can done that is proposed that is given the relationship between k s and e is that k s 0 or sub grade modulus that is equal to 1 plus mu 0 1 minus. So, this expression this expression this k s 0 is e s divided by h s 1 plus mu s 1 minus 2 mu s. So, this is also proposed by Selva Durai where e s and similarly k c 0 is e c h s 1 plus mu c 1 minus 2 mu c. So, that is so now where e c is the elastic modulus of sub soil e c is the elastic modulus of stone column material then mu c mu s is the poisson ratio of sub soil mu c is the poisson ratio of stone material. So, once we now if I express this alpha where alpha is equal to k c 0 by k s 0 then we will get where h s is the depth of sub soil and here we assume the depth of sub soil and the depth of stone column are same depth of sub soil. So, k c 0 k c is alpha we can write in this form that 1 plus mu s 1 minus 2 mu s divided by 1 plus mu c 1 minus 2 mu c into e c by e s. So, this e c by e s is called modular SEO. So, e c by e s is called modular ratio. Generally this range varies from 5 to 100. So, now these are the examples. This that expression which is from this main expression now we have to solve or determine the settlement. So, once we get this expression now we have to solve this expression by using final difference technique and then we can solve this expression and then we can determine the settlement value. So, once we get this expression. So, now then we have to solve this expression in terms of settlement and then we have to use the boundary condition also. So, suppose if this is the embankment, this is granular layer and this is the base and then this is the stone column. So, once we get this stone columns and this granular layer now what will be the boundary condition first. So, once we solve this expression. So, first we will we can derive this expression by using this soil structure or mechanical element. Then we have to non-dimensional this form of expression in terms of by using the k s s 0 and width of the beam. Then once we non-dimensional this expression then we have to solve this expression by using the final difference scheme. So, here we have this d 2 w d x 2. So, that expression we can solve by using the central difference scheme. So, that central difference scheme we can use by w 1 plus i minus 2 w i plus w i minus 1 divided by del x square. So, that means we have to divide these points into n number of. So, this is the one point another point. So, divide this total thing into n number of segments. So, each segment has a thickness of delta x. So, if it is 1 2 3 then this will be i this is i plus 1 this is i minus 1. So, once we get that then we can use this every point the same expression or we can use this equation at every point. Now, one thing is that for this as it is point is symmetric we will analyze this half portion. So, it forms 0 will start from here. So, if we apply this equation if we apply this equation expression in this 0 then we need 1 point i minus 1. So, that is here, but we do not have any node here. For example, if we are applying here this is 1 this is 2 number 2 node this is number 3 node this is number 4 number 5 and so on. So, if this is w 1 and you are using this expression the first node then this will be w i plus 1 this is w i this is w i minus 1. So, 1 will be w i 2 will be w i plus 1 and this distance is uniformly we are taking del x, but we do not have any node here, but we have to consider 1 node. So, this node is called imaginary node or fictitious node. Now, we have to determine this node here and then for that purpose we need a boundary condition. So, that boundary condition is that at this point that we will consider the slope of the settlement that is 0. So, here as a symmetric so at this point the slope will be 0. So, one slope will be 0 that is del w del x is 0 at x equal to 0. Then what will happen then again we have to use the central difference scheme if we use the central difference scheme for this one then we will get w i plus 1 minus w i minus 1 divided by 2 del x. So, that is the central difference scheme for the slope and at this x equal to 0 point it is 0. So, then at point 1 w i plus 1 is equal to w i minus 1. So, that means at the deflection at point 2 w at point 2 and w of this fictitious node 2 dash say 2 dash is same. So, this is this way by using the boundary condition we will get fictitious. But next point we can use these expressions central difference scheme all the point then the problem will arise the same when the last point there also we have if we use this expression. So, there will be w i w i minus 1 and w i plus 1. So, then we need another fictitious point if it is n then we need this is n minus 1 dash. So, this will be n minus 1. So, now we have to use another boundary condition here to solve this expression. So, that boundary condition is we can use that the here we can consider l such that the deflection of this point here also slope will be 0. So, once also where we can use this boundary condition that the slope at the x equal to 0 that is also 0 and slope at x equal to l. So, that means at slope will be 0 at x equal to 0 similarly at x equal to l here also del w del x equal to 0. Now this is one boundary condition this is another boundary condition x equal to 0 del w del x equal to 0. So, if we apply this boundary condition we can solve this we can determine the value of this 2 dash fictitious node or imaginary node if I use this boundary condition we will determine the value of this n minus 1 dash fictitious node. So, now here also the slope is 0 then the settlement of n minus 1 node will be equal to the settlement of n minus 1 node dash. Here also if we apply these things. So, we will get w n minus 1 will be w n minus 1 dash. So, by using this boundary condition. So, once we get this boundary condition then the loading condition is that if I consider there will not be any surcharge. So, q dash will be 0 if no surcharge and then beyond this if x is greater than b then the q value or the loading pattern. So, there will be 0 a q equal to 0. So, that beyond this point this no loading is applied within this point this is the q is gamma into d d is changing depending upon the x value. So, once we determine this total values then what we will get actually from this expression. So, we have some expression this is in terms of q 0 plus gamma d then we have this expression here in terms of d w d x then we have expression in terms of d square w d l x square we have the expression in terms of w also. So, we have the expression in terms of w in terms of del w and then we have the expression in terms of q. So, what will happen or what we will do, so now we will, so here we will express it in terms of w i i plus 1 and w i minus 1, this one also this is w i. So, this one also will express w i plus 1 w i and w i plus i minus 1. So, these things we will express in this form, this thing will express x plus 1 x minus 1 and this is w i. So, now we, this is q force and then we determine take w i then this coefficient, w i plus 1 then this coefficient and w i minus 1 in this coefficient. So, we will get, so once we express this term, so there must be some constant here, some constant here, some constant here, this is plus, plus, maybe plus minus anything. So, now we will express this term and then we separate all the components of w 1 in one bracket, w i plus 1 in another bracket and w i minus 1 in another bracket. So, ultimately these are the constants. So, we will get one matrix in this form, this is a 1 1 then for the first row, you will get a 1, this is a 1 1, this is a 1 1, this is a 1 1, this is