 In last class, I have discussed about that modeling of stone column supported embankment. So, before that modeling, I have discussed about what are the different load sharing mechanism if there is reinforcement is used in the embankment. So, there is basically three types of load sharing mechanism. One is because of the soil arching, one is because of the use of geosynthetic layer, another is due to the stiffness difference between the stone column and the soft soil. Now, today I will explain how those things are modeled by using this soil foundation interaction theories that we have already discussed. So, first if I go that stone column supported embankment. So, that thing that part, see assume this is the embankment, this is the and then below that use a sand pad or granular layer, then this is the existing soil which is improved by stone columns. So, this was the problems, this is the hard stratum and this portion is improved by stone column or granular pile. So, these are equals, these are the diameter of the stone column and this is the spacing between the stone columns and this is embankment. So, these are the stones, this is granular layer and this is soft soil. So, this soft soil is improved by the stone columns. So, this is the problem and suppose this is the thickness of the soft soil, this is the height of the embankment H e and H f is the thickness of the granular field. So, that means H e is the height of embankment, this is height of embankment, then H s is depth of soft soil, H s is depth of soft soil, depth is thickness of granular layer. This is the diameter of stone column is the spacing between the columns. Now, these things we have to model by using this soil interaction problem. So, first this model, this is the existing soft soil. So, this soft soils are modeled by spring dash pot system. This is spring, this is dash pot system. So, this is a spring which is stiffness is k. So, now these things again this is modeled another spring with dash pot system. This is the heart stratum. Now, in between that there is stone column. So, that those columns are modeled by non-linear springs. So, these columns are modeled by non-linear springs. So, this is the non-linear spring which is represented the stone column. So, this one is the spring dash pot system which is represented the soft soil. So, this is the spacing between the stone column and these springs are up to the region of diameter d. So, here also this is also within the d region, d is the diameter of the stone column. This way soft soil and the stone column are modeled. Then this granular layer which is basically modeled by Pasternak shear layer. So, this is used by the Pasternak shear layer. This layer is represented by this granular layer. Now, as thickness of this layer is h f and the shear modulus is g f. g f is the shear modulus of the granular layer. Next one is the embankment. This is the embankment that is also modeled by shear layer. This one is also modeled by Pasternak shear layer. Now, the thickness of this or the depth of this embankment is h e and g e is the shear modulus. So, this shear modulus of the granular material. Now, one thing we have to keep in mind. This is h s. This is h s. So, one thing we have to keep in mind that in case of h f which is uniform throughout the length. So, that means if we consider this is x direction and then the center of this thing is z direction. Then according to the z direction with respect to the j x, this h f is same. But if I consider this is symmetric, this is 0 0 point, then this h e is not same. h e is the maximum depth of the embankment or maximum height of the embankment. So, but it is up to from say center up to l c point from the center. That height of the embankment is same. But after this say this is 1 is to n slope and total width of the embankment is said twice of the b. So, this is total width of the embankment half width is the b and length of the embankment say we can extend this granular layer. So, that will be twice of. So, another spring we can spring dashpot we can consider. So, this will be twice l that twice l is the total width of the model and twice b is the width of the embankment. So, twice b is the total width of the embankment, total 2 l is the total of the model or if it is a. Now, if we consider reinforcement layer here and that we have to place at the interface. So, this is the reinforcement layer if it is reinforced embankment. So, geosynthetic layer. So, that layer is modeled by this layer is modeled by rough elastic membrane. So, that is geosynthetic layer. So, that means, these are the components generally we use. So, first soft soil is used by spring dashpot here. The reason behind this spring and the dashpot here the spring is non-linear spring and dashpot because soft soil we as we are using the stone columns. So, we have to incorporate the time effect or the consolidation effect into the model. So, that is why this time effect to incorporate the time effect we consider the dashpot and the settlement behavior to study that thing we consider the spring. So, that time effect and the settlement both things we can incorporated by using this spring dashpot. And the non-linear springs are used to idealize the stone column and pastoral shear layer is used to model the embankment and also pastoral shear layer is used to model the granular layer or the sand pad which is used. The difference between these things for the granular layer this pastoral shear thickness is uniform to the throughout the length, but in case of embankment the thickness or the depth of this pastoral layer is not same with respect to the distance. And elastic rough elastic rough membrane is used to model the geosynthetic layer. Now, we have to model the different components of this thing. So, there is basically four components one is the embankment is one component, shear layer is another component, soft soil is another component and the springs for the stone column that is another component. So, there are four components basically that and the geosynthetic layer that is also fifth component. So, first component that is embankment which is modeled by pastoral shear layer, second component is granular layer that is also both things are modeled by pastoral shear layer. Then the third component is soft soil that is modeled by spring dashpot, fourth is the stone column that is modeled by the non-linear spring and fifth one is the geosynthetic layer which is modeled by rough elastic membrane. So, these five things we have modeled by using various component or mechanical elements. So, these are the mechanical elements pastoral shear layer spring dashpot non-linear spring then the rough elastic membrane. Now, we will go and derive the expression which is available for the other case. Suppose, we are using the pastoral shear layer. So, we know that for the pastoral shear layer if we consider. So, suppose this is the pastoral shear layer and uniformly distributed load q is applied and this is the pastoral shear layer then spring are attached it is in the general pastoral shear layer model. So, that we are considering here suppose this is z and this one is x. Now, if I draw the x y z this is z this is x and this is y direction and we consider one particular element within the pastoral shear layer. So, this is pastoral shear layer. So, if I consider one particular element. So, this is one particular element of this pastoral shear layer. So, above that we are applying q. So, if I consider this portion. So, above that we are applying q and then we will get another reaction from the bottom. So, that is say for example, that reaction is q 0. So, q is applying here on the pastoral shear layer and then there is getting a reaction q 0. So, basically this q 0 if it is a spring or resting on which q 0 will be k into w where k is the spring constant or the modulus of sub grade reaction and w is the settlement. W is the settlement. So, that is the settlement in the downward direction. So, these are the now if I draw the other part of this shear layer. So, that thickness of this portion is dy. So, and this portion is dx and this is dz. Now, if I consider this cube in a bigger form. So, this is our q g q which is acting from the top and then from the bottom this is q 0 and the thickness of this for the side this is dz. So, this is q 0. So, this is q 0 this is dy and this one is dx. So, volume of this cube is dx dy dz. Now, what are the forces that is acting here? So, in the upward direction for this dz and dy side there is this force that is acting n x. Similarly, in the downward direction of this side that is n x plus del n x divided by del x into dx. Similarly, in the other side of this, this is the force dx and dz side the force that is acting that is the other side that means this ny in the downward direction of this side this is ny plus del ny divided by del y into dy. So, these are the total force which is acting on this shear layer. So, this is the side where this shear layer is acting. So, and considering this the deformation is only in the vertical direction. So, we have not considered in this analysis any lateral deformation although it is assumed that there will be a bulging in the stone column. So, those things are not modeled here. We have modeled only the vertical deformation neglecting the lateral deformation in the model. So, now, if we get the net force from this axis before that we consider the g x shear modulus in the x direction shear modulus in the y direction is same that is equal to g p or g of the pasturnal layer. So, similarly the tau x z and if I consider a two dimensional view of this figure x and z this is a three dimensional view of the model if it is a three dimensional view then for the two dimensional case. So, we will consider this shear layer is like this and the deformation at this point is same as this. Similarly, the deformation in this case will be w plus del w divided by del x into dx and shear stress this point is tau x z. Similarly, this point it will be tau x z plus del x z tau x z divided by del z into dx. So, now, if I this is this will also be del x into dx. So, similarly here also on this top the q will act here also on the bottom q 0 that will act. So, this is the shear layer that we have considered. So, now, if I consider that g x equal to g y equal to g p then tau x z that will be equal to g p shear stress into the strain in x z direction. So, this tau is the shear stress and gamma is the shear strain. So, finally, if I write that expression the tau x z that will be equal to g p and strain in x z direction. So, this is shear stress and this one is shear strain. So, similarly once we then further we can write that g p is equal to del w divided by del x. So, del w by del x is the strain shear strain that is equal to gamma x z. So, similarly tau y z is equal to g p del w del y. So, total shear per unit length of the shear postonaxial layer that is equal to n x in the x direction. So, we can see that this is the for a particular portion. So, this is the stress which is acting every. So, this side this stress is acting. So, n x is the shear force this is shear force and tau is the shear stress shear stress. So, that shear stress is acting this side. So, total shear force if I consider per unit weight that will be n x 0 to 1 unit weight tau x z into d z. So, this is tau is the shear stress which is acting along the side or along this side d x and d y d z and d y. So, on this side these forces are acting. So, we can write this is 0 to 1 tau d z according to the depth. So, that is equal to tau x z into d z z that is 1 0 to 1 unit length. So, z is 1 here consider because this is shear force per unit length this one also shear force per unit length. So, now here tau that will be equal to g p d w d x. Similarly, n y that is also g p into d w del w del y. So, g p also del w del y that is equal to w del y. So, now if I if we consider the free body diagram of this total figure. So, n x we know. So, the net force in the z direction. So, this is acting upward this one downward this is acting upward this one also acting in the upward direction this is q in the top and q 0 from the bottom. So, now if I write the force equilibrium in the z direction then we have n x plus del n x divided by del x into d x into y because this is the in the z direction total force which is acting this is force per unit length. So, now the total force is acting we have to multiply it by the length and this length is d y. So, this is the acting in this surface this is the dimension of n x is shear force per unit length. So, total force you have to multiply it by the length that length is d y. Similarly, that is minus n x into d y then plus n y plus del n y into del y d y here the length is d y. So, length is d x then minus n y into d x. Now, this two force the top one and the bottom one both are acting. So, top one this will be q into this total surface upper surface and that surface area is d x into d y and similarly q 0 d x into d y. So, now if I write that form that is plus q into d x into d y minus q 0 d x into d y that is equal to 0. So, we take all the forces in z directions and then now final form will be if this is n x d y minus n x d y both things will be cancelled out. So, from here this will be del n x del x into d x d y then plus here also n y d x minus n y d x that will cancel. So, plus n del n y divided by del y into d y into d x then plus q d x d y minus q 0 d x d y equal to 0. So, from here d x d y is not equal to 0 that. So, we can write n del n x divided by del x plus del n y divided by del y plus q minus q 0 that is equal to 0. So, this is the final expression and then finally in the further we can write that this is q x and y because in the two dimensionally that is equal to we know q 0 is k into q 0 we have written that is k into w previously. So, that will be k into w x y minus. So, now if I put this n x is equal to that n x that we have written n x n y that is g p and del w del y and n x is del w del x and n y is g p del w del y. So, now n x is g p del w del x. So, here del n x by del x that will be equal to g p del 2 w del x square. So, if I take in this side. So, this will be g p similarly n y is equal to g p del 2 w del y square. So, that is equal to g p del 2 w del y square. So, that is equal to g p del square w into del x square then minus g p del square w to del y square that is equal to 0. So, finally, we will write q x y is equal to k w x y minus if I take g p common then this will be delta square w x y where delta square is del square del x square plus del square del x square plus del square y square. So, this will be the final expression for the pastorenaxial layer in the 2 dimensional form or 3 dimensional form x and y both. Now, if we convert this expression into a 2D form then for the 2D form or plane strain condition then we can write q into x into x into x into equal to k into w x minus g p del square w del x square w in x. So, this is further simplified in the for the plane strain condition this expression and in this model we have considered the loading and the geometry is the plane strain because the as the embankment is in the plane strain condition we generally analyze the embankment in the plane strain condition. But the problem here thus will be associated that when we convert the stone column in the plane strain condition because the stone columns are generally placed in circular in diameter and it is in the triangular or rectangular grids. So, or square grid. So, if we place it in the square grid or the triangular grid and below the embankment then the problem is not actually the plane strain problem. If it is the only embankment or reinforced embankment without stone column then easily we can analyze this thing by plane strain considering it is a plane strain problem. But if it is not in if it is in the stone column supported embankment then we have to convert these things. So, generally we can use the axi-semitic unit cell approach and then but we have to convert these things for the plane strain condition. For example, the general approach for the if it is the stone column is placed in square or triangular pattern. So, generally we consider one unit cell or equivalent area. So, this is the equivalent area or cover area for this stone column. Similarly, we can consider another equivalent area for this stone column and for the another equivalent area for this stone column. Finally, we can consider. So, if this is the spacing is and this one is the diameter of the column. So, and these analysis also that means when you do the analysis this is the we will consider only this one this is stone column and then surrounding equivalent area. So, this equivalent diameter say radius is r and this one is small r. So, r is the radius of stone column and capital R is the radius of equivalent area. So, generally we analyze these things in this form. So, in the if I cut the section xx or rr then this will be the section where this is stone column this is center line this is xx section. So, this is stone column whose radius is r and this radius from the center is capital R small r is the stone column radius. So, then we analyze in this form as a axis symmetric condition, but sometime it is observed that axis symmetric condition in the that is called unit cell approach because this is called unit cell approach. So, now this unit cell approach we will get sometimes we will not get the exact result because the stone column within the center region of the embankment and the stone column the edges of the embankment their behavior are not same. So, if I consider this is the unit cell approach then sometimes we will not get the exact result. So, that is why here in this model all the stone columns are considered into the analysis. So, that means, the total embankment is analyzed here and considering the all the dimension all the stone columns are considered total embankment dimension is considered into the analysis then total things is considered into the analysis. So, that means, but here so, here so, here actually this when the consolidation expression which is available for stone column those are derived for this axis symmetric condition. So, we have to convert this consolidation expression for this plane strain condition. The problem is that when that if we do not convert this thing then what will happen that if I do the plane strain analysis the same problem then then the plane strain analysis will basically consider the diameter of the stone column here and then it is extended. So, this is plane strain model. So, that means, this is the column diameter, but actual case this diameter is only up to this portion, but if I consider the plane strain condition then it will extend in my infinite length with the same diameter d. So, that means, actually we are replacing more amount of the soil. So, to get the actual so, get the actual behavior there is two method of conversion from this axis symmetric condition to plane plane strain condition. One method is that we can keep this dimension same so, we can keep this dimension same and we can keep this radius r is equal equivalent to this equal to this spacing and then we can change the permeability and stiffness of the column material. So, there is no permeability of the soil we can change and if and the stiffness of the column also we can change. So, one method approach to convert this axis symmetric one to the plane strain one this method one keeping so, that in the or you can say this is that keep take same dimension of column and equivalent area. So, that means, that column equivalent area that means, here in the stone column we will not use the diameter we will use the d c that is the width of the stone column and here also we can use the b width of the equivalent zone. So, for the first method that we will consider that our in the first method we will take method one will take that take same diameter as width that means, d is equal to d c where d is the diameter of the column in axis symmetric condition and this is the 2 b which is the width of the stone column. Then we have to change so, because of the this method take same diameter of the width and the also we can take r that is the radius of the influence zone that is also equal to the b capital B that is the width of the equivalent area that is in the plane strain condition So, here in this analysis b is taken as s by 2 where s is the spacing this is b is the half width of the equivalent area similar to the 2 b. So, b is here taken as a s or 2 b is equal to s and then to keeping the dimensions in we have to change the properties of the soil and the stone column and we have to change this is axis symmetric real condition if I consider the properties in the real condition or the axis symmetric properties as the actual properties then we have to change those properties for the plane strain condition. So, that means we have to if I take the same diameter then change permeability soft soil and stiffness of stone column. So, that means we have to change the permeability of the soft soil and the elastic modulus or the stiffness of the stone column to convert the if I take the same width then you have to change the stiffness of the stone column to change it from the axis symmetric condition to plane strain condition. So, one expression which is available for this thing. So, that thing will be so in the next class when I will discuss about how to use this stiffness then you will give those expression how to convert this stiffness of the stone column from axis symmetric to plane strain condition, but today I will just give you the idea how what are the methods. So, if I take the same dimension then you have to change the material properties that means permeability of the soil and stiffness of the stone column to convert axis symmetric one to plane strain condition. In the second method in the method 2 so here we take same material properties that means permeability stiffness same as axis symmetric properties for both plane strain condition, but then we have to change the dimension. So, we have to change the dimension in that case we cannot take that radius of the or the diameter of the stone column in axis symmetric condition is equal to the width of the stone column is plane strain condition that we cannot change. So, that here first case we keep the dimension same and change the material properties in second method we have to keeps the same material properties, but you have to change the dimension. So, here also we have to change so that means in this case we have to reduce the width of the. So, suppose in that case if this is the axis symmetric condition in one section so this is by that thing have to convert here by reducing the width of the stone column here. So, this is for the axis symmetric and this is plane strain. So, that means the because we are keeping thus this is also continue here because we are here reducing the width of the stone column in the plane strain condition. So, these are the two methods generally we applied and the second one and the first one to convert this plane strain stone column condition from plane strain to axis symmetric condition. So, next class I will discuss this how to take this width of the stone column in different condition how for second method how to change this width and for the first method also how to change the stiffness of the stone column because then we can go for the next part or our analysis part. So, now this is the one part and now these things once we are doing in the plane strain condition. So, the final expression that we know that the from the final expression part the general expression we establish that del n x plus del n y del y that is equal to q minus q 0 is equal to 0 this is general for in the 2 D condition this expression will be del n x by del x plus q minus q 0 is equal to 0. Now, here if I condition if I consider this is the embankment and this is the granular bed then the stone columns. So, if I consider this in the above this this is the total stress which is acting above the stone column the q is the load which is acting above the stone column and this point is q and as the base of the stone column this force which is acting is q t also that means that q at the base here this is all the point this is acting q t and on the top it is acting q the force. Similarly, to the granular layer also this will act q t on the top and on the bottom it will act q s. Similarly, in the soft soil it will act q s only. So, this is embankment this is granular layer and this is soft soil. So, in this fashion it will work and so if it is a reinforcement layer is placed in between these 2 here if reinforcement layer is placed here then this q t will act on the top of the reinforcement layer and then the q b will act in the bottom of the reinforcement layer seen in the granular layer q b will act and q s will act in the base of the granular layer then the q s will act in the soft soil. So, if it is a reinforcement case then q b q t this is soft soil this is granular layer and this is reinforcement. So, these are the combination of different loads. So, if it is unreinforced embankment then q top q t at the bottom of the base of the embankment then q t at the top of the granular layer and q s at the base of the embankment and q s on the soft soil. If it is reinforced embankment then q t q b top and bottom of the reinforcement respectively q b on the top of the granular layer and q s in the bottom of the granular layer and q s on the soft top top of the soft soil. So, these are the loading distribution. So, now here if I write this for the embankment then you can write del n x del x plus q minus q t that is equal to 0 because q t is in the lower end. So, that thing is the expression for the embankment. So, now in the today's class I have discussed about this how these things are modeled by using this soil foundation interaction components and then in the next class. So, we have now derived the general expression for the embankment and then next class how to use express these things in terms of deflection and then how to what are the properties how will change the properties or stiffness of the soil to convert accessimetry to the plane strain for the method 1 and what are the width we will consider for the method 2. So, those things I will discuss in the next class. Thank you.