 In the last class I have discussed about what is soil structure or soil foundation interaction and then I have discussed about various mechanical model where the soil is idealized by spring for a and then for two parameter model soil is idealized by spring and the shear layer or spring or by tension membrane or spring and then by beam because in the first soil model that I have discussed is Winkler model where the soils are idealized by spring and those springs are linear spring discrete and there is a lack of continuity in those spring. To remove those limitations then the improved models are suggested where those continuities are removed where then the springs are connected by tension membrane or by shear layer or by beam. So, the next model of that among those improved model is the Kerr model. In the Kerr model, in the last class I have discussed about the Pasternak model where suppose this is the surface and these are the shear layer and these springs are connected by this shear layer. So, this is that shear layer and loads are applied on this layer. So, this is the model which is proposed by Pasternak. So, this is shear layer and this is spring and the expression was the shear modulus of the shear layers is g p and the spring constant of the spring are k s. So, the expression that was available for this case that q x y is k s into W x y minus g p del square W x y. So, these are the model that I have explained in the last class of the Pasternak model. In the Kerr model that suppose if this is x in this direction this is z. So, similar in the Kerr model we will get for this is x direction and perpendicular direction this is z and this normal downward direction this is z and in this direction this is y. Then first soil is idolized by once this springs and then this shear layer is placed in between two springs. Then springs are again placed to idealize the soil. So, that means here this is the shear layer with shear modulus g p this is and here again the load is applied with q x y or concentrated load can be also applied. Now, here we will get. So, this is springs with spring constant with spring constant say k k 1 k 2 and this is the spring constant k 1. So, this is this is produced by the Kerr. So, this is this is the Kerr model. So, these are the two where both the cases shear layer is used by in the first case Pasternak model shear layer is used top of the springs to connect the springs. Here the shear layers are used shear layer is used in between two spring to connect these springs. So, with different spring constant. So, here it is k 1 it is k 2 k 1 and k 2 are the spring constant and g is the shear modulus of the shear layer. Now, in that case the response function will be for the Kerr model 1 plus if I consider this is k 2 by k 1 to P or q is q x y that is equal to g by k 1 del square q x y plus k 2 w x y minus g del square w x and y. So, here this is the spring constant and this is the here another term that here the this is q x y k is w g p del square w here this terms k 1 and k 2 that will be introduced here. So, now if that k 1 is 3 times of k 2 then this expression will be 4 third q x y if I put k 1 is 3 times of k 2 then this will be 4 third q x y minus g 3 k 2 del square q x y that is equal to g 3 k 2 del square q x y that is equal to k 2 w x y minus g del square w x y. So, we will get this type of expressions which is similar to the Pasternak model. So, but here the advantage of this model over this Pasternak model instead of two parameter model here we can use the another parameter or you can use the two different types of soil with modulus of separate reaction k 1 and k 2. So, and then this is the advantage of this model. So, that we can use that additional boundary condition is also available for this model car model. So, that means here we will get two spring constant k 1 k 2. So, we will get another parameter. So, we can represent the soil behavior more correctly. So, now in this case so this is the two other difference model for where the soils are idealized by spring or shear layer or to connect those springs you can use beam or tension membrane. Now in the another limitation of the so that means in the in the last class I have mentioned that there is two limitations in the Winkler model one is the lack of continuity. So, by this improved model we can remove this limitation because here the continuity is provided by using different components. And then the next one is that in the Winkler soil model we that is the linear spring is used, but if the soil behavior is non-linear then how this linear things can be converted to non-linear things. So, that is another limitation that we have to remove there. So, now in the for that part if I consider the non-linear behavior of the soil then you will consider soil or you can consider that the non-linear model. Now first model that will use that for the non-linear part that model that the elastic plastic representation of the soil. So, that is elastic plastic. So the elastic plastic representation of a non-linear response. So, suppose you have the pressure here this is W the settlement and Q is the pressure. So, we have the Q and the W and the soil behavior is this one this is the load settlement curve of the soil. So, now here we will get a elastic and elastic elastic elastic response then we if I draw a tangent here initial part and then another on the final part. So, that is the intersection point and here we will get say the settlement is W 0. So, we will get now this non-linear response of the soil this is the actual response is now converted to a linear thing for the elastic one initial poster then the plastic that is the elastic that is the elastic response. So, now we are converting this non-linear response to a elastic elastic response. So, that means for the initial part the Q relation will be K 1 into W K 1 is the initial modulus of subgrade reaction or the initial spring constant of this initial straight line. So, that means that will be valid for any W which is less than equal to W 0. So, if W is less than equal to W 0 then the response will be Q equal to K 1 into W. Now, the next one this Q will be K 1 into W 0 into W 0 into W 0 into W 0 into W 0 that for W greater than equal to W 0. So, in the next part the Q will be K 1 into W 0 and that W 0 is constant because this is the second part that will be Q equal to K 1 into W 0 if W is greater than equal to W 0. So, this is one model non-linear model. Next non-linear model the binelier representation. In binelier representation suppose we have load settlement curve. So, this is W is the settlement Q is the pressure then we have two lines. So, two straight line this is one linear portion and this is the another linear portion of the curve. So, actually this is the again the non-linear behavior of the soil. So, this is the actual behavior of the soil is this one red one that is the non-linear behavior of the soil. So, those things we are converted to in different form. First one is the lustro-pastic representation and second one is the binelier representation this is the initial part and then the next another linear line. So, if I extend that second line. So, that value so that is the Q 0 and then we will get deflection of the intersection points a W 1. So, we will get stiffness or the K 1 is the spring constant and K 2 is another modulus of sub gate reaction. So, now we will get the response Q equal to K 1 into W that for W less than W 1 and next one Q will be Q 2 into W plus W 0. So, here this value is equal to K 1 into W plus W 0 let us say W 0. So, this is for W greater than equal to W 1. So, we will get this second response K 2 into W plus W 0 or you can put any other value. So, that is the two bilinear representation of a curve. In third model that model which is defined that is a non-linear model that is one thing is proposed by Chandra 1979 where this is soil behavior this is settlement and Q is the pressure. So, we will get a polynomial relationship that we can use Q equal to K 1 W minus K 2 W to the power Q. Now, to determine K 1 and K 2 have to plot Q by W by W minus versus Q square. So, now if I put that is Q by W this will be K 1 minus K 2 W square. So, now if I plot this Q by W versus W square then the slope of this line will be equal will give the K 1 value and the intersection intercept will give this K 1 and K 2 value. So, that means from here will this K 1 and K 2 value. So, suppose we will get this type of curve. So, if I plot W square here and Q by W then we will get this K 1 and K 2 this type of curve. So, that slope and the intercept that will give us. So, this will give us the K 1. So, intercept will give us the K 1 and slope will give us minus K 2. So, this slope will give us minus K 2. So, this slope that will give us minus K 2. So, that means slope and intercept from these two things will get the K 1 and K 2 value. So, this is another representation of the non-linear response of the soil and then and here in the previous response that here we have used this value is Q 0 W 0 and then we use this expression though that part will be. So, this is the two response one illustraplastic representation and is binular representation and then third one is the non-linear representation. So, another non-linear representation representation you can prove that is provide Kondner 1963 where also suppose this is a non-linear behavior of the soil and initial part that slope is a Q 0 K 0. So, K 0 is the initial modular coefficient of modular sub grade reaction or initial spring constant and then suppose this is again Q is the pressure W is the settlement and if I draw this parallel line from this ultimate load point then we will get this point is Q ultimate. So, this value will give Q ultimate. So, Q ultimate is the ultimate Q ultimate is the ultimate load load carrying capacity of soil. Then according to Kondner the presentation is Q K 0 or K S 0 W 1 plus K 0 W Q ultimate. So, this is one. So, this is another represent non-linear representation of the soil behavior. So, these are the few non-linear models that we can use. So, in this Winkler spring also this spring. So, you assume as a linear model. So, using this non-linear model we can remove that limitation also. So, in the next thing that is the another thing that will start that is elastic continuum model. So, in elastic continuum models that is continuous response of the soil medium seems from the work of Bushinik. So, that is first Bushinik proposed this work on this area. So, that year is 1985. So, according to the Bushinik that we analyze the problem is a semi-infinite. So, that is a semi-infinite homogeneous isotropic linear elastic solid. So, where the soil is idealized as semi-infinite homogeneous isotropic linear elastic soil solid which is subjected to concentrated force which act normal to the plane boundary. So, it is subjected to a concentrated load which act normal to the plane boundary. Now, first that we will consider that for the isotropic elastic continuum. So, suppose we have one soil surface concentrated load is applied here. So, this is x direction this is z direction. So, there will be one deformation will be in this form. So, that means here we will get this is u this is w are the two different deformation component. So, in the plane strain problem or in the plane problem a state of the plane strain exist in the x z plane. So, plane problem in the state of plane strain exist the plane strain exist in the x and z plane. So, displacement component v in the y direction is 0. So, displacement component p in y direction is 0. So, first we are going that elastic continuum models the businic first propose that. So, here the first model that we are taking the isotropic elastic continuum where in the plane strain problem that will exist suppose we apply a load p the plane strain problem exist in x z plane and displacement component v in y direction is 0. So, the non-zero stresses that is sigma x x sigma y y sigma z z and tau x y. So, these are the non-zero stresses. So, that the surface deflection deflection deflection w at x 0. So, w x at 0 point in the z direction can be. So, the surface deformation surface displacement in z direction w x that will give 2 p 1 minus mu s square divided by pi e s into log x plus c. So, the surface displacement in z direction in this plane problem that will be 2 p 1 minus mu s square pi e s log x plus c into log x c where c is an arbitrary constant. So, c is an constant. So, c is an arbitrary constant. So, now we will get this expression if it is a plane problem which is a concentrated load applied on the surface. In the next problem that if the u d l is applied on the surface suppose the same thing this is x direction here this is z direction. So, this is so one uniformly distributed load which is applied whose intensity is q and width of this loaded region say 2 b. So, in that in that case we will get another expression of the displacement of the surface of the half plane subjected to uniform distributed load u. In that case w x z will give 2 q 1 minus mu s divided by pi g s 0 to the power infinity e to the power zeta z by zeta square 1 minus mu s divided by pi g s 0 to the power plus zeta z sin zeta b cos zeta x d zeta where q is the stress intensity of the load and g s is e s divided by 2 1 plus mu s that is e s divided by 2 1 plus mu s. Shear modulus and e s is elastic modulus or young modulus mu s is the Poisson ratio. So, this is the expression for the if it is uniformly loaded region. So, the next one is isotropic elastic half plane. So, that is if I go for the isotropic then the displacement if we consider that in the surface of this loaded region. So, that means if the surface applied a concentrated load this is in the direction of r this is the z and another one applied the U D L with 2 a the loaded region this is z this one is r. So, we will get a axi-tometric condition. So, that means here for the isotropic elastic half paste then we will get the surface deflection for the concentrated load p dash and here the intensity in q dash. So, we will get w r into z r z is equal to p dash 4 pi g s r 2 1 minus mu s plus z square z square by r square. So, here similarly g s is the shear modulus and mu s is the Poisson ratio and we will get r square is equal to small r square into z square that is capital R square is small r square plus z square. So, if is the uniform loaded is applied then the deformation will be r z q star into a divided by 2 g s 0 to infinity 1 by eta to 2 1 minus mu s plus tan to z e to the power minus tau z into z 0 to z 1 eta a into d eta. So, and the surface deflection at the center of the loaded region. So, this is at the any point. So, at the center displacement at the center of the loaded region we will get w 0 0 that will be 1 minus mu s q dash a into d eta. So, this is divided by g s. So, now here this j 0 term and j 1 term 0th and first order basal function first kind respectively. So, these are the two different condition one is concentrated load another is U D L for isotropic elastic half space medium. So, in the next one that we will consider that is the third type of condition that we will consider that is the orthographic elastic continuum. So, that is your orthographic so that is orthotropic elastic continuum where we will get the expression that is available that q x z that is equal to q if it is a U D L pi k 1 minus k 2 integration L 1 2 by k 2 minus L 2 2 by k 2 minus L 2 k 2 t 2 L 1 2 k 1 L 2 2 k 1 t 1 2 d z. So, there we will get the deformation for this condition. So, where L 1 2 is mu dash 1 plus mu divided by mu 1 plus mu 1 plus mu 1 plus mu 1 plus E z L 2 2 is equal to 1 by E z 1 minus mu dash square E x by E z. L 1 1 is equal to 1 minus mu square by E x and L 4 4 is equal to 1 by z dash. 1 by z dash. Now, k 1 dash k 1 square is equal to 1 by L 2 2 into L 1 2 plus L 4 4 by 2 plus c to the power half by 2 and k 2 square that is equal to 1 by L 2 2 L 1 2 plus 1 by L 4 4 divided by 2 minus c to the power half by 2 where c is equal to 4 L 1 2 square plus L 4 4 square plus 4 L 1 2 L 4 4 minus 4 L 1 1 L 2 2. So, these are the expression by which we can determine for the orthotropic elastic continuum case what will be the deformation, surface deformation and then here we will get that e is the here we will get the e and somewhere we are using that e dash. So, we will get the e will be the Young's modulus for the plane of isotropic. So, e will be the Young's modulus for the plane of isotropic and e dash will be the again Young's modulus for directions perpendicular perpendicular to the plane of isotropic. Similarly, we will get the expression of g that g is if I consider that is e divided by 2 1 plus mu that is shear modulus for the plane of isotropic. Similarly, g dash will be shear modulus which characterizes the distortion of the angle between isotropic plane and its normal. So, we will get if we get e e dash is the Young's modulus in the plane of isotropic e dash is the Young's modulus in the direction of perpendicular to the plane of isotropic. Shear modulus g is the shear modulus in the plane of isotropic and g dash is the shear modulus which characterizes the distortion of angle between the isotropic and its isotropic plane and its normal. Similarly, we have mu and mu dash. So, similarly the value of mu is Poisson ratio ratio which which characterizes the contraction in the plane of isotropy. Basically, here x y plane when tension is applied same plane and mu dash mu dash will be the Poisson ratio which characterizes the contraction in the plane of isotropy. Tension is applied in the direction perpendicular to this plane. So, we have two mu one is mu and mu dash. So, mu dash is the Poisson ratio which characterizes the contraction in the plane of isotropy when tension is applied on the same plane. Another is Poisson ratio which characterizes the contraction of the plane of isotropy when tension is applied direction to the normal of this plane. So, on the different direction of the tension one is applied to the same plane another direction to the perpendicular to the plane then this will give you the mu and mu dash. So, these are different model that we are talking about. So, next model that we will explain that is the Vlazov model. So, in the Vlazov model so that means in the consider the state of plane strain. So, this is considered state of plane strain in the elastic in the x z plane. Then the displacement components are so u x z that is equal to 0 then w x z that will be w x into h z where h x z or that is describe the variation of displacement w x z in the z direction. So, we will have the h x z function which will show the variation of the displacement y x z in the z direction. Now, several variation of the proposed including the linear and exponential variation. So, several variation is proposed including the linear exponential variation. So, now if we have in the Vlazov model if we have the soil layer suppose it is x and here we have a z direction here then if this is the thickness of the soil layer is h and we have one components if segment if I consider if load is applied over there with intensity q x then this segment is d z then definitely we have tau x z because here the state of plane strain elastic layer in the x z this is the x z plane where the this is the elastic layer we consider and displacement components u x z 0 and w x z is w x and h z where h is a variation of the describes the variation of displacement in the z direction. So, several variations of tau x if we have the tau x here then tau x z plus del tau x z divided by del x to d x. So, we have this variation. So, now this variation of different h v value. So, h z that will be 1 minus x mu nu where nu is equal to z by h this is one variation and another one we can say this is h z that is equal to sin h gamma h minus z divided by l divided by l divided by h z by sin h gamma h divided by l. So, here gamma l r constant. So, now the expression using the principle of virtual work the expression of the q x the final response function that we will get that is k into w x minus 2 t d square w d x square this is w x. For this model we will get. So, here also this is a two parameter model where we will get k is one parameter t is another parameter. So, where k is related to e 0 1 minus e 0 s 0 to h d h d z 2 t square d z and t is equal to e 0 4 1 plus mu 0 0 to h a square z d z where e 0 is e s 1 minus mu s square and t is equal to e 0 4 1 plus mu 0 0 to h a square z d z where e 0 is e s 1 minus mu s square and t is equal to e 0 4 1 plus mu 0 0 to h a square z d z mu 0 is mu s 1 minus mu s. So, it is very important to note that that g p t and k s are directly related to e s and mu s where e s is the e m s modulus of the soil mu s is the Poisson ratio. So, these are the response that we will get for this Blasov model. Now, so this class we have discussed about various other type of model, improved model and then some non-linear model to overcome the limitation of the Inclam model and then how to determine the settlement response at various different condition of the that is also explained in this class. So, next class I will explain about the how the beams behaves if it is resting on elastic foundation. Thank you.