 Welcome to advanced geotechnical engineering course, we are going to commence module 3 lecture 1 on compressibility and consolidation. So this is module 3 lecture 1 on compressibility and consolidation. In this module the following contents are outlined. Stresses in soil from surface loads due to different types of surface loads it can be concentrated load or it can be line loads, it can be strip loads or it can be distributed loads over a certain area or irregular shaped areas loaded with certain intensity and embankment loading etc. And then after having looked into the stresses in soil from surface loads we will try to introduce ourselves to Terzaghi's one dimensional consolidation theory and application in different boundary conditions and ramp loading condition that is how the embankment construction on soft soil actually happens and methods for determining coefficient of consolidation normally and over consolidated soils, compression curves and secondary consolidation. After having discussed the one dimensional consolidation then we will try to look into Barron's theory of radial consolidation and settlement of compressible soil layers and methods for accelerating consolidation settlements. So how we can actually accelerate the consolidation settlements we will try to look into some advanced methods. So in this particular module 3 and lecture 1 we actually comments with the stresses in soil from the surface load. We all know that an important function in the study of the soil mechanics is to predict the stresses and strains imposed at a given point in a soil mass due to certain loading conditions. So always the surface is actually subjected to loading. So in that case an important function is the in the study of soil mechanics is to predict the stresses and strains imposed at a given point in a soil mass due to certain loading conditions. Basically this helps to estimate the settlement and to conduct the stability analysis of earth and earth retaining structures as well as to determine the stress conditions on underground and earth retaining structures. So once we know the stresses it helps to estimate the settlements and to conduct stability analysis of earth and earth retaining structures as well as to determine the stress conditions on underground and earth retaining structures. If you look into the idealized stress strain diagram in this particular slide the idealized stress strain diagram is shown here. The stress is on the vertical axis and strain on the x axis and the zone AB is in elastic range and zone BC is idealized as plastic. So you can see that at low stress levels the strain increases linearly with stress and that is the branch AB which is elastic range of the material. Beyond a certain stress level the material reaches a plastic state and the strain increases with no further increase in the stress. So in the idealized stress strain diagram whatever we have shown here the beyond a certain stress level the material reaches a plastic state and the strain increases with no further increase in the stress. So this is idealized stress strain diagram wherein in case at low stress levels the strain increases linearly with stress which is elastic which is within the elastic range of the material. Now let us consider the normal and shear stresses in Cartesian coordinate system. Now elemental soil mass when you look into it assume that we are having a small element having sizes of dx in x direction, dy in y direction and dz in z direction and if sigma xx and sigma yy and sigma zz these are the normal stresses acting on the planes x, y, z axis. So here it is shown sigma xx is the stress acting on y, z plane that is that dy, dz area on this over area the sigma xx acting. So sigma xx into dy dz is the force normal force acting perpendicular to that. Then there are shear stresses acting which is shown here tau xz tau xy and in this case sigma zz is shown here sigma zz is acting on dx and dy area and sigma yy is acting over dx and dz area that is x and z plane. So here we have the for convenience only we have shown only three stresses but in other phases also that is on this phase, on this phase, on this phase on the other remaining phases there will be stresses are acting. So parameter sigma xx, sigma yy and sigma zz are the normal stresses acting on planes normal to the x and y and z axis. So considered positive when they are directed on to the surface. If they are directed away from the surface over which they are acting particularly for normal stresses they are treated as in tension or otherwise the element is in compression. So if tau ij is a shear stress it means that the stress is acting on a plane normal to i axis and its direction is parallel to the j axis. So if you look into let us say tau ij that means that tau ij is a shear stress and it means that the stress is acting on a plane normal to the i axis and its direction is parallel to j axis. So for equilibrium tau xy is equal to tau yx and tau xz is equal to tau zx and tau yz is equal to tau zy. So if you look into this here for equilibrium tau xy is equal to tau yx and tau xz is equal to tau zx and tau yz is equal to tau zy. Now let us look into reducing the equation of static equilibrium. So consider the same stresses which are actually acting on x axis and y axis and z axis and we have got there is an increase in the stress because of the self weight of the element. So we have on the x axis sigma xx which is acting and other phase which is actually having sigma xx plus dou sigma xx by dou x into dx and on the z axis it is sigma zz plus dou sigma z by dou z in dz. In this phase it is sigma z. So the difference of the stress is nothing but dou sigma z by dou z into dz. So this is the rate of the change of the stress which actually undergoes because of the self weight and other reasons. So similarly the shear stresses are also shown here for this reason it is tau xy plus dou tau xy by dou z into dz. On this it is actually shown as tau zx plus tau zx by tau z into dz that is tau zx plus dou tau zx by dou z into dz. So now what we do is that we take you know equilibrium, static equilibrium in x direction for forces acting in x direction and y direction and we try to get the so called static equilibrium equations. So what we did is that now along the x direction that is that whatever the forces acting. So now we have along the x direction on one phase there is one direction sigma xx is acting and other direction it is sigma x is acting, sigma x plus dou sigma x by dou x by into dou x. So here it is referred as sigma x is equal to sigma xx, sigma y is equal to sigma yy, sigma z is equal to sigma zz. So sigma fx is equal to sigma xx minus sigma xx plus dou sigma xx by dou x into dx into dy by dz. So that is the net force acting in the x direction plus the net shear stress acting along that x direction that is tau zx minus tau zx plus dou tau zx by dou z into dz into dx dy because it is acting on dx dy plane plus dou tau yx minus dou tau yx minus tau yx plus dou tau x by dou y into dy into dx dz. So by simplifying this what we get is that as you also we know that tau yx is equal to tau xy, tau xz is equal to tau zx by using that we actually get in when we do not use any self-weight in acting in the x direction we get an equilibrium equation like this dou sigma x by dou x plus dou tau yx by dou y into dou tau zx by dou z is equal to 0. Similarly by applying and simplifying the forces in the y direction we get dou sigma y by dou y plus dou tau xy by dou x plus dou tau zy by dou z is equal to 0. So here we have two equations which we have to be satisfied in the x direction, y direction. If you consider the third direction also we get the another equation but here the static equilibrium in two-dimensional only considered for convenience. So the equations of equilibrium in two-dimensional case is that dou sigma x by dou x plus dou tau yx by dou y plus dou tau zx by dou z is equal to 0 and dou sigma y by dou y plus dou tau xy by dou x plus dou tau zy by dou z is equal to 0. Similarly in the z direction also we have taken and with that what we have got is that dou sigma z by dou z plus dou tau xz by dou x plus dou tau yz by dou y minus sigma is equal to 0. Here what has been done is that the self-weight of the element that is gamma that is the unit weight of the you know element the soil in the element into volume what we have taken is that weight force has been taken. So that is the reason why in only in the z direction it is appearing. For example if you are having some inertial forces in x direction or some body forces like seepage force which is acting then also if you are having gamma x, gamma y and gamma z then we may also get these equilibrium equations with gamma term as the one of the last terms in the static equilibrium equations. So these equations are written in terms of total stresses whatever the now we have discussed the static equilibrium equations they are you know in the in terms of total stresses. Now let us see equations of static equilibrium in terms of effective stresses. Now we know that sigma x is equal to sigma dash x plus u that can be written as sigma dash x plus gamma w h and now by differentiating this we get is that dou sigma x by dou x is equal to dou sigma dash x by dou x plus gamma w dou h by dou x. Similarly dou sigma y by dou y is equal to dou sigma dash by dou y plus gamma w into dou h by dou y dou sigma z by dou z is equal to dou sigma dash z by dou z plus gamma w dou h by dou z. Now what we do is that we know that in terms of total stress now you know we convert that into effective stresses so we can write this equilibrium equations as dou sigma dash x by dou x plus dou tau y x by dou y plus tau z x by dou z plus gamma w dou h by dou x is equal to 0. Similarly in the y axis dou sigma dash y by dou y plus dou tau x y by dou x plus dou tau z by dou z plus gamma w dou h by dou y is equal to 0. Similarly in z axis dou sigma dash z by dou z and plus dou tau x z by dou x plus dou tau y z by dou y plus gamma w dou h by dou z minus gamma dash is equal to 0. Where gamma dash is the submerged unit weight of the soil note that the shear stresses will not be affected by the pore water pressure. Now for convenience here the equations in two dimensional equilibrium are given in soil mechanics the number of problems can be solved by two dimensional problems like returning wall problem or a stiff footing for example they are called plane strain problems and a tunnel, a long tunnel, a road embankment so all those things are examples of plane strain problems where two dimensional analysis can be done. Now in the case of two dimensional with plane strain problems the main equations of static equilibrium required to be satisfied are if they are x and y direction and z direction is perpendicular to the plane of the along the length of the structure then it is dou sigma x by dou x that is y is perpendicular along the length of the y is along the length of the structure then dou sigma x by dou x plus dou tau x z by dou z is equal to 0 that is the z is the depth axis x is the horizontal axis plus dou sigma z by dou z plus dou tau x z by dou x minus gamma is equal to 0. So for weightless medium that is that if you are considering a weightless medium then the equations are reduce it to where that gamma term will get to vanish and then we have the static equilibrium equations as dou sigma x by dou x plus dou tau x z by dou z is equal to 0 plus dou sigma z by dou z plus dou tau x z by dou x is equal to 0. Now let us look into the idealization of the stress strain relationship once again. So in general what we are speaking is that the soils are non-homogeneous and they exhibit anisotropy and have highly non-linear stress strain relationships which are dependent on the stress history and the particular stress path will fall out. So in general the soils are non-homogeneous and exhibit anisotropy and have non-linear stress strain relationship which are dependent on stress history and the particular stress path fall out. So in this particular figure again a typical stress strain relationship is shown here a stress and strain and this is the actual non-linear relationship and this is the idealized relationship that is O y dash and y dash p which is this portion is the linear and this portion is the the plastic state. So linearly elastic behavior is assumed being assumed between O and y dash and that is the assumed to the yield point and followed by unrestricted plastic strain or flow at y dash p. So this is y dash p where unrestricted plastic flow is assumed where no stress increase will be there with an increase in the strain. So this is what actually this particular actual relationship is idealized to a O y dash. So the linear elastic behavior being assumed between O and y dash and then y dash p is idealized as an unrestricted plastic strain. So this is the non-hardening behavior wherein you have got this O y dash and y dash p. If only collapse condition in a practical problem is of interest then the elastic phase can be omitted and rigid plastic model rigid perfectly plastic model can be assumed. So that means that the linear elastic segment is ignored then rightly we have taken O y dash and y dash p. So this is nothing but a non-hardening behavior and rigid perfectly plastic. This is this stress strain relationship is indicated as rigid perfectly plastic that is O y dash and y dash p. So this is only if only the collapse condition in a practical problem is of interest then the elastic phase can be omitted and the rigid perfectly plastic model can be considered. Now in which the plastic strain beyond the yield point necessitate the further stress increase. So if unloading and this is particularly elastic strain hardening and softening plastic models that means that sometimes beyond the yield point there can be hardening or there can be softening. You can see that increase in the stress or decrease in the stress. If unloading and reloading work to take place subsequently subsequent to yielding in the strain hardening model then the stress at the new yield point is y double dash which is greater than y dash. Suppose if unloading and reloading work to take place subsequent to yielding in the strain hardening model that is at stress at new yield point y double dash is greater than y dash. So an increase in the yield stress is a characteristic of strain hardening. An increase in the yield stress is a characteristic of strain hardening. A further idealization is the elastic strain of softening the plastic model is represented by O y dash p dash. So this is softening model elastic strain softening model and this is elastic strain hardening model where the stress increase will happen beyond the yield point and here the plastic strain beyond the yield point is accompanied by a stress increase. The plastic strain beyond the yield point is accompanied by the stress decrease. In this case this model is actually called as the elastic strain softening plastic model. In this case this is actually called as the elastic strain hardening model. So after having seen different you know the elastic perfectly plastic and rigid plastic models you know we try to look into analysis of the stresses in soil from surface loads and further we will actually use this knowledge in when we discuss about the shear strength. So the stresses in soil from surface loads in practice the most widely used solutions are those for the vertical stress at a point below the loaded area on the surface of a soil mass. So whenever the different types of surface loads of different shapes and different because of different structures the stresses are actually transferred to the you know soil. So in fact the most widely used solutions are those for the vertical stress at a point below the loaded area on the surface of a soil mass basically for vertical stress but you can also get as we said in a element when it is subjected to loading we can also get the shear stresses in x direction, x y direction, z direction, x z direction and all in the direction in the planes on which this shear stresses are acting. So the vertical stress increment at a given point below the surface due to foundation loading is insensitive to a relatively wide range of soil characteristics such as a Trojanity, Enesotropy and nonlinearity of the stress strain relationship that is what we just discussed. The vertical stress increment at a given point below the surface due to foundation loading is insensitive to a relatively wide range of soil characteristics such as a Trojanity, Enesotropy and nonlinearity of the stress strain relationship. So accordingly the solutions from linear elastic theory in which the soil is assumed to be homogeneous and isotropic and are sufficiently accurate for use in most cases and the main exceptions are loose sands and soft place particularly where they are over lined by a relatively dense or stiff statham then you know this is some of the exceptions which are followed. However that increments of horizontal stress and of shear stress are relatively insensitive to soil characteristics. So the increments of horizontal stress and shear stress are relatively sensitive to soil characteristics. So mainly we try to determine the increase in the vertical stresses due to these surface loads. So displacement solutions from elastic theory can be used at relatively low stress levels requires a knowledge of the values of Enx modulus E and Poisson's ratio mu for the soil either for undrained conditions or in terms of effective stress. Should be noted that shear modulus G where G is equal to E by 2 into 1 plus mu where G is independent of the drainage conditions assuming that the soil is isotropic. If you assume that the soil is isotropic then G is independent of the drainage conditions and assuming that the soil is isotropic. The stresses in soil from surface loads we are discussing particularly we are now talking about the volumetric strain of an element of linearly elastic material under three principle stresses is given by delta V by V is equal to 1 minus 2 mu by E into delta sigma 1 plus delta sigma 2 plus delta sigma 3. If this expression is applied to soils for the initial part of the stress strain curve then for undrained conditions the delta V by V is equal to 0 and once you put delta V by V is equal to 0 with mu is equal to 0.5 then for undrained conditions E will be equal to 3 times G that is the G is nothing but E by 3 and if consolidation takes place then delta V by V greater than 0 then mu will be less than 0.5 for drained or partially drained conditions for drained or partially drained conditions. If consolidation takes place then delta V by V greater than 0 that means that there is some volume change occurs and the positions ratio will be less than 0.5 and for drained or partially drained conditions. So the stresses within a semi-infinite homogeneous isotropic mass with a linear stress strain relationship due to a point load on the surface were determined by Bosnes in 1885. The stresses due to surface load distributed over a particular area can be obtained by integration from the point load solutions and the stresses at point due to more than one surface load are obtained by superposition. In practice loads are not usually applied directly on the surface but the results of surface loading can be applied conservatively in problems concerning loads at the shallow depth. If we compute we assume this that the loads are applied on the surface but in practice the loads are not applied directly on the surface but the results for the surface loading can be applied conservatively in problems concerning loads at the shallow depth. If you are actually having one or two loads then the superposition principle will be used and if you wanted to get the effect due to stresses over a certain depth the integration principle is used the stresses due to surface loads distributed over a particular area can be obtained by integration from the point load solutions. Now let us consider the stresses due to point load. So point load are concentrated load, vertical load transferred to the soil from an electrical power line. So one of the practical examples is that for the point load is the vertical load transferred to the soil from an electrical power line or electrical pole. So Cq is the concentrated load and we are interested in determining the stress at a depth z and the vertical stress and sigma r the stress in the radial direction sigma t theta in this direction. Now the distribution of vertical stress sigma z with depth is given here which is actually here like a curve which takes the shape like this and it is high at the close to the surface and as you go deeper the vertical stress effect decreases. So variation of sigma z with z on the vertical through the point of application of the load cube and at any depth z1 z2 z3 where z3 greater than z2 greater than z1 if you look into it and this is something like the distribution as you go away from the load the stress decreases. So this is something like a bell shaped curve will come on both sides and as we go down the magnitude of the increase in vertical stress due to the load at the surface keeps on decreasing. Now we will see the theory behind this the variation of the vertical stress due to point load which is given in this. So the stresses at x due to point load q on the surface are as follows and these are actually obtained by taking the equilibrium in the vertical direction let us say for these vertical stress. So sigma z is equal to 3q by 2 pi z square into bracket 1 by 1 plus r by z whole square to the raise pi by 2. Similarly sigma r is given here and the sigma theta that is the stress in the radial direction and the stress then along this direction r z is equal to 3q by 2 pi into r z square into r square plus z square whole to the power of pi by 2 it should be noted that when mu is equal to 0.5 sigma theta is equal to 0 when mu is equal to 0.5 the sigma theta the stress will be 0. Now stress due to point load that the importantly vertical stress we can write it as sigma z is equal to 3q by 2 pi z square into 1 by 1 plus r by z whole square root to the raise pi by 2 and this can be written as sigma z is equal to q by z square into IP where IP is equal to 3 by 2 pi into 1 by 1 plus r by z whole square to the raise pi by 2. So this is the expression for sigma z which is independent of elastic modulus and Poisson's ratio. Now the expression for sigma z is independent of elastic modulus and Poisson's ratio. Now influence factors for the vertical stress due to point load can be given like this and this table actually shows for different values of r by z and IP values, IP or it is also called as I suffix B that is I bosonics and r by z. When r by z is equal to 0 so this is for different r by z values we can look into it this is one set of r by z this is another set of r by z this is another set of r by z. When r by z is equal to 0 IP will be 0.475 it is actually simplified as you can say that 0.478 so you can look into it so this is r by z is equal to 0 to 0.7 r by z is equal to 0.8 to 1 by 8 1.5 and r by z is equal to 1.6 to 2.6. So as we go deeper and deeper you can see that the IP value is decreasing so as r by z increases IP decreases and sigma z tends to be infinity. So influence factors for the vertical stress due to point load that means that close to the load the stresses are close to the infinity sigma z this thing but at a certain depth they also diminish. So vertical stress distribution along a vertical line this can be obtained by sigma z is equal to 3q by 2 pi z square into 1 by 1 plus r by z whole square to the raise 5 by 2. One for maximum sigma z if you differentiate this and equate it to 0 then by simplification we get r by z is equal to 0.817 when r by z is equal to 0.817 then the distribution of the vertical stress follows this type of curve where you have there is an increase and it reaches to the maximum value and that maximum value at this particular depth is sigma z max is equal to 0.0038q and this is the stress maximum stress sigma z max and then the stress again decreases for certain depth. So here you can to see that in this expression as z square is involved in the denominator of the expression for sigma z first it increases and with the depth and attain a maximum value and then decreases further with an increase in the depth. So this is how the vertical stress distribution along a vertical line not at the at its center but at a distance away from the or from this thing that is how the stress variation the distribution of the vertical stress will be like this. Now as we said that we have understood that the stresses they have whenever the surface is actually loaded with certain type of a loading then there is certain influence zone. So this is defined by a stress isobar or a pressure bulb. So the stress contour or a line which connects all points below the ground surface at which the vertical stresses is the same is called as a isobar stress isobar or stress the pressure bulb. So pressure at points inside the bulb are greater than that at a point on the surface of the bulb and the pressure at points outside the bulb are smaller than that value. So any number of stress isobars can be drawn for any applied load. So if innumerable number of stress isobars can be drawn for the applied load and the system of isobars basically indicate the decrease in the stress intensity from the inner to outer ones. So that means that as we are actually coming this is the stress bulb or a pressure bulb resembles like an onion like from the inner side inner layers will have the higher stresses and as we are traversing towards outside the stress intensity keeps on decreasing. So the mostly the stress isobars are not circular curves and they are actually classified and categorized as lemon skate curves the stress isobars are the lemon skate curves. Now let us see how we can actually get the you know plot these isobar say for example let us say that we wanted to plot sigma z is equal to isobar for 0.1 times q per unit area that is 10% isobar we wanted to plot. So now we know that sigma z is equal to sigma z is equal to q into ip by z square. So we write ip is equal to sigma z z square by q by substituting for sigma z is equal to 0.1q we can write 0.1q into z square by q so we get 0.1 z square. So when r is equal to 0 ip is equal to 0.475 we said so the depth of the pressure bulb or depth of the stress isobar can be obtained by z max is equal to root over 0.477 by 0.1 is equal to 2.185 units is the depth. Now we can actually take z and ip and r by z and r and sigma z so with that what we can do is that once you substitute say z is equal to 0.5 then we can calculate what is ip and r by z and r. So by knowing r and z we can actually plot and then the stress intensity along that particular 11 skate curve portion is 0.1q. Similarly the last point that is the z max is 2.15 and at the center where r is equal to 0 ip is 0.475 and r by z is equal to 0 and r is equal to 0 that is also intensity is 0.1q. So like this by using this procedure we can actually determine the stress isobar. So example problem let us consider a single concentrated load of 1000 kilo Newton acts at the ground surface. So we need to construct an isobar for sigma z is equal to 40 kilo Newton per meter square by making use of the Bosnesq solution. So let us consider sigma z is equal to 3q by 2 pi z square into 1 by 1 plus r by z whole square to the raise pi by 2. Now we can simplify this by taking r out r is equal to root over z into square root of 3q by 2 pi z square into sigma z to the raise 2 by 5 minus 1. So by now for q is equal to 1000 kilo Newton sigma z is equal to 40 kilo Pascal's we obtain the different values of r1, r2, r3 for different depths and different depths z1, z2, z3 so the values can be obtained. So similarly by otherwise by using the example which we have done earlier. So by putting different values we can actually calculate and this is how the isobar or influence zone. So this is very much important sometimes if you are having say isolated footing and you know if you are having the load trying to see what is the depth of extent of the influence of the zone. Suppose if you are having some weak zones then these zones can actually undergo the settlements. So this you know the prediction of this the extent of the influence is actually helps here the stress isobar or pressure bulb indicates the zone of influence. So you can see that and this is the Zmax that is 3.455 which is actually indicated here. So this is for 1000 kilo Newton's load this is how the you know the stress isobar for a sigma z is equal to 40 kilo Pascal's is given. Suppose it will be 20 here somewhere here will be there. So and mostly you know the two solutions are actually popular one is you know Bosnian solution and Westergaard solution we also compare in the Westergaard stress distribution under a point load let us say. So Bosnian assumed that the soil is elastic and isotropic and homogeneous but however the soil is neither isotropic nor homogeneous the most common type of soils are met in nature are the water deposited in sedimentary soils that means that we have alternative layers of clay and relatively compressible layers and relatively incompressible layers like sand. So we have like in alluvial soils where clay and sand, clay and sand deposits are there. So the sedimentary soils where we actually have got you know the most common type of soils are the stratified soils. So soils of this type can be assumed as laterally reinforced by numerous closely spaced horizontal sheets of negligible thickness but infinite rigidity which prevent the mass as a whole from undergoing lateral movement of the soil grains. So for this case mu or mu is equal to 0 so here you know what Westergaard is actually assumed is that the soils of this type can be assumed as laterally reinforced by numerous closely spaced horizontal sheets of negligible thickness but infinite rigidity which prevent the mass as a whole from undergoing lateral movement of soil grains. So with that you know the theory if you look into the solution proposed by Westergaard sigma z is given by q by 2 pi z square into square root of 1 minus 2 mu into 2 divided by 2 minus 2 mu divided by within square brackets of 1 minus 2 mu by 2 minus mu plus r by z whole square to the raise 3 by 2 which is indicated by q by z square into Iw, Iw is the influence factor for the Westergaard stress distribution for point load. And for the case where mu is equal to 0 the Westergaard equation for stress distribution for vertical load, vertical stress is reduced to sigma z is equal to q by pi z square into 1 by 1 plus 2 into r by z whole square root to the raise 3 by 2. So that is q by z square into Iw, so Iw is the influence factor as far as the Westergaard is also called as Westergaard stress coefficient and which is Iw is equal to 1 by pi divided by 1 plus 2 or 2 into r by z whole square to the raise 3 by 2. For r by z is equal to 0 Iw is equal to 1 by pi which is nothing but 0.3183, so you can see that this is actually 0.3183. So now here if you look into if you plot this is the I bossniks and which is influence factor of due to bossniks theory and this is the influence factor due to Westergaard theory. So values of Iw or Ib or Iw for use in the bossniks and Westergaard equation and this is the depth axis r by z. So you can see that up to 1 by z is equal to 1 there is a distinct variation is there and the value of Iw which is equal to 0.3183 at r by z is equal to 0 which is 33% less than the bossniks influence factor which is 0.478 or 0.4775. So because of this the geotechnical engineers actually prefer to use the bossniks solution as that this gives the conservative results but this Westergaard theory is also used in determining the stresses particularly for the pavements when we are actually having two layer system and three layer system and this theory is actually extended in determination of the stresses in pavement layers. Now after having considered the different types of theories, two different theories for point loads. Now what we do is that we will try to use whatever the knowledge we gained from the bossniks solution for the point load, we extend to different types of loadings. Now we assume that we have a series of line loads, a line load which is actually giving. So example of the for the line load in the real practical problem is that a long brick wall or a railroad track. If we are having two tracks, two rails then that is actually two for a broad gauge let us say is 1.69 meters separated by two line leads separated by a certain distance. So a long brick wall or a railroad track is an example for the vertical stress due to a line load. So the series of point loads which is actually given, so what we do is that along the length y you take a load intensity q per unit length into dy. So we assume that there are series of point loads which are actually there. So by using the bossniks solution we can get a stress at a depth z in terms of the Cartesian coordinates x along the horizontal axis and depth along the depth axis is z, sigma z is equal to 2q by pi into zq by x square plus z square to the raise 2 and sigma x is nothing but 2q by pi into x square z into x square plus z square to the raise 2 and tau x z is equal to 2q by pi into x z square into x square plus z square to the raise 2. So here by knowing sigma x, so this is you know when we have a line load let us say if you have got a retaining structure. So we can actually calculate at a certain depth what is the increase in the horizontal stress due to a presence of a line load or if you are having a retaining wall and you know if there is say 2 rail tracks are actually going parallel to the length of the road then you know we can actually determine what is the increase in the stress along the length of the wall that is but the sigma x. So sigma z here is obtained from the bossniks solution here what we have done is that we actually have taken q dy as the point load and then integrated over this length between minus infinity to plus infinity then with that we have got sigma z is 2q by pi into zq by x square plus z square to the raise 2. So vertical stress issued line load can be given by simplification like sigma z is equal to q by z 2 divided by pi into 1 plus x by z whole square to the raise 2. So this is again indicated by in terms of sigma z is equal to q il by z, il is the influence factor for the line load and at x by z is equal to 0 that il is equal to 0.6366 into q by z. So sigma z is equal to 0.636q by z where x by z is equal to 0. So sigma x now as I said 2q by pi into x square z into x square plus z square to the raise 2 this can be used to estimate the lateral stress pressure, lateral pressure on the earth retaining structure due to the line load on the surface of the back well. How it can be done let us look into it. So if you assume that there is a retaining structure and in the dimension form here let us assume that it is in m times h and n times h let us say that h is the height of the retaining structure and at a distance from the top of the retaining wall the load intensity is actually acting at m times h away from the that is q per unit length is actually acting. Let us assume that there is a brick wall or a boundary wall is actually existing here then we will see that what is the influence of this boundary wall on the lateral thrust ejected on the wall. So the basic by using the expression which is given for the horizontal stress obtained from the Bosnian solution we can write sigma x is equal to 2q by pi h into m square n divided by m square plus m square whole square. So however the structure will tend to interfere with the lateral strain due to the load q and to obtain the lateral pressure on relatively rigid structure a second layered q must be imagined at an equal distance on the other side of the structure. So for the two you know two line loads the lateral pressure is actually given by px is equal to 4q by pi h. So this is multiplied simply by 2 4q by pi h into m square n by m square plus m square whole square. So we can actually get the total thrust on the structure is given by capital P suffix x 0 to n that is px h into dn which is nothing but 2q by pi into divided by 1 by m square plus 1. So this is the total thrust ejected on the so by knowing m which is the cohesion which is multiplied so many times the height h and by knowing the load intensity per unit length and we can actually calculate what is the lateral thrust at a certain depth. So vertical stress due to the strip load if you look into it here strip load is the load transmitted by a structure of finite width of infinite length of the soil surface. So before looking into this let us try to look into the problem with once again with how to construct a stress isobar for you know if the point load is about 2000 kilo newtons. So let us look into this particular problem wherein we have the example problem for a single concentrated load of q 2000 kilo Newton and construct an isobar for sigma z is equal to 40 kilo Pascal's. So here the solution runs like this sigma zz is equal to sigma z is equal to Ip into q by z square that is what actually we have discussed in this lecture. Now Ip is equal to sigma zz into z square by q which is nothing but Ip is equal to sigma zz into z square by q. So what we have done is that we have rewritten this expression as Ip in terms of Ip is equal to z square into sigma z by q. Now what we do is that we know the sigma z the for which intensity we wanted to determine. So to get that what we do is that Ip is equal to 40 into z square by 2000. So here what we have done is that by putting the intensity magnitude of the point load and by putting the stress intensity for which actually we are interested in drawing the isobar we can actually get the Ip in terms of z. So Ip is equal to z square by 50 so when r is equal to 0 Ip is equal to 0.4775 by using this we can actually calculate what is the depth of the pressure bulb that is z is equal to z max at r is equal to 0. So with that we can actually get z max is equal to square root of 50 into 0.4775 with that we can actually get as 4.89 meters as the depth of the pressure bulb. Then after having obtained this like the procedure which we have discussed what we can do is that we actually take z and you take z is equal to 0.5, 1.5, 2.5, 3.5, 4 and z max is equal to 4.89 and for using Ip is equal to z square by 50 we can actually determine what is the what is the you know values of Ip. So at r is equal to 0 which is 0.4775 and r by z is equal to 0 r by r is equal to 0 the stress is again 40 kilo Pascal's. So by knowing r and z we can actually again plot the you know the by plotting this unit and on the radius axis radial axis and 0.5 meter it will be something like a you know a Levenskate curve portion wherein you can see that r is actually increasing that is 2.1 and again is dropping down. So this curve which actually takes the shape which is actually you know takes the form of a Levenskate curve and the this will be you know stress intensity for a particular Pascal's stress intensity. Suppose if you need isobar for 80 kilo Pascal's it will be inside and if it is 20 kilo Pascal's it will be outside that particular line. Now after having discussed that example problem we continue with the Wettker stress due to a strip load. So here strip load is the load transmitted by a structure of finite width. So here assume that we are having you know we have a boundary wall where it is connected with you know the foundation for that can be a strip foundation or if you are having a closely spaced columns along the length of the building and all the foundations are connected along that length that there is actually forms like a strip load with a finite width here. The width can be 2A, a strip load is the load transmitted by a structure of finite width or infinite length along this thing. So this is again you know the two dimensional you know two dimensional analysis because of the that is called again plane strain structure. So Qs is nothing but the applied uniformly applied stress over this area. So what we do is that you know for the solution for this we use again extension of both this solution and we have in the define that this is the footing which is actually running and the width is 2A and we have the loading density Qs. Now what we do is that along this is the x axis and this is the depth axis and this width is 2A that is that A this side and A this side and A that side. Now what we do is that along this x axis we assume that the Qs which is actually again Qs into dx. So we consider this like a one line load along this length and assume that that line load is actually at a distance x from here and at a depth z and so this is nothing but x and this is the depth z, x square plus z square is equal to r square and cos theta is equal to z by r and x is equal to this distance is nothing but x is equal to z tan theta tan theta. So dx is equal to z secant square theta d theta dx is equal to z by cos square theta d theta. So by using now the Bosnik solution which we obtained for line load what we can write is that small increase in the stress at a depth d sigma z z is equal to by using the line load expression for the you know vertical stress we can write d sigma z is equal to 2 but instead of Q now we write Qs dx that is we have considered that as the line load this is the intensity but it multiplied by a small horizontal distance dx, 2 Qs dx into z cube by pi into x square plus z square to the raise 2. So d sigma z is equal to 2 Qs z cube by pi r to the power of 4 z by cos square theta d theta. Now what we have done is that for dx we substituted z by cos square theta d theta and with that what we have got is that when you simplify further we have got d sigma z is equal to 2 Qs by pi cos square theta d theta with that sigma z is equal to 2 Qs by pi theta 1 to theta 2 cos square theta d theta. So theta 1 theta 2 are nothing but where we have where we have the this is the extent of this load and this is another extent of this load this is the strip load. So sigma z z is equal to Qs by pi once after simplification we can get in terms of theta, theta plus half sin 2 theta that is theta 1 to theta 2. So theta is equal to theta minus theta 1. So what we have got is that if you are having a strip load this is the expression which actually we will get if you are actually having if you are having the so called you know the increase in the stress due to a strip having width which is actually defined by a different geometry having definite breadth of 2A that is the breadth of the foundation is indicated by B is equal to 2A. So in this lecture we have actually tried to understand about the stresses caused in the soil due to some surface loads and this is as a pre you know requisite for the understanding for the compressibility and consolidation theory in soils.