 So, today we will start the first lecture on advanced foundation engineering. So, in this lecture I will be covering different design aspects of various geotechnical structures like shallow foundation, deep foundation, the retaining wall, reinforced retaining wall. Then what would be the soil structure interaction, then how we will design the different reinforced structure in geotechnical aspect. So, these things will be discussed in this lecture and before we discuss about the various design aspects of foundation and the geotechnical structure, foundation of geotechnical structures. Then we will start about the our different modules, what are the different modules that we will discuss. So, first here the course is been divided into 5 modules that is module 1 there will be soil exploration, then the module and in the soil exploration there will be 3 lectures, in then the shallow foundation that is module 2 there will be total 11 lectures, in the deep foundation module 3 there will be total 7 lectures, in the retaining structures and the reinforced earth design. So, that is module 4 there will be total 10 lectures and the soil foundation interaction then that will be module 5 there will be total 8 lectures. So, including the introduction today's lecture there will be total 40 lectures in this course. Now, the books that will be that is the reference books and this will be followed in the lectures those are the Aurora 2003 which is soil mechanics and foundation engineering that is the standard publishers distributors, then Bowels in 1997, then BMDAS 1983, then one is foundation of soil dynamics, another is BMDAS 1999 principles of foundation engineering, then Hettini's books that is Bimson elastic foundation 1979, then Quenus books for the design with geosynthetics that is 1986, then the Kramer books for the geotechnical earthquake engineering part because some part of the earthquake engineering design or design of the retaining wall under earthquake seismic condition that will also be discussed in this course. So, that part the reference book is Kramer 2003, then Ranjoon and Rao the basic and applied soil mechanics 2000, then Salpa Durai that is the elastic analysis of soil foundation interaction 1979 that is that is the reference books for the soil foundation interaction part, then N. N. Schoem and Das 2003 theory and practice of foundation engineering. So, these are the reference books which will be followed in this lectures. Now the acknowledgement is given to the professor N. Siva Kugan associate professor and head engineering college James Cook University Australia and professor B. D. Munna assistant professor department of civil engineering at Delhi and they have also given some lectures note to the which will help me to prepare these lectures. Now, first we start that the design of foundations generally requires a knowledge of factors as the load that is coming on the foundation. So, first is what is the load which is coming on the foundation, the recommends of the local building codes, the behavior of soil that will support the foundation system and geological condition of the soil. So, these things very important when we design a foundation that how much load is coming on the foundation. So, we should know how much load is coming on the foundation, then what are the required design course building course of the area where we are designing the foundation and the behavior of soil that will support the foundation system what with the soil properties. So, we should know the soil properties before we know the we start the foundation design then what is the soil properties what is their behavior because we should know before we start the foundation. So, to know that soil properties we have to go for soil exploration to and the for the laboratory as well as field testing to know these soil properties and then the geological condition of the soil that is also required. Next part is the if we start the geotechnical properties of the soil. So, before we go for the foundation aspect part. So, in the introduction lecture basically I will discuss about some geotechnical properties and soil properties also that will be required to for our design purpose. So, those are the very important properties that thing I will discuss in the introductory lecture and the for the next lecture lecture 2 I will start about the different components of this course of the foundation advanced foundation engineering. So, before we go for that part we should know the geotechnical properties of the soil. So, first is a important properties is the grain size distribution. So, grain size distribution is determined generally by the sieve analysis for the coarse grain soil and for the hydrometer analysis for the fine grain soil. So, this is a typical grain size distribution curve and so, here the x axis is the particle size and y axis is the percent final. So, 50 percent final means the 50 percent particle is 50 percent of this particle size say 50 percent finance means suppose say this is a particle size is 0.8. So, 0.8 millimeter particle is less than the 0.8 amount of the 0.8 less than 0.8 millimeter particle is 50 percent. So, that is the percent finer is x y axis. So, from this graph we can determine the 2 coefficient for coarse grain soil one is uniformity coefficient C e u another is coefficient of curvature C c. So, C e u is basically D 60 by D 10 where is D 60 denotes that the particle size which corresponding to 60 percent particle percent finance. And similarly D 10 is the particle size corresponding to 10 percent finance. So, from this curve we can determine what is the particle size for 60 percent particle percent finance and what is the particle size for 10 percent finance. So, if we know these 2 values from this graph then we will get the uniformity coefficient of the soil. Similarly, the coefficient of curvature is defined by the D 30 square divided by D 60 into D 10. So, from here you also get the 2 coefficient now the purpose of this 2 coefficient. So, for from this 2 coefficient we can identify whether the soil is the well graded soil or soil is the poorly graded soil. So, well graded soil if C e u is greater than 6 for the science and is greater than 4 for the gravels and C c must lie between 1 to 3 then that type of soil is called the well graded soil. And if it is not satisfying this condition then that of soil is called poorly graded soil. And third type of soil is graph graded soil. Now, what is well graded soil? What is poorly graded soil? What is graph graded soil? Now, well graded soil the soil where the different particle size is present in the soil in almost the equal amount. So, that there be a proper distribution of the soil. That means, here this is the proper distribution of the soil where the amount of the different sizes of particle are present in the soil and they are more or less equal amount is present in the soil. Whereas, for the poorly graded soil for a one particular range of the particle size of soil is x square. So, that means, there will be this curve will not be a flat one, it will be a straight one if it is a poorly graded soil because one particular range or for a particular particle size is excessively present in the soil. That type of soil is called the poorly graded soil. And graph graded soil means one particular range of the particles size is missing in the soil. So, there will be the graph will be like this will start from here then there is a graph then it will start from again. So, there will be a particular range of the particles is missing in the soils in that type of soil is called graph graded soil. Now, next one is the wet volume relationship of the soil. So, as we know that soil is a three phase system that is air water and solid. So, from this three phase system we can determine different soil properties and those properties are very important for the foundation design. So, if we consider the V a is the volume of the air and V w is the volume of the water and V s is the volume of the solid. Now, here this water and the water in this air they are present in the voids in between the solids. So, that is the volume of the water and volume of the air if we sum these two volume that will give us the volume of the voids. So, that is V v will be V w plus V a and total volume is V. Similarly, we can total weight of the soil if consider W then W s is the volume of the weight of the solid then W w is the weight of the water and W a is weight of the air generally it is taken as 0. So, we can neglect the air weight in the soil during the calculation. So, from this graph we can say that if it is a totally saturated soil that means all the voids are filled with water then this system become a two phase system that means the solid and the water only because that is a fully saturated soil there will be no air. Now, if similarly if the soil is a dry soil completely dry soil then this total void will be filled by the air. So, that means that will also become a two phase system because all the voids will be filled by air there will be no water. So, in that case we have to consider the weight and the volume accordingly. Now, then the different definition of this one is the void ratio E void ratio is defined by volume of void divided by volume of solid. Similarly, porosity is defined as volume of void divided by total volume. So, we can say here that the void ratio E is the volume of void divided by volume of solid. So, E value can be greater than 1, but porosity if we look at the definition of the porosity that is volume of void divided by total volume. So, the volume of void cannot be greater than the total volume. So, the porosity cannot be greater than 1. Now, similarly degree of saturation S which is expressed in percentage is defined as volume of water divided by volume of voids into 100 because it is expressed in percentage. Similarly, moisture content W in percentage is weight of water divided by weight of solid expressed in percentage. Now, the unit wave soil at any water content or any degree of saturation can be written as because what as I have mentioned that soil can be in the different stages, can be in the completely dry stage, it can be in the completely saturated stage or it can be in some normal stage with it is not it is partially saturated, this is not completely saturated. So, at different condition how we will determine the unit weight of the soil. Then if we denote unit weight of the soil is a gamma, then for gamma bulk or the unit weight at any condition is given by G S plus S E divided by 1 plus E into gamma W. So, gamma W is the unit weight of water can be taken as equal to 10 kilo Newton per meter cube G S is the specific gravity of the soil and S is the degree of saturation is the void ratio. Now, if the soil is completely dry that means, the S is equal to 0 that means, saturation is 0 the degree of saturation. Then the equation of gamma, gamma dry will be G S into gamma W 1 by 1 plus E. So, that is because here S is equal to 0. So, this equation become this one for the gamma dry and then we can write that gamma dry is equal to gamma bulk divided by 1 plus W, W is the water content of the soil. Bulk density means bulk unit weight means the unit weight at any condition with any water content. So, that is gamma bulk or gamma bulk is equal to this one. Similarly, gamma sat if I put for the gamma saturated, if the soil is completely saturated then S value will be equal to 1. So, if in that case S value is equal to 1 then this equation become this one if it is a completely saturated soil. So, for the saturated soil this equation become G S plus E gamma W divided by 1 plus E because in that case S is equal to 1. The next one is the relative density of the soil. In granular soil the degree of compaction in the field can be measured by relative density R D in percentage. So, relative density can be expressed is E max minus E natural divided by E max minus E mean into 100 because it is expressed in terms of percentage. Where E max is the void ratio of the soil in the loose state and E mean is the void ratio in the dense state and E is the in situ void ratio on the natural void ratio. So, if we know the E max E mean and E at any condition then we can determine what is the degree of relative density of the soil at that situation. The next important properties are the Atterberg limits. So, first we define the what are the Atterberg limits. So, previous the properties that we have discussed that is the well graded soil, poorly graded soil, gap graded soil. These properties are basically properties for the coarse grain soil that is for the sand mainly sand and this is the common thing and again relative density is also a very important properties for the granular soil or the coarse grain soil. Similarly, the Atterberg limits these limits are very important properties for basically for the fine grain soil or the clay soil. So, that means here we can say from this curve the soil has four different state that is the solid state, semi solid state, plastic state and liquid state. Now, we can say in this way that if we take a completely dry soil soil that means that is soil solid the soil is the solid state. Now, if we add some percentage of some amount of the water then the solid state become semi solid state. If we add more water then this semi solid state will become the plastic state. Then in the plastic state if we add more water then it will go to a state which is called the liquid state. Now, here we can say that there is a limit or there is a injunction between every state because this is that means there is four state and there is three injunctions and if I see that part that is the moisture content. So, as the moisture content increases then soil go from solid state to semi solid plastic and liquid and vice versa. If the soil is liquid state if we reduce the water content in the soil then it will go to plastic state and from the plastic state it will go to semi solid state and from the semi solid state it will go to solid state. Now, from this graph if I draw a graph which is moisture content, moisture volume in the soil water mixture. So, we can see that if we increase the if we decrease the soil water content amount then from the liquid state to it will go to plastic state and then it will go to semi solid state and as we reduce the water content the volume in the soil water mixture that will also reduce. But after a certain point or the intersection between the semi solid state and the solid state or the junction of the semi solid state and the solid state after that if we decrease the moisture content then the volume will not change volume remain constant. So, that means after this point that if we further reduce the water content but volume remain constant volume will not change. Now, let us go for the what is the this junction of this different state then the junction of plastic state and the liquid state is called the liquid limit. So, these three are the limits. So, these are liquid limits plastic limit and the sinkage limit. So, the liquid limit the limit where soil changes its states from liquid to plastic or vice versa. Similarly, the next limit is between the plastic state to semi solid state and that limit is called the plastic limit and third limit is in between semi solid to the solid and this limit is called the sinkage limit. So, we can say if we reduce the water from a liquid state of a solid then it will pass liquid limit it will go to the plastic state and then if you further reduce it it will pass the plastic limit and it will go to semi solid limit semi solid state and if you further reduce the water content then it will pass the sinkage limit and it will go to solid state. And from this graph we can say that after sinkage limit the volume of the water soil mixture will not change or does not change if we further reduce the amount of water present in the soil water mixture. So, after the sinkage limit there is no volume change. So, these four three limits are called the Atterberg limits and these are very important properties for the fine grain soil. So, next one is the hydraulic conductivity of the soil. So, what is hydraulic conductivity of the soil? Darcy proposed the following equation for the calculation calculating the velocity of flow of water through a soil and that equation is V equal to K into I where V is the Darcy velocity that is written as centimeter per second, K is the hydraulic conductivity of the soil that is centimeter per second and I is the hydraulic gradient that is del H by L. Now, what is del H by L? If water is flowing in this direction and if we take the two height of the water at two location then it will show some difference otherwise this flow will not occur. So, that means if this height difference is del H and the length of this two point A and B is L and the I hydraulic gradient can be defined as del H divided by L. So, from this expression we can calculate the velocity of a soil is the hydraulic conductivity in the hydraulic gradient. So, what is the hydraulic conductivity? Now, if we consider this I is equal to one then this V is become the Darcy velocity which is centimeter per second. So, that means K if I is equal to one K is equal to V. So, the hydraulic conductivity of the soil is the velocity of the Darcy velocity at which under the unit gradient. So, that means the hydraulic conductivity of soil is a velocity or the Darcy velocity under unit hydraulic gradient. So, that means I is equal to one. Now, in the laboratory hydraulic conductivity can be determined in the laboratory we can determine hydraulic conductivity by two methods. One is constant head method more suitable for coarse-gain soil and falling head method suitable for soil such as fine sand and sealed. So, that means it is not suitable suitable for the fine soil or that is falling head method. Now, in the field also we can determine the hydraulic conductivity of the soil. These two are the lab test that we are discussing here. Now, the type of soil and the hydraulic conductivity in centimeter per second that medium to coarse gravel the hydraulic conductivity is greater than 10 to the power minus 1. For coarse to fine sand that is 10 to the power minus 1 to 10 to the power minus 3. For fine sand silty sand that is 10 to the power minus 3 to 10 to the power minus 5. For silt clay silt and silty clay that value is 10 to the power minus 5 to 10 to the power minus 6 and for the clays the value of k or hydraulic conductivity is 10 to the power minus 7 or less than that. So, next concept or the next thing is the effective stress concept. Now, what is effective stress? Now, effective stress can be defined as the total stress minus the pore water pressure. Now, where sigma dash is the vertical effective stress and sigma is the vertical total stress u is the pore water pressure. So, that means when we apply wet or the load on a soil or the stress on a soil initially that stress is taken by the initial time is taken by the water. Now, if we do not permit any flow of the water that means this stresses will be taken by the water initial stages. Now, if we permit now the water to flow then gradually this water will flow and the stress which was taken by the water initially will be transferred to the soil skeleton. So, now as time progresses this water goes out then this stresses will be transferred to the soil skeleton. Now, soil will take the stress in that fashion. So, that means when the soil is totally dry that means there is no water. Now, if we apply the stress on a soil water mixed system then as the stress will taken by the water. So, if we apply the stress on this condition that means this water pore water pressure will be developed. Now, if the soil is totally dry that means no pore water pressure will be developed in that case u is 0 in that condition the effective stress will be equal to the total stress. So, that means if u is the dry condition u is 0 that means the sigma dash is equal to 6 sigma and if that means dry means there is no water present in the soil. Now, how we will calculate the total stress or effective stress in a soil medium. Suppose this is a soil medium where the water table is at a height at a depth of h 1 from the ground surface and this is the unit weight of the soil above the ground water table is say gamma and unit weight of the soil below the ground water table that is the saturated because it is below the water table. So, saturated unit weight of the soil is gamma sat. Now, the total stress at a point can be calculated as h 1 into gamma plus h 2 into gamma sat and pore water pressure at a depth at a point A and this A point is at a depth of h 2 from the water surface. So, at this point the pore water pressure can be determined as h 2 into gamma w where is gamma w is the unit weight of the water. Now, here if we take this is gamma h 1 into gamma plus h 2 if we take h 2 common. So, this will be gamma sat minus gamma w. So, h 1 gamma plus h 2 gamma dash this gamma dash is called as submerged unit weight of the soil that is equal to gamma sat minus gamma w. So, when you calculate the effective stress at any point of below the water table then the effective stress will be equal to the height of the soil or depth of the soil into the unit weight of the soil above the water table plus the depth of the that point from the ground water table into the submerged unit weight of the soil that is gamma sat minus gamma w. So, this way we can calculate the effective stress of the soil at any depth within the soil medium. The next thing that we will discuss about the consolidation. Consolidation is also a very important properties for the fine-grain soil. Now, there is as I mentioned that consolidation soil voids are filled by either air or water or both. So, if we want to calculate the condition to remove these voids there are two basically two methods by which we can remove these voids. One is compaction another is consolidation. So, the major difference between these two methods that by compaction we can remove the air voids, but by consolidation we can remove the water voids. So, there is there are other differences also there based on these two methods, but this is the major difference. So, here consolidation means we can remove the water voids or water present within the soil pores. So, now this consolidation once we done the consolidation this consolidation can be done in the laboratory and there some properties which are very important for our foundation design basically for the settlement because consolidation is very important properties for the fine-grain soil or the clay soil because in the sand-grain soil permeability of the soil is very high. So, once we apply the load all the soil or the water can dissipate within very short duration of the time. So, long term settlement all the settlement that we will get for the coarse-grain soil is the immediate settlement. So, there is a long term settlement is very negligible in terms of coarse-grain soil, but for the fine-grain soil or like clay where the permeability is very slow very low. So, once we apply the load there will be very negligible amount of the immediate settlement although depending upon the type of clay, but the most of the settlement will come because of the consolidation because at time progresses the water will dissipate from the clay soil slowly slowly and then the settlement will occur. So, that is the time dependent phenomenon and it will take the long term to complete the total settlement. So, that is why consolidation and that settlement is called the consolidation settlement and that consolidation settlement is very important properties for fine-grain soil during the foundation design. So, in the consolidation settlement consolidation after what we can do we can apply the load on a soil sample or apply the pressure on a soil sample and we can determine the what is the amount of void in the soil or we can determine the void ratio. So, as it is expected as we apply the more pressure the void ratio will decrease. So, here typical consolidation I mean void ratio versus log p, p means the pressure graph is presented. This is in the pressure is plotted in the logarithm scale and void ratio in the normal scale. So, this log p. So, here we can see as we apply the more pressure the void ratio decrease. This is the loading curve and this is the unloading curve. So, once the we remove the pressure so, there will be some rebound of the soil void. So, it will of further go up. Now, here the slope we can determine the compression index that is the slope of this loading curve. So, this slope of loading curve. So, the compression index C c is equal to E 1 minus E 2 divided by log p 2 minus log p 1. Now, E 1 is the void ratio corresponding to pressure p 1 and E 2 is the void ratio corresponding to pressure p 2. So, this is the basically C c is the slope of this loading path. Now, according to Schempton we can determine the C c also that is 0.009 l l is the liquid limit minus 10. In this way also we can determine what is the value of C c. Now, there is another very important thing that the over consolidated soil. Now, what is over consolidated soil? Now, if the natural pressure or the current pressure which is applied on the soil is less than the pressure which is previously subjected to the soil or the soil has experienced the more pressure which is currently applied on the soil then that soil is called the over consolidated soil. Now, for example that if the landslide is there the previously that soil before landslide is suffering more stresses or more pressure now after the landslide some soil on that point is removed and it wash away. So, that means the stresses now currently on that location. So, is less than the stress which is previously subjected to on that point. So, in that case that soil is called as the over consolidated soil. Now, how will so from this over consolidated soil? So, how will you calculate the over consolidated ratio that over consolidated ratio or the over consolidated stress. So, that over consolidated stress how we calculate that is proposed by the Casagrande in 1936. Now, from this E void ratio versus log p graph E versus log p graph also we can determine this over consolidated pressure P c. Now, this is a typical loading path and this is unloading path is a similar to this graph is a loading and unloading path. Now, from this loading path we have to identify a point O which has the smallest radius of curvature. Now, from here we can get the different curvature. Now, we will get a point O here which is has the smallest radius of curvature. Now, once you select the point O then we draw we can draw a line O A which is parallel to x axis. Now, from first select the point O which has the smallest radius of curvature. Now, from point O draw a line O A which is parallel to pressure P x axis. Now, from this point A draw a line O B such that this O B is a tangent at O. So, draw a line O B such that O B is a tangent at O then draw another line O C such that it bisects this angle A O B. Now, once you select the point O B then you once we get the O C line. Now, next step is to extend the straight portion of the loading curve. Now, extend the straight portion of the loading curve and identify the point where the straight portion of the loading curve and this bisection curve intersects. Now, identify that point and pressure corresponding to that point is called the O B. So, if this point is over consolidated pressure or maximum past effective over burden pressure. So, that the present if the present. So, this way we can determine the p c which is maximum past effective over burden pressure. Now, if present effective over burden pressure P 0 is equal to pre consolidated pressure P c then the soil is called normally consolidated soil. That means, the present effective over burden pressure is equal to this pre consolidated maximum past effective over burden pressure P c then that soil is called normally soil consolidated soil. Now, if this P 0 that means, the present effective over burden pressure is less than the P c or the past effective over burden pressure as I have mentioned then that soil is called over consolidated soil. So, this way from this log e versus log p curve also we can determine the P c value. Now, how we will calculate the consolidation settlement of a soil because as I mentioned that the foundation design if this settlement calculation is very important issue for the clay soil. Now, here how we will calculate this settlement for the consolidation settlement. So, there is two type of soil we have considered one is normally consolidated clay another is over consolidated clay. So, now in the normally consolidated clay this is the typical loading path that we will consider that is the state portion we can consider for the load normally consolidated clay. And here with the slope of this curve will give us the C c as I mentioned the slope of this curve the compression index. So, that means, this is the difference between stress is del p P 0 and P 0 plus del p that is the difference between stress and because of this stress difference this is the difference between the void ratio del e. And as a spectate as we increase the space pressure the void ratio will decrease. So, similarly then similar curve we can draw for the over consolidated clay for the over consolidated clay. So, this is the typical curve log e versus log p curve and here this is the point where we got the P c value. And the how we locate this point how we will get this P c value we can we have already explained. And then there is two parts one is the this one whose slope will give us the C c compression index. And another is this part above this P c value portion that is the C s. And here also P 0 plus P 0 plus del p there will be the del e. And here also P 0 plus del p that portion there also will get the. So, that means, here we will get the del p del e i and del e 2. So, there basically we have two different portions. Now, for the normally consolidated clay how we will calculate that for the settlement S is C c 1 plus e 0 into H log P 0 plus del p divided by P 0. So, this is the settlement calculation for typical this portion. Now, where C c is the compression index e 0 is the initial void ratio H is the thick thickness of the soil layer for which we are determining the settlement. And P 0 is the initial applied stress and del p is the stress or additional stress. That means, for example, the P 0 is equal to effective overburden pressure. And del p is the additional stress due to the applied external load on the foundation. So, later on in when you talked about the when you talk about the foundation settlement calculation you will show how this P 0 and del P 0 will be can be calculated. So, that part will be showing the later on in the during the settlement calculation of the foundation. Now, for similarly for the over consolidated clay for the case 1 if P 0 plus del p is less than P c. That means, if P 0 plus del p is less than P c in this condition in that case the settlement is calculated as C s 1 divided by 1 plus e 0 H log P 0 plus del p divided by P 0. That means, in that case instead of using C c we will use the C s where C s is the swelling index. And in the case 2 if P c in the point is in this condition that means, here P 0 is less than P c, but P 0 plus del p is greater than P c. In that condition settlement we can calculate the C s plus 1 plus e 0 into H log P c divided by P 0 plus C c divided by 1 plus e 0 into H log P 0 plus del p divided by P c. So, in this 2 conditions how we can determine the consolidation settlement. So, one is normally consolidated soil this is the expression if it is over consolidated soil if case 1 this is the expression if it is case 2 this is the expression. The next one is very important is the shear strength of the soil. Now, shear strength is generally determined by this expression the strength is coefficient of the soil coefficient plus sigma n into tan phi or sigma n this is in terms of total stress. So, the C plus sigma total normal stress on the plane of shearing. So, sigma is the total normal stress into tan phi. So, phi is the friction angle C is the cohesion of the soil. Now, these things we can draw it is the more coulomb failure criteria or more coulomb failure envelope is there. So, this is the envelope and this is the sigma axis normal stress axis and this is the shear stress axis. Now, if any more coulomb touches more circle touches this envelope that mean that failure will occur and anything. So, that means, here this expression shear strength expression is C plus sigma into tan phi phi we can determine by tan phi is the slope of this angle. So, phi if we know the slope of this angle we can determine slope of this line if we know the slope of this line we can determine the angle phi here also. So, tau f is the maximum shear stress in soil that can take without any failure under normal stress sigma. Now, this is the same expression we can write in terms of effective stress. So, in terms of effective stress this will be C bar effective cohesion then the sigma dash or tau phi dash. So, sigma dash will be sigma minus u u is the pore water pressure and similar envelope we can determine, but only this in terms of effective stress. So, tau f is the maximum shear stress the soil can take without failure under normal effective stress of sigma bar. Now, this property so that means, here we can see the C and phi these are the strength properties of the soil. So, there is a very important things for the and this C and phi these two parameters very important for the foundation design for the foundation bearing capacity design the I mean how much amount of load foundation can carry this two parameters we can determine is based on these two parameters. So, these two parameters we have to determine in the laboratory we can determine these two parameters in different by this different test one is direct shear test generally for the sand then triaxial test conducted on sand and clay both and these are three types of test depending upon the drainage condition depending that is the consolidated drain test C D consolidated undrained test then consolidated unconsolidated undrained test u u. So, depending upon the we have a nature of whether it is the consolidated or unconsolidated drained or undrained these are three types of triaxial test. Another test which is conducted is unconfined compressive strength with this is suitable for the soil whose saturated clay that means phi value is equal to 0 that mean the shear strength of the saturated clays can be calculated for this type of test star wave equal to C u C u is the undrained cohesion of the soil and here if the unconfined compressive strength is q u then if you divide it by two that means there will be the C u. So, half of the q u will give us the C u. So, that is the unconfined compressive stress divided by two that will give us the C u for unconfined compressive strength test. Now, so these are the typical sand or the clay consolidated drain test C D stress stress strength relationship during shearing. So, in the triaxial test if I take this axis this is the axial strain axis and this is the deviatoric stress then this curve will give us for the this curve is for the loose sand or normally consolidated clay and this curve is for the dense sand or over consolidated clay. Now, from these two curves we can see that for the loose sand or normally consolidated clay we will not get any prominent failure or peak of the in the curve, but in the dense sand or over consolidated clay we will get a prominent peak is peak in the stress strain curve, but ultimately we will get these two value will approaches a steady as a same condition or same level and this things after this stress that is called the residual stress. Now, so similarly if I get the axial strain and the volume change curve then for the loose sand or normally consolidated clay volume will decrease and it will decrease always as we apply the load as we apply the load volume will decrease, but for the dense sand or over consolidated clay as we apply the load initially the volume will decrease, but after that the volume will increase. So, here after certain point the volume will increase or this is due to the reason that for the loose sand as we apply the load the voids in between the soil grains that will reduce and it will reduce continuously, but in case of dense sand because this void is already in a packed condition the void is void in the soil grains between the soil grains already is in the very less amount it is in the packed condition. So, if we further apply if we further apply the stress there this volume will slightly reduce again it will become a very dense condition, but again if we apply the load then dense condition will become a loose condition that means the loose means the void in between the soil particles that will increase. So, once become very dense then after that if we apply the stress it is become again the volume will this pores volume will increase. So, that is why this volume will change for the dense sand. For similar curve we will get for the direct shear test also this is for the dense sand and this is for the loose sand and similar volume versus shear strength. So, this is a shear displacement and shear stress and previous curve was axial stress and deviatoric stress axial strain and deviatoric stress, but the similar nature of the curve will get. Now, so this in the successive lectures you will get some corrections are required. So, that in the different lectures so this is the additional slide that is put here. So, in the lecture 3 in this time 57.16 meter it is recorded the soil as well rather it should be in situ test replace this reference by this reference. So, in lecture 4 at this 19.20 minute replace refactored by refactored raise lecture 9 52.40 minute it is foundation on layer soil instead of wrap foundation. In lecture 15 unit of unit weight is kilo Newton per meter cube not kilo Newton per meter square lecture 29 it tells to discuss about a breast cut, but not discuss there at 2 minute it is reinforced foundation instead of RCC foundation as said. In lecture 30 modulus of sub grade reaction is stress per unit deflection not stress per unit area. In lecture 31 the reference of K H expression is Richard's and Elmins 1979. Lecture 34 replace Boussonic 1985 by Boussonic 1985 and the spelling is you have to change there. And in lecture 34 replace non-linear V R F soil by non-linear V V R F soil and in the V R by linear model W 0 can be replaced by Q 0. So, these are the few corrections that in the different lectures that we will follow in the come in series of lectures. So, this is the introductory lecture that I have taken. So, in the from the next lecture onward I will start the different models that the first I will start over the model 1 module 1 that means, the soil exploration then we will start module 2, 3, 4 and 5. So, in the next class I will start about the soil exploration. Thank you.