 Welcome to course on advanced geotechnical engineering, in the yesterday's lecture we have introduced ourselves to different modeling techniques in geotechnical engineering and we narrow down to a technique which is called as physical modeling or geotechnical physical modeling. This lecture is a continuation of geotechnical physical modeling topic. So this comes under module 7 lecture 2 on geotechnical physical modeling. So yesterday we have discussed that if a physical model can be model at full scale then we did not have to worry about the scale factors or scaling relationships. But if the model is not constructed at one is to one scale then we need to have some idea about the way in which we should extrapolate the observations. That means if you look from the material behavior point of view linear and homogeneous for the loads applied in the model if the material is linear or if the material behavior is linear and homogeneous for the loads applied in the model then details of the model can be predicted to prototype without much of you know worries. That means that still with the help of dimension analysis we can do that. But if the material behavior is non-linear or if the geotechnical structure has several materials which interact with each other then it requires understanding about more into dimension analysis and similitude relationships between parameters influencing model and prototype. So this dimension analysis basically it is a method for reducing elements of the form of a theoretical relationship from consideration of the variables and parameters that make up the relationship. The dimension analysis is now a method for reducing the relationship among variables which are influencing a particular phenomenon and these variables can be you know dependent variables and independent variables and dependent variable is the one where which grants the entire phenomenon and independent variables are the one which influences the phenomenon. So if any change in the independent variable then there is an influence on the independent variable. So dimension analysis of a problem leads to a reduction in the number of variables that must be studied in order to understand the problem. So by doing dimension analysis we will be able to lead to a reduction in the number of variables that must be studied in order to understand the problem. So the governing dimensionally homogeneous equation have to be deduced among the key variables influencing the special phenomenon. So for this purpose we have two methods. One is Rayleigh's method or we also call as method of product of powers. The other one is Buckingham's pi theorem. In the Rayleigh's method if you are having say variables like q1, q2, q3, q4, so qn if they are influencing a particular phenomenon by expressing as a method of product of powers and expanding the infinite series of terms and by satisfying the condition of dimensional homogeneity we can prove that the dimension of q1 is to be equal to dimension of q2 to the raise a1, q2 to the raise a2 so on to qn to the raise an. Where in n is nothing but here the number of variables. So by adjusting by writing the indical equations we can actually get the values of a1, a2, a3 so on to an. And by getting these values we will be able to get a relationship among the dependent variable and independent variable. But this method fails if you are actually having more number of variables. Let us say more than 7 variables then this method the operation involves tedious and converting one solution to another solution involves the algebraic adjustments. In such situations the Buckingham's pi theorem is a viable option and where in if you have got let us say q1 to so on to q2, q3 so on to qn then here and r dash and r double dash and r triple dash if there are these dimensionless ratios then we can actually reduce in the term of pi terms as pi1 is a function of pi2 so on to pi3 so on to pin and r dash and r double dash and r triple dash. So here what it implies is that when from the similar procedure we will be able to get this but in order to get the number of dimensionless products what we need to do is that in the Buckingham's pi theorem we have to arrange these variables in such a way that you get the non-singular matrix towards the right hand side of the matrix of the dimensional matrix. Once we solve this dimensional matrix and determine the rank of a matrix then we can actually ascertain the number of pi terms. Let us say the number of variables are say 10 and if you have got say rank of that matrix as 2 then number of dimensionless products possible for that particular phenomenon the consideration is equal to n minus r that is 10 minus 2 8. So after having obtained ascertain the number of pi terms and writing by writing linear algebraic equations and by putting pi1 to pi8 in the matrix of solutions and satisfying those equations which are reduced based on the dimensional matrix coefficients we can actually write get the pi1, pi2 so on to pi8 in the given example. So the Buckingham's pi theorem is used for reducing dimensional products and dimensionless dimensional analysis does not reveal the form of relationship between dimensional products but the correct use of dimensional products make parametric studies more efficient by revealing which variables are truly independent which variables are insignificant and also basis for the also forms the basis for extrapolating from one scale of observation to other scale of observation. So after once we get the pi terms let us say we have got say set of 5 pi terms so our similarity between model and prototype what we say is that the particular phenomenon which actually has been modeled is influenced by pi1, pi2, pi3, pi4, pi5 this implies that pi1 is a function of pi2, pi3, pi4, pi5 for similarity between model and prototype if you are actually not modeling at one is to one scale then for one is to n model small scale model for pi1 to be equal to model and prototype that is pi1 in model to be equal to pi1 in prototype we have to satisfy pi5 in model is equal to pi5 in prototype, pi4 in model is equal to pi4 in prototype so on up to pi2 in model equal to pi2 in prototype. If you are not able to achieve the complete similarity in satisfying all the pi terms then we say that the partial similarity is achieved then we need to investigate that particular effect of not modeling not satisfying this particular this thing have to be verified by experimental investigations. So let us take an example of buried flexible conduit so this is a example of a geotechnical structure which deforms under the conditions of plane strain. So plane strain in the sense that if a long body is subjected to transverse loading and it is cross section and loading do not vary significantly in the longitudinal direction a small thickness in the loaded area can be treated as subjected to plane strain. So here in this particular slide what we are seeing is a underground conduit having diameter D and H is the embedded depth and EI by B is nothing but flexural rigidity of this conduit per unit width that per B units of widths and gamma is nothing but the unit weight of the soil and ES is nothing but the soil stiffness or elastic modulus of the soil and P by B is the line load which is at a certain distance if it is at this particular point it is right on top of that and P by B is at a certain distance and X and Y are the coordinates in the X direction and Y direction and perpendicular to this is say is the Z axis. So the plane strain condition when we say that these conditions are satisfied for this type for example like retaining wall or slope of a highway embankment or railway embankment embankment dam or levee section they are examples of plane strain or a strip foundation of a footing of a strip footing of a typical foundation which extending for number of columns in one line and these are all examples of plane strain. So epsilon Z is equal to the shear strain in Z direction shear strain in ZX direction shear strain in Y Z direction will be equal to 0 that means that the strain in this direction that is ZX plane and YZ plane will be 0 and similarly the shear stress in YZ direction shear stress in the ZX direction will be 0. So only it will have in XY direction and deformations and as well as the strains will be there in these directions. So this particular structure is an example of plane strain structure so what we see is that the load P by B which actually causes in a bending moment per unit width let us say. So depending upon the loading the bending moment per unit width will per B unit width will increase or decrease. So let us say that an important parameter would be bending moment per unit width that is M by B resulting in the wall of the conduit due to construction procedures and surface traffic loading. So the example of say line load what we have considered is let us say a boundary wall and or a railway track the structural property that will influence the bending will again be a flexural rigidity per unit width that is EI by B. So in addition moments will be influenced by diameter D that is internal diameter H that is embedded depth of a conduit gamma that is the density surrounding the conduit and P by B that is the loading which is applied. So the interaction between the conduit and the soil will be influenced by the stiffness of the soil that is ES at a typical depth H that is the interaction between the conduit and the soil will be influenced by the stiffness of the soil ES at a typical depth. So what we have got is that bending moment that is M by B and we have parameters like EI by B it is a conduit parameter and loading parameters are nothing but P by B and again conduit parameter D and soil parameters are H that is the embedded depth and that is the geometry configuration and gamma is nothing but the unit width of soil and ES is nothing but the stiffness of the soil. So by solving by using the Buckingham's pi theorem and by writing dimensional matrix by influencing the variables and the solving the matrix of writing the matrix of solutions we get M by B gamma DQ is equal to a function of H by D P by B by gamma HD EI by B ESDQ P by B D square by EI by B. So if you look into this this particular arrangement has been made such a way that we can lead to some important discussion by using this problem. There can be different arrangements are possible depending upon the type of repeating variables which we have chosen but one solution to other solutions can be transformed by doing the algebraic adjustments for the pi terms because if you are having a pi 1, pi 2, pi 3, pi 4 influencing the phenomenon if these are pi terms which are say dimensionless if pi 1 is pi term pi 1 by pi 3 is also dimensionless. So depending upon the requirement without changing the number of pi terms which are used originally we can transform and get the new pi terms of choice. If you are having a particular variable and if that variable needs to be more than say if it is appearing in more than three pi terms if that variable needs to be eliminated can be eliminated by doing the adjustments what has been said but it has to be seen that the first pi term or the principal pi term used only once. So m by b by gamma d cube which is function of this thing so here there are stress and stiffness related quantities in the loading spread over the cross section of the conduit p by bd and the stress in the ground at the mid height for example here this is gamma h represents the vertical stress in the ground at the mid height of this thing and here this is the conduit stiffness and soil stiffness you know you can see that the conduit stiffness and soil stiffness are related here and so the flexural rigidity a by b and corresponding soil stiffness which is actually related. So let us discuss further you know how we can actually reduce you know different relationships. So here further the continuation of the solution for the example for buried flexible conduit now what we do is that we have said is that when you have got set of pi terms and by for similarity you know so and so pi term let us say pi 3 in model has to be equal to pi 3 in prototype. So if all these are satisfied then only the primary or principal pi term that is involving bending moment will be satisfied. So with p by b in model divided by gamma hd in model equal to p by b in prototype divided by gamma hd in prototype. So here we can actually get two situations what we said is that if you are not modeling at full scale you know then if you at one is to one that p by b in model is equal to gamma hd in model both are same you know in model and prototype are same. But when you are actually modeling at a scale different than one is to one that is say small scale model one is to n. Let us say the environment what we have is that same soil is used as that in the prototype with that we can say that gamma m is equal to gamma p by with for gamma m is equal to gamma p and one is to n small scale model what we get is that p by b in model to p by b in prototype is equal to one by n square. So that is nothing but what we have done is that by bringing this term here and bringing this term there gamma m and gamma p one hm by hp is equal to one by n dm by dp is equal to one by n by substituting that in this we get p by b in model divided by p by b prototype is equal to one by n square. So m suffix indicates model and p indicates prototype which is nothing but one by n square. Let us assume that we have got a situation where gamma m is equal to n gamma p is possible that means that there is a let us say there is a requirement which actually has come and we are able to in a physical model we are able to enhance the unit weight of the soil by n times. So that means that if you are having you know 20 kilo Newton per meter cube of a soil at by enhanced with enhanced n we get gamma m is equal to 20n and for n one is to n scale model we get by substituting in a similar way with gamma m by gamma p is equal to n we get p by b in model is equal to p by b in prototype is equal to one by n. So we can see that if you are having gamma m is equal to n gamma p we are actually getting p by b in model by p by b in prototype as one by n. Now with p by considering another pi term so what we do is that once after having obtained the line load scale factor in model in prototype with gamma m is equal to gamma p for a one is to n small scale model gamma m is equal to n gamma p for one is to n scale model by considering another pi term which actually has got p by b and e i by b terms by using this particular pi term and this relationship and for gamma m is equal to gamma p and d mm by d p is equal to one by n and we have deduced in the previous slide p by b in model is equal to p is equal to one by n square times p by b in prototype by substituting this we get e i by b in model by e i b in prototype is one by n to the power of 4. So please note that this is one by n to the power of 4 that means that whatever we are having you know the value of flexural rigidity per b widths will be one by n to the power of 4 that of in the one by n to the power of 4 times of that. So that means that if you are having a a by p p value which has to be one by n to the power of 4 times smaller similarly with gamma m is equal to n gamma p and d m by d p is equal to one by n and p m p by b in model by p by b in p is equal to one by n this is one by n with that we will be able to get a by b in model is equal to one by n cube times e i by b in prototype. So now we have understood that the flexural rigidity with p by b in model by p by b in prototype is equal to one by n with that we will be able to get this particular scale factor. Now what we do is that by using this you know term which involve flexural rigidity and flexural this soil stiffness term e i by b in model divided by e s d cube in model is equal to a b b prototype by e s d cube by prototype by with gamma m is equal to gamma p and d m by d p is equal to one by n and e i b in model by a by b in prototype is equal to one by n to power of 4 if you put this in the substitute in this what we get is that with gamma m is equal to gamma p and a small scale model and with the deduction what we reduce to here for e i by b scale factor for one is to n scale model with gamma m is equal to gamma p we get the soil stiffness in model is one by n times smaller than the soil stiffness in the prototype whereas when you have a gamma m is equal to n gamma p environment with d m by d p is equal to one by n we can actually get e i by b in model by a by b in prototype as one by n cube and by using this we get the soil stiffness in model prototype is equal to one. So this actually has got you know very high strong relevance that means that in in modeling a particular prototype behavior if you are actually having a simulation of a identical stiffness as that in the prototype the response of a particular structure for example in this case a buried flexible conduit in model so this implies that if you are actually having an environment like gamma m is equal to n gamma p and even for a small scale model we can actually maintain you know the same stiffness as that in the prototype. If you are actually having the same stiffness as that in the prototype that implies that the identical stress strain behavior of a you know in the prototype can be captured very well in a model which is actually tested in a small scale model which is reduced by one is to n tested at gamma m is equal to n gamma p. So this is very important as far as you know the physical model geotechnical physical modeling particularly with gamma m is equal to n gamma p consideration point of view. So let us now consider an example like slope in cohesive soil and assume that we actually have got a saturated clay under undrained conditions so it actually has got undrained conditions and gamma as the unit weight of the soil and d is the depth below the base that is the base layer depth and h is the height of the slope and beta is the slope inclination. So if you look into it the stability of a slope it actually depends upon the parameters like Cu, gamma, beta, h and d. So when you list of the variables we actually have got factor of safety is a function of h that is nothing but the slope height beta that is the slope inclination Cu undrained cohesion gamma and d. So by again by using this can be done by using Rayleigh's method or by using Buckingham's Pytheism and while writing dimensional matrix if you are having terms like this we can actually use force and length approach that means that force is expressed as mLt-2 the dimensions of force mLt-2 and L as L and the stress is nothing but f by L square and unit weight is nothing but f by L cube and by using this we will get two algebraic equations and by solving them and by writing matrix of solutions we can actually get relationship among the variables now here 1, 2, 3, 4, 5, 6, 6 variables are there and we get the rank of the matrix of the dimensional matrix will be 2 so 6-4 so 1, 2, 3, 4 pi terms. So in this out of these two are already dimensionless terms one is factor of safety another one is beta that is the angle or slope inclination. So here what it actually says is that the factor of safety is a function of beta d by h and c by gamma h e. Now from the similarity point of view what we say is that factor of safety for factor of safety in model and prototype to be same if at all we are modeling this particular situation of a slope in cohesive soil for factor of safety in model and prototype to be same what we say or what we have to do is that beta in model and prototype to be same that means that the slope inclination model and prototype to be same and d by h that is the ratio between the d that is the depth below the you know toe of the slope to the height of the slope d by h ratio in model and prototype to be same. If you are and then c by gamma h in model and prototype will be same if all these pi terms are same in model and prototype for a one is two n scale model then we can say that the factor of safety in model and prototype will be same. So if at all you know let us see by looking into the you know the different combinations we say that how this is actually possible. Now one thing we can actually look is that according to Taylor's 1948 theory the Taylor actually has given for slope inclination greater than 53 degrees we can say that it is independent of d by h factor depth factor then we can write factor of safety is equal to function of c u by gamma h and beta. So here the number of variables that need to be considered only two it implies that it is required to maintain the same margin of safety in model and prototype not only the geometry but also c u by gamma z should be so constant. So this implies that for similarity between model and prototype for a we actually have to you know for maintain the same margin of safety in model prototype not only the geometry that is the sloping inclination but also c u by gamma h should be constant. The c u by gamma h should be constant means how that is possible means let us say that we actually have constructed a small scale model h is reduced by h by n. If you look into it somehow let us say that the sloping inclination is achieved then the c u by gamma h in model and prototype to be constant what it implies is that for a slope which is reduced by one by n times the c u by gamma h to be same what it says that the term to be same the c u by gamma has to be reduced by one by n times. If that is reduced by one by n times then only it is possible that you know it will be able to make the similarity possible. So if you look into this if the g can be you know one of the alternatives if you see that you know by reducing the cohesion in this particular term for similar to be achieved what we can say is that by maintaining identical unit weight as that in prototype like gamma m is equal to gamma p and the c u in model and prototype let us say that it is reduced by one by n times then we can actually say by maintaining identical gamma in model and prototype we can reduce c u in the model and prototype one by n times that means that the cohesion of a soil is reduced by one by n times that in the prototype if you are able to do that when you reduce you know h by n and c u by n then they get cancelled then possibility that they will be same. But you know by you know this is you know topic to be discussed the stress strain behavior of a soil which is actually having a cohesion of one by n times the cohesion of that in the prototype and actual cohesion of say c u the both you know the stress strain behavior is drastically different for this can be explained through an example let us consider that in a prototype we are actually having a 50 kilo Pascal's of cohesion and we are actually trying to reduce this by say five times that means that a cohesion of a soil in the model is 10 kilo Pascal's. So the stress strain behavior of a soil under undrained conditions for you know yielding a 10 kilo Pascal's of cohesion and 50 kilo Pascal's of cohesion is different. So this implies that the change in soil behavior which says that this implies that the change in soil behavior and the response of a model will not be you know as similar as that in the prototype. So the change in soil behavior cannot be you know accepted but one thing you know in the in this particular you know approach the scaling down of CA implies that the change in soil behavior. But another option is that we have discussed in buried conduit example is that making gamma n times that means that the soil the self weight of the soil n times heavier that means that gamma m is equal to n gamma p and the maintaining identical cohesion as that in the prototype. If we are able to achieve a situation wherein gamma m is equal to n gamma p and Cu is identical in model under undrained. Even with that also what we are saying is that the Cu by gamma term is 1 by n times that of the Cu by gamma term is reduced. So with that we can actually say that Cu by gamma h in model prototype will be satisfied in that situation what is actually we are doing is that if you are not able to reduce the cohesion for satisfying the similarity point of view then another option another viable option it appears to be as that by enhancing the unit weight of the soil gamma m is equal to n gamma p with that we can actually maintain Cu by gamma h in model prototype identical. So this is explained in this particular slide. So what we said is that for a small scale model if h is reduced and gamma remains unchanged the strength of the soil must be reduced in the same proportion. So this implies that what we are actually discussing the change in soil behavior. Another option what we actually said is that the enhancing gamma, gamma is nothing but if you define this term gamma unit weight of the soil as rho into g where rho is nothing but the mass density let us say in this case of soil mass density of the soil and g is nothing but acceleration due to gravity. So rho which is mass density is a function of packing of particles for a soil or is a basic property of a metal and g can be changed from one celestial body to another body. By using this particular concept and assuming that you know the variation of g is possible we have examples like on the moon acceleration due to lunar gravity is about 20% of the acceleration due to gravity at the surface of the earth. Similarly the Jupiter g on the Jupiter planet is 33% more than the earth. So with that allied thinking we say that this gamma m is equal to n gamma p is possible where gamma is equal to rho into g where by maintaining identical mass density as that in the prototype that is rho m is equal to rho p and gm is actually say increased by n times we can say that gm is equal to ngp with that gamma model and prototype the scale factor for unit weight gamma m is equal to n gamma p is possible that means they provided the g sustained g is actually induced to the model which is in this example a slope in a cohesive soil it makes actually possible for us to achieve that gamma m is equal to n gamma p condition. With that what we said is that the dimensionless term Cu by gamma h in model and prototype will be identical without changing the soil properties. So that means that when you are actually having you know if any parameter which is actually influence in the model and prototype this you know it is governed by a relationship called rm, rm is nothing but physical quantity in model, rp is nothing but the physical quantity in prototype, lambda is the proportionality constant. So the relationship so the physical quantity can be time, velocity, acceleration or force or it can be stress, it can be strain so it is established by you know the relationship need to be established for each and every variable so that for a similitude to achieve between model and prototype which is not scaled which is not tested at 1 is to 1. Now let us consider another example which is you know let us say circular footing on sand. So if you look into this we have got a footing of diameter D resting on the sand having you know dry sand and gamma is the unit weight of the sand and E is the void ratio and phi is the angle of inter particle friction between sand grains that is the angle of internal friction is pi, sigma c is nothing but inter particle cohesion between the sand grains. Then the sand grains interact with each other at the inter particle cohesion generated between the sand grains is indicated by sigma c, sigma g is nothing but the crushing strength of the grain material that means that depending upon the composition of the grain they have different crushing strengths. So sigma g is nothing but the crushing strength of the grain material similarly depending upon the type of grain material we have different modulus of elasticities so E g is nothing but the modulus of elasticity of grains. Now by taking sigma c, sigma g and E g as the material properties of the you know the grains which are actually involved we can and then D g is nothing but the average particle size. So now if you look into it if you are actually having a situation of modeling this at 1 is to 1 scale and by for a given value of delta by D we and by using Buckingham we can get P by gamma D is equal to function of E phi sigma c by gamma D sigma g by gamma D E g by gamma D and D g by D. So here this is the particle size and this is a particle size to the diameter and E g by gamma D and that is nothing but the you know modulus of elasticity and unit weight of the soil multiplied by D diameter of the footing similarly we have got this terms like this. Now for P by gamma D to be model and prototype to be same what it says is that D g by D in model to be equal to D g by D in prototype similarly this pi term has to be same in model and prototype and this pi term has to be same in model and prototype and this pi term that is sigma c by gamma D in model to be same in prototype and friction angle in model to be prototype and void ratio the particular arrangement to be same in model and prototype. Now we can actually say that if you are modeling the situation in 1 is to 1 and all these pi terms will be identical. But if you are having a 1 is to n model and with gamma m is equal to gamma p condition then what we have is that let us see what will happen when you wanted to compare these pi terms. Let us assume that we are actually could able to achieve brought the same soil and void ratio and friction angle are achieved then E model in prototype and phi model in prototype are same. But being these are material properties and by identical gamma but diameter of the footing is reduced by d by n. So this term is not equal to model sigma c by gamma d model is not equivalent to sigma c by gamma d prototype which is nothing but sigma c divided by gamma d by n. So this is different from you know what we are actually need to be obtained. Then similarly sigma g by gamma d similarly E g by gamma d similarly d g by d. So if you look into is for similarity if you look in the deviation 1, 2, 3, 4 pi terms are deviating from the similarity that means that we are actually having very weak similarity as far as similarity is considered with gamma m is equal to gamma p for 1 is to n scale model. So in a similar situation let us say that by you know enhancing the gravity if you are able to impose a condition for this similar problem gamma m is equal to n gamma p. If you are able to do gamma m is equal to n gamma p for a small scale physical model which is 1 is to n is reduced by 1 by n times then what we can see is that again with the same assumption of so with sigma c sigma g and E g as material properties we can say that gamma is increased by n times and d is decreased by 1 by n times sigma c divided by n gamma divided by d by n. So they get cancelled sigma c by gamma d in model and prototype will be identical and sigma g by gamma d in model and prototype will be identical similarly E g by gamma d in model and prototype will be equivalent. But here we are going with an assumption that identical soil as that in the prototype that means that the particle sizes are constant and we have reduced in this case also d by n. So here what we have is that 1, 2, 3, 4, 5 pi terms are identical so the strong similarity is achieved except one pi term that is dg by d by n. In case of gamma m is equal to n gamma p also we are not able to in a capability not able to scale down the particles that lead to an effect what we call and we are going to discuss is called particle size effect. So what we need to do is that how this particle size effect can be eliminated that is what in the dimensionals is if any parameter or any variable or any dimensionals product deviates from the similarity then we have to see by maintaining by satisfying certain conditions how that particular parameter is insignificant in influencing the particular phenomenon let us say for making p by gamma d in model and prototype identical. So with those this discussion we understood that circular footing on sand and slope in coju soil and buried flexible conduit examples what we said is that by for a small scale model which is not tested at 1 is to 1 but which is tested at 1 is to n it implies that gamma m is equal to n gamma p condition fulfills and has actually got the superiority over gamma m is equal to gamma p and 1 is to n scale 1 is to n physical model. Now in order to enhance gamma m is equal to n gamma p there are also some attempts late 1960s by Jelikson. So this was actually introduced based on the total stress is equal to effective stress plus pore water pressure. So if you look into this you know the change in the geostatic stresses with the flow of water to the soils are possible. So introducing the hydraulic gradient symmetry method so here let us consider if water is actually flowing to the soil we know that it exerts drag forces called seepage forces on individual grains of the soil and the presence of seepage forces which causes changes in the direction of flow will cause changes in the pore water pressure and effective stresses in the soil. So by using this concept Jelikson 1969 has come out with a method called hydraulic gradient symmetry method which actually has got you know as a possibility that gamma m can be maintained as n gamma p by doing that what actually we get is that the stress and the stress strain behavior of the soil can be maintained identical as that in the prototype. So let us consider the two cases the case one is that you know hydrostatic condition when no flow takes place in this particular condition as there is no head of loss that is delta h is equal to 0 because the no flow condition then we actually have this is called hydrostatic condition where you have got you know this is the total stress and this is the pore water pressure and this is the effective stress where effective stress is nothing but gamma dash h at this particular point and this particular point it is 0. So here also in this level it is 0 and then it is 0 at this point and then here at this point the ordinate is gamma dash h so no flow head loss delta h is equal to 0 and no change in the effective stress. So this is you know the case one case two you consider where this limb is actually brought down by h and if you look into this here because of this the water flows from you know downward direction. So the water flow is actually shown here in this case and h w is the height of water column and h is the thickness of the soil sample. So we can actually write the total stress as you know nothing but sigma which is here the total stress and the depth here we can write gamma w h w and gamma w h w plus gamma sat h but here what we can actually write is that this is gamma w h w because water flows from this point to this point there is a head loss which actually takes place the pore water pressure drops by gamma if the no flow condition what we have got is gamma w is equal to h w plus h. Now but it is dropped by a term which is nothing but minus h that is nothing but gamma w into h w plus h minus h. Now by taking total stress by computing for effective stress that is total stress minus pore water pressure what we get is that gamma dash h plus h gamma w. So at this particular point we can actually write gamma dash h plus h gamma w and where the h is nothing but the head drop that is over a length h. So hydraulic gradient which is actually you know if you it will be something like a triangle which is actually having an ordinate h here and over a height of h the slope of that triangle is i, i is nothing but h by h that is small h by capital H. So for h is equal to substituting i times capital H we can write gamma dash h plus i h gamma w. So if you see the downward flow of a water increases effective stress in soil. Similarly in case 3 let us say if this limb is taken upwards then what we see is that effective stress decreases the down so we buy this in the hydraulic gradient symmetry method this particular that increase in the effective stress in the direction of flow is taken as you know as a concept and then this particular method is developed. So further developing on this method so we know that in the hydraulic in the in geotechnical engineering a small scale model test are sometimes used to study the complex nature of soil response and soil structure interaction. But it is well known that now the soil response depends upon the level of effective stress within the soil especially for granular materials. If you look into that for granular materials the soil response depends on the level of effective stress within the soil mass. So granular materials at a given density can at different stress levels behave in either as a contractive or a dilatative behavior. So they can actually have contractive behavior or they come are dilatative behavior. So the small scale model test conducted 1G often fail to reveal or represent the phenomenon that may exist at the prototype stress level. So that is what actually we have discussed it from the problems like buried flexible conduit or from the footing resting on the sand. We actually have said that the small scale model test conducted at 1G often fail to reveal or represent the phenomenon that may exist at the prototype stress levels. To overcome this it is desirable to perform small scale model test at field stress conditions and for that what we said is that gamma m is equal to n gamma p is required. So let us say that if you are actually having at a depth h we are having say unit weight of soil say gamma then the prototype we are having vertical stress at a depth h is total stress is nothing but sigma v is equal to gamma h. Now if the same depth which is actually model in 1 is to n scale model with gamma m is equal to gamma p and hm by hp is equal to 1 by n the stress there is sigma v model is equal to gamma h by n that is 1 by n times smaller than the stress which is actually there in the prototype. Now the same situation when it is say modeled with gamma m is equal to n gamma p or nothing but gamma is equal to rho ng condition with that what we can actually get is that with gamma m is equal to n gamma p and hm by hp is equal to 1 by n we get sigma v in model is equal to that is gamma m that is n gamma p into h by n which is nothing but gamma h. So what we actually say that the stress in model in prototype will be identical if you are actually having you know gamma which is actually n times the gamma in prototype. Now we will see how that is actually simulated by Jelikson. So Jelikson proposed method that employs a high hydraulic gradient within the granular soil to create a high body force and high stress levels approximating the field conditions. So we have discussed that self weight forces and the seepage forces they are treated as body forces and the based on that concept and you know the direction what we actually made sigma dash is equal to gamma dash h plus gamma w h with i is equal to h by h what we have what we can say is that sigma dash is equal to gamma dash h plus i gamma w h. So here with the sigma dash is equal to gamma dash h plus gamma w h and by substituting i is equal to h by h what we can write is that gamma dash h plus i gamma w h by dividing throughout by h we can write sigma dash by h is equal to gamma dash plus i gamma w. If you look into it the effective unit weight sigma dash by h is called as effective unit weight is increased by term i gamma w, i gamma w is nothing but the seepage force magnitude seepage it is increased by the term of which is equivalent to the seepage force magnitude in the direction of the flow. So this equation one implies for a model that is subjected to downward vertical gradient i that the effective unit weight of the soil will be increased by seepage force of the magnitude i gamma w. So this is from the basis what we actually discussed we what we have reduced we said that the sigma dash by h is written as gamma m that is the effective unit weight of the soil which is increased by the seepage force term that is i gamma w. Now by writing sigma dash by h is equal to gamma m is equal to gamma dash i gamma plus i gamma w and now we can write the unit weight scale factor and hydraulic or hydraulic gradient scale factor as gamma m by gamma p, this is gamma p. So gamma p is equal to total or submerged unit weight of the soil depending upon the ground water conditions of the prototype. So gamma p is nothing but the total or submerged unit weight of the soil depending upon the ground water conditions in the prototype and for saturated condition it can be gamma p is equal to gamma dash p. So from scaling laws the model will simulate a prototype structure scale n is to n where bm by bb is equal to 1 by n. Also from scaling laws we can say that the model will simulate a prototype structure scale n is equal to n where bm by bb is equal to 1 by n. When a 1 by n scale model test is performed under hydraulic gradient scale factor or unit weight scale factor n capital n is equal to small n the stress due to self weight of soils at homologous points of model prototype will be same. So when a 1 by n scale model test is performed under hydraulic gradient scale factor or unit weight scale factor capital n is equal to small n the stress due to self weight of soils at homologous points of model and prototype will be same. So that means that sigma v in model is equal to gamma m z m where when we substitute the gamma m that is the effective unit weight of soil as gamma dash plus i gamma w z m we can write as by using gamma m by gamma p is equal to n we can write that n gamma p into z p b n there is nothing but sigma v in model is equal to sigma v in prototype. So if the same soil is tested in the model as well as in the prototype and the same stress path is followed the strain in the model prototype will be expected to be same that is epsilon m by epsilon p is equal to 1 while the displacements of the prototype will be larger than the model by a factor n is equal to capital n. So with that what actually implies is that by maintaining a hydraulic gradient which is you know by higher with differential pressures between top and bottom of the soil we can actually ensure that identical stresses as that in the prototype. So in this slide we can actually see that a model with a soil sample of height l m say 0.3 meter is subjected to water pressure difference of say delta p is equal to 300 kilo Pascal's p t is the pressure on the top of the soil p b is the pressure on the bottom of the soil so p t minus p b delta p is equal to 300 kilo Pascal's. So if that delta p is caused due to the head difference of h then i is equal to we can write that let us say h is the head which is you know equivalent to that 300 kilo Pascal's of differential pressure and h plus l m divided by l m is nothing but the hydraulic gradient. So what we can write i is equal to 1 plus h by l m and we can write for h delta p by gamma w that is the differential pressure by gamma w is the weight of water into multiplied by l m is nothing but the length of the sample. So we can get that i is equal to 1 plus 300 divided by 10 into 0.3 where gamma w is equal to 10 kilo per meter cube and it is 101 that i hydraulic gradient is about 101 with gamma w is equal to 10 kilo per meter cube and gamma p is equal to gamma saturated is say 20 kilo per meter cube we can write and we can get n as 10 plus 101 into 10 divided by 20 with that we will be able to get 50 scale factor that is implies that if these conditions are maintained if the differential pressure of p t minus p b as 300 kilo Pascal's is maintained for a with gamma m is equal to gamma p is equal to 20 kilo per meter cube we can say that l p is equal to 50 into 0.3 that is 15 meters of the length of the soil which is actually represented in the field. So this is interesting technique wherein this is possible and similarly when with gamma p is equal to gamma dash is equal to 10 kilo per meter cube with n is equal to gamma p is equal to gamma dash that is with 10 kilo per meter cube under submerged unit weight conditions it actually says that it is 100 is equal to 100 that means it is about 50 meters. So but the relation n is equal to gamma m by gamma p if you write that gamma dash plus i gamma w in model gamma dash plus i gamma w in prototype for all practical purposes gamma w is equal to gamma dash that is you know the submerged unit weight and unit weight of water we are most identical and i is equal to 0 in the prototype then in that case we can write n is equal to 1 plus i is equal to 1 plus delta p by l m into gamma w. So this is the expression what we actually use for reducing the hydraulic gradient method. So in the hydraulic gradient similitude method which can be applied for granular soils particularly here you know the conditions are that you know where we can actually do the applications like testing of footings on sand and particularly testing of footings on sand with either concentric loads or eccentric loads or recently it has been applied for testing of piles in calculating the uplift capacity of pile in the soil embedded in sandy soil saturated soil profile where in you know so here one of the demerits is that you know the control of this particular you know delta p differential pressure need to be maintained and second thing that the surface has to be horizontal under those conditions this possible that if you are able to do then it actually simulates and satisfies the condition that gamma m is equal to n gamma p and identical stress in model prototype can be achieved. So the applications involved you know can be applied for piles in embedded in sandy soils or footings resting on sand or anchors embedded in sand particularly to the vertical or inclined pullout. So these are the possible applications of this technique and this technique actually has got merits and demerits in its application. So this you know technique with this particular limitations not extended further but the technique what we have discussed is that by manipulating the changing the G that enhancing the G and reducing for a small scale model and thus that technique what we have defined or what we have named as the centrifuge physical modeling centrifuge based in physical modeling. So the centrifuge based in physical modeling technique evolved as very powerful tool for geotechnical engineers for understanding the geotechnical behavior of structures. So centrifuge based physical modeling technique is a physical modeling technique in which a small scale model that is reduced by one is to n subjected to a rotation about a vertical axis in a horizontal plane. By doing this then identical stresses in model prototype can be maintained by with that the stress fields can be maintained as those in the prototype. This actually is possible to understand and you know the behavior of the number of geotechnical structures.