 So welcome to course on advanced geotechnical engineering module 7 entitled geotechnical physical modeling and in this module we are actually going to discuss about the centrifuge base of physical modeling mechanics of centrifuge base of physical modeling. In addition to that we will actually try to solve a problem which we have actually based on the discussion we had in the previous lecture on hydraulic gradient simulation method. So this is module 7 lecture 3 on geotechnical physical modeling. So after having discussed the hydraulic gradient similitude method we said that with the help of by maintaining differential pressures between top and bottom of the soil sample predominantly saturated granular soils there is a possibility that we can actually replicate the stresses as that in the prototype and in connection with that let us look into the example wherein uplift behavior of a pile in saturated sand profile that is that in this particular example we are considering a pile of certain length embedded in a saturated sand deposit when it is subjected to uplift. So these piles can be subjected to uplift like for example if they are happen to be the foundations of transmission line towers for example the river bed in a saturated sand profile can do exist then the piles can be subjected to uplift or what we say is also pull out. So this we would like to understand to investigate through hydraulic gradient similitude method. So for that how we need to model the uplift capacity and how we have to get the bond stress or the shear stress around the pile soil interface. So let us look into the solution like this. The problem here before us is uplift behavior of pile in saturated sand profile. So here this is the pile of diameter D this is the ground surface and this is the length of the embedment when the pile is actually subjected to uplift when the pile is subjected to uplift then the first resting force will be the self weight of the pile and on the periphery of the pile there will be shear stresses which are actually acting in the downward direction because the uplift will be in the upward direction. So along with the self weight the shear stresses will be acting in the downward direction and they will be along the length of the pile and they will be along the periphery of the pile. Now we can write by using relationship like tau z is equal to sigma h z tan delta. This sigma h z is nothing but the horizontal stress which is acting at a particular depth z below the ground surface. Sigma h z is nothing but a horizontal stress acting at a particular depth that is z below the ground surface and delta is the pile soil interface friction angle. This is for in case of uplift generally it will be on the lower side. So the delta is the interface friction angle. Now we have relationship with sigma h z is equal to k times sigma v z, sigma v z is nothing but vertical stress at a particular depth z. By with the substitution of by substituting sigma h z is equal to k times sigma v z we can actually substitute here then we get tau z is equal to k sigma v z tan delta. k is coefficient of earth pressure this depends upon the type of the pile which is installed. If it is like precast driven pile it will have a different value, it can have higher value. If it is a bored castings to pile it can have low value because the soil surrounding the pile is not stressed because of that it can have low k values. But in case of a precast driven pile which is driven into the sandy deposit there is a possibility that soil surrounding the pile will get stressed because of that there is a possibility of higher k values. So now we can write that sigma v z is equal to gamma m into z where gamma is nothing, gamma m is nothing but effective unit weight of the soil. So we have defined in hydraulic gradient symmetry method gamma m is equal to gamma dash plus i gamma w this i gamma w term is actually obtained from the so called hydraulic gradient. If you are actually having a hydraulic gradient of let us say 50 or 100 depending upon the definitions whatever we had that means that it can be possible that the gamma m can be increased by 50 times the gamma in the prototype. So by writing sigma v z is equal to gamma m into z we can write gamma dash plus i gamma w z. So by substituting this here in this we can write tau z is equal to k gamma dash plus i gamma w z into tan delta. So what we have done is that we substituted for sigma v z is equal to gamma dash i gamma w into z. Now let us see further now what we do is that in order to get simulation between model that is in hydraulic gradient simulation method and corresponding tau z in prototype what we do is that we take a ratio of tau z in model tau z in prototype is equal to from the previous expression we have k gamma dash i gamma w z tan delta. So what we have done is that we have given written in the model dimensions and prototype dimensions m's of m indicates the model and model means especially here in hydraulic gradient simulation method and p means prototype or equivalent full scale structure. So k m by k p gamma dash plus i gamma w model gamma dash plus i gamma w suffix prototype z m by z p tan delta m by tan delta p. Now with k m is equal to k p assuming that all the stress history and other aspects have been taken care and because of the reduced scale model the z m in model is z p times n that is z m by z p is equal to 1 by n and when we resemble this when we have identical stress histories and assuming that the friction angles interface friction angles in model prototype are identical that is tan delta m by tan delta p is equal to 1 and gamma dash plus i gamma w model divided by gamma dash plus i gamma w p suffix p is equal to n. So this is by virtue of the hydraulic gradient simulation method by substituting this in this particular expression what we get is that this is n times this is 1 by n times this is 1 and this is 1 with that tau z in model to tau z in prototype is equal to 1 tau z in model by tau z in prototype is equal to 1. So this indicates that the interface stresses in model and equivalent full scale structure or the pile in saturated sand profile with identical stress histories is found to be identical. Now let us see after having deduced the scale factor for interface stress in model prototype let us see how we can actually get the uplift p uplift that is nothing but the uplift force is nothing but w p plus pi d w p is nothing but the self weight of the pi d 0 to l that means that you know this varies from top of the pile to bottom of the pile tau z d z. Now by integrating this what we get is that pi d l because tau z z and then when you take the limits here pi d l, pi d l is nothing but pi d is the perimeter and l is the length of the pile into tau z. Now neglecting here as we said that one of the limitations of hydraulic gradient similarity method is that the self weight of the pile is not enhanced in the model it will be smaller but the self weight of the pile is not enhanced. So with that neglecting this term then p uplift in model and prototype we can obtain as 1 by n square with tau z in model is equal to tau z in prototype that is we have deduced just now wherein we said that by simulation of gamma m is equal to n gamma p gamma m is the effective unit weight of the soil to n is nothing but which is induced because of the simulation of you know higher hydraulic gradient and gamma p is nothing but the bulk unit weight or submerged unit weight of the soil then with tau z in model is equal to tau z in prototype and d m by d p is equal to l m by a p l p is equal to 1 by n and neglecting w p neglecting w p says that p uplift in model and prototype equal to 1 by n square that means that if you have let us say a pile which is actually taking for example 1200 kilo newtons which is an hypothetical uplift load but if you are trying to look into let us say n is equal to 50 that means that you require about 1200 newtons that is about 120 kg is sufficient for you to impose that type of higher loads. Here what we have deduced is that by using this solution we try to reduce the scale factor for the tau z which we found that it is identical as that of in the prototype then the uplift capacity of a pile in kilo newton is scaled as pi p uplift in model is equal to p uplift in prototype divided by n square so this is how you know one need to you know develop the scaling loss for the different applications of this you know in the modeling particularly the modeling which is different from a full scale that is when you do in a small scale physical modeling one is to n. Now after having discussed about the method and also discussing about you know a problem an example problem in fact the problem which we have just now discussed is after Lichensky et al 2012 and the application of the hydraulic gradient simulated method some of the selected application areas are actually given here basically to study the load settlement behavior of circular footing or square footings resting on sand and to study the response of piles in static and cyclic lateral loads embedded in sand at various stress levels. So this is in this particular study by Jan et al Jan 1990 what they have done is that they have simulated these you know these different stress levels by increasing the you know hydraulic gradients and then simulating the values from n is equal to 1, n is equal to 5, 10, 100 up to 100 they have done. So and then also they also studied the static response as well as the cyclic response of these you know piles and another application is that which is after Musso and Feralessy 2009 is the to study the collapse of a strip footing on sand or it is a dense sand under vertical eccentric loads. So to study the collapse of a strip footing on dense sand under vertical eccentric loads and as I told the recent example by this Litchensky et al is on the studying the uplifting uplift behavior of a piles embedded in sand. So like any technique we have limitations of the hydraulic gradient simulated method and these limitations some of the limitations are listed here but what need to really understand about or ensure that identical or uniform distribution of hydraulic gradient and sustained you know holding of these pressures is very important and it can be this is base can be used to model only the behavior of saturated sand and uniform fine sands or uniform silty sands tend to yield more uniform gradients across the soil layers. So if there is you know some non uniformity the ensuring the uniformity in the hydraulic gradients applied is questionable and or may not be consistent. So uniform fine sands or uniformly graded fine sands or uniformly graded silty sands tend to yield more uniform gradients across the soil layers and they are trying to give the more consistent results and the hydraulic gradient technique was found to be particularly sensitive to the distribution of hydraulic gradient in the soil profile. And more details are actually given by Yann 1990 where in they have actually discussed about the you know different issues as far as the you know with the distribution of hydraulic gradient between the soil profiles are concerned. And this hydraulic gradient method is limited to model with horizontal ground surface only mostly the geostatic stress conditions that is with the horizontal ground surface only can be done and it cannot account for the self-weight of the foundation materials like footing or pile okay. So in case of uplift particularly you know it actually matters as far as the compressive load behavior is concerned we generally not will not consider but in case of uplift one of the you know remedial measures for increasing the uplift capacity of the pile is to increase the size by increasing the self-weight of the pile. So except that you know but some selected you know limitations whatever has been highlighted here. The hydraulic gradient technique this is presented here is a viable and inexpensive scale modeling approach that may be useful for simulation of variety of geotechnical engineering problems and without you know altering you know the gravity level one can actually ensure the identical stresses in model and prototype. So in conclusion what we can say about the hydraulic gradient technique is that is a you know viable and inexpensive scale modeling technique where identical stresses as that in the model as that in the prototype can be you know simulated in the model and it is useful for a simulation of variety of geotechnical engineering problems. So the more research is warranted in this particular area where say large size models can be developed for understanding different types of problems under identical stresses as that in the prototype. So after having discussed about the different modeling techniques like we said that we have theoretical models, mathematical models and numerical models and in this particular the flow chart which is actually given here we have situation like sometimes from the mathematical model full and simplified and what we tend to do is that we tend to do the prototype actual field problem will be studied. And so it is actually like a direct modeling of prototype and if the prototype or a full scale structure is existing and which needs to be validated then one of the alternative what we said is that the scaled physical model that is which is different from a full scale physical model one of the viable options in a part from you know the hydraulic gradient simulate method what we said is that the centrifuge based on physical model. So here the centrifuge based physical model which is actually shown here so it can be helpful for modeling some prototypes or modeling some prototypes which are non-existent suppose if you are actually talking about a new phenomenon and that you know involves you know an assumption of a fictitious prototype and it actually helps us to you know understand and simplify this mathematical models and basically learning the mechanism of transport process verification and improvement of the theory. So with that what will happen the knowledge which is actually gained from this scale physical models or of the actual prototypes or the fictitious prototypes it actually helps to understand the learn mechanisms basically the mechanism of the you know behavior can be you know ascertained. So you know before going into the in-depth understanding about the centrifuge based physical modeling let us understand that you know the centrifuge based physical modeling is a physical modeling technique in which a body is rotated about a vertical axis in a horizontal plane. So when this is in the process of rotating about a vertical axis in horizontal plane there is a centripetal acceleration acts towards the center which is actually shown here for a body which is actually having mass m a centripetal acceleration acts toward the center and v is the tangential velocity and omega is the you know the angular velocity about a vertical axis. So earlier you know when investigating this when putting forward this technique many alternatives like vertical rotating about a vertical axis or rotating about a horizontal axis you know these aspects have been studied but now you know it is well established that you know this is object moving in a study circular orbit in a horizontal plane about a vertical axis is you know is the point of discussion. So here if a body of mass m like which you have shown in the previous slide is rotating at a constant radius r that is the radius from the center of the shaft to the within the body about an axis with the study speed v then in order to keep it in that circular orbit it must be subjected to a constant radial acceleration that is v square by r or it is equal to r omega square which is acting towards the center. So in order to produce this acceleration which is actually acting towards the center of the shaft or a about a vertical the axis which is rotating the body must experience a radial force that is m r omega square directed towards the axis. So by normalizing the centripetal acceleration with earth's gravity that is g and state that the body is being subjected to acceleration of n g where so what we have done is that the so called m r omega square divided by m g when we normalize we get n g is equal to r omega square. So this is the fundamental equation which we use in centrifuge based physical modeling. This helps if you are actually having a certain angular velocity omega when it is when we wanted to say this implies that if you want the higher values of n the increase the angular velocity. So by increasing the angular velocity there is a possibility that you will be able to get higher values of the n. So it is a relationship which will actually will allow us to select a required amount desired angular velocity for modeling let us say n is equal to 10, n is equal to 50 or n is equal to 100 or n is equal to some 500 and basically this is further explained it is like if you consider a string attached if you consider a string attached to the sphere having certain weight. The string experiences tension and that can be felt by the hand holding the other hand. So the tension in the string applies the centripetal force on the sphere pulling it toward the center of rotation. So the tension in the string applies the centripetal force on the sphere pulling it toward the center of rotation. So this force must be balanced by another force otherwise the sphere would move toward the center. So otherwise what will happen is that suppose if you are rotating about a certain point and at a certain radius r and it is subjected to you know the force of mr omega square central you know the centripetal force. So the force must be balanced by another force otherwise what will happen is that the sphere would move towards the center. So this is can be explained by means of a frictional or inertial force that acts on the sphere radially outward. So the force must be balanced by another force otherwise the sphere would move towards the center. So this is according to Madhavushi 2014. So this is actually explained that the force must be balanced by another force. So otherwise the sphere would move toward the center. This can be explained by means of a you know that inertial force that acts on the sphere radially towards. So you know that is how you know the so called the centrifugal acceleration what we actually get from this Newton's third law and we actually realize that in the centrifugal based physical modeling. By Newton's third law the centrifugal force and the centripetal force must be equal and opposite. The centrifugal force causes a centrifugal acceleration that acts radially outward. In the simple example of the car going around a corner we can say that the tires of the car will exert centrifugal forces on the road acting radially outward. So the centrifugal forces are exploited in a wide variety of engineering devices such as centrifugal pumps and centrifugal clutches. So this is an extension which we actually used in you know geotechnical based centrifuge modeling. So before going into that let us consider you know from the mechanics point of view rotation of a rigid body about a vertical axis. Here what we see is a vertical axis which is AA dash and this is the horizontal plane about which the P is rotating about a vertical axis in a horizontal plane. And at point P consider a rigid body which rotates about a vertical axis AA dash and this is in x direction, this is in y direction, this is in z direction and this is the angle of velocity. So omega is equal to theta dot k and this is in j vector and this is in i vector. And this point P and this vertical axis in this plane that angle subtended is pi and when it moves from this place to this place this angle which is subtended by the moving particle P or body is theta. Now when you consider the rigid body which rotates about a fixed axis AA dash and let B be the projection of P on AA that means that B be the projection of P on AA dash. What we can write is that PBE is equal to r sin pi, r is the position vector that is OP is the position vector. So we can write that PBE is equal to r sin pi, PBE is equal to r sin pi. So for a small displacement along this thing the small length delta S arc length delta S can be written as PBE into delta theta. Delta theta is the small angle which is subtended by moving from point 1, point 2, from this point to this point let us say. So delta S is the small distance along this curve. So with that what we can actually get is that delta S is equal to PBE into delta theta where V is equal to dS by dt is equal to r delta theta by delta t into sin pi. What we have done is that we have divided delta S is equal to PBE into delta theta by both sides by delta t and when delta t tends to 0 we can write V is equal to dS by dt is equal to r that is PBE and delta theta by delta t into sin pi. So this is the scalar product of a vector which is acting perpendicular to the plane of axis AA dash and OP. That means that this triangular if you take a plane and into that plane this vector tangential velocity is acting in this direction. So that means that we can write that V is equal to omega cross r, omega cross r, omega is the vector along this is equal to theta dot k is the vector along this axis and r is the position vector of this. So V is equal to omega cross r then in order to for determining the acceleration of the particle P what we do is that we differentiate this V is equal to omega cross r the cross manipulation of omega cross r then we get the dV by dt is equal to d by dt into omega cross r that means that d omega by d cross d omega by dt cross r plus omega cross dr by dt. So here we see that the acceleration is divided into two components so d omega by dt we can write it as you know this is nothing but angular acceleration rate of change of angular velocity change in angular velocity so it is nothing but alpha cross r is equal to omega cross dr by dt. So here what we can do is that this dr by dt we can write it as V and omega cross V that V can be substituted by omega cross r with that we can get is that alpha cross r plus omega cross omega cross r. So here the first component is tangential component and omega cross omega cross r is the normal component of the acceleration. So here alpha cross r which is angular acceleration and you know the position vector again it is acting in the direction of tangential velocity because when you take the projection of a a dash to the point P and with the position vector which is there and it will be perpendicular to that plane then you know it will be acting towards the you know in the direction of tangential velocity whereas the normal component where we have the central axis a a dash and the vector which is perpendicular to that omega cross r is V which is the tangential velocity vector and the perpendicular to this plane that means that it is acting towards the center that means that we have omega cross omega cross r which is actually nothing but a centripetal acceleration term. Now here as far as the centrifuge based physical modeling is concerned and if you are actually increasing the omega let us say uniformly accelerated motion then that means that the both tangential component as well as the normal component of the you know accelerations they do exist and but if you are having a constant angular velocity that means that immediately after returning certain speed and if you are able to maintain the constant angular velocity in that situations that At is equal to 0, At is equal to 0 means that the constant angular velocity has been attained. So as far as the relevance of centrifuge based physical modeling is concerned At is equal to 0 for a test which are actually conducted at constant angular velocity and for uniformly increasing G level At is present that means that if you are actually increasing that means that this can be a very short duration as far as the equipments are concerned but relevance to centrifuge based physical modeling is that if you are actually having a constant angular velocity that tangential component will not exist that is 0 and for uniformly increased G level At is present and both At as well as air will be present. So let us look into the you know simple you know after having discussed about the mechanics of you know the centrifuge modeling let us look the principle of centrifuge modeling with an example of a slope which is having saturated undrained clay. So it is to basically the principle is to generate the similitude of stresses and strains between models and the real or hypothetical prototype by simply the increasing the acceleration reveals proportional reduction in the linear dimensions that is what we have said in the previous slides that in order to increase the acceleration we have a body rotating about vertical axis in a horizontal plane. So this can be explained through a slope stability problem let us assume that we are having a stability of a slope which actually found to have the factor safety of a slope found to have dependence on let us say a slope inclination as well as you know has theta as well as the undrained cohesion as well as gamma and if it is having certain depth below the toe then d and height of the slope. So when we do the dimensionalities and when we you know apply the Taylor's theory like factor of safety is found to function of theta c by gamma h and d by h and for slopes which are actually having inclinations more than 53 degrees what we say is that the factors the slope failure only tend to occur in that case the factor of safety is found to depend only on the theta and u by gamma h term. So we said that the first simulation of model in a similarity for ensuring similarity in model prototype that means that we have identical factor of safety in model and prototype what we said is that theta in model to be equivalent to theta and prototype. So that can be achieved by somehow by means of ensuring the identical configuration of the slope but c u by gamma h to be model and prototype that means that gamma which is nothing but the rho times rho s time g rho s is nothing but the mass density of the soil. So at failure a factor of safety is equal to 1 what we can write from the Taylor stability number is that ns is equal to c u by gamma h where c u by rho s g h. So you know this you know when you expand further in order to model the slope failure for 1 is to 50 1 by 50 physical model at 1g it becomes necessary to produce a material having c u by gamma is equal to 1 by 50 for ensuring identical stability number in model prototype what it says is that when h is reduced to 1 by 50 times and in order to have c u by rho s g h in model prototype to be identical that is the stability number in model prototype will be equal means the identical factor of safety that is equal to at failure factor of safety is equal to 1. In order to ensure that it says that if you are actually talking about 1 is to 50 scale model at 1g it says that c u by gamma has to be equal to 1 by 50 that means that it implies that if you are you know if you are not making the soil weightless then you know when the alternative is to have very very low cohesions. So if you are actually having you know let us say c u of 50 kilo Pascal's in the prototype and this means that at 1 is to 50 a c u of 1 kilo Pascal's that means that almost you know above equal limit you know the soil strength is almost 0. So it is almost impossible to produce material with c u by gamma is equal to 1 by 50 and you know moment we remove the scaffolding for the what we maintain for achieving the sloping inclination there can be a failure. So for a 1 by 50 physical model at 50 g let us say and the ns will be in models stability number in model prototype are equal because here what is actually happening is that the c u by gamma is reduced by 1 by 50 but by ensuring identical c u in model prototype what has actually happened is that the gamma which is nothing but rho s into g and by is maintaining the same soil as that in the prototype and increasing the g by 50 times because we are able to induce 50 grouties what we do is that we rotate about a vertical axis in order to plane to you know induce a constant angular velocity. So with that what will happen is that for a 1 by 50 physical model at 50 g ns is equal to c u by rho s 50 g into h by 50 which actually ensures that c u by rho s g h is equal to c u by gamma h s h which is nothing but ns in model prototype ns and prototype are identical. So this is one of the classical examples which was used to you know bring out the you know merit of centrifuge modeling by using this particular example the slope stability of a slope of undrained undrained composed of undrained clay. So in this you know particular slide a slope with an inclination of beta which is actually shown here having i height h and the same when it is reduced by 1 by n times the slope height is actually is h by n times that. So this and this actually the correspondence will not be there because you know there is a vast difference of you know the because we are not able to reduce by c u by gamma by 1 by 50 the behavior is next to impossible but is the behavior is difficult to model but here in contrary to this if you see the contrary to 1g model what you can see is that by rotating a model about a vertical axis in a horizontal plane. So what you are seeing is the plan view of the you know model rotating about a vertical axis so here the model is reduced by 1 by n times but because of the increased you know centrifugal centripetal forces acting towards this by using Newton third law motion what we said that that frictional or inertial forces acting towards the model that is we take it as n is equal to ng is equal to r u omega square. So these forces they do exist when model is rotating about a vertical axis in a horizontal plane in order to keep that in circular orbit these centrifugal forces act in this direction and that is nothing but what we say ng is equal to r u omega square. So if you are actually having these constant angular elastic conditions what we can that the model slope in centrifuge would behave similar to that in the prototype that means that a correspondence between the model in particular this particular model and this prototype which is at 1g at full scale level can be achieved. So consider this with as far as the geostatic stress distribution as concerned let us say that we are having a soil profile having unit weight gamma and with a at a depth h in 1 is to 1 prototype or full scale structure the sigma v the ordinate here is gamma h where h is the you know the vertical height and for similar situations for same soil having gamma with 1 is to n 1g model where the stress is nothing but sigma v is equal to gamma into h by n. So here what we have is that very low stress compared to the stress which is actually existing in the prototype. But contrary to this when we have a model subjected to n gravities though the you know here for example the vertical ordinate is h and horizontal ordinate is gamma h. So here both vertical ordinate and horizontal ordinate have affected here in the vertical ordinate is reduced but the horizontal ordinate that is sigma v is identical as that in the prototype that means you can see that sigma v is equal to gamma n into h by n because n is now gamma is now n gamma with that what will happen is that sigma v is equal to gamma h. So this was also possible in hydraulic gradient simulated method by with the help of see page forces what we said that this can be achieved but we also discussed that you know different merits and demerits of the technique. So further we will try to explain the importance of you know the simulation of you know the stresses. So consider a shear stress shear strain behavior for loose and dense sandy soils let us say that we have got this relationship by testing a loose and dense sandy soils in a direct shear box. So this is the shear stress and this is the shear strain and this is for a dense sand and this is for the loose sand I can see that there is a hardening and then the softening behavior and here the continued hardening actually and then there will be peak will not be you know exhibited here. Now when we have got the model at small stresses and strains you can see that the stresses are very low because in the previous slide we said that the sigma v model is equal to sigma vp by n so what you can see is that the soil initially the stiffness is very high and the stresses are very low and but similarly when you look into this full scale structure behavior at large stresses you can see that the stiffness is low as well as the stresses are high. So here this when the soil is we can look into it is highly non-linear and plastic and full scale structure behavior at large stresses when large strains are mobilized. So creation of full scale stresses and strains in small scale physical models are very important and particularly when you wanted to model to capture the response of a particular full scale structure having embedded in certain type of soils it is very important to capture the identical stress strain behavior of a soil in the model as that in the prototype. So consider let us say in the centrifuge modeling let us see that you consider a model rotating about a vertical axis what we are seeing is you know the center of the shaft the model rotating about a vertical axis with a constant angle of velocity this is the top view. Now consider a small element dr having the thickness along the radial direction and this you know force acting on the area A the area perpendicular to the plane of the speaker is A and the force acting on this is F1 and F2 and F2 is more than F1 because the stress increases as we go down. So the stress increase must provide the force necessary to generate the centrifugal acceleration. So F2 – F1 that is a small df is equal to we can write it as dm into acceleration that is ar with reference to r from the center of the shaft measured from the center of the shaft to the center of the element. Further what we can write is that we can write dm as rho into dv where rho is the mass density of soil which we have used in the particular example. So where dv we can write it as dz into a that means that volume a small volume of that element having thickness dz and area A. And we can write the stress as dsv is equal to df by a where by substituting for df is equal to rho dz a by a into ar. So by using ar is equal to ng that means that when the model is rotating about vertical axis the n times gravity is when it actually exerted. So what we can say that sv is equal to when you integrate from 0 to z by n rho into ng dz when you integrate from 0 to z by n what we get is the rho dz. So that what we have discussed in the previous slide the stresses are identical at geometrically equivalent points in the prototype and centrifuge model provided the linear scale in the model is inverse of acceleration scale that is ng is equal to 1 by nL is equal to n that is also called as gl is equal to 1 condition lm by lp is equal to 1 by n and that nL is nothing but length scale factor which is lm by lp is equal to 1 by n and gm by gp that is ng is nothing but acceleration due to gravity scale factor which is gm is equal to ngp. So when you say that gl is equal to 1 that means that length is reduced to 1 by n times and acceleration due to gravity g is increased by n times. Now in this particular slide further the discussion is given wherein when you are actually having the element of soil at the surface of the earth that is in the prototype we have sigma v is equal to rho gz so similar condition is actually achieved with sigma v is equal to 0 to z by n rho ng dz that is rho gz. So consequently if the mechanical behavior of the soil is strongly dependent on stress level that such behavior could be correctly reproduced in centrifuge model that is what actually we have said is that if the mechanical behavior of the soil is strongly depend upon the stress levels then such behavior could be correctly reproduced in centrifuge based physical modeling. So in order to do this there are number of the equipments actually have been developed in the world they are actually the balanced beam and beam centrifuge and another type of equipment is drum centrifuge. In fact the initial attempts of this technique were reported in 1936 by Pokrovsky and Fedorov on the centrifuge based physical modeling a first paper the paper on this centrifuge modeling appeared in the first international conference on soil mechanics in 1936. And subsequently about after a long gap then 1970 onwards with the help of UK, USSR and other countries the further you know the work on this actually has come in and since then actually number of equipments have been developed and every 4 years you know the conferences on this centrifuge based physical modeling are held nowadays they are actually called as physical modeling in geotechnics. So we have actually two equipments one is called balanced beam centrifuge or beam centrifuge. So in this particular slide what we see is the you know a vertical shaft and this is called the arm or beam of this centrifuge and this is called the basket and initially at normal gravity when it is when the rpm is 0 the basket will be in this position but as the rpm increases as the rotation of this particular system takes place about vertical axis in the horizontal plane then the basket swings up what we call this is called swinging basket because it basket swings up and this angle from this point to this axis along the arm is actually called as the ramping angle and the time which takes to reach this horizontal level is actually called as ramping time as this is attached to the earth planet. So what we have is that virtually what we have is really the ng which is actually virtue of the rotation of the rigid body about vertical axis acting on both the sides and 1g which is actually downwards but when you look into the resultant effect of the ng as well as 1g this is nothing but the radial acceleration field is nothing but root over n square plus 1g but if you look into the higher values of n that n is equal to n dash g which is almost considered equivalent. So this is what we see is the balanced beam centrifuge where you have got two swing baskets the idea was to carry you know two different models in with a simultaneously but subsequently considering the power saving requirements and aerodynamic point of view this basket is actually placed by a counterweight and which is actually moves depending upon the weight of the model. See the purpose of this balanced beam and beam centrifuge is that the sigma mr is equal to 0 that means that if you are having a mass here m1 and r1 m2 r2 the sigma mr has to be 0. So this is you know a typical you know the balanced beam centrifuge then we also have another category of type of equipment called in the drum centrifuge but in the balanced beam centrifuge in this basket the container is placed and there are two types of containers one is called static containers other one are called you know for static test and which are rigid in nature and if you are actually doing experiments like dynamic experiments like earthquakes and other studies then there is a possibility that in order to prevent the reflection of you know the p waves and s waves there is a requirement that we have to use you know some sort of containers which actually simulate the identical motions as that in the prototype. So in the drum centrifuge in this particular slide what actually has been shown is that there will be a periphery drum and a channel will be fixed here and there will be a central shaft and a tool table. So here in the drum centrifuge what will happen is that the tool table and as well as the drum rotate about a vertical axis in a synchronized manner. So here with this what will happen is that is possible for us to do number of you know one dimensional problems and is also possible to you know simulate the elastic of space conditions in soil mechanics particularly when you take you know a pi d the perimeter which is the along the periphery of the drum then it is possible that you will be able to model you know large infinite soil extents on the surface of the soil and so here what we actually have is that a soil model which is actually created during flight either with clay or with the sand by special techniques wherein you will be able to you know get the you know this desired ground surfaces and also the stress stress and all those things can be matched and in addition also it also helps that you know in a given with identical stress stress when we have then it is possible that it can this technique this particular equipment can be used to do multiple experiments with identical stress stress. So the vertical stress in the radial acceleration field can be obtained now it is basically nothing but consider once again the rotation of the rigid body about the vertical axis where Z is the you know depth from the top surface so you can see that and consider an element having dZ in terms of dimensions Z you discuss previously we actually have said that dr and where R is the radius up till here and when you consider the Z it is nothing but RT plus Z into that element and RB is nothing but the radius up to the base of the model and RT is nothing but the radius up to the top of the model which is nothing but dr is equal to RB minus RT. So again by using this but in this is in terms of radial coordinates when we substitute here what we do is that d sigma V is equal to dF by A rho dZ A by A into AR so here by using AR is equal to r omega square what you can write is that sigma i is equal to rho omega square r top to r base into r dr so with that what we get is that rho omega square by 2 for RB plus RT into RB minus RT wherein we actually say that you know these vertical stress distribution in the radial acceleration field is not linear it is actually nonlinear that means that what is actually happening is that because of the radial acceleration field there is a certain degree of nonlinearity is actually you know uncounted. So here we can also observe that when you look into the variation of the G level with the depth the at top of the model it is RT omega square at bottom of the model it is RB omega square so if you look into this you know the way there is a variation of G level with the depth. Similarly when you have the you know at this particular point center of the model it is RT omega square at the edge of the model the gravity which is actually imposed is RT plus this offset which is say delta H omega square delta RT plus delta H omega square and so there is a possibility that the gravity level is actually varied along the horizontal distance and this with the vertical depth and if you consider the models in terms and refuse what we get is that you know there is a model which is actually caused by the identical surface curvature surface as that of the drum periphery. So in that actually the variation of the G level with the depth with horizontal distance can be eliminated but however both in beam centrifuge and drum centrifuge the variation of G level with the you know depth is existent. So the further when you look into this vertical stress in the radial acceleration field if you look into this even from with reference to the in terms of Z coordinates also the sigma v is equal to 0 to Z into rho omega square RT plus Z into dz. So this expression also indicate that there the non-linear variation of the vertical stress results due to the model rotating about a radial acceleration field. So further you know this particular slide shows about the merit of the large centrifuge which is actually this is actually having a small model small radius and this is the large radius what you can see is that when you have large radius the angle which is subtended to the center is very low and in the small centrifuge the angle is very high. So what we have is that you know here you have 45G here you have got 55G in this case what you can see is that the variation G level is you know marginally negligible here there is a variation but it is highly traceable. So you can see that here 48.9 and here 59.4 and 51.3 here 48.9 so variation in G level is very very less. So this indicates that you know the G level can vary with the depth and here also when you compare a soil layer in the prototype and corresponding centrifuge model what we said is that if you have got a in prototype with a geostatic stress condition this is the sigma v and this is sigma h for let us say for normally consolidated soil conditions. Then when you have the similar centrifuge model when it is subjected to rotation of a vertical axis in horizontal plane because of the radial acceleration field there is a small minor variation which is actually resulted because of this radial acceleration field effects. So in this particular lecture what we actually have discussed about is the example on the hydraulic gradient simulated method and then further we actually looked upon how the mechanics of the centrifuge basal physical modeling can actually look into you know the further you know understanding about the centrifuge basal physical modeling and we said that if mechanical behavior of the soil is important to capture then the you know the centrifuge basal physical modeling is one of the viable option or the tool available for geotechnical engineer to understand about the you know many facets of you know geotechnical structures.