 So welcome to course on advanced geotechnical engineering module 7 lecture number 4. This is lecture number 4 on geotechnical physical modeling. So in the previous lecture we have discussed about variation of G level or gravity level with the horizontal distance as well as with the depth. In this particular slide the merit of large centrifuge is shown. So here if you look into it there is a radius which is 1 meter from the center of the shaft to certain point within the model. Similarly here this radius is about 4 meter. So if you look into it there is variation with the horizontal distance but is limited. But in the case when you take a radius of 1 meter here it is 45 gravities and here it is 47.4 gravities. So there is a variation of G level with the horizontal distance. Similarly here it is 45 gravities at the bottom it is 55 gravities. So because here it is RT omega square and here it is RT plus 0.2 omega square but whereas if you have a large beam centrifuge the variation G level is exist variation of the G level is prevalent but however the variation is negligible. So this actually shows the merit of large beam centrifuge and here in this slide the whatever we have discussed about in the normal prototype when you have got a soil strata which is having certain thickness and because of the self weight the vertical stress is sigma v and the horizontal stress if it is normally consolidated soil the distribution is shown here. But if the model if the same prototype soil is brought and model is constructed with a reduced thickness and subjected to radial acceleration field then what we have discussed is that the variation of the vertical stress with the depth is not linear is non-linear. Then we said that this is going to cause certain degree of under stress in the upper half portion or upper portion and war stress in the bottom portion. So this variation is arisen because of the variation of the G level with the depth. So you can see that in this particular slide and this particular diagram where actually this particular line shows the distribution of vertical stress as per the full scale conditions is concerned but if you look into this there is a non-linear variation of the vertical stress in the model due to radial acceleration field. And before going further and we also have discussed about two possibilities of increasing the gamma. One is that variation of gamma by increasing the seepage forces that we said that hydraulic gradient simulator method and in the centrifuge model test we said that this can be possible by subjecting a model to high gravities this is done by rotating about a vertical axis in a horizontal plane. So if you compare the basic principle one is that in centrifuge model test the centripetal forces acts towards the center and in case of so there is in centrifuge model test centripetal forces act on the model and hydraulic gradient test seepage forces act and in the porous material soil the self weight stress is increased and in the hydraulic gradient also we have the self weight stress increase and solid structure member it can be a pile member or it can be footing member it can be a retaining wall it can be any member which is being tested. The self weight stresses also increased in case of centrifuge model test but whereas in case of hydraulic gradient test and which is not done that with that self weight stresses are not affected they remain same. And the application in the sense that they are more general problems most complicated problems the construction process can be simulated or some climatic events can be simulated like rainfall or subjected to earthquake or combination of the different loading forces can be simulated in centrifuge model test which we are going to discuss in some selected applications. And as far as hydraulic gradient test is concerned we said that this mostly used for you know granular soils or sandy soils and problems with the level ground and the self weight of the structure member is not important and wherever the self weight of the structure is not important and where you have got you know problems with the level ground like you know let us say that footing resting on horizontal surface and or problems with like uplift behavior of piles subjected to you know uplift behavior of piles embedded in sandy soils. Similarly when you have got some problems like footing subjected to eccentric loads on a level ground these things can be studied by using the hydraulic gradient similitude method. So this particular slide gives comparison of hydraulic gradient similitude method and centrifuge model test. And we said that as far as the centrifuge model test are concerned two categories of test do exist one is called variable gravity level the other one is called constant gravity level mostly the constant gravity level is adopted and in the previous discussion we have discussed that when the you know gravity reaches to a constant level that means that a particular desired gravity then when the gravity reaches to a certain level then what we can do is that you know the angular velocity is maintained constant. So the tangential component of the acceleration will be 0 and secondly if the particularly the model being tested has to be safe up to that particular gravity level is being tested then it can be subjected to some sort of loading or some sort of construction process or some sort of effect due to some forces like wave forces or some earthquake forces etc. Whereas in case if you are interested in studying some collapse behavior of a certain structure like you know some collapse behavior of a some vertical cut or some slope which is being studied wherein it can be studied with variable gravity level. So here the gravity level is increased gradually at a certain speed and the speed at which this is settled for example if you are the increase it has to be you know keep in mind with the desired conditions like undrained conditions and if you are required to have then the waiting times also have to be small. So here this is you know a certain gravity level and then this is another gravity level and then up to the collapse or failure whichever actually occurred earlier. So this is called variable gravity level but each and every gravity level you know the model actually behaves like a prototype that means that if you are actually having a sandy model sandy soil model in dry sand model the generation of pore water pressures will not be there if you are having a dry sandy soil. So up to some extent like a reinforced slope constructed with a sandy soil where if you are interested in collapse behavior of a reinforced slope constructed with sandy soil they can be studied but each and every gravity level when they are subjected to they are equivalent to equivalent prototype with respect to that particular gravity level. So this is the two categories of tests one is called variable gravity level and constant gravity level but nowadays with the development of different actuator systems and you know rainfall triggering events or earthquake triggering events the constant gravity tests are you know adopted widely. So the possibility of the geotechnical centrifuge studies include we can actually model the prototype that is that if you are actually interested in understanding a particular prototype structure which has been constructed whether you know it is safe or not or what will be its performance after 5 years or 10 years of its life can be you know studied by using you know this centrifuge base of physical modeling technique but however the size of the model depends upon the size of the equipment which is being used for the purpose. So if the equipment is reasonably large and then large prototypes can be modeled very well by simulating the majority of the you know characteristics of the prototype in the model. The another possibility of geotechnical centrifuge studies is that investigation of new phenomenon that means that a new technique or a or can be you know studied by using this centrifuge base of physical modeling wherein you know this helps to draw guidelines and then to also to assess its performance over a period of time that is before failure and at failure and also lead to you know the improvement of the or you know the phenomenon being investigated. So this is you know very interesting where a new phenomenon can be studied very efficiently by using centrifuge base of physical modeling. There another possibility is that parametric studies so if you have if we have studied for certain you know dimensions then we wanted to know what will happen to the next level see the parametric study or parametric variation is very efficient and can be studied you know efficiently in geotechnical centrifuge modeling. Then as said earlier the numerical models are required to be validated mostly they are validated by using full scale physical models in the case of you know non feasibility of constructing full scale physical models let us say in certain areas like nuclear installations or where there is no access for constructing or feasibility to construct these full scale structures then you know the numerical models are required to be validated. So one of the avenues is that to you to perform series of centrifuge base physical model test and corroborate or see how the results numerical model results are actually comparable with the centrifuge model test results. So that will actually help to validate numerical models the another important application is that use of geotechnical centrifuge for geotechnical engineering instruction that is something like called centrifuge aided learning in geotechnical engineering centrifuge aided learning in geotechnical engineering. So after having discussed the possibility of the geotechnical centrifuge studies and we said that by subjecting a model to about a vertical axis you know in a rotational plane there is a possibility of this radial acceleration field effects are there. So these effects actually what we said is that you know effect the vertical stress and then consequently the effect also the horizontal stress distribution. So the X gravity is uniform for the practical range of soil depths encountered in engineering like up to 200 meters or 300 depths of you know what we encounter for soil investigations in civil engineering the X gravity is uniform for the practical range of soil depths encountered in civil engineering. So while using a centrifuge to generate high acceleration field required for physical modeling there is a variation of acceleration through the model. So that is what actually we have been discussing while using a centrifuge to generate high gravities because we wanted to have identical stresses as that in the prototype there is a variation of acceleration through the model. So what we said is that this is attributed to the inertial acceleration field given by R omega square this is the inertial acceleration field this is due to the inertial acceleration field given by R omega square where R is the radius from the center to a certain point within the model. So this apparent problem turns out to be minor if you are able to take care of the you know the hardware that is the radius of the equipment at which the gravity scale factor N is determined and this also this operating radius that is called an effective radius we are able to select properly then we can say that that particular point the stresses in model and prototype are identical. So this apparent problem turns out to be minor if care is taken to select the radius at which the gravity scale factor N is determined. So for that what we need to do is that we have to find out you know how the variation of the gravity level is there with the depth and where it is subjected to with the vertical stress distribution where it is subjected to under stress and more stress and then comparing the errors then we can actually lead to you know the point at which we can actually you know maintain this gravity level can be a certain. So the selection of the effective radius for a you know beams and refuse like you know consider a model rotating about vertical axis which is shown here and the model is constructed with a material having row density the mass density row and this is the center of the shaft the model is rotating about a vertical axis and a capital V is the velocity in this direction and this is small element d z selected at a depth z from the top surface and RT is the radius from the center of the shaft to top surface of the model. So if you look into this what we can do is that this is the you know the stress in the this is the stress ordinate and this is the depth here in this direction. So here this is the you know the prototype stress distribution and this is the radial the vertical stress distribution in the radial acceleration field in the centrifuge model you can see that this portion there is an under stress and this portion there is an over stress. So this could be understood because of the increasing gravity level at this depth there is an over stress is possible so in order to determine this what we do is that again by following the same principle d sigma v in the small stress which we can actually find out by writing like d sigma v is equal to rho into z into d sigma v can be found out and from there by integrating from 0 to H m we can actually get rho omega square H m into RT plus H m by 2. So what we have is that this one and now here if the same point H m if you are able to find out you know what is the stress in the prototype so at this level we can actually find out it is nothing but rho into g that is the normal gravity the H m is correspondingly in the prototype is n times H m so rho g into n H m. So at the so the area of the stress diagram in the prototype is nothing but half base that is rho g n H m into H m so half rho g n H m into H m that is this is the area of the stress diagram. Now this area of the stress diagram in the model can be found out by integrating from 0 to H m. So the area of the stress diagram can be obtained like this wherein this is by writing am is equal to 0 to H m rho omega square z into RT plus z by 2 into dz. So this is actually it is nothing but 0 to H m d sigma v so by writing the rho omega square that is RT plus z by 2 omega square is nothing but the acceleration and dz is nothing but you know the thickness of the element which is under consideration. So by integrating this what we get is that and then applying the limits rho the rho omega square is equal to RT H m square by 2 plus H m cube by 6. Now by equating the two areas of stress diagrams in centrifuge model and the prototype and by using n g is equal to r u omega square this r e is called the effective radius which is measured from the center of the shaft to a point within the model where we are actually going to find out the equivalency of the stresses are there. Now what we have done is that by equating the areas of the stress diagrams in centrifuge model in area of the stress diagram in the prototype then what we get is that r e is equal to RT plus H m by 3 by simplifying and substituting for n g is equal to r u omega square we get r e is equal to RT plus H m by 3. Then for finding out the location where maximum under stress occurs what we do is that now assume that it occurs at a distance x from the surface so that means that we are now assuming at a distance x below the surface the notation is different here it is just to express in terms of function x it is written in terms of vertical distance x here. Now d sigma which is nothing but rho n g x that is nothing but the x which is nothing but the ordinate of at a particular depth x below the surface of the model that x is multiplied with n to get the prototype stress equivalent stress and g level is g and so rho g into n x minus rho omega square into x into RT plus x by 2. So this is nothing but the vertical stress in the model at a depth x in the model. Now that is nothing but the d sigma the difference between prototype stress and vertical that is that which we are trying to find out the difference between a stress here and a stress here this is the stress in the prototype this is the stress in the centrifuge model. Now by taking differentiation of this by taking this as function x rho n g x minus rho omega square x into RT plus x by 2 by taking function f dash x we can write that it is rho n g minus rho omega square RT minus rho omega square that is 2 x by 2 which reduce to rho omega square into x. So for d sigma to be maximum f dash x has to be maximum by applying the rules of maxima and minima we can say that for d sigma to be maximum f dash x is equal to 0. Then with re is equal to RT plus h m by 3 and by simplifying we get this x is equal to h m by 3 that means that and again if you look into it with x is equal to 2 h m by 3 when you substitute x is equal to 2 h m by 3 in this particular expression what you get is that d sigma is equal to 0 this implies that this term and this term will be equal and then it will be equal to 0 that means that implies that at that particular point x is equal to 2 h m by 3 sigma v in model is equal to sigma v in prototype that is where the point where the centrifuge stress the distribution of a vertical stress in the centrifuge model crosses the you know the vertical stress distribution in the prototype. So when you look back into the figure so this is that point from the surface it is actually said as 2 h by 3 2 h m by 3 and this is h m and this is actually at 50 percent of the height of this point at which this is crossing where the maximum under stress is occurring that is where actually what we have located is x is equal to h m by 3 then now we can actually find out the resulting errors after having known the location of the you know where under stress is maximum and where the sigma v in model in prototype are equal then what we can do is that after having so at x is equal to h m by 3 that is the where the maximum under stress error is possible that is defined as d sigma by sigma v prototype. So by writing the expression with x is equal to h m by 3 we can write rho h m by 3 into n into g minus rho omega square h m by 3 into RT plus h m by 6 divided by rho n g into h m by 3. So by simplifying you know this particular expression which is written and substituting for n g is equal to r u omega square we get h m by 6 re. So the percentage under stress error resulting is h m by 6 re and similarly at x is equal to h m at x is equal to h m the war stress error d sigma by sigma v p is equal to rho omega square h m into RT plus h m by 2 minus rho n g h m divided by rho n g h m which also we will get it as h m by 6 re. So as we have equated the you know areas of the we have determined the effective areas by equating the areas in the centrifuge area of this stress diagrams in centrifuge model as well as in the prototype we get the percentage under stress and percentage over stress as h m by 6 re. So this implies that using this rule we can actually find out there is exact correspondence in stress in the model and prototype at two thirds of the model depth and effective centrifuge radius should be measured from the central axis to the one third depth of the model from the top surface. So generally there are they do a number of interpretation they do exist so this actually says that re is equal to RT plus one third of the h m that is 0.33 times h m but there are also some other you know investigators have presented this re is equal to RT plus somewhere like 0.4 times h m measured from the top surface. So using this rule there is exact correspondence in stresses between model and prototype at two thirds model depth and the effective centrifuge radius should be measured from the central axis to the one third depth of the model. So and then another thing is that maximum error in the model is given by Ru is equal to R0 that is h m by 6 re. So maximum error due to either under stress or over stress is given by h m by 6 re. So for most of the geotechnical centrifuges h m by re if it is less than 0.2 h m by re less than 0.2 and therefore the maximum error in the stress profile is minor that means that if the h m by re is less than 0.2 and then the maximum error in the stress profile is minor and generally less than 3% of the prototype stress. So for most geotechnical centrifuges h m by re if it is maintained as 0.2 then what you can see is that 0.2 by 6 that is into 100 about 3.33% the stress is the variation the stress is minor. If suppose h m by re let us say is equal to 0.5 or something whether the stress will be high and is greater than 3%. So if you look into the centrifuges like about 1.5 meter radius onwards what we can see is that we will be able to ensure that except for certain models the h m by re can be maintained as less than 0.2. So this actually ensures that the variation the vertical stress is maintained within the limits and this apparent error due to radial acceleration field on the vertical stress can be minimized. So in the example problem whatever we have discussed just now an example problem is given here in a 1 by 50th centrifuge model same material is used as that of in the prototype that is that the row in model prototype identical and model of the height h m is subjected to rotation with a constant angular velocity omega about a central axis. If RT is equal to 4 meter obtain the variation of vertical stress with depth at every 50 mm in the centrifuge model and compare vertical stress in model and prototype at 50 mm, 300 mm, 150 mm, 300 mm and 450 mm depth respectively. So here what we need to do is that the RT is given that is 4 meters then height of the model is about 450 mm. So what we can do is that by the height of the model which is about 450 mm and that by knowing that what we can do is that re is equal to RT is equal to 4 plus h m by 3 0.45 by 3 so re is equal to 0.15 so plus 4 then RT is equal to 4.15 meters. And by knowing n that is equal to 50 so 50 into 9.81 is equal to 4.5 omega square you can actually find out omega you will get it in radians per second and if you would like to convert this omega there from radians per second to rpm it should be multiplied by 60 by 2 pi and for converting from rpm to radians per second multiply by 2 pi by 60. So by knowing omega now by writing the depths and for writing the stresses in model and prototype at different depths we can find out what is the variation of vertical stress in the model as well as in the prototype. So if you look into it by at the depth exactly 2 h m by 3 that is 2 into 450 divided by 3 that is around 300 mm you will see that the stresses in model and prototype are identical that is where the d sigma will be 0 and at a depth of 1 that is a depth around 450 mm you will see that the under stress error is maximum and at depth h m that is at the base of the model that is h m is equal to 450 mm you will see that the vertical stress in the model is more than the prototype stress. So we can also find out by knowing 450 that is 0.45 that is h m divided by 6 into 4.15 you can also calculate what is the percentage error due to variation the vertical stress due to radial acceleration field. So this is how the solution for this problem runs and then for calculating the stresses the appropriate units by need to be used for by while substituting rho is equal to 2000 kg per meter cube. So verification of the static equilibrium of a centrifuge model so here what we have said is that we said that now the vertical stresses variations they do exist but let us see that we have a prototype and if you are having a certain element under equilibrium so we need to satisfy or check whether the static equilibrium equations are satisfied or not. Then we can say that the model material response will be equivalent to as that in the prototype. So consider an embankment which is shown here and this dimension is length L and these are the x axis and y axis and it is actually has seepage as well as it is subjected to some body forces due to seepage as well as the sulphate forces and consider an element within the model at a certain depth as shown here. So assume that the same model is reduced by 1 by n times by maintaining this length as L by n and maintaining this gravity as ng that means that we and this all dimensions has been reduced and then here this element is actually having sizes of dx dy dz. Similarly here with the same density both models are constructed with same mass density and here the volume of the element which is considered is dx dy dz by n cube dx dy dz and here it is by reducing the 1 by n times that is dx dy dz by n cube. So this model represents the representative conditions of the prototype. Now what we do is that considering two dimensional equilibrium let us see what will happen when you wanted to derive the static equilibrium of a centrifuge model. So consider infinitesimally small element having dimensions dx dy dz in prototype that is what we have been discussing that representing homogeneous isotropic elastic continuum and where x and y which are actually shown here they are the unit weights due to body forces in x and y direction and x could be due to seepage and y could be due to sulphate of the forces. So then we have the assume that is a two dimensional element when it is subjected to the horizontal stresses and vertical stresses. So here this is the rate of change of the stress because of that there is an increase in stress. So which is nothing but sigma xx plus dou sigma xx by dou x into dx and this stress direction is sigma xs we have considered the elements in tension. Similarly the stresses in y direction here it is sigma yy plus dou sigma y by dou y into dy this is sigma y. So this x the body force acting in x direction this y body force due to sulphate acting in the y direction and this is the sign convention x is positive in this direction y is positive in this direction. Now we have shear stresses the conjugate shear stresses here dou yx plus dou yx by dou x into dx and here dou yx and here it is dou xy plus dou xy dou dou xy by dou y into dy by dou xy that is here. So when we consider now the equilibrium of you know the force equilibrium fx fy and mx and my when you take moments about the center so you will get that dou xy is equal to dou yx and by using sigma fx is equal to 0 sigma fy is equal to 0 what we get is that by for equilibrium what we for the equilibrium of the element is that we have to ensure because it is a two dimensional sigma fx is equal to 0 sigma fy is equal to 0 and sigma mx is equal to my is equal to 0. Now taking moments of the center of the element we get dou xy is equal to dou yx. Now with sigma fx is equal to 0 what we can do is that sigma fx is equal to 0. So here sigma fx is nothing but sigma xx plus tau xx tau sigma xx by tau x into dx this is you know acting on the area which is nothing but dy dz. Similarly here sigma xx stress acting on the area dy dz so the net force is nothing but dou sigma xx by dou x into dx dy dz. So if you multiply if you subtract this one similarly when you take the forces in the shear stresses in this direction and body forces in this direction and simplifying the after simplification what we get is that dx dy dz into dou sigma xx by dou x plus dou tau xy by dou y plus x is equal to 0. So this particular you know expression here we write it as dou sigma xx by dou x plus dou tau xy by dou y plus x is equal to 0 because as dx dy dz volume of the element cannot be equal to 0. So this is equilibrium equation in the x direction where it is read as dou sigma xx by dou x plus dou tau xy by dou y plus x is equal to 0. Similarly when you look into sigma fy is equal to 0 in the vertical in the y direction and simplifying the different forces what we write. So what we get is that dou sigma yy by dou y plus dou tau yx by dou x plus y is equal to 0. So this is again as dx dy dz cannot be equal to 0 what we write is that we can write this equation equilibrium equation in the so equilibrium equations for a model for the prototype is nothing but the two equations one is here in the y direction which is actually shown. So in the case of ng model considering that the small element having volume dx dy dz by n cube it yields now what we get is that dou sigma xx by dou x by n plus dou xy by dou y by n plus nx is equal to 0 and then in the y axis dou sigma yy by dou y by n plus dou tau yx by dou x by n plus ny is equal to 0. So this x and y are nothing but the body forces are equal unit weights in x direction and y direction that is the reason why they get multiplied by n. Now with that what it actually if you look into this previous expression by simplifying this the n will get cancelled n cannot be equal to 0. So dou sigma x by dou x plus dou tau xy by dou y plus x is equal to 0 dou sigma y by dou y plus dou tau yx by dou x plus ny plus y is equal to 0. So these equations whatever we have got for a prototype are analogous to the one what we have got in the centrifuge model. So this indicates that the centrifuge based physical model satisfies the static equilibrium conditions what they are actually existing in the equivalent prototype. So what we have actually reduced from this discussion that for convenience we actually have taken a two dimensional case and where we have written the forces in the x and y direction only and we have taken sigma fx is equal to 0 sigma fi is equal to 0 and by simplifying what we have reduced is that we reduced the equilibrium equations in the x and y direction and afterwards we converted them to ng model and considering a small element having volume dx dy dz by n cube it yields the equations which are actually identical to those in the prototype. So we can say that the static equilibrium conditions of the centrifuge model are satisfied. So whatever the change of boundary values is considered the solution of differential equations will not distinguish between model and prototype provided that the materials and system at corresponding points in model product can be made behave identically. So what we have noted is that whatever the change of boundary values is considered the solution of differential equations will not distinguish between model and prototype provided that the materials and systems at corresponding points in model prototype can be made behave identically. Now the limitations of you know after having discussed the like how the vertical stress can be minimized due to radial acceleration field and also how we can actually can satisfy the static equilibrium equations of a prototype in a centrifuge model. Now let us look into limitations like any modeling technique the centrifuge based on physical modeling technique also suffers from some limitations like non-homogeneity and inestropy of soil profiles is difficult to model and limitations of some modeling tools the sizes and all those things required to be considered and variation of g level with the horizontal distance and depth of the model. So this we said that water whether it is a beam centrifuge or drum centrifuge the variation of the g level with the depth cannot be eliminated but however in case of a drum centrifuge we can actually say that the variation of g level with the horizontal distance especially for small centrifuges can be eliminated because with curvature of the model but similarly even in beam centrifuges also for slope model some for example some investigators like Kumura et al in 1991 they have given the curvature for the top surface of the slopes to suit these curvature of the centrifuge and this is basically done to eliminate the variation of the g level with the horizontal distance. And then the boundary effects the models particularly especially the sandy soils they exhibit you know very severe boundary effects then nothing but the friction due to soil and the containers of the inner sides of the containers of the walls of the container. So this need to be eliminated by a special techniques particularly like applying a white petroleum grease or a lubricating agent without affecting the transparency of the front view transparent glass wall and by applying a sheets in a special way we can eliminate this friction angle up to reduce this up to you know in a big way. So the boundary effects are important need to be controlled especially for sandy soil models in case of we are having a high plastic soils there is you know this need to be arrested by applying the by lubricating the inner sides of the containers so that the boundary effects are minimized due to adhesion in case of place in case of sandy soils due to friction between the wall and the sand grains or sand grains. Then another thing is that these models they subject to scale effects particularly one of the predominance effects which we actually have discussed is that because of the our inability to not able to model the you know the grain size distribution what we said is that the scale effects are evident that these scale effects are can be minimized and also need to be found out for how they are effect in the model behavior. So this we have the one of the possible scale effects in the centrifuge based physical model means due to the grain size effects or particle size effects and which we will be discussing and how this can be solved are attended by using the same centrifuge based physical modeling technique that is nothing but what we call modeling of models. Then when we are actually reducing you know times scale factors for the time we will find that you know each and every phenomenon when it changes for example if it is for a see phase force or if it is for something like for a dynamic time or if it is for a some you know for viscous force you actually have we have different timings different scale factors for the time. So this inconsistency of the scale factors for the time is also said as you know one of the limitations sometimes you know when we have got scale factors for the time for some dynamic cases as well as for the you know some diffusion cases when they are different and if the phenomenon occurs simultaneously then we actually have to resort to some sort of you know an alternative so that the scaling considerations are satisfied. Then one of the major effects particularly the effect which is called the Coriolis effect which is actually called caused due to movement of you know particle within the model. If this suppose if that velocity with which the particle is actually moving is say V and when you are actually comparing with the velocity with which the model is rotating then we have to see that how this Coriolis effect can be you know considered. So let us look into this particular Coriolis effect in depth and what is the theory behind this Coriolis effect. So what we said is that the Coriolis effect which is nothing but it is the effect caused by the radial acceleration this is again the effect caused by the radial acceleration field when there is a movement of the model in the plane of rotation. So this movement of the model can actually occur because of certain like some certain construction process like assume that we are having a soft soil bed and with certain stresses tree and there we wanted to construct an embankment because in the construction of embankment on the soft soil happens in different stages. So let us say that we are actually simulating that so that for that what we do is that we use the sand hopper technique so inflate sand hopper is used. So when the particle is released in the when the particles are released sand particles released to construct sand embankments during flight there is a possibility that these particles are subjected to so called Coriolis acceleration or Coriolis effect. So the construction of embankments of soft soil during centrifuge test and triggering of rainfall during centrifuge test for example if you wanted to study the effect of rainfall on geotechnical structures when you are actually modeling this the rain droplets when they are actually simulated in the form of a mist or in the models and these droplets they are subjected to when they are released in the high acceleration field they can be subjected to this Coriolis acceleration that is it is basically the Coriolis effect is the effect caused by the radial acceleration field when there is a movement of the model within the plane of rotation. Then another thing is that studies on the geotechnical structures subjected to earthquakes for example when the model is subjected to earthquake perturbance due to by with the help of some actuator which can actually induce this phenomenon then what it does is that this is subjected to the particles velocities or such that they can actually get experienced with this Coriolis effect. So for some models which is actually required like construction of embankments on soft soils or let us say when we are studying rainfall effects on some geotechnical structures or geotechnical structures subjected to earthquake models it is mandatory to check whether the model is actually free from Coriolis effect or not. So in order to understand the theory of the Coriolis effect consider a model rotating about a vertical axis in a horizontal plane what we are seeing is the plan view and here this is the center of the shaft and x and y are the global coordinates at the center of the shaft and the model is rotating in this direction and this is the velocity v is equal to r omega where let us say that this is the radius which is actually up to the element a element which is considered here this is an embankment having a physical dimension length from the this axis and constructed with a material having rho. So what we need to see is that when this is subjected to the so called perturbance what will happen to the acceleration components in this. So what we need to do is that we have to first express this in the terms of polar coordinates and try to get the local accelerations at the center of the shaft and then by using the method of transformation it transform the local acceleration to the global accelerations to the local accelerations within the model that is the local axis within the model they are x dash and y dash so now what we do is that the coordinates of x and y can be expressed in radial polar coordinates as x is equal to r cos theta where r is the radius from the center of the shaft the center distance where x and y origin is there to up to the element which is actually considered in the previous slide. So y is equal to r sin theta so we have x is equal to r cos theta and y is equal to r sin theta so local velocities can be obtained by differentiating this dx by dt is equal to dr by dt cos theta minus r d theta by d sin theta similarly dy by dt is equal to dr by dt sin theta plus r d theta by dt cos theta. So let this be regenerate as one so these are the local velocities then by differentiating once again what we get is the local accelerations but these are with reference to the axis the global axis at the center of the shaft. So the local accelerations are given as d square x by dt square is equal to d square r by dt square cos theta minus 2 dr by dt into d theta by dt sin theta minus r d square theta by dt square sin theta minus r d theta by dt whole square cos theta. So similarly in the y axis d square y by dt square is equal to d square r by dt square sin theta plus 2 dr by dt d theta by dt cos theta plus r d square theta by dt square cos theta minus r d theta by dt whole square sin theta. This is let us regenerate as number 2. Now what we do by applying the transformation of from the global axis to local axis by applying the method of transformation what we can write is that the basic equation is that x dash so here you can see that one axis is getting transformed from the center of the shaft to the axis within the model which is actually rotating. So x dash is along x axis of the model that is horizontal axis of the model y dash along the vertical axis of the model. So we can write x dash is equal to C minus x sin theta plus y cos theta c is a constant y dash is equal to constant minus x cos theta minus y sin theta. So in order to get these things what we do is that to get the local accelerations with reference to axis within the model first you do the differentiation get the dx dash and dt and then after further doing successive differentiation to this what we get is the d square x dash by dt square. So the d square x dash by dt square term works out to be minus d square x by dt square into sin theta plus d square y by dt square cos theta. Similarly in the y dash direction d square y dash by dt square is equal to minus d square x by dt square cos theta minus d square y by dt square sin theta. So by we know that d square x by dt square d square y by dt square with reference to global coordinates at the center of the shaft similarly we also know here and here by substituting and simplifying what we get is that the following terms which is nothing but by simplifying the substituting for d square x by dt square and d square y by dt square what we get is the d square x dash by dt square is equal to 2 dr by dt d theta by dt plus r d square theta by dt square. So if you look into it this particular element when it is actually it has two components of acceleration one is this 2 dr by dt d theta by dt plus r d square theta by dt square and similarly we have d square by dash by dt square is equal to minus d square r by dt square plus r d theta by dt whole square. So if you look this particular element which is considered in the centrifuge model what we can write is that this particular term in the horizontal direction that is 2 dr by dt d theta by dt plus r d square theta by dt square that is this particular term. So this particular term is called as the Coriolis acceleration term. So when the model when the element is subjected to certain amount of acceleration so what we are actually acceleration field what we see is that there is some lateral acceleration which is actually acting on the element which is actually called as Coriolis term is also called as the Coriolis acceleration term which is nothing but 2 v omega ac a suffix c is indicated as Coriolis acceleration which is nothing but 2 v omega plus this term is due to inertial effect in horizontal direction because we actually assume that this model is subjected to some subparted bonds with due to some shaking let us say in that case this is actually due to inertial effect in horizontal direction. When you look into the elements here the when the vertical component d square r by dt square is due to is the inertial effect in vertical direction then r t theta r d theta by dt which is whole square which is nothing but the n g term that is nothing but r omega square term that acceleration acting towards the element acceleration acting towards the center. So here in this particular derivation by taking from global accelerations from global coordinates to the local coordinates and the local accelerations have actually expressed as we have that they have got two components in x direction two components in y direction and in the x direction what we have observed that the term which is called as 2 v omega that is the local acceleration term which is called the Coriolis acceleration term. Then we have the Coriolis effect which is actually defined as ratio of Coriolis acceleration to the radial acceleration that is ac by ar so which is nothing but we can write now ac is nothing but 2 v omega by ar let us assume that up to this ar which is nothing but like we have discussed the r can be effective radius so it is re omega square. So by simplifying using v omega is equal to v by r or v by re we can write that Coriolis effect is nothing but 2 v by v so here what we are doing is that we are actually comparing two velocities one is small v is nothing but the model velocity and capital V is nothing but the small v is nothing but the velocity of the particles within the model it can be due to seepage velocity or it can be due to some wall or wall movement the soil particles are moving or it can be due to the particle velocity which are generated because of the some seismic perturbance. And then capital V is nothing but the model velocity the velocity with which the model is rotating. So if you further you know analyze this one it is actually generally said that if the you know the percentage error due to Coriolis effect ac by ar if it is say less than 10% if that actually happens that means that if it is less than 10% then what it actually means is that the Coriolis effect can be neglected. So if the ratio of v that means that for ac by ar less than to be equal to 10% what we need to do is that the v has to be equal to 0.05 v if this is less than or equal to 0.05 v we can say that the Coriolis effect can be you know negligible that means that the model is not subjected to Coriolis effect. So there are also you know many events the tests which can actually take place let us say that we are actually investigating the effect of you know igniting a blast load on the you know performance of some structures buried structures. So when the explosive is say subject is ignited remotely then what you can see that the ejecta which is actually thrown with very very high velocities that means that this particles soil particles are thrown with very very high velocities. So in order to calculate this radius of the curvature subtended by these particles Fokhrowsky and Fedorov in 1968 they have given an empirical expression they saying that Coriolis acceleration is given by v square by RC v is the velocity with which the particle is actually is being throwing out and the RC is nothing but the radius of the curvature. So they try to compare the radius of the curvature subtended by the particles as well as with the radius of curvature which is with which the model is being rotating from which what they have reduced is that RC is equal to v square by AC and AC by substituting we get 2 v omega so we can that RC is equal to v by 2 omega so using omega is equal to v by r re what we can get is that v by 2 into v by re by taking re this side we can write RC by r is equal to v by 2 v. So it actually appears is that if RC the radius of the curvature subtended by the particles is say much much much higher than the radius of the curvature of the centrifuge then the effective radius then v is supposed to be 2 v that means that the particle will be thrown with a very high velocity and it is subjected to and then it will hit the container boundary and then falls. So in general Coriolis effect cannot be neglected for any object moving within the velocity range that means that if the velocity is within 0.05 v to 2 v then we can say that Coriolis effect cannot be neglected if the velocity is actually more than 2 v then we can say that the Coriolis effect is nonexistent that means that the model will not be affected by the Coriolis effect. So for example if you are having a model with v is equal to 30 m per second Coriolis effect can occur if the model velocity occur between so as per our this thing 0.05 v 0.05 times 30 is 1.5 m per second and this is 2 v that is so if the velocity of the moving object within the model is within 1.5 m per second to 60 m per second then we can say that the model is subjected to Coriolis effect otherwise the model is actually free from Coriolis effects. So in this particular lecture what we try to understand is that how the static equilibrium of centrifuge model can be established and also we also try to look into some aspects of how to calculate the effective radius and then we discussed about the limitations of centrifuge based on physical modeling as we have said in this particular slide where we have got the non-homogeneity and isotope of soil profile but with the advancement of the techniques this is also being you know worked out now with the modern physical modeling in geotechnical engineering and some limitations of modeling tools and boundary effects and scale effects and the inconsistence of scale factors which we are going to discuss in detail and then we actually had a discussion on the Coriolis effect.