 Welcome to lecture series on advanced geotechnical engineering course module 7 on geotechnical physical modeling lecture 7. So module 7 lecture 7 on geotechnical physical modeling. In this particular lecture we are going to discuss about modeling of capillary rise phenomenon in centrifuge and the respective scaling loss and the earthquake modeling. And how these scaling loss can be verified with an especially a technique called modeling of models. So modeling of models will be used for verifying the scaling loss or the scaling relationships which are actually deduced from the new phenomenon. So in the previous lecture we tried to understand the scaling loss for time of seepage and we also deduced based on the governing differential equation for the consolidation. We deduced also the time of consolidation and we found that the time of consolidation is 1 by n square times that of in the prototype. So this could be justified as the distance the pore water has to travel the centrifuge has to be reduced by a factor n compared to an equivalent prototype. And the pressure head driving the seepage flow is same in the prototype and centrifuge model but is applied over a distance down by a factor n. So these two combine to result in accelerating the consolidation time in the centrifuge by n square that means that this particular explanation it elucidates why the scale factor for time in centrifuge modeling is 1 by n square times that in prototype. The distance the pore water has to travel in this centrifuge model has to be reduced by a factor n compared to equivalent prototype. So we have actually reduced this distance by 1 by n times but keeping the pressure head constant. So the pressure head driving the seepage flow is same in the prototype as well as in the centrifuge model but this is applied over a short distance reduced by a factor n. So these two combine the explanation for the accelerating the consolidation time in centrifuge by n square. Now let us look into the scaling laws for centrifuge modeling of Kepler arise. The Kepler phenomenon has been very well investigated because of its relevance in environmental engineering problems and problems concerning polluted behavior of soils. So when you have got a pollutant transport in the soil and relevant to some environmental engineering problems the Kepler movement is very very important. So basically our aim is to look into the scaling laws what is the Keplerity height in a centrifuge and what will be the Kepler rising velocity that means that the rate at which the Keplerity height once in comes in contact with water and what will be the velocity of the movement of the water in the voids of you know soil and what will be the Kepler rising time in you know what is the time which actually takes to raise from a point at which the in contact with water to a certain level it depends upon we all know that it depends upon the type of soil and its gradation. So the assumptions basically we assume that prototype soil and prototype fluid are used in the model in the model fluid. So that means that we actually are using the prototype soils and prototype fluids as the model soil and model fluid and the characteristic microscopic length which is you know described by the particle size d 10 of the soil and the density of the fluid and surface tension for the fluid particle interface which seems to be independent of g level and are the same in model and prototype. That means that the detail in model and prototype that is effective particle size which is also called in environmental engineering problems is called as microscopic length, characteristic microscopic length also called as characteristic length and that is assumed to be same in model and prototype and the surface tension which is the you know for the fluid particle interfaces the fluid property and assumed to be identical in model and prototype and then you know when we use the same model fluid as that in the prototype then the mass densities of the pore fluid in the model and prototype are identical. So with these assumptions and connecting to the fundamentals of capillarity then we can actually deduce the scaling loss for the capillarity height first and then capillarity raising velocity and then time you know capillarity raising time. So the here in this particular slide it is assumed that all the grains are assumed that they are interconnected with you know thin tubes you know having certain diameter d and these diameter d is nothing but d it defines the pore size or the pore diameter and assume that this is the water level and from which the water level raises to this level and this hc is the completely saturated zone or a and in capillarity situation what will happen is that there will be a certain zone it will be completely saturated and above which there will be some fringes which are actually developed and they remain in partial saturated state up to certain distance and so we actually have two heights one is you know completely saturated you know capillarity height other one is certain zone above that is remains in a similarly partially saturated state. So whatever the scale factor we deduce we assume that you know the both the you know these terms which we said that completely the capillarity height in the completely saturated zone and capillarity height in the partially saturated zone will follow the same you know stability. So we can actually write the capillary suction as T naught which is nothing but the surface tension into pi d and then this alpha is the angle between the you know the made by the water film on the you know the surfaces of the glass tube and that is nothing but the surface of the pore surfaces but this alpha is equal to 0 if you know it is a clean tube but if a contaminated tube then alpha is not equal to 0. So for clean tubes alpha is equal to 0 so here what we can write is that capillarity section is nothing but T naught into pi d cos alpha. So this component what we have taken is the component and divided by the area that is nothing but the pi d square by 4 gives the capillarity suction. So we can write uc as 4 T naught cos alpha by d when for clean tubes or a non-contaminated soil and alpha is equal to 0 then uc is equal to 4 T naught by d or 2 T naught by r. So you can see that as you know the d pore size is actually small the uc the capillarity section will be very very high that means therefore clays the capillarity section will be very high. At equilibrium so what we do is that now when the water reaches above the water table because of this capillary phenomenon the equilibrium is nothing but the equilibrium between the surface tension forces and the self weight of the water column in the capillary tube what we assume. So this tube is nothing but you know the interconnected pore voids when we align them in a line and having approximately assuming that the diameter is d. So the T naught into pi d is equal to pi d square by 4 into hc gamma w so when you simplify this what we get is that hc is equal to T T naught by r gamma w. So this r is nothing but the pore radius which is nothing but d by 2 and with that what we get is that the capillarity raise capillary raise is equal to 2 T naught by r gamma w. So for getting the scale factor for capillarity in model and prototype capillarity height in model and prototype and we assume that hc z, hc z is the z which is actually when they when is a height you know when the water is taking when raising with depth z. So the z is actually different from here the hc z in model and prototype is equal to when we compare these terms like 2 T naught by rm rho wng because here in model it is you know it is nothing but it is n gravities and in the prototype it is which is nothing but that this is rp into rho w into g divided by T naught into m this is in p. So now we assume that T naught in model is equal to T naught in prototype and when we actually have the same soil skeleton as that in the prototype then rm the pore radius will be identical then you know by looking into this and mass densities are identical in model and prototype with that what we get is that the scale factor for capillarity height is 1 by n times of that in the prototype. That means that if you are having about 50 centimeters of capillarity height at if you are for a given soil and if you are actually testing in a centrifuge at 50 g we will actually get about 1 centimeter of capillarity height. So this particular you know hc hc zm by hc znp is equal to 1 by n so this is attributed to increase in the weight of the capillarity fluid within the capillarity tube by n. So this particular you know the reason what why we have got this one is attributed to this particular fluid becomes heavier at high gravities so because of this you know the increase in the weight of the capillarity fluid within the capillarity tube makes the capillarity rise reduced by 1 by n times in a centrifuge model. Now after Soga et al. 2003 so here the pressure and saturation profiles in prototype and centrifuge model are given. So in the prototype or a field situation if you look into it so we have got the height h and you know this d that is the depth of water table so below the water table what we see is that rho w gd that is the hydrostatic pressure here. So above this the water column you know this particular this column this is actually the suction which is nothing but rho w g into h –d so this is totally h so this is nothing but h –d. So this is you know here – rho w g into h –d so this is the in the you know suction at this level. So here what you look is that up to this is the h c z completely saturated then in here you can see that there is a decrease in the water saturation in the prototype. This also could be attributed to the series of you know the evaporation and precipitation which actually takes place. This actually also makes the water table you know the capillarity water depleted in the portion close to the at that is what we call in the Vodos zone. So this is the point where the air entry head takes place and then you know it actually depletes to this level. So if you look the similar situation in normal gravity so if you look now this so called h is now h by n times and d is d by n times. So again the hydrostatic pressure is rho w into n g into d by n so because of that here what we have got is that rho w gd which is actually identical to that of in the prototype. Similarly the suction is nothing but minus rho w into n g into h minus d by n so here again the suction is also identical. So if you see the water saturation profiles in a centrifuge model n g model and in a prototype they are analogous what you can see is that here you have partially saturated zone is actually commencing here also and this is the portion where you actually have you know the completely saturated capillarity zone and above that what I said is about the fringes development or it is also called as a fingering actually takes place in this particular zone. So if you look this into the this particular case simulation of prototype into 1 by n times in a 1 by n times but at normal gravity that is at 1 gravity. So if you look into this actually has the hydrostatic pressure which is rho w g into d by n so this is the pressure is low and again similarly the suction is also minus rho w into g into h minus d by n the suction is also low. So if you look into this here the water saturation profile in a 1g model is drastically different from you know the what we actually get in the prototype wherein this is actually due to as the less weight of water is required to be lifted so the water you know tries to have you know very high capillary raise it actually implies. So this is not the as realistic as that we observe in the prototype so the capillary model at 1g is drastically different from the one in the from the field condition. But whereas if you are able to model this with identical soil as that in the prototype and identical fluid characteristics as that in the prototype and at high gravities also so what it actually says that the water pressures as well as water saturations are you know identical to those in the prototype though there is a you know the distances are reduced by 1 by n times. So further after having reduced and discussed about the capillary height now let us look into the rate of capillary rise, rate of capillary rise that is the rate of capillary rise. So according to Landau et al 1967 and Bikerman 1970 so what actually has been assumed is that the porous medium is characterized by one pore size and is similar to a very narrow cylindrical tube so that the flowing fluid has a mean velocity. So according to Landau et al to 1967 and Bikerman 1970 porous medium is characterized by one pore size and is similar to a very narrow cylindrical tube so that the following liquid the flowing liquid has a mean velocity. So based on the you know these assumptions and further deductions the mean velocity is actually obtained as r square into delta p divided by 8ht mu w. So the r square r is nothing but the pore radius delta p is the pressure head pressure difference between the surface tension forces and weight forces of the water which is being lifted. So that u is nothing but the mean velocity that is the flow of a viscous fluid in a cylindrical tube due to the pressure difference delta p maintained at the end of the tube. So u is the mean velocity this is the this is due to the flow of a viscous fluid in a cylindrical tube due to the presence of pressure difference delta p maintained at the end of the tube. So h within brackets t is the height of liquid lifted at any instant of time t due to existing pressure difference of delta p. So because of the there is a pressure difference of delta p this delta p is due to the pressure due to self weight due to the surface tension forces as well as the self weight forces of the water being lifted in the so called the narrow cylindrical tube. If that the pressure difference is taken then that is what we get is this height of liquid lifted at any instant of time due to this particular heat the height of liquid being lifted is due to the pressure difference of delta p and which is nothing but the pressure difference due to surface tension forces and the due to the self weight of the column of water being lifted in this cylindrical tube. With those assumptions and discussion and we can actually further deduce this one. So when capillary forces cause a pressure difference delta p at the ends of an iron tube then the delta p what we said is that the difference between the pressure due to surface tension forces and pressure due to the weight of fluid lifted at an instant of time. So here the pressure due to the self weight of the fluid is given like this rho w g into h t into pi d square by 4 this is nothing but the weight divided by area, area is nothing but pi d square by 4. So this rho w g is nothing but the gamma w h t into pi d square by 5, pi d square by h this is the area into h t is the height over which the fluid is lifted. So this is nothing but rho w g h t and from the surface tension forces with h c z is h c is equal to h c z is equal to 2 t0 by r what we can deduce is that u is equal to r square into rho w g into h c z minus h t by h 8 h t mu w. So this mu w is the dynamic viscosity of water by substituting u is equal to d h t by d t and integrating time required to raise the continuous capillary zone h c z can be obtained. So by substituting for u is equal to d h t by d t and integrating this expression the resulting expression then the time required to raise the continuous capillary zone h c z can be obtained by integrating with then the time can be obtained. This time is nothing but to raise the continuous capillary zone h c z. Now so that after integration and simplification what we get is that t is equal to 8 mu w divided by r square rho w g within brackets h c z natural logarithm h c z divided by h c z minus h t minus h t. So this is situation in the prototype. Now in a centrifuge model we know that the capillary heights and any capillary raise with a time t due to the pressure difference is actually reduced by 1 by n times and the gravity level is n times g. So we can write the time in model t m is equal to 8 mu w divided by r square again the pore radius is identical as that in the prototype rho w is identical as that in the prototype. So the term is left like that and n g the g term is changed as n g and h c z was changed as h c z by n natural logarithm h c z by h c z by n minus h t by n minus h t by n. So by simplification and rearrangement of the terms we can look into this expression like t m is equal to 1 by n square by taking the common from the numerator and denominators what we get is that with 1 by n square into 8 mu w by r square rho w g and divided by into h c z natural logarithm h c z by h c z minus h t minus h t so bracket close. So with that if you look into it this particular expression is analogous to this particular expression. So only thing is that now once we take when this is substituted by t then what we get is that t m is equal to t by n square. So here also what we get like in analogous to seepage force phenomenon seepage phenomenon or the consolidation phenomenon what we get is that the time of capillary raise is also scaled by 1 by n square time that in the prototype. So in line with what we deduced in the previous lecture like time for seepage what we deduced is 1 by n square time that of the prototype similarly time of consolidation of a soil also 1 by n square that of the prototype similarly the time for the capillary raise also scaled by say similar scale factor 1 by n square then that 1 by n square times of the time of the prototype. So the t m is equal to 1 by n square into t p, t p is equal to t and which is nothing but which is in the prototype. Now after having deduced this capillary time now let us look into capillary velocity. So as we have defined in the previous in our discussion that u is equal to r square rho w g into h c z minus h t by 8 h t mu w. So what we can do is that we can write for u m as the velocity as r square rho w into n g into h c z by n minus h c t by n h t by n by 8 h t by n into mu w. So now if you compare again if you compare with prototype and model what we get is that the implies the capillary velocity u m in the centrifuge model is n times that of in the prototype. So what we get is that when you take this n out what we get is that we can write it as u, u is nothing but u or up so nothing but u m is equal to n up. So this implies that the capillary velocity in the centrifuge model is n times that of the velocity in the prototype. So what we have discussed is that capillary rise which is actually which is in fall in line with like linear similitude conditions where this scaled by 1 by n times that means that if the gravity level is actually increased by n times and the capillary height also scaled by 1 by n times. Similarly the capillary raising time is scaled by 1 by n square as that of the prototype. Similarly the capillary velocity is n times that in the prototype. So after having discussed the modeling considerations of the capillary rise and consolidation and seepage phenomenon let us look into the modeling of the earthquake in centrifuge and what are the necessary scaling considerations are required to be considered. So the need for the earthquake centrifuge modeling or the dynamic centrifuge modeling is closely associated with the nature of the earthquakes in the field and very limited quantitative data exist in the response of soil deposits or geotrache structures to the strong motion of the strong ground motions. So it actually says that you know the opportunity to study the phenomenon of earthquake induced liquefaction has been principle driving force behind the widespread interest in the generating the strong shaking on a centrifuge model. So one of the major attributes of the centrifuge modeling is that the climatic events like these earthquake can be modeled with great accuracy so that the performance of the geotrache structures to these destructive forces can be understood and it is also possible to arrive at the remedial measures and developing a theory and then guidelines so that a properly designed structures are constructed which can actually resist these destructive forces due to earthquakes. So simulation of earthquake geotrache problems in centrifuge has grown significantly in the past you know you can say 2, 3 decades and a variety of challenging problems are being tackled in various centrifuge establishments all over the world. So for this simulation of the earthquake conditions the centrifuge requires a careful consideration of modeling the base motion that means that whatever the strong motion need to be modeled and selection of the model container we will be discussing that when we subject this the base motion in the prototype we actually have the sort of primary waves and shear waves and so we need to have a special containers with non-reflecting boundaries and also you know we actually have to use appropriate fluid in the soil if you are investigating saturated sand and saturated silty sand soils subjected to earthquakes. So these scaling laws are actually deduced for a dynamic models so here this is for a case of a prototype at one gravity consider similar an embankment having the length L and assume that this embankment is subjected to shaking in this direction a dynamic shaking due to perturbance created because of the strong motion. So let this is the time axis this is the x axis let this be the amplitude variation and as a sinusoidal variation and the ap is the amplitude which is actually indicated here in the prototype so considering a motion in the prototype so we can write xp is equal to ap amplitude sin 2 pi fp tp so fp is the frequency and tp is the time in the prototype. Now by differentiating this particular term what we get is that we get dxp by dp dtp is equal to 2 pi fp 2 pi fp into ap cos 2 pi fp tp so here if you look into this this is the velocity magnitude fp into 2 pi fp into ap so then differentiating once again this term what we get is that d square xp by dtp square is equal to minus 2 pi fp whole square into amplitude in prototype into sin 2 pi fp tp so the negative sign indicates that the acceleration acting towards the center of the axis center of the rotation. So here what we have done is that we have taken motion in a prototype which is xp is equal to ap is equal to sin 2 pi fp tp and with that we have said is that by differentiating once we have got the velocity term and acceleration term by d square xp by dtp square. Then here what we have said is that the displacement magnitude from the expression whatever we have discussed is displacement is nothing but amplitude term ap in the prototype and velocity is nothing but 2 pi fp ap and acceleration is nothing but minus 2 pi fp whole square into ap so these are the displacement velocity and acceleration terms in the prototype. Considering a motion considering the motion in the model xm is equal to am sin 2 pi fm tm so this is actually the acceleration term which is obtained by d square xp by dtp square from the differentiation what we have got. So what we have got this is the acceleration magnitude term so now consider the similar embankment model with same configuration but it is reduced by 1 by n times lm is equal to lp by n and we consider an element which is actually subjected to let us say this is the with time tm and this is the xm that is the amplitude is measured in this direction so that this is actually subjected to a small amplitudes now which is am which is indicated here and it is at the ng and lm is equal to lp by n. So with that what we have got is we assume that the same motion is actually assumed where xm is equal to am into sin 2 pi fm tm so xm is equal to am into sin 2 pi fm tm. So by using the analogous expression for motion in centrifuge model the following expressions can be deduce. So again here what we have got is that displacement term am and velocity term 2 pi fm am and acceleration term the minus which is nothing but minus 2 pi fm whole square into am. So now what we do is that by the analogous expression what we have done used in the centrifuge model and with that what we have got is that displacement terms as am and velocity magnitudes as 2 pi fm am and acceleration term as minus 2 pi fm whole square am. Now here if the model in the model the linear dimensions have scale factors 1 is to n and then acceleration have scale factors 1 by 1 is to 1 by n that means that acceleration in model is n times the acceleration in the prototype in order to maintain similarity this is possible if the amplitude in model and prototype is 1 by n and frequency is fm is equal to fm is equal to n times fp then the similarity of lm by lp and acceleration in model and acceleration in prototype is equal to n can be achieved. So that means that if you look into it here if you compare absolute displacement in model am and with the displacement term in the prototype then am by ap is equal to 1 by n and then the you know in order to have the acceleration which is to be n times that in the prototype. So when you take this acceleration term in the model that is minus 2 pi fm whole square into am divided by minus 2 pi fp whole square into ap the minus minus sign will get cancelled then with that what we get is that fm by fp has to be n times that of the prototype and when the amplitude is you know 1 by n times that of the prototype with that what we get is that the acceleration will become you know n times that of in the prototype. So that means that here we have to note down that the you know the frequency has to be n times that in the prototype that means that if you are having an earthquake with the frequency of say one heads one cycle per second then the frequency in the centrifuge model at 50 g is 50 cycles per second. So when you have the frequency enhanced by n times what we get resulting time for the earthquake will be reduced by 1 by n times because frequency is increased by n times you know the time has to be reduced by 1 by n times. So if you look into the when amplitude is you know reduced by 1 by n times when the frequency is increased by n times if you look into this the velocities in model and prototype are identical velocities in model and prototype are identical that means that am by ap is equal to 1 by n and fm is equal to nfp with that what will happen is that you will get velocities identical as that in the prototype. So if you are able to have the same motion which is subjected to the structure then what we get is that for the acceleration to be n times that in the prototype the frequency has to be n times that of the prototype and amplitude then will be 1 by n times that in the prototype and for subsequently we have the time which is nothing but tm by tp is equal to 1 by n. So then when with fm by fp is equal to n and am by ap is equal to 1 by n what we get is that vm by vp is equal to 1. So the velocity in model peak particle velocity in model and prototype equal to 1. So the resulting you know these directions implies that the time in model is equal to time in prototype is equal to 1 by n. So the time taken for the earthquake to commence and you know complete will be 1 by n times that in the prototype. So the time in centrifuge is compressed by factor n and frequency is increased by factor n. So this is you know dynamic models you know it is required to be noted that the time in centrifuge is actually compressed by factor n and frequency is increased by factor n. So for example with these scale factors it can be seen that if you are having a 10 cycles of 1 hertz earthquake the duration is say 10 seconds with an amplitude of 0.1 meter in the field can be represented by a centrifuge model tested at 50 G subjected to same number of cycles the 10 cycles but of a 50 hertz frequency that means that the frequency which is 1 hertz is now increased by 50 times 50 hertz earthquake due in the duration of this is about 0.2 seconds having an amplitude of 0.1 by 50 that is about 2 mm. So these are the you know this example states that what the whatever that you know the deductions we have for the scaling loss and with these scale factors it can be seen that the 10 cycles of 1 hertz earthquake with the duration of 10 seconds and with an amplitude of 0.1 meter can be represented by a centrifuge model tested at 50 G subjected to 10 cycles of 50 hertz earthquake with a duration of 0.2 seconds so having an amplitude of 2 mm. So here the duration of earthquake also can be obtained by 1 cycle per second and 10 cycles we can actually get as 10 seconds and with an amplitude of 0.1 meter. So here also 50 cycles per second with of 10 cycles we also get here the duration as 0.2 also we can get from the time which is actually taken in the prototype like 10 by 50 is also has 0.2 seconds. So this is actually example of what we actually have based on the scaling loss for the dynamic models. So here you know based on some discussions some general summarization of scaling loss for static models was given length is actually reduced by 1 by n times area is reduced by 1 by n square then the prototype and volume is 1 by n cube times that of the prototype and pressure is that is stress is equal to 1 and strain is equal to 1 and density that is nothing but mass density equal to 1 and unit weight is n times that of the prototype mass density is identical and gravity is n times and mass is 1 by n cube force is 1 by n square and time for diffusion is 1 by n square. So please note that time for diffusion is 1 by n square and time for dynamic activity like earthquake what we have reduced just now is 1 by n. So we will have to keep in mind that we actually have two different times one for dynamic event we have got 1 by n and for diffusion event like seepage or consolidation we have got 1 by n square. So this is the scaling loss for dynamic models so these are actually summarized here with the stresses and strains are identical velocities in model and prototype are identical acceleration in model is n times that in the prototype and frequency is n times that of the prototype and the time for dynamic is 1 by n times and mass which is actually subjected to this excitation is about 1 by n cube times that is that if you are having very large chunk of mass which is subjected to strong motion in the field that particular structure can be reduced by 1 by n cube and the same structure can be subjected to equivalent to that in the field so that we will be able to get the identical response of a field structure under consideration in the centrifuge model. So if you look into now there is a conflict which actually has been addressed as far as the diffusion time as well as the dynamic events time. So in modeling stability of claim magnets that dynamic test time scale factor of 1 is to n should apply since no seepage of flow or diffusion of what occurs because if you are having a dynamic stability of claim magnets there is a possibility that we will be able to bifurcate these timings with 1 is to n that is you can say that during dynamic event we can actually take 1 by n and subsequently if a long term seepage occurs then long term diffusion occurs we can actually take 1 by n square but subsequent to earthquake any dissipation of excess pore water would be modeled using a time scale factor of 1 by n square. So however the problem arises in the study of liquefaction of saturated fine science where excess pore water pressure dissipation will occur during the earthquake event. So problem here actually occurs is that whatever you know when you look into this there is a deviation from the field event when you know before attenuation of the magnitude of the excess pore water pressure if the diffusion actually commences and this is possible for certain type of soils like sands and silty sands having non-plastic fines there is a possibility that you know the situations they happen in simultaneously in that case we have a conflict of scale factor for diffusion and dynamic events. So in that case you know it is necessary to ensure that the time scale factor for motion is same as that for fluid flow that means that we actually have to say and go with one scale factor so Tm by Tp is 1 by n and Tm by Tp in principle it actually indicates that it is 1 by n square times that in prototype. So if you look into this out of the two times factors scale factors the 1 by n square is relatively faster compared to even the dynamic scaling factor for time. So we actually have to select such a way that you know one time is actually adopted one time scale factor is adopted if the both dynamic events and diffusion events occur simultaneously so that we can actually operate on the particular model so that the results which are the results of the set of this model represent that in the prototype. So further actually two methods are actually possible for solving this conflict one is that reduction of the permeability of the soil by a factory n by crushing the particles to a smaller size. So this was actually you know tried by many investigators wherein let us say that k is equal to cd 10 square the Heisen's formula when you reduce the d10 model n prototype by 1 by root n times there is a possibility that we can actually reduce the permeability by 1 by n times and then increase the gravity by n times so that the permeability of soil is identical to that in the model prototype. But the soil constitutive behavior may get altered in the process hence not desirable. So the scaling down of the gradation is not a acceptable solution because when you scale down the sand turns out to be silt and the silt and sand behavior is actually drastically different as far as the stress strain relationship between the soil is concerned. So the reduction of the permeability of soil by a factory n by crushing the particles to a smaller size and though this avenue is there this is not possible because the soil constitutive behavior may get altered in the process. Another possibility is that increasing the viscous to the fluid in the model by a factory n that means that replacing the conventional fluid that is water by a fluid which is actually having a system of ideal characteristics with that what we can actually see that the increase in the viscous to the fluid in the model by a factory n may actually help in you know possibly to solve for this conflict. But this particular modeling changing of model pore fluid actually valid for fine sands or sands with less amount of silt and not valid for clay soils because it is not possible to you know saturate clay soils with highly viscous fluid pore fluid. Now consider here in this particular slide k is equal to k gamma by mu where mu is nothing but you know kinematic viscosity of the pore fluid which is say in the model. So now we look into this k m is equal to the k is nothing but the absolute permeability k m gamma m mu m and k p is nothing but k p gamma p and mu p. So k m by k p is equal to now if you look into this now k m is equal to n gamma p it is a gamma m changes to n gamma p and then if you are able to have in for vm as n mu p where the fluid is actually selected is n times viscous then so if you are having a water actually has 1 centi stroke of kinematic viscosity and if you are able to replace with n centi strokes that is n mu p into v p by k p into mu p. So with k m is equal to k p and by simplification what we get is that k m by k p is equal to 1 the permeabilities are identical. So by using the Darcy's law v m is equal to i m k m is equal to i p k p with that what we get is that when you have got k m is equal to k p then we get v m is equal to v p. So here in this particular case what we have used is that i m is equal to i p definition that is with h by n. So with v m is equal to v p and t m by t p is equal to now is 1 by n times that means and now this particular scale factor what we were actually getting for diffusion event because we are actually maintaining k m is equal to k p and with v m is equal to v p what we are getting is that t m by t p is equal to 1 by n. So that means that the replacement of a pore fluid with a higher viscosity is a viable option for investigating the problems when we have a conflict of scale factor for diffusion and dynamic events. So for example 100 centi stroke silicon fluid is 100 times more viscous than water but has virtually has same density. So here very very important is that whatever the pore fluid which we actually consider model pore fluid which you consider to replace the conventional pore fluid they can have it can have higher viscosity but the mass density of the pore fluid model pore fluid which is actually selected to replace the conventional pore fluid has to be identical. So this is basically required to ensure that effective stresses does not alter that is if you are actually having a pore fluid which is actually heavier then there is a possibility that the effective stresses will get actually altered. So one principle requirement is that the mass densities of this pore fluids model pore fluid and conventional pore fluid have to be almost identical. So with that what you can say that we will be able to have this particular option is valid. So thus centrifuge model using sand saturated with silicon oil and tested at 100 G would have a time scale factor for fluid flow of 1 is to 100 which is same as the time scale factor for dynamic motion given by 1 is to n since n is equal to 100. So if you are actually having this replacement of a 100 centi stroke silicon oil then there is a possibility that we will actually have the identical scale factor for the diffusion as well as this thing. So with this what we physically is happening is that by altering the pore fluid by replacing the conventional pore fluid with a model pore fluid having higher viscosity we are actually making the diffusion event slower. So with that what is actually happening is that we are able to match with the dynamic event and then we are actually going close to the field conditions when the saturated sandy deposit or deposit with sills is actually subjected to some sort of a strong motion. So what are the requirements of the ideal pore fluid in the sense that the fluid shall be like water and Newtonian fluid and it must have the same compressibility as that of water and it must be chemically polar to use along with sills and sands and if the fluid has same density and surface tension as water so that capillary effects will be properly scaled and the presence of different fluid should not alter the strength properties and damping characteristics. So that does not mean that the presence of different fluids should not increase or decrease the strength properties they have to be same and also the damping characteristics have to be same. These are the some of the selected ideal pore fluid characteristics. Now the physical explanation is actually given here in this particular chart for example when there is saturated sandy deposit subjected to the earthquake in the prototype. So you can see that how the excess pore water pressure with time takes place in the prototype. So the variation is which is like this which is shown here but when we have a centrifuge model with water as the pore fluid what actually happen is that the excess pore water raise which takes place because of the seismic disturbance but the dissipation actually commences very rapidly. So this is because of the nature of this Tm by Tp is equal to 1 by m square the diffusion event is actually faster than the which actually happens in the field. So by replacing with conventional pore fluid the schematic flow of the excess pore water pressure variation with the time is actually given here. This is the time which is in the earthquake shaking time. So you can see that this is almost identical to that in the prototype. So by replacing with conventional pore fluid with a model pore fluid having iron viscosity the centrifuge model results are represent closely as that in the prototype so which is actually explained and shown here in this particular figure. So some of the different substitute pore fluids which are actually called as silicon oil. Silicon oil actually has 80 centristokes of kinematic viscosity and row density is almost equal to that of water 4% less than that of the water. And it is also some investigators actually use glycerin water mixers and some laboratories they use model pore fluids and recently for the past one and a half decades people are actually using methyl ether or methyl solylose which actually used in the food dyeing industry as water mixable substance which is used for producing viscous fluids, high viscous fluids. So this is possible here with the methyl ether or methyl it is also called as a metallose and with that desired viscosity range can be actually obtained. The more we will be discussing with the examples from the works carried out by several investigators. So these are the, this slide actually gives that different substitute pore fluids. Now after having discussed about different modeling scaling laws but we need to have way and means of verifying this. So the modeling of models are also called as the modeling of prototype basically is a technique used in centrifuge modeling to ensure that the scaling laws derived earlier are valid. Like we have reduced that time scale factor for capillary rise is 1 by n times 1 by n square times the prototype. So that is actually done by using modeling of models. So this modeling of models is nothing but when you say that gl is equal to 1 that means that g1 l1 is equal to g2 l2 is equal to g3 l3 is equal to gl is equal to 1. That means that here what we have is that we actually have different models which can be done. So similarly in the hydrodegradient similarity method also when you actually have different pressure differences with the different thickness of soil layers there also we can actually have different models which actually represents the identical result of the let us say a particular footing resting on sand subjected to you know so called pressure difference between top and bottom. Similarly here in modeling of models the two subcategories is modeling of prototypes and modeling of models. So in this what we have seen is that where we have got the prototype dimensions are given on the log scale and gravity level are given in the log scale in the vertical axis. So here when you have a 5 meter prototype let us say a footing size of 5 meter is scaled down by 100 mm at 50g. So this is nothing but the dp is equal to 50 into 100 is equal to 5000 mm. So this is directly a modeling of a prototype into a centrifuge model. So when you have a prototype of 10000 let us say a retaining wall of 10000 mm that is 10 meters and when it is modeled by at 100g it is height of 100 mm. So 100 into 100 is about 10000 mm. So this is here we are corresponding at this is at 1g and this is at the prototype dimensions 5000 and 10000. Similarly when you try to look into so we have a when the prototype exist we try to look into the corresponding model and when other case is that when you have when the prototype does not exist what we actually do is that we actually have a three different models each model represent the model of model with each other and then they also represent overall a prototype corresponding prototype identical prototype. For example here when you have 50g at 100 that is 110 centimeter prototype dimensions it is one model. Similarly at 25g with 20 centimeter dimensions it is one model and with at 10g with 50 centimeter dimensions it is one model. But model 1 and model 2 and model 3 they actually when they are they represent a prototype having 5 meter dimensions. So in example here 5 meters. Now this dp is equal to ndm so which is nothing but n1 dm, n2 dm and n3 dm but all the one which is actually falling on this line represents the corresponding prototype in this. For example here also here what we have is that we have got you know at height of say 100 mm and at 100g to 10000 then we actually have got another model we reduced the dimension with we reduced the you know g level increases the dimension when you have increase in the you know decrease in the g level then actually we have increase in the dimensions of the model. So here the technique which is actually you know the technique of testing scaled models at very different levels with aim of verifying the same scaling loss is called this particular technique is called modeling of models. So the modeling of models is actually used for verifying the scaling loss which are actually reduced for existing phenomenon like or for the new phenomenon or anything which is actually being investigated and the modeling of modeling techniques is also used whenever we are trying to model a new physical phenomenon in the centrifuge or and the scaling loss for the one or more parameters cannot be derived easily and one of the this modeling of model technique is also used for you know checking whether the so and so model is actually subjected to the scale factors or not. So as an example let us look into you know this particular figure wherein we have the you know a thick layer of consolidation takes place in a over a 5 meter thickness and both top and bottom we have open layer that is the sand layers then this is at 1g but what it actually speak about the modeling of models is that with a thickness of 25 centimeter in the model at 20g this again this is a model of you know this particular prototype case and when we reduce this by you know another 15 centimeter that means the model dimension 10 centimeter at 50g it corresponds to again this particular prototype which is 5 meters then we reduce further to 5 centimeters increase the gravity level to 100g. So here see model 1, model 2, model 3 they are actually models of each other and but they also individually when they say they are the actual models of a particular prototype. So the idea is that all these three models should have identical time settlement behavior as that of in the time settlement behavior of the prototype. So in this lecture we actually have discussed about the capillera scaling loss and dynamic model scaling loss and then we also discussed about the technique called modeling of models.