 Welcome to lecture on advanced geotechnical engineering course. So we are in module 7 on geotechnical physical modeling lecture number 10. So module 7 lecture 10 on geotechnical physical modeling. So we have introduced ourselves to the requirement of the centrifuge based physical model testing but in this particular lecture we will try to bring out some relevance of centrifuge based physical modeling to geotechnical problems especially when we take some selected problems in geotechnical engineering and we will try to review whether the centrifuge based physical modeling is warranted or not warranted. So we are aware now that the small scale physical modeling can be performed at 1g or ng field. So physical modeling at ng requires a geotechnical centrifuge to carry out model experiments. So in order to carry out the centrifuge based physical model test we require a geotechnical centrifuge to induce high gravities. Now consider some few simple situations where physical modeling especially at small scale at 1g may be adequate and others for small scale physical modeling at ng will be required. So where it is required, where it is not required by taking some selected problems we can be reviewed. Now consider as a first example slopes in sand. So here in this particular figure a typical slope which is formed with a dry sand is shown here and the slope inclination of the sand is say beta and a certain height h. Now if we know that this phi is a friction angle and this is the friction angle the maximum friction angle whatever it can have for a stable condition is called angle of repose. So for beta greater than phi that is for slope inclination greater than phi the slope is unstable that means that the slope takes a profile which is equivalent to that same friction angle. So the slopes in sand are not stable at angles greater than the angle of repose phi in the respect to the height of the slope. So whatever will be the slopes in sand are not stable at angles greater than the angle of repose irrespective of the height of the slope. So for dry cohesionless sand the stability criterion may be stated as beta is less than phi. For example if the beta is less than phi we can say that the stability of the slope can be ensured. Let us look how this can be explained by using tau sigma plot. So consider on the y axis tau and the x axis sigma and here this is the failure envelope and this is the friction angle. So this is the friction angle whatever the soil can take and let beta be the slope inclination. Now if you look into this as long as beta less than phi the slope will be stable that is what actually we have been discussing. Suppose but beta equal to phi that means that all points in this particular line will be in contact with the failure plane. So for beta less than phi the tau is less than tau f that means that for beta less than phi you can see that for beta less than phi whatever may be the sigma 1, sigma 2, sigma 3 whatever may be the sigmas the tau will be always less than tau f. Tau f is the shear stress at failure and tau if the shear stress at a particular sigma. So for beta equal to phi then tau is equal to tau f. For beta is equal to phi moment this line is intersects with this line this line joins with this line then tau is equal to tau f. So for beta greater than phi the tau is actually greater than tau f. So slope would have already failed at all depths when tau greater than tau f that slope would have already failed at all depths. So that means that slopes cannot be raised at an angle steeper than the friction angle whether in small or full scale test. So slopes cannot be raised at an angle steeper than the friction angle whether in a small scale or in full scale when you do not use any enforcement equations. So slopes in sands when beta less than phi you can say that here for all levels of sigma tau will be less than tau f. Similarly the slope is actually stable and independent of the depth of the slope. For beta is equal to phi then tau will be equal to tau f the slope is just stable and for beta greater than phi the tau is greater than tau f the slope would have already failed at all depths. So slopes cannot be raised at an angle steeper than the friction angle whether in small scale or in the full scale. So what we need to we have understood when we have taken the slopes in sand is that it is irrespective of the height of the slope as long as beta less than phi you know as long as the beta is less than phi you know the centrifuge model tests are actually not warranted because even 2000 gravity or 3000 gravity whatever the normal stress is actually in use this is independent of the depth. So if we are actually doing 1g model test with you know the slope inclinations less than phi then you know they hold good. Similarly let us consider you know these slopes in clay but before that let us look into the undrained stability analysis by Taylor's method. So from the Taylor's method actually has been deduced from the undrained you know slope stability analysis where you know factor of safety is the lowest factor of safety obtained from the circular arc analysis so from this we can actually write weight of the you know the portion of the soil which is involved within the failure surface is a function of gamma the unit weight and h height of the slope and geometry of the failure surface. So the geometry of the failure surface can be characterized by three angles which are called alpha beta and theta and which are you know by rewriting one we can write that C by factor of safety is equal to it is indicated as C suffix r is equal to gamma h into function of alpha beta theta which is nothing but the so called the geometry of the surface is indicated like this. So CR is the required cohesion to just maintain a stable slope and function of alpha beta theta is a pure number and designated as a stability number NS. So this is actually the stability number which is put forward by the you know by Taylor in 1948. So Taylor's stability number is given as NS is equal to CR by gamma h. Now in order to get this you know for different the stability numbers the Taylor actually has given Taylor's curves wherein the Taylor's stability number is actually is on y axis and this is for different slope inclinations. Let us see for beta is greater than 53 degrees we can see that you know the you know the independent of you know the depth factor independent of depth factor and we can actually get you know different for different inclinations like 60 degrees the stability factor is about 0.191. So for 60 degrees you can see that the stability factor is about 0.191 and similarly for 90 degrees it is about for 90 degrees it is about 0.261 that is for the vertical cut. So 1 by 0.261 into C by gamma which is nothing but 3.81 Cu by gamma which is equivalent to the critical height of a slope at which is 4C by gamma. So by using this Taylor's curves one can actually obtain the stability numbers and for beta less than 53 degrees the stability number is you know found to depend upon you know the so called d by h depth factor which is also indicate as small n. So for gentle slopes the critical surface goes below the toe and always restricted above the strong layer hence depends on its location. So for beta greater than 53 degrees the stability number is found to influence only on the slope inclinations and failure surface is assumed to pass through the toe of the slope so all critical circles are pass through the toe. So this is because for such steep slopes the critical failure surface passes through the toe of the slope and does not go below the toe. So for a vertical cut beta is equal to 90 degrees and ns is equal to 0.26 under short term condition with factor of safety is equal to 1 for a critical height hc we get 3.85 into Cu by gamma. So this is actually obtained from Taylor's stability number. So in case of undrained condition either with slopes or with vertical cuts by using this we can actually get. So why this is being shown is that this is actually used in designing some of the centrifuge model test actually particularly with slope inclinations which is actually greater than 53 degrees for showing the stability of a slope under undrained conditions. So this is our example if you are actually having d by h and beta and beta which is less than say 53 degrees then this is the chart which is required to be adopted. We can see that in that case there is a possibility that the slope surface actually passes through the below the base. So it is actually called base failure. So this is depth factor is given. For depth factor d by h is equal to 1 it can be seen that beta for beta greater than 53 n is equal to 0 that is this n is equal to 0. So that indicates that is independent of the slip circle actually this n value is 0 indicates that this n value 0 indicates that the slope actually the failure surface passes through the, passes through the toe of the slope. So that is what is is again n h is nothing but the distance from the toe of the slope h is the height of the slope and d is the total height once it is given here. So by keeping d by h and beta and n can be obtained and for d by h is equal to 1 and beta greater than 53 degrees we get n is equal to 0. So that indicates that the slope surface slip surface passes through the toe of the slope. So after having taken this with slopes in sand we said that if beta less than 5 the slope can be raised to any height even at 1g or ng and the stable slope angle is going to be the same and in this case the actuating force is the body force due to gravity and the resting force is the shearing resistance due to friction. So in case of a slope in sand what we said is that if beta less than 5 the slope can be raised to any height even at 1g or ng and the stable slope is angle is going to be the same. In the case of in this case actuating force is the body force due to gravity and the resting force is the shearing resistance due to friction. So let us see with the introduction from slopes in clay particularly with undrained analysis from the given by Taylor stability chart let us look into it how it can be reduced by for slopes in clay. So consider a slope in clay where we have got Cu and gamma as the soil parameters gamma is the unit weight of the soil here in this slide the d is indicated as depth below the toe up to the firm layer and h is the height of the slope beta is the slope inclination. So we know that the factor of safety is nothing but is a function of h beta Cu gamma d by using either Buckingham's pi theorem or Rayleigh's analysis Rayleigh's method we can say that factor of safety is equal to beta function of beta d by h and Cu by gamma h. Then we said that for similarity between model prototype each and every pi term has to be identical. So in the process we discussed that the Cu by gamma has to be reduced by 1 by n times in order to if you are actually having a small scale model reduced by 1 by n times at normal gravity. In case if you are having a small scale model tested at gamma m is equal to n gamma p that is enhanced to the gravities then we say that the C by gamma need not be reduced and then automatically get reduced once gamma increases to gamma in model becomes n gamma. So but it ensures that the Cu in model prototype are identical. So Cu in model prototype identical physically means that the stress strain of the soil is retained so that ensures that there is a similarity between model and prototype. So this we have already discussed so here it implies that in order to maintain the margin of the safety same margin of safety the model and prototype on the not only the geometry but also Cu by gamma h should be same. So if g can be increased by a scale factor n that is what we actually we have been talking Cu by factor of safety rho g h is equal to function of beta into d by h. If you increase g we reduce h then the soil particles strength Cu and density rho can be kept unchanged. So based on Taylor stability chart which we have discussed just now the maximum slope height of the slope that is hc at factor of safety is equal to 1 for beta is equal to 60 degrees is given by the stability factor is 0.191 so we can write hc is equal to 1 by 0.191 into Cu by gamma. So this indicates that in order to have in order to attain in a critical height the slope actually has to raise to this height so that it will actually have a factor of safety is equal to 1. Suppose if you are having a model slope of having hm is the model height then the ratio of hc by hm will be very high and then which will actually ensure that factor of safety will be very high. So the actuating force is the body force due to gravity and the resisting force is the constant under end coefficient. So in this case the actuating force is the body force due to gravity and the resisting force is the constant under end coefficient. So here it is not possible to carry out small scale model test at 1g on slope in clay as the slope will fail only and when h reaches to hc. So what we have understood in this case particularly with slopes in clay is that based on Taylor stability chart the maximum slope height stable height of the slope is you know at factor of safety 1 is for that beta is equal to 60 degrees. So the actuating force is the body force due to the gravity and the resisting force is the constant under end coefficient. So here the slope will only fail if h tends to equal to hc. If you are actually maintaining the slope prepared reduced by 1 by n times and kept in the laboratory it will only tend to dry because it has actually has got very high factor of safety in the small physical dimensions. So here it actually says that the centrifuge model testing is actually warranted for you know understanding for a stability of a slope in under end clay and also there is a requirement that the Cu has to be maintained constant and you know it actually clearly says that you know it is not possible to carry out the small scale model test at 1g on slope in clay as the slope actually will only fail as h tends to hc. So let us see some typical you know the slope failures in centrifuge this is after Roor University Bocom a front view of the slope before you know testing is actually shown here and this is one of the traditional in a very you know earlier testing of a slope which is actually with a kaolin clay. So you can see that once the slope is actually you know subjected to failure you can see that the development of the slip surface here. So this is the formation of the slip surface and the heaving portion can be seen here the heaving portion can be seen here and the tension cracks formation can be seen here. So this is the slope which is actually formed with a saturated consolidated clay and you can see that the tension cracks. So these are the earlier you know imprints which are actually used for distant wishing you know the occurrence of the failure phase. Nowadays the advancement of the techniques like particle based velocimetry or digital image correlation techniques allow you to look into even the you know formation of shear band thicknesses and the formation of the failure plane and also you know status of the you know front elevation of the model before failure and net failure. So this is actually important information can be derived. So you can see that the classical you know the slip circle failure which is actually shown here from the testing done at Roor University Bochum, Germany. So this is a typical undrained testing which actually the slope model height of the slope is about 18 centimeters that is h is 18 centimeter. So what actually happened is that moment the height you know the gravity level increased to 33 gravities then you can see that the slope is actually undergone a slip circle failure and this shows that the validity of the similitude under the undrained conditions. So this shows that the requirement of the you know the centrifuge base of physical model test to induce failure and then here it means that the critical height of the slope is nothing but 33 into 180 mm. So that is equivalent height in meters is the you know critical height of the slope at this particular soil at failure. So this relationship between prototype and model heads at clay slopes at failure is shown here. So here hm is equal to 18 centimeter it can be seen that you know the slope is actually found to fail at you know the 33 gravities which is actually shown here is about the height of about 6 meters. So this different heights of the slopes have been tested and you can see that the here these are the large centrifuge equipment for use and this is here the small centrifuge equipment for use. So the errors due to some small centrifuge testing actually lead to some you know some discrepancies but what can be seen is that when you have a large beam centrifuge we can see that the consistency in results the modeling of the models were found to speak well and indicate the performance of a you know critical height of a failure whether it is at 5 centimeter or whether it is at 12 centimeter whether it is 18 centimeter you can see that the horizontal plate can be obtained. So this is the prototype height in meters in y axis and model height in x axis for example here it shows that when model height is you know is somewhere say 18 centimeters and if prototype height cannot be equivalent to that because this is the height at failure so it actually has you know this is at factor of safety is equal to 1. So the merit of the large centrifuge also is shown in this particular slide. So after having seen two examples of slopes in sand wherein we said that centrifuge model testing is not required as long as the sloping inclination is less than 5 that beta less than 5 and if the sloping inclination is actually greater than 5 we said that the slopes are tend to fail but when we come to slopes in clay particularly unrained condition with 5 is equal to 0 we said that the centrifuge model testing is warranted and so let us try to look some two different distinct bearing capacity problems one is the footings on sand let us say the consider a footing having width B is subjected to you know at a embedded at a depth of df is subjected to a concentric load of Q. So in the case of footings on sand the bearing capacity in addition to friction angle depends upon the size of the footing. So let us look into these are the typical failure planes we can see that this is you know the elastic wedge which is actually formed and then these are the radial shear zones and this is actually resistance generated in the embedded depth zone for a above the base of the footing. So this inclination is about 45 minus 5 degrees 45 minus 5 by 2 45 minus 5 by 2 and this size is the breadth of the footing and this is the resistance which is actually derived to counter this movement for example when the footing load is applied and this block moves this side at this blocks move this side and this is counted by the friction actually mobilized along the periphery of this failure surface these are the failure surfaces typical failure surface which are actually shown in this particular slide. So this is the effect of the embedded depth of a footing that is nothing but the gamma df that is confinement due to embedded depth. Now we can actually get for the footing on sand let net ultimate bearing capacity that is nothing but q ultimate minus gamma df is nothing but qd is equal to half gamma b n gamma plus gamma df into nq minus 1. So this is we have taken like a continuous footing so where in we have written that net ultimate bearing capacity is equal to qd is equal to half gamma b n gamma plus gamma df into nq minus 1. Now this is simplified by writing qd is equal to q n gamma by 2 plus q into nq minus 1 into df by b into b the whole bracket is multiplied by b. So this implies that the wider the footing the greater is the bearing capacity. So the wider is the footing greater is the bearing capacity further the bearing capacity footing on sand is derived from two sources one is from the frictional resistance due to weight of the sand below the level of the footing other one is the frictional resistance due to weight of the sand surrounding the surcharge or the backfill that is you know due to this and due to you know all around the footing particularly in this zone also. So one is resistance offered in this zone and resistance offered in this zone that is what actually we are talking. So this implies that a study conducted at 1g does not help in predicting the bearing capacity in the prototype because the smaller is the footing then you know we have got the less is the ultimate bearing capacity. So when b is reduced by 1 by n times you know 1 by n times the b becomes so small so the bearing capacity also will be small. So this is also you know evident we have discussed that ultimate bearing capacity of a footing is found to function of several parameters and we have said that q by gamma b is equal to function of e and friction angle sigma c by gamma b that is the crushing strength of the grains and grain to grain cohesion that is sigma g by gamma b and eg that is elastic modus of the grains eg by gamma b and then dg by b that is the dg is nothing but the average particle size to ratio of average particle size to breadth of the footing b. Now so we have said that footings on sand because of the you know the requirement whatever we have discussed in order to have the similarity the centrifuge model testing is required because where the q ultimate is actually found to a function of qd net ultimate bearing capacity is found to function of the size of the footing. Let us see in case of clay under end condition where phi u is equal to 0 and we have only the resistance from Cu that is the shear strength of the soil and weight of the wedge and shear strength of soil along the failure plane tends to resist failure. So you can see that the failure planes are distinctly different you know in case of a you know clay but when you took here in case of the you know sand you can see that the failure plane the elastic wedge is actually formed and the failure plane is actually this portion is actually shifted vertically down. So this is because of the roughness and friction of the friction caused by the material that is the surrounding soil and that is the sandy soil but in case of clay that is absent so because of that actually have you know the failure plane starts immediately from that the elastic wedge formation will not be non-existent. So you can see that this is you know the typical failure surface which is actually forms here. So for the footings on clay the normal force across the surface of this sliding can produce no frictional shear resistance on account of phi u is equal to 0. So ultimate bearing capacity can be given by Cu u is equal to Cu nc plus gamma df then net ultimate bearing capacity can be obtained by Cu d Cu suffix d is equal to Cu nc plus gamma df minus gamma df so it is actually function of only the shear strength of the soil and nc. For long and continuous footings we can write net ultimate bearing capacity is equal to 5.14 Cu and where nc is equal to 5.14 here. For rectangular footings for having dimensions breadth and length then we can write from Skempton 1951 as Cu d is equal to 5.14 Cu into 1 plus 0.2 into df by b plus 1 plus 0.2 into b by l. So these are the depth factor and shape factors which is nothing but 5.14 Cu that is nothing but as you know this is a concentrated load the inclination factor is equal to 1. So 5.14 Cu into 1 plus 0.2 into df by b into 1 plus 0.2 into b by l. So this can be seen that in case of a saturated undrained clay when we have so even if you take tau versus sigma the more circles actually will exhibit identical diameters that means that is a respect of the cell pressure we apply. It is whatever the cell pressure we apply it actually generates the corresponding sigma 1 such that the sigma 1 minus sigma 3 is constant. So that indicates that we actually get a horizontal failure envelope and which is independent of sigma. So because of this particular using this particular logic when we have got a saturated undrained clay because in the case of undrained condition there is no volume change actually occurs. So the tau is equal to Cu which is in case of a completely saturated and undrained condition. So in that situation what actually indicates is that in reality if you are actually having the similar bearing capacity problem testing on clay it says that the centrifuge model testing is not really required and 1G model testing the small scale model testing is required here and there is no term actually of the effect of the so called the breadth of the footing is independent of the size of the footing. So it means that as long as we maintain this constant cohesion and for saturated undrained conditions it actually appears that the centrifuge model testing is not warranted for footings on clay. So that is what actually has been described here. Here the shear strength is constant and equal to Cu. Hence the 1G model tests are actually valid in this case as the model size is not important and centrifuge experiments are warranted. Centrifuge model tests are not warranted. So let us take cantilever sheet pile walls in sand. Let us assume that we have got a retaining wall which is having a height retaining height h and d is the embedded depth and this is the dread surface. Now if you see that this particular ratio if d is actually small and if d by h is actually say ensures you know if d by h is small then there is a possibility that instability comes into picture. So the wall will be stable if d is greater than alpha into h where alpha is a function of friction angle phi and where h is the height of the wall retaining the soil. So here in this case the actual force, actuating force is the body force due to gravity and the resisting force is the shear resistance generated due to friction. So the scale of the model is not important you know whether we do experiment at 1G or NG provided you know if you are able to maintain d greater than alpha h. So theoretically it actually appears that the centrifuge model tests are not required but you know in the forthcoming slides we will actually look into that how the centrifuge model tests we show a different behavior. So here if you consider a cantilever sheet pile wall so here also there are two types of sheet pile walls one is very flexible and other one is say rigid sheet pile walls. In case of rigid sheet pile walls what will happen is that the wall rotates about a certain you know point that total point above the toe but in case if you are actually having very flexible sheet pile wall there is a possibility that the wall undergoes you know a failure you know buckling failure here and undergoes you know permanent hinge formation can actually occur here somewhere here where that is a point where the maximum bending moment is generated. So theoretically for cantilever sheet pile walls in sand it says that you know the centrifuge model tests are not warranted as long as d is greater than alpha h where alpha is a function of friction angle phi. Now let us consider for stability of the wall particularly if you are actually having clay. So in this case both on the dredging side that is most this is the dredge level and this is let us say a point a and this is the d embedded depth and here this is the retaining height and here and here both actually locations we have clay. So let us look the how the stability of the wall this particular expression is obtained through a detailed explanation wherein we can actually look here with the for the same example here the cantilever sheet pile wall in clay where phi is equal to 0 and is under with prevalent cohesion only. So here we can what actually happens is that the wall rotates about a you know a toe that is actually this point here and this is the dredge surface and this is the retaining height and this is actually the level at which up to which the tension cracks can occur. So Z naught is nothing but the depth of the tension cracks and now let us see sigma a is equal to active earth pressure is equal to k a sigma v minus 2 c root k a but because of the phi is equal to 0 k a is equal to 1 and k p is equal to 1 and the sigma p is equal to that passive earth pressure because here the wall moves away from the backfill and here it actually moves towards the backfill here so here the passive case actually arises so sigma p is equal to k p sigma v plus 2 c root k p. Now let us take pressure at level of point a so here what we have is that gamma h that is the so called you know the k a gamma h where k a is equal to 1 with that k gamma h minus 2 c and which I can write it like gamma h is equal to gamma h minus q u because q u unconfined comes to the strength of the soil is equal to 2 c so we can write gamma h minus 2 c so this ordinate at this point just above the dredge line is gamma h minus 2 c. Now consider the resultant of the passive and active pressure at any depth y below the dredge level. So if this is the dredge level whatever we have said and we can actually take below the dredge level that is at y is equal to 0 it is at the you know dredge surface that is at point a so net pressure is actually obtained as gamma v that is nothing but you know which is nothing but this portion plus gamma v plus 2 c gamma v plus 2 c is nothing but because y we are actually measuring from this point so k a gamma v k p gamma v so k p gamma v k p is equal to 1 so gamma v plus 2 c minus gamma h that is plus gamma y so h plus y is because this entire portion is under active case gamma h plus gamma y minus 2 c that is k a gamma h minus 2 c root k a so as k a is equal to 1 we can write gamma h plus gamma y minus 2 c by simplification the net pressure is nothing but 4 c minus gamma h this is at y is equal to 0 we also get at y is equal to 0 we get 4 c minus gamma h that means that 2 q u minus gamma h. Now by equating by this is the net pressure diagram where we have got 4 c minus gamma h acting over depth d and gamma h minus 2 c acting over depth excluding you know the so called you know the tension crack depth so earth pressure once crack actually occurs then this portion of the pressure is relieved from the earth pressure so for stability of the wall we can obtain by comparing the pressure acting this side and pressure acting this side and net pressure we can actually get 2 q by factor of safety should be equal to gamma h should be equal or equal to gamma h 2 q u nothing but 4 c u so 4 c u by factor of safety should be greater than or equal to gamma h that is what is actually written here in the third slide here of module 7 and 11 lecture 10 is 4 c u by factor of safety greater than or equal to gamma h. So this is what actually we have used in explaining you know whether the centrifuge model tests are required or not is by 4 c u by factor of safety greater than or equal to gamma h so this is actually obtained from the net pressure diagram which is active side and passive side below the ridge level so in this case the soil is saturated having constant cohesion so it actually appears that here also you know you can see that c u by gamma h actually term is coming so like in slopes in clay here also it implies that any experiment done under 1g will have to be under prototype condition and the actuating force is again body force due to gravity and resisting force is the shear strength due to under any condition. So any experiment under 1g will have to be done under prototype conditions only if you are actually doing a centrifuge model test with if you are actually doing a small scale physical model test at normal gravity this implies that you know this is not realistic and may not represent the you know the equivalent full scale model in the field. So in conclusions in relevance of centrifuge based physical modeling what we actually have did from the considering examples is that when the body force due to gravity is the only actuating force the shearing resistance due to friction is the only resting force then 1g model test would be adequate for studying the phenomenon that means that including the ignoring the dilatancy of soil when the body force due to gravity is the only actuating force and the shearing resistance due to friction is the only resting force then 1g model test would be adequate for studying the phenomenon. When actuating force is an external force and not the body force due to gravity then the resting force is the constant and then also you know 1g model test would be adequate that is actually what we have said is the bearing capacity of the footings on clay. So in all other cases like slopes in clay and slope in the retaining wall in clay what we said is that the centrifuge model based physical model tests are required. So in this particular discussion of this lecture what we have said is that you know for considering similar the simple geotechnical problems we tried to bring out the relevance of the centrifuge physical model testing then what we said is that like in agreement with whatever we have been discussing if you are having footings resting on sand it says that the centrifuge model testing is required when you are actually having footings resting on clay it says that the 1g model tests will stand good and if you are actually having you know the cantilever sheet pile walls in clay it says that the centrifuge model testing is warranted and when you are actually having you know cantilever sheet pile walls in sand it says that our analysis says that theoretically the centrifuge model tests are not warranted you know because as long as d is greater than alpha h. So now you know this particular you know discussion about whether the centrifuge model tests are warranted particularly for modeling of a cantilever sheet pile wall let us look into this with the test which are actually have been done at IIT Bombay. So here the failure modes of cantilever sheet pile walls are given so we have two typical failure modes one is the wall failure you know other one is the material failure. So in the case of the upper figure where the you know they have what actually happens is that you know the material failure the soil failure occurs the failure due to rotation about point A. So you can see that the wall undergoes a rotation and about a point A there is a point of rotation then we have got active pressure and a passive pressure and here it this portion comes towards the you know passive zone and this is active zone. So this is the net pressure where you know where you have got active and then passive and then here again passive and then active. But in case when you are actually having a failure due to formation of a plastic hinge that is the material failure material of the sheet pile wall and if H is the you know height of the retaining the soil above the dead surface D then you know we can see that this is the active pressure and this is the passive pressure and at this point this wall tries to undergo rotation and formation of a plastic hinge actually takes place here. So this particular issue was actually has was modeled by using variable gravity method and then also verified by using you know by numerical modeling. So here consider a cross section of a flexible retaining wall test setup where the wall is actually instrumented with strain gauges basically to measure the bending moments. So the bending moments are actually obtained by pasting strain gauges and calibrating by applying the known loads and once we get to be actually get the for each application of the load we will get the bending moment theoretically can be calculated and for each application of the load the response of the strain gauges to the applied load can be obtained. So based on that for each strain gauge a calibration factor of bending moment with the volts can be with output volts can be obtained. So the in the linear range if you are actually taking and that is actually gives the calibration factor for the individual strain gauges. So once during the test once it is actually subjected to let us say in this particular method what we have done is that we have taken a plane strain container where having 76 centimeters in length and 20 centimeters in the breadth and having you know retaining height of 21 centimeters and 15 centimeters is the embedded depth and the wall is 22 centimeters above the base of the container and what actually has been done is that you can see that LVDs are used to measure the surface settlements and LVDs are used to measure the lateral movements and these strain gauges are used to measure the bending moments during flight. So here the perspective view of the model prepared for mounted on the swinging basket of the large beam centrifuge facility available at IIT Bombay shown here. The wall is modeled using a thin aluminum plate having 3 mm thickness and the sand was actually a fine sand and which is having a average particle size of about 0.15 mm and there is a poorly graded sand and which is placed at 55 percent relative density purposefully to induce you know large lateral pressures. And here what has been said is that in order to observe the formation of the rupture planes colored thin colored sand lines are actually drawn both on the active side as well as towards the passive side also. So this is you know as it has been told that this particular method the testing was actually adopted is the variable gravity level testing. So the wall the model has been subjected to increased gravities where in this is actually this is at one gravity and this picture actually obtained from a camera mounted in front of the model so that it captures pictures of the desired you know area 1g and this is at 20g and this is at 30g and 35g. So you can see that as the gravity level is increased you know though we are actually though it each graph theoretically it actually says that you know the d by h ratio at 1g it is stable but it does not mean that you know as 20g the same stability is actually ensured. So that is what actually in the previous discussion when we have discussed about the relevance of centrifuge based physical modeling we said that as long as you know d is greater than alpha times h we can ensure that factor of safety that is only valid theoretically but in reality it also depends upon the you know the stiffness of the sheet pile wall and in case of when you have got a you know some flexible sheet pile walls which are actually very weak in nature then you can see that you know you have got situation like the formation of a you know the plastic hinge and then the development of the rupture lines can be seen very clearly here. So the close view of the picture at 35 gravities is actually shown here so this actually shows that the wall actually has undergone a permanent deformations and the bottom portion is not subjected to any movement and it can be seen that and this portion actually has undergone the multiple cut like you know slip planes and this indicates the so called when we take different lines and at the top also this has been observed over the length you know the width of the container because we have taken a plane strain container so because of that the entire portion get shifted downwards like a step type deformations and so this is one of the classical failure which actually observed for the deformation of a sheet pile wall embedded in sand with a very very low you know stiff wall. So this is variation of the retained soil settlements with the distance from the back of the wall so you can see that when at 20 G, 30 G, 35, 40, 50 G you can see that the settlements continue to increase so these are the crest settlements and these are the settlements away from the you know from the crest of the wall. So again so as been told here so this is actually from the front elevation can be seen that and this is the wall so these are the step type deformations are actually shown and the close view of the wall with the Rankine rupture planes is actually shown here so you can see that this one plane passing here and one plane passing here and the plane actually finally culminates at this particular point here. So from the data which is actually obtained from strain gauges and the bending moment is actually plotted so you can see that the bending moments are found to increase with G level so this is at you know at 10 G, 20 G, 30 G like that and once we have you know dropped the gravity to you know to 1 G so there is a net bending moment which is actually is shown here also. So you can see that with a during once centrifuge stress to actually reach beyond gravity level up to 35 G so you can see that you know the sharpness of this curves tend to increase and with the actually which actually happened because the wall is actually has been subjected to a bending moment which is actually more than the plastic moment capacity of the wall if that attains actually there is a attenuation of plastic hinge takes place that is what actually happened in this particular case. So here the same issues actually you know compared with you know finite element analysis of the same problem with increasing gravity by using spandine software where in when you compare here you can see that at this particular point the plastic moment capacity of a 3 mm thick wall made of aluminum having you know EG that is the E of aluminum about 72 GPa which is actually obtained as you know the plastic moment capacity is about 326.67 Newton meter per meter. So it can clearly see that this particular point actually it has crossed that plastic moment capacity and bending to our formation of a plastic hinge failure of the time. In case of another method also you can see that the sharpening of the bending moment can be observed. So this is the deformed deformation of the soil elements on the sheet pile walls at 35 G basically you can see that here also the plastic hinge formation can be seen very very clearly. So variation of the cumulative maximum bending moment with G level is actually plotted and when you see that as with an increasing G level the bending moment is actually increasing the same situation also measured in the you know FEM and this is actually level where you know the so called 326.6 Newton meter per meter that is the plastic moment capacity. So you can definitely say that somewhere between 35 to 40, 40 to 45 the wall actually attained are you know the plastic hinge formation actually has taken place, inception has taken place and developed further. So this is what actually you know we have proved by using this. So you can see that the post investigations have actually revealed that so called you know the plastic hinge actually formed at 52 mm below the you know below the dredge line and this is the 150 mm which is the depth. So you can see that this is actually point physically you know here at this point. So you can see that you know this is the point where the sharpening of the bending moment also occurred. So this is you know this is also correlated with one of the you know problem which actually happened in one of the project in one of the sites where in in order to have some diaphragm wall construction 7 meter you know behind this particular face of the wall. What actually has happened is that the wall was actually observed to deform at this level the wall was actually observed to deflect about 0.5 meter also. When the post investigation analysis actually have been carried out, test unit has been found that you know the wall which is element which is used was having inadequate suction modulus and that led to the formation of a plastic hinge. So you know this is actually the relevance of the what particular practical problem what actually has been discussed but the only difference is that in case of in this problem there is a dredge level and then occurrence of what is also there. So let us consider you know a typical you know failure study for a test at 40g. So here it same container is actually used where in we actually have got a rigid retiring wall aluminum and hinge is actually placed at this point. So this is 0.24 meter at 40g will be equal to 9.6 meter and this will be 0.4 meter at 40g will be equal to 16 meters and this is 0.36 meters this is equal to about 14.4 meters. Enter depth of the container is about 0.41 meter that is 16.4 meters at 40g. So the base layer is actually having 0.06 meter into 40 is about 2.4 meter and here this portion is actually fixed in order to prevent the sand particles entering into the hinge thin polythene sheets are actually placed here as shown in this figure. So what actually has been done is that the wall is actually propped initially with very high pressure and the centrifuge gravity is increased from 1g to 40g. In order to prevent the passive you know passive mode of failure a mechanical stopper was actually provided this side so that you know the wall will not move towards the you know towards the backfill. So here particularly we are interested in modeling the active mode of failure for a retaining wall. So once we wanted to model the active way of failure what actually has been done is that once we reach to that particular gravity so this when we withdraw the pressure at this point to 0 then what will happen is that the wall tends to move away from the backfill. So this is clearly shown in the next slide with a GIF animation which can actually shows that this is actually picture taken by a camera at 40 gravities. So you can see that the pressure which is actually released when titanium is in 40g so you can see that the formation of you know rupture planes. So this is the test which is actually we have done at 40 gravities and where you can see that the failure surface actually moves like occurs like this. So this is another typical example of you know the retaining wall mode of failure particularly active state of failure which actually can be is actually tested at IIT Bombay. So in this case consider you know let us say that we wanted currently very recently in 2013 a in flight support system was actually developed wherein what we have is that let us say we have got a reinforced soil wall or a soil nail wall or you know you have got a particular wall. So what we actually have is that we have got a wall support system which is attached to a frictionless bearings and there is a mechanical stopper and then in a pneumatic cylinder is actually placed here. Let us say that we actually have got two ports and one port allows to put pressure here other port allows to pressure put in the pressure in the reverse direction. So let us assume that initially we have got a P2 is applied and the wall is actually held you know coming toward this side by a restraining with a force adequate force and here also the care has been taken that the wall will not move by putting a mechanical stopper towards the soil wall being tested. So let us look into this how this happens once we this is at 1g status so once we go we actually go to ng that means that we have got an equilibrium and k0 condition is achieved by you know establishing forces equilibrium and forces because of this pressure applied and because this wall is actually supported on this frictionless bearings so the wall height, wall weight and all those things will not come into the picture here. The next level what we do is that we try to apply pressure P4 and remove the support so that now this particular wall at say ng if h by n is equal to is the wall height and when this actually happens at ng so its equivalent height is equal to h meters in the prototype. So let us look this in the you know the real you know demonstration wherein we can see that a 10.8 meters of wall at 40g you can see that how the wall is undergoing deformation because this particular technique allows one to test actually you know the movement of the central wall support system is also monitored by using you know you can see the LUDT which are actually placed here and they measure how much so this actually moves by about 5 centimeters in a short duration. So with that what actually happens is that we can actually see that how the wall undergoes movement and then this actually can have an impact on the you know the deformation behaviors and other aspects can be studied. So this is you know type of you know test which actually has been done by at IIT Bombay by developed in-flight wall support system at 40g. So this is a typical some reinforced soil wall constructed with marginal backfill material and where it is compacted at wet side of the backfill so you can see that there is you know the tension cracks are actually formed but a multiple number of tension cracks are formed and which actually led to the excessive deformations of a wall at the top of the in the top most zone and this is so similar situation if the wall is actually there in the field this one then you know we can see that this type of deformations can actually happen at for a 10.8 meters wall. So in this particular lecture what we try to understand is that the relevance of the centrifuge based physical model testing where we are trying to bring and then based on that we actually also try to see some selected examples. So in this module the geotechnical physical modeling where we have seen that you know how small scale physical model testing at particularly carried out at high gravities is relevant to many of the geotechnical problems and we also have brought out is that if you are actually having certain conditions like say footing resting on clay there is a possibility that you know the 1g model testing the small scale 1g model testing also holds good. So this is how the centrifuge based physical model testing is also applied for number of geotechnical problems for studying for subjected to different types of forces which actually gives the you know the behavior which is actually close to the real full scale structure.