 Welcome to course on advanced geotechnical engineering module 6 in on buried structures lecture 3. So in the previous lectures we have discussed about Marston's load theory for resident flexible pipes and different trench and projection conditions. Now let us look into what are the requirements for the minimum cover and flotation and liquefaction effects on the buried pipes. So in this lecture we are going to discuss about the what are the general loading conditions are required and then minimum cover which is required and flotation and liquefaction effects on buried pipes. So as these pipes are subjected to different types of loads one such load is live load if it is under installed below a railway embankment or a highway embankment subjected to wheel loads. So as the load is transmitted down the effective area increases and the pressure decreases. So naturally what happens is that as the load which is applied at the surface gets diminished once you know the reaches to the top of the pipe where the pressure decreases. So Bosnik's theory can be used to calculate the pressures on the different types of loads and the embedded pipes and details of live load calculations are actually connected with the minimum cover requirement. So like we can assume that wheel load is actually you know happening on the right on the top of the pipe or at happening at a certain distance away from the pipe. So one of the predominant loads which can affect the integrity of the pipes is that the live loads. So as the live load is actually transmitted down the effective area increases and the pressure basically the transmitted to the pipes will decrease but however the intensity of load transferred to the pipe due to a particular load can be calculated by using Bosnik's theory which is actually there in the stress distribution component of geotechnical engineering. The another you know effect which is need to be considered if these pipelines are installed in the seismically prone areas that the seismic loads need to be considered. So as a general the stresses induced in pipe walls due to seismic strains are quite small and do not adversely affect the design but the design codes usually allow for an increase in the allowable stress or conversely decrease in the load factors when seismic loads are induced included in a load combination. So the buried pipes that are sized to sustain other design loads usually have sufficient strength to resist the seismic imposed stresses but however there can be some effects like effects due to liquefaction and because of the or flotation of the pipes those issues need to be addressed. But as such the seismic loads if it is if the pipelines are passing through seismically prone areas the adequate consideration should be made to have the seismic loads. Then another you know important loading consideration which need to look into it is internal pressure. Basically these underground pipes or buried pipes they operate under basically there are two categories one is called high pressure pipes and low pressure pipes. The underguide pipe systems have to operate under varying levels of internal pressure. Very sewer lines normally operated very fairly low internal pressures that is the pressures in sewer lines are very low. So they can be sewer lines or can be classified as low pressure pipes whereas the water supply means an industrial process pipes operate under high internal pressures. That means that these pipelines for say cooling water running pipelines in same power plants and all they operate when very high pressure high pressures. The high pressure pipelines have to be designed for a continuous operating pressure as well as you know the short term transient pressure and these pressures also sometimes because of the rapid closure of the valve or something like that there can be possibility that you know the surges can actually happen or it is also called as a water hammer effect. So in a short pulse of time there can be a possibility that the pressures can actually increase beyond the operating pressures. So the high pressure pipelines have to be designed the high pressure pipelines have to be designed for a continuous operating pressure as well as the short term transient pressures and the short term transient pressures in the mean that it can arise because of a water hammer or water surge which can actually increase the pressure in the pipe in a short duration of a time. And the internal pressure another issue is that vacuum this vacuum can actually arise because of the you know the closure of the valves of an empty pipe. So sometimes what will happen is that certain operational events may cause a temporary vacuum in the buried pipe conduits. So one need to really concern about this particular issue in most cases the duration of application of vacuum loading is extremely short and its effects can be delineated from other live loads. The magnitude and the time variation of the transients need to be considered for both positive and negative internal pressures. So the one need to look into the magnitude and the time variation of the transients basically for both the positive and the negative internal pressures. Then another issue is that the pipe and associated contents the effects of the dead weight of the pipe wall and the fluid carried out fluid carried must be resisted by the structural capacity of the pipe. So the effects of the dead weight of the pipe wall and the fluid carried must be resisted by the structural capacity of the pipe. In practice load from these two sources are often neglected in the design of steel or plastic pipes but accounted in the design of free stress and reinforced concrete pressure pipes and concrete non-pressure pipes. So for simplicity these loads are to be added to the vertical soil loads keeping in mind that small magnitude of these loads. So these loads are actually added to the soil loads so keeping in mind of the you know as they are actually the small in the magnitude of this load is actually small compared to the soil loads. So the effect of the dead weight of the pipe wall and the fluid carried out must be resisted by the structural capacity of the pipe. So now having discussed about the different loading conditions like live loads and then we also have discussed in the previous lectures how you know the for different conditions how the soil loads can be you know calculated by using Marston's load theory. So in addition to that you know these loads due to external loads due to wheel load traffic or seismic loads are due to considered. So the minimum soil cover which is very important as far as buried conduits functioning is concerned. So in the pipeline design analysis of the minimum soil cover required are essential to protect the integrity of the buried pipe under different loading and environmental conditions. In pipeline design analysis of minimum soil cover required basically are essential to protect the integrity of the buried pipe under different loading and environmental considerations. So soil is a major component of the flexible buried pipe, the soil protects the pipe by holding the pipe in shape and then in alignment. So soil basically you know helps to hold the pipe in shape and in the alignment. So the following are the analysis of the minimum soil covers are required for the protection against wheel loads, flotation, uplift and some issues of frost are there but which are not discussed in this lecture. So first what we do is that we will try to look into wheel loads that is due to live load traffic. In case if the buried conduit pipelines are there which are embedded in a location where there is no wheel load traffic then in that case this particular live load issue will not arise. So in the pipeline design the analysis of minimum soil cover required are essential to protect the integrity of the buried pipe under different loading and environmental conditions and we will try to look about the analysis of the minimum soil covers which are required for protection against the wheel loads and due to a buoyancy of the pipe due to the high water table areas that is flotation and uplift and due to liquid action. So in this particular slide case one where you know the minimum soil cover computation is actually shown here where wheel load W directly over the top of the pipe. So here the stress distribution which is actually is on the pipe where that right at the center of the pipe where the magnitude pressure is p is equal to W by 2h square and which is actually decreases you know as it is going away from the center of the wheel load and gamma is the soil unit weight, h is the embedded depth that is the cover above the pipe and d by 2 is the at the mid height so h plus d by 2 is the mid height. So you can see that sigma v is the vertical stress which is gamma into h plus d by 2 that is actually acting at the side of the pipe and the lateral stress the resistance offered is nothing but sigma x is equal to kp into sigma y. So this passive resistance for this provided by the soil and internal pressure that is px is equal to p is actually exerted and which is counted by this passive resistance that is sigma x. So sigma y is the vertical stress that is gamma into h plus d by 2. So the passive resistance is nothing but kp into gamma into h plus d by 2. That is the passive resistance which is the passive the stress in the sigma y sigma x in the x direction. So this is the minimum soil cover requirement or computation for a case where the load is actually resting right on top of the pipe. We can also see that as the embedded depth increases the stress magnitude keeps on decreasing because let us say that in this particular figure h and capital H where capital H is greater than h. In this situation the magnitude of the live loads get reduced at the depth of the as the depth of the backfill is increased as the depth of the backfill is increased the magnitude of the live load gets decreased. So in order to in the computation of this minimum soil cover and if you are actually having an empty circular cross section of let us say having diameter d capital D and if is deflected into an ellipse. So here elliptical pattern of deformation is actually shown here. So here which actually rx and ry are the radii in x and y direction and the ratio of the radii r suffix r is equal to ry by rx. So ratio of radii of the deformed cross section in the elliptical fashion it is ratio of radii is equal to r suffix r is equal to ry by rx which this ordinate is equal to b is equal to r into 1 plus d and this is equal to a is equal to r into 1 minus d. So we can actually write ratio of radii as b by a whole cube as b is equal to r into 1 plus d. So we can write you know this is something like 1 plus d cube divided by 1 minus d cube where d is nothing but the ring deflection that is d is nothing but the ring deflection by how much actually this has actually has undergone changes. So if an empty circular cross section is described into an ellipse then px rx is equal to py ry. So px rx is equal to py ry, px is the pressure in the x direction, py is the pressure in the y direction. So px rx is equal to py ry is equal to pr and this is you know this is the you know for a empty circular cross section if it actually undergoes deflection in the form of an ellipse then you know this issue is actually is valid. Now what we need to do is that in order to calculate the minimum soil cover the horizontal pressure of the pipe on soil at spring line is nothing but px is equal to that is spring line is nothing but that mid height is equal to px is equal to py rr where py is equal to gamma h into gamma h plus w by 2h square that is due to dead load plus live load. So for d is equal to 0% by using this one d is equal to 0 then py a is equal to 1 that is ratio of radii is equal to 1 and for 2% deflection 5% ring deflection 10% ring deflection we can say that r suffix r that is ratio of radii is equal to 1.13, 1.35 and 1.83 this is 1 point r ratio of radii is 1.83 for ring deflection of 10%. So the loading is py is equal to gamma h plus w by 2h square that is due to the red light that is sulphate of the soil and plus live load and the activated, activating horizontal stress on the pipe on soil at spring line is px is equal to rr into gamma h plus w by 2h square. Now the resisting force for this soil strength and spring line is nothing but sigma x is equal to kp into gamma into h plus d by 2 where gamma is the unit rate of the soil and kp is nothing but the passive earth pressure coefficient which is actually 1 plus sin phi by 1 minus sin phi that is the Rankine's earth pressure coefficient we have taken. So here the resisting stress is actually nothing but sigma x is equal to kp into gamma into h plus d by 2 the activating horizontal stress on the pipe at soil spring line is nothing but px is equal to rr into gamma h plus w by 2h square. So this is again the same issue is actually described here the case 1 wheel load directly on the pipe where the horizontal stress applied by the pipe on soil at mid height of the pipe is given here that is the spring line and then the resisting force applied by the soil on the mid height of the pipe is sigma x is equal to kp into gamma into h plus d by 2 where d is the diameter of the pipe and phi is the friction angle of the soil and kp is the passive earth pressure coefficient w is the wheel load and h is the minimum cover and gamma is the unit rate of soil d is the deflection of pipe ring as a fraction of the pipe diameter and r suffix r is nothing but 1 plus d by 1 minus d whole cube. Now by equating this pressure applied by the you know this px which is nothing but by equating this pressure which is applied by the operated applied by you know this px and then resisted by the kx pressure applied by you know the px and then resisted by the passive earth pressure coefficient offered by the soil by equating the horizontal stress applied by pipe on soil and resisting force applied by the soil at mid height of the pipe by equating them what we get is that you know we get the expression for w is equal to 2 gamma h cube into kp by rr into 1 plus d by 2h minus 1. So if you look into this for different value of phi and for value of a typical wheel load and the different ring deflections and we can actually get the typical schematic variation of h the soil cover and with the so increase in soil cover can actually allow us to actually have accommodate higher wheel loads and let us say that you know for ring deflection of say 0% that means that the soil cover of this much 2 units is required and then where in the maximum wheel load which allowed is about 10 units of the maximum wheel load. So by equating the you know this px is equal to sigma x we have got w is equal to 2 gamma h cube into kp by rr into 1 plus d by 2h minus 1. So the schematic variation of h with w for different ring deflections actually is given in this slide. Now as the soil cover decreases the live load pressure on the buried pipe increases so look into that as the soil cover decreases the live load pressure on the buried pipe increases so the pressure transmitted to the pipe will keep on getting enhanced. So there exists a minimum height of soil cover we have to see that the minimum height of soil cover is ensured by all means. If the soil cover is less than minimum the surface live load may damage the pipe and less obvious is that is a minimum height of soil cover for dead load the weight of soil only. So is minimum height of soil cover for dead load that is weight of soil only. So now while considering now we have taken you know the wheel load right on the type now the wheel load is one of the on one of the sides of the pipe. So let us assume that we are actually having a flexible pipe and the wheel load is on the towards the left hand side which is actually shown and it actually has got load is exerted on a like a trapezium and then it is actually transferred to the type of the pipe like this. So now the deformation is caused by the punch through of a truncated pyramid of soil under the wheel. So the deformation is actually caused because of the punch through which actually happens to the truncated pyramid of soil under the wheel. If the ring stiffness is not adequate if the ring stiffness is not adequate the top of the pipe inverts and as the wheel rolls across the pipe. So the top of the pipe will get inverted and as the wheel rolls on the pipe. So you can see that the maximum moment exerted is nothing but Me is equal to 0.02 to P r square. So this deformation is actually caused by the punching mode of failure and through it is like a truncated pyramid of soil under the wheel. So the inversion is triggered by the maximum moment which is given as Me is equal to 0.02 to P r square at about 10 degrees to the right of the center line. So when the load is on the left hand side the maximum moment actually gets you know mobilized on the right hand side of the pipe and this moment causes the bending stress where sigma is equal to mc by i. So the axial stress component is neglected here because of its magnitude. So the maximum the moment m causes the bending stress sigma is equal to mc by i. So the ring compressive stress is nothing but sigma is equal to P by 2 into d by t Pd by 2t for flexible pipe where P is equal to gamma h is actually negligible. So by simplifying what we get is that P is equal to you know P is nothing but the vertical pressure on buried pipe due to the surface wheel load which is actually 30 into sigma r where sigma 30 into sigma y where sigma y is nothing but the yield stress in steel if it is a plastic pipe the yield stress in plastic respective part plastic pipe need to be used into d by t whole square. So this is for within the elastic range in the ductile hinge range then it is P is equal to 30 into sigma y into d by t whole square. So here where sigma y is nothing but the yield stress in steel and so d by t is nothing but the ring flexibility which is actually defined as you know diameter as well as the wall thickness of the plane pipe. Say for example if you are having a about 3.8 meters of pipe having the thickness of about you know something like about 20 mm of thickness then the d by t ratio is about 190 or so that indicates that you know the pipe is actually having very high you know ring flexibility and pipe is said to be flexible in nature transferred as the you know very flexible pipe. So where d by t is equal to ring flexibility which is nothing but the ratio of diameter to the wall thickness for plane pipes. So sometimes for cooling water pipelines and all where the ring flexibility will be of the order of 180 to 180 to 200 and the minimum soil cover for considerations now we are actually trying to look when the wheel load is actually on the you know the live load on the towards the left of the pipe. So what we have done is that we actually have got with you know this particular p is equal to 30 into sigma y into d by t whole square and with that after having obtained now we can actually calculate the wheel load w on the compacted granular soil punches out a pyramid with slopes of about 1 h to 1 h horizontal to 2 vertical with slope angle is equal to about 37 degrees for pressure on the pipe is approximately can be approximated as p is equal to w by b plus h into l plus h this is nothing but where h is the cover you know minimum cover. So by solving 1 and 2 so this is one where by solving 1 and 2 what we get is that the minimum cover h can be calculated. So whenever the wheel load is actually on the you know not on the right on top of the pipe then you know we need to use this particular expression and by using the equation which are actually given in 1 and 2 by solving 1 and 2 the minimum soil cover h can be determined. Now for typical granular backfill based on the analysis confirmed by the test minimum cover is about h that minimum cover need to be ensured that d by 10 that means that if you are having a d is equal to about 4 meters then minimum cover about 4 by 10 about 4 meters cover has to be ensured that is 4 by 10 that is about 0.4 meters cover. An often specified minimum allowable is nothing but h is equal to d by 6 but this implies for very perfectly flexible ring. So if you are having a flexible ring like you know we discussed about d by t is equal to about 190 or 200. So for that actually the minimum cover which is actually allowable is nothing but h is equal to d by 6. In fact the pipes have ring stiffnesses and so provide resistance to dead load collapse. Now after having discussed about the minimum cover requirement from the live load consideration point of view wherein we actually have discussed the wheel load right on top of the pipe and wheel load location on the left of the pipe. And the wheel load is located left on the pipe we said that the inversion of the pipe takes place the moment is actually mobilized on the towards the right side of the pipe. And with that actually we have calculated what is the minimum soil requirement. So where for flexible pipes we said that the minimum cover is nothing but about d by 6 need to be ensured. So the next issue we need to think about is the pipe flotation. See the possibility of the pipe flotation exists when the pipeline is constructed in areas which will be inundated and such as the stream crossings and flood plains and high ground water areas. When such conditions exist evaluate the possibility of the pipe flotation. So whenever also we need to consider the effect of the fluid pressure on the pipe due to let us say if it is in a rainfall areas. In a rainfall area if rainfall prone areas if the pipes are actually installed then we need to even consider the effect of the fluid pressure on the integrity of the pipe. So the possibility of the pipe flotation exists when the pipeline is constructed in areas which will be inundated such as the stream crossings and flood plains and high ground water table or it can be inundated due to the diversion of rainfall water into the trench which actually was used for installing the pipes. When such conditions exist the possibility of the pipe flotation need to be evaluated. So the buoyancy of the pipeline depends upon the weight of the pipe, weight of the volume of the water displaced by the pipe and the weight of the liquid load carried by the pipe and weight of the backfill. So the buoyancy of the pipeline depends upon the weight of the pipe and weight of the volume of water displaced by the pipe and the weight of liquid load carried by the pipe and the weight of the backfill. So as a conservative analytical practice what one does is that the pipeline empty will be considered. This is for two reasons one is that so that the weight of the liquid will be considered as additional safety factor and possible to the pipeline is not being in use during the period of time it also accounts for that. So as a conservative analytical practice we consider that pipeline empty for two reasons. One is that the weight of the liquid will be considered as an additional safety factor. So and the possibility of the pipeline not being in use during the period of time of maintenance will also be taken care. So in this particular slide a typical you know wedge failure which is actually shown and this is obtained based on the model test and if you are actually having a situation of water surface at this level and H is the embedded depth and this is the diameter of the pipe and this is the wall thickness and for granular soil with the shallow covers H less than 5D soils slips like a parabolic surface. So this was actually obtained from the model test data and this Z up to the mid depth of the pipe which is nothing but Z is equal to H plus D by 2. So Z is nothing but this is H and this is D so H plus D by 2 and this wedge is actually having a two vertical and one horizontal this is the wedge which actually undergoes if this Q is actually more than W then the pipe actually subjected to flotation. So we need to ensure that the W by Q is actually having at least the factor of safety of 2 so that the flotation issues can be addressed. Then D is the you know this diameter and Z by 4 is the extent of the parabolic wedge. So this area of parabola is about Z square by 6, Z square by 6. So this is the you know minimum soil cover computation for bionic soil under water to prevent flotation of an empty pipe. So here the pipe is actually considered as the empty condition. So the bionic Q on the empty pipe is the weight of the water displaced so it is nothing but Q is equal to gamma W into pi D square by 4 and D is the diameter which actually acts upward and weight of soil is the weight of the soil wedge bounded by the soil slip planes at an angle equal to the soil friction angle for cohesion less soil a reasonable soil friction angle assumed as 37 degrees for which soil slips at roughly 1 H 2 vertical this is actually based on the model that is actually this inclination is assumed to be about 37 degrees which is also assumed to be equivalent to the friction angle. So the resting force W is nothing but the bionic weight of the soil cover that is hatched cross hatched which is actually given by so the resting force is taken in this particular zone. So this rectangular portion and these small portions which are shown in the actual way and these parabolas here and these parabolas. So the resting force W is the bionic weight of soil cover that is the cross hatched is actually given by W is equal to gamma sub submerged into area. So where A is the area of the cross hatched area where D that area is given as D z D is the diameter and z that is the total rectangular area plus z square by 3 minus pi D square by 8. So that is these two areas what we have considered is that this rectangular area D z and D z we have considered and then we have to take out this area. So we have taken out this area and then we added this area and we have taken entire area and then we have taken this area we subtracted and added these two areas. So with that what we have got is that D z plus z square by 3 minus pi D square by 8. So because of that area of the area shown in the cross hatched portion is we considered and with that multiplying with gamma submerged unit weight that is let us say that if the soil is actually having 20 kilo Newton per meter cube minus 10, so about 10 kilo Newton per meter cube is the unit weight of the submerged soil into area. So a safety factor is recommended generally factor of safety of 2, so for designs purposes increase the calibrated H by a factor of 2. Suppose if you have got say 1 meter then we have to provide about 2 meters with a factor of safety of 2. So minimum cover of granular soil under water, so this is what is the minimum cover H of granular soil under water to prevent the flotation when pipe is empty. So the empty pipe floats if q exceeds y, so the q exceeds w, the empty pipe floats if q exceeds w. So equating q is equal w what we have got at equilibrium q is equal to gamma w into pi D square by 4 equal to w into gamma w into D z into z cube by 3 minus pi D cube by 8. By equating them and simplifying we get z is equal to 0.9 D, so z is equal to 0.9 D which implies that approximated as D, so this is approximated as when z is equal to H plus D by 2 when you substitute that we get H is equal to D by 2. So with a factor of safety of 2 provide that H is equal to D, so when you provide H is equal to D then the you know that is the minimum soil cover required from flotation concentration point of view from flotation concentration point of view. Now pipe uplift what is the uplift force which is experienced by the pipe, see analysis for determination of the force required to uplift the pipe in the soil is also useful. So q is the force that lifts the pipe, so the pipe uplift equation is given by sigma p is equal to k p gamma into H plus D by 2. So this is nothing but by sigma p the pipe uplift equation is nothing but which is lifted by the soil, sigma p is equal to gamma k p into gamma into H plus D by 2 and with that we can actually write by substituting and simplifying we can write that you know 3 cube by gamma D square is equal to H by D plus 2 whole square minus 3.428. So with that what we get is that we get the pipe uplift equation and if you are actually having you know the height of soil cover H is greater than 5 times D that is H by D greater than D then you know the Terzaghi model need to be adopted and with the height of soil cover is H and greater than 5 D inverted Terzaghi model for uplift force and pipe buried under high soil cover heads need to be adopted. So here in this Terzaghi by pipe inverted Terzaghi model where a bow wave of soil is formed as the pipe pours up through the soil, the shear planes that is soil slip planes do not break out at the ground surface, so the shear planes they actually do not break out at the ground surface. So from the analysis at H by D for H by D is equal to H by D greater than 5 we can actually obtain Q is equal to 20 into gamma D square. So for you know for the pipes which are actually embedded with embedded depths greater than 5 H by D greater than 5 a bow wave of soil is formed as the pipe flows up through the soil and the shear planes basically did not break out at the ground surface. So from analysis point of view when H by D greater than 5 the Q is equal to 20 gamma D square is the computed. So here in this particular figure which is actually shown here for the high H by D ratios that is H by D greater than 5 the inverted Terzaghi model is valid. So in that case the uplift force is actually calculated by using the expression given by given here it is nothing but Q is equal to 20 gamma D square and where gamma is the unit weight of soil and here as this is something like a bow wave of the soil is formed as the pipe flows through the soil and the shear planes do not break out at the ground surface. So what are the remediation measures for preventing or mitigating pipe flotation. The buoyancy effects are probably the greatest concern in areas such as flood plains and estuaries where the massive liquefaction could take place in a major earthquake also. So the following recommendations may be considered to minimize the buoyancy effects. Pipelines may be encased with concrete pipes to reduce the buoyancy effects but the increased diameter will also increase the internal lateral drag force on pipeline during the lateral spreading due to liquefaction. So pipelines may be encased with concrete pipes to reduce the buoyancy effects but the increased diameter will also increase the lateral drag force on the pipeline during lateral spreading during liquefaction. The concrete weights or gravel filled blankets can be utilized to provide additional resistance to buoyancy. So buoyancy effect can also be minimized by shallow burial of pipeline above the ground water table. So instead of taking below the ground water table the pipelines can be kept above ground water table. And where the uplift is the main concern anchors may be provided with a close spacing of about 150 meters to prevent uplift. So an appropriate anchoring system needs to be taken into account in order to reduce the buoyancy effect. So the two popular issues one is that anchors use of anchors the other one is that to keep the pipelines at shallow level itself and the other one is that encasement with concrete pipes. These are the issues which can be used the recommendations can be used for minimize the buoyancy effects on the pipeline. The another issue which we have said that when we have got the pipelines which are actually embedded in a trench narrow trenches or in embankment conditions and we said that when we are actually having these fills there is a possibility that below the water table they can be subjected to liquefaction because of the seismic perturbances. So if there is any possibility of soil liquefaction the flotation will be a major concern and additional considerations are required. So soil can liquefy and if it is saturated and shake and if the density is less than about 80% of the modified proctor density. So the soil can liquefy and then once the soil liquefies the resistance to all these uplift and all those things cannot be expected because the liquefied mass will lose its strength. So the shaking can be result of any seismic activity and if the soil is completely saturated to the ground level and the pipe is empty so there will be little resistance to flotation and the empty pipe will raise through the liquid soil. So that the onset of liquefaction what will happen is that because of the any seismic activity and if the soil is completely saturated to the ground level and the pipe is empty and there will be little resistance to the preventing the pipe from floating and then the empty pipe will raise through the liquid soil. So this actually can lead to the failure of the pipelines and many case studies actually have been reported in the failure of pipelines and manhole covers due to liquefaction particularly when they are actually embedded below the ground level and particularly in sandy type of soils and with when they are subjected to this in high water table areas when they subjected to this sort of very high magnitude of these earthquake forces the damages have been reported. So here in this particular slide where the typical pipeline which is actually subjected to you know the so called you know the liquefaction has been shown here. So here the pipeline and this is the liquefied backfill which is actually shown and H is the you know from the water table that is this is below the water table and this portion is actually above the water table. So this is the soil which actually offers the resistance and pressure distribution on pipe in a liquefied soil so buckling at the bottom is actually possible. So the distribution of the pressure at the onset of liquefaction is estimated like this which is actually shown schematically here maximum at the base and minimum at the top and the comparative base the side magnitude is less. So if the embankment liquefies when a circular pipe is empty the ring may be subjected to the hydrostatic pressure shown in this. So this is the you know the so called hydrostatic pressures momentarily will be subjected and that can lead to the damage of the pipe. So in this particular slide what we have seen is that you know this is the countering resistance and this is the hydrostatic pressure distribution at the onset of liquefaction due to some seismic perturbance. Now if somehow the flotation is prevented the catastrophic collapse may occur when they from the bottom according to the classical buckling equation. So if somehow the flotation is prevented the catastrophic collapse may lead to the buckling and this is given by prq by 3 prq by ei is equal to 3 prq by ei is equal to 3 which is nothing but h is equal to e by 4 gamma into t by r to the raise 3 this is for plane pipe that means this gives what is the height h of the water table above the bottom of the steel pipe in the embedded embankment. So lose that it can liquefy and cause the catastrophic ring collapse. So this gives actually the height you know of water table above the bottom of the steel pipe in embedded depth. So that the soil is so loose that it can liquefy and cause catastrophic ring collapse. So we can actually calculate from the combining with the buckling equation. So with that if that is you know if this depth is calculated and if we can ensure adequate factor of safety then we can actually say that the pipe can be prevented from liquefaction. So some appropriate preventive division you know the remediation measure need to be designed if it is actually prone for you know effect due to flotation effect due to liquefaction. So if somehow the flotation is prevented by catastrophic collapse may occur from the bottom according to the buckling equation which we have given. So that depth is nothing but is actually calculated is the height of water table above the bottom for a steel pipe embedded in lose sand and that can liquefy and cause catastrophic ring collapse. So here for the liquefaction so here what in this particular slide a typical forces acting on a pipe in a liquefied soil is shown here. So this is the saturated soil and this is the you know pipe and we have this is the FSP is nothing but the self weight of the pipe and FBP which is nothing but the force due to buoyancy and FEP that is the you know the force acting on the structure and this is the weight of the soil. So what will happen is that when the at the onset of liquefaction the forces acting are actually shown here wherein the shear stresses will acting downwards here and upwards here and then resistance offered by this prism in this downward direction. So this tends to go up and then this is all forces will get mobilized. So the resistance forces actually inhibiting the uplift force due to buoyancy that is FBP are provided by the weight of soil weight of the pipe the self weight of the pipe itself that is FSP which is acting downwards and weight of the overlying soil that is WSP which is acting downwards and the shear developed in the soil that is FSPP that is here the shear developed in the soil however in the case also all downward forces are shear developed in the soil and weight of the soil above the pipe and the self weight of the pipe the rest are actually in the direction of buoyancy is the pressure applied and the buoyancy thing. So the however in the event of soil liquefaction during an earthquake the shear contribution could be reduced significantly. So the FSPP FSPP component will be very very marginal. So the resistance forces inhibiting the uplift force due to buoyancy are basically provided by the weight of the pipe and weight of overlying soil in the shear developed in the soil and however in the event of soil liquefaction during an earthquake the shear contribution could be reduced significantly. So in addition the excess pore water pressure at the invert of the pipe can also contribute to the uplift force acting on the structure that is FEPP shown in the figure. So which is the excess pore water pressure which is actually the excess pore water pressure at the invert of the pipe can also contribute to the uplift force and which actually can lead to the flotation of the pipe. So when there is a positive net uplift force the pipe may float as a result. So when there is a net to positive you know uplift force acting in this direction so then there is a when the pipe can result in the flotation. So F net is nothing but FBP that is FBP which is nothing but the buoyancy force acting in this direction FEPP minus so FEPP plus FBPP minus FSP that is the sulphate of the pipe and FSPP plus FWSP. So F net pressure when it is positive then the pipe will be subjected to uplift force and the pipe may float as a result. So these case studies actually have reported that these issues are quite common when we are actually install this pilot pipes in liquefaction prone areas. So in this particular slide what we have understood is that at the onset of liquefaction if the you know if there is a positive net uplift force the pipe may float as a result. So for the two contributing forces for you know causing the pipe to lift are the one is the bayonet weight of the pipe that is FBP that is the uplift force due to buoyancy that is due to because water table is here the uplift force due to buoyancy and second thing is that the excess pore water pressure at the invert of the pipe can also contribute. So this excess pore water pressure at the onset of liquefaction momentarily will be very high and then the dissipation actually takes place but at that point of attenuation of this you know the liquefaction the excess pore water pressure is so high and then the positive net effect uplift force will be very high and will lead to the pipe flotation. So there are many methods to restrain this liquefaction the liquefaction effect on the buried pipes. So for this actually some investigators like Ling et al they are actually working on you know evolving at the use of geogrids for as a confinement the confining the gravel that means that a placement of gravel around the pipe and which is actually confined with you know a geogrid. So if some means some resistance is also considering the analysis by the geogrid which is actually offered and there are also some you know techniques which are actually tried in the literature like using the synthetic materials like geogrid as a straps at intermittently certain spacing and in preventing the you know the pipe flotation issues and all. So in this particular slide a typical mitigation technique using gravels and geogrid where a gravel confined gravel counterweight confined geogrid is used to counter the flotation at the onset of liquefaction. So how these things are actually tested is that we actually have discussed the physical modeling so these you know these situations actually have to be modeled and then you know once by understanding the phenomenon the design methods can be evolved and then can lead to the design in the field. So after having discussed about these issues now another important is that if you are actually having these steel pipes the corrosion is another issue the corrosion of the buried pipes. So three environmental agents usually exert strong influence on corrosion of the pipe wall material in buried installations particularly the water or other fluid carried by the pipe and the soil in contact with the buried pipe if it is in saline in nature then you know it can lead to the extensive corrosion and the ground water. If the ground water is actually having high chlorides and sulphates then they can also can cause a cause of concern for the corrosion. So the three environmental agents that are actually usually exerts the strong influence of corrosion which we have discussed one is water or other fluid which is carried by the pipe. So these corrosion of the buried pipes which are actually shown here and the protections are basically done by application coating and cathodic protection and cathodic protection is basically proven to be very successful in providing leak free high pressure oil and natural gas pipelines throughout the US and the other one is that coatings can be used to limit inhibit corrosion or other forms of deterioration in both concrete and steel pipes and type and extent of the coating depend upon the service environment and coal tar enamel and wrapping has been used successfully in USA for several decades. So epoxies and urethane among other you have become popular in more recent types. So in recently also in India these urethane coated you know corrosion protection systems are actually in place. This is the cathodic protection which is actually shown in this particular slide. So now we actually look into a case study where a particular buried pipe of about you know 3.8 meters diameter embedded in a soil cover at a with a soil cover about 1.5 meter that is about D by 2 and the so the loading considerations one need to consider what we have understood is that the load due to pressure generated by the flow and external pressure by the fluid if the pipe is submerged under water and external pressure generated by the weight of the earth and live loads on buried pipes and load due to thermal expansion at so load due to thermal expansion earthquakes and external pressure by the fluid if the pipe is submerged under water in the eventuality of in this particular case study we will try to look into it what will be the effect of the external pressure which is actually arisen because of the onset of rainfall. So this is you know typical installation of a flexible pipe at the site where the trench is actually made and this is the natural ground water level and the pipe is actually made with the spirals here which are actually prevented these are used for you know retaining the shape temporarily but once they are actually commissioned then these will be removed. So this is a typical strata you can see that having certain amount of fines and cohesion in it but the ground water table is actually here. So this is one of the pipes like this there are about 4 number of pipes which are actually shown here. So this is you know the other amount of pipes so you can see that the how important is the soil support so the event of changes in the you know the deformations in the pipe so the gaps can actually arise so these gaps actually can also lead to so these are actually due to some increase in pressures or deflections which actually takes place. So the ring deflections are evident actually here in this particular figure. So this is the same figure which is actually modified here. So this is a typical case here pipe 1, pipe 2, pipe 3, pipe 4 where in this area what has been done is that the rainfall actually has been simulated and then the water actually has you know has been water flow has been simulated when this has been done what actually happened is that pipe 1 and pipe 2 were found to actually under submerged water and pipe 2 is actually already with internal pressure of about 500 kilo Newton per meter square but the pipe 1 was actually found to experience distress due to high external pressure so that we will see how that actually you know can be seen through a video. So here in this particular issue the rainfall intensity which actually happened at the site is actually simulated then it can be seen that as the rainfall is actually happening or the duration the water table actually raises up to this level. So you can see that both the pipes are actually under submergence then we will see the other case this is again with a trench in this case also you can see that the pipeline actually is submerged with the water table as we go down you can see that the water table diminishes as the rainfall is simulated you can see that the water table raises and stands above the pipe 1 and pipe 2. Now we will see the third case so in this case a remedial measure is in place where in a filter layer actually has been placed at this level so this is the application of soil mechanics where in like in earthen dams a filter layer actually if it is induced what will happen is that it actually prevents the water table and maintains the water table right at the bottom of the pipe and which actually prevents reduces the external pressure and it will prevent the pipe from losing the support from the soil. So in that case what will actually happen is that the stresses have become very high and it led to very high amount of non-informed stresses and which actually has led to the failure. So this is the situation where the water table is drained or diverted with you know the so called drainage layer which is actually placed to the sooting the site condition. So the typical failure which actually has actually happened at the site here is actually shown here so here this particular pipe after excavation is actually found like this so this is because of the at the onset of you know rainfall inundation you can see that the pipe actually has undergone a buckling failure and 3.8 meter diameter pipe which actually has undergone a distress. So these issues of loading on the burial conduits is very very severe particularly this flexible pipes with flexibility ratios about D by T of the order of 200 they have to be adequate care has to be taken in the design by taking these you know appropriate loading considerations. So in this particular module what we have understood is that what are the different types of pipes there are two types flexible pipes and rigid pipes and then we also discussed the loading theories, martens loading theory for different considerations and then what are the different types of loads and what are the different type of loading conditions. Then we also discussed minimum soil cover requirement and pipe flotation particularly due to bionacy alone and due to the liquefaction effects.