 Welcome to lecture number 17 of advanced geotechnical engineering course module 2 lecture 6 under permeability and seepage. So in the previous lecture we try to understand how to construct flow nets if we are having anaestropic conditions that means the different permeabilities in horizontal and vertical direction and when we have layered soils like having stratified deposits in such situations when permeability of two soils when it is less what will happen when the water is entering from a soil having low permeability to high permeability or high permeability to low permeability how the flow nets can be constructed we have discussed and we also have discussed about how to construct flow nets for unconfined seepage conditions that is nothing but embankment dams or ethan dams and then also different construction methods for the priatic surface which is the line of seepage or is also called as line of saturation of saturation line upper most line or top most flow line which is called as priatic line namely we have discussed earlier about Dupont method and Schaffernack method and Casagrandes method we have discussed. In this lecture we will be discussing about some problems based on whatever we have discussed with numerical as well as some physical simulation which we have carried out at IIT Bombay. In continuation of that we will try to discuss how to evaluate these performance with turn without filters particularly when we have to what are the general criteria which is required for a filter design we will try to cover briefly there after we will try to calculate how to calculate the factor of safety against piping failure by both the methods Azra method and Terzaghi's method for the two cases one is when you have a sheet pile wall that is when you have a confined seepage conditions and when you have a concrete dam with a cutoff how to calculate. So this particular lecture is titled permeability and seepage number 6. So in the previous lecture we actually have discussed how to construct a top most priatic surface this is nothing but the BC is nothing but the priatic surface. So there are number of methods which are actually given and some entry and exit conditions are also given and which we are not discussing but they are required to be referred from the relevant test books. But if you look into this flow net construction for an earth dam here AE which is the dam resting on impervious stratum AE happens to be a flow line AB is an equipotential line and DE is equipotential line with a head H2. So the head drop or potential drop is nothing but H1 – H2 and BC is the flow line what we are actually discussing is the top flow line and 1, 2, 3 or you can say the 2.5 these are the flow channels they should be constructed such a way that we have the aspect ratio B by L is equal to 1 and CD which is neither a flow line nor an equipotential line this is a component of the flow normal to the CD and water flows freely down the surface of the slope. So CD are a point of conflict we can actually say that point C a point D where which is you know appear to be a point of conflict in the given example which is here. So these are the potential drops because this being here the it is easy to calculate it is generally between each equipotential line so this is the equipotential line 2, 3, 4, 5, 6. So what actually happens is that when as the pressure on the top most flow line is 0 so the total head is equal to pressure head plus elevation head when the pressure is 0 because of the atmospheric it is open to atmospheric condition then the total head is equal to elevation head. So depending upon the whatever the location is there we can divide this into say number of drops that is H1 – H2 by whatever may be the number of drops we can simply say that delta H, delta H and delta H equal potential drops can be divided as the water flows from this point to this point there is a dissipation of head which actually takes place. This is an embankment dam with a permeable filter these permeable filters are actually contained they are placed basically in the horizontal direction or they are placed at the toe in the form of a rock toe or they can be placed within the dam within the earthen dam as a chimney so it is also called as a chimney drain. So in this example the embankment dam with permeable filter construction is shown here wherein as suggested by Cassagrande the basic parambola is assumed to be start at 0.3 delta the delta is nothing but a point from this distance from this distance and beta is equal to 180 degrees and because with beta is equal to 180 degrees the delta L by L plus delta L is equal to 0 and here this is what actually I was telling about the delta H delta H and delta H these are the drops and so here this is the equipotential line with zero head this is the equipotential line with the zero head and here this is the topmost flow line and this is being equipotential line you can see that this has to be orthogonal so it actually meets this at a right angle. So here also this being the horizontal this being the flow line so the equipotential line actually meets here at the right angle right angles. So this is a you know how the flow net construction will happen or the flow happens in real conditions when we have a permeable filter. So in this particular slide the seepage through homogeneous sand beneath the base of concrete dams is shown here suppose if you are having a concrete dam which is having this configuration and there is some apron layer which is placed here and this is called as the cut-off layer basically this cut-off layer is used to divert the flow away from the in the tail water level to increase the factor of safety against piping and other conditions. And here this cut-off can be in the downstream direction or cut-off can be in the upstream direction also so in this case example where the cut-off is in the cut-off wall is provided in the downstream end and here in this case cut-off is provided in the upstream end here this is the case where there is no cut-off is provided but you can see that how the flow happens. And here for this construction the impervious blanket has been placed here and then the flow which actually cannot take place through this so which is diverted away from the face of the concrete dam so which actually we have the extended flow net which is like this. So here we have a granular filter and there is apron which is actually placed here in all the cases where the apron which is actually shown. So for this case also when you have a permeable sand layer or permeable soil layer beneath the concrete dam and how the flow net actually happens so depending upon the configuration and head of the water and high flood level requirements and all one need to have the same parameters and perform the seepage analysis and based on the results like uplift pressure and then once we have the other issue other factors like tough safety against piping and then exit gradient once we calculate which we are actually going to discuss in this lecture then it is possible for us to decide about an appropriate configuration for a particular site specific problem. In this particular slide a seepage through homogeneous dam consisting of very fine clean sand is shown here so here there is a granular filter is shown so the granular filter which is actually placed here makes it you know the flow net will be like this but this is a flow net which is actually shown for a dry weather condition. Suppose when we have a condition of continued rainstorm then the there is a possibility that the flow nets construction will actually change and leading to instability of a dam because of the seepage failure. So here because of the continued rainstorm there can be a possibility that this type of you know the complete saturation can take place and the flow net can change the way it is actually shown schematically here and that can lead to the or endanger the stability of the earthen dam or homogeneous dam under question. So in order to maintain or in order to contain the periodic surface within the you know the earthen dam body or condition with unconfined seepage condition one of the alternatives used to design appropriate filter materials when seepage water flows from a soil with relatively fine graze into a coarser material there is a danger that fine soil particles may wash away into the coarse material. So when seepage water flows from a soil relatively soil which is actually having relatively fine grains into a coarser material there is a danger that the fine soil particles will get washed away into the coarser material and then you know the what will happen is that the coarser material which is actually having high permeability will get blocked. The similar situation happens when fouling of you know railway ballast when that occurs they have they lose the you know permeability or drainage property whatever they have and also you know load sharing behavior also can get affected in case of track underlay structures. So such situation can be prevented by the use of a filter or a protective filter between the two such soils that means that when you have a soil to be protected and when you have a soil which is very hyperbole. So in order to prevent that situation of washing of fine soil particles into coarser particles there should be a filter layer then what should be the criteria for selecting this particular filter layers nowadays with the advent of materials in civil engineering some of the materials like geosynthetics preferably a non-oven geotextiles which are actually having adequate or appropriate opening sizes or recommended because there is with construction and also the timelines can be very significantly less. But moreover the performance is also superior because of the Tyler made product whatever we install but however some of the issues like clogging during the lifetime of its operation and so then this use of this synthetic materials also can lead to some sort of replacement of natural materials because the availability of the natural materials is scarce now and in such situations one of the viable options is to try out for modern materials like non-oven geotextiles or geocomposites for this application. So in this particular slide a typical flow net for an earthen dam with rock toe filter is shown. So rock toe filter is nothing but this is the soil to be protected and this is the rock toe and here this is the filter material. So this filter material what it does is that it prevents a fine soil particles to get washed out into the coarser portions and then if we are able to protect this then what will happen is that it ensures the stability to the structure and then the conditions of the dam stability can be ensured. So without filter at the toe the seepage water would wash the fine soil grains into the toe and undermine the structure so it basically undermine the stability of a structure. So this is an example what I was mentioning about a chimney drain where you have what a horizontal drain and then it is extended with a column of sand and sometimes in when we use some previous materials here there is a layer which is called as hearting is also used. So next to that hearting layer so hearting layer is nothing but a core which is having very high very low permeability and which is having high compressibility characteristics as well as high plasticity characteristics so that that remains within the along the center line of the bond and then the sand drain layer in place of for chimney drain can be constructed. But here also like you know attempts can be made to replace these layers because the constructing this layer is difficult then only if this has to be constructed we have to construct in this layer once this layer is placed and then compacted like this and then you have a this layer once it is compacted and then this material need to be placed and then compacted so the construction actually goes like this in the field. So this is a typical flow net for an earthen dam with chimney drain so here also what I mean to say is that one can think of replacing these conventional sand layer which can be used in the which is being used in the general construction now with a geosynthetic material like nano and geodistile but however the research has to address the long term performance of such systems. So before that let us look into the what are the conventional criterias which are actually available for the proper selection of the filter material. The similar criterias also have to be are there are available for the non-oven geodistile when we wanted to replace the conventional filter material that is nothing but the sand. So for the proper selection of the filter material that is the sand the are a suitable soil the size of the voids in the filter material should be small enough to hold large particles of the protected materials in place so that means that D the criteria which is actually given is that D15 of the filter ratio of D15 of the filter to D85 of the soil should be less than or equal to 4 to 5. So D15 filter is nothing but diameter through which 15% of the filter material will pass. D85 soil is nothing but diameter through which 85% of soil to be protected will pass. So this is criteria 1 and the criteria 2 is that the filter material should have a high permeability to prevent build up of large seepage pressures and hydrostatic pressure at the in the filters. So here in the filter material if the material is used is fine then there is a possibility that you know the build up of high hydrostatic pressures can happen. So for that D15 of filter to D15 of soil should be greater than or equal to 4 to 5. So D15 of the soil is nothing but diameter through which 15% of soil to be protected will pass. So once we have let us say the gradation or particle size distribution of soil to be protected, particle size distribution of filter then based on say particle size distribution of you know filter then you know we can actually see based on this criteria the ideal band for the ideal band for the band of range of particle size distribution which can be used for as a filter material. So in this particular slide a typical epthundam sections without any drainage is shown. So here this is a physical simulation which is done through centrifuge based physical modeling technique at IIT Bombay. So the model which is what we are seeing is the front elevation of the model when it is subjected to flooding in the upstream side. So as can be seen as the water on the upstream side increases and with the presence of without any drainage or let us say that we have a drain which is clogged then the possibility of its failure or under and you know the stability of a section can be seen where it got affected the moment the periodic surface reaches to the toe of the dam which is actually shown. So this type of physical simulations will allow us to understand the performance of this particular conditions in full scale level and also allows us to try out a new phenomenon or new materials where in we can actually see the response of this models with the different materials and then can lead to using the field or formulation of the guidelines for the construction of dams or the hydraulic structures in the field. So this is the physical simulation once again I am actually seeing here showing you here that this is nothing but the food dye what you can see is that as the water comes these are the telltales for telling us that this is the uppermost periodic surface which actually has developed and then you know the failure has actually occurred here and then endangered the stability of a dam section which is shown here. So let us see the numerical simulation of earthen dam the similar section without any drainage and with clogged drain. So this simulation was carried out by using geo-studio 2012 version and where in the C-PW model 2012 was used and this is basically a finite element based program which allows us to simulate or mimic the behavior whatever as close as possible to the field conditions or the physical simulation conditions which are actually shown. So let us see the video simulation of that. So here what we see is that as the head of water progresses you can see that the way it happened in the physical simulation the progress of the periodic surface a flow into the these are the flow vectors which are actually taking place can be seen and the periodic surface development can be seen. So this shows us that you know how you know the simulations can actually help us to understanding the you know the flow net behavior and the assessing the stability of the structures. Now we will try to see in this particular slide this particular condition which actually has same condition by using the same program but here what we have done is that this allowed us to do the seepage analysis for about 30 days then you know this is the condition which is actually develops here and then it undergoes failure here. So this is a condition where we actually have got again a physical simulation but in this case you can see here there is a permeable sand layer is placed that is that horizontal drain or horizontal filter is placed and the same height but which is 6 meters but now you can see as the head of water increases on the upstream side on this is the improvised layer this is the improvised layer which is actually constructed with clay layer and this is with a silty sand layer and this is with very fine sand placed at 85% density but as you can notice here that as the water progresses that this being the horizontal drain layer you can see that the water or the pre-artic surface meets this at 90 degrees closely you can see here this is how it is. So this indicates that how appropriate physical simulations can tell us about the real response of these hydraulic structures. So this is particular again similar case with transient seepage analysis during with a drain for 30 days so you can see that as the days progresses all the pre-artic surface is maintained you can see that here also from the numerical analysis by using the Cfw we could get the similar results. So we will see again the simulation of you know this by using the Cfw program but this is with a 0.6 meter drainage so you can see that now this drainage layer is indicated here and this is the drainage layer so as we observed in the physical simulation it actually meets the you know similar way and the flow is contained within the dam itself the flow is contained within the dam itself you can see here. So this is the pre-artic surface which actually develop otherwise previously we are seeing that the pre-artic surface is actually going with the here so this indicates that the importance of having an appropriate filters within the dam sections. Now let us look into some problems based on whatever we had discussed in the previous lecture. So this is a problem with for a single row of sheet pile wall structure we need to draw the flow net diagram and given that the soil is isotropic in the case 1 or case a is soil is isotropic which is having a permeability of 10 into minus 3 meter per second and second case that is case b is that soil is anisotropic where the ratio of the permeabilities kx is 6 times kz. So first case is isotropic so let us use the same Cfw program of geo studio 2012 and we will try to do it for isotropic case. In isotropic case once we get the construction so this is the FEM mesh for the sheet pile wall section so wherein we have here 15 meters of the upstream head and then where we have the here the 5 meters so the differential head here is about 10 meters and this is the permeable soil layer and the extent of this horizontal distance here is 20 meters and here is 20 meters which is taken and here also 20 meters in this problem. So this is a typical hydraulic structure constructed with a sheet pile wall and these are actually sometimes this is can be a part of a coffer dam suppose if you wanted to construct bridge pier foundations within the river or in any case any situation where you need to actually ensure the stability of these coffer dams are constructed in the form with the sheet pile walls against the instability problems. So this is a case A the isotropic section here you can say that these are the flow channels and flow net which is actually given by the Cfw program and this is the flow channel 1, 2, 3, 4, 5 and then if you take it this as 5.5 the number of flow channels which are actually involved is 5.5 and this is the you know the flow line so you can see that the equipotential lines are this being the flow line equipotential lines are actually commencing here and then dropping down at 90 degrees here so the orthogonality between the flow lines and equipotential lines can be seen and these curvilinear squares which is actually having approximately aspect ratio 1 can also be seen. So the program which actually gives this once you get this data then this particular package also this particular program also has got the capability to give the drainage or by using this we can actually calculate by number of channels and number of potential drops we can actually calculate by knowing the k we can calculate the C page or permeability of the discharge through the particular permeable sand layer where in the sheet pile wall section is embedded. Now the case B is that we are having a case which is anastrophobic section and here the program actually has automatically taken this condition and here you can see that because of the anastrophic in nature the even after having a transformed section the flow lines and equipotential lines are not orthogonal to each other. So now construct another problem so in the previous case we have seen a sheet pile wall problem now let us see a flow net for the dam section resting on two layer soil deposit with k1 is equal to 5 into 10 to the power of minus 2 mm per second and k2 is equal to 1 into 10 to the power of minus 2 mm per second. So hence we can see that when we have got homogenous soil deposit as well as when you have got say two layers what will happen to the flow nets and how these things can be used for constructing or calculating the discharges. So here we have two layers of soils one is k1 having permeability 5 into 10 to the power of minus 2 meter per second and another one is k2 having permeability 1 into 10 to the power of minus 2 meter per second. So k1 is 5 times more permeable than k2 and this is a concrete dam which is actually having a head of 10 meters and the horizontal length is about 15 meters. This is the upstream water level and this is the tail water level and in the field the seldom the situations are like this we have got a layered soil deposits. So this particular layer is indicated with this is a low color and the other one is indicated with a light green color. So when we construct the flow net by using CW program so this is the you know equipotential line here and what we can see is that this is the permeability and this is the upper layer having permeability 5 times the higher permeability than this one. So based on the conditions where what we discussed for having non-homogeneous soil deposits since here k1 by k2 is 5 the length to width ratio of flow elements in layer 2 with respect to layer 1 is 1 by 5 why because we have said that when you have got two layers then k1 by k2 ratio is equal to B2 by L2 by B1 by L1 by having you know B1, B1 is nothing but this particular width between two flow lines that is B1 and along the you know length of the equipotential line that is L1. So here you can see that the rectangular the squares are almost we have the flow net is actually having curvilinear squares so it indicates that with B1 by L1 is equal to 1 and k1 by k2 is equal to say 5 then B2 by L2 or breadth to the length of the along the flow equipotential lines equal to or L2 by B2 is equal to 1 by 5. So when you have say a permeable layer on the top and impermeable layer on the bottom then we actually have the get the flow net or the equipotential lines with having large rectangles. So that means that here the B2 by B2 by L2 is equal to 5 so the breadth will be you know let us say that if breadth is 1 unit the length will be 5 units because the permeability is 5 times. So if you are actually having say impermeable layer below and impermeable layer above and permeable layer below then the situation is that again there is a possibility that here we get the you know the small rectangles and because of the ratio of the change in the permeabilities of the layers. So here what we have seen is that when you have got non-obosinous soils the situation how the flow nets can actually change and this actually we have discussed here. So this example has given us you know clear idea how we can actually you know construct the flow net for having soil which is actually k1 by k2 is equal to 5 that is upper layer is actually having higher permeability than the bottom layer and now you know the same problem the drop in the pressure head along the length of the dam due to seepage losses can be seen here. So the B is the breadth of the dam and you can see that how the as the water flows from the upstream end to the downstream end how the you know head of water is dropping along those equipotential lines which are actually shown. Now the next example problem is that the construct the flow nets for the dam sections which are actually this is the earthen dam sections where in the first case the soil is isotropic the permeability is of the order of 5 into 10 to the power of minus 5 meter per second for the dam section and second case is that the dam is joined such a way that now here k2 is 5 times the k1. So k1 is the permeability of the soil 1 and k2 is the permeability of the soil 2. So in this particular example with case A where homogeneous dam having constructed with a soil having 5 into 10 to the power of minus 5 meter per second the FEM mesh which is actually fitted to the CW program is given here and where the upstream head of 10 meters can be seen here and this is the you know equipotential line and is given as the impervious phase. So the flow net which is actually constructed for solution for the problem with homogeneous section case A is shown here where in you can see that this is the line of C page or topmost flow line and this is the next flow line and then this is the next flow line. So here we have 1 and 2 and 2.5 it is number of flow channels are 2.5 and number of potential drops are 10. So the program actually gives like 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 these are the 0 these are the you know potential drops. So we can see that by knowing this we can actually calculate the discharge or leakage through the dam section where k into h by Nd into Nf. Now the same problem but now in this example the dam is constructed with two different materials in the layer which is on the left hand side is actually having permeability k is nothing but k1 is nothing but 1 into 10 to the power of minus 2 mm per second and here it is a 5 into 10 to the power of minus 2 mm per second. So the permeability of this layer is 5 times more than the soil 1 which is actually indicated with a green color here. So flow net for the seepage through a zone that dam this we have actually discussed for the you know as a general example but now with reference to a 10 dam the soil for the upstream half of the dam has a permeability k1 and the soil for the downstream half of the dam has a permeability say k2. So in this example we have given that it is given that k2 is equal to 5 times k1. So by using k1 by k2 is equal to B2 by L2 by B1 by L1 we can see that now with k1 less than k2 when you have a layer soil 1 with less permeability than the soil 2 then we have square elements with L1 is equal to B1 here and this is the alpha 1 and this is the boundary between the two soils having k1 and k2 then you can see that how the flow net changes and with a small rectangles here because the virtue of the change of the permeability. So for k1 by k2 is equal to 5 B2 by L2 is equal to 1 by 5. So the breadth to length ratios is actually now becomes 1 by 5 times. So whatever we actually get B1 by L1 let us say that if you are having 1 and 1 then here it actually reduces to that in the direction which is actually B2 by L2 reduced by 1 by 5 times. This we will see how the you know which we can interpret from the flow net which can be obtained by from the CW program. So here if you are having a case with k1 having permeability and k2 having soil having permeability k2 and k1 less than k2 then you can see here the flow net as we discussed in the previous slide it undergoes a change and it gets deflected here. So the similar thing can be seen in the program also. So here the number of full flow channels in soil 1 or k1 with k having permeability or nothing but 1 plus 1 and this is approximated as 2 by 3. So 2 plus 2 by 3 that is 2.67 or the number of full flow channels in soil 1. Similarly here with B2 by L2 is equal to 1 by 5 because of the ratio of k2 is equal to 5 k1. So what will happen is that this 1 will be reduced by 1 by 5 and this is also reduced by 1 by 5 and this 2 by 3 into 1 by 5 it becomes 2 by 15. So the summation of this 1 by 5 plus 1 by 5 plus 2 by 15 the number of full flow channels in soil 2 with k2 are 8 by 15. So the potential drops will remain same and though we are actually having two soil layers so we can say that the q is equal to k1 into h by nd into nf1 is equal to k2 into h by nd into nf2. So by knowing nf1 that is the number of full flow channels in soil 1 with k1. So here in this case 2.67 divided by number of potential drops which are actually here like a homogeneous section here also we have got 10 potential drops. The only difference is that because of the different soil layers you can see that changes in the flow lines here you can see this curvilinear squares and here the rectangles are actually small because of the ratio of the permeability which is actually different. For example if you are having a different material say permeability of this layer is low and this layer is high then you can see that here you will have a large rectangles. So after having seen the examples now let us try to see if you are having a sheet pile wall structure or a concrete dam structure and when there is a water flow which actually takes place we have seen that the water flows vertically down and then it actually raises upwards. So we always have to ensure for designing these structures in the water bodies we have to see that the stability of the structures against piping or some other failures. So here this particular situation which is a case for the sheet pile wall with a confined seepage problem where failure due to piping for a single row of sheet pile wall with a structure which is after Terzaghi 1922. So here a sheet pile wall structure having H1 as the upstream head and H2 as the downstream head is shown here and based on the model tests which were carried out and then it says that a zone which is within D by 2 from the D is the depth of penetration of a sheet pile wall into the permeable layer. So this particular zone is actually prone for failure. So in order to calculate the stability which is nothing but here the W dash is nothing but the weight of this prism which is nothing but if you take half gamma dash into D square into 1 meter per meter length of a sheet pile wall structure when you consider this is the plane strain structure then you can actually take it as the weight of the prism as half gamma dash is nothing but the submerged unit weight because of the presence of water. So half gamma dash D square will give me this particular weight of this prism. So what it actually has been done is that suppose if there is a uplift which is actually created so here in order to calculate this uplift pressure here we need to actually use the total head why because here in this case because it is the head which is actually dissipating from you know as you water flows from this direction to this direction. So here we actually have downward direction and here there is upward direction of the water flow. So this particular this zone is actually critical for failure and it can lead to failure. So here this calculation has to be done with by knowing the total head here by knowing the total head here let us say that can be obtained suppose if you are having one equipotential line here and one equipotential line here one equipotential line here you get H1 H2 H3 the average of those heads will actually give the average head for example here HA and HB as taken as HM is equal to HA plus HB by 2. So by the uplift pressure on the area which is d by 2 which is nothing but half gamma w into d into HM which is actually calculated as this uplift pressure. Now we take the equilibrium of this so if this uplift pressure is high there is a possibility that this gets lifted up and this actually has been found through model test that it actually instigates failure by creating a you know the prism root to the soil and it increases very drastically and you know the gap between shear pile wall and the soil will arise and then lead to the piping failure. This particular failure is actually called as piping failure. So here the average hydraulic head is considered and that is based on the total head which is actually obtained from the you know flow net diagram. So the factor of safety against piping failure or heaving can be calculated is nothing but factor of safety is given by W dash by U and for critical case for factor of safety is equal to 1 we get W dash is equal to U then we can actually calculate what should be the adequate depth of penetration of sheet pile wall in the permeable soil layer also but here half gamma dash d square divided by half gamma w d into HM will give me d gamma dash HM gamma w. So this is you know d by HM into gamma dash by gamma w. So this is also said as factor of safety is equal to the gamma dash by gamma w is nothing but IC that is the critical gradient which we have discussed that IC is equal to critical hydraulic gradient is nothing but gamma dash by gamma w into gs minus 1 by 1 plus e and IM. IM is nothing but HM by d, HM is nothing but the average total head beneath that prism and d is that depth of penetration. So which gives me the factor of safety again is the heave as IC by IM so by knowing the critical gradient critical hydraulic gradient and this particular IM which is nothing but HM by d will give me the factor of safety this should be of the order of 4 to 5 to ensure stability of the structure. Suppose if you are having a structure which is having low factor of safety one of the alternatives or options is that to provide an apron layer in the downstream side and that actually increases the resistance in the form of say w dash plus the weight due to apron say w then w plus w dash by u the factor of safety increases with the presence of a apron layer in front of the sheet pile wall structure. So to find HM find the total head within the d by 2 zone horizontally. So this is the procedure for finding the IM is given here which is nothing but HM by d which is indicated here. So as I said that we for another case with according to Hasra according to Haar he has given for structures other than single row of sheet pile walls which is having a concrete dam and with a cut off here. So for this case based on the model test has been observed that the d is the depth of penetration so the d by dash is you know which is indicated here and d by 2 at a distance the failure wedge is actually formed here but the variation of these particular criteria is still limited. Now here what we actually say that for safety of hydraulic structures again is piping failure according to Haar Haar 1935 factor of safety is defined as critical hydraulic gradient by I exit gradient I exit gradient can be obtained by you know which is nothing but delta H by L. So delta H is nothing but the last you know between the you know penultimate equipotential line in the downstream side of the hydraulic structure where that length of that flow line in that zone can be given as I exit and we can also estimate according to Haar 1962 the I exit gradient as 1 by pi into H by d and then H is nothing but the maximum hydraulic head and d is the depth of penetration of sheet pile wall. So one can determine from the flow net or one can actually also calculate from according to Haar 1962 but I exit once we know and once we know the critical hydraulic gradient this is another way of also estimating the factor of safety against piping. So here for the safety of hydraulic structures against piping suppose if you are having sometimes initially a configuration is such that you do not have any cutoffs suppose if it is found unstable against piping then there is a need for you know inclusion of cutoffs. So let us look into the example problem a stiff clay layer underlies a 12 meter thick silty sand deposit and here a sheet pile wall is driven into the sand to a depth of 7 meters and K of silty sand is 18 to 10 to the power of minus 6 meter per second the stiff clay layer can be assumed to be impervious and void ratio of the silty sand is given as 0.72 and the specific depth of the solids is given as 2.65. So what we need to do is that to draw the flow net and estimate the discharge and we need to calculate what is the pore water pressure at the tip of the sheet pile wall and factor of safety against piping failure and see whether the structure is stable against piping or not. So we have the configuration like this you have got upstream water level and downstream water level and a silty sand which is actually shown this is the depth of penetration of a sheet pile wall and this is 7 meters and the stiff clay which is the impervious layer in this case and you have got a silty sand here and this distance from the tip of this to this one is say 5 meters and this is the downstream water level where you have got 2 meters and this is 3 meters is the so this head drop is head loss is 3 meters. So this construction of the flow net works out like this so 8 potential drops and this is the potential line and this is the last equipotential line 8, 7, 6, 5 these equipotential lines are number are given and this is nothing but 8, 7, 6, 5, 4, 3, 2, 1, 0 the numbering is wrong in the slide and here this is the tip of the sheet pile wall and this is the flow channel 1, flow channel 2 and flow channel 3 or you can be approximated as 2.6 or something like that. Then we can actually calculate number of flow channels and number of potential drops and head loss is 3 meters we can calculate the discharge we need actually pore water pressure at the tip of the sheet pile wall and the datum is considered here so this elevation of this once by knowing the elevation of this we can actually calculate total head is 1.5 meters why because this is 8, 7, 6, 5, 4 so 4 by 8 into head loss is 3 meters so the total head available at the equipotential line right below the sheet pile wall is 1.5 meters. So this estimated from the flow net elevation head is minus 9 million meters below the datum so pressure head is 10.5 meters the pore water pressure is about 105 kilo Pascal's at that level. Similarly now the coming to the next example the subset of the problem is that head loss for equipotential drop that is delta H is equal to 3 by 8 so maximum exit the hydraulic gradient is nothing but 0.37 by 2.6 which is given as 0.144 and then we can actually calculate critical hydraulic gradient with Gs minus 1 by 1 plus e 0.96 so by putting critical hydraulic gradient to the I exit gradient we can actually calculate this actually as 6.7 so greater than 5. So this arrangement is found to be quite safe against with respect to the piping failure. So in this particular lecture what we actually have discussed is that how we can actually construct flow nets through a numerical simulation and we also have seen the centrifuge basal physical simulation of this particular cases with and without gradients with and without filter layers and how important is the presence of horizontal filter is also discussed and then we actually have tried to see the factor of safety against failure piping failure for a sheet pile wall structure and the criteria which is actually required the similar direction the criteria which is actually the d dash which is actually given for a concrete dam is in between db dash by d and d and then that is given by used for concrete dams which are actually having a cut off in the downstream level.