 Welcome to course on advanced geotechnical engineering, we are discussing about module 5 that is slope stability analysis. So this is the lecture number 3 on stability of slopes. In this lecture or in this lecture 3, we are going to discuss about the slope stability analysis methods, comparison of different slope stability analysis methods and some examples for evaluating at factor of safety. So in the previous lecture, we introduced ourselves to different slope stability analysis methods and we said that there are methods like ordinary method of slices and it is followed by Bishop's method of slices and then there is Morgan Stern-Prey's method and John Boo's method. So coming to the ordinary method of slices, let us recollect once again, in this method the potential surface is assumed to be circular in nature with center at O and having a radius R and the soil mass which is within the circular arc zone from the slope surface and the crust zone, it is divided into vertical planes into series of slices of width B. Preferably these slices have to be of identical width. The base of each slice is assumed to be straight line for convenience purposes. The factor of safety is defined as the ratio of the available shear strength tau f to the shear strength tau m which must be mobilized to maintain a condition of limiting equilibrium. So the factor of safety in this method is defined as the ratio of the available shear strength tau f to the shear strength tau m which must be mobilized to maintain a condition of limiting equilibrium. So in this slide and the right bottom corner a free body diagram of a typical slice is shown. The weight of that particular slice is assumed to be at the center of the slice and the normal reaction that end dash and s is the tangential force which is actually shown in a typical slice. The orderly method of slices satisfies the moment equilibrium for circular surface but neglects both inter-slice forces that is normal as well as shear forces which are actually there between the slices they are assumed to be 0 that means that the forces which are actually acting here and here and along this surface they are actually assumed to be neglected. So the advantage of this method is its simplicity in solving the factor of safety since the equation does not require any iteration process. So in summary the orderly method of slices satisfies the moment equilibrium condition and neglects inter-slice normal and the shear forces and gives the most conservative factor of safety and is useful for demonstrations or preliminary estimations. So the detailed explanation of this method is shown here in this a cross section of a slope which is ADC which is shown and with O at the center of the rotation and R is the radius and the surface ABC is assumed to be one of the failure surfaces and if a typical slice of width B if it is assumed that the soil mass within ABCD is divided into some equal number of slices having width B in horizontal direction and if it is considered the true free body diagram which is actually shown free body of the slice I which is if this is the slice I the free body diagram is actually shown here where in E1 and E2 are the normal forces inter-slice normal forces X1 and X2 are the inter-slice tangential forces and this is the normal reaction which is actually acting in the width gets oriented toward the center of rotation and this is the tangential force which is actually acting opposite the movement of the soil. So that is actually basically the resistance offered by the soil further simplifying on this we can actually get now the factor of safety we have defined already and the total weight of the slice we can give it as gamma into H the H is the height at the center of the slice into B that is the width of the slice into one unit that is the because it is a two dimension analysis we are doing parameter analysis then total normal force N is equal to sigma into L and which includes N dash is equal to sigma dash into L and U is equal to U into L depending upon the water table location we can actually calculate what is the U pore water pressure into the length of that particular straight line portion of the slice we can actually get U and multiplying by effective stress into L we get N dash. So U is the pore water pressure at the center of the base of the slice and L is the length of the base the shear force on the base can be given as tau m is the shear strength which must be mobilized to keep the slope in the equilibrium or you know endure adequate factor of safety tau m into L and the total normal forces on slice even in E2 they are even in E2 shear forces are x1 and x2. Now further considering the moment about O the sum of the moments of the shear forces T on the failure or arc AC must be equal to moment of the weight of the soil mass ABCD. So if we can actually take the moments like sigma T into r is equal to sigma wr sin alpha. So here what we have done is that we have taken the weight component into r the radius r is the lever. Now by using T is equal to tau m into L by writing tau m is equal to tau f by factor of safety into L and substituting in for sigma T r we can write tau f by factor of safety into L is equal to sigma w sin alpha. So by bringing the factor of safety term on the left hand side we can write factor of safety is equal to sigma tau f into L divided by sigma w sin alpha. So further this can be worked out like this that tau f we can actually write it as C dash plus sigma dash tan phi dash into L by sigma sigma of w sin alpha. So for an analysis in terms of effective stress we can write factor of safety is equal to sigma summation C dash plus sigma dash tan phi into L divided by sigma of w sin alpha and this can be further simplified by writing sigma dash L is equal to n dash we can write factor of safety is equal to C dash into L a where L a is the if you look into that now the summation which is removed for the coefficient term because L 1, L 2, L 3, L 4 and the summation of that is L a. So C dash L a plus tan phi dash into sigma dash n phi n dash. So sigma into n dash divided by w sin alpha within summation that is the equation one is exact but approximations are actually introduced in terms of the forces n dash. So this actually this expression gives the factor of safety in this lecture we are also going to solve some typical problems for arriving at the factor of safety of a typical slope by using manual calculation methods or also by using some software packages. The further the failures are a Swedish solution it is assumed that for each slice the resultant of the interslice forces is zero. In the Swedish method or a failures method what is assumed that for each slice the resultant of the interslice forces is zero. In the ordinary method of slices the interslice forces have been neglected but in the filious method it is more or less same as in ordinary method of slices but only difference is that for each slice the resultant of the interslice forces is assumed to be zero. So the solution involves resolving the forces on each slice normal to the base. So n dash is equal to w cos alpha minus u l because u l is nothing but the pore water pressure acting on a particular length of the slice that is l. So rewriting the equation one we can write C dash L a plus tan phi dash sigma that n dash we write it as now w cos alpha minus u l divided by sigma w sin alpha. So this is the revised expression for computing factor of safety by using the failures or Swedish method of slices. The explanation for the procedure for calculating factor of safety by using failures or Swedish method of slices is given here. Let us consider that a slope which is actually having a cross section ADC which is shown here and the slope of AD which is the question because if adequate factor of safety is ensured then there is a possibility that if the material is actually having adequate strength characteristics the slope of AD line can be maintained steeper. If the material is not having adequate characteristics and there is a need for adopting the flat inclinations for the slopes. Generally the slopes which are normally adopted for highway embankments or railway embankments they range from one vertical 1.5 horizontal to one vertical 2 horizontal. In some urban areas there is a need for steepening of the slopes in such situations one need to adopt some strengthening options for the slopes so that the areas can be maintained steeper that means that whenever there is in land acquisition problems or land availability problems the slope of AD can almost be maintained vertical and in order to make it to stand an appropriate strengthening solution need to be designed. So in this module we are going to discuss about these options for these things. Now let us assume that this particular example here the procedure we divide these slices into the entire soil mass ACD within the assumed failure surface. The failure surface which is circular arc having r radius and is assumed to be divided into let us say 7 number of slices. So this is the one slice which is a portion of the triangle and this is looking like a trapezium and this is having a certain dimension here and so there are different slices and this is the center of rotation. So first what we are going to do is that you calculate what is the area and into L which actually gives the volume into gamma gives the weight of that slice 1 that means that 1 e to calculate w1, w2, w3, w4, w5, w6, w7 and all these respective weights for each these slices will be acting in the center of the slice and let us assume that each slice having a width of small b in horizontal direction then assume that the weight is actually acting at the center. So if this is the width b, b by 2 here, b by 2 here and then this is the line joining. So once the weight is actually indicated here draw a line from to the center of the slice here and here where the weight is passing and with vertical this angle need to be recorded from graphically alpha 1, alpha 2, alpha 2, alpha 3, alpha 4, alpha 5, alpha 6 which is actually shown. So as we traverse from 1 to 7 you can see that the angle changes from positive to negative. So the components of weights w1, w2, w3 need to be find out like w cos alpha and w sin alpha can be determined from the alpha once it is obtained from graphically then the components can be obtained. For analysis in terms of torque stress the parameters Cu and phi u are used basically what we said is that that is for short term stability calculations and the value of the u will be 0 in that case factor of safety is equal to Cu into La plus tan phi u into sigma w cos alpha that means let us assume that we have got now 7 slices now. So w1 cos alpha 1, w2 cos alpha 2, w3 cos alpha 3 the summation up to w7 cos alpha divided by w1 sin alpha 1, w2 sin alpha 2 up to w7 sin alpha 7. So once we have this and then Cu into La suppose it is also possible that we can also take like Cu into L1 Cu into L2 then it is nothing but L1 is nothing but the for a slice 1 what is this particular length it is approximated as a straight line. So this graphically this length can be obtained and then it can be used for a once it is a diagram is drawn to a scale it can be obtained. So let us assume that for phi u is equal to 0 case so that is something like Cu La divided by sin alpha that means that if you are having a slope with undrained conditions where saturated clay slope then the factor of safety can be obtained as Cu into La divided by w sin alpha. Now Bishop's simplified method of slices in this solution it is assumed that the resultant force on the sides of the slices are horizontal and x1 minus x2 is equal to 0. So for equilibrium the shear force on the base of any slice is given as t is equal to 1 by factor of safety into c dash into L plus n dash tan phi dash. So resolving the forces in the vertical direction we can get w is equal to n dash cos alpha plus Ul cos alpha plus c dash L sin alpha by factor of safety n dash tan phi dash by factor of safety into sin alpha after some rearrangement of the terms and using L is equal to b secant alpha we will get a factor of safety term as factor of safety is equal to 1 by summation of w sin alpha into summation within brackets again brackets c dash b plus w minus ub tan phi dash bracket close into secant alpha divided by 1 plus tan alpha tan phi dash divided by factor of safety bracket close. So this expression actually has the factor of safety in both left hand side and right hand side. So this involves an iteration procedure initially the factor of safety assumed to be computed from the Swedish method of slices and then with that iteration value after setting number of iterations one can calculate what is the factor of safety of a slope by Bishop's method of slopes, Bishop's simplified method of slopes and this is one of the versatile method for assessing the factor of safety of the slopes. So Bishop 1955 in his paper he also showed that non-zero values of the resultant forces x1 minus x2 could be introduced into analysis by refinement has only a marginal effect on the factor of safety and so Bishop 1955 stated that how non-zero values of the resultant forces x1 minus x2 could be introduced into the analysis but refinement has only have a marginal effect on the factor of safety. So in the Bishop's simplified method the pore water pressure can be related to the total field pressure at any point by means of dimensionless pore pressure ratio which is Ru is equal to U by gamma H. Suppose if the Ru value is equal to 0 that means the slope is you can say that partially saturated when is almost dry and Ru is equal to 0.5 that indicates that the slope is completely saturated. For any intermediate value between 0 to U by gamma H it is partially saturated and Ru is equal to 0 it indicates the slope is dry. For any slope Ru is equal to U by W by B and by rewriting these terms we can write for any slice Ru is equal to U by W by B by rewriting we can write factor of safety expression by 1 by summation of W sin alpha summation into C dash B plus W into 1 minus Ru tan phi dash secant alpha 1 plus tan alpha tan phi dash by factor of safety. When Ru is equal to U then you know we have W tan phi dash into secant alpha plus 1 plus tan alpha tan phi dash by factor of safety. So here the summary in summary Bishop's simplified method considers the interslice normal forces that is X1 and X2 are the tangential forces here the tangential forces and even in E2 are the normal forces. So Bishop's simplified method considers the interslice normal forces but neglects the interslice shear forces. So kindly note here Bishop's simplified method considers the interslice normal forces but neglects the interslice shear forces. It further satisfies the vertical force equilibrium to determine the effective base normal force N dash. So it satisfies moment equilibrium for factor of safety. So in summary Bishop's simplified method satisfies moment equilibrium for factor of safety satisfies vertical force equilibrium for N for determining N or N dash considers interslice normal force. Interslice normal forces are considered and more commonly used in practice and applies mostly for circular shear face surfaces. That means that wherever there is a homogeneous soil which is used let us say for embankment construction or for highway embankment or railway embankment construction when the material is actually obtained from identical boropit area then we can say that the possible failure surface can be circular in nature. Then coming to the John Booce simplified method John Booce simplified method is based on a composite slip surface. That means basically non-circular in nature and factor of safety is determined by horizontal force equilibrium. So as in Bishop's simplified method method considers interslice normal forces but neglects the shear forces. So here also in this free body diagram of the slice which is shown for John Booce simplified method the tangential forces X1, X2 or the forces which are actually acting in this direction they are considered to be neglected and in John Booce method also like in Bishop's simplified method it considers the E1 and E2 that is these are the interslice normal forces and it satisfies both force equilibrium that is vertical force equilibrium as well as horizontal force equilibrium but it does not satisfy the moment equilibrium and considers interslice normal forces and it is commonly used for determining factor of safety for composite shear surfaces. Then coming to the another method for determining the factor of safety that is the Morgan Strang price method this satisfies both force and moment equilibrium and also assumes that the interslice function was assumed. So here the interslice forces are assumed with the form of a function and considers both interslice forces here you can see in this free body diagram of the slice which is actually shown here where E1 and E2 and E1 and E2 are forces they both are considered and they both are related in the form of a function Fx and considers both the interslice forces and they are considered assumed as a interslice force function Wx and allows selection of the interslice force function and computes the factor of safety for both force and moment equilibrium. So it computes the factor of safety for both force and moment equilibrium. The Spencer's method this is same as the Morgan Strang and price method the assumption made for interslice forces but a constant inclination is assumed for interslice forces and the factor of safety is computed for both equilibrium that is it considers again the Spencer's method very similar to Morgan Strang price method it considers both interslice forces assumes a constant interslice force function and satisfies both moment and force equilibrium conditions and computes factor of safety for force and moment equilibrium condition computes the factor of safety for force and moment equilibrium. And the free body diagram of the typical slice in Spencer's method is shown here weight is actually acting here this is the movement of the slope so opposing that this is the soil resistance which is actually offered from the mobilized shear strength and dash is the normal reaction and E1 and E2 are the normal forces of the interslice and a particular slice and T1 and T2 are the shear forces. So this is after the Spencer's method is after Spencer 1967. So let us after having discussed about the number of methods let us try to solve some typical problems by using manual calculations as well as by using some computer aided methods. So let us consider in the example 1 a 45 degrees slope is excavated to a depth of 8 meter in a deep layer of saturated clay of unit width unit weight 19 kilo Newton per meter cube. The relevant shear strength parameters are given as Cu undrained cohesion as 65 kilo Newton per meter square and 5u is equal to 0. So determine the factor of safety for the trail failure surface specified in the figure in the next slide and the cross sectional area ABCD which is within the in the failure within the zone from the top surface of the slope to the failure surface is 70 meter square. So a 45 degree slope is excavated to a depth of 8 meter in a deep layer of saturated clay of unit weight 19 kilo Newton per meter cube the relevant shear strength parameters are Cu is equal to 65 kilo Pascal's and 5u is equal to 0. So we need to determine the factor of safety for the specified failure surface there can be number of trail failure surfaces but we are actually calculating for a typical trail failure surface. So the cross section is actually shown here wherein we see that ABC is the assumed trail failure surface and AD is the slope which is inclined at 45 degrees with the horizontal and DC is the crest width and the radius is 12.1 meters and the horizontal distance OD is 4.5 meters and assume that W is actually acting right below the D vertically down and the area of ABCD portion is 70 square meters into 1 meter is the volume which is involved in the active zone. So what we call that weight is nothing but 70 cubic meters into gamma if the soil is assumed to be uniform here in this case then we get the weight. Then the height of the slope is 8 meters and the rotation which is actually taking place at a height of say 11.5 meters from the toe level which is actually at point A. So the solution is like this finding out the weight of the soil mass which is nothing but 70 into 1 into 19 that is the unit weight. So we get if it is not considered as 1 is not multiplied then we calculate weight per meter length. The centred of ABCD is 4.5 meter from O and angle which is subtended between AO and OC is 89.5 degrees. So hence the radius is actually about 12.1 meters the arc length we can actually calculate based on once we know the angle AOC we can actually calculate what is the arc length ABC and that is working out to be 18.9 meters. So the factor of safety can be given by factor of safety is equal to Cu La into R. So Cu La is nothing but the resting force into R, R is nothing but the lever arm from the distance from the center of rotation to the force along the arc surface. And W is nothing but the weight of soil mass into the horizontal distance d. So W d forms as a driving moment and Cu La R is the resting moment. So computation of the factor of safety gives 2.548 and it is to be noted that the factor of safety whatever we have computed for the trail failure surface did not be that it is a it gives the minimum factor of safety. One need to get the what is the potential failure surface and what is the minimum factor of safety we can actually can be obtained. So the minimum factor of safety can be obtained for the similar problem by using Taylor's stability charts also. So let us look into that how that Taylor's stability chart can be used to get a minimum factor of safety. So here the minimum factor of safety can be estimated by using factor of safety is equal to Cu by ns into gamma h where ns is nothing but the Taylor's stability number. So using Taylor's stability chart ns versus sloping inclination. So we Taylor's stability chart which we have discussed in the previous lectures where in we actually have seen that for beta is equal to 45 degrees the value of the assuming that d is large the value of the ns is 0.18. So by substituting these values factor of safety is equal to Cu which is 65 divided by ns which is 0.18 into 19 into 8 we get the factor of safety 2.37. Now if it is noted that this particular estimation is assumed to give the factor of safety close to the minimum factor of safety or it is a minimum factor of safety. So how the value of the ns is obtained from 0.18 can be obtained from the Taylor's chart wherein we have the sloping inclination on the x axis stability number on the y axis and for slope 45 degrees whereas here we have 45 degrees and we are actually we have assumed that the extent of the d below the toe level is assumed to be large if the d is shallow then we are actually having here only but as we assume that the extent of the d much further below then d is equal to infinity we get stability number as 0.181 it is considered as 0.18. So based on that the factor of safety minimum factor of safety is obtained and wherein we can actually calculate what is the factor of safety of a we calculated the factor of safety of a given slope. So in this example we try to determine by using the phi is equal to 0 method and we also use the Taylor's chart to get the minimum factor of safety for a potential failure surface. Now let us consider another example 2 wherein using the felonious method of slices determine the factor of safety in terms of effective stress of the slope shown in the figure for the given failure surface using peak strength parameters C dash is equal to 10 kilo Pascal's and phi dash is equal to 29 degrees. So the unit weight of the soil above and below the water table is given as 20 kilo Newton per meter cube. So in this method by using the felonious method of slices the factor of safety is determined in terms of effective need to be determined in terms of effective stress. So this is after Craig 2004. So the cross section is given here and if it is noted here the potential the given surface is shown here the given surface is shown here and these are the normal reactions which are actually passing through that and these are the tangential forces and these are the weights this force triangle what you are seeing and these are the normal reactions these are the shear force which are acting mobilized by the soil and this is the weight of the slice and the slope inclination is one vertical 1.5 horizontal. So what we are interested is that what is the factor of safety of slope for the given slope inclination which is here. Generally for factor of safety is equal to 1 means we say that is on the wedge of failure and for factor of safety is equal to 1.5 for some slopes if it is design then it is said as stable and here the r is equal to 9.5 meters and the vertical distance is about 9.15 meters. Here one approximation in felonious method is that we calculate the u value with reference to this vertical height but in reality when the water table is actually varying in line like this it is you know it is not this height in principle one need to consider this particular height that means that if you consider this reduced distance then z effective is this height so but in the felonious method we estimate the factor of safety at the center of the slice wherever the water table if the water table surface is actually here then we take that this entire area in this particular height we calculate what is the pore water pressure and multiplied by this length of this we get the ul but in reality it is this particular height. So by determining with this gw we end up on the conservative side so it is actually safe for a slope which is being designed. So that is need to be you know understand while using the felonious method or Swedish method of slices. So the computation works out like this after Craig 2004 wherein we divided into 8 number of slices and we determined weights of the slices and which are obtained as it can be here it is shown in terms of h cos alpha and h sin alpha or we can actually adopt a simplified procedure like slice number and weight computations and alpha that is from the obtained graphically then calculating w cos alpha w sin alpha and also by noting down the let h is the let us say height of the slice in the center but if you consider let us say that 3 fourth of the height is say water table is there then the gw into gamma will get the pore water pressure and then graphically if you measure what is the length then you get the u into l as the pore water pressure that is kilo meter pore water force you can say per meter length. So by using this expression factor safety in terms of effective stress we can calculate c dash la plus tan phi dash into sigma w cos alpha minus ul by w sin alpha. So the computation after simplification you get the factor safety as 1.42 so this indicates that the slope which is one vertical valenda horizontal is having a factor safety of 1.42 if one needs a factor safety more than 1.5 then it indicates that the slope needs to be flattened that means that one has to go for one vertical to horizontal from constructability point of view then this slope is tends to be ensure the higher value of factor safety. So this is again another example of felonious method of slices in this particular slice what is actually shown in this particular slide what is shown is that the slope is actually divided into portion within the failure surface is divided into 7 number of slices but there is a tension crack of certain depth. So one need to estimate in the value of the l is suppose this is the entire l and what we need to estimate is that this portion only in the participation of the tension because there exists a gap and where because of the virtue of the tension crack we cannot actually estimate what is the we cannot consider the form of a resistance. So here the center of rotation is assumed to be at height of 7 meters and the slope is actually having certain inclination and one can obtain like here height of the slope is 5 meters and the horizontal distance is 8 meters. So it actually has got inclination and the soil it is not necessarily that we get homogeneous soils we can also get the layered soils so in such situation where the portion above this is having properties of C is equal to C dash is equal to 15 kilo Newton per meter square phi dash is equal to 20 degrees and this can be a base soil where C dash is equal to 8 kilo Pascal's and phi dash is equal to 25 degrees gamma is equal to 18.5 kilo Newton per meter cube and there can be also situation of water table but in this example no water table is given. So here one need to consider like here when we are considering the soil properties which are required to be in this portion slide 6 and 7 we need to consider the soil properties for the shear strength resistance of these properties but when we are considering here particularly for slice 1 2 3 4 and 5 we need to consider these properties but for the weight computation let us say for this particular weight computation we need to consider for this portion of the trapezium the unit weight of 18.5 if this in this case unit weight is same but if this unit weight is say 19 kilo Newton per meter cube we need to take the gamma 1 and gamma 2 and calculate the actual entire weight of the composite weight of the slice. So by performing the similar exercise like identifying the normal reactions and calculating the weights and obtaining graphically so in this case you can see here alpha is equal to 0 and then alpha value changing to negative because they are actually coming towards the left side of you know moment of rotation. So the solution for this example 3 which is actually given as a failures method of slices and wherein we get the weight computation like W the dependent on the area we can actually get the unit weights by taking the unit weight concentration we get the individual weights of all the slices in this case there are 7 number of slices and C dash the resistance which are actually considered and tan phi dash you can see here up to fifth slice these properties were considered and here 6 and 7 the subsoil properties are considered and the L which is actually obtained from the graphically and then N is equal to W cos alpha T is equal to W sin alpha these are computed from weight by knowing the angle alpha then one can calculate C dash into L and UL and forgetting the UL measure the height of the height of let us assume that height is say you know is the central height of the slice in the middle and if water table is here in this case water table is also considered wherein we can actually calculate water table height so with that you can calculate what is ZW into gamma W into that L we get the UL and N minus UL we get the you know the force which is required to consider by multiplying this N minus UL into tan phi dash we get N minus UL tan phi dash the summation of this is 267.5 and the summation of C dash L is 158 summation of the T is 237.4 so the factor safety expression for by using felonious method of slices is C dash L the summation of this plus this divided by this is the driving moment so for this by using this condition with for the type of slope what we consider for the potential failure surface what is assumed then factor safety is obtained as 1.78 as it is more than 1.5 and for the type of slope configuration which is considered the slope is said to be stable. Now in this example this example basically it is an example 4 this is the using the Bishop's method of slices we need to determine the factor of safety in terms of effective stress for the slope detailed in the figure which we are going to see in the next slide the value of the RU which is actually the ratio of U by gamma H is given as 0.2 and the unit weight of the soil is 20 kilo Newton per meter cube and the shear strength parameters are C is equal to 0 kilo Newton per meter square and phi is y dash is equal to 33 degrees. So this is the slope cross section which is shown and this is the center of rotation and the factor safety is one vertical the slope inclination is one vertical to horizontal and vertical height is 7.5 if 17.5 means this horizontal distance will be around 35 meters and this is the failure surface is assumed the failure surface is assumed to entry point is here and exit point is at the top and this height is 48 meters. Now let us see how this can be solved by using Bishop's method of slices the expression is given here and which we have discussed in this lecture itself and we divided like any other method we divide the into the equal number of equal bit those slices so these are shown here and the phi dash is equal to 33 degrees and gamma RU is equal to 0.2 which is considered so estimate the weight of the slice so gamma B into H and in terms of H if you look into it is 100 H kilo Newton per meter meter and 1 minus RU into tan phi dash is estimated as 0.52 and here as it has been told because the factor safety term which is actually there in left hand side as well as right hand side you try with initial value which is say 1.1 so tan phi dash by factor safety is equal to tan 33 by 1.1 which is 0.59. So the solution for the example 4 works out to be slice number which is 8 number of slices and the heights at the center of the slices are given here and weights which are actually given in terms of gamma BH given like this similarly we calculate what is alpha and then compute W sin alpha and then compute W into 1 minus RU tan phi dash so with that we can actually get this particular term and then compute secant alpha divided by 1 plus tan alpha plus tan alpha into tan phi dash by factor safety and the product which is of these two is given as this column and this column product is given as here and the summation is 1271. So this divided by this particular W sin alpha is 1185 because in the previous slide if you look into this we have got we are actually estimating this and this in one column we estimated this particular term and in another column we estimated this term and so if you look into this here the product is this particular term the product of this column and this column is yielding the summation as 1271 divided by this summation of W sin alpha is yielding as 1185 so with that the factor of safety is computed as 1.07 so the trail value which we assumed is 1.1 so therefore that take the factor of safety is 1.07 or 08 so this indicates that the slope is for the type of soil parameters which are actually considered the slope configuration is just stable in the sense that any virtuality of for the fluctuations in the RU there can be slope can actually increase of RU there can be possibility the slope will undergo failure. So let us now after having discussed about the some examples for manual methods and now let us try to look into the comparison of the slope stability analysis methods and in this particular slide a typical slope which is actually shown how you know a failure surface is located so there are you know different options and one option is that you consider a grid of centers it is assumed that it is a perpendicular bisector of this one you know the grid of centers so this horizontal distance and vertical distances can be specified as 0.5 meter by 0.5 meter or it can range from up to 5 meter by 5 meter and more the grid of centers the more is the accuracy and also one can specify in the here what is the extension of minimum radius and maximum radius and where the circle or slip surface can proceed so based on that we can actually calculate the circular failure surfaces generally what is done is that innumerable number of failure surfaces are tried the one which actually gives so in this we can actually get the contours where factor of safety is say 1.5 or 1.3 wherever the one the center which gives the least factor of safety that is actually centered calculated or regarded as the critical factor of safety in this case you know the center of the circle and here the with entry and exit options for using for searching so different methods are there for searching the critical factor of safety in some software packages a rhombus grid is actually used in that the minimum factor minimum radius and maximum radius is specified and with that it is such as for the by trying number of slip surfaces it tries to give the the center which gives the or the slip surface which gives the least factor of safety. So a typical problem which is given in lambion bitman 1969 was considered this is an embankment which is retaining a water on the on the rear side that is actually here and the bund is constructed with a unit weight of 19.64 kilo Newton per meter cube cohesion is 4.31 kilo Pascal's and friction angle is 32 degrees and there is a drain here which is provided and this is the imperable strata and the thickness of this is a considered as 10 meters and this is the particular slope inclination which is actually shown here. So this is the schematic diagram of the slope cross section so what has been done is that different methods for actually adopted and then try to compare the factor of safety. So here firstly the ordinary method of slices was used in this the geo slope 2012 version version was used wherein first the seepage analysis was carried out and by using the CW module and wherein the periodic surface is obtained because there is a drain here and this being the equipotential line so this is the flow line wherein you can see that it actually meets at orthogonal to the equipotential line here and this portion is the periodic surface and it is considered here that this portion is the one of the failure surface which actually shown here for the this is regarded as one of the centers which actually gives the least factor of safety for based on the ordinary method of slices. So here the free body diagram of the ordinary method of slice one of the 11th slice that is that this particular slice is elaborated here so you can see that here both the lateral forces are not considered shear forces are not considered and only weight of slice is there that is actually acting downwards and this force is there that is actually acting in this direction this normal reaction which is actually shown here. So this is the force triangle for the 11th slice which is exaggerated here. So based on this ordinary method of slices for the type of problem what we considered it is actually giving as 1.161 factor of safety. Similarly by adopting for the same problem Bishop's simplified method of slices wherein similar procedure wherein we estimated the periodic surface and now we can see that we have these normal forces shear forces are again considered to be 0 so because of this difference is actually considered here. So this is the net force acting in this direction for the again the 11th slice and this is the soil shear resistance which is actually shown here and this is the weight of the slice. So this yields a factor of safety of 1.289. So when you look into the for a given slope the Bishop's simplified method gives the higher factor of safety and John Booth's method simplified method is giving a factor of safety of 1.222 wherein here you can see that this is 40.322 and 32.665 wherein you actually have these normal forces and then you actually have this shear resistance which is actually acting at the base of the slice. Now Morgenstern Price method wherein the here we have considered the horizontal normal inter slice forces that is horizontal forces as well as the you know the horizontal the shear stresses on the along this slice surfaces as well as the normal forces. So because of this force this diagram changes you can see that this is that net normal force net tangential forces and these are the normal force which are actually acting on slices and this is the weight of the slice and this is the normal reaction and this is this particular reaction. So this yields a factor of safety so this when you compare the factor of safety is actually given by Lambin Whitman and as well as computed we can see that Joe's slope 2012 computes orderly method of slices as 1.161 and Bishop's method computes 1.289 this is actually what obtained by using Joe's slope 2012 and Lambin Whitman gives 1.17 and the same analysis by for the Bishop's simplified method it gives about 1.3. So here in this particular slide the factor of safety of a slope can also be obtained by using finite element based methods or finite difference programs like FLAC. So here REL 2003 compare the limit equilibrium methods of comparison like Bishop's simplified method and John Booth's method and Morgenstern price method and whereas we compare the method by using finite element method that is by using plaxis so wherein they adopted the shear strength reduction method and here both base soil and soil in the embankment zone are found to have different soil properties and you can see that the finite element based method by using plaxis 2D gives say the band of the way the failure potential failure can actually failure surface can exist and it yields a factor of safety about 1.654. So this is actually a particular this thing failure surface which is actually obtained for Bishop's method and John Booth's method and Morgenstern price method. So you can see that because of the consideration of the interest slice forces the higher the factor of safety that indicates that more you can say that the slope is so we can actually go adopt steeper sloping inclination. Suppose let us say that we compute by using ordinary method of slices you know 1.1 factor of safety but however by considering that we tend to revise the slope and making it steeper that leads to you know a steeper adoption of a flatter sloping inclination in such situations the adoption of an appropriate method help us to arrive at computation of the factor of safety properly. So in this particular lecture what we try to discuss is that slope stability analysis method different types of slope stability analysis methods were introduced and then we try to cover some typical examples by using ordinary method of slices and failures such Swedish method of slices and Bishop's simplified solution and we compared the typical cross section example which is given by Lambien Wittman 1969 by analyzing with limited equilibrium based method for determining factor of safety by using Bishop's method John Booth's method and the values are found to be in tandem and then finally also we discussed the illustrations which are given by results reported for a given by RL 2003 RL 2003.