 So, having done this dry sand slope stability analysis now we will move on to submerged slopes and when we talk about the submerged slopes there has to be a seepage line. So, we want to understand because of seepage conditions how the stability of the slope is going to get changed alright say this is the slope and this is has a seepage line now this seepage line is inclined at an angle of I alpha normality sorry I we normally take for you know the inclination of the slope so this is alpha this is how the seepage line is seepage line means flow line is given and under these circumstances I want to compute the factor of safety of the slope yes the direction of the seepage is from top to bottom there could be a situation where seepage is from bottom to top also reverse direction bottom to top we will discuss but for a quick answer we have done this in seepage analysis if you remember the seepage force acting per unit volume of a control volume or a soil mass. Now in this case what we have to do is take the slice this is the slice which I have considered this surface is parallel to the infinite slope now tell me one thing if I ask you to find out pore the pressure at this point can you obtain what else is required the flow line is known perpendicular to this is equipotential line yes so equipotential line is going to be like this so what is the relationship between these two 90 degree equipotential line is cutting the flow line I want to find out pore the pressure at this point so what I should be doing old concepts which we discussed at this point if I insert a piezometer what is going to happen this is the height of the cut somewhere that means actually the way I have drawn is not correct because you have to be careful when you are doing this so wherever the equipotential line would cut the surface now this is the point where if I put the piezometer it will go up to this point so this is the piezometric tube and this is the height of the water column why it is so at this point the equipotential line cuts the atmosphere so the height of pore pressure or the height of water which is going to be in the piezometric tube will be hw it is okay you go back to the basics you will realize that there is a there is a seepage line and the seepage line is perpendicular to the equipotential line correct so wherever these two cut and become atmospheric the equipotential line cutting the atmospheric surface that is the value of total piezometric head so the main objective is to find out pore the pressure hw why there is a relationship between pore the pressure gamma w into hw rest is simple geometry rest is simple geometry this angle is alpha what is this angle 90 plus i is this fine and what about this angle this thing is 90 degree this whole thing this is alpha minus i so suppose this is x can I use the triangle law in triangle a b c what is the value of now can we compute x this is known this is d d is the depth of slip surface so I can say x upon sin of 90 plus i equal to yes a b that is d over sin 90 minus alpha what is the angle 90 plus i plus alpha minus i so this becomes 90 plus alpha and 180 minus so 90 minus alpha correct so what we have obtained is x equal to d sin 90 plus i cos i over cos alpha why we have obtained x because we want to obtain hw it is a good question now what is the relationship between hw and x so use triangle c b d all right what is the value of hw hw equal to x into sin of this angle this is cos of alpha minus i because this will become 90 minus alpha minus i so this will be cos of alpha minus i please check it so x is equal to d cos i over cos alpha into cos alpha minus i this is the relationship which we have got for hw hw depends upon what incidentally analyze this expression height of water or the pore or pressure at the point b is a function of depth of the point okay alpha is the seepage line perpendicular is not possible why because if alpha is equal to 90 degree what is going to happen so the analytical solution which we are doing right now is not valid for alpha equal to 90 degree why what is the reason no forget about this angle angle what is the reason the reason is any guess if alpha is 90 degree equipotential line is not going to cut the atmospheric line that means point c will not exist and if point c does not exist we cannot get hw that means I have to do some other method so this is the solution which is not valid for alpha equal to 90 degree you got this point that means hw tends to infinity okay so we have obtained hw now what else can be done with hw so in most of these analysis where we are doing slope stability you know what we want to do is we want to find out the effective shear strength or total shear strength at the slip surface so one more thing which you should realize here is that this shear force is the stabilizing shear strength the weight which is acting is destabilizing so the w sign of i term is destabilizing and w cos of i is trying to negotiate with the pressure at the acting at the base normal stress is this fine so suppose if I know u and if I know hw can I compute the value of total pressure u at base what will be equal to this equal to this is in the form of the pressure we are dealing with the forces so this has to be multiplied by the area correct so this will be equal to gamma w hw multiplied by if this is b what is the length of the inclined length of the base this happens to be b into sec i the convention is that we always take lateral directions horizontal or the width of the slice is horizontal and whatever the depths are we take them in the vertical direction for the same sake of simplicity so now this thing is getting multiplied by you know which term this is going to get multiplied by b sec i so this is the value of the ub can I write down the factor of safety term now what will be the factor of safety relationship the shear strength offered by the material so this is c equal to 0 so this is going to be n prime tan phi prime tau yes so see this is equal to 0 for dry condition dry sorry dry sands oh I am sorry it is not dry sand this is so much condition but sandy slopes so c is equal to 0 we have to compute n prime I am just coming to that once you have got ub value you can compute the n prime value and what is the stabilizing force this is a stabilizing force shear strength and what is the destabilizing force the w component which is the acting parallel to the slope so what is that now compute the value of n prime so what is the value of n prime from here we can take the vertical component vertical to this base so this is w cos i w sin i so w is gamma bd cos i minus n prime is what we are computing so pore to pressure in pore to pressure term you know what is appearing what we have written is that ub equal to the base pore to pressure is gamma w hw b sec i hw is this so this is a big expression what you have to do here this will be equal to gamma w into d cos i over cos alpha into cos alpha minus i gamma w hw so this is hw portion multiplied by b sec i this is okay n prime tan phi prime has to come in the numerator so this can be written as gamma bd cos i minus gamma w into d into b cos alpha minus i over cos alpha this is okay I can write this as gamma bd cos i minus this will become gamma w by gamma into cos alpha minus i over cos alpha so what is the factor of safety term the factor of safety term would be if I define this as now I will substitute straight away I think you can realize this so the factor of safety term will be equal to this term multiplied by tan phi prime and what is coming in the denominator gamma bd into sin i so this becomes the factor of safety term I can further simplify this as take out cos i term that is intentional so when you take out cos i term this will become 1 minus gamma w over gamma cos alpha minus i over cos alpha into cos i multiplied by tan phi prime over tan i are you happy with this what is similarity between this situation and what we have derived that means this term is a sort of a penalty or a factor which has been imposed on this term very simple to realize the situation in case of drives sandy slopes the factor of safety was tan phi over tan i submerged situation clear what is going to happen this becomes your effective stresses we have filtered out the effect of water by using this concept effective stresses normal stresses substituting over here and I get this relationship what can be done further imagine what is this term in simplified form go back to your tan plus 2 trigonometry can I convert it into tan components tan of tan of alpha tan of i try doing that yes very nice so this becomes tan alpha plus tan i term 1 plus so that means you are right so this is equal to 1 minus gamma w over gamma 1 plus tan alpha into tan i yes and of course the very well known factor that is tan phi prime over tan i so simple analysis of the logic is that whatever is happening in terms of submergence and seepage is to reduce the factor of safety is understood the submerged slopes are going to be less stable as compared to the dry slopes so the factor of safety is going to reduce i equal to 0 and tan alpha equal to 0 what is going to happen horizontal ground and seepage line is parallel to the surface and we are not interested in such type of a situation because this is a very critical situation which we are trying to solve usually gamma w upon gamma is taken as 1 by 2 another interesting situation could be when alpha is equal to i so what is going to happen i is the inclination of the slope also so alpha is in such a manner that this is just at an angle which is same as the angle of the slope parallel to the slope so in that case what is going to happen is this will become what alpha vanishes then this becomes tan square i so here I can write this as factor of safety will be equal to 1 minus gamma w over gamma and what is 1 plus tan square i tan phi prime over tan i what is the value of this bracket term this is normally written as 1 minus gamma w over gamma into 1 upon cos square i tan phi prime over tan i in this case if your alpha term is 0 what is going to happen parallel flow condition so alpha equal to 0 equal to i then what is going to happen if you put alpha equal to 0 this becomes a situation where the direction of the flow is parallel to the slope alright so what is going to happen in this case this will become i is not becoming 0 like that but because alpha is equal to i so this term will disappear this will become 1 minus gamma w over gamma multiplied by tan phi prime over tan i and usually we take gamma w over gamma as half so this is equal to 1 by 2 tan phi prime over tan i what has happened to the factor of safety it has reduced by half under submergence for the critical condition can I write that i will be equal to phi prime by 2 approximately this will remind you of your tan plus 2 geometry and the trigonometry which used to do to solve these problems. The concepts are simple just to wrap up the things what we have done is we have taken a slice we have computed the weight weight is known find out the normal stress which is acting on the surface where it is acting this is n prime and then we wanted to find out what is the pore pressure use the concept get the h w convert it into the pores acting on the base n is known n minus pore pressure is going to give you n prime this is the shear strength of the material which is stabilizing force divided by the destabilizing forces of the stresses and then this is simple analysis one of the expressions could be here I can get rid of tan alpha and I can make it 1 plus tan square i also which is more advantageous for us because i is the inclination of the slope with respect to horizontal. So factor of safety is 1 minus gamma w over gamma 1 plus tan square i multiplied by tan 5 prime over tan.