 Next part is your cantilever seed pile wall or cantilever seed pile in cohesive soil. Earlier we have finished that derivation and one example solved example in particularly cantilever seed pile in cohesion less soil. Now cantilever seed pile wall in cohesive soil. Now let us see this how the pressure distribution diagram for cantilever seed pile wall in cohesive soil it is varying. Let us say this is a and this is your d. Now if I take this is your c and this part will be completely this height is your d. How this pressure distribution diagram for particularly cohesive soil if you look at here this diagram I draw the pressure distribution diagram for cohesive soil. How this diagram has come if I write active earth pressure and passive earth pressure particularly in cohesive soil p is equal to sigma k a minus 2 c root over of k a which is equal to sigma minus 2 c when phi is equal to 0 and k a is equal to k p is equal to 1 and sigma is equal to gamma z. Similarly p p is equal to sigma plus 2 c if you look at here if you look at here your diagram this is your 4 c minus q this is your 4 c plus q. So active earth pressure intensity at point a you can say that at point a equal to minus 2 c. If you look at here active earth pressure is equal to sigma minus 2 c the minus sign is that means there will be a tension crack that will be a tension. So if I put it this is my minus 2 c because this part will be acted upon by active earth pressure and definitely this will be your p a or you can say that this is your sigma gamma z k a and that side will be your minus 2 c. If I write active earth pressure intensity that of up to dredge line up to dredge line which is equal to p a which is equal to q minus 2 c where q is equal to effective earth pressure effective means q is equal to gamma h. Then if you write at point d this is your point d on the left of seat of the pile at the dredge line the overburden pressure is equal to 0. Hence net pressure at d at point d. So left side if I write this is my left and this is your right. So left side sigma is equal to 0. So hence net pressure at d that means p p minus p a which is equal to 0 plus 2 c minus q minus 2 c which is equal to 4 c minus q. Now similarly if I go to the right side go to the right side at point c at point c point c p p minus p a which is equal to q plus gamma z plus 2 c minus gamma z gamma d instead of z I am writing because this height will be d gamma d minus 2 c which is equal to 4 c plus q. If you look at the derivation how the derivation has been made look at this is your seat pile this is your seat pile wall and it has been embedded inside cohesive soil it has been embedded inside the cohesive soil. How this pressure distribution diagram has come or arrived if you look at here this is your right side this is your left side and this right side it will retain the soil it will retain the soil as it will retain the soil soil will exert pressure on the wall that means wall will move away from the soil. So that means this part will be your active arc pressure it will be active. If I write the what is the p a active arc pressure equation it is sigma k a minus 2 c root over of k a if I say k a is equal to k p is equal to 1 for particularly case of 1 that means at rest when when phi is equal to 0 when phi is equal to 0 k a is equal to 1 minus sin phi by 1 plus sin phi if I say that for purely cohesive soil phi is equal to 0 k a and k p is equal to 1 and also sigma is equal to gamma z. So p a at the base it should be gamma z minus 2 c gamma z minus 2 c now this is your gamma z if you look at here minus 2 c once it is showing that minus 2 c what does it mean if it is a plus 2 c that means it will be somewhere else here minus 2 c means it is a tension it is tension that means 2 c part will be somewhere else here outside this tension. So here it has been made tension minus 2 c now this is all about your how the arc pressure is starting from a to d a to d for this soil for this soil. Now let us start the arc pressure from here to here arc point d if you look at at the boundary point of your d this is your point d at this surface at this surface this side particularly this side that means left side left side there is no over burden pressure because this is this is empty there is no over burden pressure that means sigma is equal to 0 if I write in terms of passive minus active p p minus p a it will be sigma plus 2 c passive arc pressure it is sigma plus 2 c as there is no soil it is blank no over burden. So it will be 0 minus 2 c minus q minus 2 c it will be 4 c minus q. So I put it exactly at this point it is 4 c minus q I put this now at point c if you look at at point c just at point at the point of d at the right side at the right side if you look at here at the right side it will be p p minus p a which is equal to q plus gamma z plus 2 c where is your q is coming this is your surcharge because here soil soil is there it is filling with the soil. So q is equal to gamma z or gamma h whatever you can write it out. So in this case the surcharge will be above this d because this is your embedded depth. So it is q is equal to gamma h plus gamma d plus 2 c plus 2 c minus minus q minus 2 c in this case q plus gamma d is your surcharge plus gamma d and minus gamma d minus 2 c means this is your p a gamma d minus 2 c gamma d minus 2 c now gamma d gamma d will go. So it will be it will be now this is minus minus this become will plus. So it will be 4 c plus q 4 c plus q now if I write it if you look at here this is 4 c plus q. So one side it is 4 c plus q other side is your 4 c minus q as I said there will be a transition it should not go in this direction there will be a transition from one point to other point. So that is why there is a smooth transition. So it is coming 4 c plus q here it is coming 4 c minus q and I draw a resultant r a is here and this is my 2 c this is all about how this earth pressure diagram has come and how it has been plotted. Now next step is your derivation from there you will have to find it out what is the depth d embedment depth d is required. Now if I take summation of if I take summation of h is equal to 0 what does it mean that means summation of horizontal forces is equal to 0. So let us put it this is my z 0 or h 0 or h 0. So summation of horizontal forces is equal to 0 that means if I take particularly this resultant force r a now I am writing r a plus z bar by 2 into 4 c plus q plus 2 c plus 2 c plus 2 c plus 2 c plus 2 c plus 2 c plus 2 c 4 c minus q minus d into 4 c minus q is equal to 0. Look at here total horizontal forces I am taking as a whole body total horizontal because it has to satisfy the equilibrium condition that means total horizontal forces is equal to 0 horizontal forces is equal to 0 means this is your r a this is your resultant force from here to here a to d r a in this direction and now this is z by 2 if you look at this this is the distance z by 2 where the resultant force will also act below this. So this is a z by 2 because somewhere else it will act somewhere else it will act it will be z by 2. So this will be z bar by 2 4 c plus q if you look at 4 c f a I am taking complete one 4 c plus q minus 4 c plus q then this will be 4 c plus q plus 4 c minus q. Now then minus d times 4 c minus q because this is a transition phase from active to passive active minus activity changing now up to whole of this this will be a active state. So 4 c minus q into d it should be deleted or neglected from here. So d into 4 c minus q it should be equal to 0 this is your total resultant forces I have taken it is to be 0 from there we can find it out z bar is equal to d 4 c minus q minus r a divided by 4 c this is my z bar that means I have to find it out a distance below the point c where the pressure intensity is changing where the pressure intensity changing it has to be first identify because where is your point b then z bar once you get it then what else is remaining because you have to find it out your movement because there are 2 unknowns you have to find it out your movement that means movement at point c considering movement is equal to 0 at point c at the base c is nothing but your at the base you can say this is the c point at the base now movement at c is equal to 0 then in this case r a y bar plus d minus d square by 2 into 4 c minus q plus z bar by 3 into z bar by 2 into 4 c minus q into 4 c minus q into 4 c minus q into 4 c plus q is equal to 0 how it has come I am taking movement at point c if you look at here r a this is your resultant force from here to here r a into y bar y bar is your distance from point d to resultant force plus d because I am taking movement at this point of the base minus d square by 2 into 4 c minus q if I take it this area this is 4 c minus q into d into because movement d by 2 so that is why 4 c minus q into d square by 2 it will be neglected means deducted then z by 3 into z by z bar by 2 if I take it like this so it will be it will be is c g will be z bar by 3 it will be somewhere else so z by 2 area is your half half this half this into this so this is your half z bar 4 c 4 c minus q plus 4 c plus q into z bar by 3 triangular distribution. This is you will be coming from z by by 3 in this equation if I put the z bar and simplify it then it is coming d square 4 c minus q minus 2 d r a minus r a this is your 12 c prime y plus r a by 2 c plus q which is equal to 0 now it is a square term now it is a quadratic equation so we can solve it so you can get it so this is our final form of the equation this is the final form of the equation this is the final form of the equation this is the final form of the equation where you can find it out the embedded depth once you will get the embedded depth once you will get the embedded depth then you can use the factor of safety and you can find it out what is your factor of safety then with 20 to 40 percent additional factor of safety you can say that with this height of h this distance from dredge line to this if this is the height is specified and the load is given then you with this you can say that this depth of embedment is suitable for particularly cantilever sheet pile wall or sheet pile in cohesive soil in cohesive soil now we will solve a problem example problem look at this even if it is not necessarily that this should be clay and this should be clay only condition is that the embedded depth below this it should be clay if above this it is not clay if it is a pile if it is a purely cohesion less soil if it is a purely cohesion less soil then pressure distribution diagram pressure distribution diagram will change for top part this is not your pressure distribution diagram this is for purely cohesion less soil cohesion less soil and c is equal to 0. But, below the embedment depth the condition to be it should be cohesive soil it should be cohesive soil. Now if I start with this what are the condition given in the example if you look at the example the example is given there are two parts first is your 3 meter second is also 3 meter then your gamma is equal to 17.3 kilo Newton per meter cube phi is equal to 30 degree and this is your water table then gamma submerged is equal to 9.5 kilo Newton per meter cube phi is equal to 30 degree phi is equal to 30 degree and this is a purely clay soil. So, that means c is equal to 57.5 kPa phi is equal to 0 degree phi is equal to 0 degree now find out this embedment depth for particularly cohesive soil. Now first step step one as I explained yesterday's last class example problem solved example problems step one is your draw the pressure distribution diagram. Now for top one this is my pressure distribution diagram because there is no c this is a cohesion less soil phi is equal to given 30 degree 30 degree 30 degree and now if this is my ra resultant force ra and this distance should be y bar at the derivation from there. Now it will start with this this is 4 c minus q this is your 4 c plus q as we have derived earlier now this is the case this is the case if I am writing this final form of equation here it will be d square 4 c minus q minus 2 d ra minus ra 12 c y 12 c y bar plus ra divided by 2 c plus q is equal to 0. Now let us start this solving the effective pressure at the dredge line if this is my dredge line with this condition what is your effective pressure q q is equal to 17.3 into 3 plus 9.5 into 3 which is equal to 80.4 kilo Newton per meter square. Now step 2 from this earth pressure diagram you can find it out if I draw the pressure distribution diagram of the top part whatever I have drawn earlier if I draw it this is my pressure distribution diagram now make it part by part this part will be your k a gamma h which is equal to 0.299 into 17.3 into 3 and this is your this is your 0.299 into 17.3 into 3 and this part is your 0.299 into 9.5 into 3 how the k a has come 0.299 5 value is given. So it will be 1 minus sin 5 by 1 plus sin 5 you can find it out what is the value of k a it is coming about 0.299 from there now with this help of this pressure distribution diagram we can find it out what is the value of r a and what distance it is acted upon from the base y bar now once now this r a is equal to r a is nothing but your resultant force acted r a is equal to you take this this part this part and this part then put the value you can find it out this will be your half h into k a gamma h then plus k a gamma h into h then plus half k a gamma prime h into h look at this for this triangular distribution pressure is equal to half base into height. So base is equal to your base is equal to k a gamma h k a gamma h this side is equal to h h is nothing but this is 3 meter this is 3 meter and for this part it will be k a gamma h into h for this part a half k a gamma prime h into your h this is your r a now putting this value now put all the value step by step you can put all the value putting this value you can find it out it will be 82.6 kilo Newton per meter this is your resultant r a it is your 82.6 kilo Newton per meter now you can find it out at what distance r a into y bar is equal to 12.78 I am just putting the value into 1 plus 45.56 into 1.5 plus 23.28 into 4. So y bar is equal to 2.13 meter now if you look at here the pressure here coming the pressure here coming the pressure because I I skip one step this pressure here coming is about 23.28 and pressure here is coming about 46.56 and here it is coming about 12.78. So how the y bar has been calculated r a into y bar at the base which is equal to 12.78 into 1 because this is 3 3 by 3 it is your 1 meter about the c g then 40 45.56 this is your 45 45 point sorry it is your 46.56 this is your 46.5 into 3 by 2 it is equal to 1.5 then 23.28 this is your 3 this will be acted by 2 by 2.3 2 by 3 into 3 no it is not 2.3 it is your 23.28 this is your half into 3 into 23.28 it is coming. So this is not 4. So this will be 3 by 2 this will be you can say that half h into base. So this is your half into 3 into 20 this much you can put it and it will be acted upon by this is the area and it will be acted upon by 2.3 of 3. So it will be 6 by 2 whatever this calculation will come from there you can find it out your y bar is equal to 2.13 meter from this you can calculate your y bar is equal to 2.13 meter. Once you get the y bar is equal to 2.13 meter then find it out the respective coefficients now I got r a and y bar now find it out respective coefficients 4 c minus q you calculate it is 149.6 then 2 r a which is equal to 165.2 165.2 then 12 c y bar plus r a which is equal to 15552.3 then 2 c plus q which is equal to 195.4 now with this values with this values because we calculated 4 c minus q take directly then 2 r a take directly then 12 c y bar plus r a is given take directly then 2 c plus q will take directly and r a is given it will come in terms of some numerical value with d with value of d if I write it it will be coming like this d square into 149.6 minus d into 165.2 then 82.6 into 155.3 by 192.6 into 192.6 25.4 which is equal to 0 now d is equal to 2.7 meter now once you get the d whatever you are getting the d whether it is correct or not that has to be checked how you will check it from the base from the base pressure distribution diagram whatever the d calculated whether it is correct or not now you can find it out what is your z bar once the d is over d is equal to 2.7 to meter I am writing d is equal to d is equal to 2.7 to meter now for check you can write z bar is equal to d into 4 c minus q minus r a divided by 4 c which is equal to 1.41 now you can calculate a r a plus z bar by 2 4 c plus q plus 4 c minus q minus d into 4 c minus q from there once you check it by taking moment about the base you see by taking moment about the base once you know all the value taking moment about the base check whether it is coming 0 or not this is our condition moment about the base. So, I have considered the taking moment about the base and how much it is coming it is coming 0.012 means almost it is equal to 0 now whatever value we are getting that is now after once you get the value d is equal to 2.7 to as I said you can take a factor of safety between 20 percent to 40 percent. So, let us say 30 percent factor of safety you have to put it. So, now d d implemented d value will be 3.54 meter with factor of safety is equal to 30 percent. So, this is all about cantilever seat pile in cohesive soil I have solved a problem if it is not let us say if it is not cohesion less cohesion less soil if it is a cohesive soil then what will happen only the pressure distribution diagram this will minus this will be plus it will change rest part will be the way I have calculated the r a with your centroid distance same calculation will be done then you find it out d and at the end you take moment at the base check whether it is equal to 0 or not. If it is coming approximately 0 that means whatever you have done the calculation that is that has been cross check and it is ok then apply your factor of safety between 20 to 40 percent as we have already discussed then consider in this case 30 percent you can take also 20 percent you can take also 40 percent do embedment depth to be implemented d is equal to 3.54 meter thanks a lot.