 In last class, I have discussed about the design methodology for the sheet pile in granular soil and cohesive soil. Now, design methodology means I have discussed about how to determine the depth of the sheet pile as the most of the resistance that sheet pile is getting that is because of the high depth that you have to provide for the sheet pile. Today, I will solve a problem where I will find the required depth of sheet pile which is in the granular soil. Now, if I start this example on a sheet pile, now for this example I have taken that is a sheet pile whose required depth we have to calculate and suppose this is the dredge level. So, this will be the d required depth of the sheet pile and this is the height of the sheet pile h above the dredge level. Now, you have to consider that the location of the water table and this is the ground level and this is dredge level. Now, this properties of this soil above this ground level water level. Now, this water level is 4 meter below the ground level dredge level is another 3 meter below the ground level below the water level and this depth we have to determine. This depth is unknown that we have to determine. Now, the properties which are taken that gamma for the soil above the water level is 19 kilo Newton per meter cube and 5 taken is 36 degree and below the water level as soil is saturated we have taken the unit weight is 9 kilo Newton per meter cube and 5 have taken the same 5 36 degree. So, here we have taken only the homogeneous soil, but that can be a layer soil. So, depending upon the type of soil you have to calculate the lateral earth pressure which is coming on the sheet pile. Now, here also same soil properties you have considered this layer that means here also gamma is equal to 9 kilo Newton meter cube phi is equal to 36 degree here also gamma dash equal to 9 kilo Newton per meter cube and phi is 36 degree. So, that means all of the same properties have taken, but that can be different. So, depending upon the type of properties you have to take that consideration in the design calculation. Now, for this sheet pile if we draw the pressure earth pressure diagram this is the dredge level this is the ground water table level. Suppose this is the top point if I consider this is this ground table is A point and this base is at B point and this dredge level is C point. Then the first we will consider this will be the earth pressure diagram up to the water table level then it will further increase then it will go this side and then from here it will go this side because that pressure distribution diagram I have already explained in previous class. Now, here the force will act in this direction but here the force will act in opposite direction here also it will act in this direction. Now the resultant force that will act here that is P A or R A. Now, this is point O point say and this now if I extend this triangular part. Now, here we can say this is the one is passive pressure 1 at B point this is also passive pressure 2 at B point. So, this is the B we have considered. Now, we consider that the this edge is at a distance of z from the base of the sheet pile and this resultant is acting at a distance of y bar from the O point where the pressure is 0 and this O point is also a distance of y 0 or y 0 or A from the dredge level and this distance of O point from the base is y capital Y and this is y bar small y bar this is capital. So, if I further if I explain what are the different components. So, this is the pressure distribution diagrams. Now, the P A or R A is the total force for this active earth pressure zone this active earth pressure zones mean if I this consider this is top T and this is C and this is C 0. So, this P A actually this P A capital P A is the total pressure or total force acting due to this T O C 0 and if you consider this is A 0 C 0 A 0 area you can say total active force that is active. So, that means total T O C 0 A 0 and T this area due to this total area this is the active pressure or active force which is acting. Now, and this force is acting at a distance of small y bar from this O point and O point is at a distance depth of y 0 or small y 0 or small A from the dredge level or C level and y and this O point is at a distance of capital Y from the base of the retaining wall sheet pile and Z is the distance from the B to this edge of the strangle. Now, first now we have to calculate from all these things have to calculate the this depth D required. Now, first we will consider if phi is 36 degree then what will be the coefficient of active earth pressure that is K A is 1 minus sin phi by 1 plus sin phi. So, that is equal to 1 minus sin 36 degree 1 plus sin 36 degree that is 0.26. Similarly, we can write K P will be 1 by K. So, this value is 3.85 that is 1 by K. Now, we will calculate the what are the forces that is acting at different level. So, that part we will first calculate. Now, if I further draw this diagram for and to show what are the forces that is acting at different level then you can see if I draw this diagram again. So, this is water table level this is dredge level this is base level. So, these are the forces here it is 0. So, you can this is the diagram the same diagram I have drawn here again to show the different force components. So, this is T point top this is A point this is at the this is O point this is C point this is C 0 this is A 0 and this is P P 1 B because this is B point and this is P P 2 B and this is the point where this distance is z. Now, this is the total force which is acting P A or R A that is a distance of capital Y a small y bar from the O point and O point is at a distance of A or small y 0 from the dredge level and this distance is capital Y from the base of the O point is at a distance of capital Y from the base of the sheet pile. Now, first if I calculate what is the pressure active pressure at A point I mean at this level then this will be as we have considered that the properties of the soil this is 19 for this top and the kilo Newton height is 4 meter this is 9 and 3 meter and then then we can write that this at A point this will be 4 into 19 into the active coefficient of the earth pressure that is 0.26. So, force at this level will be 19.76 kilo Newton per meter square that is at this level so here the force is 19.76 kilo Newton per meter square. Similarly, what the pressure active pressure at sea level the small sea level I mean this pressure because this distance from this is 3 meter water from dredge level to the water level this is 3 meter and top is 4 meter. So, we can write this will be 4 into 19 plus 9 into 3 as here that gamma is 9 and a height is this area is 3 and then the top is 19 is the gamma and this 4 and then into active earth pressure that is 0.26. So, this pressure at this point is 26.78 kilo Newton per meter square. So, the pressure at this level which is 26.78 kilo Newton per meter square. The next one we have to determine the A value or y 0 value. So, y 0 or A that is equal to p active at sea level. p point divided by gamma dash into k p minus k. This expression I have already derived in the last class. So, previous class I have already derived this expression. So, from that expression if we use that expression then we will get the A value. Now, here A value will be because p active at sea pressure active pressure at sea is 26.78. Then the gamma dash which is the lower portion gamma we are considering below the dredge level because this A or this A dash point is below the dredge level. So, here the gamma value is 9. So, that is gamma into k p is 3.85 minus 0.26. So, this A value is coming out to be 0.83 meter. So, A value that in o point is 0.83 meter below the dredge level. Now, we will calculate this p A or r a value this p A total active force due to this area is we can calculate this total for first way consider this triangle t A A 0. So, what this triangle total force is half into height is 3 meter 4 meter into the force that in 19.76. Then we will consider the force of the total this rectangular area I mean between A to C only this rectangular area we are considering. So, this rectangular area means that is 3 is the height and into 19.76. Then we will consider this triangular area between A to C. So, this rectangular area we have already considered then we will consider this triangular area between A to C. So, this triangular area will be half into 3 and then the total pressure is 26.78 from them 19.76 we have already considered. So, rest will be 26.78 minus 19.76. Then that means, this is the area. So, we have considered this triangle then we have considered this rectangle then we have considered this triangle then rest only this triangle C C 0 O that we have considered C C 0 O. So, that triangle means plus half into that distance is C to O is A that is 0.83. So, into 0.83 into the total stress at C level that is 26.78. So, that will be 26.78. So, these are the four components of this area first this triangle then this rectangle then this triangle and then the lower triangle. So, summation of this area of this forces for this total four areas that will give you the total active force acting due in this area. So, we can if I segment why if I write. So, first one will give 39.52 then next one will give 5.59. So, this is 39.28 and the third part will give 10.53 then the last one will give 11.11. So, total force that is 120.44 kilo Newton per meter. So, it is considering in terms of per meter. So, it is considering in terms of per meter. So, this is 120.44 per meter. Next we will consider so that means this P A or R A value is known. Next we will determine the what is the value of small y bar. So, small y bar that we will consider the area we will take the this with respect to o point we will determine this y bar. So, the distance of this force that is acting at different areas from the o point that we have to multiply with the force of each area and you have to divide it by the total force. So, in that case if we consider the first one then the for the first triangular portion this T A A 0 portion the total force is 39.52 which is acting as one third of this triangular area from the base. So, the distance will be that o C 0 plus C A plus one third of A t. So, O C is 0.83 and C A that distance is 3 meter then A t is 4 meter. So, one third of A t that will be one third of A t is 4 meter. So, that is the lever arm and this is the force. So, taking basically the moment with respect to this o point. Then the next one is next 59.28 for this rectangular area which is as a distance of this it is acting this half of this rectangular area. So, that means 0.83 plus this is 3 meter. So, 3 by this is 3 meter. So, 3 by next one is this 10.53 which is the force due to this triangular area only. So, that will act also one third from the base of this triangle. So, that is 0.83 plus one third of 3 meter. So, this height of this triangle is 3 meter. Then the last one is 11.11 and this is this triangle. So, that is also acting one third of the base, but from the o point that will be two third of the height of the triangle. So, that means here we can write into two third into height of the triangle is 0.83 and this total is divided by the total force which is 120.44. So, this total all the segment and then the total force. So, ultimately the y bar is at a distance of 3.05 meter from o. So, y bar have calculated that it is at a distance of 3.05 meter from o point. So, now we know the total force of this area. Now, we know the y bar then we will calculate the rest of the things. Now, again if we consider or we draw this diagram again if I draw this diagram. So, this is the water table level this is dredge level and this are the diagram and then this is the base level. So, here the total force of this area which is total force is 120.44 kilo Newton per meter and which is at a distance of 3 point a distance is 3.05 meter from o point and this o point is at a distance of 0.83 meter from the dredge level. So, this is o point a point t point and c point b point. So, this things this is the area force for this area this is p a. Now, again if I write this is p p 1 b this is p p 2 b. Now, this distance is o point is at a distance of capital Y from the base of the sheet pile and this h of this triangle is at a distance of z from the base of the sheet pile. Now, first we will calculate what is the value of p p 1 that value of p p 1 that value we can calculate this is the because that thing we have already calculated in the previous class. So, this p p 1 b the this is the net force of net pressure acting due to the passive and active this right hand side of the sheet pile this p p 1 b and the net pressure and this is the pressure at the base acting at the left side of the sheet pile that is p p 2 b. So, p p 1 b last class this has been already derived. So, this will be k p into gamma dash into h plus d minus k a into gamma dash into. Now, we will this is first we will consider this is gamma then we will apply it according to the here because as here it is d. So, it will be gamma dash. So, now you consider that this k p h and d if I replace this d and if I take this d also then we can write that k p minus k a into gamma d plus k p gamma h. So, if first for the general case we consider both are gamma. So, this is the way it we can determine this is the ultimately k. So, this thing has all previous class it has been already derived. So, that means k p minus k into gamma d plus k p into gamma into h. Now, from this figure we can see that this d is basically a plus capital Y this is a or Y 0 that is 0.83. So, a plus capital Y. So, this is the value of d. Now, if I replace this value here. So, this is k p minus k a into gamma into Y plus a plus k p gamma h. So, further if I simplify this is k p minus k a into gamma Y then plus k p minus k a into gamma into a plus k p into gamma into h. So, here we can see that this is the value of there is 3 parts and 3 gamma. So, here gamma as this is Y Y is below the dredge level. So, gamma will be 9 kilo Newton per meter cube and here also gamma and that is into a and a is also below the dredge level where the gamma is 9 kilo Newton per meter cube. But here gamma is h and with multiplying with h. And here gamma is not throughout this height of the because h is this one which is not always gamma is equal to 9 kilo Newton per meter cube because in the from top 4 meter gamma is 19 kilo Newton per meter cube and rest it is 9 kilo Newton per meter cube. So, if I put this value according to that. So, k p is 3.85 k is 0.26 here gamma is 9 into capital Y. Then the next part is plus 3.85 minus 0.26 into capital into 9 is the kilo Newton per meter cube is gamma into a is 0.83. Then the third part k p is 3.85 now here gamma into h that part will change. So, you can first we can write first gamma is 9 and there h is 4 meter. Then plus gamma is this first part gamma is 19 into 4 meter then next is 9 into 3 meter. So, finally, this is we can write this expression is 32.31 Y capital Y plus 4 23.4. So, this is the value of p p 1 at base level or B level. Similarly, we can write what would be the value of p p 2 at base level. So, p p 2 at base level that is also explained or derived in the last previous classes that is k p minus k a into gamma into Y capital Y. So, here this gamma will be 9 as it is multiplying with Y which is below the reach level. So, this will be 3.85 minus 0.26 into 9 into Y. So, that value is 32.31 Y. So, now we know this p p 2 at base level and p p 2 at base level. So, now these things will put in our general expression. So, general expression I have already derived in the last class. So, all these things we have which we are calculating here that we will put in the general expression and then we will determine the value of D. Now, the general expression which we got in the last class is 6 p a into capital Y plus small y bar plus z square into p p 1 b plus p p 2 b minus p p 2 b minus p p p 2 b into capital Y square that is equal to 0. So, this was the general expression or final expression that we got in the last class for the sheet pile and cantilever sheet pile in granular soil. Now, where z that expression it got p p b p p 2 b minus p p 2 b minus p p 1 b into Y minus 2 p a divided by p p 1 b plus p p 2 b at the base level. So, now if I put this value in the general expression value of z in the general expression then we will get 6 p a capital Y by small y bar plus p p 2 b y minus 2 p a that is whole square and then this is p p 1 b plus p p 2 b whole square into p p 1 b plus p p 2 b. So, one from the lower part this p p 1 b plus p p 2 b that one will cancel. So, we can write this whole square divided by p p 1 b plus p p 2 b minus p p 2 b capital Y square is equal to 0. Now, we will put this value 1 by 1. So, here this is 6 p a that we have calculated is 120.44 capital Y is unknown small y bar is 3.05 then plus this p p 2 b plus p p 2 b plus p p 2 b plus p p 2 b that we have calculated is 32.31 y. So, then y into y so that will be 32.31 y square minus 2 into p a is 120.44 that is whole square divided by this p p 1 b plus p p 1 b is 32.31 y plus 4 23.4 that is p p 1 b plus p p 2 b that is also 32.31 y. Then minus p p 2 b is 31.31 y square that is equal to 0. Now, if we further simplify this expression this will give us that if we multiply this 120.44 with 6 then minus p p 2 b is 31.31 y square that is equal to 0. Now, if we further simplify this expression this will give us that if we multiply this 120.44 with 6 that will give us 7 22.64 into y plus 3.05 plus 32.31 y square minus 240.88 that is whole square and divided by this is equal to 0.4. So, this will give us 32.31 y plus 32.31 y. So, that is 64.62 y capital y plus 4 23. And that is minus 32.31 capital y square that is equal to 0. So, now we have to solve this expression. Now, we if we further simplify this expression then we will get this expression. In further simplification if I simplify this expression and that means from this expression we can see that this expression is 72.64 y. So, this will be y 4 then 1 from here the y cube term will come and sorry this is this expression we can say this is p p 2 here for this expression this is 32 p p 2 b is 32.31 y. So, there is a 1 y square. So, this will be y 3. So, similarly this one will be y cube. So, that means y into y square. So, this will be y cube. So, y to the power 4 then y cube y square y and some then this form if I further simplify this expression then we can write that this expression in terms of after simplification can write that is 1 0 4 4 6 to the power y 4 plus 1 3 6 8 0 to the power y cube minus 3 1 1 3 1.33 to the power y square minus 4 4 8 3 9 1.3 to the power y square minus 4 4 8 3 9 1.3 to the power 5 y minus 9 9 1 2 1 8 that is equal to 0. Now, if I write and divide this all the terms by 1 0 4 4 point 6 then you will get y to the power 4 plus 13.1 y cube minus 29.8 y square minus 4 29.2 y minus 4 9 4 9 that is equal to 0. So, now we have to solve this simplify expression and form we will get the y value. So, this is y to the power 4 terms. So, we can solve it by trial and error method. So, that is one option. So, if I try with trial and error method. So, here I start with 6.7. So, first trial if I consider y is equal to 6.7 meter then we are getting that the term we are getting the extra term is 7 9 2.75. So, actually from the trial and error method this if I put the y value exactly then this is left hand side should be equal to 0, but here if I put y is equal to 6.7 meter the left hand side is coming plus 7 9 2.75. So, we have to reduce this y value. So, next trial we consider y is equal to 6 meter. So, then the left hand side we are getting force minus 471.4. So, that means it is in between 6 meter to 6.7 meter. So, in the third trial we consider y is equal to 6.3 meter then we are getting which is plus 15 only. So, that means it is close to 0. So, that means in the we have consider that our y value is 6 point. So, we have to further reduce this thing. So, we consider this y value is 6.29 meter. So, we consider 6.29 meter is the value of y is equal to 6.7 meter. So, we consider y by trial and error method. So, our D is y plus small a. So, y that we have consider is 6.29 meter then a is 0.83 meter. So, finally the value that is we are getting which is 217.12 meter. So, this is the D that we are getting from this after solving. So, this will be the required D of that sheet pile which we have consider. Now here also once we get the y value then we can determine what is the value of z. Then we can determine what will be the value of p p 2 b or p p 1 b that means the stresses are different level. Then one thing once we put this t value then we can check by taking summation of all the horizontal force whether it is getting close to 0 or not. If we are not getting 0 then we have to again check it with this calculation. That means there is some mistake and that means once we got this y value then we have to check whether all the horizontal or lateral forces that summation is close to 0. If it is 0 then or close to 0 then it is ok. Otherwise we have to check the calculation. Now here this D we will provide this D we are getting from the calculation. But as the last I mean previous lecture I mentioned we have to reduce it by 20 and 30 percentage. We have to increase this D by 20 to 30 percent because of the factor set. Now if I increase this D by 30 percent. So, D provided that will be 1.3 into 7.12 meter. So, this is close to 0.3 into 7.12 meter. So, we can provide D which is close to 9.5 meter or 9.3 or 9.5 meter that will ultimately provide by increase it 30 percent by. So, from the calculation we are getting 7.12 meter and we are providing this is 9.5 meter. So, we can see that our H value that means the if I again draw this figure of the sheet pile. So, this is the sheet pile. So, this is dredge level, this is ground level. So, H value that we are getting is 9.5 meter. Because this is the position of the water table and from this ground level it is dredge level ok. And this is the H and this is D that will provide or we will calculate. So, this is dredge level line. So, here this is 4 meter and this is 3 meter. So, H will be 5 meter. So, 4 plus 3 7 meter or 3 that we are providing is 9.3 or 9.5 meter. So, we can see the depth of the sheet pile below this I mean this dredge level this d value is much higher than the H value. So, that means here that is the difference between this sheet pile with the conventional retaining wall. Because here the most of the resistance this sheet pile is getting from this depth only. So, that means here required depth that we are getting is 9.5 meter whether the height of the sheet pile above this dredge level is 7 meter. So, that means the required depth is much higher than the height of the sheet pile. So, that means this is the one thing is that we have to go for very higher depth for this type of flexible structure and the that means here in the next section that I will discuss that how to design this anchor sheet pile because sometimes this required depth is very high. So, that means because here all the resistance is coming from the sheet depth because of this depth of the sheet pile that resistance itself getting from the soil. So, now providing this higher depth is sometimes very difficult. So, in that case we have to provide the anchored sheet pile to reduce this depth required depth further. So, we can reduce that depth required depth by providing anchor here. Suppose here the same this sheet pile we can reduce the depth if I suppose this is ground level. So, if I provide one anchor here and then fix it it is with anchor this one and then that means basically this is holding the this pressure is acting in this direction and this anchor is basically holding this sheet pile. So, in there that case we can reduce that required amount of the depth below the dredge level if we provide the sheet pile if we provide the anchor. So, this is our anchor and this is our flex cantilever retaining sheet pile and this is anchor sheet pile. So, where we can reduce the required depth. So, that how to design this anchor sheet piles what are the different types of design methodology. Thus those things will discuss in the coming classes or. So, in the today's class if I summarize this application part. So, here I have discussed only the design of or determination of depth of the sheet pile which is in the granular soil. So, now one thing is that this sheet pile can be in the cohesive soil. Now, in that case is in the last class I have already discussed in discussed that in the cohesive soil what would be the expression because in that case we consider the soil is purely cohesive. So, that means you have consider that soil is phi value is 0. So, in that case the final then in the similar process in that case a different for example, if I consider the same sheet pile in granular in the cohesive soil and then we will get a different value of the this different diagram. Then from this diagram we can get different value and then we can try to for suppose if I solve this same problem in the cohesive soil then what will happen. So, this is your dredge level this is the same water level and this is the distance of height depth this is 3 meter and then this required depth we have to determine. So, I have mentioned that in that case our this diagram will be this is negative this is positive then this will start from here then it will go in this form. So, here we you know that this is our dredge level this is the ground level here this value will be 4 cu minus q and here this value will be 4 cu this is plus q bar this is minus q bar. Now, force q q is the undrained cohesion of the soil and q bar is effective vertical stress at any depth. So, and this finally if I consider these things and then finally the final expression that we will get. So, similarly here also this q bar from this two pressure we have to calculate then the final expression then if I consider only the positive part then the finally the expression that we will get that is d square 4 cu minus q bar plus 2 d p a. So, this is the force which is acting that is p a then minus p a 12 cu cut small y bar plus p a divided by 2 cu plus q bar that is equal to 0. Where this q bar will be gamma into 2 cu plus q h. So, from this this if I consider same density same unit weight and height then the q bar for this case will be 19 into 4 plus 9 into 3. So, this will be the q bar for this case. Now the p a if I consider this p a is the force only the positive part and this y bar again which is acting the force from the dredge level. So, this so that means we consider this triangular part if actually if this is the rough triangular distribution in if it is a layer soil or different even in this case also if I consider in the different level because this is water is here and this water table. So, that means if it is a layer soil then we will get the different distribution, but here just we have taken a simple distribution of the to show how to calculate in case of cohesive soil, but in case of layer soil will get the different distribution. So, that in the negative part in the top then different distribution in the positive part. So, that both things we have to consider. So, here we have to we have drawn all the simple distribution and then in case of water table effect we have to consider just and then if we consider only this negative a positive part and then the total force that I have already calculated for this example in the granular soil. Similar way we have to calculate that p and then y bar. So, first we will calculate this p here and then y bar from the dredge level then these things other things are known that means p a will calculate q bar will calculate in this way and cu that is the undrained cohesion that is the property of the soil. Then all the other things we put in this expression then you will get a simplified expression in terms of d square and then once we will solve this expression then we will get the value of d from after solving the above expression d we will get after. So, once you get the d from this above expression then we will increase it by 20 to 30 percent due because of the factor set. So, in this way we can design this cantilever sheet pile in cohesive soil also in the next class we will discuss about the design of anchored sheet pile. Thank you.