 bulkhead anchors as you have finished last class about the fixed earth support. Let us solve one example anchor is there means a sheet pile is there then it has been decided to provide this anchor rod say at a distance one meter and the soil profile is saying below two meter the anchor rod this is the anchor rod below two meter of anchor rod there is a water table this is the symbol of water table and gamma above water table is eighteen kilo Newton per meter cube and phi is equal to for both the cases phi is equal to thirty degree gamma saturated is equal to twenty kilo Newton per meter cube that is below the water table and this height is three meter. Now it has been asked by means of fixed earth support method find the embedded depth d by fixed earth support find d find the d and anchor rod capacity also f a as you have discussed it can be solved by means of elastic line or by means of equivalent beam method. We will see now for the equivalent beam method how we can solve it for the case of equivalent beam method as I said earlier the steps first principle is to find it out your k a k p all other parameters. So from whatever the value of phi is given from there k a is equal to one third as it is thirty degree k p is equal to three and k dash is equal to k dash is equal to three minus one by three which is equal to two point six seven now next step once you find it out k a k p and k dash draw the pressure distribution diagram then find it out what is your pressure if I take it into two parts by means of I am solving this example by means of beam method equivalent beam method that means taking this and I am drawing this pressure distribution diagram now here is your water table here is your anchor rod. Now this is the water table and from there it is coming because at certain distance if I draw given idea how the pressure distribution diagram it is I am just drawing here in this question paper I am just drawing how it comes it comes in a way like this and this part if you remember well this part has been this part has been as it is a fixed earth method there will not be any deflection. So this part will be replaced by an equivalent equivalent load. So if you make it by means of a particularly equivalent beam method you assume up to this where it is zero up to this point of contra flexure say point of contra flexure is at a distance x up to this you consider as one part and below this you consider as a one part I am just reviewing what we have done in the last class for solving this problem now I have taken into consideration particularly here if you take it if you take it this is my point of contra flexure. So upper part from here it has been considered as a one unit and bottom part as one unit. So as assumption is that it is a hinge support for equivalent beam method with this hinge support there are two forces opposite acting to each other so that net force will be zero. So once I draw this so this is a height one meter this is your height two meter now find it out pressure distribution diagram from there you can find it out what is the value of pressure it will be twenty eight point one nine then this is your fifteen point two if I name it a b and j and here if I draw take into consideration of the bottom part take into consideration of the bottom part then this is fifteen point two and this will be your twenty seven point one and this is your r p and this is your somewhere else you can write it down this is your r b then this distance as I said this distance is your x and this distance is your a minus x a minus x then from here to here this distance is y minus y minus a which is equal to nothing but I have taken capital a and from here to here this distance is y minus x now we will solve this by step by step now let me calculate value of as you said this is the pressure distribution diagram and this part will take at a one unit this is as a one unit by equivalent beam so here you can find it out what is the pressure distribution at step by step we can find it out we can calculate then see whether this values are coming whatever I have written whether this values are coming ok or not then let us start with the top part top part distance is your say one meter if you look at here from here to here distance is one meter and here to here distance is three meter and here to here distance is three plus two plus one this is your six meter so now p a dash is equal to one third into three into eighteen plus one third into three into twenty minus nine point eight one how it has come this p a dash this p a dash it has come if you divide into small small parts this is your this this part this or you can take it as a whole your k a is equal to one third so k a is equal to one third one third into three means this is your k a this is your h this is your gamma this is your gamma if you look at here this distance this is your entire this so now as I said k a is your one third this is your three and gamma is equal to eighteen and for this part this is your k a this is your h this is your gamma submerged gamma submerged is nothing but gamma submerged gamma saturated minus gamma w gamma saturated weight is given twenty kilo Newton per meter cube and gamma w is nothing but your weight of water unit weight of water so twenty minus nine point eight one how the concept has come twenty minus nine point eight one this is the pressure at the base so it is coming twenty eight point one eight kilo Newton per meter square now next step is this is your step one you find it out pressure at the dredge line next step is find it out distance a so a is equal to p a prime divided by gamma prime k prime so from there which is equal to twenty eight point one eight divided by ten point one nine into two point two point six seven two point six seven is your k prime from where we have calculated this is your two point six seven and this is coming about one point zero four meter now then you find it out from the chart if you remember well I have given a chart last class from there you can find it out x as you know for phi is equal to for phi is equal to thirty degree thirty degree x is equal to from the chart x is equal to zero point zero eight h this is from the chart from where x is coming about zero point four eight meter that means distance below the dredge line at which you are getting point of contra flexure that means bending movement is equal to zero that distance we are getting from the chart for phi is equal to thirty degree is about zero point four eight meter now once you get this one then you can find it out what is your p prime so p prime is equal to a minus a minus x by a into p a dash which is equal to one point zero four minus zero point four eight then twenty eight point one nine by one point zero four which is equal to fifteen point two kilo Newton per meter square this is the value I have written from the pressure distribution diagram if you look at here this is eighteen kilo Newton per meter square this is twenty eight point one nine kilo Newton per meter square and this is your fifteen point two kilo Newton per meter square then once you get p a p prime then a your next step is going towards your taking what is your what is your next step will have to find it out r b will have to find it out r b from this pressure distribution diagram from the pressure distribution diagram we know all these value what is the other value is missing the other value is missing your that means we have to find it out r b force r b this force r b will come from your equilibrium conditions what is that condition this condition is movement about anchor rod is equal to zero this is your assumption earlier as I said movement about your anchor rod is zero now taking now taking movement is equal to zero of the entire free body diagram about the anchor rod about the anchor rod is equal to zero from there you can find it out half eighteen into three into one plus eighteen into three into three point five plus half into ten point one nine into three into four then plus fifteen point two into zero point four eight into five point two four then plus half twelve point nine nine into zero point four eight into five point one six minus r b prime into five point four eight let us start with this step by step how I have taken the movement half eighteen three into one if you look at here this triangle if you look at here eighteen into three eighteen into three eighteen into three half this is your area where the force is coming and it will be acted upon acted upon at a distance one below this from the movement so this is your area into one then eighteen into three eighteen into three into three point five if I look at the first case if I write it one this is your two this is your three this is your four this is your five and this is your six one part is your movement about anchor rod one now go back to second part eighteen into three into three point five if I take the into two parts this is your eighteen this distance your three this is your area and it has been acted upon at the c g this c g will be one point five and one point five plus two this is your three point five so this part is your two then come back to third third is your area half pressure here into height so pressure is your half ten point one nine into three into height is equal to four height is equal to four means this is your one third of three is one so two third of the three this is your two and this is your two so this distance is your four then come back to fourth this is the four if you take into two parts fifteen point two into x fifteen point two into x this is your zero point four eight plus five point two four this distance is your three plus two five and this is equal to point four eight divided by two this is your five point total if I add it this side is your five point two four if I am taking from the c g from here to here this side is your five point two four five point two four meter now about fifth point fifth part so this is your first second as I have written here this is third this is fourth now this part is your fifth this is your if you look at here this part is your fifth this part is your fifth now for this fifth what is your pressure half into twelve point nine nine into zero point four eight plus five point one six is your height from the c g of the triangle two up to the anchor rod these are these forces are acted upon if you look at here these forces are acting these are about the anchor rod clockwise and this is your anticlockwise so that is why this is negative this is your positive so r b prime will act at the base so that means three plus two five point four eight at a distance it is your five point four eight at a distance and this is your part six why I am solving in details if you do if you in the exam or if you do part by part one by one then chances of doing committing mistake will be less now once you get it from this after solving you can find it out r b prime is equal to sixty point five sixty point five kilo Newton now this is for once you get r b this r b is equal to equal and opposite this r b or r b prime because this is a hinge so there will not be any movement so r b prime equal and opposite force should be there so this r b prime once you get then you can very easily find it out what is the distance d or in terms of y so let us put let us put y minus a is equal to capital a now p a prime by a which is equal to twenty seven point one now taking movement now taking movement of the lower part now taking movement is equal to zero from here to here at the lower part you can find it out what is your r p value in terms of in terms of a so take this calculation r b dash into a plus a minus x is nothing but a value what is the value of a a is equal to one point zero four x is equal to what is the value of x zero point four eight one point zero four minus zero point four eight this is your a minus x so r b prime a plus x zero point five six zero point five six yeah zero point five six plus half into fifteen point two into zero point five six into a plus zero point three seven minus r p into a by three which is equal to zero now once you put the value of r b dash sixty point five in this case you can find it out r p is equal to thirteen point five five a square thirteen point five five a square and r b dash is equal to sixty point five kilo newton how do i get r p r p is coming r p is coming this is my r p half area of the triangle this into this this is your a half this and this and it will act it upon the height from there r p is coming thirteen point five five a square and r b dash is coming sixty point five sixty point five kilo newton so put this value r b dash and r p here in this equation movement equation after putting this value you will get an equation a q minus fourteen point three seven a which is equal to seven point eight five now by trial and error by trial and error a is coming about four point zero three meter now once you get the a then you can find it out y is equal to a plus a a is nothing but y minus a remember so from there you can find it out five point zero seven meter and d is nothing but your twenty to forty times more it will be one point two of y which is equal to six point zero eight meter six point zero eight meter d is equal to coming six point zero eight meter now once you get the value of d once you get the value of d then you can easily find it out what is the value of f a f a is nothing but f a is nothing but r a minus r b dash so r a is equal to r a is equal to what is your r a resultant forces of all this this this this this total resultant forces acted here this is your r a minus r b this is your f a so from there r a is coming about i have not done the calculation you can do it r a is equal to one zero six point seven minus r b dash is equal to sixty point five which is equal to forty six point two kilo Newton this is all solved example complete detail calculation i have given in this particularly sheet pile and bulk head sometimes i neglect this how the calculation has been done for you so that you can try but one example i have solved in detail detail calculation so let me just review once again these are all your soil parameter has been given a sheet pile is there and there is an anchor rod and these are all your soil parameters are there and this is the it has been asked to find it out embedded depth d by means of fixed earth support method by means of fixed earth support method and also find it out what is the force at this anchor rod so in this case as i said earlier this fixed earth support fixed earth support for particularly bulk head anchors or sheet pile with anchor head in this case there are two methods generally we adopt one is your by elastic line method other is your by beam method in beam method in beam method at the point of contra flexure at the point of contra flexure the upper part has been acted like one beam lower part has been acted open by other beam how the assumption is that at the point of contra flexure there will be a hinge there will be a hinge in the hinge that means once you assume there is a hinge that means moment is equal to zero the meaning of point of contra flexure is that at that point moment is equal to zero so once hinge has been assumed there will be two reaction force is equal and opposite so that moment is equal to zero with this beam method now next step is your draw the pressure distribution diagram of upper part and pressure distribution diagram of lower part so step one calculate your base from the phi calculate k a k p and k prime and it has been calculated phi is equal to thirty degree so k is equal to zero point three three and k p is equal to three k dash is equal to k p minus k a which is equal to three minus one by three which is equal to two point six seven and as water table is acted below three meter from the ground surface so below the water table generally we are going to consider submerged unit weight above the it will be your what is the gamma value is given it is eighteen kilo Newton per meter cube so submerged unit weight is equal to gamma saturated minus gamma w step one basic parameter you have calculated once it is over draw the pressure distribution diagram and the pressure distribution diagram has been drawn so then step one find it out p a prime pressure at this point and p dash pressure at this point so instead of finding all total what I have done it I make it into part by part one two three like this part by part from this from this so p a dash is coming about twenty eight point one eight kilo Newton per meter square so this is your twenty eight point one eight one eight kilo Newton per meter square at this point then once you get it then you can find it out what is the value of a the distance of a where this is your distance a so a has to come from p a dash by gamma prime and k prime it is equal to one point zero four meter so based on the value of five as I said earlier by means of chart from the chart value of five from the chart x is your zero point zero eight eight this is coming about zero point four eight meter once you get this is your zero point four eight meter that means point of contract lecture below the dredge line is about zero point four eight meter so we are getting zero point four eight meter value of x once you get the value of x you can find it out p prime p prime at this level what is your pressure so p prime is equal to a minus x by a into p a dash so from there it is coming about fifteen point two kilo Newton per meter square so now all these step two all the pressure distribution diagram is over all the pressure distribution diagram is over as it is an equivalent beam consider above this above the point of contract lecture this as a one body and take what is the second assumption the movement about the anchor rod to be zero so take movement about the anchor rod so that it will be zero by considering movement about anchor rod so movement has been taken into consideration so again in this case I have taken into part by part one part two part three part four part five part and six part so from there the r b value r b prime the force coming is about sixty point five kilo Newton once you get this four sixty point five kilo Newton this is about your sixty point five kilo Newton so these are equal and opposite so this value also sixty point five kilo Newton so similarly you can consider movement of these second beam this is your first assumption one unit one beam this is your second beam about the base because this is a fixed support so from there it has been done so it is in terms of r b dash and r p dash r p then r b and r p has value has been in terms of a and once you put you can get the value of a from there a is nothing but y minus a from there you can get it y is equal to five point zero seven from there d is coming about twenty percent higher twenty to forty percent so one point two times of y it is six point zero eight meter once d is coming so taking an equilibrium of the upper part r a r b dash is coming all forces at the base resultant forces is your r a so it is your r a minus r b dash force in the anchor rod is coming about forty six point two kilo Newton this is all about what I have discussed about sheet pile wall free support fixed support and sheet pile wall with your anchor rod free and fixed earth method and in this case there are two cases for each case one is your cohesion less soil another is your cohesive soil this is all case by case I have derived of how to get your embedded depth from the basic equation then each and every case we have solved one some practical example how to get d and how to get also anchor rod force this is all about we will start next class some other new topic thank you.