 last class we have finished the cantilever sheet pile wall means find it out the depth of the penetration of cantilever sheet pile wall in granular soil. The solution has been calculated how to find it out this depth of amendment of cantilever sheet pile wall. Now there is one another one method is called approximate method remember these whatever we are discussing these are for cohesion less soil or granular soil. In approximate method if this is the sheet pile wall and this is your dredge line and this distance is d let us say this part is your retaining of soil. Now this height is let us say h height in this approximate solution the pressure distribution in the back if you look at this this is my wall this is the wall and let us say a this is small d this is your c. So that means this is your dredge line up to this dredge line up to this dredge line below this dredge line there is a soil filling up the assumption for this approximate method is below the dredge line that means back of the wall this part is your soil retained soil retained that means if this is the front face of the wall and this is your back face of the wall this has been assumed that this back face will be acted upon by passive air pressure and front face will be acted upon by active air pressure. So this is not a case in the exact solution but in case of approximate solution for a quick calculation this has been made take it below the dredge line and back face acted upon by passive air pressure and front face acted upon by active air pressure. Now if you take movement at point c taking movement at point c movement at point c is equal to 0 then you can find it out half gamma k p d square into d by 3 is equal to half k a gamma h plus d whole square into h plus d by 3. Now this is a completely this is a cubic equation you can say that this is cubic equation now how to find it out this then you can start with this procedure is that first sketch the given condition sketch the wall and other conditions then find k a and k p that means active and passive earth pressures also compute distance a resultant r a compute distance a resultant r a and its location y bar if I say total forces is coming resultant is somewhere else here r a how far it is distance from this y bar of distance a means you can find it out what is the y bar then you can find it out how much is your y bar then find the in this case put the value of y instead of d you can put the value of y and find it out find it out by trial and find it out the value of what is the value of d by approximate value you can put it what is the value of d approximate method as I said it has been assumed one side of this below the d d d line this has been assumed acted by passive earth pressure that means back of the wall if this is my wall this is the wall so this is a back of the wall it is acted upon by passive pressure and front of the wall acted upon by active because it retain in the soil mass so taking moment at point c m is equal to c is equal to 0 so we are getting a cubic equation how to solve it the procedure first you draw this case the wall and other conditions that means soil parameter right c 5 what are the parameter what is the height of the wall above the d d line then once you get it find the k a and k p depending upon the 5 value you can find it out earth pressure k a and k p active as well as passive then find it out from the k a gamma d k a gamma h plus d where is your resultant earth pressure is acting and the distance of y bar then once you get y bar then put this value trial and error you put this and you will find it out d once you get the d you find it out l is equal to h plus d that means total length of your wall will equal to h plus d now these are the two methods one is your one is your whatever we solved last class one is your actual method and other is your approximate method generally design engineer they use sometimes this approximate method for quick calculation to know what is the length of this wall they are supposed to provide now we will solve a typical problem of this cantilever retaining wall in granular soil whatever solved whatever we derived last class so now we solve a typical example problem so this now this is the given conditions one is the example a wall is there in this wall this is a distance d is supposed to be find it out now one side it retained soil now water table is lying water table water table symbol generally we provide in this way water table is lying below ten feet this is your ten feet and this distance is twenty feet and this is a purely cohesion less soil gamma is equal to one one zero p c f pascal for cubic feet phi is equal to thirty degree and here gamma submerged is equal to yours sixty six zero p c f and phi is equal to thirty degree this is the condition find the length of embedment that means the d has to you have to find it out shown in figure either you can solve this by means of approximate method or whatever we derived it actual solution you can do it now first step what is your first step step one find k a k p k a dash k p dash now if you look at here you will find delta is equal to steel sheet pile and sand value of delta friction between sheet pile wall and soil is given seventeen degree now if i take a retaining wall this kind of what is the formula to find it out k and k p if this is my wall it may be inclined so this is your beta this is your row and this is alpha and this is p a this is your delta now with this what is the formula for k a and k p k a is equal to sin square alpha plus phi divided by sin square alpha sin alpha minus delta into one plus root over of sin phi plus delta sin sin square alpha sin alpha minus delta into one plus root over of sin phi plus delta sin phi minus beta divided by sin phi minus delta sin phi minus delta sin phi minus delta sin alpha plus beta alpha plus beta and this whole square now if you look at here in this conditions in this condition this is plane that means in this condition beta is equal to beta is equal to zero beta is equal to zero then what is about alpha this alpha if you look at this alpha this alpha is straight vertical that means this alpha is equal to ninety degree and what is about row what is about row value is a failure envelope row value we can we have not taken it here and what is value of delta delta is equal to seventeen degree phi is equal to thirty degree phi is equal to thirty degree now if you find it out with alpha beta if you terms are delta delta alpha beta beta now k a is to be find it out which is equal to zero point two nine nine as the soil value is same for both the sides so k a is equal to k a dash similarly for k p similarly for k p if i say to find it out the k p now i can change with this equation what is the value of k p k p is equal to sin square alpha minus phi and this is your sin square alpha sin square alpha plus delta and one plus one this is not plus this is your minus root over sin phi plus delta sin phi plus beta now sin alpha plus delta sin alpha this is not phi it is earlier alpha plus delta now sin alpha plus beta this is the value of this is the derivation of k p then putting this value we can find it out k p is equal to k p is equal to five point three eight five five point three eight five which is equal to nothing but k p dash that means first step you sketch and you find it out k a k p k a dash k p dash because these are all your r pressure r pressure values active r pressure k a and k a dash as i said earlier if the soil conditions are same soil conditions are same that means k a is equal to k a dash now k dash is equal to k p dash minus k a dash which is equal to five point zero eight six now gamma dash k p dash minus k a dash which is equal to let us say c which is equal to zero point three one now if i draw the actual pressure distribution diagram how it looks as we have explained or derived yesterday this is my wall now the there is a water table is somewhere else here so actual pressure distribution diagram will be this will go in this way then it will go somewhere else then it will go in this way now you have to find it out the area how to calculate this r pressure r pressure value unless otherwise you do not find it out the area how to calculate you know the k a and k p now we can divide into number of parts number one then put it number two then put it number three like this and this is your distance a point a at point o it rotate this is a distance a point a now next step is your step two find a r a and y bar a is your distance a is equal to distance below the dredge line where it rotate it rotate some points that is a point of rotation and now that is your point of rotation and y bar r a is equal to your resultant pressure acted somewhere else here resultant pressure acted at a distance from here this is your how much it is it is your y bar now calculate a y bar if you go back to previous whatever we have derived p a prime is equal to k a gamma h one plus k a prime gamma some must h two now this i calculate zero point one one zero into ten plus zero point zero six into ten into zero point two nine nine which is equal to zero point five one k s f now if you go back to previous what we have derived from back to last class derivation i have derived the how to get your point a value a is nothing but your p a prime divided by gamma some must k p dash minus k a dash which is equal to your zero point five one by zero point three one which is equal to one point six five now we got this is the value this is the value p a dash this is how much it is coming zero point five one the distance we got it your one point six five one point six five this is your next step now find it out what is the value of resultant r a r a is equal to zero point three three into ten by two plus zero point three three plus zero point five one into ten by two plus zero point five one ten by two into one point six five which is equal to your six point two eight k s so r a generally r a act at a distance distance y bar from point of rotation this is the point of rotation r a resultant force act at a distance y bar from the point of rotation now we will calculate your point of rotation means y bar if I say r a into y bar is equal to one point six five into fourteen fourteen point nine eight plus zero point three three into ten six point six five plus half into zero point five one minus zero point three three into ten into four point nine eight plus zero point four two into one point six five by three which is equal to y bar is equal to y bar is equal to eight point two now we get this resultant forces r a by taking this triangle by taking the area area into k a or k p you can find it out resultant force means forces at triangle one means portion one two three then some it then you can find it out what resultant forces this resultant forces at what distance it act from the point of rotation that also we calculated y bar is equal to from point of rotation this to this is your eight point two feet now point of rotation also we get it one point six five feet now next step is your find the value of find the value of d now if I say this is my p p dash what is the value of p p one dash is equal to gamma h one k p plus gamma submerged h two k p plus gamma submerged a into k from there it comes out to be nine point six six this and gamma submerged k is equal to zero point three one now p p one dash by gamma submerged k which is equal to thirty one point one six now eight r a by gamma submerged gamma prime k which is equal to hundred sixty two point zero six now six r a by gamma submerged k whole square which is equal to three ninety two point zero nine now if you look at this what is the formula generally find it out we have derived the formula to get it this is your formula now also we can get it six r a y bar p p dash plus four r a whole square by your gamma submerged k whole square this is your three twenty seven point zero zero point zero zero now if I put the equation y four plus y q this is the terminology I have calculated from here then thirty one point two minus hundred sixty two y square minus five seven eight one y is equal to thirty two seven hundred this is the formula for this has to be solved by means of trial and error means right y then y fourth then thirty one point two y q then hundred sixty two point y square then minus five point five point seven eight one y and this should be your thirty two thirty two point three seven hundred if I put the value of because this is a trial and error you can consider value start with a value initial value by assumed value see how far it is from your it should be equal because this is a fourth order equation it should be equal how far it is varying from this let us start with the value of fourteen now with this value of fourteen you can calculate y fourth y q this you can calculate approximately it is coming eleven thousand eleven thousand three forty two eleven thousand three forty two now with this eleven thousand three forty two if I take another value of fifteen y is equal to fifteen then you calculate all other parameters all other parameters as it is given so this value is coming approximately thirty two thousand seven hundred sixty how this has been calculated this has been calculated based on your approximate solutions that means you have to start with your particularly y any value of assumed value of y once you have taken is a fourteen then you can say that it is less than this then you increase to fifteen you see that how much it is coming it is slightly larger than this so it will come between fourteen point nine eight or fourteen point nine six this this range it will come now once you get the value let us say y is equal to fifteen let us say y is equal to fifteen feet now we can get it d is equal to fifteen plus one point six five which is equal to sixteen point six five feet now as I said this is the value of d we are getting sixteen point six five feet from the example you have to use the factor of seventy that means increase between twenty percent to forty percent value let us say thirty percent increase let us say thirty percent increase that means your d prime is equal to twenty one point six feet d one means as I said earlier the d value to be whatever you get the d value with that d value increase a factor of seventy twenty to forty percent we have taken thirty percent either you take factor of seventy here or you take factor of seventy in the value of k a or k p it is your one point five times you can take k a and k p one point five times so you are getting d prime is equal to twenty one point six feet that means with this value it is just stable it is just stable with value increases from twenty one point six feet it is safe environment depth to be safe or it will be more than more than safe you can say that so as I said this completely it has been calculated by means of trial and error method and this is a generally a fourth order equation first step you have to find it out draw the diagram find it out your water table where water table is lying once you get your water table find it out value of k a and k p the way I calculated once you get the k a and k p draw the pressure distribution diagram draw the pressure distribution diagram as I draw it earlier pressure distribution diagram and find it out point a is your point of rotation this is point a below this point of rotation then find it out resultant force r a and find it out the distance resultant force from point of rotation once you get r a a and y bar then find it out other terms in this y fourth y q y square these are the terms we calculated these are the terms we calculated from this based on this equation these are the terms has been calculated now once you put this value and find it out what is the this is the given value this h this is the value of h that means this is the external load applied this load is given earlier so once this is given this fourth order equation you get it now the question is how much depth I can penetrate this complete sheet file wall so once you put it the value by trial and error you can find it out what is your depth this is your depth is 16.65 then from there you put your factor safety then this is your depth if you go up to 21.6 feet then this will be stubble this will be stubble and more safe and if you provide d is equal to 15 feet this is just about to take if anything load coming beyond this it will it will fail immediately so how this this is one of the example problem how we have solved next class I will try to solve one more problem.