 Let us solve one example for cantilever retaining wall and its stability analysis check. This is the wall and this dimensions has been given. Later on we will see where there is no dimensions of wall has been given but you will have to start with this scratch level and this is part of fill and fill soil is your gamma is equal to 17.95 kilo Newton per meter square 5 equal to 28 degree C is equal to 19.12 kPa and unit weight of concrete gamma C which is equal to 23.56 kilo Newton per meter q and this dimension is given from this to this side is your 6.1 meter and this is your 1 meter this is 0.38 meter 0.38 meter then this is your 0.25 this part is your 0.25 meter and this is 1.55 meter and total is your 2.88 meter and there is a back fill at one end this is your 1.4 meter and this is your 0.6 meter and this soil gamma is equal to 19.53 kilo Newton per meter q 5 is equal to 34 degree C is equal to 35.17 if you look at here in this case in example this dimensions have been given of this retaining wall and the fill material gamma is equal to given phi is given C is given this is a C phi soil. So, there will be definitely a tension crack with a surcharge 73.9 kPa and there is a back fill also it height is also 1.4 meter and also foundation soil properties are given with this given retaining wall it has been asked check this stability earlier we have 4 stability we say that what is the condition for this stability analysis. So, check for this stability before you go that this step 1 as it is a C phi soil look at this. So, total height is your 6.7 meter and this is your of calculate and calculation 20.45 and this part is your height of tension crack h 0 which comes to be 3.54 meter. So, if you look at it q at z is equal to 0. So, but this is equal to z is equal to 0 z is equal to 0. So, q value of q will be minus 2 C tan alpha which is equal to minus 22.9 kPa. So, r z is equal to 0 q is equal to minus 2 C tan alpha which is equal to minus 22.9 minus 22.9 kPa and h 0 also you calculate h 0 is equal to 2 C by gamma which is equal to tan alpha which is equal to 3.54 meter. Now, if you look at this is a C phi soil. Obviously, there will be a tension crack there will be a tension crack. So, there are 2 ways of for solution either you remove this tension crack you take this part of this and find it out your r pressure what will happen if there is a tension crack there will be gap between soil and the wall over the period of time when the rainy season come water pore inside this then big cracks will come. So, for this kind of problem you convert into C phi soil into equivalent phi soil. So, it will be taken care of this. So, this C phi has been converted to equivalent phi soil. Now, if I convert it into equivalent phi soil. So, if you look at this q is equal to gamma h kA minus 2 C root over of kA. So, for phi is equal to 28 degree kA is equal to 0.361 for phi is equal to 28 degree from this we can find it out 17.9695 into h is equal to 6.7 into kA is equal to 0.361 minus 2 into 19.12 into 0.60 which is equal to 20.48. So, if this q is your this q is your for C phi soil if I convert this q because I will have to convert to equivalent soil of cohesion less soil. If I make it into q is equal to of a cohesion less soil this earth pressure will be gamma kA prime into h which is equal to from there I can find it out what is the value of kA prime which is equal to 20.48 divided by 6.7 into 19.95 which comes out to be 170. So, now with this if you look at this problem how the problem has been solved this problem can be take consider into two ways one is you remove the tension crack and consider the after removing the tension crack how much pressure is coming that pressure you take into consideration for analysis. In that case other case is your you take into consideration of C phi soil into equivalent cohesion less soil the reason is that for C phi soil there will be tension crack in that tension crack over the period of time what will happen there will be gap between soil back field soil as well as wall over the period of time water will be accumulated once the accumulation is there there will be a bigger gap between this wall and the soil. So, I make it into equivalent cohesion less soil. So, first I calculate what is the value of q for C phi soil this comes to be 20.48 and taking into as if without surcharge this is your q without surcharge no surcharge has been taken without surcharge then q is I make it into equivalent cohesion less soil. So, k prime is coming about 0.170 now with the help of k prime 170 with the modified value of soil we are taking C phi soil to equivalent phi soil with this k prime coming about 0.170 phi is coming about 0.45.166 degree let us take phi is equal to 45 45 degree and your k a is equal to k a prime is equal to 0.172 now with this help you find it out your p a value now step two find the value of p a that means r pressure p a is equal to 0.5 gamma h square this is because of soil p a is equal to 0.5 gamma h square this is because of your soil plus q into h this is because of your surcharge k a prime from there it is coming out to be 0.5 gamma is equal to 17.95 into 6.7 whole square plus 23.9 23.9 into 6.7 into 0.171 which is equal to 96.28 kilo Newton per meter. So, once you get the p a value active r pressure value then we can go for this stability analysis if I redraw this retaining wall is like this then your value will be it will be like this then it will be done with this surcharge now if I make it into in this way let us say this is my this is my completely whatever part of the soil retaining this is my retaining wall remember this is the wall this is the wall and this part it retain soil mass both this part if I put it into this complete part is your w 1 then this is your w 2 then this is your w 3 then this is your w 4 then from this for 4 part like part 1 1 what is the weight of the part 1.5 1.5 into 23.9 plus 6.1 into 1.5 into 17.95 which is equal to 200.09 and its lever arm or arm how much it is acting from this from the toe this distance is given about 2.130 then you can find it out also moment then for part 2 look at this part 1 part 1 if you look at this part 1 that means I am considering I am considering this part is your this is your w 1 that means this is coming about 1.5 if you look at it 1.5 into 23.9 what is 23.9 23.9 is your surcharge surcharge up to this then 6.1 this height is your 6.1 if you look at here this to this height is your 6.1 then this height 6.1 6.1 into 1.5 into weight of soil 17.95 unit weight of soil 17.95 kilo Newton per sorry meter cube so it will be your kilo Newton per meter cube so this comes out to be 200.09 this is your w 1 and how far it is from this toe because you have to find it out the moment resisting as well as this moment your then how far it is it is 1.5 divided by 2 it is your 0.75 then plus this plus this it is coming 2.13 meter from this toe now come to the second part w 2 w 2 is equal to how much is your weight of concrete weight of concrete is your 23.56 so it will be 23.56 into 0.25 into 6.1 it is equal to 35.94 and it comes out to be 1.125 look at here 23.56 is your this part if I make it into this is your 0.25 into 23.56 unit weight of concrete into height is your 6.5 so how far this distance it is acting this distance is your 1 meter 0.25 divided by 2 so this will be 1.125 meter 1.125 meter so this load is coming about these are all your kilo Newton then come to third part come to third part if you look at this third part third part is coming to be this as well as this third part is coming to me this as well as this one is your concrete other is your soil two triangles are there so if I put it into 0.13 how the 0.13 has come into picture 0.38 minus 0.25 so this part will be your 0.13 from here to here from here to here will be your 0.13 so it will be your 0.13 or 0.13 so I have taken this part into common 0.13 into 6.1 into 23.5 into 6.1 into 23.5 into 6 plus 17.95 into 0.5 plus 23.90 how it has been come 0.13 into 6.1 this side is your 6.1 into 23.56 23.56 unit weight of concrete then 17.95 this is your soil filling of this unit weight of soil into 0.5 area of the triangle will be half a into b so that is why for both the common area so it will be 0.5 will be common then 23.9 into 0.13 above these the searchers has been already acted there is a searchers 23.9 so it will be putting as 23.9 into 0.13 of the triangle will be half A into B so that is why for both the common area so it will be 0.5 will be common then 23.9 into 0.13 above these the searchers has been already acted there is a searchers 23.9 so it will be putting as 23.9 into 0.13 total is coming about to be 19.57 and the lever arm is coming 135 if you look at here this is a 1 then 0.25 1.25 and 0.13 it will be acted upon where lever arm will be acted upon where then this will be from CG to this distance will be your 1.315 meter then 4th part this is your last part at the base this is your 23.56 into 0.6 into 2.88 which is about to 40.71 23.56 is your unit weight of concrete 0.6 is your this thickness if you look at here this thickness is your 0.6 and this part is your 2.88 so this is your 2.88 this is 2.88 and this will be your 0.6 meter and this coming about this is coming about 40.71 kilo Newton and the lever arm is coming about 1.448 now if I take it so f of b total vertical force these are all my vertical forces coming out to be 296.31 kilo Newton and movement of resistance from this if I take movement of resistance movement of resistance is coming about to be 550.98 how will get movement of resistance this vertical load into this arm this distance this is giving your movement of resistance this vertical load into this arm this is giving your movement of resistance this into this this into this this into this from there movement of resistance is is coming about 550.98 this is your kilo Newton meter this is your kilo Newton meter now once you get it your first part is your earth pressure distribution diagram then we convert it into equivalent soil equivalent cohesion less soil from there we get the value of k a prime modified value of active earth pressure from there we get the value of phi then once you get the value of five modified your equivalent value of five from there we get your active earth pressure p a value then once you get it then we have to find it out how much is your self weight if this is the structure of the retaining wall with this dimensions how you have to make it into part by part so that this mistake cannot be occurred so we got this how much vertical load is coming and how far this load is from this toe so that moment of resistance we can calculate about the toe then once this calculation is over then we will proceed for next part is your stability analysis now this p a if you look at this p a it is acting how far from this y bar distance this y bar is coming about to be sixty eight point eight nine into six point seven by three plus twenty seven point three eight into six point seven by two by your ninety six point two eight which is equal to two point five five meter two point five five meter if you look at here the value of sixty eight point eight nine sixty eight point eight nine then twenty seven point three eight sixty eight point eight nine is coming from where earth pressure because of your soil I have already removed that one if I remove this one step one then it will be clear how this problem has been solved so what I have done it p a value has been calculated as zero point five gamma h square plus q h into k prime from there it is coming sixty eight point eight nine plus twenty seven point three eight that is coming about ninety six point two eight kilo Newton per meter so if you look at sixty eight point eight nine this is your earth pressure lateral earth pressure because of your soil and twenty seven point three eight because of your surcharge from this I am taking sixty eight nine with this height will be your this earth pressure will be your six point seven it will be kind of triangular so it will be acted upon by six point seven by three then twenty seven point three eight twenty seven point three eight is your surcharge because of your surcharge it will be acted upon by half of six point seven by two this is coming to be two point five five meter once you get your resistance y bar that means at what distance at what distance this active earth pressure is acted below the base this is your nothing but your y bar y bar then you find it out all of your safety first one is your overturning factor of safety overturning factor of safety so this factor of safety is equal to movement of resistance by movement of overturning this comes out to be movement of resistance is your five five zero five five zero point nine eight by your movement of overturning what is your movement of overturning ninety six point two eight this is your lateral into two point five five this coming out to be this coming out to be your value of two forty five point five eight it comes out to be two point two four which is greater than one point five this is okay this is okay now next factor of safety once the overturning factor of safety has been checked next factor of safety is your sliding sliding factor of safety next is your sliding factor of safety what is the base soil value base soil value c is equal to thirty five point one seven so c prime which is equal to zero point six seven of c so which value will be this value will be your twenty three point five six it is given adhesion factor adhesion factor has been taken as zero point six seven so modified value of c will be twenty three point five six k p a now from here for factor of safety again sliding this is for sliding now for sliding c prime is equal to twenty three point five six k p a then tan five prime is equal to tan thirty four degree which is equal to zero point six seven five so this this resist force resisting force is equal to c prime into b plus f b into tan five prime which is coming about two sixty seven point eight six now this factor of safety is equal to f r by f d which is equal to two sixty seven point eight six by your f d is equal to ninety six this is your force ninety six point two eight which is equal to two point seven eight greater than one point five greater than one point five now out what are the stability check one is your overturning movement second is your sliding third is your e should be less than equal to b by six fourth is your bearing capacity check with this two slide two checks one is your overturning and sliding the factor of safety is okay it is greater than this what is the minimum requirement for this factor of safety now we will go for after the overturning we will find it out the tension crack tension crack that means in this case net movement is equal to movement of resistance minus movement of overturning which comes out to be three zero five point four zero and x is equal to movement by forces f b which is equal to one point zero three from two that means force will be arc at a distance one point zero three from the two this is your resultant force so with this value of this e is coming about to be b by two minus x and it is about to be zero point four one meter and again you check your b by six value b by six value is zero point four eight that means e is less than b by six that means there is no tension crack now third stability analysis criteria also satisfy that means e is less than equal to b by six that means there is no tension that means q maximum and q minimum are positive then third one is fourth one is your fourth one is your you check your bearing capacity factor of safety factor of safety against bearing capacity which is equal to q na net allowable bearing capacity of soil and from this is your how much is your q maximum how to get q maximum q maximum you will get it q maximum is equal to summation of b by b into one plus six e by b b is your summation of vertical forces b is your given two ninety six point three one kilo Newton and b width is your given two point eight eight and e also give value as we got it zero point four one q maximum also you get it then you can find it out also soil pressure soil pressure value means net allowable soil bearing capacity this comes out to be c n c gamma d f n q plus zero point five gamma b n gamma the c value is given c prime is given that is your twenty three point five six from value of five five is equal to thirty four degree from this value of five from this value of five from this value of five you can find it out n c n q and n gamma n c n q and n gamma these are only your bearing capacity factor once you find it out your bearing capacity factor then you can find it out q is equal to q u is equal to c n c plus gamma d f n q plus zero point five gamma b n gamma gamma d f depth you can take it one point four meter c n c c value c prime value is your twenty three point five six gamma is given nineteen point five three d f is equal to one point four n q you will get it from the value of five and gamma b n gamma once you get q u then you can find it out net allowable pressure by dividing factor of safety you can find it out q n a once you get q n a you check your factor of safety against bearing capacity whether it is more than three or not in this case it is coming about to be more than three that means it is also satisfied this I left for the students to calculate I have solved in details step by step so last one I left it so if I summarizing to this is the example problem where the detail wall is given wall dimensions have been given we will we will see it later on there will not be any wall dimension we will start from the scratch here wall dimensions have been given with this value of gamma phi c garr c and this is your filling soil this is your foundation soil first as it is a c phi soil so what will happen there will be tension crack will occur so to overcome that we will take it into equivalent soil which will take care of your tension crack so phi has been modified to forty five degree and k a has come once you get phi and k a then find it out what is your active r pressure because of your soil as well as searcher then find it out weight of each component you divide into number of parts like w one w two w three w four I made it into weight of each component kilo newton how far this weight is acted upon from the toe from there we can find it out what is my resisting movement overturning movement is your r pressure p a so arm how we got it then we multiply force into this arm we will find it out your resisting movement then once you get it we can check there are four stability criteria one is your overburden sorry overturning second is your sliding third is your tension crack fourth is your bearing capacity all this have to check so like this like step two sliding it is more than one point five it should not be less than one point five then it is not stable once this stability analysis has been satisfied then you go for structural design we will see how this structural design has been done just in brief we will see later on on may be in the next class thank you.