 Last class, we have solved half of the problem, example problem. We have gone through this calculations of different weights coming to this wall, retaining wall. So, I have considered this dimensions has been taken as for this tentative dimensions and weight has been calculated and this is your, this is your W3 4.1 meter, 2.5 meter. I have calculated movement about A, this is also 0.4 meter, this is 0.4 meter and this is 1.3 meter. All about this tentative dimensions after satisfying this tentative dimensions, means after finding the tentative dimensions, then we have calculated last class for stability analysis. What are the forces, what is the self weight because of W1, W2 and W3 and movement about at point A. As I said earlier, you can find it out movement about this point as well as also movement about this point. So, the distance z from the point of application of resultant force at the base of the hill end, that means if this is the force coming rA and this is the rA. So, distance z, how far from A, so distance z is equal to, this is your distance z. So, z is equal to total movement about A by total vertical load. So, movement about A by total vertical load, total vertical load, this comes out to be 2826, 282641 as you have calculated in the last class, then 169, 690 which is coming about 1.67 meter, then eccentricity E is equal to z minus B by 2 which is equal to 0.42 and which is equal to less than equal to equal to your B by 6. B by 6 is equal to your 0.42, that means tension crack satisfied, there is no tension crack, that means both this maximum and minimum forces below the base of the foundation they are positive. Now, we can find it out what is the value of q maximum and q minimum, q maximum is equal to V by B 1 plus 6 E by B which is coming about to be 135750 kilo Newton and q minimum is coming about 0 because 1 minus 6 E by B, 6 E by B is nothing but E is equal to B E by B. So, 6 E into B is 1, so 1 minus this is 0, so q minimum is equal to 0. If I draw the pressure distribution below the base of the wall, below the base of the wall, suppose this is my base of the wall B, so this is the B, this is the B, the pressure distribution coming about to be this is 0 and this point is your A and this is your 135750, it is Newton per meter square, sorry this is not kilo Newton, this is Newton, Newton per meter square and this part, this part is your 0.8 meter and this is your 1.3 meter and if I name it A B C and this is D, then if I take this like this, then this is about to coming E F and this is G by means of this, if it is 0 this is this by means of simple trigonometry at point F, this value is 9231 Newton per meter square and G is equal to 7039 Newton per meter square, this value is your q max is your pressure intensity 135750 Newton per meter square. If I look at there, I have calculated q max, maximum pressure below the base of the retaining wall by means of B by B 1 plus 6 E by B, so it is coming about 135750 Newton per meter square, q minimum is coming about to be 0, I have taken this base of this retaining wall with this dimensions has been given 0.81 side and this is your 0.4, this is your 0.4 and this is 1.3, with this this pressure is your maximum, this is your q maximum 135750 Newton per meter square, minimum is equal to 0, F is equal to 9231 Newton per meter square, G is equal to 7039 Newton per meter square. By means of this, now we are going to design for this term, now there is a structural design part started, other two overturning movement has also satisfied, other two like check for your sliding factor safety against sliding and bearing capacity I have not checked we will do it later on, because this is related to your shear key. Now let us start this structural design of the third part, design of the stem, design of the stem means this part is your stem, so structural design of your stem, so actual height of stem how much it is 4.1 meter, height of stem H is equal to 4.1 meter, so actual bending moment, so bending moment is equal to 1000 into 4.10 plus 0.55 whole q Newton meter, this is the bending moment if you look at here, this is your 4.1 meter and this height is your 0.55 meter, so actual bending moment is equal to 1000, 4.1 plus 0.55 q Newton meter, effective depth consider effective depth, effective depth at the base which is equal to 400, 400 minus 40 which is equal to 360 mm. Now area of steel required A S T is equal to movement of resistance or bending moment, movement of resistance by your B D B D into T, T is equal to movement of resistance is equal to your movement of resistance is it comes out to be 100545 Newton meter, so 100545 Newton meter divided by 140 140 is your stress in steel, 140 because this is a mild steel into B is equal to q B D square q B D square into q B D into 0.87, 0.87 into 140. Sorry into 360, 360 is your effective depth D, so if I write it area of steel in terms of area of steel from this stress in steel, area of steel is coming about to be 2293, 2293 mm square. Now let us consider provide a reinforcement bar of 20 mm phi, 20 mm phi bars let us provide with this spacing is equal to spacing coming about to be 314 into 1000 by 2293 which is about 137 mm. Now provide now provide now provide 20 mm phi bars at the rate 130 mm center to center, how I have written this, this is your main reinforcement bar in the stem, because this is structural design this comes out to be 20 mm phi bar you can take it 25 or 18 depending upon the availability in the local market at the rate 130 mm center to center that means spacing is 130 mm center to center this main reinforcement has to be provided. Now distribution of steel, distribution of steel it is about 0.15 percent of main reinforcement as per your as per your IS code, so this is Indian standard code this is 0.15 by 100 into 400, 400 is your total depth D 400 into 1000 per meter which is coming about to be 600 mm square. So let us say distribution of steel let us say 8 mm phi bars which is equal to 50 into 1000 divided by your 600, so 600 is your distribution of steel total area 600, 50 is your area of this into 1000 it is coming about 83 mm say 80 mm center to center. So this shear reinforcement main reinforcement we are providing 20 mm phi bars at a spacing 130 mm center to center and shear reinforcement is coming about 80 mm we are providing at the center 80 mm center to center 8 mm phi bars. Now as you say as you see from here design of this steels particularly this stem if you look at here if you look at here this diagram this bending moment if you consider we have consider the bending moment at the base this is my stem, but this bending moment is varying at any interval from the top, so bending moment will be less here bending moment will be more and bending moment will be more at the base that means this bars are not going to provide throughout uniformly rather we need to have curtailment of bar for your better design or may be your as far as economic point of view curtailment of bar is required. Now bending moment we can say at any depth x let us say at any depth x what is your bending moment or area of steel area of steel is directly proportional to your bending moment which is directly proportional to your x plus 0.55 whole cube. Now if I take two section let us say this is x 1 and let us say this is x 2. Now I can write it area of steel at section 1 divided by area of steel at section 2 which is equal to x 1 plus 0.55 whole cube by x 2 plus 0.55 whole cube, so this is coming about to be if I if I want to curtail it at every alternative distance let us say alternative bars to be curtailed alternative bars means one bar after that other bar to be curtailed that means equal distance if I take it x 1 and x 2 is your 2 x 1. So area of steel a a s t 1 and a s t 2 which is coming about to be half and which is your x 1 plus 0.55 whole cube by x 2 plus 0.55 whole cube. So now you decide your x 2 x 2 you can take it depending upon this value you can take it so x 1 is coming about to be x 1 is coming about to be 3.1 meter from top. Now what does it mean at a distance from top 3.1 meter the bars main reinforcement bars are not supposed to be going to provide throughout that means 20 mm 5 bars at the rate of 130 mm centre to centre rather we will curtail every alternative bars. So it will become 20 mm 5 bars every alternative bars spacing will be double because you are cutting it will be 160 mm interval at the alternative bars. Now this is about your design of your stem similarly we will go for design of your tow slab this tow slab one part design of this stem for reinforcement is over that means your structural part is over now structural design of your stem. If I go for the structural design of the stem if I look at here look at this diagram this is my base slab what are the forces going to come here in this case at the base slab from here to here that means at tow one is at tow other is at hill at tow what are the forces supposed to come the force is coming to be this is your q max q maximum force is coming and here at the top may be the filling soil will come. So what kind of moment will come into picture this bar will bend like this that means compressional this side and tension sorry tension this side and compression of this side. If you look at here in this case there is a self weight of soil plus weight of surcharge and this pressure at this base at this point it will be zero. Now what kind of how the how the base slab of the hill reason how it behave it will be like this because weight will be more at the top. So that means this is your compression at this particularly if you look at here compression will be at this side compression will be this side that means in this case main reinforcement will be provided here in this case main reinforcement will be provided here. Now if I go for tow slab bending moment calculations now upward soil pressure upward soil pressure how much it is coming if I draw it once again if I draw it what I have what I have drawn it earlier let me draw it this is the part I have drawn earlier if I have drawn it earlier this is the part. So I am naming it a b c and this is my d and this will be your e e and what I have drawn earlier. So this will be f and this is g e f and g and this will be your h and this value is coming about 1 3 5 7 5 0 and this is 9 2 3 1 newton per meter square and this value is coming from 7 0 3 9 newton per meter square this values are coming from single trigonometry this total base is given and from there by single trigonometry we can find it out if this value is known other value we can find it out from this from this figure I can say that upward soil pressure c d f h c d f h upward soil pressure that is your value of c d f c d f h f c d f h because this part is your only two part. So I will consider only this part only for this many moment calculations or design of your tow slab. So I have consider into two parts base slab one is your tow slab other is your hill slab. So for tow slab I am taking into consideration of this part I am making into two parts one is your c d h f how much is your pressure intensity this is coming about to be 9 2 3 1 0 into 0.8. So this pressure at this point will be 9 2 3 1 0 into 0.8 this thickness is your this this width is your 0.8. So this comes out to be this magnitude is comes out to be 7 3 8 4 8 similarly e f h e f h part is coming about to be 4 3 4 4 0 into 0.8 by 2 it is coming about to be 1 7 3 7 6. If you look at here this is a triangular distribution this is a triangular distribution total pressure at this point is 1 3 5 7 5 0 pressure at this point is your 9 2 3 1 0. So this minus this is the pressure coming here. So this will be half this is your 0.8 into the pressure. So this is your pressure 0.8 divided by 2 half this is your load coming into picture coming into this total load coming into this. So this is your newton. Now with this upward you have to deduct it self weight because this has been made by means of a concrete r c c concrete you have to deduct by means of a self weight that means self weight of tow slab tow slab it comes out to be 0.8 into 0.4 into 25000 which is equal to your 80000 newton 0.8 is your this is your 0.8 this is your 0.4 into we unit weight of concrete that is your 25000 newton per meter square. So it is coming about 8000 newton now distance from c distance from c how much is your distance from c because you are going to find it out moment once you are designing we are going to find it out moment at this point about c. So distance from this c is about 0.4 and this part is your 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.4 into 0.53 and this part is your 0.4. Now we can calculate moment at c and this is to be coming about 29539 and 9209 and this comes about to be total coming about to be 38748 and 3289 0.0 you can check by yourself whether this calculations are correct or not this has been made by by myself. So this is coming about 35548 35548. So this is the maximum bending moment of the tow slab acted at point c and it will be in this directions because soil pressure will be more from this as you have calculated earlier from this you can find it out we can find it out what is your area of steel area of steel is coming about to be 858 mm square. So let us take let us take 12 mm 5 bars 12 mm 5 bars locally available market 8 10 12 14 these are the availability in the locally available market bars bars means steel 12 mm 5 bars. So spacing coming about to be spacing coming about to be 130 mm centre to centre you can say you can take care about 120 mm centre to centre. This calculations I have made it for structural design this is approximate because you can go in detail as per your indian standard code this is overall how this design has been made taking into consideration of structural part similarly similarly you can calculate this is for this is for your tow slab you can also go for a hill slab you can find it out hill slab in case of hill slab in case of hill slab in case of hill slab what are the things there one is your one is your soil one is your soil this is coming about 1.3 into 4.10 into 18000 if you look at this hill slab this distance is 1.3 and soil is filling up to height of 4.1 and unit weight of soil is your 18000 kilo Newton per then traffic load surcharge surcharge also given surcharge is equal to 10000 into 1.3 then you can say that weight of weight of also hill slab hill slab this comes out to be 1.3 into 0.4 into 25000 and from there you can deduct the pressure distribution up to hill slab b a g here it is 0 here it is 7039 Newton per meter square how much pressure is coming to the hill slab that has to be deducted so that is coming about 1.3 by 2 into 70590 into 0.4 into this is your this and based on that you can find it out what is your net bending movement you can find it out the distance from here from point b what are the forces what is the distance and movement about point b you can calculate so net bending movement is coming about to be I left this for your reference you can do the calculation one calculation I have done for you rest you can do it so it is coming about to be net bending movement is coming about 59378 based on the bending movement coming that you can find it out what is your area of steel area of steel comes out to be required area of steel comes out to be 59378 divided by 140 it is a stress in steel into 0.87 into 340 it is to be 1433 mm square so you can take it available either 12 mm 5 bar or 20 mm 5 bar what is there let us say 20 mm 5 bars so spacing will be coming out to be 200 200 mm centre to centre with this area of steel availability with the availability of your bars in the local market we have taken 20 mm 5 bars the centre to centre provided centre to centre provided 200 mm centre to centre that means 200 mm centre to centre means after 200 mm you provide another 20 mm 5 bars this is about your toe slab and heel slab now we will go for another most important part of the checking that is your sliding that is your sliding factor of safety against sliding as I said I have done it not done it in earlier so I just combine in part of structural design because the factor of safety against sliding is not coming within this permissible limit so total horizontal earth pressure earth pressure now this is for check for sliding earth pressure is coming about to be 18000 into 4.5 plus 0.55 4.5 how it is coming this distance is your 4.1 plus 0.4 this is your 4.5 plus this it is your 0.55 so this whole square by 2 into k a and this value is coming about to be 76507 Newton so maximum friction how much friction you are getting the friction is coming about to be mu w and this is equal to 84845 Newton so factor of safety is equal to mu w by p and this comes out to be 1.10 which is less than equal to 1.5 if you look at this sliding factor safety against sliding is about coming 1.10 which is less than is your minimum requirement of factor safety what does it mean this retaining wall will slide at the base of the foundation soil base of the foundation soil that means to satisfy this we have to provide a shear key shear key we have to provide shear key either hill or either the toe where you can provide it I am taking it the providing the shear key at the hill if I consider this is my wall this is toe and this is hill so this is a shear key so what is the intention to provide the shear key that means it will not allow it will not allow the shear key it will not allow to further proceed means it will not allow to slide it will not allow to slide this about this retaining wall about your base at the base now let us provide 1.5 times of 1.5 times of say total earth pressure say it is a p now 1.5 times of p how much now 1.5 times of p how much 1.5 times of p how much it is your 1.5 times into 7 6 5 0 7 which is equal to 1 1 4 7 6 0 newton and available friction available friction which is equal to 8 4 8 8 4 8 4 5 newton and this comes out to be 2 9 9 1 5 newton this is your unbalanced what I have done I have taken the factor of safety is equal to 1.5 1.5 instead of taking 1.10 I have taken the factor of safety is equal to 1.5 that means p into 1.5 times that means p is your total lateral earth pressure p into 1.5 times this is the thing and available friction is your mu w and unbalanced force is your 2 9 9 1 5 newton this unbalanced force has to be taken by your shear key this will be taken by your shear key now safe bearing capacity with this design with this design let us take for lateral load but for for particularly safe bearing capacity of vertical load is given 200 200 kilo newton per meter square for lateral load safe bearing capacity safe bearing capacity for lateral it takes out to be 0.7 to 0.8 into safe bearing in vertical how much it is coming it is coming about let us say 0.7 into 200 kilo newton per meter square it is coming about to be 140 thousand newton per meter square let us provide height of the key is equal to height of the key below the ground surface it is equal to y y so what will happen this 1 4 0 thousand into y y which is equal to which is equal to unbalanced force how much it is unbalanced force is equal to 2 9 9 1 5 safe bearing capacity in lateral direction which is equal to 0.7 to 0.8 times of safe bearing capacity in vertical direction safe bearing capacity in vertical direction it is given in vertical it is given 200 kilo newton per meter square because of the soil so safe bearing capacity in lateral it is coming about 1 4 1 40 thousand newton per meter square so I have taken this 140 thousand newton per meter square will be acted upon the height of y in the key so this into y is equal to total unbalanced force 2 9 9 1 5 newton from there y is about to be coming 0.21 meter 0.21 meter but as per indian standard minimum value you will have to take 300 mm this is your minimum minimum so then maximum bending moment how much maximum bending moment is coming about to be 2 9 9 1 5 into 300 by 2 it is your maximum bending moment is coming about to be unbalanced force into w l by 2 this will be acted like this so maximum bending moment will be coming about to be w l by 2 so which is your which is your 4 4 8 7 2 5 0 newton mm newton mm newton mm based on that based on that we can find it out we can find it out q b d square is equal to maximum bending moment bending moment so q is equal to for m 15 and mild steel it is equal to 0.87 into b is equal to let us say 1000 into d square which is equal to 4 4 8 7 2 5 0 newton mm and this comes out to be this comes out to be d is equal to 73 mm this comes out to be d is equal to 73 mm now minimum thickness minimum thickness as per the indian standard it should be provided is equal to 200 mm 200 mm this is your minimum minimum now if this is my this so d effective will be d effective will be your total minus clear cover of your concrete so 200 minus clear cover of concrete i have taken 60 mm so which is coming about 140 mm now the area of steel required is equal to area of steel required from this bending moment it is coming about 263 mm square which is very less very minimum so it is a very small quantity so what will happen from this heel slab alternative bars should be bent inside because the area of steel required is 263 mm square from the heel slab alternative bars or may be sorry from the toe slab the bars alternative bars should be put it inside so that it will act as a reinforcement in case of key in case of key now this completes your design now if i draw the how the reinforcement bar i will have to show in the diagram how the reinforcement bars means the reinforcement bar you have to show in the diagram as well also clear diagram you have to show you have to also give this how it looks if i take it this is my wall this is the retaining wall and this is the shear key what will happen i have taken these bars like this then i provide this way and this bar has been taken like this now other bar is coming about to be like this so these are also bars the dot points are also bars in these directions these bars are in this direction and these bars are in this direction these are vertical bars these bars are in this directions if you look at here this is my main reinforcement of this term because this main reinforcement is 20 mm 5 bars at the rate of 200 mm centre to centre and these are all my shear reinforcement in this case then what will happen so this will this will how it will why why this is a compression because this side will be go in this way so that means this part will be compression compression completely and this will be tension so in this top part if it is a compression part so this part will be given this year to 20 mm 5 bars 200 mm this is your main reinforcement and this dot is your shear reinforcement here it is a it is again these are all your 8 mm 5 bars 160 mm centre to centre this is your main reinforcement because this will bend like this as i said earlier this would bend like this so main reinforcement will come at the base in this case main reinforcement will become at the here at this side main reinforcement will be this side it has been extended it has been extended so that here at every alternative bar if you look at here 12 mm 5 bars at the rate of 120 mm centre to centre this bar has been extended as a shear key this is your 12 mm 5 bars at the rate of 240 mm centre to centre 240 mm means alternative bars has been extended to your shear key if i take a cross section in this part as i said bending moment is varying every sections from x 1 x 2 at the base bending moment will be more so at the base it is your 20 mm 5 bars at the rate of 100 mm centre to centre main reinforcement bar but above 1 meter interval it is your 20 mm 5 bars at the rate of 200 mm centre centre this means once you give this diagram this means the field engineer you will understand every alternative bars if you look at this every alternative bar this is your main bar alternative bar has been curtailed has been curtailed and this alternative bar has been curtailed because of your economy design economy design so this completes your structural design if i divide into 3 parts first one is your once again i am repeating for this retaining wall design first one is your tentative dimensions second one of this design part is your that is your stability analysis then third one is your structural design in this particularly this problem that after this tentative dimension the stability analysis there are 4 part it is satisfying one is your overturning moment it is coming i have though i have not calculated it will come within the permissible limit then e is less than equal to b by 6 that means there should not be any tension crack that also satisfy and foundation bearing capacity also has to be satisfied that also satisfy i have not done the calculation only stability analysis which is not satisfy that is your about sliding that means sliding factor of safety is coming less than equal to 1.5 it is coming about 1.2 that means we have to provide shear key at the hill or at the toe depending upon that so i have provided this shear key at the hill then what is your available on balance force has been distributed in the shear key and the distance has been coming about to be 300 mm this distance has been taken into consideration after this whole design is over this kind of diagram structural part completely showing where is your shear reinforcement where is your main reinforcement has to be shown so that field engineer as well as design engineer they can understand how the curtailment of bar where is your main reinforcement where is your shear reinforcement has been provided it has to be done this completes your design of reinforced cantilever retaining wall will start next class design of your counter fort retaining wall. Thanks a lot.