 this design of counter for retaining walls up to this dimensions, then stability analysis we have finished this stability analysis. So, then we calculated the soil pressure at the base of the base of the retaining wall and this will be if I take a, b, c and d and this term to be if I make it into I and this is my g this value will be E is equal to 1, 2, 1, 3, 9, 4 and this value comes out to be 6, 0, 6, 9, 7 Newton per meter square and this dimensions we have calculated earlier this dimensions this is your 5.5, 5.5 meter and this dimension is your 1.2 meter and this is 1 meter and this is your 3.29 meter and this completely 6.3 meter and this is your value of 6.7 meter. Now also we have checked this stability analysis criteria all the four criteria it has satisfied now we will go for a design of so first part with your dimensions tentative dimensions or may be dimensions once dimensions have been established then we have done this stability analysis geotechnical point of view stability analysis is your overturning movement factor of safety against overturning movement factor of safety against sliding and E should be less than equal to b by 6 and also bearing capacity failure check then after stability analysis once it is then whatever dimensions we have taken consideration for this design of counter for retaining wall these dimensions are now after these two steps over then we got the pressure below this retaining wall. Now third part is your your RCC design or may be structural design structural design in case of structural design so particular design of we can start with this upright slab this is your upright slab this is your upright slab this part is your heel slab and this is your toe slab this is your counter force counter force so for upright slabs bending moment bending moment we have earlier calculated it is equal to p l square by 2 l this comes out to be this comes out to be 2 5 2 0 0 Newton meter and overall thickness if you look at this thickness this upright slab of thickness of 210 mm so overall thickness is equal to t is equal to 210 mm now effective depth you can find it out 210 minus your 40 mm for your concrete that comes out to be 170 mm. Now area of this steel required area of this steel required is equal to movement of resistance that is your 25200 divided by divided by 140 a stress in steel into 0.87 this is the value of q into 170 effective depth this comes out to be 120 mm square that means for design of upright slab this is your steel reinforcement this amount of area of the steel reinforcement is required so you can take whatever the available in the market if I take 16 mm 5 bars this is available in the market then the spacing comes out to be spacing which is equal to pi by 4 into d square pi by 4 into d square this is your 201 into 1000 by 1220 this comes out to be 164.7 mm and you can take it as a at the rate of 160 mm centre to centre now this is your main reinforcement as I said earlier for cantilever retaining wall you can go for also after main reinforcement you can go for a shear reinforcement that distribution is your 0.15 percent so shear reinforcement that is your 0.15 percent 0.15 by 100 into 210 into 1000 it is coming about to be 350 mm square 50 mm square if I take 8 mm 5 bar 8 mm 5 bar the spacing comes out to be spacing is equal to your 150 mm bars 150 mm bars means 150 mm 150 mm centre to centre now this structural design this is similar same as you have done earlier particularly in case of cantilever retaining walls this will be same for upright slab as well as toe slab and heel slab first you will have to find it out maximum bending moment how much your bending moment is coming then depending upon your bending moment once you get it you can make it to into q b d square you can find it out your total depth d or total thickness d then you can find it out effective depth if you have your in this case upright slab this thickness of this upright slab from this dimensions it is given 210 so effective depth if 40 mm is your clear cover for concrete clear cover for concrete how I have done it as I said earlier if this is the cross sectional view so these are the reinforcement this cover generally your 40 mm clear cover beyond your reinforcement that is your 40 mm so this effective depth is comes out to be 210 minus 40 this is your 170 mm then if I equate with your moment to this moment is equal to your stress in steel I can find it out what is your maximum area of steel means how much reinforcement is required so reinforcement required is 120 mm square then you can take it as per the availability locally locally generally available 12 16 18 20 25 mm bars you can choose any of the bar in this case 16 mm 5 bars has been chosen and spacing comes out to be 160 mm centre to centre then as per the I score for structural design for particularly working stress method this steel reinforcement has to be provided your about 0.15 percent and from there the reinforcement comes out to be 315 mm square so now we provide 8 mm 5 bars of spacing 150 mm centre to centre the moment you go there so bending moment will be more and here it will be less so curtailment of bar will be there as I said earlier alternative bar to be curtailed so that means out of 150 mm alternative bars should be curtailed means the spacing become 300 mm centre to centre now design of the tow slab if you look at this this part I am considering only this part from here to from here to here this part is your tow slab now I will have to consider the pressure at the base and pressure above the tow slab and what is your net pressure so in this case net pressure coming about to be net pressure coming about to be more in the base than your top so we can find it out this tow slab so tow slab if I write it into this upward pressure upward pressure so like c d i f then if you look at this c d i f c d i f then if you look at this c d i f c d i f this is rectangle then i f e there are two component because this component trapezoidal component you can find it out directly upward pressure or you can make it into two component one component this and other component is your triangular component so c d i f I got this calculations one for this 107906 into one this distance is one meter and here it is here it is your 10107906 by simple distribution triangular distribution I can find it out this pressure here so this value comes out to be this value comes out to be 107906 now upward pressure of e i f e i f this comes out to be half into one into one into one into two 1 3 4 8 8 so this to this pressure is 1 2 1 3 9 4 and this to this is your 101010707906 so this part will be 1 2 1 3 9 4 minus 107906 so this comes out to be this comes out to to be this comes out to be like 6 7 4 4 so this is your total upward pressure then you can deduct then you can deduct the self weight self weight is coming downward self weight of this is coming downward with the deduct self weight which is equal to 1 into 0.4 into 25,000 this comes out to be 10,000 and distance from sea if I am taking from distance from sea distance from sea it is coming about to be 0.5 to third and this is your 0.5 and moment at sea moment at sea this comes about to be 5 3 9 5 3 then 4 4 9 6 then this is your 5000 this comes out to be 5 3 4 4 9 now this is your maximum bending moment after deducting self weight and taking into consideration of upward pressure so that means this maximum bending moment is coming to be it should coming like this it will bend like this so with this help you can find it out you can find it out this depth is given this depth is given 400 mm this part is given is your 400 mm from your dimensions calculations then you can find it out effective depth effective depth comes out to be 400 minus 60 which is equal to 360 mm now you can find it out area of steel like this you can find it out area of steel it comes out to be 1 2 9 0 mm square now you can take either 16 because you are taking the 16 mm 5 bars you can take 16 mm 5 bars so it is 16 mm 5 bars then you can find it out spacing about 150 mm centre to centre similarly shear reinforcement shear reinforcement you can take it about 8 mm 5 bars and shear reinforcement you can find it out 8 mm 5 bars 8 mm 5 bars at the rate of it comes out to be 160 mm centre to centre so the way I have done this the way I have done this design of tow slab same way you can do it this part this design of the heel slab I am not doing that you have to do it the way I have done for the tow slab particularly this heel slab you consider the part of this pressure you consider the part of the load coming what are the part of the load suppose to come particularly heel slab one is your self weight it is coming downward then weight of earth that means weight of soil it is coming downward then weight of counter fort because this is R C C it is coming downward then any any superimposed load that also you consider then afterwards then you take from here to here this part to this part the pressure distribution you deduct and see whether the upward pressure distribution is more or downward pressure distribution is more in this case downward whatever the self weight coming to the pressure because this is a retained part of the soil mass so the pressure will be more so this case how it will be bent this will be bent in this way so if you will do it like this the way I have done for tow slab you can do it for heel slab and do this calculation find it out your maximum movement you can find it out the distance from this point B and take movement about point B then your maximum movement after deducting this self weight deducting this base pressure then you find it out what is your area of steel required and main reinforcement as well as shear reinforcement if you look at this heel slab heel slab your maximum bending movement comes out to be maximum bending movement comes out to be 37577.25 Newton meter and this maximum bending movement from there you can find it out your area of steel it comes out about to be 900 mm square then you can take 12 mm 5 bars so at the rate of spacing 100 to 20 mm centre to centre then also you can provide minimum shear reinforcement also. So tow slab and heel slab as well as upright slab all this slab design has been over only remaining part is there this counter fort this counter fort structural design has to be carried out this now we will start this counter fort walls RCC design of your counter fort walls or counter fort if you look at this counter fort this top weight if I can take it whatever the weight here this weight comes out from this dimensions generally this weight will take same weight of this like 210 mm in this case so this comes out to be 420 mm that means double this stem thickness double the stem thickness it comes out to be 420 mm of width of counter fort walls then once you get the width of the counter fort walls then you can find it out then next step is you find it out earth pressure distribution or pressure distribution earth pressure if soil is retaining like this one pressure is giving soil pressure on the upright slab other way earth pressure is also giving to your counter fort walls. So this comes out to be k a gamma h square l by 2 which is equal to one third k is equal to one third gamma is equal to sixteen thousand into six point three this distance from here to here is your six point three into six point three whole square by two into l is equal to three l is l is your spacing between this counter fort this is about to three. So this comes about to be 666792 Newton meter so overall width of the counter fort at the base overall width of the counter fort at the base we can provide three point five meter why it is three point five meter three at the counter base three and three point zero eight if I if you if you look at this if I draw this diagram how this three point five comes into picture this is your zero point four two meter this is your zero point four two meter then then from there you provide this then if I take it to to to the six point three meter this comes out to be three point three point zero eight meter three point zero eight meter. So now overall width if I add it three point zero eight plus zero point four two this comes out to be your three point five meter overall width is equal to three point five meter now effective depth effective depth minus sixty mm this is for your this is for sixty your concrete this comes out to be three four four zero mm let us put it beta is your beta is your inclination of the main steel with your vertical beta is your inclination of your main steel with vertical either you to take it or you can take it so tan beta comes out to be now tan beta three point zero eight three point zero eight divided by six point three so beta is equal to twenty six degree three minute now area of steel required once you have earth pressure with this from this earth pressure this is your bending moment comes out to be so k a gamma z square by two so then from there maximum bending moment we got it from earth pressure this is your newton meter so maximum bending moment maximum bending moment comes out to be earth pressure six six six seven nine two into six point three by two this comes out to be six six sorry there is a calculation mistake here this part is your three one seven five two zero newton so this will be your three one seven five two zero so this comes out to be six six six seven nine two newton meter now you have your effective depth you have your maximum bending moment then from this you can find it out your area of steel what is your area of steel so area of steel comes out to be a st is equal to maximum bending moment seven nine two divided by stress in steel hundred forty into q is equal to zero point eight seven into effective depth which is equal to three four four four zero into sake beta or twenty six degree three minute this comes out to be your eighteen hundred mm square now we can provide eight eight numbers of eighteen mm five bars look at this for counter fort retaining wall counter fort retaining wall what you have done we need a thickness we need a thickness this thickness two ten mm is your thickness of your stem generally the counter fort retaining wall which has been constructed along with this so whatever the thickness generally we got it we make it double so available thickness means double the stem thickness this is your thickness that is your coming four twenty mm then if soil is retaining like this soil is retaining like this it will give earth pressure on this laterally one direction on your as well as laterally to your to your counter fort retaining wall from their earth pressure comes out to be k gamma h square by two into l l is equal to spacing between your earth pressure so one third into sixteen thousand into six point three whole square by two into three this comes out to be three one seven five two zero newton now overall width I have calculated if I am taking this zero point four two meter now this overall width is coming three point zero eight plus zero point four two this comes out to be three point five meter so effective depth comes out three four four four zero mm then I calculate tan beta this is your twenty six degree three minute maximum bending movement will comes out from your earth pressure diagram from this earth pressure so this is your maximum bending movement once you get maximum bending movement you can find it out your area of steel then area of steel comes out to be eighteen hundred mm square so you take what is the available locally available you can take it sixteen mm five bar twelve mm five bar or eighteen mm five bar this comes out to be eight number of bars so this is your once you get the reinforcement then you can think about your shear reinforcement and as well as also your curtailment of bars now as I said earlier as I said earlier in this case we need to have this curtailment of bars because at the base bending movement will be more at the top bending movement will be less so curtailment of bars will be curtailment of bars let h one be the depth at which let h one is your depth at which say total eight number of bars two bars let us say two bars to be curtailed I put it in even number two four six so two bar to be curtailed then I can find it out eight minus two eight is your bar minus two by eight this is your bar then h one whole square minus six divided by six minus three whole square so this comes out to be h one is equal to h one is equal to comes to be five point five meter that means below five point five meter we can curtail the bar of every two bars if there are eight bars that means two then next two bar to be curtailed then six then next two bar to be curtailed so similarly you can go for h two let us say h two is equal to another two more bar two more bar curtailed so you can find it out eight minus four by eight this comes out to be h two whole square by six minus three whole square this comes out to be h two is equal to four point four meter from top this is your from top so in this way you decide from the top at what distance two bar to be curtailed then another another distance additional two more bar to be curtailed in this way you proceed this curtailment of bars now then you will have to find it out this connection between this counter fort and upright slabs because once this counter fort retaining wall you get your main reinforcement this main reinforcement has to be connected with your counter fort retaining wall has to be connected with this upright slab as well as hill slab how this connection has to be done this has to be also again calculated I will do it next class along with this final detail reinforcement drawings thanks a lot