 I am Dr. Patil Smilkumar-Has, Professor and Head Swinging Department, Vulture Institute of Technology, Swannapur. So the today's topic of discussion is a numerical example on analysis and design of Hill's slab under cantilever rating, for a cantilever retaining wall, learning outcomes. At the end of this session, learners will be able to analyze and design Hill's slab of a cantilever retaining wall. Example, design Hill's slab of a cantilever retaining wall to retain an earth embankment with a horizontal top 3.5 meter above the ground level. Density of earth 18 kilonewton per cubic meter. Angle of internal friction 30 degree. SBC's safe bearing capacity of soil is 200 kilonewton per meter square. Take coefficient of friction between the soil and concrete as 0.5. Adopt M20 grade of concrete and 415 grade of steel. The preliminary dimensions of the retaining wall are as shown in figure 1. Now this is the figure showing the preliminary dimensions. So this is vertically is a stem and this portion of a base slab which is to slab and this portion is Hill's slab. Now here we find we have assumed the 200 mm thickness or 0.2 meter thickness at the top and we have assumed 400 mm or 0.4 meter at the bottom thickness of this particular stem slab. Now similarly, to as well as Hill's slab thickness, we have assumed it as 0.4. So the distance, the difference in level between these two heights that is ground level at downstream and up other side of the retaining wall it is 3.5 meter below the earth retained. So now first of all we are supposed to calculate the minimum depth of foundation that is D minimum which is determined by using SBC of soil divided by the density of soil multiplied by K square where K is the coefficient of active earth pressure. So that has worked out to be 1.25 meter. So now the total height from the base to the top it is 4.35 there are other 3.5 plus D minimum therefore it works out to be 4.75 meter. Now analysis of Hill's slab. Hill's slab act as a cantilever slab having a length L2 this is length L2 this is Hill's slab so this is stem slab and this is to slab of a cantilever retaining wall length L2 as shown in figure 2. So here in figure 2 we will find this particular Hill's slab. Now this is subjected to a downward pressure that is earth pressure which is weight of the earth which is retaining on Hill's slab or which is above Hill's slab and due to the horizontal pressure retaining pressure of earth and the weight of all these stem slab to slab and Hill's slab there will be a reaction P1, P2, P3, P4 so this is a reaction under the base slab. Now can you just imagine where will be the maximum bending moment produced in case of Hill's slab. Now Hill's slab is a cantilever so therefore can you guess where will be the maximum bending moment? The maximum bending moment in the Hill's slab will be at the inner face of the stem slab. If you just see the figure here you will find at this particular location you will get maximum bending moment. So this is a cantilever slab subjected to 88.3 kilo Newton per meter that is the weight of the earth then here it is subjected to a pressure bottom pressure 12.8 Newton per kilo Newton per meter and here the other value. So here we find that all these things it is subjected to a net downward force acting on this particular Hill's slab. So therefore this cantilever which will have a bending maximum at this inner face of stem slab. Now it's width is rather its length is 1.35 meter then weight of backfill it is gamma into H1 that is what's out to be 78.3 kilo Newton per meter. The self weight it is 0.4 into 1 into 25 that is 10 kilo Newton per meter self weight of the Hill's slab. So total downward load was out to be 88.3 which is as shown in figure 2. Now we will calculate maximum bending moment. So maximum bending moment is calculated. So first of all for calculation I will just show it on the figure. So this is the total downward load 88.3 so the length of cantilever it is 1.35 meter therefore 88.3 into 1.35 square divided by 2 that is the bending moment due to downward pressure. So that is the first one which we have taken 88.3 into 1.35 by 2 minus so 12.8 that means further what we have done we have divided this particular trapezoidal loading into a rectangular load and a triangular load. So P2 is a rectangular load throughout up to here. So therefore we have taken again it will be 12.8 into L2 square divided by 2 that will be the next bending moment it is upward therefore it is negative. So first one is downward therefore positive or it is clockwise positive this is anticlockwise negative ok. Then next further the triangular portion remaining. So it is one half 69.53 is the extreme pressure and minus 12.8 that is the balance the ordinate of the pressure into 1.35 this will give you the load. So again it acts at CG that is one third. The one third the ordinate therefore one third 1.35. So therefore we get 51.57 kilo Newton meter. So that is Mu will be 1.5 times M. So that was sort to be 77.35 kilo Newton meter. Now this is that's all we call it as analysis of field slab. Now we are supposed to do the design of field slab. So for design of field slab we always equate Mu with Mu limit to find out the effective depth of the slab. So required to determine and compared with the provided depth which shall be equal to or more than the day required. So here you find 77.35 into 10 to the power of 6 this is Mu it is the 0.138 fck fck is 20 b is 1001 meter d square. So we calculate d it was sort to be 167.4 which is mm so which is less than 350 mm which is provided therefore hence it is safe. Therefore it is under infor section and the area of steel is determined by using equation g point 1.1 b of is 456 2000. Now as per 1 point g point g point 1.1 b we find Mu that is 77.35 into 10 to the power of 6 is equal to 0.87 fy Astd into 1 minus Ast fy upon bdfck. So by equating we calculate the area of steel it is 636 mm square. So Ast minimum is 0.12 percent of b into d 1000 into b that is 480 mm square. Since Ast required is to be provided on the top face of the base slab and Ast minimum is to be provided perpendicular to it. So using 12 mm bars the spacing will be area of 1 bar into 1000 divided by Ast required so that was sort to be 177.82 hence provide main reinforcement 12 mm bars at 150 mm center to center. The reinforcement perpendicular to it is minimum reinforcement it is 12 mm bars spacing will be steel it is given by area of 1 bar divided by 480 that will be 235.61 say provide 12 mm distribution steel as the 12 mm bars 225 mm center to center. Now here it shows the reinforcement here you will find this is the main steel 12 mm Tor 150 mm center to center and perpendicular to that we have the distribution steel 12 mm Tor 225 mm center to center. So since this is a cantilever which is having subjected to sagging bending moment therefore steel we have provided top and here the distribution steel perpendicular to it. If we want we can even provide this particular distribution steel on both the face also so here I have provided on only one face so this is how we can design the heel slab of the cantilever retaining wall. Now these are the references used and thank you and and all for your patience link. Thank you.