 Professor and Heads Villaging Department, Valchiyan Institute of Technology, Sulapur. So today I'm going to discuss on numeric, I'm going to take a numerical example on analysis and design of TOSLAB for a cantilever retaining wall. Learning outcomes, at the end of the session, learners will be able to analyze and design a TOSLAB of a cantilever retaining wall. Example, design TOSLAB 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 of soil, 200 kilonewton per meter square, take a coefficient of friction between the soil and concrete 0.5, adopt M20 grade concrete and Fe415 steel. The preliminary dimensions of the retaining wall are as shown in figure one. See, first of all, we have to assume the preliminary dimensions. These are assumed prism preliminary dimensions based on some thumb rules. So top width is assumed as minimum 150 mm. Now we have assumed 200 mm here, 0.2 meter. And here this is taken as H by 12 where H is this 4.75 meter. So we have to determine D minimum, D minimum is nothing but it is the minimum depth of foundation below ground level. So that is given by SBC divided by the density of soil into coefficient of active earth pressure square. So then we can determine what is the actual height of stem, then further what is the total height up to the base of this particular retaining wall. So which is to be used for stability analysis of this particular structure. So without stability analysis, we cannot go for the design. So therefore, please before viewing this particular video, see my own video on stability analysis of retaining, cantilever retaining wall. Now this is analysis of TOSLAB. The TOSLAB act as a cantilever of length L1. Here you can find this is length L1. This is a TOSLAB up to the face of stem stem, right? L1 as shown in figure number two. Now this is subjected to a pressure varying pressure P1 here and at this phase it is P3 and at this here it is P2. Now we are interested in only TOSLAB. Therefore we will see only P1 and P3. So P3 is the pressure ordinate at the face of stem. Now where will be the maximum bending moment in the TOSLAB? Can you guess? So please see TOSLAB is a cantilever which is subjected to a trapezoidal pressure P1 on this side and P3 over here. So the maximum bending moment in the TOSLAB will be at the face of the stem. So that will be at this location. This is the location where we find maximum bending moment. It's a cantilever of length L1. So therefore at this particular phase we'll get the maximum bending moment. So for determining the maximum bending moment following data is required which are determined by considering one meter length of retaining ball. We consider usually one meter length of retaining ball. First we calculate what is the eccentricity? So eccentricity of the loads which are coming down and the lateral pressure. So here we take moment about the end of tone and we find out what are the stabilizing moments and what are destabilizing moments or overturning moment. Overturning moment is only moment created by the soil pressure or horizontal pressure of soil and stabilizing moment is the moment created by all weights. So weight of stem slab, then weight of base slab, then weight of soil on the earth. So which I have discussed in my previous video of stability analysis of retaining ball. So here B by two, it is 2.5 meter is B. So therefore B by two. So this is the sum of stabilizing moments and this is the destabilizing moment. The difference between these two divided by total weight downward load. So that will give us eccentricity. So P1 is the pressure at one end of the, at two end of the retaining wall. It is sigma W divided by B into one plus six E by B. So this should be in a bracket. So therefore this is 163.33 sigma W divided by B is 2.5. One plus six into E is 0.335 divided by 2.5. So this was sort to be 117.86 kilo Newton per meter. So similarly P2 it is sigma W divided by B into one minus six E by B. So that is again what's out to be 12.8 kilo Newton per meter. So all these forces are shown P1 is here, P2 is here. Then next we have to determine P3. P3 is at the face of the stem. So it is P2 plus. So I will just show you in the figure. It is up to here it is P2, right? P2 plus this particular variation. So what is this remaining portion? Up to P2 it is P2 UDL, right? Then the remaining portion, remaining portion we have to calculate by similar triangles. P2 plus P1 minus P2 divided by B into B into L1. So this is 12.8 plus 117.86 minus 12.8 divided by 2.5 into B is 2.5 minus 0.75 U is your L1. So it was sort to be 86.35 kilo Newton per meter. So P4 is the triangular load. So here you will find in the figure, this is P4. So P4 is nothing but half L1 into this particular ordinate. So that is what we have written. Half L1 into P1 minus P3. So that was sort to be 11.81 kilo Newton per meter. So bending moment M at the face of the stem is given by P3 into L1 square by 2 plus P4 into 2 third L1. So that is what sort to be 30.19 kilo Newton meter. So MU is 1.5 times M. So this completes analysis of TOSLAB. Next is design of TOSLAB. So by equating MU to MU limit, the effective depth D of the TOSLAB required is determined and compared with D provided, which shall be equal or more than D required. Now this is our maximum moment 45.285 kilo Newton meter to convert it into Newton mm multiplied by 10 to the power of six. So 0.138 FCK BD square. B is always one meter thousand, FCK is 20. So we have determined D, it was sort to be 128.09 mm, which is less than 350 mm, which is we have provided in preliminary dimension by taking 50 mm effective depth. Hence it is safe. Therefore it is under reinforcement and area of steel is determined by using G.1.1B equation of IS456 2000. Now this is MU. This is 0.87 FY AST D into one minus AST F upon FCK. A BD FCK that we have substitute, we have calculated AST, it is 366 mm square. So AST minimum is 0.12% because it's a slab, cantilever slab. So it is 0.12 by 100 into 1000 into 400. It is 480 mm square. So minimum steel can be placed on both sides or it can be placed on same side of the, only one side of the base slab. Since AST required is less than AST minimum, provide AST minimum in both directions because AST required is less. It is 366 mm square, AST minimum is 480. So therefore AST minimum I will provide in both directions at the bottom. Using 12 mm bars, the spacing of the bars is given by area of one bar divided by area of steel in 2000 that is 230 mm. Provide 12 mm bars at 300 mm centered center in both directions. The detailed arrangement of the reinforcement is as shown in figure two. So here main steel, 12 mm diameter, 200 center to center, distribution steel, 12 mm diameter, 200 center to center so that it will take care of everything. Now these are the references used for the creation of this video. Thank you one and all.