 So today I am going to deal with numerical example on analysis and design of stem slab for a cantilever retaining wall. Learning outcomes. At the end of this session, the learner will be able to analyze and design stem slab of cantilever retaining wall. An example, design the stem 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 is 18 kilonewton per cubic meter, angle of internal friction is 30 degree, safe bearing capacity of the soil is 200 kilonewton per meter square, take coefficient of friction between the soil and concrete as 0.5, adopt M20 concrete and Fe415 steel. The preliminary dimensions of the retaining wall are as shown in figure 1. So for preliminary dimensions, how to assume the preliminary dimensions for that the learners are requested to see one more video which is on stability analysis of the cantilever retaining wall because after stability analysis only we can do the design of stem. So therefore, it is necessary to see before going for this design of stem slab, we have to assume the sizes of the stem size of base slab and this particular depth of base slab will be h by 12 where h is this total depth. Then we should find out what is the minimum depth of foundation below the ground level. So that is given by the SBC of soil divided by the density of soil into K square where K is coefficient of active earth pressure and we have to determine or we have to assume this base width of the retaining wall, cantilever retaining wall. So base slab width we have to assume so it is 0.48 to 0.56 h where h is 4.75 here and then afterwards we have to this is 3.5 meter that is from this ground level to the top one. So we have to find out what should be the exact height of stem it is 4.35 now. Now the analysis of stem slab, the stem act as a cantilever of height 4.35 meter subjected to uniform varying pressure as shown in this particular figure. So the pressure at the base of the stem is K gamma into h where h is this particular height of stem. So we have to find out what is the horizontal force pH. pH is given by the area of this stress diagram that is half K into gamma h multiplied by the h that is the height that will give us the pressure at the base. So after determining the pressure at the base now we are supposed to find out the maximum bending moment. So can you just tell me what will be the where will be the maximum bending moment in stem slab. This comes under analysis of stem slab maximum bending moment will be at the base of the cantilever stem. So which is given by half K gamma h square that is the force pH into h by 3. So that will give us the value so half one sixth of K gamma h cube. So if we substitute the numericals you will get 82.31 kilo Newton meter as the maximum bending moment at the base of the stem slab. So MU it is 1.5 times M so that was sort to be 123.47 kilo Newton meter design of stem slab. So we have to equate MU limit with MU for balance section with Fe415 steel MU limit is given by MU limit is given by 0.138 fckbd square. So we have equated it to MU and we have determined what is the required depth it is 211.5 mm. So that we have taken it as 350 mm to be on safer side and overall depth D is 400 mm by considering 50 mm clear cover. Now area of steel required is obtained by using equation g.1.1b of is for 562000 where in we equate MU is equal to 0.87 fy astd into 1 minus ast fo upon bdfck. So if we substitute all the numerical values so we can find out what is the area of steel. So area of steel works out to be 1041 mm square. So that particular area of steel now we will find out what is the spacing of the bars. So we have used 12 mm diameter bars so it works of spacing works out to be 108 mm. So provide 12 mm at 100 mm center to center that is the main steel for taking this particular moment coming at the base of the stem slab. Next is distribution steel that is perpendicular to the this main steel there will be a distribution steel. So for that we have considered average thickness because it is 200 at top and 400 at bottom. So average is 300 mm. So it is 0.12 percent since it is fe415 so 0.12 percent of 1 meter that is 1000 mm into 300 mm 360 mm square. So provide 180 mm square on each face by using 8 mm bars that means this minimum steel can be placed on both the faces. Now the spacing of this the 8 mm bars it works out to be 279 and provide the bars at 270 mm center to center on tension face. And a mesh of 8 mm bars 270 mm center is given on compression face of the wall also that means on both the faces we are providing this minimum reinforcement that is distribution steel. So in this pressure diagram you find the pressure at top is 0 and the pressure at bottom is K into gamma H. So therefore this is varying so that is why the maximum bending moment goes on reducing as we go towards top it is bending going to be 0 at top. So therefore the steel reinforcement which we are providing so that is to be curtailed. Bending moment is proportional to the depth of filling and the thickness varies linearly from 200 mm at top to 400 mm depth at 4.35 meter that is at the bottom of the stand. One third of the vertical bars may be curtailed at a height of 1.5 meter from the base and another one third at a height of 3 meter from the base as shown in figure 3. Figure 3 is here. So here you can find one third of the bars are curtailed here and next one third is curtailed at 1.2.8 meter and the rest of the bar that is a one third of every third bar will go up to the top. So this is how we have to make the curtailment that means here if you see it is 12 mm diameter at 100 cm to centre at this depth and above this you will find 200 cm to centre and above that you will get 300 cm to centre the same 12 mm bars. Curtailment of vertical bars rather here it will be a check for the reinforcement. So check for shear we will be taking. So we will find out the shear force that is pH itself it is 107.17 kilo Newton. So VU that is the ultimate shear or factored shear it is 1.5 times V it is 160.75 kilo Newton. The shear stress tau V it works out to be 0.4 Newton per Newton square and if we calculate the percentage steel provided it is 0.283 Newton per Newton square and if we calculate tau C from table number 19 of highest 456 we find it is 0.4 Newton per Newton square. So therefore no shear reinforcement is required. So here you find only the vertical steel that is to take the bending moment and the nominal steel on both side in both direction that is the 0.12 percent that is minimum steel we have provided. So this is how the curtailment of reinforcement is to be done longitudinal reinforcement is to be done in case of stem of a cantilever retaining wall. So these are the references used thank you one and all.