 I am going to take a numerical example on cantilever slab. Learning outcome. At the end of the session, the learners will be able to determine the effective depth of the cantilever slab, the main and distribution reinforcement required for cantilever slab and they are able to sketch the reinforcement arrangement. This is their example, design a cantilever balcony slab projecting 1.2 meter from beam, adopt a live load of 2.5 kilo Newton per meter square, use M20 concrete and Fe415 steel, sketch the details of the reinforcement. Solution. Step 1, depth of the slab. We have to find out the effective depth of the slab. As per IS 4562000, it is one-seventh of the span for the cantilever slab. Span is 1200 mm divided by 7, it works out to be 171.4 mm. The effective depth, let us consider effective depth of 175 mm and the overall depth by considering the diameter of the bar and clear cover, it let us take it as 200 mm. The depth D may be gradually reduced to 100 mm. That means on one side it is the 200 mm, on the other side it is 100 mm. So effective depth it may be around 175 or 180 mm. Then step 2, we have to find out the design moment EMU and the design shear VU. So for calculating EMU and VU, we are supposed to find out first the self-weight of the slab. The self-weight of the slab it is given by average load taken as UDL, average load because we are having thickness on one side as 200 mm and on the other side as 100 mm. So therefore it is 0.2 meter plus 0.1 meter divided by 2 that is average thickness into 1 meter. So into 25 kilo Newton per cubic meter. So that is our density which will give us 3.75 kilo Newton per meter. It should not be per meter square it should be per meter. So finishing load is 1 kilo Newton per meter square again it is 1 into 1 it is 1 kilo Newton per meter. Now can you please tell me what will where will be the maximum bending moment for a cantilever slab. So the maximum bending moment for a cantilever slab it will be at the face of the support. It will be at the face of the support because cantilever slab is fixed at one end and or continuous at one end and on the other end it is free. Therefore maximum moment is at the face of the cantilever slab at the face of the support. Then live load 2.5 kilo Newton per meter we have already seen total load it is addition of dead load plus live load it is 7.25 kilo Newton per meter. So then we have to find out emu that is design moment it is W into L square upon 2 because it is a cantilever slab. So therefore it is 1.5 is your factor load into 7.25 that will give you W into 1.2 square divided by 2 it was out to be 7.83 kilo Newton meter that is the value of emu. Emu is design moment. Next VU is equal to W into L by L because entire load will go towards only one end because other end we do not have support it is a cantilever. Therefore it is 1.5 into 7.25 into 1.2. So 1.8 into 7.25 that will give you W into L is 1.2 it is 13.05 kilo Newton is shear force. So this is design shear is 13.05 kilo Newton design moment is 7.83 kilo Newton meter. Then we have to design the main reinforcement. So we have to find out first what is XU limit it is 0.488 for Fe 415 0.48 into 175 it is 84 mm. Then we have to find out emu limit either we can find out by equation C into lever arm this is the equation C into lever arm 0.36 fck b XU limit into lever arm is d minus 0.42 XU limit or you can even simplify this because this is already known 0.48. So therefore XU limit is 0.48 it is already known from IS for 562000. So therefore here you can if you simplify this calculation you will get a constant into fck into bd square. So that constant for the Fe 415 it works out to be 0.138 fck bd square or you can find out by using C into lever arm. So this is this works out to be 84.5 kilo Newton meter. This is emu limit the emu value which you have calculated it is less it is 7.83 only here you will find emu. So emu limit we have calculated 84.5 so the emu is less than emu limit hence it is under reinforcement. Then we have to find out the area of steel required we have to find out the area of steel required main reinforcement. So by using equation G 1.1 b of IS 456 2000. So it is emu is equal to 0.87 fy astd into 1 minus ast fy upon bd fck. So emu is 7.83 into 10 to the power of 6 equal to 0.87 into 415 into ast into d is 175 it is 1 minus ast divided by bd that is 1175 and fy is 415 fck is 20. So ast works out to be 125.8 mm square. So the minimum reinforcement also we should check it for minimum reinforcement whether this 125.8 is less than the minimum reinforcement. So if I check it for minimum reinforcement it is 0.12 divided by 100 0.12 percent of b into d b is 1000 d is 200. Please remember for minimum reinforcement you have to take overall depth it works out to be 240. So therefore this is greater than ast which you have calculated. So therefore you have to provide this particular steel minimum reinforcement you have to provide steel provide should be always equal to or more than minimum reinforcement. So therefore 8 mm diameter bar spacing area of 1 bar into 1000 divided by 240 that is 209 mm. So hence provide 8 mm diameter bar 200 mm center to center. So this is your main steel. The main steel for cantilever it is at the top then step number 4 you have to design the distribution steel you have to design the distribution steel. So distribution steel is equal to the minimum reinforcement you have to provide ast is equal to 0.12 percent 0.12 by 100 into b into d b is 1000 d is 200 it is 240 mm square. Therefore provide 8 mm diameter bars 200 center to center. So that means main steel is also 8 mm diameter bar 200 center to center distribution steel is also same. Now you have to check the anchorage length because it is a cantilever slab. So for the anchorage length you have to check whether the embedment of bar main reinforcement is up to LD or not LD is given by LD is called development length development length of the bar to transfer the tensile stress to concrete. It is 0.87 fy into diameter of bar divided by 4 tau bd as per IS 456 2000. So 0.87 fy is fy is 415 Newton per m square and diameter of the bar you can take 8 or 10 whichever we are having. So then if I take 8 it will be less than this then 4 tau bd tau bd you have to take from the table of IS 456 2000. So that is bond strength it is bond stress it is 1.2 for fe for mild steel it is 1.2 Newton per m square for mild steel if you are having HYSD bar this is to be increased by 60 percent this is to be increased by 60 percent that is therefore it is multiplied by 1.6. So we get LD as 470.12 mm. Next we have to check it for deflection you have to check it for reflection that is L by d provided is 1200 divided by 175 it is 6.85 which is which should be less than L by d max L by d max is calculated by using IS 456 2000 the cantilever for cantilever slabs L by d ratio given is 7 and which is to be multiplied by constant F1 F2 and F3 F1 is determined from figure 4 F2 is F2 and F3 are 1 for cantilever slab. So to find out F1 from figure 4 you have to find out the percentage steel it is 0.196 percent and the FS you have to find out it was out to be 240 Newton per m square and from figure 4 of IS 456 so F1 was out to be 2. So F2 and F3 are 1 so therefore 2 into 1 into 1 into 7 it was out to be 14 which is greater than L by d provided that is 6.85 hence it is safe against deflection or defense deflection control is satisfactory. Now this is the reinforcement detailing so since you have tension at top for cantilever beam this is free end and this is fixed end. So therefore the tension is at top your main reinforcement is at top and that main reinforcement should have a development length should have a development length to transfer the stresses to concrete Ld and the main reinforcement is 8 mm diameter HYD bar set 200 center to center and the distribution bar again it is perpendicular to the main reinforcement and it is again 8 8 mm diameter bar 200 mm center to center. So this is the beam and this is the development length from face of the support so maximum moment is at face therefore maximum depth at face and minimum depth at free end. These are the references used for the preparation of this particular presentation thank you Thank you one and all