 I am Dr. Patil Sunil-Kumaris, Professor and Head Civil Engineering Department to Alchin Institute of Technology, Swalapur. Today, I am going to discuss about numerical example on design of 2S lab, learning outcomes. At the end of the session, learners will be able to determine the effective horizontal span and effective thickness of slab, determine they are also able to determine the reinforcement required for the slab and sketch the reinforcement arrangement. Example design a reinforced concrete slab for a room of clear dimension 4 meter by 5 meter. The slab is supported on all 4 walls of width 300 mm. The slab is to carry a live load of 4 kilo Newton per meter square and a floor finish of 1 kilo Newton per meter square. Use M20 concrete and Fe415 steel. Assume corners are held down, sketch the details of reinforcement solution. Step 1, we are supposed to find out the thickness of the slab. So, the shorter span Lx is greater than 3.5. So, therefore, take span by depth ratio 25 that is d is equal to 1 by 25th of span by 25 that we also have to be around 160 mm. So, the overall depth be 180 mm because the from effective depth you have to add half the diameter of the bar and clear cover. So, that will give overall depth that is 180 mm. Second step you have to find out the effective span. Effective span it is equal to clear span plus effective depth. So, it is 4 plus 0.16 in x direction 4.16 similarly in y direction it is 5 plus 0.16 it is 5.16. Then you have to find out the ratio of longer span to shorter span that is Ly by Lx that is 5.16 by 4.16 that was sort to be 1.25 which is less than 2. If the Ly by Lx is less than 2 then we are supposed to design a 2-way slab. So, hence we are supposed to design a 2-way slab. Next we have to find out the design moment and shear for that we have to do the load calculation first the dead load it is world at 0.18 into 1 into 1 into 25 that is 4.5 kN per meter floor finishing load it is 1 kN per meter square into 1 meter it is 1 kN per meter live load it is 4 kN per meter square into 1 it is 4 kN per meter. So, total load is 9.5 kN per meter. So, W that is the factored load would be 1.5 times W. So, it is 1.5.9 5 times 9.5 that is 14.25 kN per meter. The slab is simply supported on all floor sides. So, the corners are held down by providing torsional reinforcement. Hence the moment coefficients are to be obtained from table 26 of IS 456 and the case here it is case number 9 that is the last case of table 26. So, we are supposed to find out the alpha x that is the coefficient for MU x and alpha y coefficient for MU y. So, alpha x from table number 26 it works out to be 0.072 that is for 1 then again you are having the next coefficient for that you have to interpolate plus 0.007 is the difference into 4 divided by 10 that was sort to be 0.0748 that is alpha x that is the interpolation between the two values of alpha x. Now, alpha y is single value it is 0.056. So, MU x is equal to alpha x into W u into L x square and MU y it is also equal to alpha y into W u into L x square only both are L x square. So, it was sort to be 13.41 V u is the support reaction it is 14.25 that is W u into R 4 upon 1 plus R 4 R is L y by L x R is L y by L x. So, R 4 upon 1 plus R 4 because it is from plate theory we have derived it by using equal deflection at a for the central point. So, into L x by 2 it is 4.16 by 2 it was sort to be 17.96 kilo Newton then design main reinforcement. So, we have to design main reinforcement. So, we have to both the reinforcement in x and y direction are main reinforcement reinforcement in x direction for that effective depth is 160 mm. So, MU limit is equal to x u max into D. So, it is 0.48 for this case of 415. So, therefore, 0.48 into 160 it was sort to be 76.8 MU limit. So, by using C into your arm this is C 0.36 fckb x u limit is C into your arm is D minus 0.42 x u limit. So, this was sort to be 70.63 kilo Newton meter. So, MU is less than MU limit hence it is under reinforcement find area of steel in x direction by using equation G 1.1 b of I s for 562000. Now, MU we have already calculated in x direction it is 18.46 into 10 to power of 6 that is to convert into kilo Newton meter into Newton mm. It is 0.87 fy into ast into D into 1 minus ast fy upon bd fck the values are substituted here. So, ast was sort to be 333.83 mm square using 10 mm bar the spacing it is area of 1 bar into 1000 divided by ast which we have calculated above. So, that was sort to be 235.3 mm hence provide 10 mm diameter bar hysd bars at 225 mm center to center. So, this is the spacing of the reinforcement in x direction and it is we provide alternate bent up bars. Next, we are supposed to calculate the reinforcement in y direction again the reinforcement in y direction is also a main reinforcement. This reinforcement will be placed above the reinforcement in x direction. So, hence the effect to depth is calculated once again it is effect to depth d is equal to effect to depth of previous case that is in x direction minus 8 that is 8 mm we are supposed to we are providing 8 mm diameter bars therefore, it will be 152 mm is the effect to depth. Now, for from the relation again g 1.1 b we are supposed to calculate the area of steel in y direction m u is equal to 0.87 fy astd into 1 minus ast fy upon bd fck m u in y direction is 13.81 into 10 to the power of 6 it is 0.87 fy is 415 ast into d is 152 into 1 minus ast fy upon bd fck. So, we get asts 260.9 mm square using 8 mm bars spacing of 8 mm bars was thought to be pi by 4 area of 1 bar into 1000 divided by this ast calculated. So, that was sort to be 192. So, therefore, provide 8 mm diameter bars at 190 mm center to center it is it specifies the prescribed maximum limit of clause that is that is 3D or 300 mm spacing should not be more than it should be less than 3D or 300 mm. Next we have to check it for shear so, tub v this is nominal shear stress it is v u upon bd v u is 17.96 into 10 to the power of 3 divided by b 1000 into 160 d it is 0.112 newton per m square. So, we have to calculate percent steel because we have to calculate first tau c tau c from table number 19 of IS 456 2000. So, in that first column is percent steel therefore, we have to calculate percent steel. So, it is percent steel is calculated it is area of 1 bar into 100 divided by 225 is your spacing which we have we are providing for 10 mm bar and this is a effective depth d. So, that was sort to be 0.1812 and 8 percent so, if you refer the table 19 we get tau c for this particular percent steel 0.218 the tau c was out to be 0.33 newton per m square for m 20 grade concrete. So, enhancement factor for the slab that means as per 40.2.1.1 so, the permissible shear carrying capacity is k tau c it is k tau c so, since the thickness of the slab is 180 mm. So, it is k tau c was sort to be 33 percent greater that is 0.33 into 1.24 the 1.24 is the value of k. So, permissible stress is 0.409 newton per m square. So, therefore, tau c max again the tau v which we have calculated it should be less than tau c it should also be less than the tau c max half the tau c max that also we should ensure. So, hence it is shear reinforcement is not required in this particular case. Next last step check for deflection L by D provided is 26 and which should be less than L by D max. So, for L by D max again we have to refer figure number 4 of IS 456 2000 so, we have to find out F 1 it is 20 into F 1 F 1 from the figure number 4 it was sort to be 1.6 for percentage steel 0.128 and we have to calculate F s first then with respect to the percentage steel for this value of F s 240 newton per m square so, it was sort to be 1.6. So, 20 into 1.6 it is 32 which is greater than 26 so, therefore, L by D max is 1.6 20 that is 32 which is greater than L by D provided hence deflection control is satisfactory. Then we have to provide a torsional reinforcement at corner the size of the mix it is L x by 5 4 1 so, it was sort to be 8 32 mm the size of the wall is to be added to it so, it is 300 mm so, 300 plus 8 32 it was sort to be 11 mm area of torsional reinforcement is three fourth of area of steel in x direction so, that is 333.33 it was sort to be 250.01 so, using 8 mm mesh we get 200 mm center to center the mesh of 8 mm diameter bars 200 mm center to center. So, then reinforcement at edge so, it will be minimum steel AST 0.12 into 1000 into D divided by 1000 it was sort to be 216 mm so, spacing of this particular steel so, this was sort to be 232 mm so, therefore, provide 8 mm diameter bar 225 mm center to center it is the reinforcement in the edge strip which is as shown in figure number 1. Now, this is the figure which shows you the reinforcement in x direction this is x direction and this is the reinforcement in y direction so, reinforcement in x direction it is 10 mm diameter bars 225 mm center to center so, alternate bar bent up alternate bar is bent up or each bar is bent up once bent up is on this side and shorter on one on right side and another bent up will be on the left side so, similarly you get the reinforcement in y direction it is 8 mm diameter bar 190 mm center to center again bent up on one side for one bar bent up on the other side for the another bar because the 50 percent of the steel should go up to end then these are the corner reinforcement mesh 8 mm diameter bar mesh at all 4 corners then this is edge steel edge strip steel in all 4 edge strips so, these are the references used for this particular presentation. Thank you. Thank you one and all.