this video is going to be very much important for you today we are going to learn how to design the reinforcement of a slab by the term reinforcement we mean the main bars and the distribution bars we are also going to check if the reinforcement provided is safe or not in addition to that we are going to check if the assumed depth of the slab is safe in deflection or not and we are also going to check whether the slab will fail in shear plus the development length of steel bars will also be checked so we have the given data a one way slab of dimension six meter into three meter is to be designed by using m20 grade of concrete and fe415 steel the depth of slab load calculation and design factored moments have already been calculated in our previous lecture which you can find in the description box of this video to design the main reinforcement we have the formula ast equal to 0.5 fck upon f5 into 1 minus under root 1 minus 4.6 mu upon fck bd square into bd where ast is the area of steel fck is the characteristic compressive strength of concrete and since we are using m20 grade of concrete fck will be 20 newton per mm square and fy will be 415 newton per mm square mu is the design factor moment which we have already calculated in our previous lecture b is 1 meter and d is the effective depth of the slab which we have already calculated as 115 mm on substituting all these values the required area of steel will be equal to 290 mm square now using 10 mm dial of steel bars spacing of the bars will be equal to area of one bar upon area of steel into thousand area 1 bar is pi by 4 d square where d is the diameter of the bar that is 10 mm and the total will be 78.5 mm square substituting this value in the previous equation spacing will be equal to 270 mm the maximum permissible spacing is 3 times d or 300 mm whichever is less and it shall be greater than 200 mm where d is the effective depth of the slab hence we can provide 10 mm steel bars at a spacing of 270 mm center to center now we need to check if the provided area of steel satisfies the criteria of minimum reinforcement as per is456 2000 for mild steel reinforcement the minimum reinforcement shall not be less than 0.15 percent of the cross-sectional area and 0.12 in case of hysd bars hence the minimum ast will be equal to 0.12 percent of area of cross section that is b into d where b is 1000 mm and d is the overall depth of the slab that is 140 mm and the total will be 160 mm square which is less than the provided area of reinforcement that is 290 mm square hence satisfying the criteria now we need to check if the effective depth of the slab satisfies the deflection criteria we have the equation fs equal to 0.58 fy into ast required upon ast provided where fy is 415 the required ast and provided astr same that is 290 mm square on substituting all these values fs will be equal to 240 newton per mm square the required percentage of steel will be equal to ast upon bd into 1000 where ast is 290 mm square b is 1000 mm and d is the effective depth of slab that is 115 mm on substituting these values the required percentage of steel will be 0.25 percent which is less than the assumed percentage of steel that is 0.4 percent hence satisfying the criteria the deflection check can be considered to be satisfied however detailed check as per the code is worked out to substantiate the above statement for 0.25 percentage of steel and fs equal to 240 the actual modification factor will be 1.5 as per this graph which is on page 38 of is 456 2000 the depth required for deflection will be span upon basic value into modification factor where basic value for simply supported slab is 20 which i have already discussed in my previous lecture on substituting these values the depth required for deflection will be equal to 100 mm which is less than the calculated effective depth of the slab that is 115 mm hence the assumed depth of the slab is safe in deflection now let's design the distribution steel 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of 1 bar upon ast into 1000 where area of 1 bar is pi by 4 d square or pi by 4 8 square on substituting these values spacing will be equal to 364 mm say 360 mm hence providing 8 mm diabolos at 360 mm center to center as per is 456 2000 the spacing of distribution bars shall not be more than 5 times the effective depth of the slab or 450 mm whichever is small 5 times effective depth is 575 mm hence maximum spacing will be 450 mm which is greater than the calculated spacing that is 360 mm hence satisfying the criteria now we need to do the check for shear to calculate the design shear we have the formula vu equal to w lx upon 2 where w is the factored load which is already calculated in the previous part of this video the value of w is 11.25 and lx is the shorter span of the slab that is 3 meter on substituting these values design share will be equal to 16.875 kilo newton tau bd is equal to vu upon bd vu is the design shear that is 16.875 b is 1000 mm and d is the effective depth that is 115 mm on substituting these values tau bd will be equal to 0.146 newton per mm square now we need to calculate the design shear strength of the concrete that is tau c for that we have to refer table 19 of is456 2000 on page 73. according to which for m20 grade of concrete and 0.25 percentage of steel tau c is equal to 0.28 newton per mm square for solid slabs the design shear strength for concrete shall be equal to tau c into k where k is equal to 1.3 as per is456 2000 on page 72 hence k into tau c will be equal to 1.3 into 0.28 which will be equal to 0.36 newton per mm square therefore tau c is less than k times dow c hence safe in shear finally we need to check the development length the development length should satisfy the following relation that is m by v plus l naught greater than l d where m is the design factored moment which is already calculated as 11.7 v is design shear which is 16.87 and l naught is number of hooks which is zero the development length is calculated by using the formula ld equal to 0.87 fy into phi upon 4 times tau vd where fy is 415 phi is the diameter of main bars that is 10 mm tau bd is the design bond stress which is 1.2 newton per mm square for m20 grade of concrete as per is456 2000 page 43. on substituting these values ld will be equal to 470.11 mm it should be noted that the value of tau bd is increased by 60 percent for deformed bars that's why we have multiplied it by 1.6 again m by v plus l naught will be equal to 11.7 into 10 raised power 6 upon 16.875 into 10 raised power 3 which will be equal to 693.33 mm which is greater than the calculated development length hence satisfying the safety criteria so this was all about this lecture do check out the courses on our android app the link will be provided in the description box of this video