 I am Dr. Patil Sunil Kumar S, Professor and Head Civil Engineering Department, Palchay Institute of Technology, Swalapur. So today I am going to discuss about a numerical example, I am going to discuss a numerical example on design of one-way slab. Learning outcomes, at the end of the session the learners will be able to determine the effective horizontal span and the effective thickness of the slab. And they are also able to determine the reinforcement required for the slab that is main reinforcement and distribution steel and sketch the reinforcement arrangement. Example, a hall has a clear dimension 3 meter by 9 meter with wall thickness 230 mm. The live load on the slab is 3 kilo Newton per meter square and a floor finish load, finishing load of 1 kilo Newton per meter square may be assumed. Use M20 concrete and Fe415 grade steel design the slab solution. The first step you have to find out the ratio of long span to shorter span L y by L x ratio. L y is 9 meter, L x is 3 meter, 9 by 3 it was thought to be 3 which is greater than 2. Hence it will be designed as one-way slab. Next step number 2, the depth of the slab. So the depth of the slab rather it is effective depth of the slab. Usually we assume between L by 22 to L by 28 but I will take L by 25 of the span. It is equal to L by 25 into span is 3000, shorter span always it is, it is 120 mm. So effective depth I will assume it as 125 mm, required minimum is 120 I will assume 125. I will assume a clear cover of 15 mm rather I will assume a clear cover of 20 mm and the diameter of the bar as 10 mm. So it is small d plus half the diameter of the bar plus clear cover so that was thought to be the capital d that is overall depth 150 mm. Then bending moment and shear force per meter width of the slab. So for that first of all we have to calculate the load distribution. So dl, dead load is equal to 0.15 that is 1 into 1 that is 1 meter into 25 kilo Newton per kilometer it is 1 meter width. So it is 3.75 kilo Newton per meter then finishing load it is 1 into 1 1 kilo Newton. Then live load it is 1 into 3 that is 3 kilo Newton per meter square. So therefore 3 kilo Newton per meter total load is 7.75 kilo Newton per meter. So w it is the partial safety factor that is 1.5 into w it is 1.5 into 7.75 kilo Newton per meter. So effective span is smaller of the two it is as per clause number 22.2 of IS 456 it is lc plus d that is 3000 plus d shorter span always it is 3.125 meter or 3000 plus width of support it is 3000 plus 230 it is 3.23 out of these two the lesser we have to take therefore effective span le is equal to 3.125 meter. Now can you please tell me what is where is the maximum bending moment in one way slab. The maximum bending moment is at center of span for one way slab it is at the mid span of the one way slab. So you have to calculate m u it is w into l square upon 8 where l is effective span. So it is 1.5 7.75 into 3.125 square divided by 8 it is 14.19 kilo Newton meter and w is equal to w into l by 2 it is 1.5 into 7.75 into 3.12 divided by 2 it is 18.164 kilo Newton. Step number 4 design for m u. So whenever you want to design first you have to find out m u limit first you have to find out m u limit for calculation of for calculating m u limit there are two methods one is you have you can find out from x u limit x u limit is 0.48 d it is 0.48 120 that is 60 mm. So m u limit is equal to this is g 0.1 1 c that is 0.36 fck bx u that is c into liver arm d minus 0.42 x u limit. So if you substitute all the values we will get m u limit as 43.114 kilo Newton meter. So you can also calculate m u limit by using direct formula a constant into fck bd square where the value of constant is 0.148 for mild steel 0.138 for fe415 and 0.133 for fe508. You will get the same value if m u is less than m u limit then it is under reinforced section. So this is must for determination of steel you should have always under reinforced section m u must be less than m u limit. Next you have to calculate m u also area of steel area of main reinforcement by using g 1.1 b of is 456 2000 it is m u is equal to 0.87 f y astd into 1 minus ast f upon bd fck a this is g 1.1 b of is 456 2000 m u is 14.19 into 10 power of 6 it is 0.87 f y is 415 ast into 125 is your effect to depth 1 minus ast divided by b into d b is 1000 it is always 1 meter width 125 415 f y divided by fck 20. So it was the ast was sought to be 333 mm square now the main reinforcement step 5 the spacing of main reinforcement I will find out using 10 mm bars the spacing s is equal to pi by 4 into d square upon ast that was sought to be 235.8 mm hence provide 10 mm diameter h y ast bars at 225 mm centre to centre please remember this should be less than spacing calculated. So the last step that is step number 6 check for maximum spacing so you have to check for maximum space maximum spacing allowed is 3 times d effect to depth 3 times 125 that is 375 or maximum is allowed is 300 the spacing provided is satisfactory because it is less than both of these two. Next step number 7 check for shear. So first we have to calculate nominal shear stress tau v it is v u upon b d v u is 18.164 into 1000 divided by b 1000 into d is 125 it was sought to be 0.145 Newton per mm square. So you have to calculate percentage steel because for calculation of tau c you require percentage steel from table 19 percentage steel area of one bar into divided by 225 that is the spacing you have provided and 125 is d and into 100 that gives you 0.279 percent. So if you refer table 19 you will get tau c from table 19 of IS 456 tau c it is for beams and slab but for both it is point it was sought to be 0.375 Newton per m square for slabs it is k into tau c tau c is given as k into tau c so k we have to take from 40.1.1 so k is 1.3 into this it was sought to be 0.487 Newton per m square. So now tau v tau c is greater than tau v tau c is greater than tau v and also tau v is less than 0.5 times tau c max tau c max is from table number 20 so hence slab is safe in shear. Next you have to check for deflection again for deflection also you require percentage steel it we have already calculated for shear in shear calculate checking it is 0.279. So you have to refer figure 4 for your constant F1 it is L by d max it is equal to 20 into F1 it is 20 into F1. So for calculation F1 you have to refer IS figure 4 from IS 456 2000. So in that particular case this is figure 4 from IS 456 2000. So FS you have to calculate first it is 0.58 into Fy into Ast required upon Ast provided so that was sought to be 240.7 Newton per mm square. So if you just use figure number 4 of IS 456 2000 you will get F1 as 1.5. So for this particular percentage steel of 0.279 then L by d max is equal to 1.5 into 20 that is 30. L by d provided is 3125 divided by 125 that is 25 which is less than L by d max hence deflection control is satisfactory. Then the distribution steel it is 0.12 percent of area of cross section 0.12 by 100 into b into d so here it is 0.12 into 1000 into 125 divided by 100 it is 150 mm square. Selecting 8 mm bars it is spacing will be pi by 4 into 8 square area of 1 bar divided by Ast into 1000 that gives you 335 provide 8 mm diameter bars at 300 mm center to center. This is distribution steel along the longer direction along 9 meter direction. Now this is the last step that is you have to sketch the reinforcement detaining which is as shown in figure in plan I have shown this is width of wall 230 mm then I have shown only the 2 bars along distribution steel and 2 bars along the x direction this is 10 mm diameter 225 centimeter center and this is 8 mm diameter 300 centimeter center. Similarly I have taken a section xx so wherein you will find bottom 10 mm diameter 225 center to center then these dotted bars along long direction 8 mm diameter 300 center to center so alternate bar is bent up so this is how you have to provide the reinforcement. These are the references used for the presentation thank you thank you one and all