 I am Dr. Patil Sunilakumaris, Professor and Head of the Civil Engineering Department, Vulture Institute of Technology, Sulapur. So today I am going to discuss about design of one-way slab. Learning outcomes. At the end of the session, the learners will be able to explain the procedure of determination of effective span and effective thickness of one-way slab and determination of main and distribution reinforcement required for the slab. And they will also be explained, they can explain the sketch, the reinforcement arrangement in one-way slab. Introduction. Any slab supported on two opposite edges in one is one-way slab as shown below. Here it is shown that this is a one-way slab, it is supported on only two edges, one is this edge and another is this edge. That is along two long edges parallel to each other it is supported and this is shortest span. So a one-way slab is the one which only bends in one direction across the span. That means this will bend only in the shorter direction. A one-way slab is one which bends in only one direction across the span and acts like a wide beam of unit width. It acts like a wide beam of one meter width or unit width. A slab is assumed to be made up of such similar strips placed side by side and acting independently. So this way we assume regarding the one-way slab. The analysis of one-way slab is same as that of analysis of beam of one meter width. The slab which is having L y by L x, L y is longer span, L x is smaller span. So that means this span is L y and this span is L x. Longer span is L y, shorter span is L x. L y by L x ratio being more than two is considered as one-way slab. That means whenever we consider any slab if we find the ratio of longer span to shorter span more than two then and then only we can design it as one-way slab. Then the design steps for one-way slab. The first step we have to find out the effective span L e of the slab as per clause number 22.2 of IS 456-2000. As per clause number 22.2 of IS 456-2000 that is regarding control of deflection. So effective span of the slab shall be lesser of the two for simply supported slabs. So effective span L e is equal to clear shorter span plus small d that is effective depth. Or effective span L e is equal to center to center distance between the supports in shorter direction. That is the clear span in shorter direction plus width of support it is. Then assume effective span of the slab effective out of these two whichever is small that we are supposed to take as effective span. So assume effective depth of the slab. Next is you have to assume the effective depth of the slab. So which is required for the sabotage. The effective depth of the slab. Next is you have to assume the effective depth of the slab. So which is required for the sabotage determination of L e. So therefore it is based on control to deflection that is 23.2. 23.2 it is control for deflection and 22.2 it is for effective span. 22.2 is for effective span and 23.2 is for control of deflection. The effective depth of the slab depends upon the bending moment and the deflection criteria. The trial depth can be obtained by using d is equal to L by 22 to L by 28. d is equal to L by 22 to L by 28. Whereas in IS 456 2000 if you refer clause 23.2 you will get effective depth of d equal to 20. So this is L by d ratio as 20 where it is we take between 22 to 28. Because if you take the constant F1 that usually work out to be 1.2 or 1.5. Therefore usually we take the L by 25 that is in between L by 22 to L by 28. So this is the question for you people. The effective span L e for a simply supported slab is given by L e is equal to clear shorter span plus d. That is effective depth or L e is equal to centre to centre distance between the supporting shorter direction. Option C maximum of A and B option D minimum of A and B. So can you please guess and tell me which is the right option. Effective span L e for a simply supported slab is given by D option that is minimum of A and B. Next you have to find the loads on the slab. The load on slab comprises of dead load floor finish and live load. The loads are calculated in kilo Newton per meter. Dead load is equal to it is overall depth of the slab in meter into 1 into 25. 25 is unit weight of concrete 25 kilo Newton per cubic meter. So this will multiplication will give you load dead load in kilo dead load of slab in kilo Newton per meter. Floor finish usually we take 1 kilo Newton per meter square it is 1 into 1 that is 1 kilo Newton per meter. And the live load W L it is assumed as 2 to 5 it is between 2 to 5 kilo Newton per meter square depending upon the occupancy of the building as per IS 875 part 2. Usually 2 kilo Newton per meter square for residential building and 3 kilo Newton per meter square for public building. If I calculate W L for public building it is 3 into 1 it is 3 kilo Newton per meter for public building. Find M U and V U. M U is equal to 1.5 times W D plus W L into L e square upon 2. And V U is equal to 1.5 times W D plus W L into L e divided by 2. Determine M U limit as per IS 456 clause number G 1.1 C. So that is given by M U limit is equal to constant into FCK that is characterization of concrete into B into D square. Where constant is 0.148 for mild steel 0.138 for Fe 4 and 5 and 0.133 for Fe 500. So one has to compare M U limit with M U. One has to compare M U limit with M U. If M U is less than M U limit it is called under reinforced section. If M U is greater than M U limit then it is called a war reinforced section which is not allowed by IS 456 2000. Therefore if you get any time M U greater than M U limit then only option with us is to increase the depth of the slab until M U is less than M U limit. Next you have to find out the area of reinforcement required as per clause number G 1.1 B of IS 456 2000. So which is given by M U is equal to 0.87 F Y E S T D into 1 minus A S T F Y upon B D F C K or you can even calculate A S T directly. It is 0.5 F C K B D divided by F Y into 1 minus square root of 1 minus 4.6 Me upon F C K B D square. So assume diameter of the bar usually 10 mm or 12 mm and find spacing of the reinforcement which is given by area of 1 bar into 1000 divided by A S T which is calculated here above. Provide the main reinforcement along the shortest direction. Please remember always we have to provide the main reinforcement along shorter direction because the slab bends in shorter direction first or load transfer is by shortest route. So therefore along the shortest direction you have to provide the main reinforcement. The slab is having spacing with a spacing less than whatever we calculate above, determined above. Find the distribution steel. This is main steel along shorter direction and along longer direction you have to provide a distribution steel. So distribution steel is also called as the steel temperature steel or it is steel in longer direction or shrinkage steel. It is also called as shrinkage steel. It is also called as temperature steel. That is in order to take the stresses induced due to the temperature or shrinkage of concrete we are supposed to provide a steel because concrete is very weak in tension. Find the distribution steel to be provided in longer direction over the main steel. Area of distribution steel is 0.15% of cross section area of mild steel and 0.12% of area of cross section for HYSD bar that is high yield strength deformed bar that is FE415 and FE500. So assume the diameter of the bar usually 6 mm or 8 mm. If it is 6 mm usually it is mild steel. If it is 8 mm it is HYSD bar and find the spacing of the reinforcement by area of 1 bar in 2000 divided by AST that is area of distribution steel calculated. Provide distribution reinforcement along the longer direction of the slab having a spacing less than the spacing determined above. Draw a neat sketch showing cross section of the slab and the reinforcement detailing. Now please remember in reinforcement detailing we will have the main reinforcement along the shorter span and the distribution reinforcement over it along the longer span. And for the main reinforcement we have to make the alternate bar bent up. So that is we require minimum 50% of the steel should go up to the end. So therefore alternate bar will be a bent up for the main steel. So these are the references used for preparing this particular presentation. Thank you. Thank you one and all.