 Sunil Kumar S, Professor and Head Civil Engineering Department, Valtran Institute of Technology, Swalapur. So, today I will discuss about design of two-way slab, learning outcomes. At the end of the session, the learners will be able to explain the procedure of determining the effective span and the effective thickness of two-way slab and the reinforcement required in short and long direction for two-way slab. When the slab is supported on all four sides and the ratio of the longer span L y to shorter span L x is less than 2. The bending moment developed in both x and y directions as shown in figure 1. This is figure 1, you will find this is longer span and this is shorter span. So, you will find it deforms in both the direction, so bending moment you will get in both the direction comparably almost nearer to each other, shorter is more and longer is less, so therefore bending moment developed in x and y direction is predominant and hence design would be made for the reinforcement in both direction. Please remember for two-way slab main steel is there on both the sides along short as well as along long direction, it is known as two-way slab, please see the deformed shape, so the moment developed depends upon the edge condition also. In building we have we come across the following boundary conditions, so all four edges discontinuous or all four edges continuous first one then one shortest discontinuous one longest discontinuous two adjacent edges discontinuous two short edges discontinuous two long edges discontinuous three edges discontinuous and one long edge continuous three edges discontinuous and one short edge discontinuous, so all these cases are shown in this figure number 2. So this is number 1 is all sides of the slab are continuous, number 2 is the one short edge discontinuous, number 3 is one long edge is discontinuous, number 4 two adjacent edges discontinuous, number 5 two short edges are discontinuous but only long edges discontinuous number 6 here two short edges discontinuous one long edge two long edges are discontinuous number 8 all three edges discontinuous and one short edge continuous, number 7 all three edges discontinuous one long edge continuous, number 9 this is all four edges discontinuous number 9 is all four edges discontinuous, number 10 here it is simply supported slab without torsionality and this is with torsionality, number 9 is with torsionality. Now design steps for 2s slab find the effective span lx of the slab as per clause 22.2 effective span 22.2 of is 4562000 effective span shall be smaller of the two that is clear shorter span plus effective depth or center to center distance between the support in shorter direction assume effective depth of the slab small d as per clause number 24.1 node 2 of is 4562000 for 2s slab of shorter span up to 3.5 meter with mild steel reinforcement the span to overall depth ratio given below may generally be you assumed to satisfy the vertical deflection limits for loading up to 3 kilo Newton per meter square this is live load. So simply supported slabs 35 continuous slabs 40 for hysd bars the above value are multiplied by 0.8 find the load on slab the load on slab comprises of dead load floor finish and live load the load can be calculated per unit area that is load per meter square. So here dead load wd it is the overall depth of the slab in meter into 1 into 25 kilo Newton per kilo meter that is density it will give you in 20 kilo it will give the value in kilo Newton per meter floor finish 1 kilo Newton per meter square into 1 it is 1 kilo Newton per meter live load assumed as 2 to between 2 to 5 kilo Newton per meter square depending upon the occupancy of the building for residential 2 and for public 3 and for library it is 5 kilo Newton per meter square find the bending moment in x and y direction for uniform distributed load of entire slab on entire slab maximum positive bending moment tension at bottom develops at the center of span and support at support negative bending moment develops in the slab with various edge conditions the maximum bending moment per unit width of the slab are given in annexure d if you refer d 1.1 you will get mx is equal to alpha x into w into lx square my is equal to alpha y into w into lx square please remember both time it is shorter span only lx square only it is as per d 1.1 of is 456 2000 alpha x and alpha y are calculated coefficients calculated from table 26 and 27 of is 456 2000 depending upon the case. So, please tell me for 2 a slab bends or deflects in shorter direction in longer direction bends in both direction or none the right answer is it bends in both the direction that is option C. Now, this is table number 26. So, depending this is case number from 1 to 9 this is case number from 1 to 9. So, you will get this is l y by lx ratio from 1 to 2 this is the coefficient for longer span this is table number 26 use for case number 1 to 9. So, please remember here you will get the positive and negative bending when negative bending is over supports and positive bending moment at mid span. Now, table number 27 it is used for case number 10 for case number 10 we use the coefficient alpha x alpha y from table number 27. So, next find the area of main steel in both the direction as per g 1.1 b m u is equal to 0.87 f y a s t d into 1 minus a s t f y and b d f c k that is g 1.1 b or you can write it as a s t is equal to 0.5 f c k b d upon f y into 1 minus square root of 1 minus 4.67 upon f c k b d square. Assume diameter of the bar usually 10 m m or 12 m m and find spacing of the reinforcement which is given by area of 1 bar divided by s t into 1000. Provide main reinforcement along short direction and long direction of the slide having spacing less than whatever we determined above. So, that means please remember both are main reinforcement then you are supposed to draw a neat sketch showing the reinforcement arrangement. So, in case of 2-way slab you will find both the steels are very important that means along short direction as well as along long direction. So, whenever you go for the interior slab if you have an interior slab interior span case then you have to first find out L y by L x. If the L y by L x let us say it is 1.5 if it is 1.5 L y by L x then you will get the negative alpha x as 0.053 and alpha x is equal to 0.041 that is for negative bending moment is 0.053 and for positive bending this is at mid span and this is at the support this is over the support. So, therefore, you will get both the coefficient both are alpha x, but this is in the native value this will give you the positive value. So, this is at mid span and this is at the over the support and similarly on the longer side you will get for any value for all values of L y by L x you have to take the same value that is over the support it is 0.032 and at mid span it is 0.024. Now please remember the negative moment is always greater than the positive bending moment. So, therefore, so over the support we have maximum bending moment for the continuous edges. So, this is how we have to refer this particular table and here you will find for 4 edges discontinuous will win only one because for simply supported over the at the support it is 0, bending moment is 0. So, therefore, you get only at mid span and that is having maximum value. Then if you are this is a table with a torsion reinforcement provided at corner will be providing mesh of bars that is torsion reinforcement to avoid lifting up of the edges. Now this table we refer when we do not assume that we are if you are not providing torsion reinforcement at corners then you have to take alpha x alpha y from this particular table. So, this is the difference between these two. So, this we have already discussed these are the references used. Thank you. Thank you and all.