 So, today let us discuss about a numerical example on design of open well staircase. Learning outcomes at the end of this session the learners will be able to determine the effective horizontal span for a open well stair and effective thickness of waste slab for each flight and determine the reinforcement arrangement of each flight and sketch the reinforcement arrangement. Example, a stair with open well is having a steps of size 280 mm by 150 mm. So, this is 280 is tread and 150 is rise. The arrangement of the stair is as shown in figure 1 arrangement is already given. So, design the stairs for a live load of 3 kilo Newton per meter square use M 20 concrete and M 4 Fe 4 1 5 steel sketch the reinforcement details. Now, this is a plan showing arrangement of steps. So, this is flight AB this is first flight here we find 1 2 3 up to 10 these are trades and 11 it is quarter landing. So, this is AB flight then this is BC flight BC flight is perpendicular to the hall. So, here we get trade 1 2 3 4 5 5 trades are there and again a landing over here and CD it is again the a third flight. So, AB and CD they are similar whereas, BC is different solution step 1 effective span for AB and CD effective span L is given by 0.15 u plus 2.8 this is going portion plus 1.4 plus 0.15 0.15 u is the thickness of the wall which they have taken here or the thickness half the thickness of beam or wall. So, whether it is here there is a beam and here there is a wall. So, half the thickness wall is taken here 300 beam is taken here 300 m width and here wall is taken 300 m width. So, half of that will be taken every time then these trades this is going portion and this is again landing portion. Now, here we get this is 4.5 meter and BC it is 0.15 1.4 1.4 1.4 plus 0.15 that is the 4.5 meter then there is the flights have effective span of 4.5 all are 4.5 here fortunately. Hence the thickness of west lab is 125th of span to 120th of span so, it is 200 plus overall thickness they have taken as 230 mm. Then loads considering 1 meter wide step load ongoing portion. So, weight of west lab is 0.23 is the total thickness into square root of 1 plus r by t square into 25s density. So, that was of a 6.523 kilo Newton per meter weight of steps half of r into t divided by t. So, into 25 that will give us 1.875 kilo Newton per meter say the finishing load is finishing including finishing load say it is 9 kilo Newton per meter with finishing load. Then live load is 3 kilo Newton per meter so, factored load it is 1.5 times dead load plus live load it is 18 kilo Newton per meter. Load on landing so, weight of landing it is 0.23 into 1 into 25 5.75 so, weight of finishing including finishing it is 9 kilo Newton per meter. So, live load 3 kilo Newton per meter so, total load 9.75 kilo Newton per meter. So, factored load is 14.625 now please see on factored load ongoing portion is 18 whereas, for landing portion is 14.625 where will be the maximum bending moment for flight A B where the maximum bending moment for flight A B is at a distance x from A where the shear force is 0. So, we get maximum bending moment where shear force is 0 therefore, we have to write the equation for shear force and equate it to 0 to find out the x. So, figure 2 shows the load on this particular flight per meter width this is flight A B so, this is R A R B so, this portion it is steps weight is there therefore, this is more and this is landing portion therefore, load is less. So, R A is calculated by taking moment about one of the rear support it is 39.6 kilo Newton then by we will take a section at distance x and if we just divide by loading on it that is the load coming that is at 2.2 meter from A we get maximum bending moment here the shear force is 0. So, maximum bending moment occurs at 2.2 meter from A hence the design moment A mu is equal to this much that is 39.6 R reaction into 2.2 minus this downward load 18 into 2.2 into 2.2 by 2. So, it is 43.56 kilo Newton meter step 4 reinforcement AST is again calculated by using relation 0.5 fck BD divided by f y into 1 minus square root of 1 minus 4.6 A mu upon fck BD square. So, by substituting the values we get AST as 646.6 mm square using 12 mm bars spacing required will be area of 1 bar into 1000 divided by 646.6 that is 174.94 mm. So, therefore, provide 12 mm bars at 170 mm centre to centre then distribution steel so this is perpendicular to the slide that is the perpendicular to the main steel we are having AST is 0.12 percent of the cross sectional area 0.12 by 100, 1000 is width and thickness is 200. Therefore, we get 240 mm square. So, using 8 mm bars spacing required will be area of 1 bar divided by 240 into 1000 it is 209 mm. So, provide 8 mm bars at 200 centre to centre this is regarding distribution steel. Next is for design for flight BC flight BC loading is different flight AB it is different therefore. So, we have to design AB and BC separately. So, for flight BC the loading diagram is shown here the this is landing portion therefore having lesser load and this is going portion having more load and again landing portion having lesser load. So, due to symmetry here the bending moment will be at the mid span. So, therefore, we calculate RA or AB reaction then we calculate maximum bending moment which is at mid span. So, maximum bending moment works out to be 41.4 kilo Newton meter. So, the reinforcement again determined by using 0.5 fckbd upon f y into 1 minus square root of 1 minus 4.6 mu upon fckbd square. So, if we substitute all the values we get AST is equal to 612.2 mm square. So, the spacing of bars it is area of 1 bar 1000 divided by 612.2 that is 184.7 mm. So, let us provide 12 mm torque 180 mm centre to centre in this particular portion then distribution steel again 8 mm diameter 200 centre to centre. Now, this is reinforcement detailing. So, this is reinforcement detailing for the first flight that is AB here there is a beam here there is a wall. So, therefore, we find the bottom steel bottom steel is main steel that is 12 mm diameter bars at 170 mm centre to centre it is provided bottom. And above that you get dots these dots are nothing but this is distribution steel it is 8 mm diameter bar h ysd bar at 200 mm centre to centre. So, here at the we get fixity therefore, this bottom main steel is taken up to the top. So, that is again you get top and bottom steel here and distribution steel again in this portion where you have top and bottom steel distribution steel is provided see in a staggered manner. So, next is the flight BC for flight BC we are having landing here and landing at both ends and in between that we are having a going portion. So, therefore, here you get 12 mm diameter bars at the bottom that is your main steel. So, at 180 mm centre to centre here it is 170 here it is 180 mm centre to centre and this particular thing again this is taken up to top and again we are having bottom x star bar z bottom. So, then here again distribution steel staggered and here the distribution steel 8 mm dot 200 mm centre centre that. So, this is how we are supposed to show the two sections in case of open well staircase. So, one for the two flights on either side and one for for the central span that is central flight. So, these are references used for preparing this particular presentation. Thank you.