 Patil Sunil Kumar is Professor and Head Severe Engineering Department, Valchainstoratakumar Solapur. So, today we shall discuss about design of open well stairs. Learning outcomes, at the end of session, the learners will be able to explain the procedure for determining the effective horizontal span of a dog laid stair and effective thickness of waste lab for each flight, then determine the reinforcement required for each of the flight and the sketch or reinforcing the reinforcement detailing. Open well staircase, open well staircase is one which have open space in the middle portion and flights running its along its periphery. When all steps are difficult to arrange in two flights, a short third flight having three to five steps may be provided along the direction perpendicular to the hall. This type of staircase is also called, this staircase is also used when the lift duct has to be provided within the staircase. Figure number one shows the typical plan of open well staircase. So, this is a open well staircase. Here we have the first flight, the A from A to B it is first flight, so this first flight is resting on a beam here, below the first riser, a beam here, so it is resting over here. So, these are the number of threads 1 to 10 and 11th is the landing, quarter landing. Then again here 12, 13, 14, 15, 16 these four five steps, so they are running perpendicular to the hall and here again second flight, so from 17 to from 18 to 27, there are all the trades and this is again quarter landing and this is the landing at the top. So, design of open well staircase, so height of each for each flight is equal to number of risers into the rise. Number of threads in each flight, it is number of risers in each flight minus 1. Effective span, the stair slab span longitudinally perpendicular sorry depending upon the supports the effective span is the horizontal distance between the center to center of supports. So, here for example, I will show you for the first flight, so the span will be from here to here. That means from the beam where your first riser is resting, the center of beam to all these risers and landing and it is resting over here or the beam or over the wall over here and for the second flight that is BC, so it is the center of this particular wall to the center of this wall. So, this is landing portion B and C and these are the going portion this one, so likewise and this is similar to the first flight. Then load calculations to determine the dead load of the waste slab, it is necessary to assume the thickness effective thickness of waste slab which is given as by it is between 120th to 125th of the effective span. The effective span is already determined though you can find out the thickness of waste slab. The overall thickness T is effective thickness plus effective cover, effective cover is from the center of reinforcement to the edge. To find the total load per meter horizontal width of stair, we we have to find out we have to take this total thickness of the slab waste slab then afterwards square root of 1 plus r by T square into density of concrete. So, weight of steps is equal to 1 half r into T divided by T this is to bring it to horizontal portion into density of concrete. So, dead load on the going portion is equal to weight of slab plus weight of steps plus weight of finishing the dead load on the landing portion is weight of slab plus weight of finishing. So, assume the weight of finishing maximum as 1 kilo Newton per meter square. The live load is assumed as per IS 875 part 2 that is depend depending upon the whether it is residential building public building what type of load we are coming about. So, we will get the live load from IS 875 part 2 total load is equal to dead load plus live load factored load is 1.5 times total load. The projected loading pro diagram for the total factored load for flight AB and DC is shown in figure 2 and for BC in figure 3 load distribution diagram of flight AB and BC are same or they are different. So, can you please guess whether they are same or they are different the load distribution diagram for factored load on flight AB and DC are different why they are different because here you will find from the first riser this is landing portion and this is going portion. So, therefore, here load is more here load is less and this is for flight AB. If you see the for flight BC so, this is landing here and again going portion and landing here here load here is less due to landing portion here due to steps then going portions load is more. So, this will show you the load distribution for flight AB and next one is for flight BC. So, flight AB as well as flight DC are similar so, therefore, it is same for both the things. Then determination of design moment the maximum bending moment will be at the center of the flight so, that is for BC not for both. Find maximum support reactions and determine the maximum bending moment MU that is 0.36 if see it is equated to 0.36 MU limit is 0.36 FCK B into XU limit into D minus 0.42 XU limit so, that is for under reinforced section XU limit is equal to 0.42 D for FE 4 and 5 0.46 D for 500 0.53 D for smile steel compare MU limit with MU if MU is less than MU limit it is under reinforced section find the area of longitudinal steel. Find the area of longitudinal steel using formulae point AST is equal to 0.5 FCK BD upon FY into 1 minus square root of 1 minus 4.6 MU upon FCK BD square. Find the spacing from the formula SV is equal to 0.487 FY ASVD divided by VUS where VUS is the total it is VU minus tau CVD find the area of distribution steel it is 0.2 percent of BND and the spacing using above formulae is determined for this also draw a needs detailed sketch of the stair showing reinforcement detailing which is in figure number 4 now this is figure number 4. So, this is flight AB here below this particular riser there is a beam so, which supports the flight. So, therefore, your span it is from this center of the beam up to the end or wall or beam whatever it is there. So, this is your the horizontal span projected span for flight AB. So, this is a waste slab and above that you are having steps. So, therefore, in this portion this going portion you get more load and this is landing portion where you get lesser load because no steps are there on this. So, you are having dead load of waste slab plus finish and plus live load here dead load of waste slab plus steps plus finish and plus live load. So, now in this particular case to find the maximum meaning moment it is not at the mid of the flight. So, to find out the maximum meaning moment we have to take a section at a distance x from here and we have to find out what is the shear force at that section. So, bending moment is maximum at a section where shear force is 0. So, therefore, by equating that particular shear force to 0 we find out the distance x and we find out the maximum meaning moment at x. So, by using that we call it as m u then we compare it with m u limit and we find out the steel. So, here you get main steel at the bottom and this dotted portion dotted one which is perpendicular to this slide it indicates the distribution steel and for hugging moment since due to partial fixity or full fixity we get by hugging moment at the top. So, therefore, here also there is a main steel which is extended further here and here we have again at the top steel. So, that is for if any particular hugging moment is there. Now, this is for flight BC. For flight BC we are having landing here and we are having landing here landing on both the sides and in the middle portion it is going portion this is going portion wherein we have steps. So, therefore, here you will get the bottom main steel and distribution steel perpendicular to it. So, these are the we are supposed to draw the section for the first flight and section for the second flight. So, for the first and third flight it is same for the second flight it is different. Therefore, two sections are to be drawn for a dog like to stare sorry open well staircase and these are the references which are used for this particular presentation and thank you one and all.