 Dr. Patil Sunil-Kumaris, Professor and Head of the Civil Engineering Department, Volchian Institute of Technology, Sulappur. So, today I am going to discuss about a numerical example on design of dog leg restaure. Learning outcomes. At the end of session, the learners will be able to determine the effective horizontal span and effective thickness of the waist slab for each flight. Then, they are also able to determine the reinforcement required for each flight and sketch the reinforcement arrangement. This is an example. Design a dog leg restaure for an office building in a room measuring 2.8 meter by 5.8 meter clear. Vertical distance between the floors is 3.6 meter. Width of flight is to be 1.25 meter. Allow a live load of 3 kilo Newton per meter square. Sketch the details of the reinforcement. Use M20 concrete and Fe415 steel. Assuming the stairs are supported on 230 mm walls at the end of outer edge of landing slab. Solution. The first step is to assume a rise. The minimum rise required for a public building is 150 mm. So, floor to floor height is 3.6 meter. A dog leg stair is always having two flights. Therefore, so 3.6 meter is divided by 2 to get flight height of each particular flight. It is 1.8 meter that is 1800 mm. So, number of risers in one flight is 1800 divided by rise that is 150 mm that is 12. So, hence the number of trades required it is 12 minus 1 that is 11. As the stair is 1.2 meter width therefore, the minimum width of the landing required is also 1.25 meter. So, for 11 trades we need a length of going. Going length of 11 into T. So, selecting the trade T as 300 mm. The step arrangement are shown in the plan below here in figure number 1. So, here you will find these 11 number of stairs that is 11 trades and this is a landing. This is again a landing. So, again the size is again same which is given over here 2.8 by total it is 5.8 meter. So, next second step is to find the effective span. The stairs slab is spanning longitudinally effective span is the center to center distance. That means, here you can see it is the center to center distance between this wall to this wall. So, therefore, it is 1.25 plus 3.3 going plus 1.25 plus half the thickness of wall here half the thickness of wall here that is again total is 0.2 through 5 that is 0.23 that is totally 6.03. So, loads thickness of waste slab with the thickness of waste slab is to be assumed from 120 to 125th of this span. So, that is it works out to be 300 mm to 240 mm. Let us take a thickness of the waste slab as 250 mm and the overall depth will be 280 mm. Let us find the load per meter horizontal width of the stair. The weight of waste slab is equal to 0.28. It is the thickness into square root of 1 plus r divided by T square into 25. 25 is the density of the concrete. So, it works out to be 7.83 kilo Newton per meter. So, weight of step that means, we are having a triangular step therefore. So, one half the 0.15 into 0.25 0.15 is rise 0.25 is thread and 0.25 that is to make it to horizontal. So, multiplied by 25 it is 1.875 kilo Newton per meter. So, the load total is 9.7 kilo Newton per meter that is the lead load of waste slab plus dead load of step. So, in going portion the finished load let us consider it as 10.5 kilo Newton per meter. Now, in landing portion there are no steps triangular step therefore, there it is 0.25 into 1 into 25 that is 6.25 kilo Newton per meter. So, with a finishing material it may be taken as 7.25 kilo Newton per meter that is 1 kilo Newton per meter I have assumed finishing material load. Then live load is 3 kilo Newton per meter square the factor load on the going per meter horizontal width is given by 1.5 it is a load factor into 10.5 plus 3 that is 20.25 kilo Newton per meter and the loading on the slab per meter width it is 1.5 into 7.25 plus 3 that is 15.375 that is on this landings. So, here in the landing portion we have 15.375 kilo Newton per meter and in going portion of 3.3 meter we have 20.25 kilo Newton per meter and again in mid landing it is 15.375 kilo Newton per meter. So, now based on this particular sketch can you guess where is the maximum bending moment in a fly? The maximum bending moment will be at the mid span because it is having a symmetrical loading and span is also symmetrical therefore, we have at the mid span the maximum bending moment will be at the mid span. So, first of all we have to find out RARB that is the reactions from the downward loads which we have which we are getting. So, it is 54.4 kilo Newton each particular load then maximum moment occurs at the mid span and its value will be we have to calculate the what is the maximum bending moment at mid span that was out to be 87.5 kilo Newton per meter that is 54.4 that is the reaction into 6.03 by 2 that is 1 is the clockwise positive therefore, next is anticlockwise minus 15.37 into 1.365 that is that particular load and again this is the distance it is it acts at the center of the UDL. So, that is taken minus again another 33.3 by 2 that is going portion by 2 into 20.2 that is the load coming on it and it is acting at the center of that. So, that will give you bending moment 87.5 kilo Newton per meter. Then we have to find out MV limit 0.36 fck b into x u limit into d minus 0.4 to x u limit. So, x u limit for Fe 4 and 5 is 0.48 d. So, if we substitute the values fck 20 b is 1000 mm that per meter run and it was out to be 172.5 into 10 to the power of 6 Newton mm that is your MV value MV limit value. So, hence the section can be designed as singly reinforced section. So, because the bending moment which we have calculated it is 87.5 however, the MV limit is 172.5 therefore, it can be considered as a singly reinforced section. Then step number 5 that is reinforcement. So, let AST be the area steel required for the reinforcement required as per IS G 0.1.1 b of IS for 562000. So, MV is 0.87 fy AST into 1 minus AST fy upon bdfck. So, if you substitute the MV value over here we get AST value as 1063 mm square. So, if we calculate the spacing of the 16 mm diameter bars spacing was out to be area of 1 bar pi by 4 into 16 square. So, into 1000 divided by 1063 it was out to be 189 mm center to center hence provide 16 mm bars at 180 mm center to center. Now, the distribution steel. So, distribution steel is always calculated based on percentage basis 0.12 percent of cross sectional area. So, 0.12 divided by 100 into cross sectional area b is 1 meter that is 1 meter we have we have considered and 280 mm is the gross thickness. So, this was out to be 336 mm square. So, using 10 mm bars spacing required will be equal to area of 1 bar into 1000 divided by AST required. So, that will give you 230 give us 233 mm provide 10 mm bars at 230 mm center to center. So, the reinforcement is provided in the other flight also in the same manner because both are similar and identical. Now, this is the reinforcement details the details of the reinforcement are shown in figure number 3. This is a sectional elevation of the flight and the how the please note you should carefully note the manner in which the reinforcement is to be provided. So, here you will find the mean reinforcement it starts from here it will be going up to the top and again it is going into the over the wall. So, this is main reinforcement that is 16 mm HYSD bars at 180 mm center to center. And here you will find the vertical portion it is called as rise that is 150 mm and the horizontal portion is called as thread. So, that is called as it is 300 mm and here we find the distribution steels distribution steel is perpendicular to this slide here you will find the distribution is steel is provided the it is the 10 mm torque 230 mm center to center. Now, here we are supposed to provide a top steel here also and here also here a bottom steel here because. So, here we will be having the bending moment the hogging bending moment at the edges due to the partial fixity. So, that is why we are supposed to provide a top steel here. So, this particular the distance up to here. So, this is up to L by 4 from the edge. So, therefore, so in this particular way that means here we find this particular reinforcement it is having an arrangement of seizure over here. So, therefore, it may it is popularly known as the kind she is you know on the field. So, therefore, so the reinforcement arrangement is to be done very carefully and this this is nothing but a inclined slab staircase is nothing but a inclined slab. So, wherein so we are providing slab in inclined portion and the loading we are considering per square meter horizontal plan. So, that is why we have shown the loading over here on a horizontal projected that is why I have said here the projected load distribution on one flight. So, this is a projected load distribution. So, for this particular loading what we find we have shown here actually the simply supported. However, we will be having a fixity because here we will be having a wall on the this particular above the landing also. So, therefore, the we consider here the partial fixity and we provide the steel at top also and here we find the loading less because here we do not have triangular step loading here we are having waste slab loading plus triangular step loading in the going portion. So, this 3.3 meter it is called going portion and the 1.25 meter that is called as the 1.365 this is by taking up to the centre. So, this is called as mid landing this is landing portion. So, in this particular way shown in figure 3 carefully note the manner in which the reinforcement is to be provided at the inner edge of the landing slab. So, here you will find this is the main reinforcement along longitudinal reinforcement which goes up and it is goes into the above the wall and this is 16 mm HD bars at 180 mm centre to centre and these dots indicate the perpendicular that is distribution steel 10 mm diameter bars at 230 mm centre to centre and here due to partial fixity we have the hogging bending moment at the ends. So, therefore, we provide top steel also here. So, here also we provide top steel. So, this is at up to L by 4 from the edge of the beam and wherever we have top and bottom steel we provide the distribution steel. So, alternatively and here it is at for the sagging moment bending moment therefore, it is at the over the longitudinal steel. Now, these are the references which I have used for preparation of this particular the power point presentation and thank you one and all.