 So, today I am going to discuss a numerical example on design of square footing. Learning outcomes at the end of this session the learners will be able to determine the area of footing, then the effective and overall depth of footing and they are also able to determine the reinforcement required for a footing and they can sketch the reinforcement arrangement in footing. This is a design example design a square footing for a short actually loaded column of size 300 mm by 300 mm carrying 600 kilo Newton load use M 20 concrete and Fe 4 and 5 steel. The safe bearing capacity of the soil is 180 kilo Newton per meter square sketch is the details of the reinforcement solution. The first step we are supposed to find out the size of footing. The size of footing for that we have to consider the load on the column P is the load on the column 600 kilo Newton and the self weight of the column is usually taken 10 percent of the load on the column. So, that is 600 divided by 10 percent of the 600 diverse out to be 60 kilo Newton. So, total load it is 660 kilo Newton. So, the SBC of the soil that is safe bearing capacity of the soil is 180 kilo Newton per meter square. So, if we divide the total load that is 1.1 times P by SBC we get the area of footing as 3.667 meter square. The size of this footing if you take root of it is 1.91 meter by 1.91 meter. So, let us provide a 2 meter by 2 meter footing. Second step so, we have to find out the reaction soil reaction from the factored load. Now, please remember while finding area of the footing we have not taken the factored load we have taken given load as it is with a self weight of footing because the this is a serviceability condition. So, since it is a serviceability condition the factor for serviceability is only 1. So, therefore, here we are not multiplied load by 1.5 we have taken as it is. So, whereas, when we go for design we require a factor load. So, for design soil reaction from the factored load we calculate factor load 1.5 times P it is 1.5 times 600 divided by B square that is the area of the footing it is 2 meter by 2 meter we get 0.225 Newton per mm square as the upward pressure. Now, we shall we should find out the depth of footing from the depth of footing is calculate done from three consideration from the consideration of single shear. The depth will be found and this particular depth will be checked for 2A shear and this is also checked for bending. So, the critical section is at a distance d from the face of the column for as shown in figure for a single shear now this is a plan of the footing. So, here the critical section for the single shear is at a distance d from the face of the column. Now, d is the effective depth of the footing and B is the breadth of the footing which is 2000 mm small b is the width of the column that is 300 mm. Now, v u is the soil pressure from the shaded area. Now, the shaded area is shown in figure here. So, this is the shaded area. So, total distance that is this is 2000 from that I have to 2000 minus I have to make this particular the size of the column which I have considered and then afterwards. So, we have to find out what is this particular shaded area. So, the shaded area is calculated. So, it is 2000 minus 300 divided by 2 minus this particular d that is the distance from the face of the column. So, therefore, here we will get q u into b that is b is the total width of the footing then b minus b divided by 2 minus d. So, that will give you the total this particular soil pressure from the shaded area it was out to be 450 into 180 minus d. So, assuming 0.2 percent steel for m 20 concrete the tau c is determined from table 19 of IS 456 2000 tau c was out to be 0.32 newton per mm square. So, minimum depth required is tau c bd is equal to v u. So, tau c bd if we equate to v u we get minimum depth required will be equal to 351 mm provide a effective depth of 360 mm. So, therefore, so now let us check that particular depth whether it is safe or not from bending moment consideration this is checked from bending consideration. So, m u limit is 0.36 fck b into x u limit into d minus 0.4 to x u limit. So, this is as per g 0.1 0.1 c that is maximum moment carrying capacity of the footing in bending. So, x u limit by d is 0.48 for fe 4 and 5. So, therefore, m u limit was out to be 0.138 fck bd square. So, if we substitute 0.138 fck is 20 b is 2000 d square is 360 square. So, it was out to be 815.2 into 10 to the power of 6 Newton mm. So, m u is calculated at the face of the column as shown in figure 2. Now, where will be the maximum bending moment for the footing slab? Now, footing slab is nothing but a two way cantilever. So, therefore, can you tell me where will be the maximum bending moment at the footing slab? The maximum bending moment of the footing slab is at the face of the column here at the face of the column it is maximum because it is a cantilever. So, therefore, at the face of the column we have a maximum bending moment. Now, this is a critical section for maximum bending moment. So, m u is calculated it is moment of the forces under shaded area about x x. It is q u into b into b minus b by 2 into b minus b by 4 it is at the center of this b minus b by 2 is this particular distance. So, it is q into b into b minus b square divided by 8. So, that will give us a bending moment of 162.563 into 10 to the power of 6 Newton mm which is less than m u limit which we have calculated. What is m u limit we have calculated? So, m u limit we have calculated is 815.2 into 10 to the power of 6. So, which is greater than this particular value of 162.563. So, therefore, depth provided is sufficient. So, now, let us check for Tuasier. The critical section for Tuasier is at a distance d by 2 from the face of the column. So, it is shown on the in the figure over here figure 3 shows critical section for Tuasier. And the critical section for Tuasier is this one. So, this is critical section for Tuasier. So, here it is at a distance d by 2 from the face of the column here, here also it is at a distance d by 2. So, therefore, the perimeter or the each side of a critical section works out to be b plus d. So, this is also b plus d and this side is also b plus d. Now, this particular to find out what is the shear force along this particular critical section. So, we have to find out the upward pressure in the shaded area for Tuasier stress we have to calculate. The Tuasier stress is given by upward pressure in the shaded area. So, divided by area of the critical section. So, area of the critical section consists of perimeter of the critical section into the depth effective depth. So, it is this is the perimeter that is 4 into b plus d. So, that was what we 26440 mm. So, 2640 mm into effective depth is 360. So, that will give us the this is the perimeter of the critical section into the depth effective depth that will give us the area resisting this particular shear critical shear. So, critical Tuasier is the area shear force in this in this shaded area. Now, that we will calculate so, 0.225 into 2000 by 2000 this is this will give us the total downward load minus 660 into 660 that is the area of the this particular breadth of the inner portion. So, that will give us the total downward load divided by this perimeter into depth that will give us the Tuasier stress as 0.844 Newton per mm square. So, the maximum shear stress permitted for as per to with respect to I is 4562000 is 0.25 times root fck. So, 0.25 times root fck is 20 it is 1.118 Newton per mm square whereas, we are getting 0.844 as Tuasier. So, therefore, this is depth 360 mm is sufficient from consideration of Tuasier. So, now, Mu we have to you use this particular with equation g 0.1.1 b from is 4562000 to find out the area of steel we equate Mu root to the this AST into BD into D into 1 minus AST into Fy upon BD fck. So, we equate it we find out area of steel it is 1299 mm square. And using 12 mm bars spacing of the bars it is area of 1 bar divided by area required into 2000 it was sort to be 174 mm centre to centre. So, provide 12 mm bars at 170 mm centre to centre. So, that is the main steel and since it is a 2 AST provided the percentage steel provided was sort to be 0.185. So, Tuasier as from is 456 Tuasier was sort to be 0.308 Newton per mm square and Tuvi it is VU upon BD it was sort to be 0.306 which is less than the Tuasier max or Tuasier therefore, therefore, no we can provide the reinforcement. The development length also we should check so, which is less than Tuasier 0.3606 is less than Tuasier therefore, development length Tuvi D it is M 20 concrete and Fe 4 and 5 it is 1.2 times 1.6 1.92 so, LD provided it is 564. So, this much length is available beyond the phase of the column hence it is ok. Now, this is the reinforcement arrangement this is the main steel and the perpendicular to it we find the another that is also main steel because it is a 2 way cantilever. So, both are 12 mm torque HYSD bars 170 mm centre to centre both way these are references used for the preparation and thank you one and all.