 the design of slab base, we will see the design procedure as well as we will see one example, so how to design our slab base. If you want to see the theoretical background of the slab base, we have I have another video on this, so you can refer that video for the theoretical background of the slab base. Learning outcome of the is that at the end of the session students will be able to design the slab base. We have discussed all these procedure in our previous videos, just now I will just we will take that and overview all step overview of the steps. You can find all these steps in my previous videos also previous video also. First of all we will assume our grade of concrete then we will find out the bearing strength of concrete which is equal to 0.45 fck, we get the value of bearing strength of concrete which is a stress bearing stress bearing strength, now we will get the value of area which is equal to p divided by the bearing strength of concrete this is 0.45 fck, so load from column divided by the bearing strength of concrete will get the required area. These are the various projections of the slab base length width depth of column longer projection smaller projection design procedure. If we assume the square base plate we will find out the value of l and b which is equal to under root a and b to achieve economy we will keep both the projections equal and we will find out the value of a which is equal to b by this equation d plus 2b into bf plus 2a which is equal to a we will find out a and b by this equation. Now intensity of pressure w can be find out by p divided by a1 where a1 is the area of the base plate provided in mm what actually we are providing is a1, now the minimum thickness of the slab base is calculated from equation number 6 it should not be less than the thickness of the column flange. Then nominal we are providing holding down bolts which are 2 or 4 in numbers of 20 mm when we are when the base plate is subjected only to the axial compressive load will provide only 2 bolts will be it will be enough we will see one example the question is design a slab base for a column section is hb 350 at 710.2 Newton per meter subjected to an factored axial compressive load of 150 kilo Newton load is transferred to the base plate by direct bearing of column flanges the base rest on concrete pedestal of grade m20. And solution is first of all we will assume grade of steel which is equal to Fe410 and for Fe410 Fe which is equal to ultimate capacity of the steel which is equal to 410 mega Pascal yield strength equal to 250 mega Pascal and for m20 grade concrete bearing strength of concrete as per code is equal to 0.45 fck which is equal to 9 Newton per mm square. As per is 800 2007 the partial safety factor for material is equal to 1.1 and for weld it is equal to 1.25 for shop welding. Now from is handbook we will take down the properties that is properties of is hb 350 thickness of flange tf is equal to 11.6 mm thickness of web tw is equal to 10.1 mm depth of section 350 mm width of flange bf equal to 250 mm square. Now first of all we will calculate the area of the slab base a is equal to load in Newton divided by the bearing pressure in Newton per mm square we will get the value in mm square we will convert it into meter square. And let us provide a rectangular base plate of length of size l by b and we will assume a equal to b equal overhang. Now we have to find out this overhang we have got this value area of the base plate then we will get l from our handbook 350 that is the depth of column plus 2 a is unknown we are going to find this a multiplied by b is the width of flange 250 and we assume a equal to b so a equal to 55.24 mm we will round off it to 60 mm and we will get length as 470 mm and width as 370 mm. Now the bearing pressure of concrete is determined by w is equal to p by a1 which is equal to 150 multiplied by 1000 and actual area what we have provided. Now the value comes out to be 8.62 Newton per mm square which is less than 9 Newton per mm square which is alright. Okay what is mean by this 9 what is 9 we have given the grade of concrete as 20 Newton per mm square and for 20 Newton per mm square the bearing strength of concrete comes out to be 9 Newton per mm square which is equal to 0.45 fck 0.45 into 20 which is equal to 9 Newton per mm square which is greater than 8.62 so it is safe. Now to find out the thickness T s which is equal to under root 2.5 w into a square minus 0.3 b square into gamma m naught upon f y this equation w is the bearing pressure 8.62 is the smaller projection b is the longer projection but here in case we have kept the projections equal 60 and 60 gamma m naught 1.1 f y 250. So 2.5 8.62 60 0.3 60 1.1 250 the thickness comes out to be 15.458 mm we will provide it as 16 mm. First of all we have seen that clause it should be greater than the thickness of flange. Now what is the thickness of flange for I s H b 350 11.6 greater than thickness of flange therefore safe. Hence provide a base plate of 470 by 370 by 16 mm size here I have provided a 470 by 370 60 60 I have provided 2 volts of 20 mm y you will see here. In the question we have given the load is transferred to the base plate by direct bearing. This implies that the column end and base plate have been machined for perfect bearing. Direct bearing we have taken perfect bearing. Also we have not given any bending moment in the question therefore connection of the column with base need not be designed. However to keep the column in position we need some cleat angles. So we will provide 2 cleat angles of nominal size 55 by 55 by 8 you can take another size also it is a nominal size 55 by 55 by 8 is provided for connecting the column flanges with the base plate as shown in the given figure this is the cleat angle. We are providing we have to connect this flange to the base plate by the help of this cleat angle. We are provided with 2 volts of 20 mm size. Now if there are some review question you can pause the video and answer these questions. The very first question is the bearing strength for M25 grade of concrete is just now we have seen for M20 grade concrete now we have to say what is the bearing strength of concrete for M25 grade of concrete. And next question is calculate the area required for the slab base to carry an axial compressive load of 1000 kilo Newton we take grade of concrete as M20. So you have to calculate the area these are the answers the bearing strength of concrete for M25 grade is equal to 11.25 Newton per square how it comes which is equal to 0.45 fck 0.45 into 25 area is nothing but load divided by 0.45 fck. So load is 1000 into 1000 divided by 0.45 into 20 we will get it as 0.1111 meter square these are my references thank you.