 Hello friends. Today we are going to see design of compression member. A little brief introduction about myself, myself, Kaji, Syed, Sujad Ali. Currently working as assistant professor in civil engineering department, Valjan Institute of Technology. So, the learning outcome of this session is at the end of this session students will be able to design the members which are subjected to axial compression. What is compression member? Basically the member which is generally subjected to compressive force is nothing but compression member. So, this is the definition of compression member members which are primarily subjected to compressive force are nothing but compression members. This diagram represents one there is one short column and another one is long column. What is mean by short column? A column whose cylindriness ratio is less than 12 and what is long column? The column whose cylindriness ratio is greater than 12. Basically short column will fail by crushing and long column will fail by buckling. Now, we will see there are various nomenclatures for the compression members. The same member which carries the vertical compression compressive force in building is called as columns or posts. The same compression member in trusses is called as struts. The same compression member in words of crane is called as boom. How to select the shape of compression member? Basically the member to basically will select the member in such a way that will get the maximum movement of inertia. Ideally for the same cross sectional area the circular sections are the best sections. After that we are having square tubing after that we are having rectangular section. So, for the design of beams will select an i-section and for the design of struts the members which carry compressive force in trusses are nothing but the struts will select single angle section as well as double angle section either of the two. But we will select the equal angle, equal single angles because we will get same radius of gyration along both the axis. Ideally the member for same cross sectional area is the best is our pipe section because we are getting the maximum movement of inertia for the pipe section. But due to the difficulty in connection because we get some difficulty in connecting this pipe section we will select square tubing or rectangular tubing but due to the advancement in building nowadays we can select these sections suitably. Now how to select the effective length of compression member? This table is given in IS 800 2007. It is given in table number 11. So, for different boundary conditions we are having the different values of effective length. At one end we are having translational restrain, rotational restrain. At the other end we are having translational restrain free rotation free. This represents a one end fix other end free type of sub that column whose effective length will be equal to 2L. Next we will see the design steps for compression members. The following are the usual steps in the design of compression member. First of all we will assume the design stress in the compression member. Then we will calculate effective sectional area. Now how to calculate effective sectional area? It is given by the formula that is area will be equal to load divided by stress. The load will be given to us. We will get the value of load and we will assume that design stress and with the help of this we will calculate area. Then we will select a section from IS handbook number 1 and we will calculate the value of r minimum. Then depending upon the boundary conditions we will calculate the effective length and by the use of this effective length and this r minimum we will calculate slenderness ratio kL by r. General formula for slenderness ratio is kL by r. kL is the effective length and r is the radius of gyration. Then we get the value of this slenderness ratio with the help of this slenderness ratio by using the IS table number 9abc. Depending upon the buckling class we will calculate the value of design stress and with the help of this design stress we will calculate the value of the load carrying capacity of the section. Then if the load carrying capacity is less than the factored load that is the given load our section means our section fails or if the load carrying capacity is very much higher than this design load the section will become uneconomical then we have to revise the section. This is how we have to assume the slenderness and this is how we have to assume the design stress depending upon the slenderness ratio for the angle stress the value is assumed to be around 90 mega Pascal. We will see a small numerical example for the design so that we will get a clear understanding of the design of compression member. The question is design a single angle strut connected to the gusset plate to carry 180 kilo Newton factor load the length of the strut between center to center connection is 3 meter. Given the data is goes somewhat like this we have given the single angle section then we have given the factored load after that we have given the support condition center to center connection is 3 meter. Basically first of all we will assume the value of fcd which is equal to 90 mega Pascal then with this with the help of this fcd we will calculate the area required which is equal to 180 kilo Newton divided by the stress we will get around 2000 mm square with the help of is handbook number one or the steel table we will select a suitable angle section. Here we here I have selected 90 by 90 by 12 which is having an area of 2019 mm square from that I got the value of rvv which is equal to 17.4 mm. For the continuous member we will get the value of effective length will be equal to 0.85 into 0.85 L which is equal to 0.85 into 3000 which is equal to 2550 mm. This is given in is 800 2007 for the angle strut. Then we calculate the value of cylinder ratio this effective length divided by r will get the value of cylinder ratio. Now by the use of this is table number 9 c will get the value of fcd. Basically we are using single angle section that is why we are using the buckling class c this is given in is table number 10 which buckling class we have to select then we get the value of fcd which is equal to 61.62 MPa then we calculate pd which is equal to a into fcd but this value comes out to be less than 180 then we have to revise the section. Now we will select some higher sections we will try here 130 by 130 by 8 we have this area we have this rvv we get cylinder ratio similar steps. Now the value of fcd as per table number 9 c of is 800 2007 equal to 107 Newton per mm square the carrying capacity load carrying capacity is equal to 216 kilo Newton which is greater than 180 kilo Newton therefore it is safe. Now you these are some review question you can pause the video and answer these questions the answers for the questions are okay. So these are my references thank you.