 Hello friends, I am Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Singapore. So, today I am here to explain you about the expression for the crippling load when both ends of the columns are hinged using Euler's column theory. The learning outcomes of today's lecture is the students will be able to study the concept of long column and short column and the assumptions made in Euler's column theory. They will also be able to derive the crippling load of the column with both ends hinged. So, what is compression member or column or a strut? So, these are the member in a structure which are subjected to axial compressive load. So, the column will be the structure which is vertically and with both ends fixed and subjected to axial compressive load. Whereas, strut will be the member which is not vertical and one or both of its ends are hinged or pin jointed which is also subjected to axial compressive load. The difference between long column and short column. A column is said to be long when the length upon least lateral dimension is greater than 12. So, least lateral dimension means if you take the cross section of the column, so it will be breadth and depth. So, whichever is the least in that that will be taken as least lateral dimension. So, whereas short column will be when the length divided by least lateral dimension is less than 12. So, generally the long column fails by buckling. So, it is subjected to buckling stresses whereas short column fails by crushing and it is subjected to crushing stress. The radius of gyration in case of long column is lesser than that of short column whereas load carrying capacity is lesser for the long column and greater for the short column. Now, you can see the failure mode in case of long column and short column. So, in case of long column the failure takes place by buckling. So, this is the how it buckles whereas in short column the failure takes place by crushing. So, here the crushing occurs and in case of brittle material a fracture takes place, sudden fracture will take place. So, the assumption made in Euler's column theory is that the column is initially perfectly straight and load applied is axial. The cross section of the column is uniform throughout its length. The column material is perfectly elastic, homogeneous, isotropic and obeys Hooke law that is stress is proportional to strain within elastic limit. The length of the column is very large compared to its lateral dimension. So, that we have seen that length upon least lateral dimension it is greater than 12. The direct stress is small compared to the bending stress and the column will fail by buckling alone. The self weight of column is negligible. Now, how the failure of column takes place? So, in case of short column the failure takes place by direct compressive stress as it is not a long column it fails by crushing. So, the crushing stress or compressive stress is compressive load divided by area whereas in long column the failure takes place by buckling stress. So, buckling stress sigma b is equals to p into e by z where e is the eccentricity of buckled column and z is i upon y that is section modulus whereas in between the long column and short column the failure may takes place by combined effect of direct compressive and buckling stresses. Sign convention a moment which tends the column to have a convexity towards its initial center line is taken as positive and the moment which tends to bend the column we having concavity towards the initial center line is taken as negative. Now, what is crippling load? Here pause the video and try to write answer on a paper. The load at which the column just buckle is known as crippling load or it is also known as critical just or it is also known as buckling load. Now, for the expression of crippling load when both ends of the columns are hinged. So, we will assume or consider a column a b having a length of l and cross sectional area a which is hinged at a and b and the p is the crippling load which will just buckle the column and due to this crippling load the shape of the column will be deflected shape of the column will be a c and b. Now, we will consider a section at a distance of x from the end a and at that point the deflection is y. So, the moment due to the crippling load at this section is p into y. So, minus p into y. So, negative sign because this concavity is towards the center of the column. So, we are giving your negative sign, but the mathematically moment is given by e i d 2 y by dx square. So, equating this both moment. So, what we will get e i d 2 y by dx square is equals to minus p into y. Therefore, e i d 2 y by dx square now taking this term on the left hand side it will become positive plus p y is equals to 0. Therefore, dividing this whole term by e i I will get d 2 y by dx square plus p upon e i into y is equals to 0. So, this equation can also be written as d 2 y by dx square plus alpha square y is equals to 0 that is equation a where we are considering alpha is equals to p upon e i. Therefore, alpha square is equals to p upon e i. Therefore, alpha is equals to under root p upon e i. So, the solution of this equation a is y is equals to c 1 cos alpha x plus c 2 sin alpha x. Therefore, y is equals to now putting the value of alpha. So, y is equals to c 1 cos into brackets. So, x term is as it is and alpha is under root p upon e i plus c 2 sin and x term is as it is under root p upon e i we have replaced this alpha with under root p by e i. So, this is equation number 1. So, now we know the end condition. So, at x is equals to 0. So, deflection y is 0 at a. So, putting this in equation 1 we get. So, now y is equals to 0. So, c 1 cos. So, now we have put x is equals to 0 means this whole term will become 0 that is cos 0 plus c 2 sin again this term will become 0. So, now this is equals to 0 is equals to c 1 cos 0 is 1 plus c 2 sin 0 is 0. So, we are getting c 1 is equals to 0 that is equation number 2. Now, again we know one more end condition that is at x is equals to l at b point the deflection is again 0. So, putting that in equation 1 we get 0 is equals to c 1 cos. Now, instead of x here I have put l into under root p by e i plus c 2 sin l under root p upon e i. So, now as c 1 is 0 this term will become 0. So, 0 is equals to 0 plus c 2 sin l under root p by e i that is equation number 3. So, now from this equation we know that for the left hand side to be 0 either c 2 must be 0 or sin l under root p i should be 0. So, as c 1 is equals to 0 and if we again put c 2 is also 0. So, from equation 1 we get y is equals to 0 this means that bending will be 0 in this column. So, which is not true because. So, there is a deflection of the column. So, therefore, the term second term that is sin into bracket l under root p a upon e i should be equal to 0. So, now when this term will become 0 if sin 0 sin 0 it is equals to 0 or for sin phi pi is equals to 0 or sin 2 pi it is 0 or sin 3 pi. So, for this value the sin is 0. Therefore, this term l into under root p upon e i it should be either 0 or pi or 2 pi or 3 pi. Now, taking the least practical value from this. So, l under root p upon e i is equals to pi. So, now squaring this and solving we get p is equal to 0. So, this is equal to pi square e i upon l square. So, this is the crippling load for the column with both ends hinged. So, these are my references which I have referred. Thank you. Thank you very much for watching my video.