 I am Prashant Vishwanath Dinshati, assistant professor, department of civil engineering from Walchand Institute of Technology, Singapore. Today I am here to explain you about the expression for the crippling load when one end of the column is fixed and other is free using Euler's column theory. The learning outcome of today's lecture is student will be able to study the failure of long columns. They can derive the crippling load for column when one end is fixed, other end is free. The compression member, so the compression member may be called as column or a strut, so which is subjected to axial compressive load. So we can have this column is a structure which is vertical and both its end fixed which is subjected to axial compressive load. Now how the failure of the column takes place? So either it takes place by direct compressive stress which is occur in case of short column that is sigma c is equals to p c upon a that is compressive stress is equals to compressive load divided by cross sectional area whereas long column fails by buckling stress that is sigma b is equals to p into e by z where e is the eccentricity and z is equals to i by y. So now some column may be failing either with the combined effect of direct compressive and buckling stress when we will call it as a long column features of long column. So long column is the column which is having length to least lateral dimension ratio greater than 12 and it generally fails by buckling and it is subjected to buckling stress whereas radius of gyration is less for long column and load carrying capacity is lesser. So the failure mode is by buckling. Sign convention, so a moment which will bend the column with it convexity towards its initial center line of the column is taken as positive and the moment which will bend the column with the concavity towards the initial center line is taken as negative. Failure of column, how does the failure occurs in long column and short column? Here pause the video and try to write answer on a paper. Long column fails by buckling whereas short column fails by crushing. The expression for crippling load when one end of the column is fixed other end is free. So now we will consider a column of ab having the length l and cross sectional area a which is fixed at end a and free at end b and p is the crippling load applied. So after applying the crippling load the column will deflect in this shape that is ab dash and we will consider a section at a distance of x from a where the deflection of the column is y. Now the moment due to the crippling load, so now this crippling load will be acting at this point, so this load into this distance will be the moment. So p into a minus y, so the moment is also given by this formula e i d 2 y by dx square. So now equating this both moment we will get e i d 2 y by dx square is equals to p into bracket a minus y, so this multiplying this into the bracket we get p into a minus p into y. So now taking this negative term on the left hand side I will get e i d 2 y by dx square plus p y is equals to p a. Now this can this is divided by e i, so what I will get is d 2 y by dx square plus p upon e i into y is equals to p upon e i into a. Now this equation can be written as d 2 y by dx square plus alpha square y is equals to alpha square a where alpha square is equals to p upon e i therefore alpha is equals to under root p by e i. The solution of this equation is y is equals to c 1 cos alpha x plus c 2 sin alpha x plus alpha square a upon alpha square. So now putting again the value of alpha we will get y is equals to c 1 cos x. So now instead of alpha here I am putting its value that is under root p upon e i plus c 2 sin x under root p upon e i plus a. So this we will treat it as equation number 1. Now we know the boundary or end condition at x is equals to 0 y is also 0 means deflection at a point is 0. Putting this in equation number 1 we get. So now y is equals to 0 and now x term is 0 means this term will become 0 and this term will also become 0. Therefore 0 is equals to c 1 cos 0 plus c 2 sin 0 plus a. Now cos 0 is 1 and sin 0 is 0 so this term will become 0. So from this we get c 1 is equals to minus a that is equation number 2. Now one more condition we know that at x is equals to 0 dy by dx slope is also 0 at fixed end. So this is the equation. So we have to find now dy by dx. So from equation 1 differentiating we will get dy by dx is equals to c 1 minus sin x under root p upon e i into under root p by e i plus c 2 cos x under root p upon e i into under root p by e i. Now plus 0 the differentiating or derivative of constant is 0. So dy by dx is equals to minus c 1 under root p upon e i sin into bracket x under root p upon e i plus c 2 under root p upon e i into cos x under root p upon e i. So this we will treat as equation number 2. Now putting x is equals to 0 and dy by dx is equals to 0 in this equation number 2. So we will get 0 is equals to minus c 1 under root p upon e i sin 0 as x term is 0 this will become 0 plus c 2 under root p upon e i cos 0. Now as c 1 is equals to minus a and now here 0 is equals to minus c 1 under root p upon e i into sin 0 plus c 2 under root p upon e i into cos 0. Now cos 0 is 1 sin 0 is 0 so this term will become 0 therefore c 2 will so for this term now left hand side is 0 now the remaining term is c 2 under root p upon e i. So now here either c 2 must be 0 or under root p upon e i must be 0 but for crippling load you are having a value of p therefore under root p upon e i cannot be equal to 0 therefore c 2 will be equals to 0. Now again substituting c 1 is equals to minus a and c 2 is equals to 0 in equation number 1 we get y is equals to minus a cos into bracket x under root p upon e i plus a this is equation number 3. Now again one more condition we know that at x is equals to l the deflection y is equals to a. Now putting this in equation number 3 so here instead of y I am putting here a is equals to minus a cos l under root p upon e i plus a. So now here the x value is replaced by l so now this positive a I will take on the left hand side so a minus a it will become 0 is equals to minus a cos into bracket l under root p upon e i. So now for this term to be a 0 either a should be 0 or cos into bracket l under root p upon e i should be 0 but a cannot be 0 because due to the crippling load there is a deflection at the free end that is a. So therefore a cannot be equal to 0 therefore the second term that is cos l under root p upon e i should be equal to 0. Now this is equal to 0 where for the value of cos pi by cos pi by 2 or cos 3 pi by 2 or cos pi by 2. So therefore this term l under root p upon e i should be either equals to pi by 2 or 3 pi by 2 or 5 pi by 2. Now taking the least practical value of this so we will get l under root p upon e i is equals to pi by 2. So now taking this l on the right hand side we will get this squaring both side. So we are getting this equation p upon e i is equals to pi square upon 4 l square therefore p is equals to pi square e i upon 4 l square this is the crippling load. Now what is the effective length of column with one end fixed and other end free? Now if you see when both ends of the column are hinged so actual length p is equals to pi square e i upon l square. So if we are using effective length so then also it is pi square e i upon l effective square. Therefore l effective is equals to l. Now for the condition one end fixed and other end free now just now we have derived p is equals to pi square e i upon 4 l square whereas by Euler's it is p is equals to pi square e i upon l effective square. So here l effective is equals to 2 l. So these are my references which I have referred. Thank you. Thank you very much for watching my video.