 Hello friends, 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 both ends of the column are fixed using Euler's column theory. The learning outcome of today's lecture is that students will be able to derive the crippling load when both ends of the columns are fixed. So, the compression member it is a member of a structure which is subjected to axial compressive load. Column is the structure which is vertical and both its ends are fixed rigidly while subjected to axial compressive load. How the failure of column may occur? The column failure may occur by direct compressive stress, so this case will be majorly found in case of short column where length upon least lateral dimension is lesser than 12. So, it fails by crushing and whereas long column it fails by buckling if the length upon least lateral dimension is greater than 12 it is called as long column and it will fail due to buckling stresses that is sigma b is equals to load into eccentricity divided by z. So, in between the column long and short they may fail by combined effect of direct compression and buckling stress. Sign convention a moment which will bend the column with its convexity towards the initial position of the column is taken as positive whereas the column or the moment which will bend the column with its concavity towards the initial centre line is taken as negative. Now, what is column and strut? So, explain the difference between column and strut here pause the video and try to write answer on a paper. So, as we know that column is a compression member which is vertical and both its ends are fixed rigidly while subjected to axial compressive load. Now, what is strut? So, strut is also compression member but is not vertical and one or both of its ends are hinged or pinned while subjected to compressive load. Now, to find the crippling load when both ends of the column are fixed. So, we will consider a column of AB having length L with uniform cross sectional area A which are fixed at and A and B and a crippling load P is applied to the column which just buckles the column. So, due to that buckling the curvature of the column will be A, C and B. So, as the both ends are fixed it will be subjected to fix and moment that is M 0 at and A and B. So, this will be acting anticlockwise and here it will be acting clockwise moment. Now, we will consider a section at a distance of x from A which is having a deflection or y at this point. So, the moment due at this point due to the applied crippling load and moment is so M 0, M 0 is the moment due to fixity. So, this is clockwise and minus P into y. So, this crippling load P into this distance y it is acting anticlockwise. So, I am giving here a negative sign. So, again we can have the moment as E I d 2 y by dx square. So, now equating this both the moment we get E I d 2 y by dx square is equals to M 0 minus P into y. So, now I will take this negative term on the left hand side. So, I will get E I d 2 y by dx square plus P into y is equals to M 0. So, now 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 M 0 upon E I. So, now this equation can be written as d 2 y by dx square plus alpha square into y is equals to M 0 upon E I where alpha square is P upon E I or alpha is equals to under root P by E I. So, now the solution of this equation is y is equals to C 1 cos alpha x plus C 2 sin alpha x plus M 0 upon E I into alpha square. Now putting the value of alpha in this we will get y is equals to C 1 cos into bracket x under root P upon E I where alpha is equals to P upon under root P upon E I plus C 2 sin x under root P upon E I plus M 0 upon E I into P by E I. Now this E I E I will get cancel. So, we will get this equation that is y is equals to C 1 cos x under root P I plus C 2 sin into bracket x under root P upon E I plus M 0 upon P. So, this will be your equation 1. Now depending upon the end condition so we know that as and a is fixed at x is equals to 0. So, deflection at this point is 0 and also the slope that is d y by d x is also 0. Similarly, when x is equals to l at point b x is equals to l y is equals to 0 and d y by d x is also 0. So, now substituting the value that is x is equals to 0 y is equals to 0 y is equals to 0 in equation 1. So, I get y is equals to 0 C 1 cos now x term is 0. So, this whole term will become 0 it is cos 0 plus C 2 sin and this term it will become 0 sin 0 plus M 0 by P. So, cos 0 is having value 1 and sin 0 is having value 0. So, putting that I will get 0 is equals to C 1 plus M 0 by P therefore, C 1 is equals to minus M 0 by P this will be equation 2. So, now again for the slope I want d y by d x. So, this equation 1 differentiating I will get d y by d x is equals to C 1 minus sin x under root P upon E I into under root P by E I plus C 2 sin x under root P upon E I into under root P upon E I plus derivative of this constant is 0. So, what I will get is d y by d x is equals to minus C 1 sin into bracket x under root P upon E I into under root P by E I plus C 2 cos into bracket x under root P upon E I bracket close into under root P upon E I. Now substituting the value now we know that x is at x is equals to 0 d y by d x is also 0. So, putting this in above equation what I will get is 0 is equals to minus. So, now this sin 0 sin 0 is 0. So, I am putting here 0 and cos of 0 is 1. So, 0 is equals to minus C 1 plus C 2 into 1 under root P upon E I. So, this term as it is multiplying by 0 it will become 0. So, I will get 0 is equals to C 2 under root P by E I. Now if we see this equation. So, the left hand side is 0. So, when it will be 0 when either C 2 value should be 0 or under root P upon E I value should be 0. So, but for a crippling load of P the value of under root P upon E I cannot be 0. Therefore, naturally C 2 term must be equals to 0. So, putting C 2 is equals to 0. Now substituting C 1 is equals to minus M 0 by P and C 2 is equals to 0 in equation 1 we get y is equals to minus M 0 by P cos into bracket x under root P upon E I plus now 0 because C 2 term is 0 here plus M 0 by P. So, I will get y is equals to minus M 0 by P cos into bracket x under root P by E I plus M 0 by P. So, this is equation number 3. Now, again we know the end condition at x is equals to l y is again 0. So, now putting this value in equation number 3. So, what I will get is y is equals to 0 and here x is again l. So, instead of x I am putting here l. So, 0 is equals to minus M 0 by P cos of l under root P by E I plus M 0 by P. So, now this negative term I will take it on the left hand side. So, I will get M 0 by P into cos of under bracket l under root P by E I is equals to M 0 by P. So, now this M 0 by P here. So, I am taking on the right hand side this P will get multiplied and M 0 will get divided. So, putting that value cos of what I will get is cos into bracket l under root P by E I is equals to 1 because M 0 M 0 and P P will get cancelled. So, I will get 1 here. So, the cos of l under root P by E I is equals to 1. So, when this cos value will be 1. So, for cos 0 you will be having value 1 or cos 2 pi you will be having 1 cos 4 pi you will be having 1 or cos 6 pi like this you will be having 1. So, now putting this term that is l under root P by E I it should be either equals to 0 or 2 pi 4 pi 6 pi. Now, taking the least practical value that is 2 pi. So, equating it l into under root P by E I is equals to 2 pi. Now, squaring both the sides what I will get is l square into P upon E I is equals to 4 pi square. So, from this I will get P is equals to 4 pi square E I upon l square. So, now this is the crippling load when both ends of the columns are fixed. Now, what will be the effective length of the column with both end fixed? So, now if you see the condition when both ends are hinged. So, the crippling load is P is equals to pi square E I upon l square whereas, Euler's formula is also P is equals to pi square E I upon l effective square. So, now comparing this relation we get l effective is equals to l. Now, when both ends of columns are fixed. So, this P is equals to 4 pi square E I upon l square whereas, Euler's load that is P is equals to pi square E I upon l effective square. Now, comparing this what we will get l effective is equals to l by 2. These are my references which I have referred. Thank you. Thank you very much for watching my video.