 I am Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Singapore. Today I am here to explain about the expression for the crippling load when one end of the column is fixed and other is hinged using Euler's column theory. The learning outcome of today's lecture is student will be able to derive the expression for the column when one of its end is fixed and other is hinged. Now, what is a compression member? The compression member may be a column or a strut depending upon its position, so which is subjected to an axial compressive load whereas column is a structural member held vertically with its both ends fixed rigidly subjected to axial compressive load. The failure of column may occur due to direct compressive stress, buckling stress or combined effect. So, when the column is a short column, when the length upon least lateral dimension is less than 12, it is treated as a short column and it fails generally by direct compressive stress which is given by sigma c is equals to p c by a that is load in compression. And whereas long column if length upon least lateral dimension is greater than 12, it is called as a long column it generally fails by buckling stress where sigma b is equals to p into e where small e is the eccentricity divided by section modulus whereas in between the column may fail by the combined effect of direct compressive stress and buckling stress. Sign convention a moment which will bend the column with its convexity towards the initial position is taken as positive and the moment which will bend the column with concavity towards the initial position is taken as negative. End condition for column state the four types of end conditions for the column here pause the video and try to write answer on a paper. The first end condition is both the ends of the columns are hinged. One end is fixed, other end is free, third is both the ends of the columns are fixed, fourth is one end is fixed and other is pinned or hinged. The crippling load when one end of the column is fixed and other is hinged, here you can see a column which is having length l and cross sectional area a which is fixed at end a and hinged at b and it is applied with the crippling load p. Due to this crippling load the column will deflect like this and as the end a is fixed there will be a fixed end moment that is m 0 which will try to bring back the slope of the column to 0 at a. Now to balance this moment m 0 we require a force horizontal force h which is shown here. So this is a horizontal force which will counter balance this moment. Now we will see the section at this point. So let it be at a distance of x from a and the deflection or lateral displacement b y. So the moment at this point will be minus p into y because this p is now it is anticlockwise. So I am taking negative for anticlockwise positive for clockwise and plus h into l minus x. So this horizontal force h into this distance that is l minus x and this forms a clockwise moment. So this moment as mathematically the moment is given by e i d 2 y by dx square. So equating this moment we will get e i d 2 y by dx square is equals to minus p into y plus h into bracket l minus x. Now I will take this minus 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 h into bracket l minus x. Now dividing all by e i so I will get d 2 y by dx square plus p upon e i into y is equals to h into bracket l minus x upon e i. Now putting p upon e i is equals to alpha square. Therefore alpha is equals to under root p by e i. So now putting this alpha square and alpha I will get this equation that d 2 y by dx square plus alpha square into y is equals to h into bracket l minus x upon e i. So the solution of this equation is y is equals to c 1 cos alpha x plus c 2 sin alpha x plus h into bracket l minus x upon e i alpha square. So again now putting the value of alpha is equals to under root p by e i and alpha square is equals to p upon e i. So I will get this equation. So here cancelling this e i and e i and putting the value of alpha I will get y is equals to c 1 cos into bracket x under root p upon e i plus c 2 sin into bracket x under root p upon e i plus h into bracket l minus x upon p. So this we will treat it as equation number 1. Now we know the end condition at A as it is fixed. So at distance x is equals to 0 the deflection is 0 that is y is 0 and also the slope dy by dx is also 0 and when x is equals to l it is hinged. So there the deflection is again 0. Substituting the value of x is equals to 0 y is equals to 0 in equation number 1 which I have just now we have derived. So putting this value x is equals to 0 so this term will get 0 and again here putting this x is 0 so this term also becomes 0 and here again putting x is equals to 0. So I will get 0 is equals to c 1 cos 0 plus c 2 sin 0 plus h l upon p. So now cos 0 is 1 sin 0 is 0 so the second term will become 0 so I will get c 1 plus h into l upon p is equals to 0. Therefore c 1 is equals to minus h upon p into l so this we will treat it as equation number 2. So now again we want the slope to be found out here so again I am differentiating this equation number 1. So dy by dx is equals to c 1 so cos of this term is minus sin of this term into this value constant value. So it is c 1 into minus 1 sin x under root p upon e i bracket close into under root p by e i plus now the derivative of sin is cos x plus c 2 cos into bracket x under root p upon e i into root p upon e i minus h upon p. So now here h l upon p that will be a constant it will be 0 then minus h into x upon p so it will be minus h upon p. So now substituting the value of at x is equals to 0 dy by dx is equals to 0 in this equation. So again now x term this term will get 0 and again this term will also get 0. So 0 is equals to minus c 1 into so sin of 0 is 0 and plus c 2 so this cos of 0 is 1 into under root p upon e i minus h upon p. So 0 is equals to c 2 under root p upon e i minus h by p. So I will take on that side so it will become positive c 2 under root p upon e i is equals to h by p therefore I will get c 2 is equals to h upon p under root e i upon p. Now substituting the value of c 1 is equals to minus h upon p into l and c 2 is equals to h upon p under root e i by p in equation 1. So I will get this equation this equation number 1 so now instead of this c 1 I will put this value that is c 1 what I have found out. So y is equals to minus h upon p into l cos into bracket x under root p upon e i plus h upon p under root e i by p into sin into bracket x under root p upon e i bracket close plus h upon p into bracket l minus x. Putting the second end condition y is equals to 0 at x is equals to l in this above third equation. So now substituting this value y is equals to 0 and instead of x here I am putting the value l so I am getting this equation. So now putting this value I will get 0 is equals to minus h upon p into l cos into bracket l under root p upon e i plus h upon p under root e i by p into sin bracket l under root p upon e i plus 0. Now this term will become 0 into 0 this value it will become 0. So now again I am taking this negative terms on the left hand side so I will get this h upon p l cos into bracket l under root p upon e i is equals to h upon p under root e i by p into sin l under root e i by p into root p upon e i. So now again I am taking this term on the left hand side so it will become multiplication so denominator will go into numerator so this p p will get cancel h h will get cancel so what I will get is l under root p upon e i cos into bracket l under root p upon e i is equals to sin l under root p upon e i. So now I will bring this cos term here so it will become in the denominator so sin upon cos it will become tan so l under root p upon e i is equals to tan into bracket l under root p upon e i. The solution of this above equation the tan value at 4.5 radians it will be equal to same that is so l upon under root p upon e i is equals to 4.5 radians. So now squaring both sides l square p upon e i is equals to 4.5 square so it is 20.25. So 20.25 is nearly equal to 2 pi square so putting that value I will get p is equals to 2 pi square e i upon l square which is the crippling load. So these are my references which I have referred thank you thank you very much for watching my video.