 Hello friends, I am Prashant Vishwanath Dinshiti, assistant professor, department of civil engineering from Walchand Institute of Technology, Singapore. Today I am here you to explain about the Rankine's formula for the crippling load. The learning outcome of today's lecture is student will be able to derive the crippling load of column by using Rankine's formula. The compression member it may be a strut or a column which is subjected to axial compressive load whereas column is a structural member which is vertical and both its ends are fixed while subjected to axial compressive load. The failure of columns may be due to direct compressive stress if the column is a short column. The short column is the column when the length to its least lateral dimension is lesser than 12 and the long column will fail by buckling stress. So long column will be the column when the ratio of length to least lateral dimension is greater than 12. So the column in between long and short it may fail by combined effect of direct compressive stress and buckling. Crippling stress in terms of effective length and radius of gyration. The moment of inertia can be returned as i is equals to a k square where k is the radius of gyration and a is the cross sectional area, i is the least moment of inertia hence k will be also the least radius of gyration. We know the Euler's formula for the crippling load it is given by p is equals to pi square e i upon L effective square. So now in place of i we will write as a k square writing that what we will get is pi square e i into a k square upon L effective square. So if I take this k in the denominator what I will get p is equals to pi square e into a upon L by k bracket square. Now the crippling stress is equals to crippling load divided by area so putting that value in p by a what we will get here is pi square e a upon L upon k bracket square. So as a a will get cancel and L upon k is the slenderness ratio it is the ratio of length of column to least radius of gyration. So where k will be taken as under root i minimum upon area. So limitation of Euler's formula so we know the Euler's formula it is given as pi square e a upon L effective upon k bracket square. So considering both ends hinge L effective is equals to L hence crippling stress is pi square e a upon L upon k bracket square where L by k is the slenderness ratio. So if L by k is small then the crippling stress would be very high as it is a denominator if it is very small so you will get the crippling stress very high that may be greater than the crushing stress also. So but for material the crippling stress cannot be greater than the crushing stress. So when slenderness ratio is less than a certain limit Euler formula gives a value of crippling stress which is greater than crushing stress that cannot be accepted. Now for example if we consider a mild steel column with both ends hinged having crushing stress 330 Newton per mm square and modulus of elasticity 2 into 10 raise to 5 Newton per mm square. So now if we equate the crippling stress to the crushing stress pi square e upon L upon k bracket square is equals to 330 so putting the value of E so we will get L upon k square is equals to pi square into 2.1 into 10 raise to 5 divided by 350. So taking the square root of both the terms L upon k is equals to 79.27 say nearly equals to 18. So now hence the slenderness ratio if it is less than 80 then the mild steel column with both ends hinged then Euler formula will not be valid. So it will give the crippling load greater than the crushing load. Now column type and failure so when we will consider column as a long column or short column and when the failure of long column will occur or how will it to occur and failure of short column occurs by here pause the video and try to write answer on a paper. A column having the ratio of length to least lateral dimension greater than 12 is called as long column. A column having ratio of length to least lateral dimension less than 12 is called as short column. Failure of long column occurs by buckling whereas failure of short column occurs by crushing. Now Rankine's formula so Euler formula gives correct result only for long column. So when the column is short or the column is not very long the formula will not give the correct results. So based on experiments performed by Rankine he established an empirical formula which will be applicable for both types of column. So the Rankine's formula is given by 1 upon P is equals to 1 upon PC plus 1 upon PE. So we will treat it as equation number 2 where P is the crippling load by Rankine's formula and PC is the crushing load and PE is the Euler's load or crippling load by Euler's formula. So sigma C will be the ultimate crushing stress and A will be the cross-sectional area. Now we know the Euler's load that is crippling load by Euler's it is pi square A upon L effective square. So here sigma C is constant and area is also constant. So the PC will be constant because PC will be crushing stress into area as both these are constant this will be constant. The value of PC is constant hence the value of P majorly depends upon this value of P that is the Euler's crippling load formula. Now we will consider case 1 when the column is short so PE that is Euler's crippling load is given by pi square A upon L effective square and the Rankine's formula is given by 1 upon P is equals to 1 upon PC plus 1 upon PE. So if L effective is small then the value of PE will be large so L effective it is in the denominator so when it is very small this PE will be large when this PE will be large then 1 upon PE will be small. So 1 upon PE in Rankine's formula it will be small. So it is negligible as compared to PC hence when the column is short so 1 upon PE is proportional to 1 upon PC therefore the crippling load will be nearly will be equal to crushing load. Hence crippling load by Rankine's formula is approximately equals to crushing load. Now case 2 when the column is long so PE is equals to again pi square A upon L effective square and Rankine's formula 1 upon P is equals to 1 upon PC plus 1 upon PE. So here if the column is long L effective is large as it is in the denominator so this PE crippling load by Euler's will be small. So when it is small 1 upon PE 1 upon PE will be large enough when compared with 1 upon PC hence we can neglect 1 upon PC in equation number 2 then we will get 1 upon PE is equals to 1 upon PE or PE is equals to PE hence the crippling load by Rankine's formula is approximately equal to crippling load by Euler's formula. Now again we will take 1 upon PE is equals to 1 upon PC plus 1 upon PE so now for all lengths of column so we will cross multiply it so PE plus PC is divided by PE into PC is equals to 1 upon PE. Now taking the reciprocal to both the sides and dividing numerator and denominator by PE so what we will get 1 upon PE is reciprocal is PE and here PE into C divided by PE I will get PC here again PE divided by PE into V1 and PC divided by PE here it will be as it is. So now again but we know that PC is equals to sigma C into A and PE is equals to pi square EI upon L effective square. So now putting this value in this above equation we will get PE is equals to sigma C into A upon 1 plus sigma C divided by into bracket pi square EI upon L effective square here I is equals to AK square so if we put I is equals to AK square so we will get PE in terms of radius of gyration also PE is equals to sigma C into A upon 1 plus sigma C upon pi square E into bracket L effective upon K bracket square. So now this term sigma C upon A 1 plus alpha into bracket L effective upon K square is equals to PE is equation B. So here alpha will be equal to sigma C upon pi square E where alpha is known as Rankine's constant and this formula gives the crippling load by Rankine's formula. So these are the references which I have referred thank you thank you very much for watching my video.