/Problem on Calculation of Slenderness Ratio for Columns – Columns and Struts – Strength of Materials

Problem on Calculation of Slenderness Ratio for Columns – Columns and Struts – Strength of Materials

Video: Problem on Calculation of Slenderness Ratio for Columns – Columns and Struts – Strength of Materials

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Hello Friends here in this video we will see a problem on calculation of slenderness ratio for a rectangular section column here we have a question a steel bar of rectangular section I'll write this in the data here the section is a rectangular 40 mm into 50 mm so that will be the area for us 40 mm into 50 mm next it is pinned at each end pinned at each end then it is subjected to axial compression next the bar is 2 meter long so length of the bar is given as 2 meter which is mm mm determine the buckling load and the corresponding axial stress using Euler's formula so here we have to find the buckling load by Euler's formula so that would be called as Euler's buckling load and the corresponding axial stress so we have to find out the value of stress also calculate slenderness ratio slenderness ratio I'll denote it by lambda if the proportional limit of the material is 200 Newton per mm square that is the compressive stress which is given take young's modulus the value is given this will be 4 steel 2 into 10 raised to 5 Newton per mm square so now with the data available and we have understood what are the unknowns here let us try to get the solution to this problem now the first thing they have mentioned is that it is a rectangular section 40 mm into 50 mm so I will draw the section here now this rectangle will have width equal to 40 and depth will be 50 so this will be the width of the column which is 40 mm and here we have the depth of the column which is 50 mm these are x and y axis respectively and here I'll draw the section now this is the rectangular section of the column 40 mm into 50 mm next it was given that it is subjected to axial compression and the column was pinned at each end she pinned means pinned it is another word for hinged so here we have the column which is hinged at both the ends so I will draw the diagram to explain it so here is the column which is hinged at both the ends this diagram indicates the column is hinged and this will be the total length of the column now when load is acting over this column and it is a compressive load from both the sides so under the action of this compressive load this column will Bend and it will start bending from the hinged end and it would be completed up to the hinge end it means the total length is taking part in the bending so this length would be called as the effective length of the column so now the first thing which I will write here is since both the ends of the column are pinned or hinged so therefore effective length else of X e will be equal to the length of column so here we have effective length is equal to the total length and the column length is mm mm so this is the effective length now after getting this effective length I'll see the first question the first thing which we have to calculate here is the Euler's crippling load or we can say euler's buckling load so now since euler's crippling or we can say buckling load is given by now the formula I will write it onto the next page it is given by P suffix e euler's load it is equal to pi square e I minimum upon effective length whole square so I will keep this as first equation now effective length value we have capital e is given in the problem the only unknown here is I minimum that is the minimum moment of inertia so for that we will calculate the moment of inertia about x axis and then about Y axis and then we are going to compare which value is minimum that we are going to select so here I will say that therefore mi that is moment of inertia for given rectangular section about x axis is given by that would be IX X is equal to we know the formula of a rectangle moment of inertia about x axis that is BD cube by 12 here B is 40 D is 50 so 40 into 50 cube divided by 12 therefore I xx value will be it is 416 0.

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67 into 10 raise to 3 mm raise to 4 in a similar manner we will calculate about y-axis so therefore am i forgiven rectangular section about y-axis is given by I YY is DB cube by 12 moment of inertia for rectangle about y-axis so it is 15 into 4 T cube divided by 12 so I YY comes out to be 266 point 6 7 into 10 raised to 3 mm raised to 4 now after getting the I x6 and I have ever values so now after getting ixx and iyy here I will compare that is I x6 value was 416 point 6 7 into 10 raised to 3 i YY to sixty six point six seven so I YY is less so I will say that therefore iy y is less than I X 6 so hence I minimum is equal to iyy in the value is 266 point 6 7 into 10 raised to 3 mm raised to 4 so this value we are going to put in the Euler's formula I will say that therefore put all values in equation number 1 so we have euler's load is equal to PI square into here it is PI square e I minimum upon effective length square the value of e is given as 2 into 10 raised to 5 into I minimum which is 266 Oh point 6 7 into 10 raised to 3 divided by effective length square so effective length we have found out it was 2,000 so therefore it is 2,000 square hence the euler's load it comes out to be 130 1.

6 0 into 10 raised to 3 Newton or it is 130 10.6 0 kilonewton so this will be our first answer that is we have found out what is euler's crippling load now the next question was to get the stress value that is how much is the corresponding axial stress because of the load now we have found out the load value because of this load how much will be the stress which is induced in the column so I will say that therefore axial stress induced in the column because of Euler's crippling load that will be stress I will denote it as Sigma and it is equal to euler's load upon the cross section area EULA's load is 130 10.6 0 into 10 raised to 3 Newton divided by the cross-section area is 14 to 15 because 40 is the width and 50 is the depth so from here we'll get the value of stress it comes out to be 65 point seven nine Newton per mm square this is the second answer now after getting the stress the next thing is we have to find the slenderness ratio which is denoted as lambda so here I'll again use the formula that since euler's load is given by again I am using the Euler's load formula and that is P suffix E is equal to pi square e I minimum upon effective length square now after adding this formula here I can write it in another way i minimum the minimum moment of inertia it can be written as a square into K square divided by effective length square now how I have written this here I will explain it that since K it is called as radius of gyration and that radius of gyration K is equal to root of I upon a that is the moment of inertia divided by cross section area so now here I want I so I will keep I on one side this would become when I remove the root sign here so that would be K square it would be K square into a square so we arrive I will write it down as K square into a square so that is it could be your since this is the root sign so first of all when it goes on to the other side that is that would be K square and this a will remain as it is so it is K square into a so here I'll write down that it is K into K square is a K square because here I have explained this that since the minimum moment of inertia will be equal to a K Square and here K is also called as minimum radius of gyration so instead of I minimum I can write down a K square now after this the euler's load can since it is a load and we know that stress is equal to load upon area so therefore load is equal to stress into area so here I will write down instead of load stress into area and here this stress will be the limiting stress given in the problem that is if the proportional limit of the material is this is the stress which is Sigma C the compressive stress given in the problem so I will use this compressive stress into area which is the euler's load and that is equal to PI square e a and this K square I will send it into the denominator so that becomes le upon K whole square I am sending this K square into the denominator of effective length so here if I see from both the sides area and area will get cancelled out and Here I am left with PI square e upon now effective length upon K that is nothing but the slenderness ratio which is denoted by lambda I will write down since slenderness ratio lambda is equal to effective length upon K so now here we have as shift lambda square on one side so lambda square goes on one side is equal to PI square e and this Sigma C will be shifted in the denominator so after this since we want this lambda I will keep it on one side and bring the root sign on to the other side so removing the square here we would have root onto the other side so root of pi square e upon Sigma C from this we are going to get the slenderness ratio so putting the values here therefore lambda is equal to root of pi square into capital e is 2 into 10 raised to 5 divided by Sigma C is given as 200 so here I will get the answer as 99 point 3 5 so this is the third answer so here we have completed this problem if we see in this question there were three things asked first was EULA's load and that euler's load we had calculated it was 130 1.

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60 kilonewton so this was the first answer next the second question was to calculate the axial stress that we have found out sixty five point seven nine Newton per mm square and the last thing was slenderness ratio whose value is ninety nine point three five and after finding all three unknowns we can say that the problem has been completed.