Hello, I am going to discuss the calculation of the compressive strength of acompression member. So in last lecture we have seen the strength calculation formula of thecompression member as per the IS code and we have seen this strength depends on threemain factors one is the yield stress of the material that means the material properties, then thelength of the member from which we can find out the slenderness ratio and because ofbuckling the reduction of the strength occurs and then another is imperfection factor whichdepends on the buckling class of the member.So three factors are introduced in the design strength calculation of the compressive member,these three factors are the slenderness ratio, then buckling class and the material propertiesthat is yield strength. So based on that the formula has been derived which is basically similarto the British code. And I am going to solve a workout example through which we willunderstand that the formulas that are used in the IS code, how to make use of those whilecalculating the compressive strength of a compression member.
Example: Determine the design axial load on the column section ISMB 400, given that theheight of the column is 3.5 m and that it is pin-ended. Also assume the following:
That means from this what we could see that ISMB 400 members are used. So from this wecan find out what is the buckling class where ISMB 400 we know the thickness of themember that means thickness of the flange, width of the flange and the depth of the crosssection those things we know, so from this we can find out the buckling class then anotherthing we can find out that is the slenderness ratio, slenderness ratio here the length is given3.5 meter.So we know the radius of gyration of the section ISMB 400, so from that, we can find outslenderness ratio ofcourse the slenderness ratio to find out slenderness ratio we have to knowthe effective length. So effective length depends on the end connections, here ends areconnected by pin-end joint, so effective length here will be will not be reduced that will be 1into 3.5 because here K will be 1 as per the IS code, right.So one is buckling class then slenderness ratio and grade of steel fu means fe250. So fy valueis 250 it was given. So based on these three we can find out the value of fcd.
So using the formula we can find out the design compressive strength of the member of thegiven member and we could see that about y-y axis it is going to fail first that means y-y axisis the weaker section and the same can be made by the use of table 9, means in place ofcalculation of all these by fcd we simply can find out using table 9 from where fcd we canget.
Example: Calculate the compressive strength of a compound column consisting of ISHB [email protected] 54.7 kg/m with one cover plate of 300×16 mm on each flange (as shown in the figure) andISHB 250Cover plate-300×16z zy y125 mm16 mmhaving a length of 4 m. Assume that the bottom of the column is fixed and the top is hingedand = 250 N/mm2
Therefore we do not need to calculate fcd value for both the direction, because the bucklingclass is the same for both the direction. So we will consider r minimum in which direction itis coming, the r minimum will be the failure criteria. So we will straight calculate the rminimum value and then we will find out λ and then according to buckling class we will findout the fcd value and once we get fcd value we can find out the value of compressive forcewhich can be carried by that particular member.
So this is how we can calculate the compressive strength and compressive force of acompression member using IS 800:2007, ok this is all about the calculation of compressivestrength of the member, thank you.
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