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Module 1: Compression Members

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Video 1
In this lecture we will be focused on compressive strength calculation of strut angles. Weknow when a member is subjected to compressive force and used in a bracing system or in aroof truss are called struts. This strut basically takes light load and the effective length will becomparatively less. So for such cases we generally use angle section to take care thecompressive load.
Now this compressive load may act concentrically on the angle section or may act through itsone leg which will be eccentric and the strength calculation of such type of angle sections aregiven in code in clause 7.5 of IS 800:2007. Now in clause 7.5.1 of IS 2007 it is told that thecompression in single angles may be transferred either concentrically to its centroid throughend gusset or eccentrically by one of its leg to a gusset or adjacent member.
That means an angle sections is loaded with the axial compression member and either thecompressive load will be transferred through this Cg of this angle section by the use of gussetplate or it may be transferred through one of its leg may be through one of its leg.So when the angle section is transferring the load through its one leg, then the eccentricitywill come into picture and because of this eccentricity three types of things will happen one istorsional buckling, torsional buckling means basically flexural torsional buckling, another isflexural buckling which comes for all the members which is common, and another istorsional buckling.So the combination of flexural torsional buckling can be calculated from a formula which is given in the clause 7.5.
Now for concentric loading when the angle is concentrically loaded this can be calculatedthrough the clause 7.1.2 mean in case of concentrically loaded in compression the designstrength may be evaluated as per clause 7.1.2 of IS 800:2007. This clause 7.1.2 we havediscussed already that means it is a concentric loading what we have designed earlier in caseof compression member, which may be channel section, maybe I section, maybe other type ofsection.
However if this is not concentric then we have to go for another clause which is given inclause 7.5.1.2 when the angle section is loaded by the axial compressive load through its oneleg. So in clause 7.5.1.2 it is told that when the single angle is loaded concentrically throughone of the its leg the flexural torsional buckling strength may be evaluated using anequivalent slenderness ratio, λe , which is given by
Where k1 , k2 , k3=¿ constants depending upon the end condition as per Table 12 of IS 800-2007. Depending upon end condition means whether it is fixed or it is hinged, whether it isconnected by one bolt, two bolt or welded depending on the end condition end fixitycondition the value of k1, k2, k3 is defined in table 12.
Where, l = centre to centre length of the supporting memberrvv = radius of gyration about minor axisb1 , b2 = width of two legs of the anglet = thickness of the legε = yield stress ratio,
Now for a particular grade of steel this π 2 E/250 is constant, so that can be calculatedbecause E is constant for a particular grade of steel, therefore ε is also constant. So if wecan calculate λvv and λϕ and if we find out the value of k1, k2, k3 from table 12 then Ican find out the equivalent slenderness ratio λe
Now we have used rvv in the previous expression which is the radius of gyration aboutminor axis, which will be the minimum radius of gyration about the minor.So for a particular angle section we can find out the properties from the geometry, from SP: 6we can find out the value of rvv for a particular angle section then we know what is the widthof the legs of the angle b1, b2 and thickness of the leg and also if we know the grade of steelthen we can find out the yield stress ratio ε that also we can find out.And the constants k1, k2, k3 that can be found in table 12 of IS 800:2007. Now this value of k1,k2, k3 depends on number of bolts at the end of the member as well as the connecting memberfixity that means gusset or connecting member fixity means what type of fixity is therewhether it is fixed, or hinged depending on that and whether number of bolts are more than orequal to 2 or 1. So depending on that we can find out k1, k2, k3.Say for example if I have a fixed connection, if fixity is there and if the number of bolts are 2or more than 2 then I can take the value of k1 as 0.2, value of k2 as 0.35 and value of k3 as 20.So similarly I can consider different type of member fixity, if it is hinged then k1, k2, k3 valuecan be found, similarly if number of bolts are 1 at the end then depending upon the fixity Ican find out the value of k1, k2, k3.
Video 2
 
Now the theory whatever we have discussed theory means the expression for finding outλe value has been discussed, so for such type of members how to find out the designstrength that can be understood if I go through this example.Example: An ISA 150×150×12 used as a strut has the effective length as 3 m. Calculate thestrength when it is connected bya) One bolt at each endb) Two bolts at each endc) Welded at each endSolution:Basically what changes we will find out for different cases that is the equivalent λ which isλe. So λe depend on k1, k2, k3 so that k1, k2, k3 can be found from table 12 and depending onthe type of fixity and number of bolts the k1, k2, k3 value will change and accordingly thestrength will be going to change. So through this example we will try to understand.For ISA 150×150×12, A = 3459 mm2, [Table III, SP:6(1)-1964]rvv = 29.3 mmFor angle sections, Buckling curve ‘c’ is used. [Table 10, IS 800:2007]Imperfection factor, α = 0.49 [Table 7, IS 800:2007]
Now coming to third case when it is connected by weld at each end, so for this case what wecan consider that this will be similar to the earlier. So in case of weld connection we canassume it will be fixed at both the end and as we have calculated the k1, k2, k3 valueconsidering two bolts for this case also will become same.This case will be exactly similar to earlier case, i.e., Connected by two bolts at each end.Therefore,Pd=¿ 410.9 kNSo what we have seen here that Pd value for first case we got 316.1 kilonewton and forsecond case we got Pd is equal to 410.9 kilonewton and in third case Pd value we got 410.9kilonewton, that means when two bolts or weld connections are used the design strength isbecoming higher compared to the one bolt connections, right. So if we want to increase thestrength of the member we should choose for higher number of bolts means where one bolt issufficient we will choose two bolt with smaller diameter. So that we can accommodate morestrength on the member with same member size member size will be same but strength willbe more because λe is going to change, right.