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Video 1
This lecture we will focus on laterally unsupported beam. So design strength of laterally unsupported beam will be calculated based on the codal provisions, which is given in clause 8.2.2 of IS 800-2007. Now in case of laterally unsupported beam, the lateral torsionalbuckling will play an important role and because of lateral torsional buckling, the full plasticity of the section will not be developed that means the member will fail before at any it’s (())(0:59) full bending stress of the section, the it will fail due to laterally torsional buckling and this lateral torsional buckling happens. Unlike RCC section or the stocky section, their lateral torsional buckling does not come intopicture, but in case of steel rolled section where we have made economic design in terms of the material, we face this type of lateral torsional buckling and if we do not provide lateral support then such type of buckling will occur. So this buckling occur, because of the cross sectional shape, support conditions and effective length. So depending on all these bending strength of laterally unsupported beam will be calculated.
The design bending strength for laterally unsupported beams isMd=βb Zp f bdWhere,Zp = Plastic section modulus of the cross-sectionβb = 1.0 for compact & plastic sections= Ze/Zp for semi-compact sectionsfbd = design bending compressive stress given by,f bd=Χ¿f yγm0 ΧLT = bending stress reduction factor to account for lateral torsion bucklingNow bending stress reduction factor to is calculated by,Χ ¿=1 ∅¿+√∅¿ 2−λ¿ 2 ≤1.0Where, ∅¿=0.5[1+α¿(λ¿−0.2)+ λ¿ 2]
α LT = imperfection factor for lateral torsional buckling of beams= 0.21 for rolled steel sections= 0.49 for welded steel sectionsSuppose, if we use plate to make a I section with the use of welding, then for such type of section, we can use α LT as 0.49 otherwise for the rolled section we can α LT as 0.21λLT = non-dimensional slenderness ratio given by,λ ¿=√βbMZp crf y ≤√1.2 ZMe fcry ¿√ffcry, b
Where,Mcr = elastic lateral buckling moment (Cl. 8.2.2.1) is given by,M cr=√{(π(2LE¿ )I 2 y)[G It+ π(2LE¿)I2w]}¿ βb Zp f cr ,bIt = torsional constant ¿∑ bi ti 3 /3 for open sectionIw = warping constant Iy = moment of inertia about weaker axisry = radius of gyration about weaker axisL¿ =effective length for lateral torsional buckling (Clause 8.3)hf = centre-to-centre distance between flangestf = thickness of flangeG = shear modulus
fcr,b is the extreme fiber bending compressive stress and is given by,fcr,b = extreme fiber bending compressive stress corresponding to elastic lateral bucklingmoment and is given byf cr, b= 1.1 π2 E(rLy¿)2 √1+ 20 1 (L r htfy¿ f )2For different values of KL/ry & hf /tf corresponding values of fcr,b is given in Table 14, IS800:2007. Values of fbd can also be found from Table 13(a) and 13(b), IS 800: 2007corresponds to different values of fcr,b and fy
The following simplified equation may be used in case of prismatic members made of standard rolled I-sections and welded doubly symmetric I-sections, for calculating the elasticlateral buckling moment, McrMcr = π2 E Iy hf 2L¿2 √1+ 20 1(L r htfy¿ f)
However, Mcr for different beam sections, considering loading, support condition and nonsymmetric section, shall be more accurately calculated using the method given in Annex E ofIS: 800-2007.
Video 2
So in today’s lecture what we can see that laterally unsupported length of the member, there is a chance of lateral torsional buckling. Because of lateral torsional buckling before developing the full bending stress, the member may fail and therefore, we have to find out what is the lateral torsional buckling moment. So in next lecture, we will go through one example, then the detail methodology of the design will be clear. Thank you.