Beams and lateral-torsional buckling
A steel I-beam is efficient in bending about its strong axis, but that efficiency has a weakness. The compression flange behaves like a strut that wants to buckle sideways. Because it is joined to the tension flange through the web, it cannot simply bow out; instead the whole section moves sideways and twists together. This is lateral-torsional buckling, and it can govern long, unbraced beams well below their full plastic moment.
If the compression flange is held sideways at close intervals, by a concrete slab, purlins or bracing, the beam is laterally restrained. Its strength is then the plastic or elastic moment of the section, depending on section class: plastic and compact sections reach Zp fy / γm0, semi-compact sections only Ze fy / γm0. Shear in the web is checked separately, and high shear reduces moment capacity.
For laterally unsupported beams, the elastic critical moment Mcr depends on the unbraced length, the section's lateral stiffness, torsional and warping constants, the moment distribution and where the load is applied. Loads applied on the top flange are worse than loads at the shear centre. The code converts Mcr to a design strength through a slenderness and imperfection factor, in the same style as column curves. Practical cures are bracing the compression flange, choosing wider flanges, or using box sections. Deflection and web bearing under concentrated loads complete the checks.
Restrained, plastic or compact: Md = βb Zp fy / γm0λLT = √(βb Zp fy / Mcr)φLT = 0.5 [1 + αLT (λLT − 0.2) + λLT²]Md = βb Zp fbd, fbd = χLT fy / γm0
Problem. A simply supported floor beam spans 6 m with a slab that fully restrains its compression flange. Factored UDL is 40 kN/m. Using E250, find the required plastic modulus.
- Mu = 40 × 6² / 8 = 180 kNm.
- Required Zp = Mu γm0 / fy = 180 × 10⁶ × 1.10 / 250 = 792 000 mm³.
- Choose a section with Zp above 792 × 10³ mm³ and plastic or compact class, check that Md does not exceed 1.2 Ze fy / γm0 (simply supported beams), then check shear and deflection.
- Without the slab, the 6 m unbraced length would need an LTB check and a heavier section.
Answer. Zp ≥ about 792 × 10³ mm³
Quick quiz
1Which flange of a beam tends to buckle?
The compression flange.
2Is a top-flange load better or worse for LTB than a load at the shear centre?
Worse.
3Name one way to prevent LTB.
Brace the compression flange at close intervals (slab, purlins, bracing).
- IS 800 (India): General construction in steel: limit state design of steel members and connections.
Clauses and documents to read
- IS 800:2007 cl. 8.2.1 (laterally supported beams)
- IS 800:2007 cl. 8.2.2 (laterally unsupported beams)
- IS 800:2007 Annex E (elastic lateral torsional buckling moment)
- IS 800:2007 cl. 8.4 (shear)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- N. Subramanian, Design of Steel Structures, Oxford University Press, latest edition.
- C. G. Salmon, J. E. Johnson, F. A. Malhas, Steel Structures: Design and Behavior, Pearson, 5th edition.
- W. T. Segui, Steel Design, Cengage, latest edition.
Free and official sources
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