Rolled I or H section in bending to SIA 263

The code governs which inputs appear and how every value is derived.

Member

Material properties, shared by every verification. The cross-section is not shared: SIA 263 classifies it from the internal forces, so it belongs to each verification and appears with it.

About this check

clauses, worked example and questions

Checks a rolled I or H section in bending at the ultimate limit state, to SIA 263:2013: section class, bending and shear resistance, lateral-torsional buckling between lateral supports, and the deflection. Every step shows its formula and the clause it comes from.

What it checks

What you enter and what you get

You enter fy,k and τy,k from table 1 of your copy of SIA 263, E and G (clause 2.2), the profile dimensions, γM1, MEd, VEd and whether the internal forces come from an elastic or a plastic analysis. For lateral-torsional buckling: lD, the end moments of the segment, the moment diagram, the load position and the warping of the end sections. For the deflection: the static system (simply supported, or continuous over 2 to 5 equal spans), the span, the service load and the limit. Vibration is an option.

You get the section properties, the class, every resistance and each utilisation with a pass or fail and its clause, and a drawing of the profile to scale. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Worked example

A rolled section with h = 114 mm, b = 120 mm, tw = 5 mm, tf = 8 mm and r = 12 mm in fy,k = 235 N/mm², under MEd = 20.8 kNm and VEd = 24.9 kN, with lateral supports every 1,250 mm and end moments of 15.6 and 14.3 kNm of the same sign, and a deflection check over three continuous 5.00 m spans under 3.0 kN/m against l/350. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Characteristic yield strength fy,k235 N/mm²§ t.1
Characteristic shear yield strength τy,k135 N/mm²§ t.1
Modulus of elasticity E210000 N/mm²§ 2.2
Overall depth h114 mm
Flange width b120 mm
Web thickness tw5 mm
Flange thickness tf8 mm
Root radius r12 mm§ t.5b
Resistance factor γM11.05§ 4.1.3
Design bending moment MEd20.8 kNm
Design shear force VEd24.9 kN
Internal forces found byElastic analysis
Lateral-torsional bucklingConsidered§ 4.5.2
Distance between lateral supports of the compressed flange lD1250 mm
Larger end moment of the segment |Mmax|15.6 kNm
Smaller end moment of the segment |Mmin|14.3 kNm
The two moments haveThe same sign at both ends of the segment
Moment diagram between the lateral supportsUniformly distributed load, simply supported
Load applied atShear centre
Warping of the end sectionsNot prevented — kφ = 1,0
Shear modulus G81000 N/mm²§ 2.2
DeflectionConsidered
Static systemContinuous over equal spans — uniformly distributed load
Number of equal spans n3
Span l5000 mm
Service load, uniformly distributed q3.0 kN/m
Deflection limit, span divided by350
VibrationNot considered
Calculation: Common values
StepValueRef.
Characteristic yield strengthfy,k = 235 N/mm²§ t.1
Characteristic shear yield strengthFrom SIA 263 table 1, not derived as fy/√3.τy,k = 135 N/mm²§ t.1
Reduction factorε = √(235/fy). The class limits are scaled by it.ε = 1.0000§ t.5a
Resistance factorγM1 = 1.05§ 4.1.3
Gross areaA = 2b·tf + (h − 2tf)·tw + (4 − π)·r², the plates and the four fillets.A = 2,534 mm²
Second moment of areaFlanges, web and fillets about the major axis. No resistance check uses it — the deflection is inversely proportional to it.Iy = 6,060,890 mm⁴
Elastic section modulusWel = 2Iy/hWy,el = 106,331 mm³
Plastic section modulusTwice the first moment of the half section about the major axis, fillets included.Wy,pl = 119,491 mm³
Second moment of area, minor axisIz = 2,308,338 mm⁴
Radius of gyration, minor axisiz = √(Iz/A). It sets the critical length between lateral supports: for the same section resistance an IPE resists lateral-torsional buckling markedly less well than an HEA, HEB or HEM.iz = 30.18 mm
Uniform torsion constant2/3·(b − 0,63tf)·tf³ + 1/3·(h − 2tf)·tw³ plus the fillet junction term — the expression the section tables use.K = 59,942 mm⁴
Warping constantIω = tf·b³/24·(h − tf)², the flanges' lateral inertia on the lever between them.Iω = 6,471,936,000 mm⁶
Compressed flange — flat widthc = (b − tw)/2 − r, measured to the toe of the fillet.c = 45.50 mm§ t.5b
Compressed flange — slendernessLimits 9.0 · 10.0 · 14.0 for class 1, 2 and 3.c/tf = 5.69§ t.5b
Web in bending — flat widthh − 2tf − 2rbw = 74.0 mm§ t.5a
Web in bending — slendernessLimits 72.0 · 83.0 · 124.0 for class 1, 2 and 3.bw/tw = 14.80§ t.5a
Section classThe least favourable of the two compressed parts. Every IPE, HEA, HEB and HEM profile is class 1 in bending except the HEA 180 to 320 in S 355.– = Class 1§ t.5a
Calculation methodDecided by the class: class 1 admits the plastic resistance, so EP — elastic analysis, plastic resistance — the general case.– = EP§ t.5a
Calculation: Check 1 — bending resistance
StepValueRef.
Governing section modulusPP and EP both take the plastic modulus.Wy,pl = 119,491 mm³
Ultimate bending resistanceMR = fy·Wy,plMR = 28.08 kNm
Design bending resistanceMR/γM1 = 26.74 kNm§ 4.1.3
Section modulus requiredW ≥ γM1·MEd/fy — the sizing form used to pick a profile out of the tables in the first place.Wreq = 92,936 mm³
Bending resistanceMEd ≤ MR/γM1η = 0.778 · Pass78 % of the capacity used
Calculation: Check 2 — shear resistance
StepValueRef.
Effective shear areaAv = A − 2b·tf + (tw + 2r)·tf. The last term hands back the web-to-flange junction and both fillets.Av = 846 mm²§ (42a)
Ultimate shear resistanceVR = τy·Av.VR = 114.2 kN
Design shear resistanceVR/γM1 = 108.7 kN§ 4.1.3
Shear resistanceVEd ≤ VR/γM1η = 0.229 · Pass23 % of the capacity used
Calculation: Lateral-torsional buckling — not decisive
StepValueRef.
Moment ratio over the segmentψ = Mmin/Mmax with signs, Mmin the smaller in absolute value.ψ = 0.9167
Critical length ratiolcr = 2,7·iz·(1 − 0,5ψ)·√(E/fy), for EP.lcr/iz = 43.72
Critical lateral-torsional lengthlcr = 1,320 mm
If the moment were pushed up to the resistanceA profile taken from the tables usually has a little more resistance than the moment needs, and a first approximation for the extra spacing that buys is lcr·√(MR/(γM1·MEd)), on the assumption of a second-degree relation between the two. NOT IN SIA 263. The alternative is the full lateral-torsional buckling check.not in SIA 263lD,max = 1,496 mm
Lateral-torsional bucklinglD = 1,250 mm ≤ lcr = 1,320 mm: lateral-torsional buckling is not decisive and the bending resistance of the section stands alone. This is information, not a check.lD / lcr = Not decisive — 0.95
Calculation: Check 3 — deflection
StepValueRef.
Static system– = Continuous over equal spans — uniformly distributed load
Support momentFrom the three-moment equations for 3 equal spans. For three spans this comes out at exactly −0,100.M/(q·l²) = -0.10000
Deflection coefficientw = k·q·l⁴/(E·I), that is 1/145.26. Obtained by integrating the bending moment twice over the span and scanning for the largest deflection, so it is derived rather than read off.k = 0.0068842
Deflectionw = 10.14 mm
Limitl/350. SIA 260 annex A gives indicative values; others may be agreed.wlim = 14.29 mm
Deflectionw ≤ wlimη = 0.710 · Pass71 % of the capacity used

How this module is verified

Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.

Questions

Why is τ_y,k entered and not derived from f_y?

SIA 263 tabulates τy, and the tabulated values are not fy/√3. For S 235 the table gives 135 N/mm² where the division gives 135.7.

When is lateral-torsional buckling decisive?

When lD exceeds lcr. Below it, the section resistance stands alone. In the worked example, lD = 1,250 mm ≤ lcr = 1,320 mm.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

Related

Norm: SIA 263:2013 · Page updated: 26 October 2026