Castellated and cellular steel beam 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 steel beam with web openings, castellated or cellular, by the Vierendeel model to SIA 263:2013: the chords, the montant between two openings, lateral-torsional buckling 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, the type of openings, the parent profile (b, tf, tw, r), the opened beam (h, a, h0, e1, e0) and γM1 (clause 4.1.3). The internal forces come from the span l and the design load qEd of a simply supported beam, or are entered directly, with the distance xD to the first chord section. For lateral-torsional buckling you enter E (clause 2.2), G, the distance lD between lateral supports of the compressed flange, the loading, the load position za and the warping of the end sections. For the deflection you enter the service load and the limit.

You get the derived geometry, the class, the forces in the chords and the montant, and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Worked example

A castellated beam cut from a parent profile 200 mm deep (b = 100 mm, tf = 8.5 mm, tw = 5.6 mm), opened to h = 480 mm with a = 60 mm, h0 = 200 mm, e1 = 120 mm and e0 = 240 mm, in fy,k = 235 N/mm², simply supported over 6.24 m under qEd = 16 kN/m with lateral supports every 3.00 m, and a service load of 7 kN/m against l/350; the chord, montant and deflection checks pass, but lateral-torsional buckling does not (η = 1.43). 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
OpeningsPolygonal openings — castellated
Flange width b100 mm
Flange thickness tf8.5 mm
Web thickness tw5.6 mm
Root radius r12 mm
Overall depth of the opened beam h480 mm
Depth of a chord at the opening a60 mm
Height of the intermediate flat h0200 mm
Width of a montant e1120 mm
Clear width of an opening e0240 mm
Resistance factor γM11.05§ 4.1.3
Internal forcesSimply supported under a uniformly distributed load
Span l6240 mm
Design load qEd16 kN/m
Distance from the support to the first chord section xD240 mm
Lateral-torsional bucklingChecked, on the section at the opening
Modulus of elasticity E210000 N/mm²§ 2.2
Distance between lateral supports of the compressed flange lD3000 mm
Shear modulus G81000 N/mm²
LoadingUniformly distributed load, simply supported
Load applied at za from the shear centre za0 mm
Warping of the end sectionsNot prevented — kφ = 1,0
DeflectionConsidered
Service load, uniformly distributed q7 kN/m
Deflection limit, span divided by350
Calculation: Common values
StepValueRef.
Characteristic yield strengthfy,k = 235 N/mm²§ t.1
Characteristic shear yield strengthτy,k = 135 N/mm²§ t.1
Resistance factorγM1 = 1.05§ 4.1.3
Height of an openingBetween the two chords. With flats it is two sloping sides of 80 mm and a vertical side of h0 = 200 mm.h − 2a = 360 mm
Rise of one sloping side(h − 2a − h0)/2 = 80.0 mm
Run of one sloping side(e0 − e1)/2 = 60.0 mm
Angle of the cutMeasured from the horizontal. It is the angle of the zig-zag the parent web was cut along.α = 53.1 °
Depth of the parent profileNot entered: it follows from the others, since the cut runs between two levels a from each flange. Compare it with the profile you started from — if it does not match, one of h, a or h0 is not what you think it is.d = (h − h0)/2 + a = 200.0 mm
Gain in depthCastellation alone typically gives about 1,5; flats take it further.h/d = 2.40
Area of one chordA1 = b·tf + tw·(a − tf), the fillets neglected.A1 = 1,138.4 mm²
Static moment at the openingS = b·tf·(h − tf) + tw·(a − tf)·(h − tf − a)S = 519,452 mm³
Lever arm of the chord forcesaN = S/A1 — the distance between the centroids of the two chords. The whole external moment is turned into a couple on this arm.aN = 456.30 mm
Web area of one chordA1w = (a − tf/2)·twA1w = 312.2 mm²
Plastic modulus of one chordAbout the axis that cuts the tee into two equal areas — 5.69 mm from the outside of the flange.W1,pl = 10,250 mm³
Horizontal section of a montantA2 = e1·twA2 = 672 mm²
Plastic modulus of a montantW2,pl = tw·e1²/4, the montant being a plain rectangle.W2,pl = 20,160 mm³
Pitch of the openingse0 + e1 = 360 mm
Web slenderness at the openingLimits 9.00 · 10.00 · 14.00. The EP figure of 10 for S 235 and 8 for S 355 is the class 2 outstand limit 10ε — the web stub above an opening is an outstand, free along the cut edge.(a − tf − r)/tw = 7.05§ t.5b
Class of the chord webA castellated beam is generally designed by EP, which needs class 2 or better.– = Class 1§ t.5b
Support reactionREd = 49.92 kN
Largest moment, at midspanq·l²/8MEd = 77.88 kNm
Chord force at midspanN1 = M/aN. The moment is carried almost entirely by a couple of axial forces in the two chords; what is left over is carried by their local bending, M1.N1,Ed = 170.7 kN
Moment at the first chord sectionMD,Ed = 11.52 kNm
Chord force thereN1D,Ed = 25.25 kN
Chord shear thereV1 = V/2 — the vertical shear is shared equally between the two chords.V1D,Ed = 23.04 kN
Moment at the next chord sectionME,Ed = 27.07 kNm
Chord force at the next sectionN1E,Ed = 59.33 kN
Chord shear at the next sectionV1E,Ed = 20.16 kN
Calculation: Check 1 — chord at midspan, axial force
StepValueRef.
Plastic axial resistance of a chordN1,pl = fy,k·A1: a uniform stress over the whole chord, because at midspan the shear is nil and there is no local bending.N1,pl = 267.5 kN
Chord under axial force aloneN1,Ed ≤ N1,pl/γM1η = 0.670 · Pass67 % of the capacity used
Calculation: Check 2 — chord at section D, shear
StepValueRef.
Plastic shear resistance of a chordV1,pl = τy,k·A1wV1,pl = 42.1 kN
Chord under shear aloneV1,Ed ≤ V1,pl/γM1η = 0.574 · Pass57 % of the capacity used
Calculation: Check 3 — chord at section D, axial force, shear and local bending
StepValueRef.
Shear stress in the chord webτ = V1,Ed/A1wτEd = 73.8 N/mm²
Reduced yield strengthfyr = √(fy,k² − 3τ²). The reduction is applied over the WHOLE chord for simplicity, even though the flanges could in fact reach the full fy,k.fyr = 197.2 N/mm²
Local bending moment in the chordM1 = V1·e1/2, the lever being half the montant width — section B-B lies there, measured from the mid-opening hinge at A. This moment is the part of the general bending that the couple of chord forces does not take.M1,Ed = 1.382 kNm
Normal stress at the governing sectionσ = N1,Ed/A1 + M1,Ed/W1,plσEd = 157.0 N/mm²
Chord under axial force, shear and local bendingσEd ≤ fyr/γM1η = 0.836 · Pass84 % of the capacity used
Calculation: Check 4 — montant, horizontal shear
StepValueRef.
Horizontal shear in the montantV2 = ΔN1: the change in chord force from one opening to the next has to be handed across by the montant.V2,Ed = 34.08 kN
Plastic shear resistance of the montantV2,pl = τy,k·A2V2,pl = 90.7 kN
Montant under horizontal shearV2,Ed ≤ V2,pl/γM1η = 0.394 · Pass39 % of the capacity used
Calculation: Check 5 — montant, axial force, shear and bending
StepValueRef.
Axial force in the montantThe difference between the chord shears on either side.N2,Ed = 2.88 kN
Bending moment in the montantM2 = (V1D + V1E)·(e/2) − (N1E − N1D)·(aN − h0)/2. The horizontal shear bends the montant, so it carries tension on one side and compression on the other.M2,Ed = 3.408 kNm
Shear stress in the montantτ2,Ed = 50.7 N/mm²
Reduced yield strengthfyr = 218.0 N/mm²
Normal stress in the montantσ = N2,Ed/A2 + M2,Ed/W2,pl — the same form as for the chord, applied to the montant.σEd = 173.3 N/mm²
Montant under axial force, shear and bendingσEd ≤ fyr/γM1η = 0.835 · Pass84 % of the capacity used
Calculation: Check 6 — deflection
StepValueRef.
Second moment of area at the openingComputed from the geometry, not entered: the two tees and nothing between them, Iy = 2·[b·tf³/12 + b·tf·((h − tf)/2)² + tw·(a − tf)³/12 + tw·(a − tf)·((h − a − tf)/2)²]. The flanges give 79.4 % of it. Fillets neglected, as for A1 and S; the flats sit in the montants, not in this section.Iy = 119,038,137 mm⁴
Bending deflection of the equivalent beam5·q·l⁴/(384·E·Iy), on a beam of constant inertia equal to that at the opening.wM = 5.53 mm
Total deflectionThe bending deflection raised by about 20 % to allow for the shear deformation, which in a beam full of holes is no longer negligible. That factor is an estimate from the literature, not a norm value.w = 6.63 mm
Limitl/350wlim = 17.8 mm
Deflectionw ≤ wlimη = 0.372 · Pass37 % of the capacity used
Calculation: Check 7 — lateral-torsional buckling
StepValueRef.
Torsion constant at the opening(2·b·tf³ + 2(a − tf)·tw³)/3. Only the two chord webs are there.K = 46,971 mm⁴
Torsion constant between the openingsThe full web. The openings cut K by 31 % — and they cut Iz by only 0.4 % and Iω not at all. It is K that the openings take away, which is exactly why the check is made on the section at the opening.K = 68,045 mm⁴
Elastic critical moment at the openingWith Iz, K and Iω of the OPENED section, β = 0, kφ = 1.0. The solid section would give 99.5 kNm — 4 % more, and using it would be unconservative.McrD = 95.6 kNm
Bending resistance of the opened beamfy,k·A1·aN — both chords at yield on the lever aN, which is the same couple model the chord forces use.MR = 122.1 kNm
Relative slendernessλ̄D = 1.1297
Reduction factorαD = 0.49, taking the beam as a WELDED I — which it is, being cut and rewelded. It is treated as a welded plate girder throughout.χD = 0.4687
Lateral-torsional bucklingMEd ≤ MD,Rdη = 1.429 · Fail143 % 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 lateral-torsional buckling checked on the section at the opening?

The openings cut the torsion constant K. In the worked example they reduce K by 31 %, leave Iz within 0.4 % and Iω untouched. The solid section would overstate McrD by 4 %, which is unconservative.

Where is the governing chord section?

It has to be found by iteration. On a simple beam it lies at roughly a quarter of the span. The module checks only the sections it is given, so move xD along the beam and watch the utilisation.

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