Checks a steel beam with web openings under a moment and an axial force, compression or tension, by the Vierendeel model to SIA 263:2013: the chords, the montant, lateral-torsional buckling, the member in compression and the deflection. Every step shows its formula and the clause it comes from.

What it checks

Geometry of the opened beam: the chord area A₁ = b · t_f + t_w · (a − t_f), the static moment S and the lever arm a_N = S/A₁ between the chord centroids. Fillets are neglected.
Class of the chord web, table 5b: (a − t_f − r)/t_w, the web stub at the opening being an outstand free along the cut edge.
Chord at midspan: N_1,Ed = M_Ed/a_N + N_Ed/2 ≤ N_1,pl/γ_M1, with N_1,pl = f_y,k · A₁.
Chord at section D, shear: V_1,Ed ≤ V_1,pl/γ_M1, with V_1,pl = τ_y,k · A_1w.
Chord at section D, axial force, shear and local bending: σ_Ed = N_1,Ed/A₁ + M_1,Ed/W_1,pl ≤ f_yr/γ_M1, with M₁ = V₁ · e₁/2 and f_yr = √(f_y,k² − 3τ²).
Montant: the horizontal shear V_2,Ed = ΔN₁ against V_2,pl = τ_y,k · A₂, then σ_Ed = N_2,Ed/A₂ + M_2,Ed/W_2,pl ≤ f_yr/γ_M1.
Lateral-torsional buckling, as for a welded plate girder (α_D = 0.49) on the section at the opening: M_Ed ≤ M_D,Rd, from M_crD and M_R = f_y,k · A₁ · a_N.
Flexural buckling about both axes, clause 4.5.1: N_K,y,Rd and N_K,z,Rd, with the area and inertias at the opening and the curves of a welded I.
Stability, bending and compression: N_Ed/N_K,Rd + M_Ed/((1 − N_Ed/N_cr) · M_Rd) ≤ 1, with ω = 1.0 and M_D,Rd as the bending resistance.
Deflection: 5 · q · l⁴/(384 · E · I_y) with I_y at the opening, raised by about 20 % for shear deformation, against w_lim.

What you enter and what you get

You enter f_y,k and τ_y,k from table 1 of your copy of SIA 263, the type of openings, the parent profile (b, t_f, t_w, r), the opened beam (h, a, h₀, e₁, e₀), γ_M1 (clause 4.1.3), and N_Ed with its sense. The moments and shears come from a simply supported beam under q_Ed or are entered directly, with the distance x_D to the first chord section. For the member you enter E (clause 2.2) and the buckling lengths, from the end conditions and intermediate supports or directly; for lateral-torsional buckling l_D, G, the loading, z_a and the warping of the end sections; for the deflection the service load and the limit.

You get the derived geometry, the chord and montant forces, the buckling resistances 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

A cellular beam is checked as an equivalent castellated beam.
Compression brings the member check in.
Only sections D and E, one pitch e₀ + e₁ apart, and the montant between them are checked. Move x_D to find the governing chord section.
Local buckling of the montant is not checked. Buckling of the compressed chord over e₀ is covered while the beam is held laterally.
The opened section is used over the whole length for M_crD, N_cr and N_K,Rd, which is conservative.
The 20 % allowance for shear deformation is an estimate, not a norm value.
Short spans, concentrated loads and continuous beams need particular attention.

Worked example

A castellated beam from a profile with b = 100 mm, t_f = 8.5 mm and t_w = 5.6 mm, opened to h = 480 mm with a = 60 mm, h₀ = 200 mm, e₁ = 120 mm and e₀ = 240 mm, in f_y,k = 235 N/mm², simply supported over 6.24 m under q_Ed = 14 kN/m and N_Ed = 40 kN in compression, held against buckling about z and laterally every 1,560 mm, with a service load of 7 kN/m against l/350. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Characteristic yield strength f_y,k	235 N/mm²	§ t.1
Characteristic shear yield strength τ_y,k	135 N/mm²	§ t.1
Openings	Polygonal openings — castellated
Flange width b	100 mm
Flange thickness t_f	8.5 mm
Web thickness t_w	5.6 mm
Root radius r	12 mm
Overall depth of the opened beam h	480 mm
Depth of a chord at the opening a	60 mm
Height of the intermediate flat h₀	200 mm
Width of a montant e₁	120 mm
Clear width of an opening e₀	240 mm
Resistance factor γ_M1	1.05	§ 4.1.3
The axial force is	Compression
Design axial force N_Ed	40 kN
Internal forces	Simply supported under a uniformly distributed load
Span l	6240 mm
Design load q_Ed	14 kN/m
Distance from the support to the first chord section x_D	240 mm
Modulus of elasticity E	210000 N/mm²	§ 2.2
Buckling lengths	From the end conditions and intermediate supports
Ends, buckling about the strong axis y	Held against sway — both ends pinned
Intermediate supports against buckling about y n_y	0
Ends, buckling about the weak axis z	Held against sway — both ends pinned
Intermediate supports against buckling about z n_z	3
Lateral-torsional buckling	Checked, on the section at the opening
Distance between lateral supports of the compressed flange l_D	1560 mm
Shear modulus G	81000 N/mm²
Loading	Uniformly distributed load, simply supported
Load applied at z_a from the shear centre z_a	0 mm
Warping of the end sections	Not prevented — k_φ = 1,0
Deflection	Considered
Service load, uniformly distributed q	7 kN/m
Deflection limit, span divided by	350

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 235 N/mm²	§ t.1
Characteristic shear yield strength	τ_y,k = 135 N/mm²	§ t.1
Resistance factor	γ_M1 = 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·t_f + t_w·(a − t_f), the fillets neglected.	A₁ = 1,138.4 mm²
Static moment at the openingS = b·t_f·(h − t_f) + t_w·(a − t_f)·(h − t_f − a)	S = 519,452 mm³
Lever arm of the chord forcesa_N = S/A₁ — the distance between the centroids of the two chords. The whole external moment is turned into a couple on this arm.	a_N = 456.30 mm
Web area of one chordA_1w = (a − t_f/2)·t_w	A_1w = 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.	W_1,pl = 10,250 mm³
Horizontal section of a montantA₂ = e₁·t_w	A₂ = 672 mm²
Plastic modulus of a montantW_2,pl = t_w·e₁²/4, the montant being a plain rectangle.	W_2,pl = 20,160 mm³
Pitch of the openings	e₀ + 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 − t_f − r)/t_w = 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 reaction	R_Ed = 43.68 kN
Largest moment, at midspanq·l²/8	M_Ed = 68.14 kNm
Chord force at midspanN₁ = M/a_N + N_Ed/2. 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.	N_1,Ed = 169.3 kN
Moment at the first chord section	M_D,Ed = 10.08 kNm
Chord force there	N_1D,Ed = 42.09 kN
Chord shear thereV₁ = V/2 — the vertical shear is shared equally between the two chords.	V_1D,Ed = 20.16 kN
Moment at the next chord section	M_E,Ed = 23.69 kNm
Chord force at the next section	N_1E,Ed = 71.91 kN
Chord shear at the next section	V_1E,Ed = 17.64 kN

Calculation: Check 1 — chord at midspan, axial force
Step	Value	Ref.
Plastic axial resistance of a chordN_1,pl = f_y,k·A₁: a uniform stress over the whole chord, because at midspan the shear is nil and there is no local bending.	N_1,pl = 267.5 kN
Chord under axial force aloneN_1,Ed ≤ N_1,pl/γ_M1	η = 0.665 · Pass66 % of the capacity used

Calculation: Check 2 — chord at section D, shear
Step	Value	Ref.
Plastic shear resistance of a chordV_1,pl = τ_y,k·A_1w	V_1,pl = 42.1 kN
Chord under shear aloneV_1,Ed ≤ V_1,pl/γ_M1	η = 0.502 · Pass50 % of the capacity used

Calculation: Check 3 — chord at section D, axial force, shear and local bending
Step	Value	Ref.
Shear stress in the chord webτ = V_1,Ed/A_1w	τ_Ed = 64.6 N/mm²
Reduced yield strengthf_yr = √(f_y,k² − 3τ²). The reduction is applied over the WHOLE chord for simplicity, even though the flanges could in fact reach the full f_y,k.	f_yr = 206.7 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.	M_1,Ed = 1.210 kNm
Normal stress at the governing sectionσ = N_1,Ed/A₁ + M_1,Ed/W_1,pl	σ_Ed = 155.0 N/mm²
Chord under axial force, shear and local bendingσ_Ed ≤ f_yr/γ_M1	η = 0.787 · Pass79 % of the capacity used

Calculation: Check 4 — montant, horizontal shear
Step	Value	Ref.
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.	V_2,Ed = 29.82 kN
Plastic shear resistance of the montantV_2,pl = τ_y,k·A₂	V_2,pl = 90.7 kN
Montant under horizontal shearV_2,Ed ≤ V_2,pl/γ_M1	η = 0.345 · Pass35 % of the capacity used

Calculation: Check 5 — montant, axial force, shear and bending
Step	Value	Ref.
Axial force in the montantThe difference between the chord shears on either side.	N_2,Ed = 2.52 kN
Bending moment in the montantM₂ = (V_1D + V_1E)·(e/2) − (N_1E − N_1D)·(a_N − h₀)/2. The horizontal shear bends the montant, so it carries tension on one side and compression on the other.	M_2,Ed = 2.982 kNm
Shear stress in the montant	τ_2,Ed = 44.4 N/mm²
Reduced yield strength	f_yr = 222.1 N/mm²
Normal stress in the montantσ = N_2,Ed/A₂ + M_2,Ed/W_2,pl — the same form as for the chord, applied to the montant.	σ_Ed = 151.7 N/mm²
Montant under axial force, shear and bendingσ_Ed ≤ f_yr/γ_M1	η = 0.717 · Pass72 % of the capacity used

Calculation: Check 6 — deflection
Step	Value	Ref.
Second moment of area at the openingComputed from the geometry, not entered: the two tees and nothing between them, I_y = 2·[b·t_f³/12 + b·t_f·((h − t_f)/2)² + t_w·(a − t_f)³/12 + t_w·(a − t_f)·((h − a − t_f)/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.	I_y = 119,038,137 mm⁴
Bending deflection of the equivalent beam5·q·l⁴/(384·E·I_y), on a beam of constant inertia equal to that at the opening.	w_M = 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/350	w_lim = 17.8 mm
Deflectionw ≤ w_lim	η = 0.372 · Pass37 % of the capacity used

Calculation: Check 7 — lateral-torsional buckling
Step	Value	Ref.
Torsion constant at the opening(2·b·t_f³ + 2(a − t_f)·t_w³)/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 I_z by only 0.4 % and I_ω 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 I_z, K and I_ω of the OPENED section, β = 0, k_φ = 1.0. The solid section would give 335.2 kNm — 1 % more, and using it would be unconservative.	M_crD = 330.6 kNm
Bending resistance of the opened beamf_y,k·A₁·a_N — both chords at yield on the lever a_N, which is the same couple model the chord forces use.	M_R = 122.1 kNm
Relative slenderness	λ̄_D = 0.6077
Reduction factorα_D = 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.	χ_D = 0.7808
Lateral-torsional bucklingM_Ed ≤ M_D,Rd	η = 0.751 · Pass75 % of the capacity used

Calculation: Common values — flexural buckling
Step	Value	Ref.
Buckling length, strong axisk·l with k = 1.00 — held against sway — both ends pinned.	l_K,y = 6,240 mm
Buckling length, weak axisl/(n_z + 1) = 6,240/4: the member buckles between its supports.	l_K,z = 1,560 mm
Relative slenderness, strong axis(l_K,y/i_y)/λ_E, i_y = 228.7 mm, λ_E = π·√(E/f_y,k) = 93.91.	λ̄_K,y = 0.2906
Relative slenderness, weak axis(l_K,z/i_z)/λ_E, i_z = 25.0 mm.	λ̄_K,z = 0.6656
Buckling resistance, strong axisχ_K·A·f_y,k/γ_M1, curve b, χ_K = 0.9675.	N_K,y,Rd = 493.0 kN	§ 4.5.1
Buckling resistance, weak axisχ_K·A·f_y,k/γ_M1, curve c, χ_K = 0.7459.	N_K,z,Rd = 380.1 kN	§ 4.5.1
Elastic critical load, strong axisπ²·E·I_y/l_K,y². It sets the amplification 1/(1 − N_Ed/N_cr) of the moment.	N_cr,y = 6,336 kN
Elastic critical load, weak axis	N_cr,z = 1,208 kN

Calculation: Check 8 — stability, bending and compression
Step	Value	Ref.
Bending resistance of the opened beamThe lateral-torsional buckling moment found above.	M_D,Rd = 90.8 kNm
AmplificationWith I_y at the opening — the weaker section — over the whole length.	1/(1 − N_Ed/N_cr,y) = 1.0064
Axial termOver the smaller buckling resistance, with the area and inertias AT THE OPENING (two tees), curves of a welded I.	N_Ed/N_K,Rd = 0.1052
Moment term, amplifiedω = 1,0: the beam carries transverse load.	ω·M_Ed/M_Rd = 0.7554
Stability, bending and compressionN_Ed/N_K,Rd + M_Ed/((1 − N_Ed/N_cr)·M_Rd) ≤ 1	η = 0.861 · Pass86 % 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

How does the axial force reach the chords and the montant?

Each chord carries N_Ed/2 on top of its share of the moment, M_Ed/a_N. Compression adds to the compressed chord, and the chord checks take the more loaded one. In the montant the two halves cancel, so V₂, N₂ and M₂ are unchanged.

Which bending resistance enters the stability check?

M_D,Rd, the lateral-torsional buckling moment of the section at the opening. In the worked example it is 90.8 kNm, against M_R = 122.1 kNm.

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.

Norm: SIA 263:2013 · Page updated: 28 October 2026

Castellated or cellular beam in bending and axial force to SIA 263

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