Checks a rolled I or H section, or another standard section type, under a centred axial force at the ultimate limit state, to SIA 263:2013: section class, flexural buckling about both axes with its own buckling curve, section resistance and the slenderness limit. Every step shows its formula and the clause it comes from.

What it checks

Section properties from the profile dimensions, root fillets included: A, I_y, I_z, i_y, i_z and N_pl = A · f_y,k.
Section class in uniform compression: the flange outstand c/t_f, with c = (b − t_w)/2 − r (table 5b), and the web c/t_w (table 5a), with ε = √(235/f_y,k). The whole web is compressed, so it takes 33, 38 and 42 · ε, not the 72, 83 and 124 · ε of bending. The least favourable wall sets the class (table 5a).
Buckling curve for each axis from the section type and the direction of buckling. For a rolled I or H section: curve b, α_y = 0.34, about the strong axis; curve c, α_z = 0.49, about the weak axis.
Slenderness: λ_K = l_K/i, λ_E = π√(E/f_y,k) and λ̄_K = λ_K/λ_E, about each axis.
Flexural buckling, clause 4.5.1: χ_K computed rather than read off the chart, N_K = χ_K · A · f_y,k, and N_Ed ≤ N_K/γ_M1 on the governing axis.
Section resistance, clause 4.1.3: the gross section yielding, A · f_y,k/γ_M1, and the net section fracturing, 0.9 · A_net · f_u,k/γ_M2; the smaller governs.
Slenderness limit: λ_K ≤ 250 for bracing and secondary members, ≤ 200 for main structural members, with a separate category for structures subject to fatigue.
Force on each intermediate support: F_Ed = 0.01 · N_Ed, across the member in the plane it holds.

What you enter and what you get

You enter f_y,k and f_u,k from table 1 of your copy of SIA 263, for the grade and the plate thickness, E (clause 2.2), the section type and its dimensions h, b, t_w, t_f and r, and the member length l. The buckling lengths come from the end conditions and the number of equally spaced intermediate supports about each axis, or are entered directly. You also give the inclination α, N_Ed as a magnitude, γ_M1 and γ_M2 (clause 4.1.3), any holes or openings deducted, and the member category.

You get every intermediate value with its clause, the governing axis, the utilisation of each check with a pass or fail, and a drawing of the section to scale. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

A centred axial force only. Compression with bending about either axis, and flexural-torsional buckling, are each their own calculation.
The section drops to curve d when t_f exceeds 100 mm for a rolled profile or 40 mm for a welded one.
Holes filled by bolts need not normally be deducted; oversize or slotted holes and empty openings do.
An intermediate support counts only if it is tied to a bracing.
For a member that is not vertical, bending from its own weight may be neglected only while its deflection stays within l_K/3500.

Worked example

A rolled I or H section with h = 171 mm, b = 180 mm, t_w = 6 mm, t_f = 9.5 mm and r = 15 mm in f_y,k = 235 N/mm², 6.00 m long with pinned ends held against sway, three intermediate supports about the weak axis, under N_Ed = 640 kN. 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
Modulus of elasticity E	210000 N/mm²	§ 2.2
Type of section	Rolled I or H
Overall depth h	171 mm
Flange width b	180 mm
Web thickness t_w	6 mm
Flange thickness t_f	9.5 mm
Root radius r	15 mm
Length of the member l	6000 mm
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
Inclination of the member to the horizontal α	90 °
Design compressive force N_Ed	640 kN
Resistance factor γ_M1	1.05	§ 4.1.3
Section resistance	Considered
Characteristic tensile strength f_u,k	360 N/mm²	§ t.1
Resistance factor of the net section γ_M2	1.25	§ 4.1.3
Reduction of the section	None — the gross section
Member category	Main structural members — λ_K ≤ 200

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 235 N/mm²	§ t.1
Reduction factorε = √(235/f_y,k).	ε = 1.0000
Elastic limit slendernessλ_E = π√(E/f_y,k).	λ_E = 93.91
Gross areaFrom the dimensions, root fillets included.	A = 4,525 mm²
Second moment of area, strong axis	I_y = 25,101,340 mm⁴
Second moment of area, weak axis	I_z = 9,244,525 mm⁴
Radii of gyrationi = √(I/A).	i_y · i_z = 74.5 · 45.2 mm
Plastic axial forceN_pl = A·f_y,k.	N_pl = 1,063.4 kN
Flange outstandc = (b − t_w)/2 − r. Limits 9.0 · 10.0 · 14.0 for class 1, 2, 3.	c/t_f = 7.58	§ t.5b
WebIn UNIFORM compression the whole web is compressed, so it takes 33/38/42·ε and not the 72/83/124·ε it gets in bending, where only half of it is. Limits 33.0 · 38.0 · 42.0 for class 1, 2, 3.	c/t_w = 20.33	§ t.5a
Section classIn uniform compression, the least favourable wall.	– = Class 1	§ t.5a
Curve about the strong axisCurve b. Buckling about the STRONG axis puts the flange tips — which carry residual compression — near the neutral axis, where they matter least, so curve b applies. Buckling about the WEAK axis puts them at the extreme fibre, where they matter most, so curve c applies. It is the same section either way; what changes is the direction it deflects in.	α_y = 0.34
Curve about the weak axisCurve c.	α_z = 0.49

Calculation: Common values — strong axis (y–y)
Step	Value	Ref.
Buckling lengthk·l with k = 1.0 — held against sway — both ends pinned.	l_K,y = 6,000 mm
Slendernessλ_K = l_K/i.	λ_K,y = 80.6
Relative slendernessλ̄_K = λ_K/λ_E.	λ̄_K,y = 0.8578
Buckling coefficientCurve b, computed rather than read off the chart.	χ_K,y = 0.6882	§ 4.5.1
Ultimate buckling resistanceN_K = χ_K·A·f_y,k.	N_K,y = 731.8 kN	§ 4.5.1
Design buckling resistanceN_K/γ_M1.	N_K,Rd,y = 697.0 kN	§ 4.5.1

Calculation: Common values — weak axis (z–z)
Step	Value	Ref.
Buckling lengthl/(n_z + 1) = 6,000/4: the member buckles between its supports.	l_K,z = 1,500 mm
Slendernessλ_K = l_K/i.	λ_K,z = 33.2
Relative slendernessλ̄_K = λ_K/λ_E.	λ̄_K,z = 0.3534
Buckling coefficientCurve c, computed rather than read off the chart.	χ_K,z = 0.9217	§ 4.5.1
Ultimate buckling resistanceN_K = χ_K·A·f_y,k.	N_K,z = 980.2 kN	§ 4.5.1
Design buckling resistanceN_K/γ_M1.	N_K,Rd,z = 933.5 kN	§ 4.5.1

Calculation: Check 1 — buckling
Step	Value	Ref.
Governing axisThe buckling lengths differ, so both axes are checked; the smaller resistance governs.	– = Strong axis (y–y)
BucklingN_Ed ≤ N_K/γ_M1	η = 0.918 · Pass92 % of the capacity used	§ 4.5.1

Calculation: Check 2 — section resistance
Step	Value	Ref.
Gross section, yielding	A·f_y,k/γ_M1 = 1,012.8 kN	§ 4.1.3
Net areaNo reduction: the gross area.	A_net = 4,525 mm²
Net section, fracture	0,9·A_net·f_u,k/γ_M2 = 1,172.9 kN	§ 4.1.3
Section resistanceN_Ed ≤ N_Rd	η = 0.632 · Pass63 % of the capacity used

Calculation: Check 3 — slenderness limit
Step	Value	Ref.
Slenderness limitλ_K ≤ 200	η = 0.403 · Pass40 % of the capacity used

Calculation: Common values — intermediate supports
Step	Value	Ref.
Force each intermediate support must resist1 % of the axial force, F_Ed = 0,01·N_Ed, across the member in the plane it holds — for each of the 3 supports about z, and passed on by them to a bracing.	F_Ed = 6.40 kN

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 do the two axes take different buckling curves?

The curve follows where the residual stresses sit relative to the direction of buckling. Buckling about the strong axis puts the flange tips near the neutral axis; about the weak axis it puts them at the extreme fibre. Using curve c for both axes is conservative on the strong axis; using curve b for both is not conservative on the weak one.

Why does the strong axis govern in the worked example?

Only because its buckling length is the longer: three supports shorten the weak-axis length to 1,500 mm. Those supports must exist and be tied to a bracing.

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: 27 October 2026

Rolled steel section in compression to SIA 263

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