Checks a doubly symmetric rolled I or H under a torque together with axial force, bending about both axes and shear, to SIA 263:2013: the uniform and warping torsion along the member, then the elastic comparison stress at every point where the stresses peak. Every step shows its formula and the clause it comes from.

What it checks

Limit: f_y,k/γ_M1, equation (6). No plastic reserve is used: with torsion the whole calculation is elastic (clause 4.3.5.4).
Section properties: A = 2b·t_f + (h − 2t_f)·t_w + (4 − π)·r², I_y, I_z, S_y, S_y,w above the toe of the fillet, the torsion constant K with the fillets, the warping constant I_ω = t_f·b³·(h − t_f)²/24, ω_max = b·(h − t_f)/4 and S_ω = b²·t_f·(h − t_f)/16.
Torsion along the member: the parameter χ = L·√(G·K/(E·I_ω)), then at the section the uniform torsion T_v,Ed = G·K·φ′, the warping torsion T_ω,Ed = −E·I_ω·φ‴ and the bimoment M_ω,Ed = −E·I_ω·φ″, with the largest of each along the member and where it occurs.
Shear stresses: τ_v,f = |T_v|·t_f/K and τ_v,w = |T_v|·t_w/K; the warping shear τ_ω = T_ω·S_ω/(I_ω·t_f); τ_Vy = V_y·b²/(8·I_z) and τ_Vz = V_z·S_y/(I_y·t_w).
Direct stresses: σ_N = N_Ed/A, the bending gradients about both axes, and the warping stress σ_ω = M_ω·ω_max/I_ω at the flange tips.
Class of each plate: every flange outstand to table 5b, with its own stress ratio ψ and the edge that is more compressed, and the web to table 5a.
Comparison stress, equation (6): σ_g,Ed = √(σ_x,Ed² + 3·τ_Ed²) ≤ f_y,k/γ_M1 at the flange tips, at the flanges over the web, in the web at the toe of the fillet and in the web at the neutral axis, each in its own check, and a scan along all the walls for anything higher in between.

What you enter and what you get

You enter f_y,k from table 1 of your copy of SIA 263, γ_M1 (clause 4.1.3), h, b, t_w, t_f and r, and the signed forces at the section: N_Ed, M_y,Ed, M_z,Ed, V_z,Ed and V_y,Ed. For the torsion you choose how the torque is carried: uniformly only, with the section free to warp; computed along the member from its supports and torques; or split as entered, with T_v, T_ω and M_ω from a frame program. To compute it you enter E and G (clause 3.2.2.1), the length L, the support at each end (fork, built in or free), a concentrated torque T_Ed at its distance a, a distributed torque m_x,Ed and the position x of the section verified.

You get the section properties, the split of the torque, every stress, the classes of the plates, and the utilisation of each check with a pass or fail.

Assumptions and limits

A doubly symmetric rolled I or H. For a welded or folded section enter r = 0.
A fork prevents twist but leaves the flanges free to warp; a built-in end prevents both, as a welded end plate on a stiff support or continuity does.
Where warping is prevented, part of the torque is carried by warping, and next to the restraint the whole of it.
Enter the forces with their signs. Where a sign is not known, try both: the module takes the worst point for the signs given.
A section check only. Flexural, lateral-torsional and shear buckling, distortion of the cross-section and fatigue are not checked here.

Worked example

A rolled section with h = 490 mm, b = 300 mm, t_w = 12 mm, t_f = 23 mm and r = 27 mm in f_y,k = 355 N/mm², a cantilever of 5 m built in at the left end and free at the right, with a torque T_Ed = 20 kNm at its free end, verified at the built-in end under M_y,Ed = 200 kNm and V_z,Ed = 40 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	355 N/mm²	§ t.1
Resistance factor γ_M1	1.05	§ 4.1.3
Overall depth h	490 mm
Flange width b	300 mm
Web thickness t_w	12 mm
Flange thickness t_f	23 mm
Root radius r	27 mm
Axial force N_Ed	0 kN
Bending moment about y M_y,Ed	200 kNm
Bending moment about z M_z,Ed	0 kNm
Shear force along z V_z,Ed	40 kN
Shear force along y V_y,Ed	0 kN
How the torque is carried	Computed along the member — supports and torques
Modulus of elasticity E	210000 N/mm²	§ 3.2.2.1
Shear modulus G	81000 N/mm²	§ 3.2.2.1
Length of the member L	5000 mm
Support at the left end	Built in — twist and warping prevented
Support at the right end	Free end
Concentrated torque T_Ed	20 kNm
Its distance from the left end a	5000 mm
Distributed torque m_x,Ed	0 kNm/m
Section verified — distance from the left end x	0 mm

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 355 N/mm²	§ t.1
Resistance factor	γ_M1 = 1.05	§ 4.1.3
Limit of the comparison stressNo plastic reserve is used: with torsion the whole calculation is elastic.	f_y,k/γ_M1 = 338.1 N/mm²	§ (6)
Gross area2b·t_f + (h − 2t_f)·t_w + (4 − π)·r².	A = 19,754 mm²
Second moment of area, strong axis	I_y = 869,731,765 mm⁴
Second moment of area, weak axis	I_z = 103,654,513 mm⁴
First moment of the half section	S_y = 1,974,428 mm³
First moment above the toe of the filletEverything above z = 195.0 mm.	S_y,w = 1,746,278 mm³
Torsion constant⅔(b − 0,63t_f)t_f³ + (h − 2t_f)t_w³/3 + 2(t_w/t_f)(0,145 + 0,1r/t_f)·D⁴, D the circle inscribed at the fillet.	K = 3,092,726 mm⁴
Warping constantt_f·b³·(h − t_f)²/24.	I_ω = 5,643,052,875,000 mm⁶
Sectorial coordinate at the flange tipb·(h − t_f)/4; negative at the top right and bottom left tips.	ω_max = 35,025 mm²
Sectorial static moment at the webb²·t_f·(h − t_f)/16.	S_ω = 60,418,125 mm⁴
Torsion parameter of the memberSmall (below about 1): warping carries the torque over the whole member. Large (above about 10): uniform torsion does, except near a restraint.	χ = L·√(GK/(EI_ω)) = 2.299
Uniform torsion at xG·K·φ′ at x = 0 mm.	T_v,Ed = 0.000 kNm
Warping torsion at x−E·I_ω·φ‴. T_v + T_ω is the torque at x.	T_ω,Ed = 20.000 kNm
Bimoment at x−E·I_ω·φ″.	M_ω,Ed = -42.632 kNm²
Largest bimoment along the memberAt x = 0 mm, the section verified.	max \|M_ω\| = 42.632 kNm²
Largest warping torsion along the memberAt x = 0 mm.	max \|T_ω\| = 20.000 kNm
Largest uniform torsion along the memberAt x = 5,000 mm.	max \|T_v\| = 16.025 kNm
Uniform torsion shear, flange\|T_v\|·t_f/K, opposite on the two faces.	τ_v,f = 0.00 N/mm²
Uniform torsion shear, web\|T_v\|·t_w/K.	τ_v,w = 0.00 N/mm²
Warping shear, flange over the webT_ω·S_ω/(I_ω·t_f): towards −y in the top flange when positive, +y in the bottom one.	τ_ω = 9.31 N/mm²
Shear from V_y, flange over the webV_y·b²/(8·I_z), in both flanges.	τ_Vy = 0.00 N/mm²
Shear from V_z, web at the neutral axisV_z·S_y/(I_y·t_w).	τ_Vz = 7.57 N/mm²
Warping stress at the flange tipM_ω·ω_max/I_ω, sign at the top left tip.	σ_ω = -264.60 N/mm²
Direct stress from N aloneN_Ed/A.	σ_N = 0.00 N/mm²
Bending stress gradientsσ = σ_N + a·y + c·z + M_ω·ω/I_ω: a·I_yz + c·I_y = −M_y, a·I_z + c·I_yz = M_z.	∂σ/∂y, ∂σ/∂z = 0.000 / -229.956 N/mm² per m
Class of the flange, top rightClass 1 to 3: c/t = 117.0/23.0 ≤ 83.35 (21ε√k₂, k₂ = 23.8000), ψ = -1.000, ε = 0.814, supported edge the more compressed.	c/t = 5.09	§ t.5b
Class of the flange, top leftClass 1 to 3: c/t = 117.0/23.0 ≤ 12.52 (21ε√k₁, k₁ = 0.5366), ψ = 0.169, ε = 0.814, free edge the more compressed.	c/t = 5.09	§ t.5b
Class of the flange, bottom rightClass 1 to 3: c/t = 117.0/23.0 ≤ 13.54 (21ε√k₁, k₁ = 0.6280), ψ = -0.255, ε = 0.814, free edge the more compressed.	c/t = 5.09	§ t.5b
Class of the flange, bottom leftNo compression: no local buckling.	– = in tension
Class of the webClass 1 to 3: c/t = 390.0/12.0 ≤ 100.89 (62ε(1−ψ)√(−ψ) [ψ ≤ −1]), ψ = -1.000, ε = 0.814.	c/t = 32.50	§ t.5a
Governing point	– = flange tip, σ_g = 320.9 N/mm²	§ (6)
Scan along all wallsNothing between the named points is higher: point 1 governs.	max σ_g = 320.9 N/mm²

Calculation: Check 1 — Comparison stress at the flange tips
Step	Value	Ref.
Direct stress at the flange tipsWorst of the 4: top left. σ_N + σ_M + σ_ω = 0.0 + -56.3 + -264.6.	σ_x,Ed = -320.94 N/mm²
Shear stress at the flange tipsτ_v,f = 0.00, taken in full at the edge; no flow reaches a free edge.	τ_Ed = 0.00 N/mm²
Comparison stress at the flange tipsσ_g,Ed ≤ f_y,k/γ_M1	η = 0.949 · Pass95 % of the capacity used	§ (6)

Calculation: Check 2 — Comparison stress at the flanges over the web
Step	Value	Ref.
Direct stress at the flanges over the webWorst of the 2: top. σ_N + σ_M + σ_ω = 0.0 + -56.3 + -0.0.	σ_x,Ed = -56.34 N/mm²
Shear stress at the flanges over the web\|τ_Vy − τ_ω\| + τ_v,f = \|0.00 − 9.31\| + 0.00.	τ_Ed = 9.31 N/mm²
Comparison stress at the flanges over the webσ_g,Ed ≤ f_y,k/γ_M1	η = 0.173 · Pass17 % of the capacity used	§ (6)

Calculation: Check 3 — Comparison stress at the web at the toe of the fillet
Step	Value	Ref.
Direct stress at the web at the toe of the filletWorst of the 2: top. σ_N + σ_M + σ_ω = 0.0 + -44.8 + -0.0.	σ_x,Ed = -44.84 N/mm²
Shear stress at the web at the toe of the fillet\|τ_Vz,w\| + τ_v,w = 6.69 + 0.00, z = ±195.0 mm.	τ_Ed = 6.69 N/mm²
Comparison stress at the web at the toe of the filletσ_g,Ed ≤ f_y,k/γ_M1	η = 0.137 · Pass14 % of the capacity used	§ (6)

Calculation: Check 4 — Comparison stress at the web at the neutral axis
Step	Value	Ref.
Direct stress at the web at the neutral axisMid-height. σ_N + σ_M + σ_ω = 0.0 + 0.0 + -0.0.	σ_x,Ed = 0.00 N/mm²
Shear stress at the web at the neutral axis\|τ_Vz\| + τ_v,w = 7.57 + 0.00.	τ_Ed = 7.57 N/mm²
Comparison stress at the web at the neutral axisσ_g,Ed ≤ f_y,k/γ_M1	η = 0.039 · Pass4 % of the capacity used	§ (6)

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 no plastic resistance used?

With torsion the check is elastic, clause 4.3.5.4: the comparison stress must not exceed f_y,k/γ_M1 anywhere in the section.

How much does warping add in the worked example?

At the built-in end the whole torque is carried by warping, T_ω,Ed = 20 kNm, with a bimoment of 42.6 kNm². Of the −320.9 N/mm² at the governing flange tip, −264.6 N/mm² is the warping stress and −56.3 N/mm² the bending. η = 0.949.

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: 14 November 2026

Rolled I or H section under torsion, axial force, bending and shear to SIA 263

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