Timber beam in bending in the fire situation to SIA 265

Checks a rectangular solid-timber or glulam beam in bending about y-y in the fire situation, to SIA 265:2021: the residual section after charring and lateral torsional buckling on that narrower section. Every step shows its formula and the clause it comes from.

Basic SIA 265:2021

Part of the Basic plan

This module is included in the Basic plan and above, with its calculation note.

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About this check

clauses, worked example and questions

What it checks

Charring: notional charring rate β_n (clause 4.5.2.3, table 13), charring depth d_char,n = β_n · t (clause 4.5.2.2, equation (50)) and effective loss of thickness d_ef = d_char,n + d_red (equation (49)).
Residual section: b_fi and h_fi, each reduced by d_ef at every exposed face, and W_fi = b_fi · h_fi² / 6 (clause 4.5.2.2, figure 13).
Lateral torsional buckling in fire: λ_rel,m,fi from equation (42) of clause 4.2.9.3, formed with h_fi and b_fi, and the factor k_m,fi (clause 4.2.9.2, equation (39)).
Resistance in fire, clause 4.5.2.5, equation (51): f_Rd,fi = 1.8 · f_m,d · η_w · k_m,fi. k_h and k_sys do not apply.
Bending in the fire situation: σ_m,d,fi = M_d,fi / W_fi ≤ f_Rd,fi.
For comparison only, the resistance with the factor k_fi of EN 1995-1-2: 1.7 · f_m,d · η_w · k_fi · k_m,fi, with k_fi = 1.25. It is not part of SIA 265.

What you enter and what you get

You enter the member (wood type, strength class, service class, b, h), the strengths f_m,d, f_m,k and E_0,05 from your copy of SIA 265, how the compression edge is held (continuously, or by lateral supports at a spacing a), the design bending moment in fire M_d,fi, whether the beam was protected at the start of the fire, the material for charring, the exposure time t, the zero-strength layer d_red, and the exposed faces.

You get the charring depth, the residual section, the slenderness and k_m,fi, the resistance and stress in fire, and the utilisation with a pass or fail, each with its clause. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Rectangular sections; the charring model of SIA 265 clause 4.5.2.
A floor beam is usually exposed on three faces, the top being protected by the deck; tick the faces that the fire reaches.
For category A and B floor areas, M_d,fi = 0.6 · M_d may be assumed.
With fire on several faces, the module warns when the width before the fire is below the minimum for R60 (140 mm, clause 4.5.2.4); heat accumulation must then be considered. It is flagged, not calculated.
Only the fire situation is verified here. The ultimate limit state of the same beam is its own module.

Worked example

The beam of the ultimate-limit-state example (C24, b × h = 80 × 120 mm, service class 1, lateral supports every 2.00 m), unprotected, charring as solid softwood, after 30 minutes of fire on the bottom and both sides, with d_red = 7 mm and a design moment in fire of 1.593 kN·m (0.6 · M_d). In this example the bending check in fire does not pass: the residual width is 18 mm and lateral torsional buckling reduces the resistance on that section. The module also warns that the section is narrower than the minimum width for R60 with fire on several faces. These are the module's own results, computed when this page was built.

Inputs
Input	Value	Ref.
Wood type	Solid
Strength class	C24
Service class η_w	1	§ 3.2.1.3 t.4
Section width b	80 mm
Section depth h	120 mm
Design bending strength f_m,d	14.0 N/mm²	§ 3.3.2.1 t.8
Characteristic bending strength f_m,k	24 N/mm²	§ Annexe B
Modulus of elasticity, 5% fractile E_0,05	7300 N/mm²	§ Annexe B
Compression edge	Lateral supports at spacing a	§ 4.2.9
Spacing of lateral supports a	2000 mm	§ 4.2.9.3
Design bending moment in fire M_d,fi	1.593 kN·m
Protected at the start of the fire?	No
Material for charring β_n	Softwood or beech — solid, finger-jointed, reconstituted	§ 4.5.2.3 t.13
Fire exposure time t	30 min
Zero-strength layer d_red	7 mm	§ 4.5.2.2
Exposed faces	bottom, left, right

Calculation: Fire situation
Step	Value	Ref.
Notional charring rate	β_n = 0.80 mm/min	§ 4.5.2.3 t.13
Charring depthd_char,n = β_n · t	d_char,n = 24.00 mm	§ 4.5.2.2 (50)
Effective loss of thicknessd_ef = d_char,n + d_red	d_ef = 31.00 mm	§ 4.5.2.2 (49)
Residual widthb_fi = b − d_ef at each exposed face	b_fi = 18.0 mm
Residual depthh_fi = h − d_ef at each exposed face	h_fi = 89.0 mm
Residual section modulusW_fi = b_fi · h_fi² / 6	W_fi = 23,763 mm³	§ 4.5.2.2 f.13
Relative slendernessλ_rel,m = 1.15 · (√(a·h_fi) / b_fi) · √(f_m,k / E_0,05)	λ_rel,m,fi = 1.5455	§ 4.2.9.3 (42)
Lateral torsional buckling factor	k_m,fi = 0.4186	§ 4.2.9.2 (39)
Design resistance in firef_Rd,fi = 1.8 · f_m,d · η_w · k_m,fi. k_h and k_sys do not apply.	f_Rd,fi = 10.550 N/mm²	§ 4.5.2.5 (51)
Design stress in fireσ_m,d,fi = M_d,fi / W_fi	σ_m,d,fi = 67.037 N/mm²
Bending in the fire situationσ_m,d,fi ≤ f_Rd,fi	η = 6.354 · Fail635 % of the capacity used	§ 4.5.2.5 (51)
Alternative including k_fif_Rd,fi* = 1.7 · f_m,d · η_w · k_fi · k_m,fi, k_fi = 1.25Not part of the norm. Utilisation would be 538.3%.	f_Rd,fi * = 12.455 N/mm²	EN 1995-1-2

How this module is verified

Verified clause by clause against SIA 265:2021 and against independent calculations used on real projects, before release.

Questions

Why does lateral torsional buckling matter more in fire?

The slenderness is formed with b_fi and h_fi, not with b and h. In the worked example the width falls from 80 to 18 mm and the depth from 120 to 89 mm, so λ_rel,m,fi rises to about 1.55 and k_m,fi falls to about 0.42.

Is the Eurocode value used?

No. The EN 1995-1-2 resistance with k_fi is shown beside the result for comparison only; the verification follows SIA 265.

Is this module free?

The fire situation is part of the Basic plan, with the calculation note of every module. The ultimate limit state is free.

Norm: SIA 265:2021 · Page updated: 21 October 2026