Structural Engineering Tools

Timber beam in biaxial bending in the fire situation to SIA 265

Checks a rectangular solid-timber or glulam beam bent about both axes in the fire situation, to SIA 265:2021: the residual section after charring, lateral torsional buckling, and the interaction of the two bending stresses. 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.

About this check

clauses, worked example and questions

Checks a rectangular solid-timber or glulam beam bent about both axes in the fire situation, to SIA 265:2021: the residual section after charring, lateral torsional buckling, and the interaction of the two bending stresses. Every step shows its formula and the clause it comes from.

What it checks

  • Charring: notional charring rate βn (clause 4.5.2.3, table 13), charring depth dchar,n (clause 4.5.2.2, equation (50)) and effective loss of thickness def = dchar,n + dred (equation (49)), on each exposed face.
  • Residual section: bfi, hfi and the section moduli Wy,fi, Wz,fi.
  • Lateral torsional buckling in fire: λrel,m (clause 4.2.9.3, equation (42)) and km (clause 4.2.9.2, equation (39)) on the residual section.
  • Resistance in fire, clause 4.5.2.5, equation (51): fRd,m,fi = 1.8 · fm,d · ηw · km,fi; kh and ksys do not apply.
  • Biaxial bending in fire, clause 4.2.3.1, equation (14): the two utilisations add linearly.
  • For comparison only, the interaction with the factor kfi of EN 1995-1-2 (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 fm,d, fm,k and E0,05 from your copy of SIA 265, the lateral support spacings, the design moments in fire My,d,fi and Mz,d,fi, whether the member was protected at the start of the fire, the material for charring, the exposure time, the zero-strength layer, and the exposed faces.

You get the residual section, every intermediate value with its clause, the utilisation with a pass or fail, and figures of the residual section and of the charring depth against time. 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.
  • For floors in categories A and B, the module notes that Md,fi = 0.6 · Md may be assumed; otherwise enter the moments of the accidental combination for fire.
  • With fire on several faces, narrow sections are flagged (minimum width before the fire for R30 and R60, clause 4.5.2.4); heat accumulation is flagged, not calculated.
  • The ultimate limit state at normal temperature is its own module.

Worked example

The beam of the ultimate-limit-state example (GL24h, b × h = 680 × 280 mm, service class 2), unprotected, after 60 minutes of fire on all four faces, with design moments in fire of 94 and 1 kN·m. In this example the check does not pass: the residual section is too small for these moments. These are the module's own results, computed when this page was built.

Inputs
InputValueRef.
Wood typeGlulam
Strength classGL24h
Service class ηw2§ 3.2.1.3 t.4
Section width b680 mm
Section depth h280 mm
Design bending strength fm,d16 N/mm²§ 3.4.2.2 t.9
Characteristic bending strength fm,k24 N/mm²§ Annexe B
Modulus of elasticity, 5% fractile E0,059600 N/mm²§ Annexe B
Lateral support spacing, bending about y-y ay6820 mm§ 4.2.9.3
Lateral support spacing, bending about z-z az6820 mm§ 4.2.9.3
Design moment about y-y in fire My,d,fi94 kN·m
Design moment about z-z in fire Mz,d,fi1 kN·m
Protected at the start of the fire?No
Material for charring βnSoftwood or beech — glulam§ 4.5.2.3 t.13
Fire exposure time t60 min
Zero-strength layer dred7 mm§ 4.5.2.2
Exposed facestop, bottom, left, right
Calculation: Fire situation
StepValueRef.
Notional charring rateβn = 0.70 mm/min§ 4.5.2.3 t.13
Charring depthdchar,n = βn · tdchar,n = 42.00 mm§ 4.5.2.2 (50)
Effective loss of thicknessdef = dchar,n + dreddef = 49.00 mm§ 4.5.2.2 (49)
Residual widthbfi = 582.0 mm
Residual depthhfi = 182.0 mm
Section modulus about y-yWy,fi = bfi · hfi² / 6Wy,fi = 3,213,028 mm³§ 4.5.2.2 f.13
Section modulus about z-zWz,fi = hfi · bfi² / 6Wz,fi = 10,274,628 mm³
Relative slenderness in fire, y-yλrel,m,y,fi = 0.1101§ 4.2.9.3 (42)
Relative slenderness in fire, z-zλrel,m,z,fi = 0.6294§ 4.2.9.3 (42)
Lateral torsional buckling factor in fire, y-ykm,y,fi = 1.0000§ 4.2.9.2 (39)
Lateral torsional buckling factor in fire, z-zkm,z,fi = 1.0000§ 4.2.9.2 (39)
Bending resistance in fire, y-yfRd,m,y,fi = 1.8 · fm,d · ηw · km,y,fi. kh and ksys do not apply.fRd,m,y,fi = 23.040 N/mm²§ 4.5.2.5 (51)
Bending resistance in fire, z-zfRd,m,z,fi = 23.040 N/mm²§ 4.5.2.5 (51)
Bending stress in fire, y-yσm,y,d,fi = My,d,fi / Wy,fiσm,y,d,fi = 29.2559 N/mm²
Bending stress in fire, z-zσm,z,d,fi = Mz,d,fi / Wz,fiσm,z,d,fi = 0.0973 N/mm²
Biaxial bending in fireσm,y/fm,y + σm,z/fm,z ≤ 1η = 1.274 · Fail127 % of the capacity used§ 4.2.3.1 (14)
Alternative including kfieach resistance taken as 1.5 · fm,d · ηw · kfi with kfi = 1.15Not part of SIA 265. Shown for comparison only.interaction * = 1.3294EN 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 the resistance rise to 1.8 · f_m,d in fire?

SIA 265 equation (51) converts the design strength to the fire situation with this factor; the residual section carries the load. kh and ksys are not applied.

Is the Eurocode value used?

No. The EN 1995-1-2 interaction with kfi 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.

Related

Norm: SIA 265:2021 · Page updated: 20 October 2026