Structural Engineering Tools

Timber bar in tension and biaxial bending in the fire situation to SIA 265

Checks a rectangular solid-timber or glulam bar under axial tension and bending about both axes in the fire situation, to SIA 265:2021: the residual section after charring, lateral torsional buckling and the interaction of the three 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 bar under axial tension and bending about both axes in the fire situation, to SIA 265:2021: the residual section after charring, lateral torsional buckling and the interaction of the three 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 = βn · t (clause 4.5.2.2, equation (50)) and effective loss of thickness def = dchar,n + dred (equation (49)).
  • Residual section: bfi and hfi, each reduced by def at every exposed face; Aef = bfi · hfi (clause 4.5.2.2, figure 13), Wy,fi = bfi · hfi² / 6 and Wz,fi = hfi · bfi² / 6.
  • Lateral torsional buckling in fire: km,y,fi and km,z,fi (clause 4.2.9.2, equation (39)).
  • Resistances in fire, clause 4.5.2.5, equation (51): fRd,T,fi = 1.8 · ft,0,d · ηw; fRd,m,y,fi = 1.8 · fm,d · ηw · km,y,fi, and the same about z-z with km,z,fi. kh and ksys do not apply.
  • Tension with biaxial bending in fire, clause 4.2.4.1, equation (21): σt,0,d,fi / fRd,T,fi + σm,y,d,fi / fRd,m,y,fi + σm,z,d,fi / fRd,m,z,fi ≤ 1.
  • For comparison only, the interaction with each resistance taken as 1.5 · f · ηw · kfi, with kfi = 1.15, from EN 1995-1-2. 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 ft,0,d, fm,d, fm,k and E0,05 from your copy of SIA 265, the lateral support spacings ay and az, the design forces in fire Nd,fi, 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 t, the zero-strength layer dred, and the exposed faces.

You get the charring depth, the residual section and its moduli, the resistances and stresses in fire, and the utilisation of equation (21) 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.
  • The connection loss is not deducted again from the residual section Aef.
  • For category A and B floor areas, Nd,fi = 0.6 · Nd may be assumed.
  • Enter a spacing of 0 where the compression edge is held continuously, which gives km = 1.
  • Only the fire situation is verified here. The ultimate limit state of the same member is its own module.

Worked example

The section of the ultimate-limit-state example (GL24h, b × h = 280 × 440 mm, service class 2, compression edge held about y-y, lateral supports every 5.15 m about z-z), unprotected, charring as solid softwood, after 10 minutes of fire on all four faces, with dred = 7 mm, Nd,fi = 150 kN and My,d,fi = Mz,d,fi = 2.655 kN·m. 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 b280 mm
Section depth h440 mm
Design tensile strength ∥ grain ft,0,d12.8 N/mm²§ 3.4.2.2 t.9
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 ay0 mm§ 4.2.9.3
Lateral support spacing, bending about z-z az5150 mm§ 4.2.9.3
Design tensile force in fire Nd,fi150 kN
Design moment about y-y in fire My,d,fi2.655 kN·m
Design moment about z-z in fire Mz,d,fi2.655 kN·m
Protected at the start of the fire?No
Material for charring βnSoftwood or beech — solid, finger-jointed, reconstituted§ 4.5.2.3 t.13
Fire exposure time t10 min
Zero-strength layer dred7 mm§ 4.5.2.2
Exposed facestop, bottom, left, right
Calculation: Fire situation
StepValueRef.
Notional charring rateβn = 0.80 mm/min§ 4.5.2.3 t.13
Charring depthdchar,n = βn · tdchar,n = 8.00 mm§ 4.5.2.2 (50)
Effective loss of thicknessdef = dchar,n + dreddef = 15.00 mm§ 4.5.2.2 (49)
Residual widthbfi = b − def at each exposed facebfi = 250.0 mm
Residual depthhfi = h − def at each exposed facehfi = 410.0 mm
Notional residual sectionAef = bfi · hfi. The connection loss is not deducted again here.Aef = 102,500 mm²§ 4.5.2.2 f.13
Section modulus about y-yWy,fi = bfi · hfi² / 6Wy,fi = 7,004,167 mm³
Section modulus about z-zWz,fi = hfi · bfi² / 6Wz,fi = 4,270,833 mm³
Lateral torsional buckling factor, y-ykm,y,fi = 1.0000§ 4.2.9.2 (39)
Lateral torsional buckling factor, z-zkm,z,fi = 1.0000§ 4.2.9.2 (39)
Tensile resistance in firefRd,T,fi = 1.8 · ft,0,d · ηwfRd,T,fi = 18.432 N/mm²§ 4.5.2.5 (51)
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 = 1.8 · fm,d · ηw · km,z,fifRd,m,z,fi = 23.040 N/mm²§ 4.5.2.5 (51)
Tensile stress in fireσt,0,d,fi = Nd,fi / Aefσt,0,d,fi = 1.4634 N/mm²
Bending stress in fire, y-yσm,y,d,fi = My,d,fi / Wy,fiσm,y,d,fi = 0.3791 N/mm²
Bending stress in fire, z-zσm,z,d,fi = Mz,d,fi / Wz,fiσm,z,d,fi = 0.6217 N/mm²
Tension with biaxial bending in fireσt/ft + σm,y/fm,y + σm,z/fm,z ≤ 1η = 0.123 · Pass12 % of the capacity used§ 4.2.4.1 (21)
Alternative including kfieach resistance taken as 1.5 · f · ηw · kfi with kfi = 1.15Not part of the norm. Shown for comparison only.interaction * = 0.1282EN 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

Is the section loss at the connection deducted in fire?

No. The tensile stress in fire is taken on the notional residual section Aef = bfi · hfi; the module does not deduct the connection loss again.

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: 21 October 2026