Timber beam in shear to SIA 265

The code governs which inputs appear and how every value is derived.

Member

Material and cross-section, shared by every verification. Change the verification above and these stay as they are.

About this check

clauses, worked example and questions

Checks a rectangular solid-timber or glulam beam in shear at the ultimate limit state, to SIA 265:2021, including stress perpendicular to the grain and a notched support. Every step shows its formula and the clause it comes from.

What it checks

What you enter and what you get

You enter the member (wood type, strength class, load duration, service class, b, h), the design shear strength fv,d from your copy of SIA 265, whether the piece contains the pith, the system factor ksys (clause 5.7.2.2) and the design shear forces Vz,d and Vy,d. For the transverse stress you choose tension, compression or none, and enter its magnitude with fc,90,d and ft,90,d. For a notch you enter its side, hef, the distance c from the support reaction to the notch corner, VEd and whether it is cut on a slope.

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

Assumptions and limits

Worked example

A glulam beam GL24h, b × h = 140 × 700 mm, in service class 1 with ksys = 0.8, under 172 kN of shear, a transverse compression of 1.96 N/mm², and a notch on the tension side leaving 440 mm of depth at 140 mm from the support reaction, with 140 kN at the notch. In this example none of the three checks passes. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Wood typeGlulam
Strength classGL24h
Load duration ηtShort to permanent§ 2.2.6
Service class ηw1§ 3.2.1.3 t.4
Section width b140 mm
Section depth h700 mm
Design shear strength fv,d1.8 N/mm²§ 3.4.2.2 t.9
Piece contains the pith?No§ 3.3.2.1 t.8 n.6
System factor ksys0.8§ 5.7.2.2
Design shear force, z direction Vz,d172 kN
Design shear force, y direction Vy,d0 kN
Transverse stress at the sectionCompression perpendicular to the grain§ 4.2.7.1
Transverse stress, magnitude |σ90,d|1.96 N/mm²
Design compressive strength ⟂ grain fc,90,d1.9 N/mm²§ 3.3.2.1 t.8 / 3.4.2.2 t.9
Design tensile strength ⟂ grain ft,90,d0.15 N/mm²§ 3.3.2.1 t.8 / 3.4.2.2 t.9
Loads within h of the support neglected?No§ 4.2.7.2
Notch at the supportOn the tension side§ 5.2.2
Reduced depth hef440 mm§ 5.2.2.1
Design shear force at the notch VEd140 kN
Distance support reaction to notch corner c140 mm§ 5.2.2
Notch cut on a slope?No§ 5.2.2.2
Calculation: Common values
StepValueRef.
Load duration factorηt = 1.00§ 2.2.6
Moisture factorηw = 1.00§ 3.2.1.3 t.4
Design shear resistancefRd = fv,d · ksys · ηt · ηwfRd = 1.440 N/mm²§ 4.2.5.1 (23)
Calculation: Check 1 — shear parallel to the grain
StepValueRef.
Shear stress, z directionτxz,d = 1.5 · Vz,d / A, the maximum at the neutral axis of a rectangleτxz,d = 2.633 N/mm²§ 4.2.3
Shear parallel to the grainτd ≤ fv,dη = 1.828 · Fail183 % of the capacity used§ 4.2.5.1 (23)
Calculation: Check 2 — shear with stress perpendicular to the grain
StepValueRef.
Correction coefficientkt,90 = 4.0 for transverse compression, 1.0 for transverse tensionkt,90 = 4.0§ 4.2.7.1
Transverse stress, signedcompression is introduced as a negative valueσ90,d = -1.960 N/mm²
Compressive resistance ⟂ grainfc,90,d · ηt · ηwfc,90,d = 1.900 N/mm²
Tensile resistance ⟂ grainft,90,d · ηt · ηwft,90,d = 0.150 N/mm²
Shear component usedThe interaction of §4.2.7.1 is written for shear in one direction. The z component is used, since a reaction applied across the grain acts together with that component.τxz,d = 2.633 N/mm²§ 4.2.7.1
Shear with stress perpendicular to the graininteraction ≤ 1η = 1.412 · Fail141 % of the capacity used§ 4.2.7.1 (27)
Interaction expression[(fc,90,d + σ90,d)/(fc,90,d + kt,90·ft,90,d)]² + (τd/fv,d)²·[1 − (fc,90,d/(fc,90,d + kt,90·ft,90,d))²] ≤ 11.4124§ 4.2.7.1 (27)
Calculation: Check 3 — shear at the section reduction
StepValueRef.
Notch depthΔhef = h − hefΔhef = 260.0 mm
Reference notch depth25 mm for softwood solid timber, 45 mm for softwood glulamΔh0 = 45 mm§ 5.2.2 t.16
Reduction factorkred = √(Δh0/h) / [√(α − α²) + 0.8·(c/h)·√(1/α − α²)] ≤ 1, α = hef/hkred = 0.3852§ 5.2.2 t.16 (58)
Reduced shear resistancefRd = kred · fv,d · ksys · ηt · ηwfRd,notch = 0.555 N/mm²§ 5.2.2.1 (57)
Shear stress on the reduced sectionτd = 1.5 · VEd / (b · hef)τd,notch = 3.409 N/mm²§ 5.2.2.1 (57)
Shear at the section reductionτd ≤ kred · fv,dη = 6.145 · Fail615 % of the capacity used§ 5.2.2.1 (57)

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 module ask whether the piece contains the pith?

For pieces containing the pith the design strength is halved (clause 3.3.2.1, table 8). Drying cracks radiate from the pith and open up the shear plane.

How is shear in two directions treated?

SIA 265 gives no model for it. Where Vy,d is not zero, the module uses the resultant of DIN 1052 and labels it in the results.

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.

Related

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