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Timber beam design to EN 1995-1-1

Timber is strong for its weight, but its strength depends on how long the load acts and how wet the timber gets: that is the factor k_mod. A deep, narrow joist can also roll over sideways before it reaches its bending strength (lateral torsional buckling). And timber creeps a lot, so the long-term deflection is often what decides the size. Every number below comes from the free STATIX timber beam calculator with the inputs in step 1.

1. The problem

Input	Value
Supports	Simply supported
Span L	4.0 m
Load combinations	SANS 10160-1 (1.2G + 1.6Q)
Resistance	EN 1995-1-1 (kmod, γM)
Width b	75 mm
Depth h	300 mm
Plies	1
Material	Solid timber
Grade fills the properties	EN 338 C24 (traceable)
Service class	1 — interior, dry
Permanent gk	2.0 kN/m
Imposed qk	2.5 kN/m
Imposed duration kmod	Medium-term

2. Design strengths

Each characteristic strength (bending, shear, bearing) is multiplied by k_mod (load duration and service class) and divided by the material factor γ_M. Small, deep sections get a size bonus k_h in bending.

[EN 1995-1-1 2.4.1, 3.1.3, 3.2] k_mod = 0.80, k_h = 1.00 f_m,d = 14.77, f_v,d = 2.46, f_c,90,d = 1.54 MPa

3. Bending

The bending stress is the design moment over the section modulus; it must stay below the design bending strength (reduced for buckling in the next step).

[EN 1995-1-1 6.1.6] M_d = 12.80 kNm, W_y = 1,125 × 10³ mm³ σ_m,d = 11.38 MPa

4. Lateral torsional buckling

The compression edge of a beam that is not held sideways can buckle out of plane and twist. The critical stress depends on the effective length between restraints (longer when the load sits on top); it gives a relative slenderness and a reduction factor k_crit on the bending strength.

[EN 1995-1-1 6.3.3] effective length 4,200 mm, σ_m,crit = 25.77 MPa λ_rel,m = 0.965 → k_crit = 0.836 σ_m,d / k_crit f_m,d: 11.38 ≤ 12.35 MPa OK

5. Shear

The shear stress at the neutral axis is 1,5 V / (b_ef h); cracks in the timber reduce the width that works (b_ef = k_cr × b).

[EN 1995-1-1 6.1.7] V_d = 12.80 kN, b_ef = 50.3 mm τ_d / f_v,d: 1.274 ≤ 2.462 MPa OK

6. Bearing at the supports

Timber is weak across the grain. The reaction spread over the bearing length (plus a small allowance each side) gives the bearing stress perpendicular to the grain.

[EN 1995-1-1 6.1.5] effective bearing length 130 mm σ_c,90,d / f_c,90,d: 1.313 ≤ 1.538 MPa OK

7. Deflection, now and later

The instantaneous deflection is from the dead and imposed loads. Creep then adds a part of it again (k_def, larger for wetter service classes), mostly on the permanent load. The final deflection is usually the limiting check for timber.

[EN 1995-1-1 2.3.2.2, 7.2] u_G = 3.90, u_Q = 4.88 mm u_inst = 8.78 mm → u_fin = 11.99 mm (with creep)

The calc sheet shows the deflection limits and every load combination.

Run this example yourselfOpen the free STATIX timber beam calculator and press Load the example (75 × 300 C24 joist, 4 m) under "How it is calculated". Change any value and every step updates, with a calc sheet.
Open the timber beam calculator

A worked example for learning, not a design. Check every value against the current code and your own project, and have the design checked and signed by a registered professional engineer.

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