Timber beam design
A sawn-lumber beam is designed by adjusting a published reference stress through a chain of factors and then checking bending, shear, deflection and bearing against it. This runs the whole NDS allowable-stress chain, including the beam stability factor that applies whenever the compression edge is free to move sideways.
| Quantity | Value | Note |
|---|---|---|
| Species and grade | Douglas Fir-Larch No. 2 | NDS Supplement Table 4A, No. 2 |
| Reference Fb | 875 psi | 12-inch-deep, normal duration, dry |
| Load duration CD | 1.00 | Table 2.3.2 |
| Size factor CF | 1.10 | Table 4A, by nominal depth |
| Volume factor CV | 1.000 | beams 5 in and deeper |
| Repetitive factor Cr | 1.00 | three or more members at 24 in or less |
| Beam stability CL | 1.000 | compression edge braced |
| Adjusted Fb | 963 psi | the product of all of the above |
| Section modulus S | 21.39 in³ | bd²/6 |
| Moment of inertia I | 98.9 in⁴ | bd³/12 |
| Moment M | 8640 lb-in | wL²/8 |
| Bending stress fb | 404 psi | 42% of allowable |
| Shear V | 240 lb | wL/2 |
| Shear stress fv | 25.9 psi | 25% of allowable |
| Deflection | 0.118 in | limit 0.400 in |
| Bearing stress | 46 psi | 7% of Fc⊥ |
| Allowable uniform load | 81 plf | governed by bending |
The stability factor is the one people skip. It only applies when the compression edge is unbraced, but when it does apply it can cut the allowable bending stress by half — which is exactly the case for a beam whose top edge is free to move sideways.
fb = M/S ≤ Fb′ · Fb′ = Fb·CD·CF·CV·CL·Cr
fv = 3V/2bd ≤ Fv′ · Parabolic distribution, maximum at mid-depth.
Δ = 5wL⁴/384EI · L/360 for floor live load, L/240 for total load.
CL from FbE = 1.20E′min/(Rb)² · NDS 3.3.3, with Rb = √(le·d/b²).
A clear path from inputs to results.
A sawn-lumber beam is designed by taking a published reference stress, multiplying it through a chain of adjustment factors, and then checking bending, shear, deflection and bearing against the result. This runs the whole NDS allowable-stress chain, including the beam stability factor that applies whenever the compression edge is free to move sideways.
How to use
- Choose the species and grade, which set the reference stresses.
- Enter the member size, span and uniform load.
- Pick the load duration and say whether it is a repetitive member.
- Set the unbraced length of the compression edge and the bearing length.
- Read the four utilisation checks and the allowable load.
The formula
Reference: NDS for Wood Construction (AWC), §3.3 and Tables 2.3.2, 4A and 4.3.1; reference values from the NDS Supplement.
Worked example
2×10 Douglas Fir-Larch No. 2 over 12 ft at 40 plf
- 1.5 × 9.25 in member, 12 ft span, 40 plf uniform load.
- Douglas Fir-Larch No. 2, normal load duration.
- Compression edge continuously braced, 3.5 in bearing.
- Deflection limit L/360.
- 1
Section modulus is 1.5 × 9.25² / 6 = 21.4 in³ and moment of inertia is 98.9 in⁴.
- 2
The simple-span moment is 40 × 144 × 12 / 8 = 8,640 lb-in, and the shear is 240 lb.
- 3
Size factor CF for a 9.25 in depth is 1.1; the volume factor is 1.0 at this span; the member is braced so CL is 1.0.
- 4
Adjusted Fb is 875 × 1.0 × 1.1 = 963 psi, and the applied bending stress is 8,640 / 21.4 = 404 psi — 42% of allowable.
- 5
Deflection is 0.118 in against a limit of 0.400 in, so it uses 29% of its allowance.
- 6
Bending governs, and the allowable uniform load is 81 plf against the 40 plf applied.
42% bending · 29% deflection · 81 plf allowable · bending governs.
Interpreting the result
The governing check is the one to fix. Here bending governs, so a deeper member or a shorter span helps immediately; if deflection governed, a stiffer species or a deeper member would be the only levers. The stability factor is the one people skip — it only applies when the compression edge is unbraced, but when it does it can halve the allowable bending stress.
Assumptions & limitations
- Visually graded sawn lumber, No. 2 grade, dry service, normal temperature.
- Reference values are from the NDS Supplement Table 4A and apply to a 12-inch-deep member under normal-duration load.
- Emin is taken as 0.35E, the published ratio for this grade, for the stability calculation.
- Simply supported, uniformly loaded, with the load applied to the top edge.
- Wet service, fire-retardant treatment, incising and elevated temperature each carry their own factor and are not modelled.
- Deflection is bending deflection only; shear deflection is negligible at these span-to-depth ratios.
Frequently asked questions
Why is the allowable bending stress higher than the reference value?
Because of the size factor. A 2×10 is stronger per square inch in bending than a 12-inch-deep reference member, and CF captures that. The volume factor and the stability factor then pull it back down for long or unbraced members.
When does the beam stability factor apply?
When the compression edge is not continuously braced against lateral movement — a beam whose top edge is free to translate sideways. It does not apply to a joist with decking screwed to it, which is why floor joists rarely use it and headers often do.
Can I use this for engineered lumber?
Not directly. LVL, glulam and I-joists have their own reference values and their own adjustment factors, published by the manufacturer. Use this for solid sawn lumber only.
What does the load duration factor do?
Wood carries more load for a short time than for a long one. Snow gets 1.15, wind and seismic 1.6, and permanent load 0.9. It is the reason a roof can be designed for a snow load that would fail it if applied every day.
Should I design to L/360 or L/240?
L/360 for live load under floors, L/240 for total load, and L/180 where finishes are not sensitive. The limit is about what the finishes above can tolerate, not about safety — a beam can pass bending and still crack plaster.
Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.