How to calculate the U-value of a timber frame wall, and why the studs matter
A U-value tells you how quickly heat passes through a wall, roof or floor. For a wall made only of flat, continuous layers, working it out is simple arithmetic. Very few real walls are like that. In a timber frame wall the insulation sits between studs, and the studs let heat through about six times faster than the insulation beside them. Allowing for them correctly is the difference between a U-value that reflects the wall and one that flatters it.
This guide works through a real timber frame wall, layer by layer, using the method set out in BR 443 Conventions for U-value calculations and BS EN ISO 6946.
Step 1: the resistance of each layer
Every layer resists the flow of heat. Its thermal resistance, R, is its thickness in metres divided by its thermal conductivity, λ (BR 443, section 1):
R = thickness ÷ λ
So 120 mm of insulation with a λ of 0.022 W/mK has a resistance of 0.12 ÷ 0.022 = 5.455 m²K/W. Use the conductivity the manufacturer declares for the product you're specifying, from its datasheet or certificate.
Air spaces are given a resistance directly rather than a conductivity. A typical unventilated cavity in a wall is 0.18 m²K/W (BR 443, section 4.7.1). A cavity next to a foil-faced board resists more, because the low-emissivity surface reduces the heat radiated across it.
Step 2: add the surface resistances
The thin film of air against each face of the wall adds a little resistance too. BR 443 gives the values to use (section 4.1):
| Element | Heat flow | Inside, Rsi | Outside, Rse |
|---|---|---|---|
| Wall | Horizontal | 0.13 | 0.04 |
| Roof | Upwards | 0.10 | 0.04 |
| Floor | Downwards | 0.17 | 0.04 |
Figures in m²K/W. A roof steeper than 60° is treated as a wall.
For a wall of continuous layers, the total resistance is the sum of everything, and the U-value is one divided by the total:
U = 1 ÷ (Rsi + R₁ + R₂ + … + Rse)
Step 3: allow for the timber
Where insulation sits between studs, heat has two routes through that layer: through the insulation, or through the timber. Timber has a λ of about 0.13 W/mK against 0.022 for PIR insulation, so a disproportionate share of the heat takes the timber route.
How much of the wall is timber. BR 443 sets a default timber fraction of 15% for timber frame walls, based on 38 mm studs at 600 mm centres plus the plates, dwangs (or noggings elsewhere in the UK) and extra studs a real frame has. A lower 12.5% can be used only if the frame meets specific conditions, such as a single top plate and no mid-height dwangs (section 4.4.1). For other elements, the fraction is the timber width divided by the spacing, plus 1% for extra pieces: ceiling joists of 47 mm at 400 mm centres give 13% (section 4.5).
The combined method. BS EN ISO 6946 works out two estimates of the wall's total resistance and takes their average (BR 443, section 1):
- The upper limit imagines the heat travelling straight through the wall along separate paths: one through the insulation, one through the stud, and so on. It works out the resistance of each complete path, then combines the paths in proportion to the area each covers.
- The lower limit works layer by layer instead. Each mixed layer is given one combined resistance, as if heat could spread sideways freely within it, and then the layers are added up as usual.
The true figure lies between the two. The method is valid only while the upper limit is no more than 1.5 times the lower. Beyond that, or where metal passes through the insulation (apart from point fixings like wall ties), a detailed numerical calculation is needed instead (BR 443, section 2).
Worked example: a timber frame wall with a block outer leaf
This is a real wall from a recent project, listed from the inside out:
| Layer | Thickness | λ (W/mK) | R (m²K/W) |
|---|---|---|---|
| Inside surface, Rsi | 0.130 | ||
| Plasterboard | 12.5 mm | 0.19 | 0.066 |
| Service void, between 47 mm battens (8.8% timber) | 25 mm | air: given R / timber 0.13 | air 0.665 / timber 0.192 |
| PIR insulation board, foil-faced | 25 mm | 0.022 | 1.136 |
| PIR insulation between studs (15% timber) | 120 mm | 0.022 / timber 0.13 | insulation 5.455 / timber 0.923 |
| OSB sheathing | 9 mm | 0.13 | 0.069 |
| Breather membrane | (negligible) | ||
| Cavity, unventilated | 50 mm | 0.180 | |
| Medium density blockwork | 100 mm | 0.51 | 0.196 |
| Render | 15 mm | 1.0 | 0.015 |
| Outside surface, Rse | 0.040 |
Two layers are bridged: the service void by battens, and the main insulation by studs. Everything else is continuous, and together those continuous layers and the two surfaces add up to 1.832 m²K/W.
Upper limit. With two bridged layers there are four paths through the wall. Each path's share of the area is its share of the first bridged layer multiplied by its share of the second:
| Path | Share of area | Total R of the path |
|---|---|---|
| Air void, then insulation | 91.2% × 85% = 77.52% | 1.832 + 0.665 + 5.455 = 7.952 |
| Air void, then stud | 91.2% × 15% = 13.68% | 1.832 + 0.665 + 0.923 = 3.421 |
| Batten, then insulation | 8.8% × 85% = 7.48% | 1.832 + 0.192 + 5.455 = 7.479 |
| Batten, then stud | 8.8% × 15% = 1.32% | 1.832 + 0.192 + 0.923 = 2.948 |
The paths are combined as parallel routes for heat: divide each share by its path's resistance, add the results, and take one over the total. That gives an upper limit of 6.581 m²K/W.
Lower limit. Each bridged layer gets one combined resistance in the same way: one over (share ÷ R for the air, plus share ÷ R for the timber).
- Service void: 1 ÷ (0.912 ÷ 0.665 + 0.088 ÷ 0.192) = 0.547
- Insulation between studs: 1 ÷ (0.85 ÷ 5.455 + 0.15 ÷ 0.923) = 3.141
Added to the continuous layers: 1.832 + 0.547 + 3.141 = 5.521 m²K/W.
The result. The limits are within 1.5 of each other (6.581 ÷ 5.521 = 1.19), so the method is valid. The total resistance is their average, (6.581 + 5.521) ÷ 2 = 6.051 m²K/W, and the U-value is 1 ÷ 6.051 = 0.165 W/m²K, which rounds to 0.17 W/m²K.
The common shortcut, and why it's wrong
Some simple calculators handle a bridged layer by giving each material its share of the layer's resistance and adding them up: 85% of the insulation's 5.455 plus 15% of the stud's 0.923, and likewise for the battens. It looks reasonable, but it treats the wall as if heat had to pass through the insulation and the stud one after the other, when in fact the stud is a short cut around the insulation.
For this wall the shortcut gives a total resistance of 7.231 m²K/W and a U-value of 0.138 W/m²K, about 16% better than the wall's true 0.165. Leave the timber out altogether and it looks better still, at 0.126 W/m²K.
On a wall well inside the limit, that may not change the outcome. On a wall close to the limit, it's the difference between a design that complies and one that doesn't, and between a heat loss calculation that matches the building and one that doesn't.
Pitched roofs work the same way
A roof with insulation between and below the rafters is calculated the same way, with the roof's surface resistances (0.10 inside) and the rafters as the bridging timber. In a dormer roof from the same practice, with 150 mm of PIR between rafters at a 12.75% timber fraction, 70 mm of PIR below them and a battened service void, the combined method gives 0.109 W/m²K.
Ventilated cavities behind cladding and tiles
Where there's a well-ventilated cavity, such as behind tile-hanging, timber cladding or the battens under roof tiles, the air in it is at outside temperature. The cavity and everything outside it are left out of the calculation. Because the cladding still shelters the wall from the wind, the outside surface resistance is increased to match the inside one: 0.13 for a wall and 0.10 for a roof (BR 443, section 4.7.3).
Not every vented cavity counts as well ventilated. BR 443 notes that a timber frame wall cavity vented to NHBC standards, with an open joint about every 1.2 m, has enough openings to count as slightly ventilated, and is calculated by the procedure for that case in BS EN ISO 6946 (section 4.7.1). Weep holes in a masonry outer leaf don't usually count as ventilation.
What this calculation doesn't include
- Corrections. BS EN ISO 6946 adds small corrections for metal fixings through the insulation, for gaps in the insulation, and for rainwater flowing under the insulation of an inverted roof. They were all zero in this example, but check whether they apply to yours.
- Ground floors. A floor on the ground loses heat through the ground beneath it, and its U-value depends on the floor's size and shape. Ground floors are calculated with BS EN ISO 13370 instead.
- Junctions. Heat lost at corners, around windows and doors, and where floors meet walls is dealt with separately, as linear thermal bridges. The extra timber there doesn't count towards the 15% (BR 443, section 4.4.1).
Sources
- BRE: BR 443, Conventions for U-value calculations (2019 edition), sections 1, 2, 4.1, 4.4.1, 4.5, 4.7.1 and 4.7.3.
- BS EN ISO 6946, Building components and building elements: Thermal resistance and thermal transmittance: Calculation methods, as described in BR 443.