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Construction & DIY

Bridged U-Value Calculator (Timber Stud or Joist Wall, ISO 6946 Combined Method)

U-value of a wall, roof or floor where timber studs or joists interrupt the insulation, by the ISO 6946 combined method, with the heat loss the studs add — drawn against the stud share.

Results

U-value with the studs counted 0.294 W/(m²·K)
U-value if there were insulation only 0.224 W/(m²·K)
How much the studs raise the heat loss 31.6 %
Upper limit of the total resistance R′T 3.483 m²·K/W
Lower limit of the total resistance R″T 3.313 m²·K/W
Total resistance RT 3.398 m²·K/W
Is the combined method suitable? Yes — the two limits are within 1.5 of each other
How it changes
U-value with the studs countedShare of the area taken by studs or joists 0 – 40. U-value with the studs counted: 0.223714 → 0.405384; U-value if there were insulation only: 0.223714 → 0.223714U-value with the studs countedU-value if there were insulation only00.10.20.30.40.5010203040Share of the area taken by studs or joists (%)

The dashed line marks the value you entered.

What this tool does

Insulation between studs is not the whole story: the timber conducts heat about four times better, so a quoted U-value of the insulation alone is too optimistic. ISO 6946 brackets the answer between two limits — one treating each path (through studs, through insulation) separately, the other assuming the heat spreads out sideways at each layer — and takes their average. A frame with 15 % studs, 140 mm of 0.035 insulation and some boards loses roughly 30 % more heat than the insulation alone suggests, and continuous insulation over the frame is how that is reduced. The method suits the cases where the two limits are within a factor of 1.5.

Formula

R′T = 1 ÷ [f ÷ R_stud path + (1 − f) ÷ R_insulation path] · R″T = Rsi + 1 ÷ [f λ_t/d + (1 − f) λ_i/d] + R_other + Rse · RT = (R′T + R″T) ÷ 2 · U = 1 ÷ RT (ISO 6946)

Variables

SymbolMeaningUnit
frShare of the area taken by studs or joists%
ddThickness of the layer with studs and insulationmm
liThermal conductivity of the insulationW/(m·K)
ltThermal conductivity of the studsW/(m·K)
roResistance of all the other layers togetherm²·K/W
rsiInside surface resistance Rsim²·K/W
rseOutside surface resistance Rsem²·K/W
UU-value with the studs countedW/(m²·K)
U0U-value if there were insulation onlyW/(m²·K)
UPHow much the studs raise the heat loss%
RUUpper limit of the total resistance R′Tm²·K/W
RLLower limit of the total resistance R″Tm²·K/W
RTTotal resistance RTm²·K/W
OKIs the combined method suitable?—

Worked example

  • Share of the area taken by studs or joists15 %
  • Thickness of the layer with studs and insulation140 mm
  • Thermal conductivity of the insulation0.035 W/(m·K)
  • Thermal conductivity of the studs0.13 W/(m·K)
  • Resistance of all the other layers together0.3 m²·K/W
  • Inside surface resistance Rsi0.13 m²·K/W
  • Outside surface resistance Rse0.04 m²·K/W
  • U-value with the studs counted0.294 W/(m²·K)
  • U-value if there were insulation only0.224 W/(m²·K)
  • How much the studs raise the heat loss31.6 %
  • Upper limit of the total resistance R′T3.483 m²·K/W
  • Lower limit of the total resistance R″T3.313 m²·K/W
  • Total resistance RT3.398 m²·K/W
  • Is the combined method suitable?Yes — the two limits are within 1.5 of each other

Limitations

  • The result is an estimate based only on the values you type. Real situations often include factors this calculator does not know about.
  • For work that must comply with a standard or be signed off, check the result against the applicable code and have it reviewed by a qualified engineer.