The possibility of a thermal or cold bridge occurring in a specific location can
be appraised by calculation. Alternatively, the calculations can be used to
determine how much insulation will be required to prevent a cold bridge. The
composite lintel of concrete and steel shown below will serve as an example ~Wall components, less insulation (steel in lintel is insignificant):Resistances of above components:Resistances of surfaces:To achieve a U-value of say 0„27 W/m^2 K,total resistance required = 1...0„27 = 3„703 m^2 K/W
The insulation in the cavity at the lintel position is required to have a
resistance of 3„703 † 0„613 = 3„09 m^2 K/W
Using a urethane insulation with a thermal conductivity () of 0„025 W/mK,
0„025¾3„09 = 0„077m or 77 mm minimum thickness.
If the cavity closer has the same thermal conductivity, then:
Summary of resistance = 0„613 † 0„180 (Ra) = 0„433 m^2 K/W
Total resistance required = 3„703 m^2 K/W, therefore the cavity closer is required
to have a resistance of: 3„703 † 0„433 = 3„270 m^2 K/W
Min. cavity closer width = 0„025 W/mK¾3„270 m^2 K/W = 0„082 m or 82 mm.
In practice, the cavity width and the lintel insulation would exceed
82 mm.
Note: data for resistances andvalues taken from pages 469 to 471.102.5 mm brickwork outer leaf, = 0.84 W/mK
100 mm dense concrete lintel, = 1.93 ..
13 mm lightweight plaster, = 0.16 ..Brickwork, 0.1025...0.84 = 0.122 m^2 K/W
Concrete lintel, 0.100...1.93 = 0.052 ..
Lightweight plaster, 0.013...0.16 = 0.081 ..
Internal (Rsi) = 0.123 ..
Cavity (Ra) = 0.180 ..
External (Rso) = 0.055 ..
Summary of resistances = 0„613 ..Thermal Bridging