SCOTTISH EXECUTIVE

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Building Regulations: Technical Standards

Appendix A: The proportional area calculation method for determining U-values of structures containing repeating thermal bridges

A1 Full details of the calculation method are given in CIBSE Design Guide A: 1980, Section A3. However, two examples are given in this Appendix which illustrate the method as applied to frequently encountered designs.

A2 If the element design does not include a continuous cavity all the element layers have to be analysed together. Where the design does incorporate a continuous cavity, however, the section should be divided into two parts along the centre of the cavity and the parts analysed separately. The resistances of the two parts should be determined with half of the cavity resistance assigned to each. The thermal resistances can then be added together to obtain the total resistance of the element.

A3 Thermal bridging may be disregarded where the difference in thermal resistance between bridging and bridged material is less than 0.1m²K/W. For example, normal (i.e. non-lightweight) mortar joints need not be taken into account in calculations for brickwork but must be taken into account for lightweight insulating blockwork.

A4 In timber framed walls, for the purposes of demonstating compliance with the requirements for the conservation of fuel and power, only the repeating timbers i.e. the vertical studs and, if they penetrate the full depth of the insulation, horizontal dwangs between the studs, should be included. All other timbers penetrating the wall, such as those at the perimeter of the wall and those around openings, should be ignored.

A5 In some cases, the joists in timber roof and floor constructions will project beyond the surface of the insulation. For the purposes of demonstrating compliance with the requirements for the conservation of fuel and power, however, the calculations should take the depths of the joists to be the same as the depth of insulation, hence ignoring the effect of the projections. Joists which are wholly beneath insulation can also be ignored.

Example 1

What is the U-value of the proposed wall construction shown below?

diagram

The resistance of the outside surface of the wall is 0.06 m²K/W and the inside surface resistance is 0.12 m²K/W.

With this construction the thermal bridging of the insulation by the timber studs must be taken into account as follows:

Consider the wall as inner and outer leaves with the boundary between leaves at the centre of the cavity.

Resistance of inner leaf

Combination of these resistances:

Fractional area of timber stud:

formula

Fractional area of insulation: F_ins = (1 - F_t) = 0.937

The resistance of the inner leaf is then obtained from:

formula

Resistance of outer leaf

This is treated as an unbridged structure (the difference in resistance between brick and mortar is less than 0.1 m²K/W).

Total resistance of wall

The total resistance of the wall is the sum of the resistances of the inner and outer leaves:

R_inner + R_outer = 2.36 + 0.27

Hence total resistance = 2.63 m²K/W

U-value of wall

formula

Example 2

If the proposed wall construction is now that shown below, there are two thermally bridged layers:

a. that of the blockwork by the normal mortar joints; and

b. that of the insulation by the timber studs

diagram

Consider the wall as inner and outer leaves with the boundary between leaves as halfway through the cavity.

Resistance of inner leaf

In this case there are two bridged layers and the precise location of the bridge in one layer with respect to the bridge in another layer is generally unknown (or is not readily determined). There are four different combinations of paths through the blockwork and insulation and it is therefore assumed, because pitch centres do not coincide, that heat flows through them in proportion to their relative areas. The average resistance of the leaf is determined as follows:

1. RESISTANCE OF NON-BRIDGED LAYERS

2. RESISTANCE OF BRIDGED LAYERS

Heat flow paths:

The two bridged layers create four paths:

block/insulation
block/timber
mortar/insulation
mortar/timber

Material resistances:

Resistance of heat flow paths:

Fraction of face area of materials:

formula

Fraction of face area of heat flow paths:

Sum of parallel resistances:

The sum of resistances in parallel is given by the formula:

formula

The figures in this case are:

formula

The resistance of the inner leaf is therefore: formula

Resistance of outer leaf

Total resistance of the wall

The total resistance of the wall is the sum of the inner and outer leaf resistances:

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