Energy Source Builder

ASHRAE Sets Thermal Correction Factors for Metal Framing

Steel framing has several strengths for home building. It's lightweight, uniform, dimensionally stable, easy to install, cost competitive and resistant to fire and decay. A major drawback--and perhaps a fatal flaw--is thermal performance. While this problem is well known, steel industry boosters often remain silent when the issue of heat loss comes up. A trade journal even stated that a steel-framed wall assembly had a higher R-value than a wood-framed one.

Steel studs offer the possibility of thicker walls and thicker insulation. However, heat loss from a wall assembly is affected by framing as well as insulation. Steel conducts heat over 300 times faster than wood. The picture is far from complete, but current information shows that steel framing has a large negative impact on the energy performance of residential buildings. (A recent study sheds more light on this issue. See Issue #38.)

ASHRAE Correction Factors

"Metal studs significantly reduce the overall thermal performance of a wall system," says Merle McBride, Ph. D., research associate with Owens-Corning. "The only remaining question is the degree of thermal degradation." The negative impact of metal framing on wall R-value is reflected in correction factors published in the most recent revision of American Society of Heating, Refrigeration and Air-conditioning Engineers (ASHRAE) Standard 90.1. Dr. McBride is the chairman of the ASHRAE Standard 90.1 Committee and guided development of the table.

"One reason we developed the table was to discourage people from ignoring the effect of steel framing," says Dr. McBride. The correction factors also provide a simple and consistent calculation method for thermal performance of metal-framed walls.

Hot Box Tests

The correction factors are based on several calibrated hot box tests for 2x4 walls. Solid, 16- to 18-gauge, C-channel steel studs were used for the tests. Only straight wall sections were tested. The "framing factor" or the percent of the wall occupied by framing was 12 percent for walls with studs spaced 16 in. on-center and 9 percent for 24 in. on-center spacing. Correction factors for other wall configurations were extrapolated from those results.

For comparison, take two walls with 2x8 studs and framing spaced 24 in. on-center. With wood studs, the wall would have an insulating value of R-23.8. The example below shows a wall with steel studs at R-12.3.

Other Factors

Several factors that affect thermal performance were not considered in preparing the table. The impact isn't known for other parts of a building, such as roof and ceiling assemblies, walls with corners and the wall-to-roof and wall-to-floor connections. Screws and other fasteners probably promote heat loss.

Several factors not considered might reduce heat loss. These include: thermal breaks in the studs, punch outs in the web, special stud facings and wider stud spacing.

Other factors would increase heat loss. These include: thicker gauge steel and framing factors higher than 12 percent of wall area.

Foam Sheathing

Adding foam insulation sheets to a wall certainly improves the thermal performance, but it's not clear how much.

"There is conflicting information about the exact benefits of foam sheathing," says Dr. McBride. Some people think the benefit might exceed the rated R-value of the material, because foam blocks the massive thermal bridging effect.

"Our tests suggest that the additional R-value may be somewhat less than the rated value," he says. He believes that fasteners and foil facing on the foam boards helped promote heat loss in these tests.

Many of the remaining questions could be answered by additional research. ASHRAE is planning to conduct large scale tests on wall sections using a wide variety of materials, including steel framing. The American Iron and Steel Institute finished its own study earlier this year and plans to release the results soon.

ASHRAE Correction Factors for Metal Wall Framing


To use the correction factors from the table, simply multiply them by the insulation value in the cavity. That number is then added to the insulating values of the other wall components. Here's an example for a wall using 8-inch C-channel studs 24 inches on-center plus R-25 batt.




Component                                     R-value
Outdoor air film (7.5 mph)                       0.25
Lap siding                                       0.81
1/2-inch plywood sheathing                       0.62
Cavity and metal stud (R-25 x 0.38 = 9.5)        9.50
1/2-inch gypsum board                            0.45
Indoor air film (still air)                      0.88

Total                                           12.31

If you want to calculate heat loss, you'll need to convert the R-value to U-factor by dividing the R-value into 1. 1 / 12.31 = U-0.81

This article appeared in Energy Source Builder #36 December 1994
©Copyright 1994 Iris Communications, Inc.