# Thermal Bridging

How does one quantify thermal bridging when it comes to the dimensions of the area that is a thermal bridge?

For instance; if there is a concrete area that is 6" wide x 12" tall, how does one calculate that area into the whole R-Value of the structure?

Asked by Peter L

Posted Thu, 06/12/2014 - 20:00

Posted Thu, 06/12/2014 - 20:00

## Other Questions in Energy efficiency and durability

### Insulating a cathedral ceiling

In Energy efficiency and durability | Asked by jon cypert | Nov 25, 11

### poly under slab and over sill ?

In Energy efficiency and durability | Asked by Jocelyn Smith | Nov 25, 14

### Spray foam insulation

In Energy efficiency and durability | Asked by Danny Mac | Nov 17, 14

### Mold on inside of plywood sheathing

In Energy efficiency and durability | Asked by Aaron Saucier | Nov 25, 14

### Designing a passive solar / net zero house in NC Mountains

In General questions | Asked by bill Moseley | Nov 23, 14

Even concrete has a thermal resistance. To answer your question we'll need to know the thickness of the concrete between the warm and cold sides to estimate the thermal resistance of the concrete which is about r 0.169 / " To compute the equivalent r value for the structure we'll need it's r value and area then (area of concrete/r of concrete + area of structure / r value of structure) / (area of structure+area of concrete) = 1/r equivalent.

Posted Thu, 06/12/2014 - 23:20

The concrete area would be around 12" wide and 12" tall and the concrete beam would transition into the interior of the home. I hope that is enough info...

Posted Fri, 06/13/2014 - 00:41

Peter,

The short answer is that one can perform a back-of-the envelope calculation of the effect of the thermal bridge by calculating the areas of the different wall assemblies; converting the R-values of each area to U-factors; and coming up with an average U-factor for the entire wall by applying the area percentages to the various U-factors.

The more accurate answer is that you need a software program like THERM to make these calculations.

Posted Fri, 06/13/2014 - 06:43

Petrer,

If the distance between the warm area and cold area is 6" then the concrete will have a thermal resistance of r=1. If the rest of your house is effectively r30 then the one square foot of concrete is losing as much heat as 30 square feet of house.

Posted Fri, 06/13/2014 - 09:56

Jerry,

Wow, that seems like a lot of loss. So if there was 2 square feet of concrete, would that equal to 60 square feet of house?

In the winter the loss would be heat and in the summer it would be a loss in cool conditioned air but are both calculated the same way?

Posted Fri, 06/13/2014 - 10:22

"So if there was 2 square feet of concrete, would that equal to 60 square feet of house?"

Sort of, except that the rest of the house also has thermal bridges, so the chances of the walls being R30 are very remote. That one foot of concrete also doesn't lose heat to air leakage the way the rest of the house does and may well benefit from being below grade. Very, very roughly the ratio is probably closer to 1:10.

Posted Fri, 06/13/2014 - 10:33

Peter.

Despite Malcolm's attempt to confuse you (sorry Malcolm)! What i said is the very definition of thermal resistance and not subject to interpretation! The same heat flow will occur, at the same temperature difference, where the product of area and r value is equal! So yes you are correct in the statement "So if there was 2 square feet of concrete, would that equal to 60 square feet of house?"

Posted Fri, 06/13/2014 - 10:48

The R value of concrete varies with density, with higher density concrete being half that of the value Jerry cited. Here is a table that shows the effect:

http://archtoolbox.com/materials-systems/thermal-moisture-protection/24-...

I imagine the R value shown in that table are just typical, with the actual value for concrete varying considerably according to the mix and degree of aeration.

There is hardly any use to having an "average structure R value," other than for wall sections framed in the same way so that a "framing factor" can be applied. In the end, you want to know the expected rate of heat loss of the structure at design outside temperature (heating climate). That is simply the sum of the losses from the various parts of the shell, and this what the various heat loss programs do, by having you enter data for framed walls, foundation walls, slabs, ceilings, windows, etc. A concrete beam is just another element, with size and R value, to add to the total.

Posted Fri, 06/13/2014 - 15:44

How's the house coming Jerry?

Posted Fri, 06/13/2014 - 23:27

Malcolm,

Thank you for asking. We're looking at lots, floor plan about settled, design still evolving. Here is a file of the current details

Posted Sat, 06/14/2014 - 09:20