There are two ways to calculate R value of a wall, for simplicity sake, I will label them Type 1, conventional and Type 2, ASHRAE.

Here is the wall assembly in question.

Conventional 2X6 stud wall construction with 5” spray foam insulation in cavity.

Code requires this wall assembly to have 2 layers of gypsum board on the interior

This wall has thermal bridging at 2X6 members – this is the issue which causes problems

Wall assembly – 2X6 with Spray Foam

R values of components

Air film – inside                          .68

2 layers gypsum board            1.12

5 1/2” spray foam                    37.4

2X6 stud                                 6.875

Sheathing                               1.3

Air film – outside                     .17

Calculation using Type 1, conventional method:

R value through cavity – add components

Air film – inside                         .68

2 layers gypsum board            1.12

5 1/2” spray foam                    37.4

Sheathing                               1.3

Air film – outside                     .17

Total R                                    41.1

R value through stud – add components:

Air film – inside                                     .68

2 layers gypsum board                        1.12

2X6 stud                                 6.875

Sheathing                               1.3

Air film – outside                     .17

Total R                                    10.6

Calculate total R of the wall assembly – multiple the above by the percentage of each in the wall.

This will assume that the wall has 18.75 percentage stud area

10.645 X .1875 + 41.17 X .8425 = 36.68

Calculation using Type 2, ASHRAE method:

Calculate R value of components not in cavity:

Air film – inside                       .68

2 layers gypsum board          1.12

Sheathing                               1.3

Air film – outside                     .17

Total                                       3.77

Calculate the wall cavity using the weighted average of the U values of the components:

((1/37.4) X (.8125)) + ((1/6.875) X (.1875)) =.049 = 1/.049 = R20.4

Add the two together 3.77 + 20.4 = 24.17

Difference in methods

Two equations to consider:

2 + 2 =4

½ + ½ = 1/1 = 1

AND

2 X .1875 + 2 X .8125 = 2

(1/2 X.1875) + (1/2X.8125) = .5 = 1/.5 = 2

But with unequal numbers things change:

4+2= 6

¼ + ½ = ¾ = 1/.75 = 1.33

AND

2 X .1875 + 4X.8125 = 3.6

1/2X.1875 + ¼ X.8125 = .296875 = 1/.296875 = 3.368

The long and short of this is that using the ASHRAE method of the weighted average of U values, weights the smaller number over the bigger number. That’s why ASHRAE does it, as more heat will go through the wall where there is less resistance holding that heat back.  In other words, more and more heat with go through the stud as more and more resistance is put in the cavity, without increasing the heat on the inside.

Technical ASHRAE information

The fractional area method used above is found in ASHRAE Fundamentals 2009  Equation   in

chapter 10 (pages 25.7-8), as shown below. This equation shows the addition of the thermal transmittances (U-value), based on their surface-weighted fractions of the space. The inverse of the summation of these values equals the R value.

Conclusion : Thermal bridging is a big deal, so either use out insulation or double studs in the wall.