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Does plywood Add R-value?

So the R-value of the ceiling is 22.04 ft2 oFh / BTU. ... Total R-Value of Composite Wall. Material R-Value (ft2 o Fh / BTU) 1/2 inch Gypsum Board (Drywall or plasterboard) 0.45 Wood Siding, 1/2 inch 0.81 Plywood, 3/4 inch 0.94 Insulating sheathing, 3/4 inch 2.06 8 more rows

e-education.psu.edu - Composite Wall R-Values | EGEE 102 - E-education.psu.edu

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Generally, walls are made up of several layers of different materials. The R-value of a composite wall is calculated by adding the effective R-values of each of the layers of the wall. For example, the image below shows a wall made up of four layers—½ inch drywall inside for aesthetic purposes, real insulation in between the studs, ¾ inch plywood sheathing outside, and wood siding as the final exterior finish. Together, the layers of the wall are preventing heat loss.

Diagram of a wall

We can calculate this wall’s composite R-value by adding the R-values of each layer.

Plasterboard (1/2 inch)

+ Fiberglass (3.5 inches @ 3.70 per inch)

+ Plywood (3/4 inches)

+ Wood Siding (1/2 inch)

-----------------------------------------

Total R-Value of Composite Wall

We can find the R-values for the walls by using the table from the page about R-Values (the table is repeated below): Building Materials and their R-Values Material R-Value (ft2 o Fh / BTU) Plain glass, 1/8 inch 0.03 Stone per inch 0.08 Common Brick per inch 0.20 Asphalt Roof Shingles 0.44 1/2 inch Gypsum Board (Drywall or plasterboard) 0.45 Wood Siding, 1/2 inch 0.81 Plywood, 3/4 inch 0.94 Insulating sheathing, 3/4 inch 2.06 Fiberglass, per inch (battens) 3.50 Polystyrene per inch 5.00 Polyurethane Board per inch 6.25 Cinder Block (12 inches) 1.89

Plasterboard (1/2 inch) = 0.4524

+ Fiberglass (3.5 inches @ 3.70 per inch) = 12.95

+ Plywood (3/4 inches) = 0.94

+ Wood Siding (1/2 inch) = 0.81

-----------------------------------------

Total R-Value of Composite Wall = 15.15 ft 2 ° Fh Btu

Examples

Example 1 A ceiling is insulated with 0.75" plywood, 2" of polystyrene board, and a 3" layer of fiberglass. What is the R-Value for the ceiling? Solution: The ceiling consists of three layers, and all three layers together prevent the heat loss. So, we need to add the R-values of all three layers 3/ 4 " of plywood has an R-value of 0.94 2" of polystyrene at 5.0 per inch will have an R-value of 10.00 3" of fiberglass at 3.7 per inch will have an R-value of 11.10 So the R-value of the ceiling is 22.04 ft2 oFh / BTU. Pink insulation in the ceiling Example 2 Please watch the following 1:31 video presentation about Example #2. A wall consists of 0.5” wood siding (R = 0.81), 0.75” plywood (R = 0.94), 3.5” of fiberglass (R = 13.0), and 0.5” plasterboard (R = 0.45). What is the composite R-value of the wall? Click here for a transcript of Composite Wall R-Values Problem # 1 video. Lesson 7b, Screen 7: Composite Wall R-Values Example 10 This problem, we need to calculate composite R-value. The wall consists of four layers. One half inch wood siding, and its R-value is given straight away for half inch as .81. And we have three-quarter inch plywood and this plywood’s R-value is also given as .94, this is for 3/4". Whereas the fiberglass, each inch has an R-value of 3.7. We are using 3 and a half inches. So, 3.5 times 3.7 would give us about 13.00 R-value. And the last layer would be one half inch plaster board which is also drywall and its R-value is given as .45. 1/2" wood siding: 0.81

3/4" plywood: 0.94

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3 1/2 Fiberglass (3.7/inch): 13.00

1/2" plaster board: 0.45

Total: 15.2 So when you add these up you get 15.2 which means the composite R-value of this wall is 15.2 degrees F, foot squared, hour over BTUs. = 15.2 ° F f t 2 h B T U Example 3 Please watch the following 1:18 video presentation about Example #3. What is the R-value of a wall that is made up of wood siding (R = 0.81), 5” of fiberglass (R = 3.70 per inch), and a layer of 0.5” drywall (R = 0.45)? Click here for a transcript of Composite Wall R-Values Problem #2 video. Lesson 7b, Screen 8: Composite Wall R-Values Example 11 What is the R-value of a wall that is made up of wood siding (R= 0.81), 5” of fiberglass (R=3.70 per inch), and a layer of .5” drywall (R=0.45)? Here we have, again a wall made up of three different layers. The first layer is wood siding which is outside, obviously, and its R-value is given as 0.81, whatever the thickness might be of that wood siding. And we have a second layer of fiberglass. And this fiberglass thickness is given as 5” and its R-value is given as 3.7 per inch. So we are using 5 inches so 5 times 3.7 would be 18.50. And we have a third layer of a drywall and this drywall has an R-value of 0.45, half inch drywall. Wood siding: 0.81

5" fiberglass 3.7/inch: 18.50 (5x3.7)

Drywall: 0.45

Total R-value: 19.76 When you add all these three layers up, you get a total R-value of 19.76. So the answer is 19.76 degrees F, foot squared, hour over BTUs. This is the composite R-value. C o m p o s i t e R − v a l u e = 19.76 ° F f t 2 h B T U Example 4 Please watch the following 1:44 video presentation about Example #4. A wall is made up of 8” of stone, 3” of polyurethane board, and 0.75” of plywood. Calculate the composite R-value for the wall. Click here for a transcript of Composite Wall R-Values Problem #3 video. Lesson 7b, Screen 9: Composite Wall R-Values Example 12 A wall is made up of 8” of stone, 3” of polyurethane board, and 0.75” of plywood. Calculate the composite R-value for the wall. This is a problem where we have three layers for a wall and those three layers are made up of stone, polyurethane board and plywood. The first one is stone and its thickness is 8” and each inch of stone wall will provide an R-value of 0.08. Therefore, all these 8” together would provide 0.64. And the second layer is made up of 3” of polyurethane and each inch provides an R-value of 6.25, therefore, together, all 3” would provide an R-value of 18.75. The third layer is again three quarters inch plywood, and it provides an R-value of 0.94. So all these three together would have an R-value of 20.33 or the composite R-value is 20.33 degree F, foot squared, hour over BTUs. 8" stone (0.08/inch): 0.64

3" polyurethane (6.25/inch): 18.75

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0.75" plywood: 0.94

Total: 20.33 C o m p o s i t e R − v a l u e = 20.33 ° F f t 2 h B T U

Insulation Needs By Region

The United States map below shows insulation needs by region, as indicated by color and numbers. Instructions: Click on the “zone” buttons below the map to see the U.S. Department of Energy’s Recommended Total R-Values for new construction of houses. Note that insulation R-values are different for the ceilings, walls, floor, etc. U.S. Department of Energy recommended total R-Values for new construction houses, by regions and by various parts of the house.