Thermal Calculations: ===================== Since the building has underslab insulation, the detailed method of FPSF calculation (as opposed to the simplified method) is required. These are those calculations, with the text below taken directly from the controlling HUD document. The following steps outline the detailed design approach for heated buildings. Step 1: Fn = Site design Air Freezing Index ------------------------------------------- Select the 100-year return period design air freezing index, Fn100, from Figure A1 or Table A3. This information was prepared by the National Oceanic and Atmospheric Administration's National Climatic Data Center specifically for use in FPSF design. The Fn values are conservative because they are not adjusted for the insulating benefit of a normal snow cover on the ground. A lower return period value may be used for less important structures or those that are resilient to infrequent ground freezing. See Table 3 for Fn values at return periods less than 100 years. From Table A3: .............. Location Mean Annual Temperature 100-Year-Return Fn =========================================================== Spokane WA 47.2 1232 Step 2: Rf = Floor-Slab Cross-Section R-value --------------------------------------------- Calculate the thermal resistance of the design floor slab, Rf, considering all insulating materials in the cross- section including any floor coverings. When determining Rf, dry condition R-values, as presented in Table A2, shall be used for all materials, including insulation. If the floor cross section and resulting thermal resistance of the floor slab varies over its area, calculate Rf as the average over the perimeter 39 inches (1 m) of the floor. In super- insulated slabs (where the calculated Rf value exceeds R28), the designer must follow the design procedure for unheated buildings, since the heat from the building is substantially blocked from moving into the ground and protecting the foundation. Calculation (R values from Table A2): ..................................... Concrete, 6 inches: 6 x 0.05 = 0.3 Gravel, 6 inches: negligible 0 XPS Type IV, 3 inches: 3 x 5.00 = 15.0 Vapor barrier: negligible 0 ===================================================== Rf 15.3 (Rf << 28 so heated-building method valid.) Step 3: Rv = Required Vertical Wall Insulation R-value ------------------------------------------------------ Determine the minimum required thermal resistance of the vertical wall insulation, Rv, from Table A4 given h from Figure 5, Fn from Step 1, and Rf from Step 2. From Table A4: .............. Fn 15.0 < Rf < 28.0 hó12" h=24" ==================================================== 750 8.5 11.4 1500 8.5 11.4 From Figure 5, h (shown there to be the height of above-grade foundation perimeter) is 4" << 12". Therefore Rf=8.5. Step 4: Select Thickness of Vertical Wall Insulation ---------------------------------------------------- Based on the required Rv value from Step 3, select an adequate thickness of vertical XPS or EPS insulation using the following effective resistivities: Type II EPS - 2.4 R per inch; Type IX EPS - 3.2 R per inch; and, Types IV, V, VI, VII XPS - 4.5 R per inch. Common nominal thicknesses are 1", 1-1/2", 2", and 3". The insulation shall extend from the bottom of the footing to the exterior wall envelope as shown in Figure 5. Using Type IV XPS requires thus 2 inches of insulation: 8.5/4.5 = 1.9 < 2 The design calls for 4 inches: 4 >> 2. Step 5: Foundation Depth or Horizontal Wing Insulation ------------------------------------------------------ Horizontal wing insulation is placed below ground extending outward from the vertical wall insulation as shown in Figure 5. For climates where Fn is less than 2,250, wing insulation along the walls is not required and the designer may proceed to Step 7. Step 6: not required - see Step 5 --------------------------------- Step 7: Foundation Depth or Corner Wing Insulation --------------------------------------------------- Since more heat loss occurs at building corners than through mid-wall sections of heated buildings, additional frost protection in the form of horizontal wing insulation or a deeper foundation is required for more severe climates (Fn > 2250). Fn=1232 << 2250 so no corner horizontal insulation required either. Conclusion: ----------- The detailed FPSF design procedure for this foundation as designed requires 2 inches of Type IV XPS along the complete external foundation perimeter. The depth below grade may be the minimum allowed, 12 inches. The design specifies 4 inches of Type IV XPS perimeter insulation for the full height of the foundation perimeter, materially exceeding the requirement; the footing depth below grade is 12 inches, meeting the requirement. Structural Calculations: ======================== Shear In South (Windowed) Wall ------------------------------ 0.25 x W x A F = -------------- L where-- F is shear Force in pounds per foot, W is wind pressure in pounds per square foot, A is the Area of wind surface in square feet, and L is the Length of plywood shear bracing in feet (for a sheet extending the full wall height). The wind pressure is assumed to be 20 psf, described in one source as "the highest wind velocity you'll ever encounter other than on the top of a bald mountain or in southern Florida." (Also, less poetically, as prescribed by Code). The Area is that of the east or west wall, which with a 10-foot height, a 32-foot wall, a 36-foot truss extension width, and a 7.5-foot (over the 10-foot wall) height (5:12 pitch) is-- (10 x 32) + (2 x 1/2 x 18 x 7.5) meaning Area = 455 square feet. The plywood sheet Length on wall is 8 feet (two 4-foot sections). The shear force per foot in the shear wall is thus-- 0.25 x 20 x 455 F = ----------------- = 284.375 8 To resist 285 pounds a foot with an 8-foot shear wall of wall-height CDX plywood requires a precisely calculable nailing schedule listed in, among other places, the UBC. The specifications for this house call for 1/2" plywood (15/32" in the tables) nailed "with 8d annular-ring nails to the framing at 4-inch intervals around the perimeter and 6-inch intervals over the intermediate stud; nail penetration shall be a minimum of 1-1/2 inches." That supplies a code-accepted shear resistance of 340 pounds a foot, about 20% above requirements. Shear in the north wall need not be figured, since the it is essentially the south wall with several times as much shear wall in it. The east and west walls also need not be figured: though the wind-exposure area is doubled, the shear-wall length is materially more than doubled. The south wall controls. (These calculations and the related analysis will be submitted to a Structural Engineer for review and approval.) {end}