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.)
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