HOUSES

We’ve been dealing with commercial-type space, but let’s switch over to houses. There are times when plumbers are the ones who must figure the min­imum requirements for houses. This is especially true in very rural areas. Fig­ure 9.5 provides the information needed to compute the fixture requirements for a typical, single-family home. The same table can be used to figure the fix­ture requirements for an apartment building. Check back to Figure 9.2 for ex­planations of the numbers noted in the headings of the table. There is no big secret to this table. Each home is required to have a minimum of one toilet, one lavatory, one bathing unit, one kitchen sink, and one connection for a washing machine.

if you look at the table closely, you will see that the basic minimums are required for each dwelling or dwelling unit. This means that each apartment

”I

to

 

Подпись: PLUMBER’S AND PIPE FITTER’S CALCULATIONS MANUAL

40 sq ft

Restaurants617 per person

Persons (total)

Male

Female

1-50

1

1

51-100

2

2

101-200

3

3

201-300

4

4

For each

additional 200

persons over

300 add

1

2

 

Persons (total)

Male

Female

Comply with board of health requirements

1-150

1

1

151-200

2

2

201-400

3

3

For each additional 200 persons over 400 add

1

1

 

FIGURE 9.4 ■ Minimum fixtures for restaurants. (Courtesy of Standard Plumbing Code)

Building or occupancy2

Occupant content2

Water closets3

Lavatories4

Bathtubs, showers and miscellaneous fixtures

Dwelling or Apt.

Not Applicable

1 for each dwelling

1 for each dwelling

Washing machine

House

or dwelling unit

or dwelling unit

connection per unit5. Bathtub or shower—one per dwelling or dwelling unit. Kitchen sink-one per dwelling or dwelling unit.

Подпись: CALCULATING MINIMUM PLUMBING FACILITIES ■ 173FIGURE 9.5 ■ Minimum fixtures for homes and apartments. (Courtesy of Standard Plumbing Code)

in a building must be equipped with the same minimum requirements that would be found in a home. Of course, local codes may offer a different ruling, so always check your local code requirements before designing or installing plumbing systems.

Underlying Layer’s Maximum Temperature

The limitation of minimum temperatures of SMA laydown has been commonly understood and adopted; however, it is quite a different matter when laying down the SMA mixture on an underlying layer that has an excessive temperature, as in the following examples:

• On a recently placed intermediate course that has not yet cooled

• While hot recycling in situ (or in place) with the simultaneous placement of a new wearing course, so-called hot remix plus

• During the execution of the so-called Kompaktasphalt—the laydown and compaction of two courses (intermediate and wearing) at the same time with one passage of a paver (see Section 10.5).

As a general rule (see Section 10.4), rollers compacting an SMA layer operate directly behind the paver. This practice is correct in typical circumstances since the mixture cools quickly. But when the underlying layer is warm, such a method may create some problems, with the most frequent one resulting in the binder bleeding under the rollers and creating fat spots.

The following comments and observations deserve mention:

• The rolling operation cannot be started until the spread mixture has reached the optimum start temperature related to the viscosity of the applied binder; therefore placing the mixture on an underlying course that is still warm— with temperature exceeding 50-70°C (e. g., on a course made several hours earlier)—should be accompanied by ongoing temperature control.

• Because of the warm underlying course, the SMA does not cool as fast, consequently giving more time for compacting and enabling a greater dis­tance between the rollers and the paver.

• In cases of fat spots being squeezed out by the rollers, rolling should be stopped until the mixture temperature in the course drops to a point where rolling does not overcompact the mat; the time (or temperature) at which roll­ing can resume should be determined experimentally for specific binders.

Unfavorable air temperatures (too low) and an excessively warm underlying course cause various problems. Having an excessively warm underlying layer cre­ates more favorable laydown conditions because this allows for more time to work with the mix.

A JIG FOR MARKING PLUMB CUTS IN RAFTERS

2. Then nail on a 1×2 fence.

1. Mark cut lines with a small rafter square.

A JIG FOR MARKING PLUMB CUTS IN RAFTERS

 

lx stock, the same width as rafters

 

A JIG FOR MARKING PLUMB CUTS IN RAFTERS

If you need to make a series of plumb cuts to match and align rafter tails along an eave, make a jig like the one shown here. Use a rafter square to mark the plumb-cut angle of tie roof (4-ІП-12 or 5-in-12) on a short length of 1×4 or 1×6, depending on the rafter size. Cut a parallelogram-shaped template with
identical plumb-cut angles, then nail a short 1×2 fence to the upper edge of the template. Once a chalkline has been snapped across all the rafter tails, use this template to mark the plumb cut on them so that they can be cut to length.

Подпись: over first board. Nail through miter joint, into end of truss.

MITERING FASCIA BOARDS

CUTTING AND NAILING GUIDELINES FOR GUT­TER AND FASCIA BOARDS. Use long, straight stock for gutter or fascia boards. Both types must butt together over the solid backing of a rafter tail. Gutter boards can meet in a simple butt joint; fascia boards should meet with a mitered joint, as shown in the bottom illustra­tion at left.

Nail gutter and 2x fascia boards to truss or rafter ends with 16d, hot-dipped galvanized nails. Nail fascia boards with a finish hammer, being careful not to miss the nail or you’ll leave hammer tracks in the wood. I lave a partner hold the other end of the board. Hold the board down a bit on the rafter tail so that the roof sheathing just skims over it (see the illustration on the facing page). To line up

Подпись: INSTALLING GUTTER AND FASCIA BOARDS Подпись: Position the gutter board on the truss or rafter end so the sheathing overlaps the top edge.

everything correctly, tack a piece of scrap wood on the top edge of a rafter and let it project downward, the way the roof sheathing will. Drive the topmost nail high enough so it will be covered by the drip edge that will be installed later.

At the bottom corner of the roof, the cave trim meets the barge rafter. If the exterior trim will be covered with cladding, as on this house, plumb-cut the barge rafter and nail it to the plumb-cut end of the gutter board.

Install the gutter board so that it runs long, just like the lookout boards, then mark the correct overhang distance on the gutter board and on the topmost lookout. Be sure to take the barge rafter’s thickness into account. Snap aline between these two marks, then cut the gutter board and lookouts in place. Now you’re ready to cut and install the barge rafter.

The procedure is similar if you’re installing fascia instead of gutter boards; however, miter the end of the fascia board to lit a mitered plumb cut on the barge rafter (see the photo onp. 131). It’s easier to make both miter cuts before nailing the boards. Miter and install the fascia first, then snap the cutoff lines on the lookouts as described previously.

Plumb-cut barge rafters

Select straight, clear stock lor the barge rafters, and start with a board that’s a little longer than you need. Make the bottom plumb cut on the ground. This is a mitered plumb cut if vou’re working with finished trim. With one or two helpers holding the hoard above, nail the barge rafter to the lookouts and to the gutter or fascia board. I like to make the top plumb cut with the board in place, judging by eye where the top of the cut should be. When you install the opposite barge rafter, mark the top plumb cut against the rafter you’ve already cut to ensure a light-lit ting joint.

STEP 7 Sheathe the Roof

Sheathing a roof is much like sheathing a floor, except that the deck is sloped, not flat, and you work at least 10 ft. off the ground. Anyone who has worked in the construction trades for more than a few years has heard stories about someone who has fallen from a roof—or has seen it happen. 1 was sheathing a roof many years ago on a cool morning with a bit of frost on the ground. The sunny side of the roof had no frost. Without thinking, I stepped onto the shady side of the roof and was immediately careening down a fairly steep pitch, heading for doom. Fortunately, I had my straight-clawed hammer in my hand. Instinctively using it like an ice ax, I stink the claw into the sheathing, stopping myself just short of the roof’s edge—a thrilling wav to

O C* /

Loads

Introduction. Loads acting on buried structures include the dead load of the struc­ture itself, the dead load of the earth cover over the structure, the weight of the fluid within the structure, live loads from vehicles, and, under certain circumstances, exter­nal hydrostatic pressure from groundwater.

The structure dead load is only significant for rigid structures. Flexible structures are manufactured from plastic or metal. In each case, the weight of the material is insignificant when compared with the total load on the structure. For rigid structures, however, because the material is generally concrete and because the pipe wall thick­ness is considerable, the weight of the material should be included in the determination of the total load applied on the structure. For concrete pipe, the pipe weight, Wp, can be estimated using the following equations:

Circular:

Подпись: (5.23a)Подпись: (5.23b)Подпись: (5.23c)Wp = 3.3h (Di + h) in U. S. Customary units Wp = 74 X 10-6h (Di + h) for SI units

Arch or horizontal elliptical:

Wp = 2.8h (Si + h) in U. S. Customary units Wp = 63 X 10-6h (Si + h) for SI units

Vertical elliptical:

Wp = 4.2h (Si + h) in U. S. Customary units Wp = 94 X 10-6h (Si + h) for SI units

where Wp = pipe weight, lb/ft (kN/m)

Di = inside pipe diameter, in (mm)

Si = inside horizontal span, in (mm) h = pipe wall thickness, in (mm)

(See Concrete Pipe Technology Handbook, American Concrete Pipe Association, 1994.)

Earth Load. The first detailed studies of the loads on buried pipes were conducted by Anson Marston at the Iowa State University in the early 1990s. These studies resulted in the Marston load theory for rigid pipes. The theory provides a methodology for determining the loads on buried pipes in almost any installation condition.

Marston theorized that a pipe in a trench was in static equilibrium. Therefore, the summation of vertical forces was zero. He also concluded that the pipe and backfill would settle relative to the in situ trench walls. He then went about determining the different forces acting on the pipe. Represented pictorially in Fig. 5.31, these are the weight of the soil in the trench, the resisting vertical force at the bottom of the trench, and the shear forces present at the interface of the backfill and the native trench wall.

Loads

FIGURE 5.31 Forces acting on a buried pipe as presented by Marston. (From Soil Engineering, 4th ed., HarperCollins, 1982, with permission)

Through a mathematical transformation of the equilibrium equation, Marston arrived at the following equation for a rigid pipe in a trench:

W£ = CdlBl (5.24)

with

Подпись: (5.25)Cd =

d 2Kp

where WE = earth load, lb/ft (kN/m)

7 = soil unit weight, lb/ft3 (kN/m3)

H = height of cover, ft (m)

Kp! = frictional coefficient
Bd = trench width, ft (m)

Marston also investigated the loads on rigid pipes in embankment conditions. However, since there are no trench walls, it was necessary to determine the relative movement of the pipe and soil directly above the pipe to the fill material adjacent to the pipe. The soil directly above the pipe is called the soil prism (see Fig. 5.32). This gave a measurement of the shear forces at the interface of the embankment and soil prism. Marston then used similar procedures for determining the loads on pipes in embankments as he used for pipes in trenches. He set the system in static equilibrium and summed vertical forces. Using this procedure, he derived formulas for several embankment installation conditions.

Подпись: FIGURE 5.32 Illustration of prism load acting on pipe. (From A. P. Moser, Buried Pipe Design, McGraw-Hill, 2001, with permission)

Marston’s student, M. G. Spangler, expanded the previous work of Marston in determining a method for relating the strength of an installed rigid pipe to the strength

of a pipe in a three-edge bearing test. Marston originally introduced the concept of a bedding factor for this purpose. Spangler refined the method by introducing the 0.01-in (2.5-mm) crack as a laboratory performance limit for equating the in-field perfor­mance to the three-edge bearing test performance. Later, as use of corrugated metal pipe increased, Spangler noticed that the Marston load theory did not provide satisfac­tory results for flexible pipes. The load that a flexible pipe was able to support was much greater than what was predicted using Marston load theory and bedding factors. Spangler completed a series of field and laboratory tests to investigate the loads on flexible pipes. His analyses resulted in the now famous Iowa formula, which was sub­sequently revised by R. K. Watkins.

More recently engineers have attempted, through the use of computers and finite element analysis, to better represent soil-structure interaction and the resultant loads on buried pipes. They have met with varying degrees of success. The works of both Marston and Spangler are still widely used in engineering practice. The calculation of pipe deflection by the Iowa formula is given in Art. 5.8.6. For a complete discussion of the Marston load theory and Spangler’s Iowa formula, see A. P. Moser’s Buried Pipe Design, 2d ed., McGraw-Hill, 2001.

Loads Подпись: D Подпись: (5.26)

Representations of earth loads are gradually moving away from the use of the Marston loads. In lieu of the Marston loads, the earth load is represented as a proportion of the soil prism load. The soil prism load is the weight of the column of soil directly above the pipe:

where Wc = prism load, lb/ft (kN/m)

у = soil unit weight, lb/ft3 (kN/m3)

H = height of cover, ft (m)

Do = outside pipe diameter, ft (m)

This is depicted graphically in Fig. 5.32. Depending upon the pipe type (stiffness) and the relative quality of the soil envelope, the effective earth load on the pipe may be greater than, equal to, or less than the soil prism load. This modification of the soil prism load is made via an arching factor. Therefore, the total vertical earth load acting on the structure, WE, is

Подпись: (5.27)W£ = VAF (Wc)

where Wc = prism load

VAF = vertical arching factor

Live Load and Impact. Culverts are usually designed for the live load generated by an AASHTO HS 20 truck. The controlling loading for culverts consists of two axles spaced 14 ft (4.3 m) apart, each weighing 32 kip (145 kN), with wheels on the axle spaced 6 ft (1.8 m) apart transversely. The 16-kip (73-kN) wheel load is the same as for an H 20 load­ing. The live load applied to underground structures under load factor criteria is either a standard HS truck, or a live load lane.

Where the culvert has a span of 20 ft (6.1 m) or greater, it is classified as a bridge and must be investigated for an alternate military loading of two axles 4 ft (1200 mm) apart with each axle weighing 24 kip (107 kN). The live load lane consists of a uniform load applied in conjunction with a concentrated load. The concentrated load is distrib­uted across the design lane of 10 ft (3000 mm), and is the uniform load. Because of this and because of the relatively short spans associated with culverts, the standard HS truck usually controls as the critical loading.

Loads

FIGURE 5.33 Distributed load area for single dual wheel. Conversions: 1 in = 25.4 mm, 1 ft = 0.305 m. (From Concrete Pipe Design Manual, American Concrete Pipe Association, 2007, with permission. )

Where the fill over a culvert is 2 ft (600 mm) or more, the wheel live load of 16 kip (73 kN) is applied as a concentrated load acting on the wheel print area and uniformly distributed over a rectangle with sides increasing at a rate of 13/4 times the depth of cover. This is represented pictorially in Fig. 5.33. If areas from several concentrated loads overlap, the total load is uniformly distributed over an area as defined by the outside limits of the individual areas.

Rigid structures with less than 2 ft (600 mm) of cover use a different method for dis­tributing the live load. AASHTO code requires that in this case the live load be distributed using the same method as is used in distributing live load in a concrete slab. This method is generally only applied to reinforced concrete box culverts or three-sided culverts.

An impact factor is added to the highway live loading. The factor is equal to 30 percent of the live load for a soil cover of 1 ft (300 mm) or less and decreases to 20 percent for a cover up to 2 ft (600 mm) and to 10 percent for a cover up to 2 ft 11 in (875 mm). There is no impact applied when the cover is equal to or greater than 3 ft (900 mm).

Install step flashing at roof – wall intersections

Use metal step flashing to waterproof the inter­section where a roof butts into a wall. Usually made from aluminum or copper, metal step flashing is bent to form a series of elongated, L-shaped pieces that are lapped over each other in successive shingle courses as well as upward along the wall. The step flashing we used on this house is 10 in. by 6 in., and each leg is 3 in. wide.

Подпись:

Install step flashing at roof - wall intersections

The illustration on p. 144 shows how step flashing is installed on each course of shingles that runs into a wall. (It’s also used where shingles meet a chimney or a skylight curb.) Each time a regular shingle is laid down, a step shingle is placed under it—on the part of the regular shingle that will be covered. Lap step shingles by about 2 in., one on top of the other. One nail above the tar strip should be enough to

Подпись: I III III II III III III III III III III III II III III III III III III III III II III III III III III III III II III III III III III III III III II III III III III III III III III II III III III III III III III III II III STEP FLASHING Install step flashing at roof - wall intersectionsПодпись: When a roof intersects a wall, metal step shingles are used to flash the intersection. The step shingles are covered by both the three-tab shingles on the roof and by the siding on the sidewall.Подпись: Ready for rain. Finishing the roof is a big step forward. High-quality roof shingles should last 25 years or more and require little or no maintenance.

hold each step shingle in place. As each succes­sive course of shingles is nailed in place, a por­tion of the step shingle on the roof is covered. Later, when the wall is covered with siding or stucco, the vertical leg will be covered as well.

Water and Soil Storage

As physico-chemical and biological reactions occur in the soil and the water, sam­pling periods of short duration are recommended. Some chemical variables should be measured in-situ in a sub-sample (temperature, pH, redox potential, and electrical conductivity) whereas the main water sample is preserved to prevent reduction or loss of target analytes, and transported to the laboratory without delay and kept cool until further treatment. Preservation stabilizes analyte concentrations for a limited period of time. Some samples have a very short holding time (from few hours to some days).

Each analytical method available will have its own requirements for specimen preparation. The most appropriate sampling method specifications for each para­meter can be found in many textbooks (e. g. SMEWW, 1998). They should consider, for each chemical parameter, the bottle type (glass/plastic, dark), preservative (acid­ification, cold), typical sample volume, the need of filtration, and maximum storage time.

RESTAURANTS

Подпись: ✓ fast code fact If you are a residential plumber, be aware that code requirements for commercial work can be very different from what you are accustomed to working with. Be sure that you are using the proper section of the code for the type of work that you are doing. There are a lot of restaurants in society. This is a common type of building for plumbers to work with. Finding the number of water closets and lavatories re­quired in a restaurant is no more difficult than the other examples that we’ve been working with. However, there are additional requirements for restaurants.

Essentially, you must check with your lo­cal code office and comply with minimum requirements that are established by the Board of Health.

Before we do a sizing example for a restaurant, let’s discuss two alternative op­tions. You will notice if you look at the headings in Figure 9.4, that the number 6 and the number 17 are next to the heading for restaurants. We’ve already discussed the option of number 6; it is the one where one bathroom might be allowed for use by both sexes. The option per­taining to number 17 are that if alcoholic beverages will be served, the estab­lishment must meet facility requirements as set forth for clubs or lounges.

Now, let’s do a sizing example. Assume that the restaurant we are work­ing with is rated for 250 people. How many toilets are needed? Four water closets are required in each restroom. What is the required number of lavato­ries? The building calls for three lavatories in each restroom. Do you notice a difference in the ratings for restaurants, compared to the other types of build­ings we have done thus far? If you review the tables, you will see that restau­rants required, in almost all cases, the same number of fixtures for males as for females. In previous examples, female restrooms required more fixtures. This is not a big issue, just something i wanted to point out.

Make the Attic Accessible

CODE REQUIRES AN ACCESS HOLE

for the attic. This allows workers to get into the attic to install insulation and wiring. You may also want to get up there some day to check on a roof leak or just to see how the spi­ders are doing.

The standard attic access hole has a rough 2x frame that is 22/ in. wide by 31 in. long. This rectangular opening fits nicely between ceiling joists installed on 2-ft. centers. Usually, the hole is located in a closet or along a hallway. Once you select a spot for access to the attic, cut a pair of 2x$ to lit between the ceiling joists and nail them 31 in. apart. The bottom edges of the blocking should be flush with the bottom edges of the joists. Take the time to measure the hole diagonally to

make sure its square. After V*-in. dry – wall is nailed in place, you will have a 21^-in. by 30-in. access hole.

If you are using blown-in insula­tion to insulate the ceiling, build a plywood or OSB curb around the access hole in the attic (see the illus­tration at location). Make sure the curb is tall enough (at least 12 in.) to keep the insulation from falling on your head every time you open the lid to the attic. The lid can be cut from

drvwall and should be insulated with

/

rigid foam board.

Instead of a simple lid for the access hole, you may want to install a pull-down stair kit. If you’re using the attic for storage and need to go up there on a regular basis, a pull-down staircase is the way to go. The stairs
are factory-made and designed to tit in the 22’A-in. space between the joists. The rough opening must be longer, however—54 in. is common. Check the manufacturer’s specifica­tions before you frame the opening.

Create attic storage. There is some usable space in most truss-roof attics. To make use of it, don’t lay sheathing directly on the joist chords of roof trusses, since trusses are not engineered for this. Instead, frame a storage platform above the joist chords, making sure that there’s room for insulation beneath the platform. You can support the platform’s joists on short (8-in. to 12-in.-long) cripple studs nailed to the double top plates of nearby walls. Sheathe the platform with l/’-in. plywood or OSB.

Make the Attic Accessible

CREATING USABLE SPACE IN THE ATTIC

Make the Attic Accessible

Some usable storage space can be created in the attic, even if the house has a truss roof, by framing a floor that is supported by interior walls and sits above the trusses’joist chords.

 

Make the Attic AccessibleПодпись:

than with trusses, the ridge board can also run long at this stage. This process ensures a straight barge rafter. It also lets you straighten a bowed gable rafter. Hoist a supply of 2×4 lookout stock up to the roof, set each lookout in its notch, and secure the end of each one against the face of the first inboard truss (or rafter) with two 16d nails. Keep the top sur­face of the lookout flush with the top edge of the truss. Now, before driving a pair of 16d nails into each notch, move the gable-end rafter in or out to make the framing member as straight as possible. Nail all the lookouts in their notches, and leave them like this for now.

I Make truss ends straight

If the truss tails have been set straight, begin to cut and install the gutter or fascia boards. If they’re not straight, snap a line across the top or bottom edges, then mark plumb cuts to

I lineup the truss ends in the same plane. This is another exception to the “never cut a truss" rule. As long as you’re onlv making a small cutoff at the end of the truss (well away from joints and gussets), there’s no chance of struc­tural damage. Check the cave overhang called

O’ о

for in the plans. Measure this distance on the gable-end trusses at opposite ends of the house. Then snap a line across all the trusses, going from one end of the house to the other. Mark the cut lines on the truss ends and make the cuts. This technique also works when building plans call for exposed rafter tails that must be cut in the same plane. Use a iig or pattern, such as the one shown in the top illustration on p. 132, to mark identical plumb cuts on the rafter tails. You can make the cuts from above, while standing on the top plate of the wall, or from below, while standing on a ladder or seal Inkling.

Install the gutter or fascia boards

Although both gutter and fascia boards are nailed to the rafter tails, the distinction between the two is that gutter boards are later

covered with trim (aluminum or vinvl clad-

/

ding or lx finish material), fascia, on the other hand, is a finished surface. Some builders install lx fascia boards over 2x gutter boards or subfascia, while others use the 2x stock as the finished fascia.

Obviously, smooth, accurate cuts are required to install fascia boards, while gutter boards are less demanding. On this house, we installed gutter boards that were latei clad with coil aluminum (see chapter 7).

STEP4 Secure the Trusses to the Interior Walls

On small houses, trusses are generally engi­neered to obtain their support from exterior walls without needing further support from interior walls. Still, its not uncommon for trusses to cross over and bear on interior walls. Inmost regions, these trusses can usually be nailed directly to the interior wall with two 16d toenails on one side and one on the opposite side. T his is not the case, however, if you live in a part of the country where the weather may be freezing one day and boiling the next. In areas with extreme temperature fluctuations, trusses must be able to expand and contract freely. Otherwise, drvwall ceilings

nailed to these trusses tend to crack. Check with your building department for the code requirements in your town or city. In addition, ask area builders what the local practice is.

To secure a truss to a wall while still allow­ing it to adapt to fluctuations in temperature and humidity, use a truss clip, as shown in the illustration above. These clips, which are nailed both to the wall plates and to the truss, feature a slot that allows the truss to move up and down as it expands and contracts— just make suie (lie nail is slightly loose in the joist chord.

INSTALLING BACKING FOR DRYWALL

 

Ceiling joists or joist chords

STEP4 Secure the Trusses to the Interior Walls

Dry wall nails

 

2×6

wall

 

STEP4 Secure the Trusses to the Interior WallsSTEP4 Secure the Trusses to the Interior Walls

STEP4 Secure the Trusses to the Interior Walls

Подпись: SNAPSHOTSSTEP4 Secure the Trusses to the Interior Walls

Doing carpentry on the ground is easier than doing carpentry on top of the house. . .

You really have to watch your step. . .

We handle the trusses carefully. . .

Plenty of bracing is required to tie the trusses together and to the rest of the house. . .

When all the trusses are up, the house looks like an upside-down ship with the hull removed.

STEP5 Add Drywall Backing

At this point, you need to add backing (some­times called deadwood) to walls that run parallel to the joist chords. The backing pro­vides a nailing base lor ceiling drywall (see the illustration on the facing page). It’s easier to put the backing in now rather than after the roof is complete and hampered by sheath­ing. This is a good place to use knotty, crooked 2x stock. Nail the stock, one 16d nail every 16 in., to the double top plate of parallel walls so that the 2x backing overhangs the double top plate by at least 1 in. on both sides. Some builders use drywall clips instead of solid backing, but I like to fasten nails into some­thing solid. (Details on drywall installation are covered in chapter 9.)

This process is different for trusses that are allowed to expand and contract. In this case, nail flat 2×4 “ladder” hacking between the joist chords on a parallel wall. Use truss clips to attach the backing to the double top plate of the wall, as explained on p. 127. This allows the trusses to expand and contract without stressing the drywall.

STEP6 Install the Barge Rafters and the Fascia Boards

As explained earlier in this chapter, barge rafters extend beyond the end of a building, creating an overhang at the gable ends. The rafters are supported by 2×4 lookouts that fit into notches cut in the gable-end rafters. The lookouts run back to the first inboard rafter (or roof truss). At the bottom corner of the roof, a barge rafter meets a gutter or fascia board, which extends along the eave and is fastened to the ends of the rafter tails.

There are a few tricks to making sure that these exterior trim details are done correctly. If you plan to cover the trim with aluminum cladding, as we did on this house (see chapter 7), the culling and installation work is a little easier. On the other hand, if the barge rafter and fascia will be exposed as finished trim, youTl need to cut tight-fitting miter joints where the boards meet. Both types of installa­tion will go more smoothly if you follow the order of the steps described below.

Let the lookouts run long

I like lo install lookouts long, then snap a line from eave to ridge and cut the lookouts in place. If you’re framing with rafters rather

ІШІ

 

STEP4 Secure the Trusses to the Interior WallsSTEP4 Secure the Trusses to the Interior WallsSTEP4 Secure the Trusses to the Interior Walls

Some One-Dimensional Numerical Integration Formulas

Подпись: I = Подпись: b f (x) dx a Подпись: (4A.1)

This appendix summarizes some commonly used numerical formulas for eval­uating the following integral:

Detailed descriptions of these and other numerical integration procedures can be found in any numerical analysis textbook.

4A.1 Trapezoidal rule

For a closed integral, Eq. (4A.1) can be approximated as

1 = hif 1 + 2 Щ fi + (4A.2a)

where h is a constant space increment for discretization, n is the number of discretization points over the interval (a, b), including the two end points, and fi is the function values at discretized point, xi.

For open and semiopen integrals, Eq. (4A.1) can be computed numerically as

1 = h (эf2 + 2£ fi + 3fn-1^ (4A.2b)

4A.2 Simpson’s rule

For closed integrals, one has

h

1 = 3 [ f 1 + 4( f 2 + f 4 + f 6 + ■ ■ •) + 2( f 3 + f 5 + f 7 + •• •) + fn] (4A.3a)

For open and semiopen integrals, one has

Подпись: 1 = 12[27f2 + 13( f 4 + f 6 +•) + 16( f 5 + f 7 + ■ ■ •) + 27f n-1] (4A.3b)

4A.3 Gaussian quadratures

Equation (4A.1) can be expressed as

n

1 = wif (Xi) (4A.4)

i=1

where wi is the weight associated with the ith abscissa xi in the discretiza­tion. The weight wi is related to orthogonal polynomials. Table 4A.1 lists some commonly used orthogonal polynomials and their applied integral range, ab­scissas, and weights. Definitions of those polynomials and tables of abscissas and weights for different Gaussian quadratures are given by Abramowitz and Stegun (1972).

TABLE 4A.1 Some Commonly Used Gaussian Quadratures

Some One-Dimensional Numerical Integration Formulas

Gauss Range (a, b) Abscissas Xf Weight wi

in which lkj and akj are elements in matrices L and A, respectively, and K is the size of the matrices. In terms of akj’s, lkj’s can be expressed as

k-1

Подпись: (4B.4)lkk — I akk ^ ^ lkj

Some One-Dimensional Numerical Integration Formulas Подпись: for k — j + 1,..., K (4B.5)

j—1

Computationally, the values of lkj’s can be obtained by solving Eqs. (4B.4) and (4B.5) sequentially following the order k — 1,2,…, K. Numerical examples can be found in Wilkinson (1965, p. 71). A simple computer program for the Cholesky decomposition is available from Press et al. (1992, p. 90). Note that the requirement of positive definite for matrix A is to ensure that the quantity in the square root of Eq. (4B.4) always will be positive throughout the computation. If A is not a positive-definite matrix, the algorithm will fail.

For a real, symmetric, positive-definite matrix A, the Cholesky decomposition is sometimes expressed as

A — L Л Lt (4B.6)

in which L is a unit lower triangular matrix with all its diagonal elements having values of ones, and Л is a diagonal eigenvalue matrix. Therefore, the eigenvalues associated with matrix A are the square roots of the diagonal el­ements in matrix L. If a matrix is positive-definite, all its eigenvalues will be positive, and vice versa.

In theory, the covariance and correlation matrices in any multivariate prob­lems should be positive-definite. In practice, sample correlation and sample covariance often are used in the analysis. Owing to the sampling errors, the resulting sample correlation matrix may not be positive-definite, and in such cases, the Cholesky decomposition may fail, whereas the spectral decomposition described in Appendix 4C can be applicable.