Building Construction Principles, Materials, & Systems 2009 UPDATE Medan Mehta Solution Manual

Online Instructor’s Manual to accompany

Homework and Classroom Assignment Manual 2009 Update First Edition

Madan Mehta University of Texas at Arlington

Diane Armpriest University of Idaho

Solutions: Homework and Classroom Assignment Manual

to accompany

Building Construction: Principles, Materials, and Systems 2009 Update by Madan Mehta Walter Scarborough Diane Armpriest

A NOTE TO INSTRUCTORS This document, provides answers to questions in the Homework and Classroom Assignment Manual. Most questions in Assignments 1 through 9 (related to Part I of the textbook—Principles of Construction) have definitive answers. However, this is not true with assignments related to Part II of the text (Assignments 10 through 30), particularly those that require the drawing of construction details. Although most details provided in the Solution Manual are based on accepted principles and practices, some variation is possible due to regional differences or the aesthetic goals. The instructor may wish to address this issue during the course of class and review and (or) alter the response provided as necessary.

Different line and fill colors have been used in details to improve their readability. For the same reason, different colors have been used in the text part of the solutions.

As stated in the introduction to the Assignment Manual, answers to many questions cannot be obtained directly from the textbook. The students should be reminded that developing appropriate solutions to many of the questions requires the ability to think creatively and look beyond rote answers. In some situations, references to later chapters of the text and (or) resources outside the textbook are also required.

The Instructor may provide the students with these references, or require that they find them on their own.

Assignment 1 Name

Chapter 1

1.1(a)

The current version of the MasterFormat (MF04) was introduced in 2004 and includes major changes from the 1995 version. Obtain a copy of the 1995 version of MasterFormat (MF95) from your college library or online and list at least three major changes that were made in 2004.

Note: An online version of MF95 is available from: www.ctrlspecbuilder.com/csi. MF04 may also be obtained from there by following the link: Standard & Formats.

1. MF04 has 50 divisions in place of 16 divisions in MF95.

2. Division 2 (Site Construction) in MF95 has been removed and its contents distributed over three divisions in MF04: Division 31 (Earthwork), Division 32 (Exterior Improvements), and Division 33 (Utilities). Division 2 in MF04 deals with Existing Conditions.

3. Division 15 (Mechanical) in MF95 has been removed and its contents distributed over three divisions in MF04: Division 21 (Fire Suppression), Division 22 (Plumbing), and Division 23 (HVAC). Division 15 in MF04 has been reserved for future expansion.

4. Division 16 (Electrical) in MF95 has been moved to Division 26 in MF04.

1.1(b)

Which divisions of MF95 did not change in MF04?

Division 00 (Procurement and Contracting); Division 01 (General Requirements); Division 03 (Concrete); Division 04 (Masonry); Division 05 (Metals); Division 06 (Wood, Plastics, and Composites); Division 07 (Thermal and Moisture Protection); Divsion 08 (Openings, called Doors and Windows in MF95); Division 09 (Finishes); Division 10 (Specialities); Divsion 11 (Equipment); Division 12 (Furnishings); Division 13 (Special Construction); and Division 14 (Conveying Equipment, called Conveying Systems in MF95) remain essentially unchanged.

1

1.2

You are the project architect for a new office building conducting a routine visit to the construction site. You observe that the concrete in the floor slab of the building is being excessively consolidated. When you return to your office, you send an email to the concrete subcontractor whom you met at the site to place your observation on record. Have you done the right thing? Explain.

No, because the architect’s construction observations should be communicated to the general contractor (GC) with a copy to the owner. The architect’s communication link is with the GC, not the subcontractor.

1.3

During a routine visit to the construction site, the project architect observes that the opaque parts of the exterior wall were being constructed before installing the glazed elements (which were available at the site). The architect was concerned that this sequence of wall construction would compromise the aesthetics of the wall and discussed the issue with the general contractor, who disagreed. In your opinion, should the architect instruct the general contractor to install the glazed elements first and then construct the opaque parts around the glazed elements? Explain your answer.

It is fine for the architect to raise the issue with the GC during a site visit. However, due to the architect’s role as a construction observer, he/she has no authority to instruct the GC as to the sequence of construction. The architect should place his/her observations and the concerns on record with a copy to the owner.

1.4

You have been retained as the facilities manager for an independent school district (ISD) that administers a large number of existing schools and formulates policies for the construction of new ones. Because of its experience, the ISD has a short list of architects and general contractors who are highly professional and competent. The ISD board has requested your recommendation for the project delivery system they should adopt for the construction of the new facilities. What is your recommendation to the board? Explain.

Because the ISD, based on its own experience, has a short list of highly competent architects and general contractors, the traditional design-bid-build method of project delivery is not well suited in this situation. Two project delivery methods that should be recommended are: • Negotiated contract method • CMAR method In both methods, a selected general contractor (from the ISD’s list) may be brought on board during the design phase.

2

Assignment 2 Name

Chapter 2

The following answers are with reference to IBC (2009) and IRC (2009).

2.1

2.2

Obtain the latest edition of the International Building Code (IBC) or a copy of the current building code of your city from your college library. List the chapter numbers and chapter titles that deal with: (a) Administrative provisions of the code

Chapter 1: Administration

(b) Occupancy classification of buildings

Chapter 3: Use and Occupancy Classification Chapter 4: Special Detailed Requirements Based on Use and Occupancy

(c) Types of construction

Chapter 6: Types of Construction

(d) Allowable heights and areas of building

Chapter 5: General Building Heights and Areas

In which chapter of the IBC (or the building code of your city) are provisions relating to the building permit, certificate of occupancy, and stop work order provided?

Chapter 1: Administration

2.3

With reference to the IBC (or the building code of your city), under what conditions and in which format should a stop work order be issued by the building official?

When the building official (or his/her designated official, i,e., the building inspector) notices, during an inspection, that any work at the construction site is in gross violation of the code, he/ she should issue a stop work order. The violation may be noticed either during a code-mandated inspection or during an additional inspection deemed necessary by the building official.

(Note: Exercising his/her discretion, the inspector may not issue a stop work order but direct its immediate correction and subsequent verification if the violation is minor.)

3

2.4

The IBC (or the building code of your city) references a number of construction standards. (a) How many standards organizations are referenced?

A total of 17 standards are referenced by IBC (2006). (b) Which organization provides the maximum number of referenced standards?

The International Code Council (ICC) with 10 referenced standards.

2.5

Obtain a copy of the latest edition of the International Residential Code (IRC) from your college library and answer the following questions: (a) Does IRC apply to all residential occupancies? Explain.

IRC applies only to one- and two-family dwellings.

(b) Compare the total number of pages in the IRC and IBC documents and comment on why there is a large (or small) difference in the number of pages.

The total number of pages in both IBC (2006) and IRC (2006) are approximately the same although IRC refers only to one- and two-family dwellings while the IBC refers to all types of buildings including one- and two-family dwellings. Note that IRC is a comprehensive code that regulates all aspects of residential design and construction including all services required in a typical residence, such as mechanical, electrical, plumbing, lighting, energy efficiency, swimming pools, etc. The IBC regulates only building construction. Building services are regulated by respective speciality codes.

4

Name

2.6

With reference to Table 2.1 in the text (or the Occupancy Classification chapter of the IBC or your city’s building code), determine the occupancy classification of the following buildings. (a) A university physics and chemistry laboratory building.

Occupancy B (b) A concert hall with a seating capacity of 2,000.

Occupancy A, Division 1, i.e., Occupancy A-1 (c) A 20-story apartment building with a floor area of 10,000 sq ft on each floor. The ground floor of the building contains offices (e.g., apartment management offices) and shops (e.g., dry cleaning pickup, barber, coffee shop, etc.).

Apartment units are Occupancy R-2, and shops and offices belong to occupancy B. Therefore, it is a mixed occupancy building. However, if the area of offices, shops, etc., at the ground floor is less than 10% of the total ground floor area (i.e., less than 1,000 sq ft), the building may be treated as belonging to its major occupancy—R-2. (see text).

(d) A school of architecture building with faculty offices, studio spaces, an auditorium (200seat capacity) and a number of classrooms (each for 75 students).

A school of architecture building belongs to Occupancy B. However, classrooms and auditorium will be classified as Occupancy A-3 because their occupant load (seating capacity) is not less than 50. It is, therefore, a mixed occupancy building comprising Occupancies B and A-3.

(d) A day care center (for a maximum of 100 children, 2-1/2 years or older).

Occupancy E

5

2.7

A 3-story (45-ft-high) city library building (Occupancy A-3) is constructed using Type II(B) construction and is fully sprinklered. The building’s footprint measures 200 ft x 150 ft as shown in the following site plan. Using the information provided in “Principles in Practice” in Chapter 2 in the text, determine if the building conforms with the allowable area and height requirements of the building code. Show calculations.

Property line

40 ft

Line of road

18 ft

80 ft

City Library

40 ft

150 ft

200 ft

80-ft-wide right of way

Property line

Summary of Findings Occupancy A-3 Type of construction: Type II(B) From Table 1, page 46 of the text: At = 9,500 sq ft Maximum allowable number of stories = 2 Maximum height = 55 ft Because the building is sprinklered, allowable maximum number of stories = 3,and allowable maximum height = 55 + 20 = 75 ft.

Thus, the number of stories and height provided are both OK. Area Increases F = perimeter of building that fronts an accessible open space ≥ 20 ft wide = 2(200) + 150 = 550 ft P = perimeter of entire building = 2(200) + 2(150) = 700 ft F = 550 = 0.79 700 P From Eq. (2): If =

W = 30 because all perimeter dimensions of building included in F front an open space width > 30 ft F - 0.25 W = (0.79 - 0.25) 30 = 0.54 30 30 P 6

Name From Eq. (1): Area increase due to frontage = At(If) = 9,500(0.54) = 5,130 sq ft Is = 2.0. Hence, from Eq. (3): Area increase due to sprinklers = At(Is) = 9,500(2) = 19,000 sq ft Allowable Area and Area Provided From Eq. (4): Total allowable area per floor = Aper floor = 9,500 + 5,130 + 19,000 = 33,630 sq ft Area provided per floor = 200(150) = 30,000 sq ft < 33,600 sq ft. OK From Eq. (5): Aall floors allowed = 3.0(Aper floor) = 3.0(33,600) = 100,800 sq ft Aall floors provided = 3(30,000) = 90,000 sq ft < 100,800 sq ft. OK

2.8(a) A four-story (55-ft-high) office block without an automatic sprinkler system is of Type II(A) construction as shown in the following drawing. (a) Using the information provided in “Principles in Practice” in Chapter 2 in the text, determine if the building conforms with the allowable area and height requirements of the building code. Show calculations.

Note: Because the width of open spaces (W) is different on different faces of the building, the value of W is the weighted average over all four faces of the building. If the width of an open space is less than 20 ft, it is ignored (see Principles in Practice) and if it is greater than 30 ft, it is set equal to 30 ft. The weighted-average value of W for this building is 25.54 ft, but you must verify it and show calculations.

Notes on Frontage

15 ft

Property line

80-ft-wide right of way

200 ft

80 ft

Office Block

22 ft

25 ft

180 ft

Line of road

Property line

An important difference between this problem and the previous problem is that in this problem, some open space widths (Wvalues) lie between 20 ft and 30 ft.

Therefore, an average value of all W-values that are ≥ 20 ft must be determined. These W-values are: 25 ft, 22 ft and 160 ft (set = 30 ft).

Note that the road width is included in W.

Wav = 25(180) + 30(180) + 22(200) = 25.54 180 +180 + 200 Summary of Findings Occupancy B Type of construction: Type II(A) From Table 1, page 46 of the Text: At = 37,500 sq ft 7

Maximum allowable number of stories = 5 The number of stories provided (4) and height provided (55 ft) are both OK. Because the building is not sprinklered, no increase is allowed in the number of stories and the height of building.

Area Increases F = perimeter of building that fronts an accessible open space ≥ 20 ft wide = 200 + 2(180) = 560 ft P = perimeter of entire building = 2(200) + 2(180) = 760 ft F = 560 = 0.74 P 760 From Eq. (2): If =

F - 0.25 W = (0.74 - 0.25) 25.54 = 0.42 30 30 P

From Eq. (1): Area increase due to frontage = At(If) = 37,500(0.42) = 15,750 sq ft Is = 0 (no sprinklers).

Allowable Area and Area Provided From Eq. (4): Total allowable area per floor = Aper floor = 37,500 + 15,750 = 53,250 sq ft Area provided per floor = 200(180) = 36,000 sq ft < 53,250 sq ft. OK From Eq. (5): Aall floors allowed = 3.0(Aper floor) = 3.0(53,250) = 159,750 sq ft Aall floors provided = 4(36,000) = 144,000 sq ft < 159,750 sq ft. OK (b) If this building is provided with an approved automatic sprinkler system throughout, can its height be increased to 8 stories? Explain.

No, because only one story may be increased beyond the basic allowable number of stories, which in this case is 5. (c) If this building is provided with an approved automatic sprinkler system throughout, what should be its minimum construction type if it is: •

10 stories high?

Type I(B), which will allow up to (11 + 1 =) 12 stories.

•

20 stories high?

Type I(A), which will allow an unlimited number of stories, subject to the limitations imposed by the city’s zoning code.

•

40 stories high?

Type I(A), which will allow an unlimited number of stories, subject to the limitations imposed by the city’s zoning code.

8

Assignment 3 Name

Chapter 3

3.1

Estimate the approximate total floor live load in your bedroom (dormitory room) by adding the estimated weights (in pounds) of all movable items (including the probable maximum number of occupants that might be present in the room even for a short duration). Divide the total load by the floor area of the room (in sq ft). Compare your result with the minimum floor live load for residential occupancies given in Table 3.1 in the text.

Weight of bed

400 #

Weight of other furniture items (tables, chairs, book racks, cabinets, etc.). List them and give their approximate weights and finally sum them.

Table 1 Table 2 Chairs (2) Book rack inc. books

100 # 150 # 120 # 400 #

Total

770 #

Weights of people

8* people with an average weight of 180 # each = 1,440 #

Load of other items

Clothes, shoes, computer system, etc. = 500 #

Total live load in room

400 + 770 + 1,440 + 500 = 3,110 #

Area of room

140 sq ft

Live load in room (in psf)

3,110 = 22.2 psf 140

Minimum prescribed live load in Table 3.1 Comment on your findings.

40 psf

The actual live load is well below the minimum design live load prescribed in the code. (Note: in most occupancies, the actual maximum possible occupancy load is less than that stipulated by building codes or standards.)

* This assumption is based on the fact that the room area of 140 sq ft will accommodate a maximum of 8 persons at a time.

9

3.2

Using the International Building Code or the building code of your city, answer the following questions: (a) In which chapter of the code is information about the floor live loads included?

Chapter 16: Structural Design (b) Provide the floor live load table number and caption.

Table 1607.1: Minimum uniformly distributed live loads and minimum concentrated live loads (c) From Table 3.1 in your text, provide the floor live load values for the following occupancies.

If the occupancy listed below is not included in the text table, refer to the code table.

•

Office room in a bank building

50 psf

•

Outdoor deck in a house

40 psf (Table 1607.1, IBC 2009)

•

Restaurant floor

100 psf ( Table 1607.1, IBC 2009)

•

Second floor corridor in a school building

100 psf ( Table 1607.1, IBC 2009)

(d) Which of the above live load values is the highest and why?

Live loads in school corridor and restaurant floor are highest because of the large number of people that may congregate in these spaces.

3.3

In Example 2 of “Principles in Practice” in Chapter 3 in the text, the dead load of a typical contemporary wood frame residential floor was determined to be 9.5 pounds per sq foot (psf).

What is the dead load of the same floor if the floor joists are: (a) 12 in. on center (o.c.)? Show calculations.

From the text, weight of 1 ft long floor joist = 4.1 # Weight of floor joist that contributes to 1 sq ft of floor = 4.1 # Hence, the total dead load of 1 sq ft of floor = 6.4 + 4.1 = 10.5 psf 10

Name (b) 24 in. o.c.? Show calculations.

From the text, weight of 1 ft long floor joist = 4.1 # Weight of floor joist that contributes to 1 sq ft of floor = (4.1/2.0) = 2.05 #, say 2.1 # Hence, the total dead load of 1 sq ft of floor = 6.4 + 2.1 = 8.5 psf

3.4

In Example 2 of “Principles in Practice” in Chapter 3 in the text, the dead load of a typical contemporary wood frame residential floor was determined to be 9.5 pounds per sq foot (psf).

If the subfloor consists of 1-1/8 in. thick tongue-and-groove plywood panels (no underlayment) and the floor finish consisted of 1-1/2 in. thick terrazzo, determine the floor’s dead load. Show calculations.

Note: You will need to refer to Table 1 in “Principles and Practice” in Chapter 3 in the Text.

From Example 2 (of the text), weight of 1/4 in. thick hardboard underlayment and 3/4 in. thick plywood subfloor = 3.2 psf. This is replaced by 1-1/8 in. thick plywood subfloor, which weighs = 9(0.4) = 3.6 psf (Table 1 of Principles in Practice of the text), resulting in an addition of (3.6 - 3.2) = 0.4 psf.

Weight of 1-1/2 in. thick terrazzo floor = 19 psf (Table 1) Therefore, the weight of the floor = 9.5 + 0.4 + 19 = 28.9 psf, or say 30 psf.

3.5

The dead load of a typical residential roof with a light covering (such as asphalt shingles) is generally assumed to be 10 psf. If the roof covering consists of clay tiles, what is the dead load of the same roof? Show calculations.

From Table 1, weight of clay roof tiles = 12 to 18 psf. Assume 18.0 psf on the conservative side (in absence of the precise information from the manufacturer). Asphalt shingles weigh 2.0 psf (Table 1 of Principles in Practice of text). Therefore, the additional dead load on this roof = 18.0 - 2.0 = 16.0 psf. Thus, the roof dead load = 10 + 16 = 26 psf.

11

3.6

Using the International Building Code, the International Residential Code, or the building code of your city, determine the approximate ground snow load in your city, and then comment on the minimum approximate live load for which a roof in your city should be designed.

The International Building Code (2009) gives ground snow load contours superimposed on the map of the U.S. Assuming that the roof snow load is the same as the ground snow load (a conservative assumption), the roof should be designed for the snow load if the snow load is > 20 psf. If the snow load < 20 psf, the roof should be designed for 20 psf.

3.7

Refer to Figure 3.11 in the text and estimate the approximate basic wind speed for your city.

How does it compare with the maximum and minimum basic wind speeds for the continental United States?

The maximum and minimum wind speeds for the continental U.S are 85 mph (West Coast) and 150 mph (South Florida) respectively.

12

Assignment 4 Name

Chapter 4

4.1

A concrete cylinder (6 in. diameter, 12 in. high) failed in compression at a load of 142.5 kips.

Determine the strength of concrete. Show calculations.

(Ultimate) strength of concrete, fc =

Failure load, P Cross-sectional area of cylinder, A

P = 142.5 k (“k” is the symbol for kips) A = π(3)2 = 28.27 in.2 (Note: radius of cylinder = 3 in.)

Hence, fc =

4.2

142.5 k 28.27 in.2

= 5.041 k = 5.041 ksi = 5,041 psi in.2

(“ksi” is an acronym for kips per square inch and “psi” is an acronym for pounds per square inch)

Is it easier to break a concrete beam by subjecting it to bending or (axial) compression?

Explain the reasons for your answer.

It is easier to break a concrete beam by bending it rather than by subjecting it to axial compression. In bending, one half the beam’s cross-section is subjected to tension, and the other half is subjected to compression. In axial compression, the entire cross-section of the beam is subjected to compression. Because concrete is much weaker in tension than compression, the beam’s failure occurs in tension. The other reason is that in bending, the stresses are concentrated at the edges (extreme fibers) of the cross-section (Fig. 4.17). In axial compression, the stresses are uniform over the entire cross-section (Fig. 4.14).

The stress-strain relationships of two concrete mixes are shown in the sketch at right and are marked as 1 and 2. Determine which mix:

1 Strength of Mix1

(a) has greater strength.

Strain

Mix 2

Ultimate strain of Mix 2 Ultimate strain of Mix 1

13

Strength of Mix 2

2 Stress

4.3

(b) has higher modulus of elasticity and why.

Mix 2, because the the initial part of the stress-strain curve of Mix 2 has a greater slope (see Section 4.5 and Fig. 4.13 of the text).

(c) is relatively more ductile and why.

Mix 1 is more ductile, because its ultimate strain is greater than that of Mix 2 (see page 84 of the text).

4.4

Your structural engineer has advised the use of 50 ksi steel for the structural frame of the building you are designing. What does 50 ksi represent?

50 ksi (kips per square inch) represents the yield strength of steel.

4.5

Elastomeric joint sealants are commonly used to fill the joint space between building components. Because the sealant is subjected to frequent expansion and contraction due to the movement of components, your partner has advised the use of a sealant with a low modulus of elasticity (called low-modulus sealant). Is your partner correct? Explain.

The partner is correct in recommending the use of a low-modulus sealant because a sealant with a low modulus of elasticity will be subjected to lower stresses. Remember: Modulus of elasticity, E =

Stress, f Strain, ε

Thus, Stress = (Modulus of elasticity) x (strain), or f=Eε Therefore, for the same deformation (strain), a low-modulus sealant will be subjected to a lower stress. This is particularly significant for a sealant because it is subjected to fatigue caused by repeated expansion and contraction that may lead to premature failure.

14

Assignment 5 Name

Chapter 5

Following the format of the example given below, sketch in plan the following components/ materials and then (using Table 5.3 in the text) determine their R-values. Sketches must use the standard graphic symbols (see Appendix at the end of this manual) and the dimensions shown must be actual dimensions. Ignore inside and outside surface resistances for determining the R-values of components/materials (in Problem 5.1) but include them for the Rvalues of assemblies.

7-5/8 in.

Example: An 8 in. nominal (7-5/8 in. actual) brick wall. Scale: 1” = 16”; i.e., each grid division = 4 in.

8” thick brick wall

From Table 5.3: R-value of 1” thick brick wall = 0.2.

Therefore, R-value of 7-5/8” (= 7.625”) thick brick wall = 0.2(7.625) = 1.53.

(a) An 8-in. nominal brick wall with 5/8 in. thick plaster on one side. Scale: 1” = 16”.

Material 5/8” (0.625”) plaster 7-5/8” (7.625”) brick wall Total R-value (Rt)

8-1/4”

8” nominal thick brick wall with 5/8” thick plaster on one side

R-value 0.35(0.625) = 0.22 0.2(7.625) = 1.53 = 1.75 i.e., an R-2 wall

(b) A 5-1/2-in. thick fiberglass (batt) insulation. Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

5-1/2”

5.1

R-value = 3.5(5.5) = 19.25, i.e., R-19 insulation

5-1/2” thick fiberglass insulation

15

4”

(c) A 4-in. thick extruded polystyrene board. Scale: 1-1/2” = 1’.

R-value = 5.0(4) = 20.0, i.e., R-20 insulation

4” thick extruded polystyrene insulation

(d) A 1/4-in. thick glass sheet (see Chapter 28). Scale: full-size.

R-value = 0.14(0.25) = 0.035, i.e., R-0. In other words, the R-value of a glass sheet (without considering the surface resistances) is almost zero.

1/4” thick glass sheet

(e) A 1-in. thick insulating glass unit (IGU) consisting of two sheets of 3/16-in. thick glass with 5/8 in. air space between panes (see Chapter 28). Scale: full-size.

1”

3/16” th. glass 5/8” air space

1” thick IGU

3/16” th. glass

Material 3/16” (0.19”) glass Air space 3/16” (0.19”) glass Total R-value (Rt)

R-value 0.14(0.19) = 0.03 1.00 0.14(0.19) = 0.03 = 1.06 i.e., R-1

(Thus, disregarding the interior and exterior surface resistances, the R-value of an IGU with an air-filled cavity equals approximately1.0. See Fig. 28.19 for the cross-section of an IGU.)

5.2

Sketch in plan the wall assembly described below and use Table 5.3 in the text to determine the assembly’s R-value and U-value. Follow the format given in Example 3 in Chapter 5 and label all elements of the assembly.

2 x 6 nominal (1-1/2 in. x 5-1/2 in. actual) wood studs spaced 24 in. on center with 5/8-in. thick gypsum board on the interior, 1/2 in. thick oriented strandboard (OSB) sheathing toward the exterior, air-weather retarder, 1 in. air space followed by 4 in. nominal (3-5/8 in. actual) brick veneer. The space between studs contains fiberglass insulation. Approximate scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

16

Name Note: In determining the R-value of the stud-insulation layer of the assembly, ignore the presence of studs and assume that the layer consists entirely of the insulation. Assume the R-value of OSB = the R-value of plywood.

Refer to Chapter 6 in the text for a description of an air-weather retarder and Figure 14.1(d) for a threedimensional sketch of the assembly.

5/8” th. gypsum board 2 x 6 wood stud 5-1/2 th. fiberglass insulation

Wood frame wall clad with brick veneer

1/2” th. OSB Air-weather retarder 4” (nominal) brick veneer

1” air space

Material Inside surface resistance 5/8” gypsum board 5-1/2” th. fiberglass insulation 1/2” th. OSB Air-weather retarder 1” air space 3-5/8” thick (4” nom.) brick veneer Outside surface resistance Total R-value

5.3

R-value = 0.20 0.6(0.625) = 0.38 3.5(5.5) =19.25 0.9(0.5) = 0.45 = 0.0 = 1.0 0.2(3.625) = 0.73 = 0.70 = 22.71, i.e., R-23 assembly

Determine the R-value of the assembly in Question 5.2, but for the OSB sheathing, substitute 1/2 in. thick plywood laminated with aluminum foil sheathing and provide an air-weather retarder. A sketch is not required but show calculations.

The R-values of OSB and plywood are assumed to be identical. However, because the plywood is laminated with aluminum foil, the R-value of a vertical air space = 2.5 (see Notes on pages 110 and 111 of the text). Therefore, the R-value of the assembly = 22.71 + 2.5 -1.0 = 24.21, i.e., R-24 assembly.

17

5.4

Sketch in plan the wall assembly described in Question 5.2, but substitute 1/2 in. thick (average) wood siding for air space and brick veneer. Approximate scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

Note: Refer to Figure 14.1(a) in the text for a three-dimensional sketch of the assembly.

5/8” th. gypsum board 2 x 6 wood stud Wood frame wall clad with wood siding 1/2” th. OSB Air-weather retarder Wood siding 5-1/2” th. fiberglass insulation Material R-value Inside surface resistance = 0.20 5/8” gypsum board = 0.38 5-1/2” th. fiberglass insulation =19.25 1/2” th. OSB = 0.45 Air-weather retarder = 0.0 1/2” (av.) wood siding = 0.9(0.5) = 0.45 Outside surface resistance = 0.70 Total R-value = 21.43, i.e., R-21 assembly

5.5

Does the wall of Question 5.4 comply with the insulation requirements of the International Energy Conservation Code (IECC) for Climate Zone 7 (see text)? Give reasons for your answer.

From Figure 5.22 (text), the minimum R-value of the insulation in the cavity spaces between wood studs of a single-family dwelling in Climate Zone 7 is R-21. Because the R-value of cavity insulation in Question 5.4 is R-19, it does not comply with the prescriptive provisions of the IECC. Note that the prescriptive provisions of IECC do not consider the overall R-value of the assembly.

18

Name

5.6

Determine the overall R-value and U-value of the exterior walls a building with the following properties: The opaque parts of the exterior walls are constructed using the assembly in Question 5.2.

Windows (with an R-value = 1.8) constitute 30% of the wall area.

The U-value of the opaque portion of the wall, Uop =

1 = 0.043 23

The U-value of the glazed portion of the wall, Ugl =

1 = 0.556 1.8

From Eq. (6) of text, Uo =

Aop(Uop) + Agl(Ugl)

Therefore, overall R-value =

5.7

Aop + Agl

=

70(0.043) + 30(0.555) = 0.197 70 + 30

1 = 5.0. Hence, it is an R-5 wall assembly.

0.197

Does the wall of Question 5.6, if used in a single-family dwelling of wood frame construction, comply with the insulation requirements of Climate Zone 3 of the International Energy Conservation Code? Explain.

It complies, because it meets the requirement for minimum R-value of cavity insulation. The provided cavity insulation is R-23, and that required for Climate Zone 3 is R-13.

5.8

A single-family wood light frame dwelling in Climate Zone 6 has an unconditioned basement.

Determine the minimum thickness of fiberglass insulation required in the ground floor assembly (floor immediately above the basement).

From Figure 5.24 (text), the minimum R-value of floor insulation is R-30. From Table 5.3 (text), the R-value of 1 in. thick fiberglass insulation is approximately 3.5. Therefore, the minimum thickness of fiberglass insulation = (30/3.5) = 8.6” 19

5.9

Using online resources, list three major North American manufacturers of extruded polystyrene board insulation (XEPS) and the colors used by each to distinguish their products.

From “www.xpsa.com” (“Selected Websites” on page 131 of the text), three major manufacturers of extruded polystyrene insulation are: Dow Corp. (Styrofoam brand insulation—blue in color) Pactive Corp. (GreenGuard brand insulation—green in color) Owens Corning (Foamular brand insulation—pink in color)

20

Assignment 6 Name

Chapter 6

6.1

Using Table 6.1 in the text: (a) Determine the water vapor pressure difference between the inside and outside air across an assembly under the following conditions: outside air temperature = 90 oF, RH = 70% and inside air temperature = 70 oF, RH = 45%.

Air temp.

Outside 90 oF Inside 70 oF

RH 70% 45%

Vap. press.

100.5(0.7) = 70.4 psf 52.5(0.45) = 23.6 psf

Vap. press. difference = 46.8 psf

(b) Diagram the assembly and show the direction of vapor drive through it (see Figure 6.8 of the text for an example).

INSIDE OUTSIDE Vap. press. 23.6 psf Vap. press. 70.4 psf Vapor drive direction Assembly

6.2

Using the straight-line method illustrated in Example 2 of “Principles in Practice” in Chapter 6, determine whether the inside air will condense on a wall assembly that consists of a 1/4 in. thick glass sheet under the following conditions: (a) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 40 oF (b) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 20 oF (c) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 0 oF R-value (inside air film) R-value (glass) R-value (outside air film)

0.7 0.0 0.2

Dew point of inside air (70 oF and 45% RH) = 47.5 oF (Table 1, Principles in Practice, Ch. 6).

21

As the diagram shows: (a) The dew point (47.5 oF), occurs outside of the glass sheet. Therefore, inside air will not condense. (Note: glass is impervious to water vapor).

Inside air film (0.7) Glass sheet

Outside air film (0.2)

70 oF

(b) The dew point (47.5 oF), occurs inside. Therefore, inside air will condense on glass sheet.

50 oF 47.5 oF 30 oF

(a) Dew point temp.

(c)

40 oF

(b)

20 oF 10 oF

(c) Same as (b)

0 oF INSIDE

OUTSIDE

Note: The thicknesses of layers of the assembly in this diagram are proportional to their R-values.

6.3

Using the straight-line method illustrated in Example 2 of “Principles in Practice” in Chapter 6, determine whether the inside air will condense on the interior glass sheet of the insulating glass unit in Question 5.1(e) under the following conditions: (a) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 40 oF (b) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 20 oF (c) Inside air temperature = 70 oF, RH = 45%; outside air temperature = 0 oF R-value (inside air film) R-value (glass) R-value of air space R-value (outside air film)

0.7 0.0 1.0 0.2

Dew point of inside air (70 oF and 45% RH) = 47.5 oF (Table 1, Principles in Practice, Ch. 6).

As the diagram (following page) shows, inside air will not condense in cases (a) and (b) because the dew point occurs within the air space, where the air and vapor cannot reach because an insulating glass unit (IGU), being a fully sealed unit, is impervious to both. In case (c), inside air will (just) condense on the interior glass of the IGU. 22

Name Note: The thicknesses of layers in this diagram are proportional to their R-values

Inside air film (0.7)

Outside air film

Air space (1.0) Glass sheet

(0.2) Glass sheet

70 oF (a)

50 oF 30 oF

40 oF

(b)

Dew point temp.

(c)

20 oF

10 oF 0 oF

INSIDE

6.4

OUTSIDE

What general conclusions can you draw by comparing the results of Questions 6.2 and 6.3?

By increasing the R-value of an assembly, it is possible to prevent the condensation of inside air. Thus, in case of the IGU in Question 6.3, if we increase the R-value of the space between the glass sheets of the IGU (air space in this case), the inside air will not condense even in case (c). (See Chapter 28 of the text for the R-value of an IGU.)

6.5

In which situations, would you: (a) recommend the use of a vapor retarder under a slab-on-grade?

A slab under a conditioned space should be provided with a vapor retarder to prevent the moisture from under the slab from migrating into the interior of the building. (b) recommend not using a vapor retarder under a slab-on-grade?

A slab exposed to unconditioned space (e.g., a slab-on-grade in a parking garage, pavement or road, etc.) does not require a vapor retarder.

23

24

Assignment 7 Name

Chapter 7

7.1

Determine the type of construction of a 2-story building whose structural and envelope components have the following fire ratings (FR). Refer to Table 7.1 in the text. (a) Structural frame: steel columns and beams FR = 15 min Floor: open-web steel joists, steel deck, and concrete topping FR = 25 min Roof: open-web steel joists, steel deck, and rigid board insulation FR = 15 min Exterior non-loadbearing walls: concrete masonry FR = 2 hr All components of the building are constructed of noncombustible materials. Therefore,the building is of Type I or Type II construction. From Table 7.1 of the text, the fire ratings of assemblies of this building do not meet the requirements of Type I(A), Type I(B) or Type II(A), but meet those of Type II(B) construction. Hence, the type of construction is Type II(B).

(b) If the structural frame of the building is replaced by bearing walls with a FR = 2 hr, what is the construction type? Explain.

Replacing the structural frame of the building by 2-hr rated loadbearing walls will not alter the type of construction because the ratings of the floor and roof are less than 1 hr. This assumes that the bearing walls are made of noncombustible materials.

7.2

What is the construction type of the single-story office building described below?

Roof and ceiling: light-gauge steel with gypsum board ceiling Exterior and interior load bearing walls: light-gauge steel and gypsum board

FR = 1 hr FR = 1 hr

All components of the building are constructed of noncombustible materials. Therefore,the building is of Type I or Type II construction. The fire ratings of the assemblies of this building meet the requirements of Type II(A) construction (Table 7.1 of the text).

25

7.3

Determine the type of construction of a 20-story office building with fully-glazed exterior walls and the following fire ratings: Structural frame: reinforced concrete beams and columns FR = 3 hr Floors: reinforced concrete FR = 2 hr Roof: reinforced concrete FR = 2 hr Exterior bearing walls: no exterior bearing walls provided.

All components of the building are constructed of noncombustible materials. Therefore,the building is of Type I or Type II construction. The fire ratings meet the requirements of Type I(A) construction (Table 7.1 of the text).

7.4

As mentioned in Chapter 7 in the text, a first source of information for architects and engineers about an assembly’s fire rating is the International Building Code (IBC). Obtain a copy of the IBC (or the building code of your city) and provide the chapter number and title that includes this information.

Chapter 7: Fire-Resistance-Rated Construction (International Building Code 2009)

7.5

A ceiling tile has a flame spread index (FSI) of 20 and smoke developed index (SDI) of 400. To which class of interior finish does the tile belong?

The ceiling tile belongs to Class A because FSI < 25 and SDI < 450 (Table 7.3 of the text).

26

Assignment 8 Name

Chapter 8

8.1

The sound absorption coefficients of a ceiling tile have been measured as listed below. Determine the tile’s noise reduction coefficient (NRC).

Frequency (Hz) Absorption coefficient 125 0.25 250 0.42 500 0.55 1000 0.75 2000 0.88 4000

0.85

α + α500 +α1,000 + α2,000 NRC = 250 From page 174 of the text, NRC is the average 4 sound absorption coefficient, α, in frequencies 0.42 + 0.55 + 0.75 + 0.88 from 250 Hz to 2 kHz. Hence, = 0.65 or, NRC = 4

8.2

You are the architect of a multistory office block [of Type 1(A) construction)], whose owner requires that the walls separating the corridor from the office spaces must render speech sounds originating in the office spaces inaudible in the corridor. What minimum STC value would you recommend for the walls?

STC 60 (see Table 8.1 of the text).

8.3

Using the information provided in Table 8.1 in the text: (a) Sketch in plan a wood stud wall assembly to provide a minimum STC of 60.

From Table 8.2 of the text, the following wood stud wall assembly is an STC 61 assembly.

2 layers 1/2” th. gyp. board Fiberglass insulation

1” space between studs

1/2” th. gyp. board Wood stud

Double wood stud wall with fiberglass infill STC 61 27

(b) Would you recommend using this wall to separate the offices from the corridor in the building of Question 8.2? Explain.

No, because the building of Question 8.2 is of Type I(A) construction, which requires that the partition walls in the building must be of noncombustible materials, i.e., wood studs must be replaced by metal (galvanized steel) studs. However, the STC of the above wall with metal studs may not be 61. Relevant STC data must be obtained for such a wall.

28

Assignment 9 Name

Chapter 9

9.1

Explain whether the following buildings require building separation joint(s). If the joint(s) is (are) required, indicate its (their) location(s) on the plans provided and give your reason for the location. (a) Office Building and Parking Garage 50 ft

Building separation joint

2-story reinforced concrete frame parking garage

100 ft

150 ft

200 ft

50 ft

6-story reinforced concrete frame office building

Width of separation joint = 2” Reasons for the location of separation joint: The heights of the two buildings are quite different

(b) Furniture Warehouse 600 ft

600 ft

Building separation joint 30-ft high, 1-story furniture warehouse. Structural system consists of steel columns, steel joists, and joist girders.

Reasons for the location of separation joint: The dimension of 600 ft in both directions is too large to be without separation joints. Because it is a warehouse, the joints may slightly exceed the recommended distance of 250 ft.

Width of seperation joint = 2” 29

(c) High School Activities Center 100 ft

30-ft high (two30-ft high (onestory) aerobics story) gymnasium. center, weight Roof structure rooms, and consists of offices. Strucprecast concrete tural system double-tees consists of steel supported on columns and masonry walls beams, steel floor and roof deck, and non-loadbearing exterior masonry walls

Building separation joint

200 ft

150 ft

100 ft

Note: Refer to Chapter 20 for a description of double-tees.

Width of separation joint = 2” Reasons for the location of separation joint: The structural systems of two parts of the building are quite different

(d) Hospital In-Patient Room Block Building separation joint 0 16 ft

60 ft

220 ft

100 ft

Width of separation joint = 2” Reasons for the location of separation joint: The overall length of the building is too large to be without joints. The shape of the building suggests the location of separation joint as shown.

30

Name In a stick-built aluminum-glass curtain wall, the aluminum mullions (vertical sticks) extend from floor to floor over several floors with a spliced expansion joint at each floor level, as shown in the following sketch (see also Figure 30. 3 in the text). (a) If the floor-to-floor height = 13 ft, determine the minimum width of a spliced joint required to accommodate thermal movement between the connecting mullions.

Mullion

From Eq. (1), Ch. 9 of the text, δt = α L (∆t) L = 13 x 12 = 156” ∆t = 80 oF, and From Fig. 9.12 of the text: α = 13 x 10-6 (oF)-1

Top of floor Joint splice

Rail

13 ft

9.2

Glass

Hence, minimum width of joint splice, δt = (13x10-6)(156)80 = 0.16”

Top of floor

(Part) Elevation of an Aluminum-Glass Curtain Wall

(b) Is the total width of the spliced joint between connecting mullions a function of thermal movement only? If not, which other factors must be considered to arrive at the minimum total width of the spliced joint?

Note: As shown in Figure 30.3 in the text, each mullion is anchored to the spandrel beam.

Another important consideration is the elastic deflection of spandrel beams (due to live loads).

This must be obtained from the project’s structural engineer and added to δt. If the building’s structural frame (columns and beams) are of reinforced concrete, creep in beams and columns must also be accounted for. Additionally, consideration must be given to tolerances to account for dimensional variations and uncertainties in our calculations.

31

(c) In this stick-built aluminum-glass curtain wall, the rails (horizontal sticks) are supported by the mullions. If each rail is 5 ft long, determine the width of the joint at each end of the rail to allow thermal movement. Should this joint also provide for any other movement? Explain.

From Eq. (1), Ch. 9 of the text, δt = α L (∆t) L = 5 x 12 = 60”, ∆t = 80 oF, and from Fig. 9.12 of the text: α = 13 x 10-6 (oF)-1 Hence, minimum width of joint splice, δt = (13x10-6)(60)80 = 0.06”, which is approximately 1/16”, requiring 1/32” between mullion and rail on both sides.

No other movement is to be considered.

9.3

A movement joint between two components has been estimated as 0.42 in. If the joint is sealed with a Class 25 sealant, what is the minimum total width of the joint?

From Eq. (4), Ch. 9 of the text: Joint width of 100 = (joint movement) + tolerances sealed joint sealant class Thus, Joint width of 100 = (0.42) + tolerances = 1.68” + tolerances sealed joint 25 The amount to be added for tolerances depends on the component’s material, and the quality of workmanship. If the above cmponent is of concrete or masonry, a joint width of 2.0” should be adequate in most circumstances.

32

Assignment 10 Name

Chapter 11

10.1

Draw the cross-sections of the following lumber members: 2 x 4, 2 x 6, 2 x 8, 2 x 10 and 2 x 12, and provide their actual sizes. Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

Note: Use appropriate graphic symbols and draw all cross-sections in one row, keeping the bottom of the cross-sections aligned. Show dimensions using extension lines and dimension lines.

1-1/2” 1-1/2” 1-1/2”

2x8

2 x 10

2 x 12

Draw the following wood members in cross-section and provide their actual overall dimensions: (a) 4 x 8, (b) two 2 x 8 nailed together, (c) two 2 x 8 nailed together with 1/2 in. thick intervening blocking, and (d) two 2 x 8 nailed together with 1/2 in. thick intervening OSB. Scale: 2” = 1’.

Note: Show dimensions using extension lines and dimension lines.

4x8

3-1/2”

7-1/4”

7-1/4”

3-1/2”

2 - 2 x 8 with 1/2” intervening blocking

2-2x8

33

7-1/4”

3”

3-1/2”

7-1/4”

10.2

7-1/4”

5-1/2”

3-1/2”

2x6

2x4

9-1/4”

1-1/2”

11-1/4”

1-1/2”

2 - 2 x 8 with 1/2” intervening OSB

10.3(a) The following drawing shows the detail of an outdoor wood deck. (a) Complete the detail by using appropriate graphic symbols for the finished grade and each wood member. Also provide the missing dimensions.

Decking (2 x 6 lumber with 1/8 in. space between members)

11-1/4”

Guard rail bottom plate (2 x 4 lumber)

Beam (3-2 x 12 lumber)

2’4-3/4”

Guard rail verticals (2 x 4 lumber)

1-1/2”

3’6”

Handrail (out of 2 x 6 lumber)

Joist hanger

3 in. diameter pipe column 6 ft o.c.

1/4 in. thick U-shaped steel plate welded to column top

Steel base plate Concrete footing Finished grade

Outdoor Deck

34

16 in.

2 x 12 lumber joists 16 in. o.c.

Name (b) Which type of lumber would you specify for: decking? guard rail verticals? joists? beam?

10.4

Pressure-treated softwood lumber for decking. Southern Pine or Douglas fir may be selected for its greater strength, and (hence) greater durability. Heartwood of Cedar or Redwood may be used in lieu of pressure-treated wood (refer also to Question 10.7 of this assignment for alternative decking materials).

For guard rail verticals, joists and beam—Southern Pine or Douglas Fir. Lumber grade must meet structural considerations.

Draw a three-dimensional, freehand sketch of the detail in Question 10.3, showing the beam and the decking in cross-section.

Handrail of guard unit

2 x 4 guard unit verticals @ 5-1/2” o.c.

2 x 4 bottom plate 2 x 6 decking Beam

Joist

Note: Maximum code-allowed clear space between guard unit verticals = 4”

Pipe column Base plate Finished grade

35

10.5

Your partner has recommended the use of quarter-sawn lumber for the decking in Question 10.3. Explain why you agree or disagree.

Agree, because quarter-sawn lumber provides greater abrasion resistance, durability and resistance to warpage. (Cost and availability of quarter-sawn lumber should, however, be checked.)

10.6

Assume that the outdoor deck in Question 10.3 measures 18 ft x 24 ft. Draw its framing plan and determine the total amount of lumber required (in board feet) for the project in beams, joists amd decking (ignore guard unit). Assume that the beams (on both sides of the joists) are 24 ft long and the end joists are doubled. Scale: 1” = 5’; i.e., 1 grid division = 15

18’

in.Note: Show each joist by a single line, doubled joists by two closely spaced lines, and the beams by a heavy Beam 24’ line.

Doubled joist

2x12 joists @ 16” o.c.

Lumber in beams (2 beams) =

2[3(2x12)]24 12

Lumber in joists (21 joists) =

21[(2x12)]18 = 756 BF 12

Pipe column

= 288 BF

Lumber in bottom plate (2x4 members, total length = 2(24 +18) = 84 ft =

84[(2x4)] = 56 BF 12

Lumber in decking (38 members, 2x6, approximately 23.5 ft long =

38[(2x6x23.5)] = 893 BF 12

Total lumber quantity = 288 + 756 +56 + 893 = 1,993 BF = (approx.) 2.0 MBF

36

Name

10.7

The outdoor deck in Question 10.3 consists of solid lumber, which is being increasingly replaced by synthetic materials. Using online resources, describe the advantages and disadvantages of synthetic materials for an outdoor deck compared to lumber. Determine which components might be replaced with synthetic materials. Also provide names of at least two manufacturers of these materials.

Synthetic decks (decking and guardrail caps), made from wood waste and recycled plastics are available. They are preservative-free. However, their long-term durability is less certain because of their brief history. Several manufacturers make synthetic decking that look like real wood.

An online resource for all types of decks is: “www.decks.com”, which also provides information about synthetic deck manufacturers, such as: CorrectDeck, EverGrain and LockDry.

10.8

Your intern has specified the following grades for various sizes of lumber for a wood frame apartment building. Determine if you would be able to obtain the lumber as specified.

Lumber Grade Yes/No 2x8

Construction

No

2 x 10

Standard

No

2x4

No. 1

Yes

2x6

Stud

Yes

2 x 12

Utility

No

2x8

No. 4

No

2x4

No. 2

Yes

37

38

Assignment 11 Name

Chapter 12

11.1

(a) If you were to replace the beam in the outdoor deck (of Question 10.3, whose sketch is reproduced below) with a glulam beam, would you specify a balanced or an unbalanced beam?

Explain.

Balanced beam, because the beam is continuous over several supports. (Remember, an unbalanced glulam beam can only be used on a single span.)

Decking (2 x 6 lumber with 1/8 in. space between members)

Beam (3-2 x 12 lumber)

Joist hanger 2 x 12 lumber joists 16 in. o.c.

3 in. diameter pipe column 6 ft o.c.

Outdoor Deck 39

(b) In addition to 2-by solid lumber and glulam, which other wood-based materials can be used for the beams of this deck.

Laminated veneer lumber (LVL) beams is another alternative.

11.2

The sketch of the deck in Question 11.1 shows prefabricated sheet metal joist hangers. Using a three-dimensional sketch, show how you would detail the connection between the beam and the joists without the joist hangers.

Beam

2 x 4 ledger nailed to beam

2 x 10 joists @12” o.c.

11.3

Give your recommendations for the type of (manufactured) wood product and its exposure rating for the following components of a typical wood frame building: (a) exterior wood siding

Plywood (Exterior exposure rating)

(b) floor sheathing

OSB (Exposure 1)

(c) roof sheathing

OSB (Exposure 1)

(d) wall sheathing

OSB (Exposure 1)

40

Name

11.4

An extremely important part that contributes to the structural integrity of a wood frame building are the nails. The use of power-driven nails has not only increased the efficiency of nail installation, but also to more objective method of specifying them. Increasingly, therefore, for all framing connections, nails are being specified by their length and shank diameter instead of the penny size.

In this question, you are being asked to provide the nail length and shank diameter of Common nails and Box nails for the four commonly used nail sizes (6d, 8d, 10d and 16d) used in the structural framing of wood buildings. (Note: Refer to Table 1 of publication “ESR-1539” of the International Staple, Nail, and Tool Association (ISANTA), entiltled “Power-Driven Staples and Nails for Use in all Types of Building construction”. You can access this publication by going to “www.isanta.org” and looking for ESR-1539.)

Penny size

11.5

Common nail

Box nail

Nail length

Shank dia.

Nail length

Shank dia.

6d

2.0”

0.113”

2”

0.099”

8d

2.5”

0.131

2.5”

0.113”

10d

3.0”

0.148

3.0”

0.128”

16d

3.5”

0.162”

Not listed

Georgia Pacific is one of the major manufacturers of Wood I-joists, which manufactures Ijoists in three series–GPI series of joists, WI series of joists and WO series of joists. By accessing Georgia Pacific web site (www.gp.com) and using sketches, explain the essential differences between the three series of joists.

LVL flanges, widths: 1-3/4”, 2-5/16”, 2-7/16”, and 3-1/2”

Lumber flanges, widths: 2-1/2”, 2-7/16”, and 3-1/2”

9-1/2”, 11-7/8”, 14” and 16”

GPI Series

Lumber flanges, width: 2-1/2”

9-1/2”, 11-7/8”, 14” and 16”

WI Series 41

16”

WO Series

11.6

Your partner has specified GPI 20 joists, 14” deep, spaced 16” on center for a floor above two bearing walls that are 20 ft apart. The floor is a residential floor (live load = 40 psf and dead load = 10 psf). Using Georgia Pacific literature, determine if your partner’s specification of joists is structurally adequate. Explain your answer.

As per the load tables provided by Georgia Pacific, GPI 20 joists, 14” deep, spaced 16” o.c. will span a maximum of 21’-3” for the above load and span conditions. Because the span of joists is 20 ft < 21-3”, the selection is OK.

11.7

What is the maximum length in which Gergia Pacific I-joists are available?

Georgia Pacific joists are available in standard lengths of 24 ft to 48 ft in increments of 4 ft. thus, the maximum standard length is 48 ft.

11.8

You are the designer-builder for the house of a client who wants to avoid the use of all materials that emit VOCs. The plan of the house has a few large floor and roof spans for which solid lumber is inadequate. Which wood materials would you specify for: (a) floor framing?

Trussed floor joists (with nail plates)

(b) floor sheathing?

1-1/2x6 tongue-and-groove (T&G) lumber board subfloor laid diagonally over joists.

(c) roof framing?

Roof trusses (with nail plates)

(d) roof sheathing?

1x6 T&G lumber board roof sheathing laid diagonally over rafters.

(e) wall sheathing?

1x6 lumber plank wall sheathing laid diagonally over studs.

42

Assignment 12 Name

Chapter 13

12.1

The following sketch shows the ground floor plan of a two-story wood light frame forest lodge with living room and kitchen on the ground floor and one bed room and bath at the second floor. It is constructed entirely of sawn lumber. Using the plan as an underlay, draw the floor framing plan (layout of the floor joists and supporting beams, if any) for the second floor of the building (refer also to the second floor plan shown later in this assignment).

Use a single line (preferably in blue) to show the joists and a heavy line for a beam. Every joist need not be shown as long as the overall floor framing is clear.

30’-0”

14’-0”

2x8 Southern Pine # 2 joists 16” o.c.

21’-6”

Living Room

Beam

Beam

Kitchen Doubled 2x8 Southern Pine # 2 as header to support stair stringers

2x8 Southern Pine # 2 joists 16” o.c.

12’-0”

GROUND FLOOR PLAN 43

18’-0”

Up

12.2

Based on the framing plan, determine the preliminary size and spacing of the joists. Provide this information below as well as on on the framing plan.

Note: Refer to Appendix B in the text. For more accurate size and spacing information, refer to the tables provided in the International Building Code or the International Residential Code.

12.3

12.4

Minimum 2 x 8 Southern Pine #2 @ 16” o.c.

From Table R502.3.1(1) of the International Residential Code, 2 x 8 Southern Pine # 2 @ 16” o.c. will span a maximum of 14’-2”. Because the joist depth must remain constant for the entire floor, the same joist size is used for the entire floor. Note that a larger depth of joist (2 x 10 SP #2 @16” o.c.) may be used to provide a stiffer floor.

Suggest the preliminary size and spacing of the studs, and lumber grade for the walls of the building of Question 12.1. (Note: Refer to Appendix B in the text or the building code tables.)

Size and spacing

Minimum requirement: 2 x 4 @ 16” o.c. (Table R602.3(5), International Residential Code).

Grade of lumber

Stud grade or Construction grade

Using the second floor plan, shown on the following page as an underlay, draw the roof framing plan of the building with hip roofs at the ends (as shown in the isometric sketch of the roof). Using appropriate tables from the text (or the building code), provide the preliminary size, spacing and the lumber grade for rafters.

Maximum span of rafters = 21.5/2 = 10.75 ft = 10’-9” From Table R802.5.1(1), International Residential Code,2 x 6 (Southern Pine #2 or Sprucepine-fir #2 or Douglas-fir #2) @ 16” o.c. will be adequate. [Dead load = 10psf, live load = 20 psf, ceiling not attached to rafters.]

12.5

Draw in plan the detail of the arrangement of studs you would recommend at P and Q (see second floor plan). Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

2 x 4 stud 2 x 4 stud Fiberglass insul. in stud cavities

DETAIL P

DETAIL Q 44

Name 2 ft wide roof projection

Hip rafter

P Living Room Bedroom Ridge board

Q Down

Ridge board Bath

Hip rafter

Valley rafter

Wardrobe

SECOND FLOOR PLAN Hip rafter

Hip rafter Common rafters and jack rafters 2x 6 Ridge board, hip rafters, valley rafters 2 x 8

Hip

Hip Valley Hip

45

12.6

As stated in Section 12.9 of the text, most connections between the framing and sheathing members of a wood light frame building are made using nails. For a conventional wood light frame building, the number (or spacing) of nails, size of nails and the type of nailed connection (face-nailed, toe-nailed or end-nailed) required for a particular connection have been standardized and are provided in the building codes. Using the appropriate table of the International Building Code or the International Residential Code, sketch the following connections and give the size, number (or spacing) of nails and type of connection.

(a) Between studs (double studs)

(b) Between top plates (double top plates)

10d common nails @ 24” o.c.

Face nailed

10d common nails @ 24” o.c.

Face nailed

(c) Between rim joist and top plate

8d common nails @ 6” o.c.

Toe nailed

(d) Between stud and sole plate

3-8d common nails Toe nailed OR 2-16d common nails End nailed

(e) Between stud and top plate

2-16d common nails End nailed

(f) Between floor joist and top plate or beam

3-8d common nails Toe nailed

46

Assignment 13 Name

Chapters 13 and 14

13.1

The following drawing shows a partial floor plan of a 2-story student dormitory building constructed using wood light frame construction. (The upper floor plan is the same as the lower plan.) (a) Suggest the floor framing plan (layout of the floor joists) for the second floor of the building. Treat the floor plan as an underlay and use sawn lumber only. Use a single line (preferably in blue or red) to show the joists. Every joist need not be shown as long as the overall floor framing is clear.

60 ft 6 ft

20 ft

8 ft

6 ft

20 ft 2x6 floor joists 16” o.c

Bath Room

Room

15 ft

2x10 floor joists 16” o.c.

Bath

Room Bath

Bath Room

15 ft

Corridor

Span of joists in corridor = 8 ft.

Use double 2x6 SP#2, 16” o.c. (see note on the following page.

Room

15 ft

Span of joists in rooms = 15 ft Use 2x10 Southern Pine (SP) #2 @ 16” o.c.

(As per the International Building Code, 2x10 SP#2, 16” o.c. spans up to 16’-1”.)

Double 2x10 SP #2 end joist

Bath Room Bath

Second Floor Plan

47

13.2

Based on the framing plan, determine the preliminary size and spacing of the joists. Provide this information on the framing plan drawn on the previous page. Use sketches and notes to illustrate any framing aspects that you are not able show in the framing plan.

(Note: Refer to Appendix B in the text for size and spacing information. For greater accuracy, refer to the tables provided in the International Building Code.)

2x10 floor joists in room

Double 2x10 end joist

Joist hanger Corridor wall

Double 2x6 floor joists, 16” o.c. in corridor

While the dormitary rooms receive a live load of 40 psf, the corridor must be designed for 100 psf. The use of double 2 x 6 joists in corridor serves two purposes: (a) strengthen corridor floor to be adequate for 100 psf live load, and (b) provide an increased clear height in the corridor. Thus, if the ceiling height in rooms is 9 ft, and that in the corridor is 8 ft, a clear space of approximately 16 in. is available between corridor ceiling and the floor above for utility and sprinkler lines.

To provide access to utility lines, the corridor ceiling should consist of removable ceiling tiles supported on framing (perimeter angles and inverted tees) attached to corridor walls.

Span of joists in rooms = 15 ft Use 2x10 Southern Pine (SP) #2 @ 16” o.c.

(As per the International Building Code, 2x10 SP#2, 16” o.c. spans up to 16’-1”.)

Span of joists in corridor = 8 ft.

Use double 2x6 SP#2, 16” o.c. (based on structural calculations).

48

Name

13.3

The following drawing is a section through the building in Question 13.1. Based on your answer to Question 13.1 and assuming that the roof framing consists of prefabricated wood trusses bearing on the exterior walls, determine which walls of the building are loadbearing walls. Highlight these walls in the following (first floor) plan.

5

Prefabricated wood truss

12

R 9’-0”

2’-6”

T

5’-0”

1’-10-1/2”

9’-0”

S

Concrete slabon-ground

10’-0” 4’-0” 4’-0”

Section A-A

2 x 6 studs @ 16” o.c.

2 x 6 studs @ 16” o.c.

2 x 6 studs @ 16” o.c.

A

A

First Floor Plan

Bearing walls are highlighted in blue color. Size and spacing of studs are provided in response to Question 13.4.

49

13.4

The following questions of this assignment are about the building in Question 13.1. (a) Suggest the preliminary size and spacing of the studs for the bearing walls. (Refer to Appendix B in the text.) Also suggest lumber grade for the studs.

Size and spacing

Exterior walls supporting roof trusses: 2 x 6 @ 16” o.c.

Interior walls supporting floor joists: 2 x 6 @ 16” o.c.

Grade of lumber

Preferably “Stud” grade. “Construction” grade may be used

The use of 2 x 6 studs for interior walls is based on structural as well as acoustical considerations to accommodate 5-1/2 in. thick fiberglass between studs. Additional measures such as the use of a resilient channel on the walls and ceilings may also be needed (see Chapter 8 of the text). (b) Draw Detail R in Section A-A. For the truss, assume that the bottom and top chords are made of 2 x 6 lumber, the exterior wall cladding is beveled wood siding, the interior finish is 5/ 8 in. 1-1/2” gypsum and =the Scale: = 1’, i.e.,board 1 grid division 2 in. roof cover is asphalt shingles (see Chapter 32).

Asphalt roof shingles over roof underlayment (see Chapter 32)

Truss top chord

Nail plate 2x4 blocking Truss bottom chord Metal drip edge (see Chapter 32) 5/8” thick Type X gypsum board

Finished wood fascia 2x6 continuous rough fascia 2x6 soffit joist 2x6 ledger

Trim

Soffit vent

Detail R

Note: Insulation and vapor retarder in ceiling and wall, and air-weather barrier in wall are not shown for clarity

Approx. scale: 1” = 1’

50

Name (c) Draw Details S and T in Section A-A. Assume 1-1/8 in. thick tongue-and-groove OSB subfloor at the second floor, a 6 in. thick reinforced concrete slab with 18 in. x 30 in. perimeter beams. Scale: 1-1/2” = 1’.

5/8” Type X gypsum board over vapor retarder T&G OSB subfloor

Wood siding, see Chapter 14

2 x 10 rim joist 1/2” OSB sheathing

2x4 blocking as nailer

Air-weather retarder Wood siding, (see Chapter 14)

Pressure-treated sill plate

5/8” Type X gypsum board over vapor retarder Finished ground

Detail S

10 mil thick vapor retarder 4” sand subbase, see Chapter 19 Concete slab-on-ground foundation with perimeter and interior (ribs) beams (see Chapter 19)

Detail T (d) If the floor-to-ceiling height of the building is 9 ft, what length of studs would you use?

8 ft-85/8 in. long precut studs (Chapter 13 of the text) (e) Using the information from the previous questions and assuming that the rough opening size of each window is 4 ft x 5 ft, draw on the following page a partial elevation of the 2story exterior wall that illustrates the framing around a window. Assume that the lintel header is made of 2 x 12 lumber and is supported on one jack stud at each end. Label each member of the frame. Scale: 1/2” = 1’; i.e., each grid division = 6 in.

51

Double top plate

16”

16”

T&G OSB subfloor

16”

16”

16”

16”

Double top plate

Bottom plate

16”

16”

Rim joist

Cripple stud

Lintel header Jack stud King stud

Rough opening height = 5’-0”

8ft-85/8 in. long precut studs

Rough opening width = 4’-0”

Rough sill Cripple stud

52

Assignment 14 Name

Chapters 13 and 14 cont.

The adjacent sketch shows the ground floor plan of a two-story weekend house constructed using wood light framing. The foundation system consists of 8 in. thick concrete masonry walls with a crawl space.

Draw the ground floor framing plan of the house using the foundation plan (shown below) as an underlay. Provide the size and

Balcony above

Kitchen-dining

Living Wall above

Laundry

Bath

spacing of the floor joists.

Up

Note: Use a single line to show the joists, rim joists,

Ground Floor Plan

and blocking; two closely spaced lines for doubled joists, and a heavy line for beams. Refer to Chapter

2x12 floor joist 16” o.c.

21 for masonry wall foundation system.

Pier

20’-0”

14.1

Full-depth blocking

Pier

8” thick CMU foundation wall

WLF wall above

Double 2x12 joists supported on 8”x 8” reinforced masonry piers capped with termite shield. Provide joist hangers on both sides of double joist to support floor joists.

Double joist to support wall above

Ground Floor Framing Plan

Full-depth blocking to support wall above Foundation Plan 15’-0”

20’-0”

53

Using the first-floor plan of the weekend house (shown below) as an underlay, draw the second floor framing plan and provide the size and spacing of the floor joists.

2x10 floor joist continuous from Beam A to outer face of balcony

In-line blocking

2x10 floor joists 24” o.c.

2x12 joist 24” o.c., nailed to 2 x 10 joist 2x12 floor joist 24” o.c.

2x12 floor joist 24” o.c.

10’-0”

Double end joist

Beam A Joist hanger

Balcony above

ny lco ang a B erh ov Exterior wall

10’-0”

Balcony above

Wall above

2x10 floor joist 24” o.c. (see 3-d sketch)

20’-0”

14.2

15’-0” Glulam or LVL Beam A (11-1/4”) deep

Glulam or LVL Beam B (11-1/4”) deep Glulam or LVL Beam C (11-1/4”) deep 2x12 floor joist 24” o.c.

Second Floor Framing Plan 15’-0”

20’-0”

First Floor Plan

54

Name

14.3

The adjacent sketch shows a three-dimensional view of the roof of the weekend house. Using the second floor plan (shown below) as an underlay, draw the roof framing plan and give the size and spacing of roof members.

Fascia rafter

Study

Bedroom Fascia rafter

A

A

Glulam or LVL ridge beam Bath

2x12 rafters 16” o.c.

Second Floor Plan

Note: Structurally, 2 x 12 rafters are oversized (for a span of 20 ft), but this size allows 9-1/2” (R-30) fiberglass insulation between rafters with about 2” space under roof sheathing for ventilation in this cathedral ceiling, see Question 14.4.

Roof Framing Plan

55

14.4

Draw section A-A (scale: 1” = 5 ft) and then draw details at the eave and the foundation (scale: 1-1/2” = 1’). Assume 2 x 6 studs for exterior walls, OSB sheathing, fiber-cement siding, tongue-and-groove OSB subfloor, and fiberglass insulation as needed.

Hold back sheating by 1-1/2” on each side of ridge beam to provide vent, covered by continuous ridge vent

Glulam or LVL ridge beam 2 x 12 rafters

Cont. soffit vent

Blocking 2 x 12 floor joists 2 x 12 floor joists Double 2 x 12 beam

Glulam or LVL beam

Fully grouted 8” x 8” reinforced concrete masonry pier

Section A-A

56

Name

Fiber cement siding Air-weather retarder 1/2” OSB sheathing Treated wood starter strip Treated wood sill Termite shield Fully grouted 8” reinforced masonry foundation wall 9-1/” (R-30) fiberglass insulation Portland cement plaster

2 x 12 rafters Asphalt shingles No. 30 organic felt underlayment

Detail at Foundation Level

Ice dam protection membrane up to 24” beyond exterior wall 1/2” OSB sheathing Galvanized steel edge flashing

Fiber cement siding

Vapor retarder behind gypsum board

Rough fascia Finished fascia 2 x 4 soffit joist Screened vent

Detail at Eave

14.5

Examine the plan of the staircase shown in Question 14.2. Assuming that the riser height is 7-1/2 in., determine the floor-to-floor height of the first story. Show calculations.

Note: Refer to the first few pages in Chapter 33 of the text.

Total number of risers = 16. Total floor-to-floor height = 16(7.5) = 120” = 10’-0”

57

58

Assignment 15 Name

Chapter 16

15.1

As stated in the text, wide-flange (W) sections, channel (C) sections, angles (L), and hollow structural sections (HSS) are among the more commonly used cross-sectional profiles in steel frame buildings. Each of these profiles is available in various standard dimensions.

Structural engineers and architects routinely access their dimensional properties, which are tabulated (along with the structural properties) in the Steel Construction Manual (SCM) published by the American Institute of Steel Conctruction (AISC).

(If your curriculum requires you to take a course in steel design, you will become familiar with the use of the SCM. If not, it is still worth your while to browse through the SCM and look at the kind of information it provides.)

Using the information given in Section 16.2 in the text or the SCM, neatly draw the following sections (showing rounded corners and sloped edges, if any). Provide all important dimensions using leader and dimension lines. (a) W 14x132 and W 14x43. Because W-sections have two (horizontal and vertical) axes of symmetry, draw these sections with their horizontal axes of symmetry aligned. Scale: 3” = 1’, i.e., 1 grid division = 1 in.

8”

5/8”

1/2”

5/16”

12-5/8”

14-5/8”

1”

W 14x132

W 14x43 59

13-5/8”

14-3/4”

• What is the weight of each W-section? W 14 x 132 (132 #/ft); W 14 x 43 (43 #/ft) •

Which dimensions are equal between them? 12-5/8” (inside of flanges)

(b) C 10x30 (flange width = 3.0 in., flange and web thickness are 7/16 in. and 11/16, respectively); C 10x25 (flange width = 2.90 in., flange and web thickness are 7/16 in. and 1/2, respectively); and C 10x20 (flange width = 2.75 in., flange and web thickness are 7/16 in. and 3/8, respectively).

Scale: 3” = 1’.

3”

2-7/8”

2-3/4”

11/16”

1/2”

3/8”

C 10x30

L

10”

10”

L

C

C

L

C

C 10x25

C 10x20

• What is the weight of each C-section?

C 10x30 (30 #/ft); C 10x25 (25 #/ft); C 10X20 (20 #/ft)

• Which dimensions are equal between them?

60

Depth of section = 10”

Name (c) L 6x6x1/2 and L 6x4x1/2. Scale: 3” = 1’.

1/2”

6”

6”

1/2”

1/2”

1/2” 4”

6”

L 6x4x1/2

L 6x6x1/2

• What is the weight of each L-section?

L 6x6x1/2 From Steel Manual, cross-sectional area of angle = 5.77 sq. in. Therefore, voume of 1 ft long angle = 5.77/144 = 0.04 c.ft. Weight = 490(0.04) = 19.6 #/ft. (Density of steel = 490 #/c.ft, see p. 74 of text.)

L 6x4x1/2 Area = 4.75 sq. in. Voume of 1 ft long angle = 4.75/144 = 0.033 c.ft. Weight = 490(0.033) = 16.2 #/ft.

• Which dimensions are equal between them?

Vertical leg (= 6”) and thickness of legs (= 1/2”)

61

15.2

Using freehand sketches, illustrate three methods currently used to produce hollow structural sections (HSS).

Note: You can access the information using online resources by typing “Dimensions and Section Proerties of HSS,” which will lead you to a free downloadable PDF document. You can also access this document through www.steeltubeinstitute.org and following the link to hollow structural sections.

A hollow structural section (HSS) is manufactured from a steel plate, heating it to make it more plastic, progressively bending it to shape (round, rectangular or square) through a set of rollers and then welding the opposite ends of the plate by electric resistance welding (without the addition of any weld material). This is one of the three methods of manufacturing. For other two methods, refer to the references provided.

15.3

Using the document referenced in Question 13.2: (a) Sketch three different shapes of HSS produced by the manufacturers.

Square HSS

Rectangular HSS

62

Round HSS

Name

(b) List the largest and the smallest sizes of square HSS. Largest: HSS 32 x 32 x 5/8 (i.e., 32” x 32” x 5/8”) Smallest: HSS 1-1/4x1-1/4x1/8 (i.e., 1-1/4” x 1-4” x 1/8”)

(c) List the thicknesses in which 9x5 HSS are available.

HSS 9x5x5/8 HSS 9x5x1/2 HSS 9x5x3/8 HSS 9x5x5/16 HSS 9x5x1/4 HSS 9x5x3/16

63

64

Assignment 16 Name

Chapter 16 cont.

The Steel Joist Institute (SJI) publication “Standard Specifications, Load Tables and Weight Tables for Steel Joists and Joist Girders” is available as a free download. Go to www.steeljoist.org/sjipdf/ and open the link “Entire Catalog” to answer the following questions. (a) Sketch in section a 12K3 steel open-web joist. Assume top and bottom chord members are made of 1-1/4 in. x 1-1/4 in. angles. Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

12”

16.1

Note: “12” specifies the depth (in inches) of the joist and “3” , called section number, gives its relative weight and slenderness. A 12K3 joist is lighter and more slender (with a smaller load-carrying capacity) than a 12K4 joist; both have the same depth = 12”

Joist 12K3

(b) If this joist spans 20 ft, what is its total weight? Show calculations.

From Steel Joist Institute (SJI) Publication: “Standard Specifications—Load Tables and Weight Tables, 42nd Edition”, page 28, weight of 12K3 = 5.7 #/ft. Therefore, weight of 20 ft long joist = 5.7(20) = 114 #.

(c) List various K-series joists that are 18 in. deep.

18K3, 18K4, 18K5, 18K6, 18K7, 18K9, 18K10

65

(d) Which of the above 18 in. deep joists can carry the highest load?

18K10 (e) Which of the above 18 in. deep joists can span the longest distance?

18K10 (f) K-series joists are generally used as top-chord bearing. Illustrate the commonly used support condition of a top-chord bearing K-series joist on: • a wide-flange beam

Open-web steel joist Wide-flange beam support

• a masonry wall Wall continues above. Void filled with grout after joist is welded to bearing plate (see Fig. 24.13(b) of text)) Open-web steel joist Steel bearing plate anchored to bond beam (see Chapter 24) Bond beam Concrete masonry wall

66

Name (g) Describe a situation where you would recommend bottom-chord bearing of K-series joists? Illustrate using sketches.

Bottom chord bearing is generally used where the joist cantilevers from a supporting beam (see page 7 of SJI Publication referred to in Question 14.1).

Wide-flange beam support

(h) What is bridging of steel joists and how is it provided?

Bridging provides lateral stability to joists by connecting the joists together using (top and bottom) horizontal rows of angles (Figure 16.13 of Text and page 8, SJI Publication). In each row, one angle is welded to top chord and the other angle is welded to the bottom chord. The number of rows is a function of the joist span and joist’s section number (SJI Publication, pages 18 and 19).

Horizontal rows of angles may be replaced by diagonal (cross) bridging angles connecting the adjacent joists. Even with horizontal rows, one central diagonal bridging set is generally required at the center of the joists.

67

The adjacent sketch shows the plan of a typical floor of a 6-story steel frame office building.

Using the plan as an underlay, draw the floor framing plan, and provide the preliminary sizes of W-shape (primary and secondary) beams and columns.

Note: Refer to Chapter 16 and Appendix B in the text.

60 ft 30 ft

25 ft

30 ft

W 16x ... beam

25 ft

W 16x ... beam W 18x ... girder

25 ft

100 ft

Elev

Rest Room

P 25 ft

16.2

Run the beams in the long direction to try to equalize the depth of beams and the supporting girders. Using 2 intermediate beams per 25 ft span, as shown. Beam span = 30 ft. Approximate beam depth = (30 x 12)/22 = 16.4 in., say 16 in. (see page 959 of the text).

Girder span = 25 ft. Approximate depth of W-shape beam = (25 x 12)/16 = 18.75 in., say 18 in. Note: the depth of next W-shape available (greater than 18”) is 21 in. (see Steel Manual) which is excessive for a 25 ft span.

68

Name

16.3

Draw a detail section through a spandrel beam in Question 16.2 showing the deck, pour stop, and the floor concrete. Assume 1-hour fire-rated floor. Indicate the location of the detail on the framing plan in Question 16.2. Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

3”

Pour stop

Composite floor deck with 3” concrete topping W 18 x ... girder W 16 x ... beam

Angle shop welded to beam and field bolted to girder

Detail at P

69

Repeat Question 16.2 using K-series joists instead of W-shape secondary beams.

30 ft

25 ft

30 ft

Joist

25 ft

16.4

Joist girder

Joist girder

Run the joists in the long direction to somewhat equalize the depth of joists and the supporting beams. Typical spacing of floor joists is 2 to 4 ft (see page 959 of the text). Use a spacing of 3’1-1/2” to provide 8 spaces in 25 ft span, as shown. Joist span = 30 ft. Approximate joist depth = (30 x 12)/20 = 18 in.

70

Assignment 17 Name

Chapter 18

17.1

Due to environmental concerns, manufacturing plants are generally located away from human settlements. Is there any environmental concern specific to lime manufacturing that further necessitates the distance? Does the same concern apply to portland cement manufacturing? Explain.

Lime is manufactured by heating limestone (calcium carbonate). This process produces carbon dioxide(Section 18.1 of text), considered a major environmental pollutant. Because approximately two-thirds of the raw material used in manucturing portland cement is limestone, the manufacturing of portland cement also produces carbon dioxide.

17.2

Portland cement is manufactured in various types. Which type would you recommend for use in: (a) a concrete foundation?

Type I/II portland cement

(b) a concrete foundation with sulfate exposure?

Type V portland cement

(c) an elevated concrete slab?

Type I/II portland cement

(d) a precast concrete beam?

Type III portland cement

(e) a terrazzo floor in the entrance lobby of an office? Type I/II white portland cement (f) a tilt-up concrete wall?

Type I/II portland cement or Type III

(g) an approximately 6 ft thick concrete wall?

Type IV portland cement

(h) cold-weather conditions?

Type III portland cement

71

17.3

While increasing the ultimate compressive strength of concrete, fly ash slows the rate of strength gain in concrete. Is there any particular portland cement type with which fly ash should not be used? Explain.

Type III is high early-strength portland cement. The use of fly ash with Type III portland cement reduces the latter’s ability to gain early strength. Therefore fly ash is not used with Type III portland cement.

17.4

Concrete structures are designed based on concrete’s 28-day strength. Therefore, the cylinders are tested 28 days after casting. Is there any reason to test the cylinders at earlier intervals of time? Explain.

On a fast-track construction project, where formwork must be removed early (in some cases, in less than 24 hours after concrete’s placement), a general contractor (GC) must obtain test results of concrete’s early strength to ensure that it is safe to remove the formwork. These tests are in addition to the normal 28-day strength tests.

17.5

A concrete slab-on-ground measures 40 ft x 70 ft with an average thickness of 9 in. How much concrete would you order for the slab? Show calculations.

Total quantity of concrete = 40 x 70 x (9/12) ft3 = 2,100 ft3.

Because 27 ft3 = 1 yd3, the above quantity = (2,100)/27 = 77.8 yd3 Therefore, order 80 yd3 to allow for some wastage.

17.6

What are the differences between (a) insulating concrete, (b) foamed concrete, and (c) structural lightweight concrete? Explain the differences and provide examples of situations in which you would use each type.

Insulating Concrete and Foamed Concrete: Both insulating concrete and foamed concrete are nonstructural concretes, and are primarily used as low-slope (flat) roof deck insulation. With approximately 12 in. slump, they both are free-flowing, and can easily provide slope to roof drains (see Chapter 31). Their approximate compressive strength is between 150 to 500 psi, and a density of 30 to 50 pounds per cubic foot 72

Name (pcf). Their main ingredients are portland cement and water. The difference between them, however, is that the insulating concrete contains expanded perlite or vermiculite as aggregate. Foamed concrete does not have any aggregate, but contains a foaming agent which, when foamed, creates air bubbles within cement-water slurry. The bubbles do not collapse but become part of the hardened concrete, which increases the R-value of concrete.

Structural Lightweight Concrete: Structural lightweight concrete is a structural material. Apart from portland cement, sand and water, it contains expanded shale as coarse aggregate. The resulting concrete weighs between 85 to 115 pcf, as compared with approximately 140 pcf for normal weight concrete. Structural lightweight concrete can be produced in fairly high strength (up to 6,000 psi), and is commonly used in the floors and roofs of multistory concrete structures. Being lighter in weight, it reduces the dead load of the structure, reducing structural member sizes. The higher cost of the coarse aggregate is more than recovered by the smaller member sizes. When used in precast concete structural members, it also reduces the cost of transporting them to the site.

17.7

A concrete mix has the following ingredients per cubic yard: Portland cement (PC) 600 lb Water 350 lb Admixture None If the mix described above is changed as shown below, indicate on each line the effect of the change of ingredients on concrete’s workability and strength. Insert the appropriate term, increase,, decrease decrease,, or (almost) unchanged in the space provided. increase PC 600 600 600 640 600 600 600 600

Water 380 350 325 350 350 250 350 350

Admixture None WRA None None HRWR HRWR AEA Set retarder

Workability Increase Increase Decrease Decrease Increase Increase Increase Unchanged

Strength Decrease Unchanged Increase Increase Unchanged Increase Decrease Unchanged

Note: “WRA” implies water-reducing admixture, “HRWR” implies high-range water reducer, and “AEA” implies air-entrainining admixture.

73

17.8

The consolidation and finishing of concrete results in a certain amount of bleeding. Excessive bleeding is discouraged because it adversely affects concrete’s performance. Is there any advantage to an acceptable amount of bleeding? Explain.

Bleeding is the rise of water to the top surface of concrete as the concrete is consolidated and finished. Excessive bleeding implies over consolidation resulting in the separation of concrete ingredients, resulting in a weaker concrete. Certain amount of bleeding, inherent in concrete placement and finishing, is helpful, however, because the surface water evaporates faster, reducing the water-cement ratio in the mix, that increases the strength of concrete.

17.9

Steel is sold by its weight. If a certain number of No. 4 reinforcing bars cost $1,000.00, what is the cost of the same number of No. 8 bars?

Note: You will need to refer to Table 18.3 in the text.

From Table 18.3, cross-sectional areas of one No. 4 bar and one No. 8 bar are 0.20 in.2, and 0.79 in.2 respectively. Because the weight of bars is proportional their cross-sectional areas, one No. 8 bar is (0.79/0.20), i.e., 4-times heavier than one No. 4 bar. Thus, the same number of No. 8 bars will weigh 4-times more than #4 bars, costing $4,000 instead of $1,000.

17.10

(a) Sketch in three dimensions a welded wire reinforcement mat with the following specifications, 4 x 8-D 6.0 x D 8.0, giving all important information on the sketch.

Transverse deformed wire (crosssectional area = 0.08 in.2) @ 8” o.c.

Longitudinal deformed wire (crosssectional area = 0.06 in.2) @ 4” o.c.

(b) What is the difference between the mat described above and that designated as 4 x 8W 6.0 x W 8.0?

The only difference is that the wires in this case are plain (undeformed wires).

74

Assignment 18 Name

Chapter 19

18.1

An exterior concrete slab-on-ground is 6 in. thick and is reinforced with No. 4 bars 12 in. on center in both directions. Draw a section through the slab giving all important dimensions and label all parts. Scale: 1 in. = 1 ft.

Note: You will need to refer to Table 19.1 in the text.

No. 4 rebar @ 12” o.c.

No. 4 rebar@ 12” o.c.

6” thick concrete slab 2” cover Compacted earth as subgrade

Section Through an Exterior Slab-on-Ground

18.2

The following sketch shows the section through a doubly reinforced concrete beam. Assume that the tension steel consists of 8 No. 8 bars in two layers. Calculate the clear space between bars in the top (compression steel) and also in the bottom (tension steel), given that beam width = 18 in., No. 3 stirrups, and the concrete cover is the minimum required for an interior beam.

18”

Clear space between (top) compression steel bars (Note: “No. 8” is also written as “# 8”) Diameter of # 8 bar = 1.0” , see Table 18.3.

Clear cover (as measured from the outside of stirrups) = 1.5”, see Table 19.1—interior beam (i.e., not exposed to weather).

Compression steel Cover Cover (refer to the text)

Closed-loop # 3 stirrup Tension steel (#8 bars) in two layers

Stirrup diameter = 0.375”, see Table 18.3 Clear space = 18 - 2(cover) - 2(stirrup dia.) - 3(bars) 2

Doubly-reinforced beam

= 18 - 3 - 0.75 - 3 = 5.63” 2 Note: “2” in the denominator indicates two spaces between bars

75

Clear space between five (lower layer) bottom, tension steel bars Clear space = 18 - 2(cover) - 2(stirrup dia.) - 5(bars) 4 18 - 3 - 0.75 - 5 = 2.31” 5 Clear space between three bottom (upper layer), tension steel bars = 5.63” (same as for compression bars)

18.3

(a) A rectangular reinforced concrete grade beam supports a wood floor with a crawl space below. The beam measures 15 in. x 24 in. and is reinforced with 4 No. 6 bars at the bottom, and No. 3 closed-loop stirrups 12 in. on center throughout the length of the beam. Draw a cross-section through the beam giving all important dimensions and label all parts. Scale: 1 in. = 1 ft.

Note: See Figures 21.36 and 21.37 in the text to visulize the construction described in this question.

15” # 3 rebar as hanger bar, see Figure 19.14 24” Reinforced concrete grade beam

1-1/2” cover

# 3 closed loop stirrup @ 12” o.c.

Note: Steel rebar dowels that connect the beam with piers have not been shown for clarity

4 - # 6 bars 10-mil thick polyethylene sheet moisture-vapor retarder over crawl space ground

Reinforced concrete drilled pier

(b) From the sketch above, determine the clear distance between the bottom bars in the beam.

Clear space 15 - 2(cover) - 2(stirrup diameter) - 4(bar dia.) between bars = 3 = 15 - 2(1.5) - 2(0.375) - 4(0.75) = 2.75” 3 76

Name (c) Determine the area of steel as a percentage of the total cross-sectional area of beam.

From Table 18.3, crossectional area of one # 4 bar = 0.44 in.2. Area of 4 bars = 4(0.44) = 1.76 in.2.

Ratio of steel area and total beam cross-sectional area = (1.76)/(15 x 24) = 0.005 = 0.5% (Note: In determining steel area as a percentage of total beam area, we include only longitudinal steel, ignoring hanger bars and stirrups)

18.4

Draw in three dimensions the two alternatives of closed-loop stirrups shown in the text.

18.5

Explain the difference between the functions of stirrups in concrete beams and ties in concrete columns.

Concret is a brittle material. It is weak in tension, and hence, weak in shear (refer to Section 4.7 of the text). Concrete, therefore, has a limited capacity to withstand shear. The additional shear capacity in a reinforced concrete beam is provided by stirrups.

Because a column is primarily in compression, the primary purpose of ties in a reinforced concrete column is to prevent the buckling of (slender) rebars due to compression. However, ties also increase the shear resistance of columns subjected to bending by gravity and (or) lateral loads.

77

18.6

An 18 in. x 18 in. interior concrete column is reinforced with 8 No. 9 bars. Assuming No. 3 ties, draw a cross-section that shows the arrangement of ties in the column.

Scale: 1 in. = 1 ft.

Note: Refer to Figure 19.27 in the text.

18” Clear space between longitudinal bars 18 - 2(1.5) - 2(0.375) - 3(1.128) 2 = 5.43” < 6”. Therefore, the central longitudinal bar of the column does not need lateral support from a tie corner, see Figure 19.27 of the text. (Note: 1.128 in the above calculation is the diameter of a #9 bar, see Table 18.3.) =

# 3 ties 18”

# 9 longitudinal bar

The center-to-center spacing between ties (along the height of a column) is provided by a simple rule-of-thumb provision in the ACI (American Concrete Institute) Code.

18.7

Explain the difference between shoring and reshoring.

Shoring refers to the vertical support of formwork in a cast-in-place concrete structure. Both formwork and shores are temporary elements that provide structural support to wet concrete during construction. In a single story structure, shores and formwork are removed after the horizontal concrete elements (beams and slabs) have gained sufficient strength so that they can sustain their self load and construction loads. If we follow this procedure for removing formwork and shores in a multistory building, excessive shoring material is required. Therefore, after the concrete above a given floor has gained sufficient strength, the formwork and shores under that floor are removed, and that floor is reshored (reshoring is partial shoring). Precations are taken that a reshored floor is not subjected to construction loads. Generally, with weekly placement of concrete at each floor, one floor is shored and two lower floors have reshores. A floor below these three floors can generally be without any support, see Figure 19.12.

78

Name

18.8

Draw in three dimensions a typical chair and a typical bolster to support steel reinforcement.

Plastic coated feet

A typical bolster Several other profiles are available.

A typical plastic rebar chair.

Several other proprietary chairs are available.

18.9

An interior reinforced concrete slab-on-ground is 65 ft x 65 ft and is 6 in thick. (a) Draw the plan of the slab and show the location of control joints that you recommend.

Scale: 1/16 in. = 1 ft.

13’

Control joint

Edge of slab

13’

13’

13’

From Section 19.11, center to center spacing of control joints is approximately 24 x slab thickness = 24 (0.5) = 12 ft. therefore, we provide control joints at 13 ft o.c., giving 5 equal spaces.

79

(b) Draw a section through a typical control joint. Scale 2 in. = 1 ft.

6”

1-1/2” Width of control joint = 1/8” approx.

80

Assignment 19 Name

Chapter 20

19.1

The following diagrams show the plans and three-dimensional sketches of reinforced concrete floors. Determine in each case whether the slabs are one-way or two-way slabs and provide reasons. Also draw sections at locations A, B, and C and provide the approximate dimensions of slab and beam. Scale: 1 in. = 1 ft.

24 ft

20 ft

24 ft

A

Slab measures 20’ x 48’. Long to short span of slab = 48/20 = 2.4 > 2.0; hence, it is a one-way slab. Slab span = 20’. From Section 20.1, approximate slab thickness = (span/24) = [(20 x 12)/24] = 10”

Beam

Section A

24 ft

20 ft

24 ft

10” thick slab

B

These are two slabs. Each slab measures 20’ x 24’.

Long to short span of slab = 24/20 = 1.2 < 2.0; hence both are two-way slabs. Slab span = 24’ (the larger of the two spans). From Section 20.1, approximate slab thickness = (span/36) = [(24 x 12)/36] = 8” 81

8” thick slab

Beam

Section B

24 ft

20 ft

24 ft

C

These are four slabs. Each slab measures 10’ x 24’. Long to short span of slab =24/10 = 2.4 > 2.0; hence all are one-way slabs. Slab span = 10’. From Section 20.1, approximate slab thickness = (span/24) = [(10 x 12)/24] = 5”

5” thick slab

Beam

Section C As described in Chapter 20 in the text, a number of floor systems are used in contemporary concrete frame buildings, for example: • • • •

Beam-and-girder floor Band beam floor One-way joist floor Two-way joist floor (waffle slab)

• •

Flat plate floor Flat slab floor

The suitability of each system is generally a function of column spacing and layout. In the next few questions, you are asked to recommend the floor system that best suits a given column layout, and to sketch the floor framing plan and detailed sections with member dimensions. Note that for some layouts, more than one system may be suitable. Reference Chapter 20 and Appendix B in the text.

82

Name

19.2

(a) The column layout plan shown below is typical of multistory concrete frame buildings.

Which floor system(s) is(are) not suitable for the layout? Give reasons.

This column spacing (approximately 30 ft in both directions) is very commonly used in office buildings with a cast-in-place concrete structural frame. The most suitable system for this layout is a one-way joist floor with beams running in the 30 ft dimension and joists in the 35 ft dimension (see margin box in Section 20.4 of the text). Another system (generally less economical) is beam and girder system with beams running along 30 ft dimension. With three beams in a 35 ft bay, the slabs will be one-way slabs (see Question 17.3(a)). All other systems will generally be less economical.

30 ft

35 ft

35 ft

35 ft

Beam Joist

Beam

30 ft

30 ft

A

B

This beam and the one on the opposite side do not support the joists, but are required Column Layout Plan to support the wall load and provide lateral load resistance. (b) Using the plan above as an underlay, draw the framing plan of the most suitable floor system for the layout.

83

(c) Draw a section through the floor at location A and provide approximate dimensions of the members. Scale: 1 in. = 1 ft.

3.5” Slab Beam 20” Line of joists

Section A

36”

Assume 1-hour fire-rated floor. From Table 20.1, minimum slab thickness = 3.5”.

We do not yet know the width of beams supporting the joists. Assume 36” wide beams (to be verified later). Hence, clear span of joists = 35 - 3 = 32’. From Appendix B, the approximate depth of joists = span/18 = (32 x 12)/18 = 21.3”. With a 3.5” thick slab, the joist depth below the slab = 21.3- 3.5 = 17.8”. From Figure 20.8, select 20” deep, 30” wide pans. Thus, beam and joist depth = 23.5”. Approximate beam width = 1.75(21.3) = 37”. Thus, our initial assumption of 36” wide beams is reasonably OK. (d) Draw a section through the floor at location B and provide approximate dimensions of the members. Scale: 1 in. = 1 ft.

Standard 1:12 taper to match the joists

Slab 3.5”

12” x 23.5” edge beam

20”

Joist

6”

30”

12”

Section B Joist spacing = 36” o.c.

We have used standard module one-way joists. If the required fire rating were 2-hours, requiring a minimum 4.5” thick slab (see Table 20.1), it would have been more economical to use a wide-module one-way joist system.

84

Name

19.3

(a) Using the column layout below, repeated from Question 19.2, as an underlay, draw the floor framing plan of an alternative system that can be used for the same layout.

30 ft

35 ft

35 ft

35 ft

Girder

Girder

Beam

Beam

30 ft

30 ft

C

D

The above floor framing plan shows a girder and beam system with girders running along the shorter span. Three beams have been used in each bay so that no beam delivers its load at mid span of the girder.

85

(b) Draw a section through the floor at location C and provide approximate dimensions of the members. Scale: 1 in. = 1 ft.

We do not have column dimensions.

Therefore, we will use center-to-center span for determining girder dimensions.

From Appendix B, approx. girder depth = (30 x 12)/12 = 30”; width = 18”.

18” 4.5”

30”

18” x 30” girder

22.5” Clear span of beam = 33.5’. Therefore, its approximate depth = (33.5 x 12)/15 = Line of 27”; width = 16”. beam Clear span of (one-way) slab = 10’ - 16” = (8.67 x 12)/24 = 4.5” Note that we could post-tension the girder to reduce its depth to 27” so that beams and girders align at the bottom.

Section C

(c) Draw a section through the floor at location D and provide approximate dimensions of the members. Scale: 1 in. = 1 ft.

16”

27”

4.5” thick slab 16” x 27” beam Line of girder

Section D Column

19.4

(a) Using the column layout of the multistory reinforced concrete office building shown below as an underlay, draw the framing plan of the most suitable floor system for the layout.

86

Name 45 ft

45 ft

45 ft

Joist 5’ o.c.

45 ft

45 ft

45 ft

E

48” x 28.5” beam

Beam (15” x 28.5”)

(b) Draw a section through the floor at location E and provide approximate dimensions of the members. Scale: 1 in. = 1 ft.

48” The most economical system for this floor is a one-way joist system, Assume 45” wide beams (to be verified later). Clear 4.5” span of joists = 41.25’. Hence, joist depth = (41.25 x 12)/18 = 27.5”. Using a 4.5” thick slab to provide a 2-hour fire rating, the depth of joists below slab = 23.0”.

Therefore, use 24” deep pan form. 4.5” thickness of slab is large for a standard 24” module system. Therefore, use wide module sysem with 7” wide joists at 5’ o.c. (i.e., 53” wide pans), Beam depth = 28.5”. Beam width = 1.75(27.5) = 48” approx.

48” x 28.5” Note: Post-tensioning the beam beam will reduce its dimensions.

87

19.5

Using the column layout of the multistory reinforced concrete apartment building (shown below) as an underlay, draw the framing plan of the most suitable floor system for the layout.

20 ft

20 ft

20 ft

16 ft

12” x 16” perimeter beam (see next page)

16 ft

16 ft

8” thick flat plate floor

For this column layout, the most economical reinforced concrete floor system is a flat plate floor—a beam-less floor so that the slab rests directly on columns (see Section 20.3). Spacing between colmns = 20’. From appendix B, the approximate depth of slab = (20 x 12)/30 = 8”.

Perimeter beams are genedrally required, however, to carry wall loads. Perimeter beam depth = (20 x 12)/15 = 16”. Provide a 12” x 16” perimeter beam.

88

Assignment 20 Name

Chapter 21

20.1

Soil samples must be tested in a soil testing laboratory for an accurate analysis. However, many soil properties depend on the type of soil, which can be approximated from a visual examination of the soil. The International Building Code divides soils in a few broad categories (see Table 21.1 in the text). (a) If in a soil sample, you observe that: • most particles are less than 1/4 in. diameter • the soil feels gritty and coarse when the particles are rubbed between your fingers • when the sample is wetted, it cannot be rolled into a thread text)) is: this soil type (according to Table 1, Principles in Practice of Chapter 21 in the text Sandy soil (SW/SP/SM/SC) (b) If in a soil sample, you observe that: • most particles are less than 1/4 in. diameter • when the particles are rubbed between your fingers, the soil feels smooth • when the sample is wetted it rolls into a thread this soil ((Table Table 1, Principles in Practice of Chapter 21 in the text text)) is: Fine-grained soil (ML/CL/MH/CH) (c) If in a soil sample, you observe that: • a large percentage of particles are greater than 1/4 in. diameter the soil ((Table Table 1, Principles in Practice of Chapter 21 in the text text)) is: Gravelly soil (GW/GP/GM/GC)

20.2

(a) If a 3 ft x 3 ft square isolated column footing rests on soil described in Question 18.1(a), what is the maximum load the footing can carry as determined from the presumptive bearing capacity given in Table 21.1?

Note: See Figure 21.27(b) for a sketch of isolated column footing.

Allowable presumptive bearing capacity (BC) of soil = 2.0 ksf Allowable load on column = Area of footing x BC = 9.0 x 2.0 = 18.0 k 89

(b) The drawing in Question 20.3 shows a typical foundation detail of a wood light frame wall. If the footing rests on soil described in Question 20.1(a) and its width is 22 in. (as shown), what is the maximum load that the footing can carry per foot length as determined from the presumptive bearing capacity given in Table 21.1?

Allowable load on wall = (Width of wall in ft) x BC = (22/12) 2.0 = 3.7 k/ft

20.3

(a) The cross-sectional detail shown below is incomplete in many respects. Complete the detail by adding missing features and provide all labels. (b) The depth below grade of the bottom of the footing in the detail above is 21 in. Which conditions must you examine to ensure that the 21 in. depth is acceptable?

1. Adequate bearing capacity of soil 2. Bottom of footing must be below the frost line of the location Wall insulation OSB subfloor Floor finish as needed, not shown

2 x 4 wood stud wall 1/2 in. OSB sheathing

Wood siding

Insulation in floor with vapor retarder below subfloor Vented crawl space

Parging on wall

Note: Parging is a thin stucco (portland cement based plaster) applied on a masonry wall to hide imperfections and improve dampproofing.

8 in. thick CMU wall

Plastic batten bar to retain vapor retarder Vapor retarder

Liquid waterproofing over parging

Existing grade

Fully grouted reinforced masonry foundation wall

21 in.

Perforated pipe drain to daylight or storm sewer

22 in.

Wall Footing Detail 90

Name

20.4

(a) A 10 ft deep basement is constructed with reinforced concrete walls on a large unobstructed site. Although the final choice of excavation support system is up to the general contractor (GC), your opinion, as an experienced construction contract administrator for the project, has been sought by the GC, who is relatively new to the field. What is your suggestion to the GC as the most suitable excavation support system? Explain.

Open excavation (sloping or benched), as shown in Figures 21.7 and 21.8 of the text.

(b) If the basement described above exists on a relatively restricted site, which excavation support system would you suggest? Explain.

Soil nailing or sheet piling. Soil nailing is, however, more comonly use because of the ease with which it can be done. (c) If the basement is 50 ft deep below the ground, which excavation support system would you suggest? Explain.

System comprising soldier piles, lagging and tiebacks.

20.5

(a) Assuming that the basement in Question 20.4(a) is a simple rectangle in plan with few openings and pipe penetrations, which waterproofing system would you recommend for the walls? Give reasons for your choice.

Self-adhering single-ply membrane.

(b) If the basement consists of a complex geometry with several corners, protrusions, and penetrations, would your recommendation change? Explain.

Liquid-applied waterproofing.

91

(c) Which waterproofing system would you recommend for the basement in Question 20.4(c)? Explain.

Blind side waterproofing which is sandwiched between the excavation support system and the reinforced concrete wall.

92

Assignment 21 Name

Chapters 22 and 23

21.1

Masonry mortar types are classified as M, S, N and O. Which mortar type is most commonly specified and why?

Type N mortar is the most commonly used. That is why it is referred to as the general purpose mortar. With its approximately 1:1 portland cement and lime proportions, it balances mortar strength (provided by portland cement) with other necessary characteristics required of mortar such as watertightness of masonry in the hardened state of mortar, and workability and water retentivity of mortar in its plastic state (provided by lime).

21.2

What would you recommend for the mortar type (M, S, N, or O), type of brick–facing brick (FBS, FBX, or FBA), or building brick–and the grade of brick (SW, MW, or NW) for the following situations?

Location of bricks (a) Exterior brick veneer (in running bond) in New York (b) Exterior brick veneer (in running bond) in southwest California (c) Exterior brick veneer (in stack bond) in southwest California (d) Interior exposed brickwork in an office lobby (e) Brickwork in a loadbearing wall with stucco finish on the exterior and drywall facing on the interior in southern Arizona

21.3

Brick type FBS

Brick grade SW

Mortar type N

FBS

SW

S

FBS

SW

S

FBX Building brick

MW NW

N N or S depending on the required strength of the wall

Complete the following drawings by showing in plan the layout of bricks and mortar joints in two adjacent courses of a single-wythe, L-shaped brick wall in running bond. Use modular bricks. Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

Plan of course 1

Plan of course 2 93

21.4

Complete the following drawings by showing in plan the layout of bricks and mortar joints in two adjacent courses of a two-wythe brick wall in English bond using modular bricks.

Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

Plan of course 1

Special unit (half header)

21.5

Plan of course 2 Complete the following drawings by showing in plan the layout of bricks and mortar joints in two adjacent courses of a two-wythe brick wall in Flemish bond using modular bricks.

Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

Plan of course 1

Special unit (half header) Plan of course 2

21.6

Complete the drawing on the right by showing in section the layout of bricks and mortar joints in a two-wythe brick wall in Common (American) bond using modular bricks.

Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

94

Section

Name

21.7

Complete the following drawings by showing in plan the layout of 8 in. (nominal) CMUs, mortar joints, and vertical reinforcement in two adjacent courses of the wall. Assume the use of 2-cell CMUs (with 1-1/2 in. thick face shells and 1-1/4 in. thick webs) and the wall is reinforced with No. 5 rebars at 16 in. o.c.

Scale: 1-1/2” = 1’; i.e., 1 grid division = 2 in.

Mortar Grout

# 5 Rebar

Plan of course 1

Plan of course 2

21.8

Name the natural stone you would specify in the following situations and provide reasons for your choice. (a) Exterior monumental entry stairs to a city hall With the amount of foot traffic that an entry stair would have, granite is the best choice, being the strongest and most durable of stones. (b) A residential kitchen countertop

Granite because it is more acid resistant than other stones.

(c) A residential bathroom

Marble because of its durability, good appearance and lower cost than granite.

(d) A living room in a middle-income apartment

Marble or slate. Both take good polish, and are available in several colors.

(e) Exterior wall cladding of an 8-story office building Limestone—more economical than marble or granite.

95

21.9

Estimate the approximate fire rating of the wall of: •

Question 21.5

4 hours (Table 23.3)

• Question 21.7 4 hours. Because it is a partially grouted wall, we need to determine its equivalent solid thickness to determine a more exact value of its fire resistance (see note below).

21.10

From what source would you obtain more precise values of the fire rating of the walls in Question 21.9? Explain.

A commonly-used source for obtaining fire resistance rating of assemblies is the International Building Code (Chapter 7). A more authoritative source is the Fire Resistance Design Directory published annually by the Underwriters Laboratories, Inc. (UL). Masonry associations—Brick Industry Association (BIA) and National Concrete Masonry Association (NCMA) also provide fire rating of their products.

Note to the Instructor: This question, as the following explanation shows, may be too difficult for the students, unless the Instructor gives them hints and help.

The fire resistance rating of a CMU wall is (i) a function of equivalent solid thickness of wall and (ii) the type of coarse aggregate used in the manufacture of the unit. Lightweight aggregate (e.g., expanded clay, shale or slate) gives a higher fire resistance for the same quivalent thickness of wall than normal weight aggregate (e.g., crushed limestone).

Equivalent Solid Thickness The equivalent solid thickness of the wall is obtained by subtracting the space occupied by ungrouted cells from the total space occupied by the wall. In Question 19.7: Thickness of both face shells = 2(1.5) = 3.0” Thickness of three webs = 3(1.25) = 3.75” Hence, the area of each cell = (7.625 - 3.0) x [(15.625 - 3.75)/2] = 4.625” x 5.94” = 27.5 sq in.

The total area occupied by one CMU = 15.625 x 7.625 = 119.1 sq in.

Because only one cell is ungrouted, the total solid area of one CMU = 119.1 - 27.5 = 91.6 sq in.

Thus, the percentage solid area of CMU wall = (91.6/119.1)x100 = 77%. Hence, the equivalent solid thickness of wall = 0.77(7.625) = 5.9”.

Fire Rating From Table 720.1(2) of the International Building Code, a CMU wall with an equivalent thickness of 5.9” (with normal weight aggregate) has a fire rating of 4 hours.

If the coarse aggregate was expanded shale in CMU, an equivalent solid thickness of 5.1” would give 4-hour fire rating.

96

Assignment 22 Name

Chapter 24

22.1

The owner of the two-story weekend house, who had earlier decided to use wood light frame construction (see Assignment 14 in this manual), now wants to use CMU walls and wood light frame floor and roof. The project architect has advised the owner that the dimensions of wall elements in plans and sections need to be redeveloped to eliminate site-cutting of CMUs. On the architect’s instructions, you (as her intern) have redrawn the first-floor plan as shown below, utilizing 8 in. CMU for the exterior walls and 4 in. CMU for the interior walls.

The completion of the plan requires that you provide exact dimensions of wall elements and rough openings. (a) Replace the approximate dimensions shown below and provide the missing dimensions.

(Refer to Section 24.6 in the text.)

A 75/8” 6”

7’ 43/8” 8’0”

75/8” 6” 6’ 83/8”

5’ 115/8”

6’-9” 20’-0” 19’ 115/8”

6’-0”

19’ 115/8” 20’-0”

B

5’ 43/8” 5’-6”

3’ 43/8” 1’ 35/8” 3’-3” 1’-3”

25’ 35/8”

8’-9”

9’ 35/8” 9’-3”

1’ 35/8” 1’-3”

35’ 35/8” 35’-3”

8’ 83/8”

Note: Dimension of an opening = multiple of masonry unit (8”) + 3/8” Dimension of a masonry panel = multiple of masonry unit (8”) - 3/8”

15’-3” 15’ 35/8”

First Floor Plan 97

(b) Draw below the detail sections at locations A and B [Question 22.1(a)], showing the junction of the wall with the floor. Scale: 2” = 1’; i.e. one grid division = 1-1/2 in.

The three-dimensional sketch is provided here as reference.

2x12 floor joist terminates over 2x8 plate 2x10 floor joist continuous on both sides of 2x8 plate.

2x10 and 2x12 joists nailed together (see Question 12.2, Assignment 12)

Threshold under sliding door Lightweight concrete Single-ply membrane waterproofing over subfloor to extend over blocking and under threshold

Blocking 3x12 ledger bolted to bond beam 16” deep bond beam 2x12 floor joist 2x8 plate

Reinforced masonry wall.

Vertical reinforcement not shown

1/2” th. gypsum board

EIFS cladding. Thickness of expanded polystyrene insulation as needed Liquid air-weather retarder on masonry (see Ch. 27)

2 x 1 wood furring strip

Lintel unit with one additional course above (fully grouted) to provide a 16” deep lintel beam

2x6 wood nailer

Sealant and backer rod Single-hung aluminum window with thermally broken frame (see Figures 28.23 and 28.4 of Text)

Section B 98

Section A

Name (c) Draw below the detail section at location A showing the junction of the wall with the roof. Scale: 2” = 1’.

Asaphalt shingle roof No. 30 organic felt underlayment Peel-and-stick ice dam protection membrane to extend 24” beyond exterior wall (toward the interior) 16” deep bond beam 2 x 12 (oversized) rafters to allow 9-1/4” thick insulation with 2” air space above Edge flashing

2 x 4 soffit joist 2 x 4 Ledger EIFS cladding

Finished fascia board Rough fascia Soffit Screened vent

99

22.2

A highrise apartment complex is to be constructed using CMU bearing walls and precast concrete hollow-core planks. (a) In the schematic design stage of the project, what nominal thickness of CMU walls would you assume if the apartment building is: •

6 stories high?

8 in. thick (see Appendix B)

•

12 stories high?

10 in. thick for the first few floors, and 8 in. thick in upper floors

(b) Which alternative construction system(s) would you examine during the schematic design stage for this project? Explain.

Loadbearing cast-in-place reinforced concrete walls and floors using tunnel forms

22.3

A CMU loadbearing wall system is often used to enclose long-span spaces such as gymnasiums and auditoriums. Apart from the economics, what are other important reasons for the choice of the system?

The walls of a gymnasium and an auditorium must have high fire resistance (being assembly occupancies). They also must have sufficient strength to support loads from long-span roof truss, and must be impact-resistant, particularly the gymnasium walls. The only two suitable options are reinforced concrete and reinforced masonry walls. The latter is generally more economical, and that is why it is commonly used.

(Note: The other fire-resistive construction, gypsum board on light-gauge steel wall framing, can provide adequate fire resistance, but does not have sufficient structural strength or impact resistance.)

22.4

As the architect for a single-story commercial development, you have decided to use concrete tilt-up wall construction. The floor-to-roof height is 22 ft. What approximate thickness of the wall will you assume in the schematic design stage?

From Section 24.9 of the text: Approximate wall thickness = 22 x 12 = 5.5 in., or say 6 in.

48 100

Assignment 23 Name

Chapter 25

23.1

The detail of the insulated metal panel shown in Figure 27.51 of the text is reproduced below (left-hand image). (a) Does the factory-installed seal exemplify rain-screen principles? Explain.

The factory-installed seal is provided so that the joint between panels will function as a rain screen. The seal provides the air barrier, a necessary component of rain screen. Consequently, the air space in the joint wil have pressure equalization at all times, and will function as a pressure-equalized, drainage-capable joint. Because the seal is on the interior side of the joint, it is (relatively) protected from weathering.

Factory-installed seal

Site-installed backer rod and sealant

101

(b) If the seal is brought forward (as shown in the right-hand image above) and is siteinstalled, how would the performance of the detail change? Explain.

With the seal in this exterior location, the joint functions as a barrier-type joint. Because the seal is exposed to weathering elements, its ability as a barrier will deteriorate over time. As it deteriorates, water from wind-driven rain can be sucked through the joint when the outside air pressure on the face of the wall is higher than the inside air pressure.

23.2

(a) What would you change in the following window sill detail so that it performs based on the rain-screen principles?

Insulating glass unit

Interior gasket

Exterior gasket Glazing stop

Sealant Backer rod

Weep holes

Provide a continuous seal between the glass and the interior side of the frame before installing the interior gasket to provide an air barrier. With weep holes and some air permeability of exterior gasket, pressure equalization in the glazing pocket should occur, making the glazing pocket more water resistant.

102

Assignment 24 Name

Chapter 26

24.1

The following drawing shows the elevation and plan of a multistory brick veneer facade. Using the drawing as an underlay, locate vertical and horizontal expansion joints on the elevation.

Vertical expansion joint 5 ft

Horizontal movement joint at floor level

Top of floor

Top of floor

Top of floor

Top of wall foundation Finished grade

Elevation (Approximate scale: 1/12” = 1’; i.e., 1 grid division = 3 ft)

15 ft

5 ft

5 ft

5 ft

15 ft

Plan

103

5 ft

5 ft

5 ft

24.2

If the building in Question 24.1 is a reinforced concrete frame structure, draw a detail section at the foundation level. Show all components and label them. Assume that: • the brick veneer is backed by an 8 in. thick CMU wall.

• 2 in.-thick rigid foam insulation between the CMU wall and brick veneer (modular bricks).

• 16 in. thick, 24 in. deep reinforced concrete wall foundation.

• 6 in.-thick ground-supported slab isolated from the foundation wall.

• the building is located in Madison, Wisconsin.

Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

Note: You will need to refer to Chapters 26, 19, and 5 in the text.

2” air space Liquid air-weather retarder on CMU wall Galvanized steel furring section

2” extruded polystyrene insulation

5/8” gypsum board

3-5/8” brick veneer

Reinforced CMU backup wall

Masonry tie

Masonry wall reinforcing

Flashing termination bar Mortar capturing device

1/2” isolation joint with asphalt-saturated fiber board and topped with sealant

Two-piece flashing (stainless steel with hemmed edge and self-adhering single-ply membrane flashing) Weep holes

Concrete slabon-ground

10 mil vapor retarder

Stainless steel Zsection cover 1/2” fiber cement board

4” sand layer

Extruded polystyrene insulation

Compacted ground

Detail at Foundation Level

104

Name

24.3

For the building in Question 24.2, draw a detail section of the wall assembly at an intermediate floor (one-way joist system, shown below). For the shelf angle, assume 1/4-in.-thick equal-leg angle. Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.)

Note Items not identified in this detail are the same as on Page 84.

Slab

3-5/8” brick veneer Galvanized steel shelf angle 2-piece flashing Sealant and backer rod at movement joint

Spandrel beam Joist

Dovetail restraint anchor (see Figure 26.22 of the text)

Detail at an Intermediate Floor Level

105

24.4

Repeat the detail in Question 24.3, but assume that: • brick veneer is backed by a light-gauge steel stud wall (stud size 600 S 162-43) • fiberglass insulation in the stud wall cavity • 5/8 in. thick interior drywall • 5/8 in. thick gypsum sheathing, and • 1 in. thick rigid foam insulation over sheathing Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

5/8” glass mat faced gypsum sheathing 5/8” gypsum board Air-weather retarder wrap on sheathing and spandrel beam

Light-gauge steel stud back-up wall Slab

1” extruded polystyrene insulation Mortar capturing device 2-piece flashing Weep holes Sealant and backer rod Brick veneer

Deep leg and regular leg nested tracks

106

Spandrel beam Joist

Name

24.5

The following drawing shows a partial elevation of an exterior wall of an office building that is to be clad with precast concrete curtain wall panels. Using the drawing as an underlay, design the wall’s elevation showing precast concrete wall panels, surface finishes, and any relief joints you would like to include. Interior daylighting considerations require a minimum 35% of the wall area to be glazed.

36 ft

Relief joint

13 ft

13 ft

13 ft

Spandrel beam Column

36 ft

Joint between panels

Partial Elevation Approximate scale: 1” = 12’; i.e., 1 grid division = 3 ft

107

Relief joint

24.6

Using three-dimensional, freehand sketches, show how you would explain your ideas to an intern architect as to the construction of the wall in Question 22.5 (including connection details).

Precast concrete panel Dead load support for panel is located as close to columns as possible. Only two dead load supports per panel.

Lateral load support (tieback) for welding to bottom of spandrel beam

Column Floor slab

Dead load support Steel embed to which tieback is welded Pocket in spandrel beam to provide recessed bearing for dead load support

Tieback for welding to steel embed on top of spandrel beam. Number of tiebacks a function of lateral load.

108

Assignment 25 Name

Chapter 27

25.1

The following drawing shows the partial elevation of a multistory residential building with a reinforced concrete structure. The exterior wall consists of portland cement stucco backed by light-gauge steel studs. Using the drawing as an underlay, show the location of joints in stucco.

30 ft

30 ft

Spandrel beam Column

Partial Elevation

Expansion joint

Approximate scale: 1” = 10’, i.e., 1 grid division = 2.5 ft

Control joint

Red lines represent expansion joints provided at the bottom of each spandrel beam. Blue lines represent control joints (see answer to Question 25.2).

109

25.2

Draw the detail of the portland cement stucco-clad wall assembly in Question 25.1. The structural frame of the building consists of a reinforced concrete one-way joist floor as shown below. Assume: • 600 S 162-43 studs • fiberglass insulation in stud cavity • 5/8 in. thick interior drywall • 5/8 in. thick gypsum sheathing.

Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

7/8” portland cement stucco finish Stucco, self-furring lath backed by asphalt-saturated paper

Light-gauge steel stud wall with cavity insulation

Sheet air-weather retarder on sheathing Glass mat faced gypsum sheathing 1” in. extruded polystyrene insulation Galvanized steel Z-section anchored to beam

2-piece galvanized steel expansion joint accessory Backer rod and sealant Spandrel beam Joist

Deep leg and regular leg galvanized steel nested tracks

110

Name

25.3

The following drawing shows the partial elevation of a multistory residential building with a reinforced concrete structure. The exterior wall consists of EIFS backed by light-gauge steel studs. Using the drawing as an underlay, show the location of joints in EIFS.

30 ft

30 ft

Spandrel beam Column

Partial Elevation Approximate scale: 1” = 10’; i.e., 1 grid division = 2.5 ft

111

Expansion joint at bottom of each spandrel beam (see Question 25.4)

25.4

Draw the detail of the wall assembly in Question 25.3. The structural frame of the building consists of reinforced concrete one-way joist floor as shown below. Assume: • 600 S 162-43 studs • fiberglass insulation in stud cavity • 5/8 in. thick interior drywall • 5/8 in. thick gypsum sheathing Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

Glass mat faced gypsum sheathing Liquid air weather retarder on sheathing and spandrel beam

Light-gauge steel stud wall

EIFS finish lamina Expanded polystyrene insulation

Spandrel beam Joist

EIFS finish over backwrapped mesh on both sides of sealant and backer rod

Galvanized steel deep leg and regular leg nested tracks

112

Name

25.5

(a) Using online resources, list three manufacturers of EIFS in North America.

Dryvit Systems Inc. STO Corporation Senergy Inc. (b) Select one of the EIFS manufacturers from the above list and, using sketches and notes, describe two important EIFS systems it manufactures.

Liquid air-weather retarder on glass mat faced gypsum sheathing Adhesive in vertical notched trowel configuration Insulation Base coat Standard mesh Base coat Finish coat

Vent and drainage track Dryvit Infinity (Pressure Equalized, Rainscreen) Vent and drainage System with Standard Impact Mesh assembly

The two systems are similar except the additional, thicker mesh (Dryvit’s “Panzer mesh”)

Dryvit Infinity (Pressure Equalized, Rainscreen) System with High Impact Mesh 113

114

Assignment 26 Name

Chapters 28 and 29

26.1

The solar heat gain coefficient (SHGC) for a 1/4-in.-thick clear glass sheet is 0.86 and its U-value = 0.98. If the same glass is tinted, which of the two properties, the SHGC or the Uvalue, will be more affected? Explain.

If the glass is tinted, the SHGC of the same glass will be reduced. U-value, which measures air-to-air heat transfer will not be affected.

26.2

How is the SHGC of glass related to its U-value? Explain.

SHGC measures the amount of solar heat that is transferred to the interior through a glass.

U-value is the amount of heat gained or lost through an element because of the difference between the inside and outside air temperatures. The two properties are, therefore, not related to each other. However, if the glass has a reflective coating (which is a thin metallic coating), it slightly decreases the U-value of glass because of the coating’s low emissivity.

And because the coating reflects solar beam, it also reduces the SHGC value of glass.

26.3

Determine the light-to-solar gain (LSG) index of glass with a visible transmittance (VT) of 0.78 and SHGC of 0.58.

LSG =

26.4

VT = SHGC

0.78 0.58

= 1.35

Among the three insulating glass units (IGUs) shown below, select the one that most suits windows in the following climates: (a) Hot, sunny climate IGU 2 (b) Cold cimate, requiring passive solar heating through windows IGU 3 (c) Hot, humid climate with generally overcast skies IGU 1

Low-E coating

Low-E coating

Low-E coating

Reflective coating

OUTSIDE

INSIDE

IGU 1

OUTSIDE

INSIDE

IGU 2

115

INSIDE

OUTSIDE

IGU 3

26.5

A 1/4-in.-thick annealed glass is 50 in. x 40 in. What is the maximum wind load it can resist for a breakage probability of 8/1000? Provide your answer in Pa and also psf.

3.0 kPa (62.7 psf), see Table 28.28 of the text.

26.6

A 1/4-in.-thick annealed glass is 50 in. x 80 in. What is the maximum wind load it can resist for a breakage probability of 8/1000? Provide your answer in Pa and also psf.

1.8 kPa (37.6 psf), see Table 28.28 of the text.

26.7

You are designing an aluminum-glass curtain wall for an office building in New York City. The interior relative humidity would be maintained at approximately 45%. What minimum value of condensation resistance factor (CRF) would you specify for the curtain wall?

26.8

From Table 1 in Principles in Practice, Ch. 28, for 40% RH, CRF = (57 + 48)/2 = 52.5. For 45% RH, CRF should be approx. 60. (a) Does an interior drywall emit any radiation? Yes, all objects emit radition at all times. (b) Give a few examples of objects that emit longwave radiation.

All terrestrial objects emit longwave radiation (road surfaces, building surfaces, interior objects, etc.). (c) Give a few examples of objects that emit shortwave radiation. Sun, high temperature lamps.

26.9

26.10

In a window with nailing flanges, are the windows installed before or after applying the exterior wall cladding (e.g., siding or masonry veneer)? Windows with nailing flanges must be installed before applying exterior wall finish.

Draw a section (to a suitable scale) through a jamb of such a window assembly. Identify each component and explain their roles.

Fiber cement siding

Polyurethane thermal break (to reduce heat conduction from outside to inside)

Nailing flange (fin) to anchor window

Air-weather retarder Preglazed, single-hung aluminum window with thermal break and nailing fins

OSB sheathing

Inulating glass unit Insulation NOTE: The window crosssection shown is conceptual, but gives all its major features.

The actual cross-section is generally more complex.

Gypsum board

Jamb Detail

Line of stool 116

Name

26.11

In the following cross-section (at the sill level of an aluminum window): (a) label the components with leader lines; (b) explain the function of the parts designated as A, B, C, and D, and also label them; and (c) show which is the “inside” and “outside” of the assembly.

Spacer

Gasket

Double-sided tape topped with sealant

Snap-on stop Setting block INSIDE

A OUTSIDE

C Thermal break

B

Sill pan

D

“B” and “C” are screw splines meant to connect horizontal window frame member with the vertical frame member. “A” and “D” are weep holes.

26.12

A preglazed window is labeled as F-C35 60 x 60. (a) Interpret the label fully.

F-- fixed window C--performance class of window is “Commercial”. Therefore, it meets or exceeds the performance requirements for this grade for air leakage, water leakage, and wind load resistance.

35-- The maximum wind load resistance of the window is 35 psf.

60 x 60-- the window has been tested to meet the requirements of its performance class (“C”) to a maximum size of 60” x 60”. (b) Which other performance rating label can this window have? Explain.

The window will also have a label from the National Fenestration Rating Council (NFRC) giving its thermal performance ratings (U-value and SHGC).

117

26.13

Complete the surrounding wall around the following window head and sill details. Assume modular brick veneer cladding with a light-gauge steel stud backup with 4-in.-wide studs, fiberglass insulation, 5/8-in.-thick interior gypsum board and 5/8-in. glass mat faced gypsum sheathing, air space, and brick sill. Label all components. Scale: 1/2 full-size.

Air-weather retarder wrap Gypsum sheathing Lintel angle Light-gauge steel header

Brick veneer Self-adhering rubberized asphalt flashing and weep holes Stainless steel drip edge Stainless steel drip edge 12-ga. galvanized steel angle as nailer

Head Detail 12-gauge galvanized steel angle as nailer Wood stool Wood apron (trim)

Sealant and backer rod Brick sill

Gypsum board Self-adhering rubberized asphalt flashing and weep holes

Sill Detail 118

Name

26.14

Complete the surrounding wall around the following windowsill detail. Assume everything is the same as in Question 24.13, but replace the brick sill with a precast concrete sill. Label all components. Scale: 1/2 full-size.

12-gauge galvanized steel angle as nailer Sealant and backer rod Wood stool Precast concrete sill

Wood apron (trim)

Gypsum board

Self-adhering rubberized asphalt flashing and weep holes Air-weather retarder wrap on gypsum sheathing Brick veneer

Sill Detail

119

26.15

Complete the surrounding wall around the following window head and sill details. Assume portland cement stucco on a light-gauge steel stud exterior wall with 6-in.-wide studs, fiberglass insulation, 5/8-in.-thick interior gypsum board, and 5/8 in.-thick glass mat faced gypsum sheathing, and an appropriate sill. Label all components. Scale: 1/2 full-size.

Light-gauge steel header

Head Detail

Self-adhering rubberized asphalt flashing

Wood stool

Corner bead

Wood apron (trim) Light-gauge steel stud material fastened to track

Portland cement stucco with selffurring lath and building paper Air-weather retarder wrap

Gypsum board

Gypsum sheathing

Sill Detail 120

Assignment 27 Name

Chapter 30

27.1

Complete the following sill detail of a multistory glass-aluminum curtain wall at an intermediate floor. Show part of the spandrel beam and the floor. Label all components.

Scale: Approx. 1/2 full-size.

Alumimum clip for snap on connection with stool

Vision glass (IGU)

Aluminum stool

Semi-rigid mineral wool insulation

5/8” th. gypsum board

Metal pan cover over insulation Spandrel glass

Insulation attached with an impaling clip Light-gauge steel stud wall with fiberglass insulation

2” air space Outside face of structural frame

Glass-aluminum curtain wall system

Sill Detail

121

27.2

(a) Complete the following head detail of the glass-aluminum curtain wall in Question 25.1.

Show part of the ceiling. Label all components.

Scale: 3/4 full-size.

Fireseal (firestop) between curtain wall and spandrel beam

Metal pan cover Semi-rigid mineral wool insulation

Spandrel beam

Spandrel glass

Impaling clip

Suspended ceiling

Head Detail

Glazing stop Vision glass (IGU)

Insulation attached with an impaling clip

(b) Is this curtain wall inside or outside glazed? What are the characteristics of the detail that provide this information?

It is an inside glazed curtain wall because: (a) it has ain interior glazing stop at the top of a glass panel (b) it does not have any pressure plates

122

Assignment 28 Name

Chapters 31 and 32

28.1

You have specified a gravel-covered, 4-ply built-up roof membrane on a low-slope roof. The general contractor (GC) has been advised by the roofing subcontractor that the membrane is ready for inspection prior to laying the flood coat and gravel. After his inspection, the GC has requested that you observe the roof. (a) How would you verify that the membrane is indeed a 4-ply membrane?

By measuring the exposure of felts, which for a 4-ply built-up roof = 8-1/2”.

(b) If this roof area measures 30 ft x 60 ft, approximately how many rolls of felt have been used?

The total area of roof = 30 x 60 = 1800 sq ft = 18 roof squares.

Each felt roll covers 1 roof square for one ply.

Because it is 4-ply roof, the total number of rolls required = 18 x 4 = 72 (c) If this building is in a relatively warm climate, which roofing asphalt would you specify for the flood coat?

Type III roofing asphalt (d) Calculate the total weight of the flood coat on this roof.

For the flood coat approximately 60 # of asphalt per roof square is needed. Hence, the total weight of asphalt in the flood coat = 18 x 60 = 1,080 # (e) Calculate the total weight of aggregate on this roof.

Generally, 400 pounds of aggregate per roof square is needed in a built-up roof. Therefore, the total weight of aggregate = 60 x 400 = 24,000 #. (f) If this building is in a warm, hurricane-prone coastal area, which part(s) of built-up roof specifications would you change and why?

The codes do not allow a gravel (or aggregate) covered built-up roof in a hurricane-prone region (basic wind speed greater than 120 mph). Even in a non-hurricane prone region, the codes do not allow such a roof beyond a certain building height. Therefore, if a built-up roof is specified in a hurricane-prone region, it must be covered with a granule- or foil-faced capsheet instead of flood coat and aggregate.

123

28.2

The drawing below shows a section through a steel roof deck supported by open-web steel joists. The joists bear on a CMU wall with brick veneer cladding. Complete the detail assuming 3 in. thick rigid foam insulation and a single-ply roof membrane. Label all components.

Approximate scale: 2” = 1’; i.e., one grid division = 1-1/2 in.

Continuous galvanized steel cleat Metal coping

Waterproofing sheet under coping

Galvanized steel counter flashing

Treated plywood and treated lumber nailers

Termination bar to anchor flashing to wall Bond beam

Single-ply membrane flashing Single-ply roof membrane anchored to treated wood nailers at roof edge Two-layer polyiso board insulation

Bond beam

124

Name

28.3

The roof detail shown below is an alternative to the detail shown as Figure 31.27 in the text.

In which situation would you recommend this detail in place of Figure 31.27 and why?

Roof detail shown in Figure 31.27 of the Text is similar to the detail shown here in Question 26.2. This detail is appropriate for low height parapets ( preferably not exceeding 24”). If the parapet is high, the membrane flashing is likely to sag when hot. Therefore, two details are generally used. One of these is the detail shown here, using a through-wall sheet metal flashing and counter flashing. The other detail, which generally performs better, uses ribbed metal panel cladding on the inside face of the parapet wall.

Sheet metal flashing to extend from bed joint of masonry up to bottom of parapet nailer Sheet metal reglet Removable sheet metal counterflashing Top of membrane sealed with roofer’s cement Membrane flashing

125

28.4

Your structural engineer has advised you that the low-slope roof of the bank building you are designing will be subjected to a maximum wind uplift of 50 psf. The building is located in a region that is subjected to substantial hailstorms. What performance criteria would you specify for the roof in terms of wind uplift, hail impact damage, and safety against exterior fire.

Note: Factory Mutual Research Corporation (FMRC) classifies wind uplift ratings of roofs in increments of 15 psf, starting at 30 psf; i.e., roof wind uplift ratings are: 30 psf, 45 psf, 60 psf, and so on.

Wind uplift: Because the factor of safety for a low-slope roof’s wind uplift design is 2.0, a wind uplift resistance of 100 psf is needed. The FMRC wind resistance classification are in steps of 15 psf starting at 30 psf. Therefore, this roof system must have a rating of 105 psf.

Hail Impact: There are two FMRC ratings, severe hail impact resistance (SH) and medium hail impact resistance (MW). Specify SW since the region gets substantial hailstorms.

Fire: Specify Class A—the most fire-resistive system, commonly used for commercial roofs.

28.5

The drawing on the following page shows the detail at the eave of the roof of a wood light frame office building you have designed in Boston, Massachusetts. (a) The owner wants to use asphalt shingles as roof cover and has asked you to provide the following information: • •

Roof slope 6.5:12 (see leader lines on facing page) Type of roof underlayment Because this is a standard slope steep roof, one-layer No. 15 organic felt is the minimum required. Either one layer No. 30 or two layers of No. 15 felts is preferrable.

•

Type of ice dam protection membrane A proprietary peel-and-stick ice dam protection membrane.

•

Type of valley underlayment No. 15 organic felt; No. 30 felt is preferred

126

6.5

12

(b) Which of the two, roof underlayment or valley underlayment, will be installed first?

Valley underlayment. (c) What is the difference between valley flashing and valley underlayment? Explain.

Valley flashing covers the valley underlayment. In an open valley, valley flashing is exposed. Valley flashing is the first line of defense against water penetration through the valley, valley underlayment is the second line of defense.

127

(d) Complete the detail shown on the following page and label all components. Assume asphalt shingle roof cover (12 in. x 36 in., 3-tab shingles, see Figure 32.5 in the text). Note that shingles are laid in double coverage with 5 in. exposure and standard 2 in. head lap.

Rafter ends are to remain cut as shown with fascia and soffit added. Add missing materials of your choice in a fully finished detail.

Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in.

Asaphalt shingles No. 30 organic felt underlayment Peel-and-stick ice dam protection membrane Galvanized steel drip edge

Insulation Vapor retarder if needed Screened vent Finished fascia board Rough fascia

Gypsum board

Wood trim Corrugated tie Brick veneer

Soffit Air-weather retarder wrap Glass mat faced gypsum sheathing

Completed Eave Detail

128

Assignment 29 Name

Chapter 33

29.1

The monumental exterior entrance steps to a historical building have weathered substantially, requiring complete replacement. The total rise from the existing grade to the entry level of the building is 5 ft 3 in. It has been decided to keep the same existing riser height for each step, which is approximately 4 in. Determine the number of risers and treads and give their dimensions. Show calculations.

Total height of steps = 63” With 4” riser dimension, number of risers = 63/4 = 15.75. Use 16 risers, with riser height = 63/16” = 3-15/16” rise.

Let (2R + T) = 25”. Therefore, [2(63/16) + T)] = 25. giving T (tread width) = 17”.

29.2

The following drawings show the first-floor and second-floor plans of a staircase in a 2-story single-family residence. (a) If the finished floor to finished floor height is 10 ft 1 in., determine the riser height of the stair and comment on whether it meets the building code requirement.

16 1

A

A Up 3Up

10

P First Floor Plan

Second Floor Plan

Number of risers = 16. Riser dimension = 121/16 = 7.56” < 7.75” (maximum allowed).

Therefore O.K. (b) If the tread width of this stair is 10 in. and the width of each landing is 3 ft 4 in., determine the dimension marked “P” in the above drawing. Show calculations.

P = 3’4” + 7(10”) + 3’4” = 12’6”

129

(c) Does the 10 in. tread width meet the building code requirement? Explain.

Yes, it meets building code requirement for a residential stair provided a nosing projection of 3/4” to 1-1/4” is provided. (d) Assuming a wood stair, draw a section through the stair, showing only one tread and two risers. Assume treads made out of 1-1/8 in. thick OSB with a carpet as the finish material.

Line of finished Scale: 2” = 1’; i.e., 1 grid division = 1-1/2 in. stringer 11-1/4” 10”

Rough tread

7.56”

Carpet applied in pieces.

Each piece extends from bottom of nosing to top of riser, fastened to tackless strips provided at bottom of each riser and back of each tread. Additional fastening provided by staples at top of riser and underside of nosing.

Rough riser Rough stringer

Tackless strips

Detail Section of Stair

(e) Draw section A-A through the entire stair in Question 27.2(a) on the following page.

Show handrail and other important features. Walls surrounding the stair are framed with 2 x 4 studs.

Scale: 3/8” = 1’; i.e., 1 grid division = 8 in.

130

Name

Landing platform

Landing platform

Section A-A

131

132

Assignment 30 Name

Chapters 34 and 35

30.1

Draw a partial floor plan and section of the tile floor assemblies and applications described below. Include all components required for proper installation including dimensions, type of setting material, and grout. (a) Bathroom floor: 1 in. x 1 in. ceramic mosaic tile over a wood subfloor.

Scale: 1/4 full-size; i.e., 1 grid division = 1 in.

Mosaic tiles with joints between tiles filled with grout Latex-modified (thin) setting material 1/2” thick cement backer board glued and screwed to subfloor with joints taped 1” thick T&G OSB subfloor

(Partial) PLAN SECTION (b) Entryway: 12” x 12” slate tiles over a 6 in. thick concrete slab.

Scale: 1/4 full-size.

12” x 12” tile Grout

12” x 12” ceramic tiles with joints between tiles grout filled Latex-modified (thin) setting material 6” thick (suspended) concrete floor slab with high spots ground smooth

(Partial) PLAN

SECTION 133

30.2

Specify the type of fiber and carpet construction for the following carpet applications. (a) An economical, soil-resistant carpet (in a muted color) to be used on a ground-level floor of an office building.

Fibers—polypropylene; Carpet construction—fusion bonded (b) A multicolored rug in shades of deep red and blue for use in a residential living room.

Fibers—wool; Carpet construction: woven (c) A very durable, bright, fade-resistant carpet for a commercial waiting room.

Fibers—nylon; Carpet construction: cut pile

30.3

(a) Draw a section that shows 4 adjacent planks of residential wood flooring over a 1-1/8 in. thick tongue-and-groove (T&G) OSB subfloor. Show and label all important details.

Scale: 3/8 full-size; i.e., 1 grid division = 2/3 in.

3/4” thick, 4” wide T&G wood plank flooring blind nailed to subfloor Cleavage plane (building paper or vapor retarder) 1-1/8” thick T&G OSB subfloor

134

Name (b) What criteria would you use to select the appropriate species of hardwood for this application?

A large number of softwood and hardwood species are used for strip or plank flooring. The two most important properties of wood for flooring are: density of wood and dimensional stability. The greater the density, the greater the floor’s resistance to wear and tear. Greater dimensional stability ensures that the floor will be less affected by the changes in relative humidity of air. Among the hardwoods, maple and oak being denser species, are considered good flooring materials.

30.4

You are selecting finish floor materials for several different projects listed below. What would you specify as the appropriate floor covering? Why? (a) Living room in a home for persons with multiple chemical sensitivities.

Natural stone or ceramic tiles on thick-set portland cement mortar bed. Being made of inorganic compounds, these flooring systems do not have any emissions. Other alternatives are portland cement terrazzo flooring and solid wood flooring. (b) A medical research laboratory.

Sheet vinyl flooring. It is easy to clean and is joint free. (c) Entry lobby for a stone manufacturing plant.

Natural stone panels on thick-set mortar bed—A mosaic of various stones made by the plant. (d) Circulation areas in an airport.

Epoxy terrazzo floor. It is long-wearing and can be laid in almost any pattern and color.

(e) Bathroom floors in a multistory apartment complex.

Marble or ceramic tiles.

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30.5

(a) Draw (to a suitable scale) a section through a direct-hung suspended ceiling system with a narrow-face grid in a steel frame structure.

Floor slab comprising composite metal deck and concrete fill Top flange of steel beam Steel hanger wire tied to clip angle anchored to floor slab

Wall angle Perimeter wall

Inverted tee cross runner

Inverted tee main runner

9/16” Narrow face

Lay-in ceiling tile

(b) Is this system appropriate for use with a suspended plaster ceiling? Explain.

The system shown above is suitable for lightweight, lay-in (acoustical) ceiling tiles. It is not suitable for a suspended plaster ceiling, which is much heavier.

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Name

30.6

(a) If the suspension system in 28.5(b) is not appropriate for a plaster ceiling, what would you recommend?

A suspended plaster ceiling requires a more robust supporting grid. The wet plaster is applied to a metal lath attached to the grid, as shown in the following sketch.

(b) Sketch (to a suitable scale) the system you would recommend for this application.

Floor slab

16-gauge, 2” x 1” galvanized steel channels as main runners 48” o.c. suspended with hanger wires 16-gauge, 1” x 1” galvanized steel channels as cross runners 16” o.c. tied to main runners with pre-bent wire clips Galvanized steel lath screw fastened to cross runners Gypsum plaster or portland cement plaster ceiling

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